Putnam
inequalities
geometry
trigonometry
triangle inequality
inequalities unsolved

Take any point to be the center and draw a circle of radius 1.This circle will enclose more than half the number of points.What is the minimum radius of the circle that will enclose all N points?(This circle doesn't have to be centered on one of the N points.)

 :D . I hope it is a right forum to post.

Any attempt yet? :D
The solution is $\frac{2}{\sqrt{3}}$

A masterpiece is still there. No one prove it? :(

[quote="K81o7"]I don't really understand the problem.  What properties does this set of points have?[/quote]

  You have a set of point. You chose one arbitrary point (any points) in that set, and draw the circle (the center is the point you chose) with radius 1 that enclose more than half of the number of other points. The question is what is the minimum radius of the circle that will enclose all N points?(This circle doesn't have to be centered on one of the N points.)

  For instance, you have 3 points (A,B,C). You chose point A as the center and draw the circle with radius 1. The circle that you draw will enclose point B (or point C or both B&C)(more than half of 3 points means that in the circle should have at leat $[\frac{3}{2}]=2$ points). Then you chose point B as the center and draw the circle radius 1....so on.
  So what is the minimum radius of the circle that will enclose 3 points A, B,C? (That circle doesn't need to have the center at A,B, or C).
  I wish it helps.

It is one year already. I wish I could see the solution  :lol: . I will immediately call s.o who could prove this problem genius.
 :D

I still don't understand the question... but I do know of a similar sounding problem that was in a putnam preparation book that I think involved the number $\sqrt{2}/3$ and was about lattice points enclosed by a circle...

Well, now that this has been brought back, I believe I see a way to do it...

Hint: Prove that any two points are within a distance $2$ of each other  :)

[hide="elaboration"]For a point $P$, call the set of points within a distance of $1$ of $P$ $S_{P}$ (including $P$).  For two points $P$, $Q$, the two sets $S_{P}$ and $S_{Q}$ each have greater than $\frac{N}{2}$ elements.  It then follows that they intersect in some element.  Then there exists a point $T$ such that $PT$ and $QT$ are both at most $1$, meaning, by the triangle inequality, that $PQ \le 2$.[/hide]

Edit: And btw, this should probably be in Combinatorics...

Assumption: no 3 points are collinear.

This can be proven using Helly's theorem.

We can easily show that distance between any two points is <= 2.

Now around each point draw a circle of radius $\frac{2}{\sqrt 3}$

If we can show that any three circles have a common point, then all have one common point (Helly's theorem for 2 dimensional space), which would imply that all points can be enclosed in a circle of radius $\frac{2}{\sqrt 3}$ (centered at the common point of all circles)

Consider any 3 points ABC, which form a triangle, where each side is $\leq 2$. We can show using elementary geometry and trigonometry that the circum radius of this triangle is $\leq \frac{2}{\sqrt 3}$

Now we can apply Helly's theorem and we are done.


1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]

1.  [hide] $n(n+1) - n^2 = n.$  Thus, the sum of the first k terms of the first series subtracted from the first k terms of the second series is 1 + 2 + 3 +... + k.  So k (k+1)/2 = 210.  k=20.[/hide]


\[ (1 \plus{} x\sqrt {x^2  \plus{} y^2 } )dx \plus{} ( \minus{} 1 \plus{} \sqrt {x^2  \plus{} y^2 } )ydy \equal{} 0\]

Rearranging, we have

$ 1\plus{}\left( x\plus{}y\frac{dy}{dx}\right) \sqrt{x^{2}\plus{}y^{2}}\minus{}y\frac{dy}{dx}\equal{}0$

$ \Rightarrow \frac{d}{dx}\left( x\plus{}\frac{1}{3}\left( x^{2}\plus{}y^{2}\right) ^{3/2}\minus{}\frac{1}{2}y^{2}\right) \equal{}0$

$ \Rightarrow x\plus{}\frac{1}{3}\left( x^{2}\plus{}y^{2}\right) ^{3/2}\minus{}\frac{1}{2}y^{2}$ $ \equal{}C$


1.The number of ways in which we can choose 2 distinct integers from 1 to 100 such that difference between them is atmost 10 is????

2.Number of integers between 1 and 1000(both inclusive) which are divisible by either of 10,15 and 25 is???

3.The number 3 can be written as 3,2+1,1+2 or 1+1+1 in four ways.In how many ways can the number n be written??

4.The no. of ways of selecting 2 nos. from the set {1,2,3,4...,12} whose sum is divisible by 3 is??? :?:

5.Also,The no. of times the digit 0 will be written when listing the integers from 1 to 1000 is?????   :?:  :?:   :(

[hide="1"]
Number of ways to pick two numbers with a difference of one: $ (1,2),(2,3)...(n,n \plus{} 1)$ where $ n \plus{} 1 \equal{} 100$
So there are $ 99$ ways
Similarly for numbers with a difference of two: $ (1,3)...(n,n \plus{} 2)$

So the answer we are looking for is

$ (100 \minus{} 1) \plus{} (100 \minus{} 2)...(100 \minus{} 10)$
$ \equal{} 10.(100) \minus{} \frac {10}{2}.(10 \plus{} 1) \equal{} 945$

So there are $ 945$ pairs with a difference $ \le 10$ [/hide]

[hide="2"]
Number of inigers between $ 1$ and $ 1000$ divisible by $ 2(5)$ or $ 3(5)$ or $ 5(5)$

The intigers divisible by $ 5$ are $ 5k$ where $ k \equal{} [1,200]$ so the number of intigers that are divisible by $ 2$ or $ 3$ or $ 5$ are
$ \lfloor \frac {200}{2} \rfloor \plus{} \lfloor \frac {200}{3} \rfloor \plus{} \lfloor \frac {200}{5} \rfloor \equal{} 206$
But we have over counted the ones that are divisible by $ 2(3), 2(5), 3(5)$ and $ 2(3)(5)$
$ 206 \minus{} \left ( \lfloor \frac {200}{6} \rfloor \plus{} \lfloor \frac {200}{10} \rfloor \plus{} \lfloor \frac {200}{15} \rfloor \plus{} \lfloor \frac {200}{30} \rfloor \right ) \equal{} 206 \minus{} 72 \equal{} 134$[/hide]

[hide="4"]
Split the set into numbers $ \mod 3$
$ (1,4,7,10) \equiv 1 \mod 3$
$ (2,5,8,11) \equiv 2 \mod 3$
$ (3,6,9,12) \equiv 0 \mod 3$

So we need the sum of two inigers to add to $ 0$ or $ 3 \mod 3$.
The answer is $ 4(4) \plus{} \binom{4}{2} \equal{} 16 \plus{} 6 \equal{} 22$ [/hide]

[hide="5"]
$ \lfloor \frac {1000}{10} \rfloor \plus{} \lfloor \frac {1000}{100} \rfloor \plus{} \lfloor \frac {1000}{1000} \rfloor \equal{} 111$
There are also numbers in the form $ a0b$ where $ a,b \equal{} [1,9]$
So the total it
$ 111 \plus{} 9(9) \equal{} 192$[/hide]

[quote="rohitarura"]
3.The number 3 can be written as 3,2+1,1+2 or 1+1+1 in four ways.In how many ways can the number n be written??[/quote]

[hide]Define $ a_n$ to be the number of ways in which $ n$ can be written as the sum of positive integers when order matters. Therefore, $ a_1 \equal{} 1$, $ a_2 \equal{} 2$, and $ a_3 \equal{} 4$.

Using recursion, we see that if we start with $ 1$, then it becomes the number of ways in which $ n\minus{}1$ can be written. Similarly for $ 2,3,\dots,n\minus{}1$, and starting with $ n$ just yields one way, simply writing $ n$. Therefore, the number of ways to write $ n$ is just the sum of the number of ways to write all the numbers below $ n$, plus 1 for just $ n$. So $ a_n \equal{} 1 \plus{} \sum_{k \equal{} 0}^{n \minus{} 1} a_k$. So $ a_4 \equal{} 8$, $ a_5 \equal{} 16$, and you can see quickly that $ a_n \equal{} 2^{n \minus{} 1}$ when $ n > 0$, and 0 when $ n \equal{} 0$.[/hide]

The answer to the 2nd question is 146.
Please solve it

Problem #2 is the same as how many numbers from 1 to 200 is divisible by at least one of 2, 3, or 5. (I just divided everything by 5, since clearly if $ 5 \not| n$ then $ \{15,20,25\} \not| n$) So just basic PIE ($ \#(\psi)$ is just the number of elements satisfying condition $ \psi$):
\begin{align*} \#(2|n \vee 3|n \vee 5|n) & = \#(2|n) + \#(3|n) + \#(5|n) - \#(2|n \wedge 3|n) - \#(2|n \wedge 5|n) - \#(3|n \wedge 5|n) + \#(2|n \wedge 3|n \wedge 5|n) \\
& = \left \lfloor \frac {200}2 \right \rfloor + \left \lfloor \frac {200}3 \right \rfloor + \left \lfloor \frac {200}5 \right \rfloor - \left \lfloor \frac {200}6 \right \rfloor - \left \lfloor \frac {200}{10} \right \rfloor - \left \lfloor \frac {200}{15} \right \rfloor + \left \lfloor \frac {200}{30} \right \rfloor \\
& = 100 + 66 + 40 - 33 - 20 - 13 + 6 \\
& = 206 - 66 + 6 \\
& = 140 + 6 \\
& = \boxed {146} \end{align*}

[quote]2

Number of inigers between and divisible by or or

The intigers divisible by are where so the number of intigers that are divisible by or or are
\lfloor \frac {200}{2} \rfloor + \lfloor \frac {200}{3} \rfloor + \lfloor \frac {200}{5} \rfloor = 206
But we have over counted the ones that are divisible by and
206 - \left ( \lfloor \frac {200}{6} \rfloor + \lfloor \frac {200}{10} \rfloor + \lfloor \frac {200}{15} \rfloor + \lfloor \frac {200}{30} \rfloor \right ) = 206 - 72 = 134[/quote]

I disagree...
In fact, The intigers divisible by 5 are 5k where k=[1,200]. The number of intigers that are divisible by 2 or 3 or 5 and takes the form 5k, are, in fact, 206, as in your solution.
And we have over counted the ones that are divisible by 2(3), 2(5), 3(5), that is, the new number obteined is:
206 - (33 + 20 + 13) = 140. However, after all those procedures, we have under count the ones that are divisible by (2)(3)(5), that is, the right number are 140 + 6 = 146.

Forgive me for not knowing how to use latex, as well as other features of the forum.

the first is 1 and 11 and the last is 80 and 100, so if we count, the answer would be 80-1=79...


Suppose that the centre of nine-point circle of triangle lies on the incircle of the triangle. Show that $ R \equal{} 4r$

Have:
1). $ |HI|^2\equal{}2.r^2\minus{}4.R^2.R^2.cosA.cosB.cosC$ and $ |OH|^2\equal{}R^2\minus{}8.R^2.cosA.cosB.cosC$
$ \Longrightarrow |OH|^2\minus{}2.|HI|^2\equal{}R^2\minus{}4.r^2.$
2). If N=the center of nine-point circle, then: N=the midpoint of [OH], and:
$ \Longrightarrow |IN|^2\equal{}\frac{2.(|IO|^2\plus{}|IH|^2)\minus{}|OH|^2}{4}\equal{}\frac{2.|IO|^2\minus{}(|OH|^2\minus{}2.|IH|^2)}{4}\equal{}
\\
\equal{}\frac{2.(R^2\minus{}2.Rr)\minus{}(R^2\minus{}4.r^2)}{4}\equal{}\frac{R^2\minus{}4.Rr\plus{}r^2}{4}\equal{}\frac{(R\minus{}2r)^2}{4}\Longrightarrow \\
\Longrightarrow |IN|\equal{}\frac{R\minus{}2r}{2}.$
3). The centre of nine-point circle of triangle lies on the incircle $ \Longleftrightarrow
 |IN|\equal{}r \Longleftrightarrow \frac{R\minus{}2r}{2}\equal{}r \Longleftrightarrow R\equal{}4r.$

Alternatively,

It's well known that the nine-point circle is tangent to the incircle (and easy to prove by inversion).  Then if the nine-point center lies on the incircle, the radius of the nine-point circle is $ 2r$.  It then follows that $ R\equal{}4r$.

I hope that his one is also possble solution.

Denote by $ F$ Feurbach point (point where the nine-point circle and incircle are tangent to each other), by $ I$ center of incircle and by $ N$ center of nine-point circle.
Let $ p$ be line that contains $ F$ and touches incircle and nine-point circle
(that is in fact common tangent to nine-pint circle and incircle).

Draw lines $ IF$ and $ NF$.

Both these lines are perpendicular to line $ p$ so it follows that $ N$, $ I$ and $ F$ are collinear.
Since $ N$ is on incircle and $ N$, $ I$ and $ F$ are collinear we can conclude that:
$ \frac{R}{2}\equal{}NF\equal{}NI\plus{}IF\equal{}2r$ so it follows that $ R\equal{}4r$.


What's the english name for it? I mean the blue line $k$, which goes through tangency points

This is the polar line of  the intersecting points of the tangents. and the point is the pole of its polar


Fie $x_{n}=\sum_{k=2}^{n}\frac{1}{klnk}-\int_{2}^{n}\frac{1}{xlnx}dx$ pt $n\geq 2$. Demonstrati ca $x_{n}$ este marginit si determinati $\lim_{n\to\infty}\frac{1}{ln(lnn)}\sum_{2}^{n}\frac{1}{klnk}$.

[hide="Soluţia clasică."]Fie $f : \left[ 2, \infty \right) \to \left( 0, \infty \right)$ astfel încât $\lim_{x \to \infty}f(x) = 0$. Avem $0 \leq f(k)-\int_{k}^{k+1}f(x) \, dx \leq f(k)-f(k+1)$, deci $0 \leq \sum_{k=2}^{n-1}f(k)-\int_{2}^{n}f(x) \, dx \leq f(2)-f(n)$, i.e. $f(n) \leq \sum_{k=2}^{n}f(k)-\int_{2}^{n}f(x) \, dx \leq f(2)$.
Dacă $f$ nu este $L^{1}$, atunci $\lim_{n \to \infty}\left. \left( \sum_{k=2}^{n}f(k)\right) \right\slash \left( \int_{2}^{n}f(x) \, dx \right) = 1$.
[/hide]


topology
advanced fields
advanced fields theorems

Please download following file and see at red line. I have one question : What is canonical topology?

Looks like Neukirch, chapter 3.1 ;)
The second factor is just some $ \mathbb R^n$ (with $ n$ the number of infinity primes), so you can either use the standard topology of $ \mathbb R^n$ or the product topology of $ n$ times $ \mathbb R$. But both topologies are the same, so this doesn't matter, and this is the standard/canonical topology.

Thanks!  :lol: That text is belong to Chapter: Riemann-Roch Theory in that ONE!


geometry
rectangle
combinatorics solved
combinatorics
permutations

How many permutations pi = {q_1,q_2,...,q_n} of the numbers 1, 2, ...,n are there such that for each k = 1, 2, ...,n the following condition is satisfied: │k - q_k│  \leq  1 ?

Let u be the number of such permutations and 'a' such a permutation.

a(n) = n or a(n) = n-1.

. if a(n) = n then there are u(n-1) ways to put the a(1) .. a(n-1)
. if a(n) = n-1 then a(n-1)=n and there are u(n-2) ways to put the a(1) ..
a(n-2)

So u(n) = u(n-1) + u(n-2)

u(0) = 1, u(1) = 1, u(2) = 2, ....

u is Fibonacci sequence.

Even it doesn't look like it on the first sight, I think it's similar to this problem:

In how many ways can we tile a rectangle 2xn with 2x1 rectangles? We have a similar idea for that one, although it's another type of prb! It's amazing how we can find links between such different problems!


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

i've tried searching for a similar problem, but my wording didn't match with any existing thread on it...

so, suppose you have $ E\subset F$ a field extension, and $ A\in M_n(F)$ a square matrix with characteristic polynomial which lies in $ E[x]$: is it true that $ A$ is similar to a matrix in $ M_n(E)$?

of course, if you found a "particular counterexample", just try to put some more hypothesis (e.g. the fields to be char 0, the extension to be algebraic, finite, separated..) and see if you can prove anything.

No. Let $ c$ be an element of $ F\setminus E$, and let $ q$ be its minimal polynomial over $ E$. If $ M(p)$ is the companion matrix for the polynomial $ p$, let $ A\equal{}\begin{bmatrix}M((x\minus{}c)q(x))&0\\0&M(\frac{q(x)}{x\minus{}c})\end{bmatrix}$ in block form. The characteristic polynomial is $ q^2$, which has coefficients in $ E$. The minimal polynomial is $ (x\minus{}c)q(x)$, which doesn't.

That's essentially the only kind of counterexample you'll get; look up rational canonical form for more detail. You can't really exclude it without trivializing everything.


geometry
3D geometry
trigonometry
vector
puzzles

Diagonals are drawn on a cube that meet at a single vertex. What is the angle made between them?

[hide]use the dot product.  The angle in question is \[ \cos^{-1}{\frac{(0, 1, 1) \cdot (1, 1, 0)}{2}}=\cos^{-1}{\frac{1}{2}} \] or $60^{\circ}$.[/hide]

Another way:

[hide]Draw another diagonal which joins the other two vertices. This forms an equilateral triangle, thus the angle is 60 degrees.[/hide]

[quote="mathnerd314"]Hexaditidom's solution was the one I saw in the book. What's the dot product?[/quote]

Essentially the same way, but the dot product of two vectors is the product of the magnitudes times the cosine of the angle between them.  You can show that the dot product of two vectors, $(a_1, a_2, a_3, ..., n_1) \cdot (b_1, b_2, b_3, ...b_4)=a_1b_1+a_2b_2+...+a_nb_n$.

I assume they're face diagonals?

[hide="Answer"]Since they're both face diagonals, then by definition we have that the line connecting the other endpoints of the two diagonals is also a face diagonal, and we have an equilateral triangle. Therefore, the angle between the lines is $\boxed{60^\circ}$.[/hide]

[quote="mathnerd314"]Hexaditidom's solution was the one I saw in the book. What's the dot product?[/quote]
Let's say we have 2 vectors $\vec{a}$ and $\vec{b}$.  By the traingle law, we have $\vec{b}-\vec {a}$ as there sum.
Now, by the cosine law,
$|\vec{b}-\vec{a}|^2=|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|\cos\theta$
$\\|\vec{b}-\vec{a}|^2=(\vec{b}-\vec{a})\cdot (\vec{b}-\vec{a})=|\vec{b}|^2-2\vec{a}\cdot\vec{b}+|%Error. "arrow" is a bad command.
{a}|^2\\ \implies |\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|\cos\theta=|\vec{b}|^2-2\vec{a}\cdot\vec{b}+|\vec{a}|^2\\ \implies -2|\vec{a}||\vec{b}|\cos\theta=-2\vec{a}\cdot\vec{b}\\ \cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$

EDIT: The defintion of the dot product is as follows.
If we have 2 vector $\vec{a}=\left< a_1,a_2,a_3,...,a_n\right>$ and $\vec{b}=\left< b_1,b_2,b_3,...,b_n\right>$in $V_n$, then
$\vec{a}\cdot\vec{b}=\sum_{i=1}^n a_ibi$.
The above formula solves the problem. :)

Masoud Zargar


geometry
trapezoid
geometric transformation
homothety
rotation
ratio
geometry proposed

[color=darkblue]Let $ ABCD$ be a trapezoid with $ AB\parallel CD$. Exterior equilateral triangles $ ABE$ and $ CDF$ are constructed. Prove that lines $ AC$, $ BD$ and $ EF$ are concurrent.[/color]

Let's name $ \{M \} \in AC \cap BD$, see that $ \frac {AM}{MC} \equal{} \frac {AB}{CD}, \leftarrow \equal{} \rightarrow \frac {AM}{MC} \equal{} \frac {AE}{CF}$ and, with $ AE \parallel CF$ we are done.

Best regards,
sunken rock

[quote]with $ AE \parallel CF$ we are done. [/quote]
Sorry, but you still haven't explained why from this fact the statement follows.

[hide="My solution (pretty similar)."]Denote by $ O$ the intersection point of $ AC$ and $ BD$. Clearly we have $ AE\parallel CF$, $ BE\parallel DF$, $ BO\parallel DO$ and $ AO\parallel CO$. Hence $ OBEA$ is homothetic to $ ODFC$. Thus $ O$ is the centre of homothety. But in this homothety $ E$ is homothetic to $ F$ - that is why $ EF$ goes through the center of homothethy - $ O$. So $ EF$, $ AC$ and $ BD$ are concurrent at $ O$.[/hide]

[hide="Generalization:"]If $ A_1B_1\parallel A_2B_2,B_1C_1\parallel B_2C_2,C_1A_1\parallel C_2A_2$, then $ A_1A_2,B_1B_2,C_1C_2$ are concurrent.[/hide]

[hide="Image:"][geogebra]f1c407f6328fc364fa50050992154ffedd4cf2f4[/geogebra] [/hide]

[hide="Proof:"]Let $ M$ be the intersection of $ A_1A_2$ and $ C_1C_2$. It is clear that $ B_2A_2C_2$ is similar to $ B_1A_1C_1$, and $ A_2C_2M$ is similar to $ A_1C_1M$. So $ A_2B_2: B_2C_2: C_2A_2\equal{}A_1B_1: B_1C_1: C_1A_1$ and $ C_2A_2: A_2M: MC_2\equal{}C_1A_1: A_1M: MC_1$. Therefore $ A_2B_2: B_2C_2: C_2M: MA_2\equal{}A_1B_1: B_1C_1: C_1M: MA_1$. This implies that $ A_2B_2C_2M$ and $ A_1B_1C_1M$ are similar. The spiral similarity which brings $ A_2B_2C_2M$ to $ A_1B_1C_1M$ brings $ B_2M$ to $ B_1M$. The transformation takes $ A_2M$ to $ MA_1$, so the transformation contains a rotation of $ 180^{\circ}$. Thus, $ B_1,M,B_2$ are collinear. So $ A_1A_2,B_1B_2,C_1C_2$ concur at $ M$.

Q.E.D.[/hide]

Sorry for late reply: it follows that EF divides AC at the same ratio as BD does.
Alternately, one may use Desargues...

Best regards,
sunken rock


AMC
AIME
modular arithmetic
number theory
relatively prime

I understand the first part of the solution:

The label  will occur on the $ \frac12(1993)(1994) \pmod{2000}th$ point around the circle. (Starting from 1) A number  will only occupy the same point on the circle if $ \frac12(n)(n \plus{} 1)\equiv \frac12(1993)(1994) \pmod{2000}.$

But why can't you multiply both sides by two to obtain $ n(n \plus{} 1)\equiv (1993)(1994) \pmod{2000}$ and thus $ n(n\plus{}1)\equiv 42\pmod{2000}$ and $ n \equal{} 6 ?$

2 is not relatively prime to 2000, so you can't multiply by it in mod 2000. For example: 0=1000 mod 2000 turns into 0=2000 mod 2000 when multiplying by 2, so does 0=1000 mod 2000?

Oh I see, thanks a lot! Need to review my mod rules....

Actually, that's not correct. You can multiply by any number in any mod - it's division that gets tricky. Say $ a\equiv b\pmod m$. Then $ a \equal{} b \plus{} mk$. We can multiply this by $ c$: $ ac \equal{} bc \plus{} mck$, or $ ac\equiv bc\pmod{mc}$ or by extension $ ac\equiv bc\pmod m$.

The reason Brut3Forc3's example doesn't work is that $ 0\not\equiv1000\pmod{2000}$.

Sorry, I meant that you can't multiply by a number and use the statement you get to deduce the truth of the original statement. (It's parallel to starting with 1=2 and multiplying by 0 to get 0=0, a true statement.) So I guess that's dividing.

To put another way, multiplying by a number not relatively prime to the modulus is not a reversible step.

Mods only give you filters on what cannot work. So if you get a solution through mods, remember that it's only filtered. You have to check to make sure it actually does work.

In your example, you have at least two solutions - $ n\equiv1993\pmod{2000}$ and $ n\equiv6\pmod{2000}$.


geometry
IMO
circles
angles
midpoint
IMO 1983

Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. One of the common tangents to the circles touches $C_1$ at $P_1$ and $C_2$ at $P_2$, while the other touches $C_1$ at $Q_1$ and $C_2$ at $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2=\angle M_1AM_2$.

Let B be the other intersection of the circles $C_1, C_2$ and $H \equiv P_1P_2 \cap Q_2Q_2$ the intersection of their common external tangents, identical with their external homothety center lying on the center line $O_1O_2$.  Since the points pairs $(P_1, P_2), (Q_1Q_2)$ are centrally similar with the homothety center H and coefficient $\frac{r_1}{r_2}$, where $r_1, r_2$ are radii of the circles $C_1, C_2$, the midpoints $(M_1, M_2)$ are centrally similar with the same homothety center and coefficient. Let the line HA meet the circles $C_1, C_2$ at points $R_1, R_2$, respectively, different from the point A. Again, the points pairs $(R_1, A),\ (A, R_2)$ are centrally similar with the same homothety center and coefficient. Therefore, the triangles $\triangle M_1R_1A \sim \triangle M_2AR_2$ are centrally similar with the same homothety center and coefficient. Thus their corresponding sides $M_1R_1 \parallel M_2A,\ M_1A \parallel M_2R_2$ are parallel. Let B' be the intersection of the lines $M_1R_1, M_2R_2$, forming a parallelogram $M_1AM_2B'$. The radical axis AB of the circles $C_1, C_2$ cuts the tangent length segments $P_1P_2, Q_1Q_2$ at their midpoints, hence, it also cuts the segment $M_1M_2$ at its midpoint N. As $M_1M_2$ is one tdiagonal of the parallelogram $M_1AM_2B'$, the line AB is the other diagonal line of this parallelogram. Since AN = BN, the vertex B' of this parallelogram is identical with the point B and the parallelogram is a rhombus (perpendicular diagonals or all sides equal). As a result,

$\angle M_1AM_2 = \angle M_1BM_1 \equiv \angle R_1BR_2 = 180^\circ - (\angle AR_1B + \angle AR_2B) =$

$= 180^\circ - \frac{\angle AO_1B + \angle AO_2B}{2} = 180^\circ - (\angle O_2O_1A + \angle O_1O_2A) = \angle O_1AO_2$

Little more general:

Let  $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. Let $S$ be such point on line $O_1O_2$ so that tangents on $C_1$ touches it at $P_1$ and $Q_1$ and tangents on $C_2$ touches it at $P_2$ and $Q_2$. Let  $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2 = \angle M_1AM_2$.

[b]Proof[/b]: Since $S$ is image of $M_1$ under inversion wrt circle $C_1$ we have: \[ \angle O_1AM_1 = \angle O_1M_1'A'= \angle O_1SA  \] Since $S$ is image of $M_2$ under inversion wrt circle $C_2$ we have:\[ \angle O_2SA= \angle O_2A'S'= \angle O_2AM_2  \] Image of $A$ is in both cases $A$ itself, since it lies on both circles.
Since $\angle O_1SA = \angle O_2SA$ we have: \[ \angle M_1AO_1=\angle M_2AO_2  \] Now: \[ \angle O_1AO_2 = \angle M_1AM_2-\angle M_1AO_1+\angle M_2AO_2 = \angle M_1AM_2  \]

[quote="orl"]Let $ A$ be one of the two distinct points of intersection of two unequal coplanar circles $ C_1$ and $ C_2$ with centers $ O_1$ and $ O_2$ respectively. One of the common tangents to the circles touches $ C_1$ at $ P_1$ and $ C_2$ at $ P_2$, while the other touches $ C_1$ at $ Q_1$ and $ C_2$ at $ Q_2$. Let $ M_1$ be the midpoint of $ P_1Q_1$ and $ M_2$ the midpoint of $ P_2Q_2$. Prove that $ \angle O_1AO_2 \equal{} \angle M_1AM_2$.[/quote]
I think this has a similar idea as [b]yetti[/b]'s solution, but a bit different:
[hide="Solution"]
Let $ P_1P_2$ and $ Q_1Q_2$ meet at $ R$.  Let $ RA$ meet $ C_2$ at $ B$.  Now, it is well-known that $ O_1O_2$, $ P_1P_2$, and $ Q_1Q_2$ are concurrent at $ R$, the center of homothety between $ C_1$ and $ C_2$.  Now, it is well-known that $ O_1O_2$ bisects $ \angle P_1RQ_1$.  Since $ P_1R \equal{} Q_1R$, we have that $ O_1O_2R$ meets $ P_1Q_1$ at its midpoint, $ M_1$, and $ P_1Q_1$ is perpendicular to $ O_1O_2R$.  Similarly, $ O_1O_2R$ passes through $ M_2$ and is perpendicular to $ P_2Q_2$.  Since $ O_2P_2\perp P_2R$, we have that $ RA\cdot RB \equal{} RP_2^2 \equal{} RM_2\cdot RO_2$, which implies that $ ABM_2O_2$ is cyclic.  Yet, since $ A$ and $ B$ lie on $ C_1$ and $ C_2$ respectively and are collinear with $ R$, we see that the homothety that maps $ C_1$ to $ C_2$ about $ R$ maps $ A$ to $ B$.  Also, $ O_1$ is mapped to $ O_2$ by this homothety, and since $ M_1$ and $ M_2$ are corresponding parts in these circles, $ M_1$ is mapped to $ M_2$ by this homothety, so $ \angle O_1AM_1 \equal{} \angle O_2BM_2 \equal{} \angle O_2AM_2$, from which we conclude that $ \angle O_2AO_1 \equal{} \angle M_2AOM_1$. [/hide]

Let $B$ be the other intersection point of these two circles. Let $P_1P_2$, $Q_1Q_2$, $O_1O_2$ meet at $Q$. Let $AB$ meet $P_1P_2$ at $P$.
Clearly, $P_1Q_1$ and $O_1O_2$ are perpendicular at $M_1$; $P_2Q_2$ and $O_1O_2$ are perpendicular at $M_2$.
Since $\triangle O_1P_1M_1 \sim \triangle O_2P_2M_2$, \[\dfrac{O_1P_1}{O_1M_1} = \dfrac{O_2P_2}{O_2M_2} \Rightarrow \dfrac{O_1A}{O_1M_1} = \dfrac{O_2A}{O_2M_2} \qquad{(*)}\] Since $P$ is on the radical axis, $PP_1 = PP_2$, so in the right trapezoid $P_1P_2M_2M_1$, $AB$ is the midsegment. So we have $M_1A = AM_2$ and $\angle AM_1O_1 = \angle AM_2O_1$.
Let $M$ be a point on $O_1O_2$ such that $\triangle AM_1O_1 \cong \triangle AM_2M$ (which means $AM=AO_1$ ve $MM_2 = M_1O_1$). So, from $(*)$, we get \[\dfrac{AM}{MM_2} = \dfrac{O_2A}{O_2M_2}\] This means, $AM_2$ is the angle bisector of $\angle MAO_2$. So, \[\angle O_2AM_2 = \angle M_2AM= \angle M_1AO_1 \Longrightarrow \angle M_1AM_2 = \angle O_1AO_2.\]

It is similar to all but slightly shorter...Let, $P_1P_2,Q_1Q_2,O_1O_2$ concur at $Q$.Then a homothety with centre $Q$  that sends $C_1$ to $C_2$.Let $QA \cap C_1 =B$.Under the homothety $A$ is the image of $B$.So, $\angle M_1BO_1 =\angle M_2AO_2$ and $\triangle QP_1O_1 \sim \triangle QM_1P_1 \implies \frac{QO_{1}}{QP_{1}} = \frac{QP_{1}}{QM_{1}}$.so $QO_1.QM_1 = Q.P^2_1 = QA .QB$ Hence  $A,M_1,B,O_1$ are concyclic.so $\angle O_1BM_1=\angle O_1AM_1 \implies \angle O_1AM_1= \angle O_2AM_2 \implies \angle O_1AO_2=\angle M_1AM_2 $

I'm sorry that I can't use LaTeX because I am new user and don't have permission to add images to posts.
It can be easily solved with this lemma:
Let A-symmedian of triangle ABC intersect its circumscribed circle u at D, P on AD satisfying AP = PD. Let us define center of u as O. Then B, C, P, O are concyclic.
Now let S be intersection of tangents to circles from the problem, AS intersects  C_1 at B. One can prove that AP_1BQ_1 is harmonic qudrilateral, so P_1Q_1 is symmedian of triangle AP_1B. By lemma we get that B, A, O_1, M_1 are concyclic, thus angle O_1AM_1 is equivalent to angle O_1BM_1. By homotety we get angle O_1BM_1 = angle O_2AM_2. Q. E. D.

[color=#0ff]To $\LaTeX$ this,[/color]
It can be easily solved with this lemma:
Let $A$-symmedian of $\triangle ABC$ intersect its circumscribed circle $u$ at $D, P$ on $AD$ satisfying $AP = PD$. Let us define center of $u$ as $O$. Then $(B, C, P, O)$ are concyclic.
Now let $S$ be intersection of tangents to circles from the problem, $AS$ intersects $ C_1$ at $B$. One can prove that $\square AP_1BQ_1$ is harmonic qudrilateral, so $P_1Q_1$ is symmedian of $\triangle AP_1B$. By lemma we get that $(B, A, O_1, M_1)$ are concyclic, thus $\angle O_1AM_1$ is equivalent to $\angle O_1BM_1$. By homothety we get $\angle O_1BM_1 = \angle O_2AM_2$.
$Q. E. D.$

Let $P$ be the exsimilicenter of $C_1$ and $C_2$, and let $\infty$ be the point at infinity on line $l=O_1O_2$. Then the pairs of points $(P, \infty)$, $(M_1, O_2)$ and $(O_1, M_2)$ are pairs under an involution on line $l$ (this involution is inversion on $P$). That means that $(AP, A\infty)$, $(AM_1, AO_2)$ and $(AO_1, AM_2)$ are also pairs under an involution on the pencil of lines through $A$. Since by inversion
$$\angle PAM=\angle PO_2A=\angle O_2A\infty,$$
this involution must be a reflection over the angle bisector of $\angle MAO_2$, which solves the problem. $\blacksquare$


calculus
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domain
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3D geometry

Evaluate the two closed chain integrals =) Wasn't very fun from class.

$ a) \int \limits_{[1, \minus{} 1 \plus{} i, \minus{} 1 \minus{} i]}\frac {dz}{z}$
 [hide="Answer"]$ 2\pi ?i$[/hide]

$ b) \int \limits_{[i, \minus{} 1, \minus{} i,1]}\frac {dz}{z}$

And of course it's easier to first find a formula for $ \int_{[a,b]}\frac {dz}{z}$ and $ 0 \notin [a,b]$.

No one enjoy complex analysis? XD

I am sort of unsure about how to really do these, and I'd like to get them done before class tomorrow so I can keep up.
Could someone help out? =)

Not sure if we can just use Cauchy's residue/integral formulas and say they're both $ 2\pi i$.

Well, if I understood your notations correctly (I mean the notation for your contour of intergation) then, of course, you can just use Cauchy residue theorem. The function $ f(z) \equal{} \frac {1}{z}$ has one singularity (pole of the the first order)  inside domain and everything is fine. :)

Ah the notation is supposed to be line contours from some point to the next. The contour image for the first one is a sideways cone-ish shape.
The second one is a diamond. And then integrating both counter-clockwise.

So to use the residue I just multiply $ f(z)$ by $ z$? Since at the pole it would then by $ > 0$ but still finite. So the integral for both is $ 2\pi i$ because its $ 2\pi i R(z_0)$
where $ R(z_0) \equal{} \lim _ {z \to z_0} (z \minus{} z_0)f(z)$ which is a simple pole for both (same function) ? Edit: and $ z_0 \equal{} 0$

Suppose we wanted to integrate over the line $ [1,\minus{}1\plus{}i]$ for the function $ \frac 1z$ and we used the antiderivative as $ \log(z)$. Our teacher said something about principle branches and said we'll get to it in a few weeks. Could someone explain this to me?
Thanks =)

$ log(z)$ can only be defined over a simply connected domain (otherwise its multivalued because going around an angle of 2pi around the origin doesn't give the same value), so we essentially need to chop off a ray going away from $ 0$. Such a choice is called "picking a branch".


The expression $ 1.25 \times 10^{\minus{}14}$ can be written as $ \frac{1}{2^m \cdot 5^n}$. What is the value of $ m\plus{}n$?


1. Two different  teams  with  four  students each  were training  for  an  ineternationa math  contest.  The  students  were  given an  assessment  exam and  their  scores  were  ranked.  What  is  the  probability  that  the  four  best  scores  come  from only  one  of  the  teams  assuming  that  no  two  students  got  the  same  score?

2.  Let  p(x)  =  x^2  + bx  + c , where  b  and  c  are  integers.  If  p(x) is  a  factor  of  both   x^4 + 6x^2 + 25  and  3x^4  +  4x ^2 + 28x + 5  , what  is p(1)?

1.  [hide]this (i think) would just be (8c4)/8!(8 people, how many ways can you choose 4 to be on the top team.)which is 1/4!4! which is 1/576.  then multiply by two because it could be on either team.  so the answer is 1/288.  I probably got it wrong.[hide][/hide][/hide]

2) [hide]Factor x4 + 6x2 + 25:

x4 + 6x2 + 25

= x4 + 10x2 + 25 - 4x2

= (x2 + 5)2 - 4x2

= (x2 - 2x + 5)(x2 + 2x + 5)

So b =  :pm: 2 and c = 5. Long division yields that (x2 - 2x + 5) is a divisor of the second polynomial, so b = -2, c = 5. So p(1) = 1 - 2(1) + 5 = 4.[/hide]

Scrambled wrote:1.  [hide]this (i think) would just be (8c4)/8!(8 people, how many ways can you choose 4 to be on the top team.)[/hide]

Just a question, why do you need to [hide] divide by (8!)[/hide]?

Scrambled -- I think you have a serious problem with using ordered possibilities for the denominator but unordered possibilities for the numerator.

karen canete wrote:What  do  you  think  should  be  the  answer  for  number  one  problem?

[hide]The answer is 2/(8c4) = 2/70 = 1/35.

There are 2 "good" possibilities.  Either all 4 are from team A or all four are from team B.  There are 8c4 possibilities total.

[/hide]

Just a note.  Any time you talk about probability, you are assuming some underlying probability distribution.  Here I'm guessing you were assuming that all rank orderings of the 8 students were equally likely.  To be technically correct, you need to say that.

A slightly more natural way to phrase the problem (while still specifying the probability distribution) is to say that the students were split into two teams of 4, with each split equally likely.  Then what is the probability that the four top scores were on the same team?

--Dan

Alternatively, there are [hide]4!*4!*2 good orderings (4! for the members of one team, 4! for the members of the other team, and 2 for the orders of the teams) compared with 8! orderings for the students overall, which gives you (luckily enough) the same answer.[/hide]


A 6x6 rectangle is tiled by 1x2 dominos. Prove that there exists a [i]fault line[/i], that is, there is some way to make a straight line cut through the rectangle that does not cut through any domino.

Can you generalize as to what dimension rectangles will always have a fault line?

[hide]Each line must have an even number of dominoes crossing it because there are an even number of spaces on each side. Suppose there are no vertical fault lines. That means a minimum of $2*(6-1)=10$ horizontal dominoes. Suppose there are no horizontal fault lines. That means a minimum of $2*(6-1)=10$ vertical dominoes.There are $18$ dominoes.

This argument is extendable to even rectangles such that $2*(v+h-2)<vh$[/hide]


So apparently there were an infinite number of solutions for this?

I don't get how this works...

I did $f(x)=(x-(a-bi))(x-(a+bi))=((x-a)+bi)((x-a)-bi)$
$=(x-a)^{2}+b^{2}=x^{2}-2ax+a^{2}+b^{2}$

Sum of coefficients is $1-2a+a^{2}+b^{2}=(a-1)^{2}+b^{2}=18 \implies |a-1|=3,|b|=3$

Solutions set is $\{ (4,3), \; (4,-3), \; (-2,3), \; (-2,-3) \}$

Where's the flaw in logic?

If $b=0$, then the other root need not be its conjugate, I think.

The question was:

A certain quadratic equation with real coefficients has one root $a+bi$ for integers $a$ and $b$, and the sum of the coefficients is 18. Find all possible pairs $(a,b)$.

Your analysis is correct; the answer should be something like $\{\textrm{Zuton's 4 answers}\}\cup \{(a, 0) \mid a \in \mathbb Z\}$.


When fatty acids are transformed to say stearic acid or any other acid of substance. Is it fair to say that the acid has transformed to such a degree so that its considered a new substance chemicaly? Is such a substance "chemicaly" considered a new substance even though its derived from the former fat/wax/oil?

The fat/wax/oil is normally an ester. Typical would be triglycerides, which are esters of glycerol (a trivalent alcohol) with three fatty acids. An ester is not the same chemical substance as an organic acid.

The basic reaction for creating an ester would be

alcohol + acid $\longrightarrow$ ester + water

ROH + R'COOH $\longrightarrow$ R'COOR + water

Hydrolyzing the fat would mean breaking the ester bond by reversing this reaction. Strong alkalis are one crude chemical way to do this (old-fashioned lye soap), but the result then would not be the free acid R'COOH but rather the salt.

Yes, we are very definitely talking about chemical reactions.


algebra
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combinatorics

Problem: In how many ways can you cover a $ (1 \times n)$ board with blue, yellow and green $ (1 \times 1)$ squares so that the number of blue and yellow squares is odd, and the number of green squares is even?

I've tried to solve this with recurrence relations but I always keep coming short.

[hide="So this is my attempt at solving."]It is obvious that $ n$ has to be even because the total number of squares is $ 2k \minus{} 1 \plus{} 2l \minus{} 1 \plus{} 2m \equal{} 2(k \plus{} l \plus{} m \minus{} 1)$. Suppose $ a_n$ is the number of such coverings. Therefore, $ a_n \equal{} 0$ for odd $ n$.

The smallest such board is for $ n \equal{} 2$, and it has two solutions, BY and YB.

For the even $ n$, we can create $ a_n$ from $ a_{n \minus{} 2}$ by adding two blue, two green, or two yellow squares. So it should be $ a_n \equal{} 3a_{n \minus{} 2}$. But we can also create $ a_n$ from $ a_{n \minus{} 4}$ by adding $ 1\times 4$ cells with even number of greens, yellows or blues. We can color a $ 1 \times 4$ cell with no more than 2 colours, so we pick the first color and the positions for it, and then pick the second color. But we must exclude the combinations when the last two colors are the same: we can pick the fourth color in 3 ways, the third in 2, the second in 2, and the first in 1 way, so in total 12 ways for coloring a $ 1 \times 4$ cell. From this I concluded
$ a_n \equal{} 3a_{n \minus{} 2} \plus{} 12a_{n \minus{} 4}$

We can also construct $ a_n$ from $ a_{n \minus{} 6}$ by adding a $ 1 \times 6$ cell which has exactly two of green, blue and yellow cells (otherwise, it falls within the first two cases). We must kick out combinations that were already incuded with $ a_{n \minus{} 2}$ and $ a_{n \minus{} 4}$. Those are when the last 2 colors are the same and when last four positions contain exactly 2 colors.  We count positions when the first two colors are the same - that means that the last four positions contain exactly 2 colors: first we pick one of three colors. Then we pick the sixth color in 2 ways, and the fifth in one way. The fourth one we pick in two ways, and the third in one way. In total, 12. combinations. Now we count combinations when the last two colors are the same: we pick the last color in three ways. The fifth in one way. Now we choose two out of 4 positions for the second color, 18 ways. Now the total of combinations that needs to be kicked out is 12+18=30, so there are 60 new coverings with $ a_{n \minus{} 6}$. (Not really convinced that my counting was good)

Therefore,

$ a_n \equal{} 3a_{n \minus{} 2} \plus{} 12a_{n \minus{} 4} \plus{} 60a_{n \minus{} 6}$[/hide]
Any help would be appriciated.  :blush:  :)

Your recurrence relation isn't working out so well; there's a lot of overcount and undercount happening everywhere, and once this starts happening you'll want to start from the beginning and rethink. I think recursion works, but not that particular sequence, and I haven't tried thinking of one that does.

[hide="Try..."]selecting the green tiles first, then dividing up the rest between blues and yellows. There's a binomial coefficient identity that will simplify the result, though I can't get it to a closed form yet.[/hide]

I would approach the recursion as follows:

Let $ a_n$ be the number of ways to color a $ 1 \times 2n$ board with the desired properties.  Then $ a_n$ is also the number of ways to color a $ 1 \times 2n$ board with an even number of blue but an odd number of green and yellow, or with an even number of yellow but an odd number of blue and green.  Let $ b_n$ be the number of ways to color a $ 1 \times 2n$ board with an even number of each tile.  Then verify (by considering the possible colors for the last two squares) that $ b_{n \plus{} 1} \equal{} 3b_n \plus{} 6a_n$ and $ a_{n \plus{} 1} \equal{} 2b_n \plus{} 7a_n$.  This is enough to compute -- it's also possible to solve for a recursion involving just $ a_n$, e.g., by getting $ b_n \equal{} \frac{a_{n \plus{} 1} \minus{} 7a_n}{2}$ from the second equation and substituting into the first.  The resulting equation is $ a_{n \plus{} 2} \equal{} 10a_{n \plus{} 1} \minus{} 9a_n$, with closed-form solution $ a_n \equal{} \frac{9^n \minus{} 1}{4}$.

An alternative way to approach this problem is using generating functions.  We have
\[ (x\plus{}y\plus{}z)^n\equal{} \sum_{colorings} x^{g}y^{b}z^{y}\]
where $ g$ is the number of green squares in the coloring and so on.  Call this $ f(x,y,z)$.  We are looking for the sum of the coefficients corresponding to $ g$ being even and $ y, b$ being odd.

If this were a one variable polynomial, we could extract the sum of the even coefficients by looking at $ \frac{f(1)\plus{}f(\minus{}1)}{2}$, and the sum of the odd coefficients by looking at $ \frac{f(1)\minus{}f(\minus{}1)}{2}$.  In this case, we have a three variable polynomial, so we'll take an average over the vertices of a three dimensional cube.  We assign $ \pm 1$ to the vertices of the cube in such a way that the sign remains unchanged along the $ x$ axis (we want $ g$ to be even), but changes along the $ y$ and $ z$ axis (we want $ b$ and $ y$ odd), and then add up $ f$ over all the vertices of the cube using those signs.  Doing so gives that the expression we want is equal to
\[ \frac{1}{8}(f(1,1,1)\minus{}f(1,\minus{}1,1)\minus{}f(1,1,\minus{}1)\plus{}f(1,\minus{}1,\minus{}1)\plus{} \\ \plus{}f(\minus{}1,1,1)\minus{}f(\minus{}1,\minus{}1,1)\minus{}f(\minus{}1,1,\minus{}1)\plus{}f(\minus{}1,\minus{}1,\minus{}1))\\\equal{}\frac{1}{8}(3^n\plus{}(\minus{}3)^n\minus{}1\minus{}(\minus{}1)^n),\]
which is equal to 0 if $ n$ is odd and $ \frac{1}{4}(3^n\minus{}1)$ if $ n$ is even.

Another alternative method:  Clearly the total is $ 0$ for odd $ n$, so let us assume that $ n$ is even, say $ n \equal{} 2k$.  Let us consider all of the $ 3^n$ colorings of the board with three colors.  Divide the board into $ k$ consecutive $ 2 \times 1$ dominoes, and consider the operation $ \sigma$ defined as follows:

Scan the board from left to right until the first domino which is not entirely green is encountered, and replace that domino by a new domino according to the operations $ bb \rightarrow by \rightarrow gb \rightarrow gy \rightarrow bb$ and $ yy \rightarrow yb \rightarrow yg \rightarrow bg \rightarrow yy$.  (e.g. if the first non-green domino is Green/Blue, replace it by one that is Green/Yellow).

$ \sigma$ has the following two useful properties:  First, applying $ \sigma$ a total of $ 4$ times brings you back where you started.  In other words, $ \sigma$ divides that $ 3^n \minus{} 1$ boards which are not entirely green into $ \frac {1}{4}(3^n \minus{} 1)$ quadruples $ (B, \sigma(B), \sigma^2(B), \sigma^3(B) )$.  Additionally, in each quadruple there is exactly one board containing both an odd number of blue squares and an odd number of yellow squares (and therefore an even number of green squares).


vector
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complex numbers

1. An intramural quarteback throws a football with an intial speed of 9.7 m/s at an upward angle of 38 degrees above a horizontal playing field.

a) How long will the ball reamin in the air before falling to the ground if it leaves his hand at a point 1.5m above the ground?

b) What horizontal distance will the ball travel before striking the ground?

2. Five vectors of equal magnitude radiate outward from thecenter of a regular pentagon, one vector pointing toward each vertex. Show by a graphical treatment that the sum of all five vecotrs is zero.

Alright for 1a, I have to find t when Y is 0. I tried putting the equation and solving for t, but i get a - value. Can someone please give me a correct euqation.

For 1b, i have to find x(range) but the equation also gives me a -.

[b]* Please not that I'm using -9.8 m/s for g.[/b]

For 2, I don't know how to prove it. I drew a pentagon and 5 vecotrs pointing to the center and I don't know how it proves the area to be 0. Can someone please instruct me in how I do this.

Thank You.

You could think about the sum of the five vectors to be zero, if the sum of the complex numbers at these points add to zero. But you can rotate and shrink your axis until one of the vectors is at 1 in the complex plane. Then the five point you want to sum are roots of $x^{5}-1=0$, and since we know the sum of the roots of a polynomial is equal to minus the co-efficient of the term 1 less than the degree of the polynomial, which in this case is 0, we are done. More generally, it dosnt have to be 5 vectors, but any amount! This is the first proof that came into my head, no doubt theres simpler


I  have A*x_r=B*x_r, for r=1 to R where x_r are the orthonormal vectors. Here, A and B are n^2*n^2 Hermitian matrices, A is full-rank positive definite matrix and it has only n distinct eigenvalues. That is the multiplicity order of eigenvalues for A is n. I do not have further information about B except that some of its eigenvalues are positive.

In this case, how can I determine the value of R?

Alternatively, is it possible to determine the null-space dimensionality of (A-B)?

Many thanks

This question would have a better chance of being answered in the Linear Algebra forum
 http://www.artofproblemsolving.com/Forum/index.php?f=349
but I can't move it there.


When I create a color tag like this:

[color=#444444][color=green]this is suppose[/color]d to be green[/color]

It shows up as:

[code][color=#444444][color=green]this is suppose[/color]d to be green[/color][/code]
when my intention is:
[code][color=green]this is supposed to be green[/color][/code]

[u][b]WHAT SEEMS TO BE THE PROBLEM:[/b][/u]
Suppose message length = N
(for example, my above intent has N=28)
Count N characters from the intended color tag over. Place a [/color] tag there. Place a [color=#444444] tag at the beginning.

[color=green]Not a problem for me.[/color]

[color=blue]Is there anything else odd about your system?[/color]

I am using Chrome, and XP.

Not a problem in other forums, and it seems that some others are also experiencing the same problems.

You are embedding color tags inside other color tags which is not allowed.

I know. The problem is, it is automatically done for me, which I do not desire.

What do you mean it's automatically done for you?

Please specify exactly what you are typing and what you are clicking.

[hide="color test"]
[color=red]test[/color]
[color=#4488FF]test[/color]
[color=green]this is supposed to be green[/color]
Everything seems OK
[/hide]

[color=#444444][color=red[/color]]color test[/color]

Typed:

c,o,l,o,r,[space],t,e,s,t

Then, I highlighted the string, "color test"

Then, I press the drop-down arrow for font colour. Pick red.

It looks fine right now.

I click on this box to type more stuff. The #444444 color tag comes in.

I have the same problem for a long time, when I use the option highlight - select from drop down

Solution? Type [ color = red ] text [ / color ] without spaces

Example with drop down: [color=#444444][colo[/color]r=red]12345[/color]
Example with typing: [color=red]12345[/color]

[quote="james4l"]I click on this box to type more stuff. The #444444 color tag comes in.[/quote]
I don't understand what you mean by this. What box are you clicking on?

He means on the textbox (where you write post).

levans, try to do this and you'll understand:

1)Type AoPS

2)Highlight it (like you want to copy that text)

3)Select font color from the drop down menu red or any.

4)Click on the textbox and you'll see

[color=#444444][col[/color]or=red]AoPS[/color]

When I preview, the text is all black, the code shows
[code][color=red]AoPS[/color][/code]
In the textbox, the code shows
[code][color=#444444][col[/color]or=red]AoPS[/color][/code]

Wow, OK, I was able to duplicate this in Chrome, but not Firefox or IE. Looks like some sort of bug or incompatibility in V8 (Chrome's JavaScript implementation). If you are using Chrome, please just type the color tags in directly.

Edit: It seems to be an issue with the focus DOM method being used within the change event handler which Chrome deals with differently than other browsers. Basically, what it comes down to is that the change event handler is being called twice within Chrome but only once in the other browsers as the other browsers are able to handle the focus() method, but Chrome keeps the focus on the font color selection.

[color=#444444][color=re[/color]d]Red text.[/color] Does this work?

No, it doesn't. Safari is also affected.


Prove that $n(n+1)(n+2)$ cannot be a square.

[color=red]Edit by Megus: I've reposted question, realslow - don't delete statement after the problem is solved !!  :mad: [/color]

It was surely posted many times and even general but it's faster to give a reply.

as $(n+1,n(n+2))=1$ we have $n+1=m^{2}$ for some $m$ and also $n(n+2)=m^{4}-1$ has to be a square which is impossible.

Can someone repost the statement please ¿

[thanks Megus :) ]


AMC AIME function algebra polynomial Duke college

Here is another sample problem that I couldn't figure out on the AMC site:

Find the sum of all positive integers n for which n:^2: - 19n + 99 is a perfect square.

Any suggestions?

Fierytycoon

Theres a nice little trick you can use for this.

Firstly, (n-10):^2:=n:^2:-20n+100 is always less than n:^2:-19n+99. (Check the algebra).
Then, for all n>18, (n-9):^2:=n:^2:-18n+81 is bigger than n:^2:-19n+99. check algebra again.

So for all n>18, we have our number between two consecutive perfect squares, so it can't be a square itself.
Checking these small values of n give the only solutions as n=1, 9, 10 and 18 (the formula is sort of symmetrical about n=9.5 so you only need to check the smaller numbers) and their sum is 38.

Thanks for the solution. I think I understand now. By the way, how could one tell from the function (without actually plugging in the values of n) that it was symmetrical about 9.5? Admittedly, I do not know much about symmetrical functions.

Hmm, after taking a look back at the problem, I don't think that it is such a good one for an AIME contest. I mean, if I ever encountered a problem like this one while taking the real AIME, and couldn't figure out how to do it, I would just plug in the reasonably small values of n, and, in this case, still get the right answer without having to think.

Fierytycoon

The above idea is a good one.  One more might be to say that n^2 - 19n + 99 = n^2 -19n +90 + 9 = (n-9)(n-10) + 9.  There aren't going to be many perfect squares that fit that mold and the trivial solutions are obvious.

Manipulating polynomials can shed so much light on number theory problems like this :)

I did it in a different way. I said n^2 - 19n + 99 = x^2
So, n^2 -19n + (99 - x^2)  = 0.

For n to be an integer the discriminant has to be a perfect square. That is,

19:^2: - 4(99 - x^2) = k^2
4x^2 - 35 = k^2
4x^2 - k^2 = 35
(2x + k)(2x - k) = 35

The integer solutions to this follow from the fundamental theorem of arithmatic. And once you find the values of k then n = (19 :pm: k)/2.

Hint: a similar method to this works for that challenge problem that Roman gave. The one with the (a^2 + b)/(a + b^2) = n.

[quote="Fierytycoon"]Thanks for the solution. I think I understand now. By the way, how could one tell from the function (without actually plugging in the values of n) that it was symmetrical about 9.5? Admittedly, I do not know much about symmetrical functions.
Fierytycoon[/quote]
OK well I sort of cheated, I plugged in the values! But once I saw that, just looking at
n(n-19) + 99
you can see that plugging in 5 gives 5 x -14, plugging in 14 gives 14 x -5. Etc.

Here's another problem, from the Duke Math Meet 1998, that utilizes a similar idea:

Find all integers n such that n - 3 divides n :^2:+ 2.

[quote]Find all integers n such that n - 3 divides n:^2: + 2.[/quote]

Well, the method (at least the one I use here) is slightly different. We can express n:^2: + 2 as n:^2: - 9 + 11 = (n + 3)(n - 3) + 11. As n - 3 clearly divides (n + 3)(n - 3), we need to find all n - 3 that divide 11. That would simply force n = 14 + 11k, where k is 0, 1, 2...

Fierytycoon

Actually it's the other way around, so there are only a finite number of solutions. Also, don't forget that negative numbers can have this property also.

[quote]Actually it's the other way around, so there are only a finite number of solutions.[/quote]

Oh, silly me. Right, so n = 14, 4, 2, and -8 (symmetry about 3), leading to 11, 1, -1, and -11, all of which divide 11.

Fierytycoon

First I would like to applaud Complex Zeta on his insightful reply [see above].

Now to a more interesting idea... One the same idea of number theory, what books do you reccomend for a nice introduction without being slammed into the floor with meaningless theorems and their proofs. I own a copy of A FRIENDLY INTRODUCTION TO NUMBER THEORY which is very nice for the man who doesn't have the balls do go straight into such a vast sector of mathematics.

Thank you.

What are you specifically looking for? If you want to know all the ``olympiad'' theorems, I shall shamelessly advertize my website at http://www.albanyconsort.com/simon. I have a section with all the useful theorems you might want to know for olympiads with proofs (well, at least all the ones for which I could find simple and short proofs). I wanted to put the fundamental theorem of arithmetic up there, but the only ``simple'' proof I knew used Liouville's Theorem from complex analysis, and I didn't want to put up a proof of that too because that could lead to a big mess of unproven theorems. I apologize that it's a bit disorganized. I just put the theorems up there in the order that I thought of them, but perhaps later I'll organize them to make them more useful.

If you want a book, the the Art of Problem Solving books are really good.

I don't know what you mean by "meaningless theorems and their proofs", but some good elementary number theory books that I would recommend are (listed in increasing order of difficulty):

[u]Excursions in Number Theory[/u] by Ogilvy and Anderson - Very informal, introductory book on Number Theory. Not a whole lot of proofs, very expository. Appropriate for either middle school or early high school students.

[u]The Higher Arithmatic[/u] by H. Davenport - An excellent and classic book on Number Theory, very good for high school students and has lots of nice historical commentary and bibliography information.

[u]Number Theory[/u] by George Andrews - Another good and very inexpensive book too. probably geared towards advanced high school students. Has a very combinatorial approach to it and has a nice chapter on Partition Theory.

[u]A Course in Number Theory[/u] by Rose - Mostly geared towards advanced undergraduates, but about half of it is accessible to high school students. The chapters come in pairs, the first introducing a basic concept followed by a second which goes deeper into that concept. Introduces some more advanced and modern concepts like RSA cryptography and elliptic curves in an accessible way.

[u]An Introduction to the Theory of Numbers[/u] by Hardy and Wright - This is one of the Holy Books of Number Theory. Geared towards advanced undergraduates and graduate students. But even if you don't understand all of it, it will expose you to a lot of great information and beauty of number theory.

And here is a page with a lot of Number Theory books - http://www.ericweisstein.com/encyclopedias/books/NumberTheory.html

Zeta, did you mean the Fundamental Theorem of Algebra?

Gauss, which one of those are best for the olympiad?

Hmm. I would have to say none of them particularly. They all have exercises and problems in them, but most of them go outside of the scope of what the regular Olympiads cover. The exercises from Davenport and Andrews are probably easier than your typical Olympiad problems (though the ones with stars next to them may be somewhat similar), while the ones from Rose and Hardy et Wright are in some cases of similar difficulty but not typical of the kinds of problems you will see there. So I wouldn't recommend any of them *specifically* for preparing for the Olympiads. But knowing more about Number Theory and doing problems, even these, never hurts.

For a list of books that are probably better for preparing for Olympiads I would look at the Resources section of this site, or here - http://www.ericweisstein.com/encyclopedias/books/MathematicsProblems.html

Yes, there isn't really any purely algebraic proof of the Fundamental Theorem of Algebra, beccause it's not really an "algebraic" theorem. It's really more of a theorem of analysis in that it relies essentially on the metric completeness of the field of real numbers.

The proof using Liouville's Theorem is the shortest, but the most elementary that I have seen is one using winding numbers. I *believe* there is a proof along those lines in Rudin's book, Principles of Mathematical Analysis, but I can't find it right now to see.

There is a short but much more advanced proof in Ian Stewart's book, Galois Theory, which shows that the fundamental theorem of algebra applies to any field K[i], i:^2: = -1, where K is an ordered field in which any positive element has a square root and every odd degree polynomial has a root. This applies to the field of real numbers because of the intermediate value theorem (note again that it relies on a theorem from analysis).

Here are some links to some further discussion of the FTA -
http://www.math.niu.edu/~rusin/known-math/01_incoming/FTA
http://www.math.niu.edu/~rusin/known-math/98/fta

Oh yeah! I just remembered. There is an excellent and very readable exposition on the Fundamental Theorem of Algebra in a Springer-Verlag book called [u]Numbers[/u] by Ebbinghaus et. al (Chapter 4 by R.Remmert). It gives a detailed history of the Theorem and several different proofs of it.


The probability that a student will pass in maths is 3/5 and that he will pass in English is 1/3. If the probability that he will pass in both is 1/8. what is the probability that he will pass in at least 1 subject?
a.	97/120
b.	87/120
c.	53/120
d.	120/297

isn't the solution supposed to be 1 - 2/5 x 2/3 => 1-4/15 => 11/15 ???? Am i missing something here???? :o

Actually,it should be 3/5+1/3-1/8=97/120
Try to use a venn diagram

Fizsle, your solution is almost correct. You used the technique of complementary counting. However, you forgot the fact that he had a 1/8 chance of passing both tests. If you add 1/8 to your solution at the end, you would get the right answer.

As Nerdy_girl_number1 said, the problem is alot easier with a Venn-Diagram.

[quote="fizsle"]isn't the solution supposed to be 1 - 2/5 x 2/3 => 1-4/15 => 11/15 ???? Am i missing something here???? :o[/quote]

Yes -- you assumed the two probabilities were independent.  However, we know that they aren't (because we know the probability that the student passes both is $\frac{1}{8}\neq \frac{3}{5}\cdot \frac{1}{3}$), so your assumption (and thus the calculation which resulted from it) was flawed.

Thanks everyone that was very helpful....


Suppose that two triangles $ ABC$ and $ PQR$ have the following properties:

(i)  The points $ A$ and $ P$ are the midpoints of the segments $ QR$ and $ BC$, respectively.
(ii)  The lines $ QR$ and $ BC$ are the internal angle bisectors of the angles $ \angle BAC$ and $ \angle QPR$, respectively.

Prove that $ AB \plus{} AC \equal{} PQ \plus{} PR$.

This has been discussed many times before.

Thanks, found it.  :blush:

http://www.mathlinks.ro/Forum/viewtopic.php?t=17687
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6530
http://www.cut-the-knot.org/htdocs/dcforum/DCForumID6/472.shtml

 :D


How did the AMC give the USAMO in the olden days without fax machines and the internet?

By mail would take really long, though, wouldn't it?

The AMCs and the AIMEs are sent via mail

[quote="Anaxerzia"]How did the AMC give the USAMO in the olden days [b]without fax machines[/b][/quote]
Fax machines have been around since the nineteenth century.  The first USAMO was in the early 1970s. :wink:  http://www.ideafinder.com/history/inventions/fax.htm

I don't know if the AMC always used fax for the USAMO, but it isn't impossible.  Also, overnight mail is still an option for sending in USAMO solutions.

When I took it in the late 90s, they wanted solutions faxed to them. I think the problems were sent by mail.


Find all positive integers $ n\geq 3$ with the following property:
There exists a positive integer $ M_{n}$ such that for any $ n$ positive numbers $ a_{1}$, $ a_{2}$, ..., $ a_{n}$ the inequality
$ \frac{a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}}{\left(a_{1}a_{2}...a_{n}\right)^{1/n}}\leq M_{n}\cdot\left(\frac{a_{2}}{a_{1}}\plus{}\frac{a_{3}}{a_{2}}\plus{}...\plus{}\frac{a_{n}}{a_{n\minus{}1}}\plus{}\frac{a_{1}}{a_{n}}\right)$
holds.

Привет,

In fact, for any $n\geq 4$, there is no constant M such that the inequality

$\frac{a_1+a_2+...+a_n}{\left(a_1a_2...a_n\right)^{1/n}}\leq M\left(\frac{a_2}{a_1}+\frac{a_3}{a_2}+...+\frac{a_n}{a_{n-1}}+\frac{a_1}{a_n}\right)$

holds for any n positive reals $a_1$, $a_2$, ..., $a_n$. In fact, take some positive x, and define $a_i=x^{i-1}$ for every i. Then, the inequality becomes $\frac{\left(x^n-1\right)/\left(x-1\right)}{x^{\left(n-1\right)/2}}\leq M\left(\left(n-1\right)x+\frac{1}{x^{n-1}}\right)$ which is wrong for large enough x, since the left hand side grows asymptotically faster than the right hand side. ([url=http://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=34203]Thanks to Buri aka Burmeister for this counterexample.[/url])

For n = 3, the inequality holds with M = 1; in fact, for any three positive reals a, b, c, we have

$\frac{a+b+c}{\sqrt[3]{abc}}\leq\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$.

In fact, by AM-GM, we have $\frac{b}{a}+\frac{c}{b}+\frac{c}{b}\geq 3\sqrt[3]{\frac{b}{a}\cdot\frac{c}{b}\cdot\frac{c}{b}}=3\sqrt[3]{\frac{c^2}{ab}}$, what rewrites as

$\frac{b}{a}+2\cdot\frac{c}{b}\geq 3\frac{c}{\sqrt[3]{abc}}$; similarly,
$\frac{c}{b}+2\cdot\frac{a}{c}\geq 3\frac{a}{\sqrt[3]{abc}}$;
$\frac{a}{c}+2\cdot\frac{b}{a}\geq 3\frac{b}{\sqrt[3]{abc}}$.

Summing up these three inequalities, we get $3\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\geq 3\frac{a+b+c}{\sqrt[3]{abc}}$, what becomes $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\geq\frac{a+b+c}{\sqrt[3]{abc}}$, qed..

  Darij

Привет!!!
Thank you very much!!!  :)
Pobol'she by takikh umnykh liudei!!! ;)


search
ratio
superior algebra
superior algebra theorems

Long primes are primes $p$ where $10$ is a primitive root, I search informations about this subject since I have an oral for the ENS of Lyon in 3 months. I would know how Artin find the ratio (number of long primes)/(primes) which is:
\[C_{artin}=\prod_{p\in\mathcal{P}}1-\frac{1}{p(p-1)}\]

Thanks

Have a look here

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

Hey, I hope they won't give such things at oral ENS... In fact, who knows, after all nobody stops them.  ;)  :D . By the way, nice to be sure that you have an oral examination in 3 months, I still have some worries for the next year in written examinations... :blush:  :blush:

It's the "second concours", I will show a personnal work and I think I will speak of long primes and primitive roots

Ok, the famous TIPE... I know, I also have some troubles with this these holidays since we must present something in the first year...

Not exactly :)
I'm in university, the "second concours" is for students and this oral exam is like a TIPE but I think it's harder since I'm completely alone to do it, maybe one teacher I know, will see what I've done and correct me.
I guess it's good you start your TIPE at first year because you will have more ideas and a better analyse of your subject.


The perimeter of a hexagon is 300 units. One side of the hexagon is 45 units long and the lengths of the other sides are in the ratio $ 1: 2: 3: 4: 5$. Find the positive difference between the lengths of the longest and shortest sides.

Let the side of the shortest side be $ x$. We then have $ 45 \plus{} x \plus{} 2x \plus{} 3x \plus{} 4x \plus{} 5x \equal{} 300$, so $ x \equal{} 17$. Thus the smallest side is 17 and the largest is 5(17)=85, giving us a positive difference of 68.

well we know that 1 side is 45

and there is the ratio of the sides

1:2:3;4:5

using this critical information we get the equation

45+x+2x+3x+4x+5x=300

solving we get x=17

so our answer is 17 or $\boxed{17}$

i always sub 1 with x. so we have $15x+45=300$
then
$300-45=15x$
$255=15x$
$x=\frac{255} {15}$
$x=17$
the shortest is $17$ and the longest is $85$
the positive difference, therefore, is $68$


find the smallest natural integer which has the time;somme of 9 consecutive natural integers;somme of 10 consecutive natural integers;somme consecutive of 11natural integers

I'm not absolutely sure what you're saying.  I think the question is to find the least positive integer expressible as a sum of $9,10$ and $11$ consecutive integers, in which case I think the answer is $495$.

Let the nine numbers be $n-4, n-3,\cdots,n+3,n+4$, the ten be $k-4,k-3,\cdots,k+4,k+5$ and the eleven be $m-5,m-4,\cdots,m+4,m+5$.  Then adding them, we get that the sum must be a multiple of $9$ and $11$, and of the form $10k+5$ - i.e. an odd multiple of five. So we must have an odd multiple of $5,9$ and $11$, and obviously the least example is $495$, which gives $n=55$, $k=49$ and $m=45$.


inequalities
trigonometry
inequalities proposed

Prove that :
 $ \frac{m_a.cosA}{sinB.sinC} + \frac{m_b.cosB}{sinA.sinC} + \frac{m_c.cosc}{sinB.sinA} \geq 3R $

What are the $m_{a} $...???? :(  :blush:

:huh:  $ m_a, m_b , m_c $ are the medians of the triangle $ ABC $

Your inequality is equivalently with $\sum am_a\cos A\ge 3S\ or\ \sum \frac{\cos A}{\sin {\phi_a}}\ge \frac 32$, where $\phi_a,\ \phi_b,\ \phi_c$ are the measures of the angles between the median and the side coresponding to its.

We know: $\sum \cos A=1+\frac rR;\ 6sr\le \sum am_a \le 2s(R+r);\ a.s.o.$


inequalities
combinatorics proposed
combinatorics

find the least n so that in any partition of (1,2,3,4,5,.........,n) in two sets, there are tree distinct integers a,b,c in the same set such that ab=c.

What about givin upper and down bounds for the number n if the set 1,2,3...,n is going to be partitioned into m sets?

I think maybe there is a conection with Shur bounds for a similar problem!

Well, to be rigourous Schur's numbers $s(k)$ are not relevant for that because Schur was not dealing with pairwise distinct integers.

For each integer $k\geq 2$, let $n(k)$ be the least integer $n$ satisfying the following property :
For each partition of $\{1,2, \cdots, n \}$ into $k$ classes, one of the classes contains two distinct numbers together with their sum.

Clearly, $n(k) \geq s(k)$ but this lower bound is probably far from best.

Concerning your problem, the least desired integer $N(k)$ satisfies :
\[N(k) \leq 2^{n(k)}\]
since any partition of $\{1,2,3, \cdots, 2^{n(k)} \}$ into $k$ subsets induces a similar partition of $\{2^1,2^2, \cdots, 2^{n(k)} \}$ and then a partition of $\{ 1,2, \cdots, n(k) \}$. From the definition of $n(k)$ there exists $a,b,c$ pairwise distinct which belong to the same classe such that $a+b=c$. In that case, the three numbers $2^a,2^b,2^c$ are pairwise distinct and belong to the same class of the initial partition with $2^a \times 2^b = 2^c$.

As far as I know, the best upper bound for $n(k)$ is (by me :D ) : $n(k) \leq [n \cdot (n!) \cdot e] + 1$,
which in turn leads to $N(k) \leq 2^{[n \cdot (n!) \cdot e]+1}$.

I have no idea about the fact that $N(k) = 2^{n(k)}$ or not.

Note that few values of $n(k)$ are known. According to Sierpinski, we would have :
$n(2) = 9, n(3) = 24, n(4) = 67, n(5) = 197$.

Pierre.


find the maximum value of $ k$ such that: the sequence given by:
$ u_n \equal{} (1 \plus{} \frac {1}{n})^{n \plus{} k}$ be an increase sequence

Hi victory
the solution of
(8/9)k>243/256 is maximum value of k


vector
geometry
3D geometry
sphere
integration
calculus
symmetry

Here's a problem that shouldn't be too difficult but I can't quite get it.

What is the expected magnitude of the dot product of 2 vectors randomly chosen from the d-dimensional unit sphere?

I can do the d=2 case which I believe is
$ \frac {1}{2\pi} \int_0^{2\pi} |\cos(x)|dx \equal{} \frac {2}{\pi}$

but I can't get the general case. Any help would be appreciated, thanks!

I also believe your answer is true. In general, evaluating the integral $ I_n \equal{} \frac{\iint_{S^{n\minus{}1}\times S^{n\minus{}1}}| \mathrm{u} \cdot \mathrm{v}| \, d\mathrm{u}d\mathrm{v}}{\iint_{S^{n\minus{}1}\times S^{n\minus{}1}} \, d\mathrm{u}d\mathrm{v}}$, I obtained $ \frac{\Gamma( \frac{n}{2})}{\sqrt{\pi} \, \Gamma(\frac{n\plus{}1}{2})}$ in the case of dimension $ n$. This gives

\[ \begin{array}{|c|ccccccccc|}
\hline
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & \cdots \\
\hline
I_n & 1 & \frac{2}{\pi} & \frac{1}{2} & \frac{4}{3\pi} & \frac{3}{8} & \frac{16}{15 \pi} & \frac{5}{16} & \frac{32}{35 \pi} & \cdots\\
\hline
\end{array}\]

ah, thanks! interesting answer. and actually, I think knowing the variance would also be helpful. So, with

$ E_n \equal{} \frac {\Gamma(\frac{n}{2})} {\sqrt{\pi} \Gamma(\frac{n\plus{}1}{2})}$

how would you evaluate the following integral?

$ I_n \equal{} \frac {\int\int_{S^{n\minus{}1} \times S^{n\minus{}1}} (|u \cdot v| \minus{} E_n)^2 du dv}  {\int\int_{S^{n\minus{}1} \times S^{n\minus{}1}} du dv}$

or just

$ \frac {\int\int_{S^{n\minus{}1} \times S^{n\minus{}1}} |u \cdot v|^2 du dv}  {\int\int_{S^{n\minus{}1} \times S^{n\minus{}1}} du dv}$

would be enough as well. Is there a straightfoward way to evaluating these integrals?

Note- my $ n$ is your $ d$, and the surface of the $ n$-dimensional ball is actually the $ (n\minus{}1)$-dimensional sphere in the usual convention.

The expected value of the square is downright easy- it's $ \frac1n$ by a symmetry argument. You can get the variance from that once you have the average. It's the absolute value in the original average that's ugly.

An alternate interpretation of the answer: that's the ratio $ \frac{V_{n\minus{}2}}{V_{n\minus{}1}}$, where $ V_k$ is the volume of the $ k$-dimensional unit ball. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=45050]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=26191]here[/url] for a calculation of these volumes.
Is there a more direct way of getting this as a ratio of volumes/areas of balls/spheres or something similar?


ν.δ.ο. η εξισωση  $ (p\minus{}1)!\plus{}1\equal{}p^m$ με $ m$ φυσικο και $ p>5$ πρωτο ειναι αδυνατη.

[quote="Athinaios"]ν.δ.ο. η εξισωση  $ (p \minus{} 1)! \plus{} 1 \equal{} p^m$ με $ m$ φυσικο και $ p > 5$ πρωτο ειναι αδυνατη.[/quote]

Mporei na kano kai lathos, alla auto den einai to theorima Liouville?

Oχι δεν κανεις λαθος. Γι αυτο εγραψα και οτι ειναι καπως γνωστη. Για οποιον ομως δεν το εχει υποψιν του πιστευω οτι ειναι ενα σχετικα καλο θεμα.

Λημμα: Αν $ q^m$ ειναι η μεγιστη δυναμη ενος πρωτου που διαιρει τον $ a \minus{} b$ και $ q^k$ η μεγιστη δυναμη του ιδιου πρωτου που διαιρει τον $ n$, τοτε $ q^{k \plus{} m}$ ειναι η μεγιστη δυναμη αυτου του πρωτου που διαιρει τον $ a^n \minus{} b^n$.

Ειναι $ 1 < (p \minus{} 1)! \plus{} 1 < (p \minus{} 1)!p \equal{} p! < p^{p \minus{} 1}$ (γιατι; ), αρα $ m < p \minus{} 1$. Αν ο $ m$ διαιρειτε με ολους τους πρωτους $ p_1, p_2, ..., p_j < p$, τοτε εχουμε $ m \plus{} 1 < p$ και ο $ m \plus{} 1$ δεν διαιρειτε με κανεναν πρωτο μικροτερο του $ p$ το οποιο δινει $ m \equal{} 0$, ατοπο. Αρα υπαρχει πρωτος $ q < p$ και προφανος $ q < p \minus{} 1$ που δεν διαιρει τον $ m$, οποτε απο το παραπανω λημμα, αν $ q^k$ ειναι η μεγιστη δυναμη του $ q$ που διαιρει τον $ p \minus{} 1$, τοτε $ q^k$ ειναι και η μεγιστη δυναμη του $ q$ που διαιρει τον $ p^m \minus{} 1 \equal{} (p \minus{} 1)!$, απο οπου προκειπτει οτι ο πρωτος $ q < p \minus{} 1$ δεν διαιρει το $ (p \minus{} 2)!$ το οποιο ειναι παλι ατοπο. Αρα η παραπανω εξισωση δεν μπορει να εχει λυση.

αλλη λυση,
[hide]
εχουμε αφου $ p>5 \equal{}> p\minus{}1 > \frac{p\minus{}1}{2} > 2$
αρα,
$ (p\minus{}1)^2 | (p\minus{}1)! \equal{} p^m\minus{}1 \equal{} (p\minus{}1)(p^{m\minus{}1}\plus{}p^{m\minus{}2}\plus{}...\plus{}1)$
δλδ,
$ p^{m\minus{}1}\plus{}..\plus{}1 \equiv 0 (mod p\minus{}1) \leftrightarrow m \equiv 0 (mod (p\minus{}1)) \equal{}> p\minus{}1 \leq m$(1)
ομως λογω της (1) παιρνουμε,
$ p^m\minus{}1 \geq p^{p\minus{}1} \minus{}1 > (p\minus{}1)!$. σχεση που ολοκληρωνει την αποδειξη.[/hide]


Find the sum of all prime numbers between $ 1$ and $ 100$ that are simultaneously $ 1$ greater than a multiple of $ 4$ and $ 1$ less than a multiple of $ 5$.

$ \textbf{(A)}\  118\qquad \textbf{(B)}\  137\qquad \textbf{(C)}\  158\qquad \textbf{(D)}\ 187 \qquad \textbf{(E)}\ 245$

[hide]Since the number must be 1 less than a multiple of 5, the units digit must be 9. Since it also must be 1 more than a multiple of 4, then we can represent the number as $4x+1$. We know that $x=2,7,12,17, \text{or } 22$, because these give units digits of 9. These x values give 9, 29, 49, 69, and 89, but only 29 and 89 are prime. The sum of these is 118. [b]A[/b][/hide]

I got 189 for some reason...
ooooh
59 doesnt work nvm... 130...
ok ok wow my bad... 29 + 89 = 118...


function
logarithms
combinatorics unsolved
combinatorics

[quote="galois"] consider all permutations of the first n integers.take a particular permutation.then the no of runs of the particular permutation is calculated as shown below:
example n=6
then let 1 3 2 4 6 5 be a permutation.
then 1->3 is increasing
3->2 is decreasing
2->4->6 is increasing
6->5 is decreasing
therefore the different monotone sequences are counted.and the no of runs of the above permutation is 4.similarly for others [/quote]
The horrible generating function I posted in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29440]permutations and monotonic sequences[/url] thread allows to show that, for $n\geq 4$, the number of permutations with even number of runs is the same as that with odd number of runs. I wonder whether anybody can find a nice bijection between the corresponding two sets.

I might be wrong, but it looks like tha parity of the number of runs depends only on the first two and the last two numbers. If the permutation begins with $a,b$ and ends with $c,d$, the number of runs is even iff $a<b,c>d$ or $a>b,c<d$. We can find a bijection by taking each permutation to the permutation in which the last two terms are reversed.

Is this wrong? :?

It's perfect, grobber. Thank you. Another question I cannot answer: is the sequence $U(n,k)$ of numbers of permutations with $k$ runs unimodal in $k$ for fixed $n$ (or, maybe, even $\log$-concave)? I computed $U(n,k)$ for $n\leq 8$ and it is true in that range...


trigonometry
series summation
algebra
Trigonometric Identities
IMO Shortlist
IMO Longlist

Prove the identity \[\sum\limits_{k=0}^n\binom{n}{k}\left(\tan\frac{x}{2}\right)^{2k}\left(1+\frac{2^k}{\left(1-\tan^2\frac{x}{2}\right)^k}\right)=\sec^{2n}\frac{x}{2}+\sec^n x\] for any natural number $n$ and any angle $x.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Well, at first, the problem has two typos (one from the original source and one that occured during the LaTeXing). Here is the correct formulation:

[color=blue][b]Problem.[/b] Prove the identity

$\sum\limits_{k=0}^n\binom{n}{k}\left(\tan\dfrac{x}{2}\right)^{2k}\left(1+\dfrac{2^k}{\left(1-\tan^2\dfrac{x}{2}\right)^k}\right)=\sec^{2n}\dfrac{x}{2}+\sec^n x$

for any natural number n and any angle x.[/color]

For the [i]solution[/i], denote $A=\tan^2\dfrac{x}{2}$. Then,

$A+1=\tan^2\dfrac{x}{2}+1=\dfrac{\sin^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}}+1=\dfrac{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}}=\dfrac{1}{\cos^2\dfrac{x}{2}}=\sec^2\dfrac{x}{2}$

and

$\dfrac{1+A}{1-A}=\dfrac{1+\tan^2\dfrac{x}{2}}{1-\tan^2\dfrac{x}{2}}=\dfrac{1+\dfrac{\sin^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}}}{1-\dfrac{\sin^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}}}=\dfrac{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}-\sin^2\dfrac{x}{2}}=\dfrac{1}{\cos x}=\sec x$.

Thus,

$\sum\limits_{k=0}^n\binom{n}{k}\left(\tan\dfrac{x}{2}\right)^{2k}\left(1+\dfrac{2^k}{\left(1-\tan^2\dfrac{x}{2}\right)^k}\right)=\sum\limits_{k=0}^n\binom{n}{k}\left(\tan^2\dfrac{x}{2}\right)^{k}\left(1+\dfrac{2^k}{\left(1-\tan^2\dfrac{x}{2}\right)^k}\right)$
$=\sum\limits_{k=0}^n\binom{n}{k}A^k\left(1+\dfrac{2^k}{\left(1-A\right)^k}\right)=\sum\limits_{k=0}^n\binom{n}{k}A^k+\sum\limits_{k=0}^n\binom{n}{k}A^k\dfrac{2^k}{\left(1-A\right)^k}$
$=\sum\limits_{k=0}^n\binom{n}{k}A^k+\sum\limits_{k=0}^n\binom{n}{k}\left(\dfrac{2A}{1-A}\right)^k=\left(A+1\right)^n+\left(\dfrac{2A}{1-A}+1\right)^n$
$=\left(A+1\right)^n+\left(\dfrac{1+A}{1-A}\right)^n=\left(\sec^2\dfrac{x}{2}\right)^n+\left(\sec x\right)^n=\sec^{2n}\dfrac{x}{2}+\sec^n x$,

and the problem is solved.

  darij


Prove exist $ a$ is belong to R satisfy $ \frac{1}{3}\leq\,$ {a^n} $ \leq\frac{2}{3}$ with every $ n$ is belong to $ N*$ :o

It's been posted:  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174077


lol I didn't think about that! ns  :)
try to find more solutins!
[hide="hint"]I used induction[/hide]
p.s: u forgot to prove that binary works!  ;)


let $ a,b,c\ge 0$, such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
Prove that : $ 9 \plus{} ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \plus{} 3abc\geq 6(ab \plus{} bc \plus{} ca)$

my Solutuion right or wrong
let $ a,b,c\ge 0$
CMR ([b]prove that[/b]) thì phải
$ \sum (\frac {a}{b \plus{} c})^3 \plus{} \frac {9abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq \sum \frac {a}{b \plus{} c}$

$ x \equal{} \frac {2a}{b \plus{} c}$
$ y \equal{} \frac {2b}{c \plus{} a}$
$ z \equal{} \frac {2c}{a \plus{} b}$

$ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 4$
..............................
please help me by pqr

q=xy+yz+zx
p=x+y+z
r=xyz
my solution
$ \sum (\frac {a}{b \plus{} c})^3 \plus{} \frac {9abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq \sum \frac {a}{b \plus{} c}$
$ < \equal{} > p^3 \plus{} 12r \plus{} 3pr\geq 16p$
$ < \equal{} > p^3 \plus{} (3r \minus{} 16)p \plus{} 12r\geq 0$

$ f(p) \equal{} p^3 \plus{} (3r \minus{} 16)p \plus{} 12r$
$ f'(p) \equal{} 3p^2 \plus{} 3r \minus{} 16$
$ f'(p) \equal{} 0 < \equal{} > p \equal{} \sqrt {(16 \minus{} 3r)/3}\ge \sqrt {13/3}$
$ \equal{} > f'(p) > 0 with p\in [3;4]$
$ \equal{} > Min f(p) \equal{} min f(3)$

If $ p\geq 4$
we have$ r\geq \frac {16p \minus{} p^3}{4p \plus{} 9}$

[quote="fesha3t"]can you help me?!
i think it's very easy if you Prove $ (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(a \plus{} b \plus{} c)^2\geq 27$
easy[/quote]

Try:$ a\equal{}1;b\equal{}c\equal{}\frac{1}{2}$

[quote="caubetoanhoc94"][quote="fesha3t"]can you help me?!
i think it's very easy if you Prove $ (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(a \plus{} b \plus{} c)^2\geq 27$
easy[/quote]

Try:$ a \equal{} 1;b \equal{} c \equal{} \frac {1}{2}$[/quote]

no , $ a^2\plus{}b^2\plus{}c^2\equal{}3$

[quote="fesha3t"][quote="caubetoanhoc94"][quote="fesha3t"]can you help me?!
i think it's very easy if you Prove $ (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(a \plus{} b \plus{} c)^2\geq 27$
easy[/quote]

Try:$ a \equal{} 1;b \equal{} c \equal{} \frac {1}{2}$[/quote]

no , $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$[/quote]
Okie.It's easy.
Setting:$ p\equal{}a\plus{}b\plus{}c;q\equal{}ab\plus{}bc\plus{}ca;r\equal{}abc$=>$ p^2\minus{}2q\equal{}3$
Inq
$ <\equal{}>p^2q \ge 27r$
$ <\equal{}>p^2(p^2\minus{}3) \ge 54r$
We have:
$ 54r \le 3p(5q\minus{}p^2)\equal{}3p(\frac{3}{2}p^2\minus{}\frac{15}{2})$
We need prove that:
$ p^2(p^2\minus{}3) \ge 3p(\frac{3}{2}p^2\minus{}\frac{15}{2})$
$ <\equal{}>(2p^2\minus{}3p\minus{}15)(p\minus{}3) \ge 0$
It's true by $ p \le 3$  :P


conics ellipse geometry calculus LaTeX USAMTS MATHCOUNTS

Just curious, but since like I was the only person in ARML without a TI-someting, I felt cheap. I guess I really want one.
Just list your opinions.

1) stop making multiple accounts
2) find a calculator with a full featured, extensible, object oriented programming language.  wait; i forgot. there isn't one
3) i was going to vote for nspire (quite an improvement in ram, and multiple shades of gray; great display/graphics), actually nspire cas if i could, but i accidentally voted other.  oops.
4) don't use the word "suck" lightly

Also: as a message to both you and mathand... : Don't make fake email addresses on Yahoo! and GMail upon will.  The emails are eventually going to be recycled, but fake emails are a gateway to spam, and there are soo many conditions that need to be met before an email can be recycled (spam filters improved, etc.).  If you want to create tons of email addresses, do it elsewhere.  There are MANY honest email providers out there, some even dedicated to junk mail.  Don't spam up the ones that were not intended to be that way.

the above: well said :)

I've had my 89T for.. 2.5 years now? I'm happy with it. I guess the TIN-spire (LMAO.. actually that was pretty lame) might be better with all the non-hacked grayscale and stuff (cough games) but I think it's waaay too big to carry around.. even for a graphing calculator. The stuff sticking out the sides to accommodate the larger screen are just ugly anyway :P

and about RAM.. you probably don't need too much of it unless you're going to install a buttload of apps (i.e. cabri geometry) or games (BOMBERMAN W00T). 180k on the 89T is enough.

sorta irrelevant conclusion: until you get to calculus, use your brain as a portable calculator and use mathematica (or free alternative like sage) at home.

i only have TI84 silver...
i want a TI89 or Nspire cuz it can do exact values

I guess since the TI-NSPIRE was just recently released, nobody's really got to use it that much. :|
The full potential of it hasn't been released yet. (Extra upgrades.)
mathcrazed said that he got a NSPIRE at mathcounts nats
He said it was amazing. (I don't really believe him) :wink:

NSPIRE CAS or just NSPIRE?

[quote="logic812"]NSPIRE CAS or just NSPIRE?[/quote]You DO know what a CAS is, right?  The only reason you would want a regular nspire instead is for backward compatibility and to be able to use it on whatever test which doesn't allow calculators with CAS.

I personally think you shouldn't worry about getting a graphing calculator right now.  I would wait a year or two first.  By then there would be more support (games :) ) as well as simply a reason to use it.  Don't get a graphing calculator because other people have one.  Get one when you are going to use it.  The calculator you have now should be more than enough.

I completely agree with undefined's conclusion: just use something like Sage (ooooh, it supports python scripts) or WIRIS when you can. (Or Mathematica at school)

I agree, but you everyone except me and Yushi had at least a TI-84 Silver...
I wonder what they needed those for...
(some people did nothing during team with a TI-89 in LAYING IN FRONT OF THEM!!! :!:  :censored: )
Yeah, but Mathematica sucks at our school... sighs.
(I don't even know how to operate it) :lol:
The only reason I put NSPIRE vs NSPIRE CAS was because with NSPIRE, you could use the
TI-84 keyboard, I've heard that the regular keyboard sucks.

[quote="logic812"]I agree, but you everyone except me and Yushi had at least a TI-84 Silver...
I wonder what they needed those for...
(some people did nothing during team with a TI-89 in LAYING IN FRONT OF THEM!!! :!:  :censored: )[/quote]Don't worry about that.  Just because they have one, doesn't mean you need to get a TI-89.  I guess this kinda shows how much use it has for you atm.[quote="logic812"]Yeah, but Mathematica sucks at our school... sighs.
(I don't even know how to operate it) :lol:[/quote]Mathematica is great (the greatest) software (it's built on Maxima I think), just unfortunately expensive.  How can you say it's bad if you don't know how to use it?[quote="logic812"]The only reason I put NSPIRE vs NSPIRE CAS was because with NSPIRE, you could use the
TI-84 keyboard, I've heard that the regular keyboard sucks.[/quote]Thus the reason I mentioned backward-compatibility.  The NSPIRE even has a full TI-84 emulation software, I think.

[quote] Mathematica is great (the greatest) software (it's built on Maxima I think),[b] just unfortunately expensive[/b]. How can you say it's bad if you don't know how to use it? [/quote]

If you win an award in USAMTS (http://www.usamts.org/) , you can get mathematica for free!!! so this is how it works, the people who received a gold medal get to pick first, and the remaining prizes get to be chosen be the silver medal people, and lastly the bronze medal people. i am gonna try to get gold, but most likely get silver, so i have a chance to get mathematica for free.  :lol:

I'm pretty annoyed now: both SAGE as well as Maxima have Emacs plugins, but no Vim plugin (well, Vim can have python syntax highlighting for SAGE files, but that's about it). :(

I should see if I can get a crack for the expired copy of Mathematica that we got from school :rotfl: jk

(oh yeah just to clear things up: TI-81(?)<85<86<83<83+<84<84silver<89<89titanium<NSPIRE err something like that imo)

I <3 my nspire. I like the nspire keyboard (would you call it that?) better than whatever other one comes with the nspire.

[quote="undefined117"]YES #H34N1, MY FELLOW SWEDE[/quote]lol, I don't have either so I should stay out of this discussion, shouldn't I?

[quote="undefined117"]maybe.. the nspire is becoming the windows vista of TI calculators :O[/quote]oh noes! :o

TI-89 Titanium, pwns the n-spire. (obviously)

Yet some (crazy, whacko) experts estimate that [i]eventually[/i] :wink: , the TI-NSPIRE would pwn the TI-89.
I wonder when that day will come.

Wow, I'm the 1000th viewer here!
I wonder why nobody ever polls or posts here?

[quote="logic812"]I wonder why nobody ever polls or posts here?[/quote]
Because some people on this forum talk like this:
[quote="logic812"]Wow, I'm the 1000th viewer here![/quote]

Nah, I'm just kidding (somewhat).  But don't feel offended, seriously.

Back on topic...:
But what do you mean by "experts estimate"?  If you mean that they'll make newer versions of the TI-NSPIRE, then of course they will be better than the TI-89 Titanium.  But no matter how much time passes, the fact remains that the TI-89 Titanium exceeds the TI-NSPIRE.
Hey... I just might buy a TI-89 Titanium in the next day or so (or perhaps I already have).  2.7 MB is pretty impressive for a non QWERTY-keyboard calculator.

You do realize the Casio (or HP? I forgot) has an SD card slot available so you can expand it.. I don't like them though, since they make loud beeps =P not good in a testing environment.

TI isn't always better: [url=http://en.wikipedia.org/wiki/Casio_ClassPad_300]The Casio Classpad[/url]

[quote="MustItAlwaysBeMe"][quote="#H34N1"]The 89 titanium has a faster proc. than teh nspire[/quote]What processor, exactly, does the NSpire have?  The 89 obviously has a Motorola 68K.  Also keep in mind that the NSpire has 32 MB (i think) of permanent flash memory. (but then again, we're talking about a calculator here :) )[/quote]
NSPIRE: ARM @ 150 MHz (unconfirmed)
TI-89 Ti: Motorola 68000 @ ≤16 MHz

Source:Wikipedia

Anyway, looking at this data, the NSPIRE seems faster than the 89 Ti up to this point.

Serious speed doesn't really matter. My TI can calculate fibonacci faster than eclipse (using basically the same program -- same for loop and iteration). That might be because the 43rd number overflows a long though.

[quote="MustItAlwaysBeMe"][quote="logic812"]I read some newspaper (online) article and some experts estimated that the NSPIRE would be a better calc overall than the 89 Ti[/quote]And what exactly is that supposed to mean? They [i]estimated[/i]?  How did they estimate?  And who were the "experts"?[/quote]

Here is someone who understands what I'm complaining about.  I have already expressed that I don't understand how the "experts" did this.

Anyways, have you decided which calculator to buy yet, logic812?  Because if you haven't, then you must really highly value our opinions...

Perhaps we should start a "TI-89 vs TI-NSPIRE" topic if you've already bought a calculator...

I kind of find it hard to buy a calc while I'm in Korea, but if I had a choice, I would defenitely (not :lol: ) buy the N-SPIRE.
Serioiusly, I think most of the people that have posted in this thread would agree with me, although 16 shades of gray sounds really appealing.  :(

Oh well...


A chord, perpendicular to the circle's radius, bisects the radius.  Find the ratio of the smaller area the chord splits the circle into to the larger area.

$ AIME$

[hide="solution"]
Assume wlog that $ r\equal{}2\sqrt{3}$. The radii that extent to the edge of the bisector-chord make an angle of $ 2\cos^{\minus{}1}\frac{1}{2}\equal{}\frac{2\pi}{3}$

We know that the area of the sector is $ \frac{1}{2}\cdot r^2(\theta^r \minus{} \sin \theta^r)$

$ \Rightarrow \frac{4\pi \minus{} 3\sqrt{3}}{8\pi \plus{} 3\sqrt{3}}$[/hide]

I thought you were only allowed to answer with integers between 0-999 on the AIME?

[hide="Solution w/o Trig"]
[asy]defaultpen(fontsize(12pt));
pair O = origin;
pair A = dir(150);
pair B = dir(90);
pair C = dir(30);
pair D = intersectionpoint(A--C,O--B);

draw(Circle(O,1));
draw(O--A);
draw(O--B);
draw(O--C);
draw(A--B);
draw(B--C);
draw(A--C);
draw(rightanglemark(O,D,A,3));

label("$A$",A,NW);
label("$B$",B,N);
label("$C$",C,NE);
label("$O$",O,S);
label("$D$",D,NE);[/asy]

Let the radius of $ \odot O$ be $ r$. From the diagram, $ OA \equal{} OC \equal{} r$. We are given that $ BD \equal{} OD \equal{} \frac{1}{2}r$. We also know that $ AD \equal{} DC$ because the radius of a circle bisects a chord that is perpendicular to it. Because $ AD \equal{} DC$, $ BD \equal{} DO$ and $ \overline{AC} \perp \overline{OB}$, quadrilateral $ OABC$ is a rhombus. It follows that $ AB \equal{} BC \equal{} r$.

Now we notice something interesting: in $ \triangle OAD$, the hypotenuse is twice that of one of the legs. This makes the triangle a $ 30\minus{}60\minus{}90$ right triangle. We have that $ AD \equal{} DC \equal{} \frac{\sqrt{3}}{2}r$ and $ \angle ODA \equal{} \angle ODC \equal{} 60^\circ$. Now we can find the area of the circular segment $ ABC$ and find the ratio that we desire.

$ [\text{circular segment }ABC] \equal{} \frac{1}{3}\cdot\pi r^2 \minus{} 2\cdot \frac{\frac{1}{2}r\cdot \frac{\sqrt{3}}{2}r}{2}\\
\equal{} (\frac{1}{3}\pi \minus{} \frac{\sqrt{3}}{4})r^2$

All the information we have now is adequate to find the ratio.

$ \frac{(\frac{1}{3}\pi \minus{} \frac{\sqrt{3}}{4})r^2}{\pi r^2 \minus{} (\frac{1}{3}\pi \minus{} \frac{\sqrt{3}}{4})r^2} \equal{} \frac{\frac{1}{3}\pi \minus{} \frac{\sqrt{3}}{4}}{\frac{2}{3}\pi \plus{} \frac{\sqrt{3}}{4}} \equal{} \boxed{\frac{4\pi \minus{} 3\sqrt{3}}{8\pi \plus{} 3\sqrt{3}}}$
[/hide]

In the actual AIME they ask you to find the remainder when $ a*b*c*d*e*f$ is divided by $ 1000$, where your answer is of the simplified form $ \frac {a \minus{} b\sqrt{c}}{d \plus{} e\sqrt {f}}$.  I didn't feel like typing that though so I just put the main portion of the problem: to find the ratio.


Functional Analysis
advanced fields
advanced fields unsolved

I'm looking for a simple proof of the following fact:

"Every weakly convergent sequence in $ l^1(\mathbb{N})$ converges strongly."

This should be quite elementary. My current proof seems to be much too complicated. Any suggestions?

I am too lazy to write down a proof of my own, but the standard proof can be found for example [url=http://books.google.com/books?id=bjiigCL7T5AC&printsec=frontcover&dq=martin+functional+analysis+schechter&lr=#v=onepage&q=&f=false]here[/url] at pp 194-195.

Thank you, that is the kind of proof I was looking for.


With all these competitions and challenges going around, lets have a sticky with the names of the contests and the leaderboard or winner of each contest.  Just an idea.

and a good idea at that.  I think we should too

I just came across this and think that it's a great idea. Could a moderator of this forum post a sticky that does this?

[quote="h78reg"]perhaps there should be a more challenging all-star contest for the top 3 of each contest. :)[/quote]

   If we do implement this idea it should be by age.


inequalities
induction
real analysis
real analysis unsolved

[url=http://bildr.no/view/509277][img]http://bildr.no/thumb/509277.jpeg[/img][/url]

Denoted by red line.

Thanks

I fully agree the way the proof is written in your book is misleading. Since the author of the problem is Claudiu Raicu, a good friend of mine, I feel obliged to render justice and clarify it.

We want to build that sequence of intervals inductively. Start with any nondegenerate $ I_0 \equal{} [a_0, b_0]$. Clearly $ f(I_0) \subseteq [0, \infty)$, since $ f$ is positively defined. (If your question was about the notation, in general $ f(X) : \equal{} \{f(x) \ \mid \ x \in X\}$).

Assume intervals $ I_k$, $ 0\leq k \leq n\minus{}1$ have been built, with the stated properties. Now take $ x_n \in \mathbb{Q} \cap I_{n\minus{}1}$ with $ f(x_n) \geq n$ (possible, since $ f$ is unbounded on any interval).
Then we take a neighborhood $ I_n \equal{} [a_n, b_n] \subseteq I_{n\minus{}1}$ of $ x_n$ such that for all $ x \in I_n$ to have $ f(x) \geq f(x_n)$ (possible because $ x_n$ is a local minimum). Putting the inequalities together, we also get $ f(x) \geq f(x_n) \geq n$, hence $ f(I_n) \subseteq [n, \infty)$.

Thus we have proven the induction step, and we can build and consider the sequence of intervals $ (I_n)_{n\geq 0}$. From here on, read from your book - it's correct.

[quote="mavropnevma"]I fully agree the way the proof is written in your book is misleading. Since the author of the problem is Claudiu Raicu, a good friend of mine, I feel obliged to render justice and clarify it.

We want to build that sequence of intervals inductively. Start with any nondegenerate $ I_0 \equal{} [a_0, b_0]$. Clearly $ f(I_0) \subseteq [0, \infty)$, since $ f$ is positively defined. (If your question was about the notation, in general $ f(X) : \equal{} \{f(x) \ \mid \ x \in X\}$).

Assume intervals $ I_k$, $ 0\leq k \leq n \minus{} 1$ have been built, with the stated properties. Now take $ x_n \in \mathbb{Q} \cap I_{n \minus{} 1}$ with $ f(x_n) \geq n$ (possible, since $ f$ is unbounded on any interval).
Then we take a neighborhood $ I_n \equal{} [a_n, b_n] \subseteq I_{n \minus{} 1}$ of $ x_n$ such that for all $ x \in I_n$ to have $ f(x) \geq f(x_n)$ (possible because $ x_n$ is a local minimum). Putting the inequalities together, we also get $ f(x) \geq f(x_n) \geq n$, hence $ f(I_n) \subseteq [n, \infty)$.

Thus we have proven the induction step, and we can build and consider the sequence of intervals $ (I_n)_{n\geq 0}$. From here on, read from your book - it's correct.[/quote]

Thank you very much!


For what integers $m > 1$ is there an ordering $a_{1}, a_{2}, ..., a_{m}$ of the numbers $1, 2, ..., m$, such that the sums $a_{1}, a_{1} + a_{2}, ..., a_{1} + a_{2} + \cdot\cdot\cdot + a_{m}$ leave different remainders when divided by $m$?

Are you sure that m should be among the ordered numbers?
EDIT: I thought there wo'nt be any solution then. That's false if we put $a_1=m$.

For n odd, there's no solution because $a_1+a_2+...+a_m\equiv m(m+1)/2\equiv 0\equiv m\equiv a_1 (mod m)$.

You could set $a_1=m$, then there should be no problem.


prove that for all $ a,b,c$ positive reals
$ \sum_{cyclic} (1 \plus{} \frac {a}{b} ) \ge 2 \plus{} \frac {2(a \plus{} b \plus{} c)}{(abc)^\frac {1}{3}}$

$ \sum_{cyc}(1 \plus{} \frac {a}{b}) \equal{} 2 \plus{} \sum_{cyc}(\frac {a}{b}) \plus{} \sum_{cyc}(\frac {a}{c})$
From AM-GM: {see proof, http://www.mathlinks.ro/viewtopic.php?t=117266 }
$ \sum_{cyc}(\frac {a}{b}) \ge \frac {a \plus{} b \plus{} c}{(abc)^\frac {1}{3}}$
$ \sum_{cyc}(\frac {a}{c}) \ge \frac {a \plus{} b \plus{} c}{(abc)^\frac {1}{3}}$

$ \sum_{cyc}(1 \plus{} \frac {a}{b}) \equal{} 2 \plus{} \sum_{cyc}(\frac {a}{b}) \plus{} \sum_{cyc}(\frac {a}{c}) \ge 2 \plus{} \frac {2(a \plus{} b \plus{} c)}{(abc)^\frac {1}{3}}$


limit
trigonometry
calculus
calculus computations

1.
$ 0 < U_n < 1$

$ cosU_n \equal{} U_n^n$

so $ U_n \equal{} \sqrt [n]{cos U_n}$

so $ lim U_n \equal{} 1$
2.
$ 0 < U_n < 1$

$ cosU_n \equal{} U_n^n$

so $ cos(lim U_n) \equal{} 0$

so $ lim U_n \equal{} \frac{\pi}{2}$

What is False and why ?

Use \lim and \cos.  Any argument that has a sequence of numbers in $ (0, 1)$ approaching $ \frac {\pi}{2}$ should be suspect.  If we assume $ U_n \to L \neq 1$ then $ (U_n)^n \to 0$ and so $ \cos U_n \to 0$, but $ \cos$ has no zeroes on $ [0, 1]$, a contradiction.  It follows immediately that if $ U_n$ converges, it must converge to $ 1$.  The argument by contradiction can actually be modified to show that really $ U_n \to 1$ (e.g., choose $ N$ such that $ (1 \minus{} \varepsilon)^n < \cos \frac{3}{2}$ to see that for $ n \geq N$ we must have $ U_n \in [1 \minus{} \varepsilon, n]$), or those inclined to intuitive rather than formal arguments could just draw the picture.