geometry
3D geometry
conics
parabola
calculus
integration
calculus computations

How do I solve questions of the following type:

Find the bounded volume between:
The plane z=0,
the plane x/4+y/2+z/4=1,
and the hyperbolic cylinder x=2y^2

I want to know what is the way to solve these kind of questions. How do I draw it? What are the steps for solving such a question?

In addition, since I have a big problem with drawing, I will be very graeful if someone can tell me how to draw a cube inside a spere. What are the steps?

Any kind of help will be extremely appreciated.

Thank you in advance.

I think this is correct.  Perhaps, someone else here can verify.

Step 1:  Find the region of the solid projected onto the xy plane (i.e. the z=0 plane).
When we set $ z \equal{} 0$, the plane $ x \plus{} 2y \plus{} z \equal{} 4$ intersects the $ z \equal{} 0$ plane on the line $ x \plus{} 2y \equal{} 4$.  The x and y intersects of this line is $ (4,0)$ and $ (0,2)$, respectively.
Also the trace of the cylinder $ x \equal{} 2y^2$ on the $ z \equal{} 0$ is the curve $ \displaystyle y \equal{} \pm\sqrt {\frac {x}{2}}$.
Next, we need to find the points of intersection of the curve $ \displaystyle y \equal{} \pm\sqrt {\frac {x}{2}}$, and the line $ x \plus{} 2y \equal{} 4$.
You will find that these points are $ (2,1)$ and $ (8, \minus{} 2)$.
Therefore, the region, $ R$ of the solid projected onto the xy plane (z=0 plane) is the bounded region between the parabola $ x \equal{} 2y^2$ and the line $ x \plus{} 2y \equal{} 4$.

N.B. The projected region of the solid onto the xy plane tells you the boundaries around the solid.  We also need to know what bounds the the roof of the solid.  For this solid, its height is determined by the plane $ x \plus{} 2y \plus{} z \equal{} 4$ or $ z \equal{} 4 \minus{} 2y \minus{} x$.

Step 2:  Find the volume using the triple integral $ \displaystyle\int\!\!\int\!\!\int_R dV$ over the region $ R$ which in this case is the area bounded by $ x \equal{} 2y^2$ and $ x \plus{} 2y \equal{} 4$.  Thus, the volume of the solid is given by
$ \displaystyle V \equal{} \int_{x \equal{} 0}^{x \equal{} 2}\int_0^{y \equal{} \sqrt {x/2}}\int_0^{z \equal{} 4 \minus{} 2y \minus{} x}dzdydx \plus{} \int_{x \equal{} 2}^{x \equal{} 4}\int_0^{y \equal{} (4 \minus{} x)/2}\int_0^{z \equal{} 4 \minus{} 2y \minus{} x}dzdydx \plus{} \int_{y \equal{} 0}^{y \equal{} \minus{} 2}\int_{x \equal{} 2y^2}^{x \equal{} 4 \minus{} 2y}\int_0^{z \equal{} 4 \minus{} 2y \minus{} x}dzdxdy$

Triple integral problems like this can be a bit tricky so the first thing you should always try to do is draw a picture.  It really does help.  If you can see what you are doing, you can get a feel for what your integrals need to look like.   The algebra is made a lot easier if you can see what you're doing so you have an intuitive sense of what is right before you even compute an integral.


calculus
integration
geometry
perimeter
inequalities
area of a triangle
Heron\u0027s formula

A triangle with integral sides has perimeter 8. The area of the triangle is

$\text{(A)} \ 2 \sqrt 2 \qquad \text{(B)} \ \frac{16}{9} \sqrt{3} \qquad \text{(C)} \ 2 \sqrt 3 \qquad \text{(D)} \ 4 \qquad \text{(E)} \ 4 \sqrt 2$

[hide]I think the only triangle with a perimeter of 8 is a 2,3,3 triangle.

Using Heron's formula, $\sqrt{4(2)(1)(1)}=2\sqrt{2}$.

So $\boxed{\text{A}}$[/hide]

[hide="Answer"]By triangle inequality, sides must be $(2,3,3)$. By Heron's, $A=\sqrt{4(2)(1)(1)}=2\sqrt{2}\Rightarrow \boxed{A}$.[/hide]

[hide]

the side is 3,3,2 which is an isosceles triangle . So the height is $\sqrt{3^2-1}=2\sqrt{2}$ and base = $2$ .
Hence the area is $2\sqrt{2}$
$\boxed{A}$
[/hide]

[hide]The sides must be 2, 3, 3, or otherwise it isn't a triangle.

Drawing the altitude, pythagorean theorem tells us that the altitude is $2\sqrt2$, so the area is $2\sqrt2(\frac{1}{2})(2)=2\sqrt2$. [/hide]


calculus
derivative
function
quadratics
calculus computations

let $ a,b,c$ be different reals and
\[ f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-a)(x-c)}{(b-a)(b-c)}+\frac{(x-c)(x-b)}{(a-c)(a-b)}\]

Determine $ f'(x)$

$ f(a) = f(b) = f(c) = 1$, but $ f(x)$ is quadratic in $ x$.  Hence $ f(x) = 1, f'(x) = 0$.


[quote="shobber"]$x>0,y>0$, prove:
\[(x+y)^n > x^{n-1}(x+ny)\]
for all $n \geq 2$.[/quote]

Binomial expansion shows more positive terms on the left.
Thats it

[hide]
the RHS is just the first two terms in the expansion of the LHS

so we just have to show that the other terms in the expansion of the LHS is $>0$ which is obviously the case since both $x$ and $y$ are positive

[/hide]

just an extension of Bernoulli's inequality right???


calculus
integration
calculus computations

[hide=" answer"]

$ \int_{0}^{3} \frac{3}{\sqrt{1\plus{}t}} \vec{i} \; dt \minus{}\int_{0}^{3} 9t^{2} \vec{j} \; dt \plus{} \int_{0}^{3} \frac{18t}{(1\plus{}t^{2})^{2}} \vec{k} \; dt$

For the first integral, let $ u \equal{} 1\plus{}t \iff du \equal{} dt$. Thus, we get:

$ \int^{3\plus{}1}_{0\plus{}1} \frac{3}{\sqrt{u}} \vec{i} \; du \equal{} 6u^{\frac{1}{2}}\vec{i} \; \Bigg|_{1}^{4} \equal{} (12 \minus{} 6)\vec{i} \equal{} 6\vec{i}$

We don't even need to bother with the second integral, since the multiple choice gives us the same coefficient in front of j for each answer.

For the third one, let $ u \equal{} 1 \plus{} t^{2} \iff du \equal{} 2t \; dt \iff \frac{du}{2} \equal{} t \; dt$ to get:

$ \int_{0^{2} \plus{} 1}^{3^{2} \plus{} 1} 9u^{\minus{}2} \vec{k} \; du \equal{} \frac{\minus{}9}{u} \vec{k} \; \Bigg|_{1}^{10} \equal{} \minus{}\left( \frac{9}{10} \minus{} \frac{9}{1} \right) \vec{k} \equal{} \frac{81}{10} \vec{k}$

Thus the answer is d.
[/hide]

this is solved..moderator can close my topic

PrimeFactorization, why are you so heavily invested in doing other people's homework for them?   :P

Because I'm bored. At least I showed my work, so hopefully this dude will learn, but meh. I don't really care.


algebra
polynomial
modular arithmetic
algebra proposed

If $ a_0,a_1,\dots,a_{50}$ are the coefficients of the polynomial
\[ \left(1\plus{}x\plus{}x^2\right)^{25}\]
show that $ a_0\plus{}a_2\plus{}a_4\plus{}\cdots\plus{}a_{50}$ is even.

[quote="Leon"]If $ a_0,a_1,\dots,a_{50}$ are the coefficients of the polynomial
\[ \left(1 \plus{} x \plus{} x^2\right)^{25}\]
show that $ a_0 \plus{} a_2 \plus{} a_4 \plus{} \cdots \plus{} a_{50}$ is even.[/quote]

Let $ P(x)\equal{}\left(1 \plus{} x \plus{} x^2\right)^{25}$. Then $ a_0 \plus{} a_2 \plus{} a_4 \plus{} \cdots \plus{} a_{50}\equal{}\frac{P(1)\plus{}P(\minus{}1)}{2}$ $ \equal{}\frac{3^{25}\plus{}1}{2}$

And, since $ 3^{25}\equal{}(\minus{}1)^{25}\equal{}\minus{}1\pmod 4$, we get that  $ \frac{3^{25}\plus{}1}{2}$ is even.

Q.E.D.


algebra
polynomial
calculus
derivative
algebra proposed

For all integers $n \geq 2$ and a complex number $c \in C$, the following equation:
$x^n + cx + c = 0$
has a root of modulus at most $2$.

[quote="Raluca"]If $\mid x_1\mid,...,\mid x_n\mid>2$ => $\mid x_1x_2...x_n\mid>\mid x_1\mid+...+\mid x_n\mid$ (see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60904 )
But $\mid x_1\mid+...+\mid x_n\mid \geq \mid x_1+...+x_n\mid$.
So  $\mid x_1x_2...x_n\mid>\mid x_1+...+x_n\mid$.
Using Viete relations we get $\mid c\mid>\mid c\mid$, which is false.[/quote]

R u sure? ;)
Viete relations look like:
$x_1+...+x_n = 0$
$x_1 x_2+...+x_{n-1} x_n = 0$
.......
$x_1 x_2...x_{n-2} + ..... = 0$
$x_1 x_2...x_{n-1}+...+x_2...x_{n-1}x_n = - x_1x_2...x_n$ or:
$\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}+1=0$
Don't worry there is no ... sugar :)

Heeey, I had already noticed my mistake, that's why I deleted my post ! :oops:

Actually, I solved another problem: $x^n+cx^{n-1}+c=0$

Anyway, thanks a lot for teaching me the Viete relations.  :lol:

No sugar at all, you were right!  :)

We can use

[u]A consequence of Lucas's root theorem*[/u]
If all complex roots of a polynomial in $\mathbb C[X]$ are represented inside a convexe closed curve, then so are all complex roots of its derivative (that is inside [b]the same [/b]convexe closed curve).

Denote $P(x)=x^n+cx+c$ with roots $x_1,...x_n$.
If $c=0$ the conclusion is obvious.
If $c\neq 0$, let $y=\frac{1}{x}$. We get the new equation
$cy^n+cy^{n-1}+1=0$
Let $f(y)=cy^n+cy^{n-1}+1$ with roots $y_1,...y_n$.
Suppose, by reductio, that $\mid x_1\mid,...,\mid x_n\mid>2$. Then
$\mid y_1\mid,...,\mid y_n\mid<\frac{1}{2}$, so $y_1,...y_n$ are inside a circle of radius $\frac{1}{2}$ and it follows that all roots of $f '(y)$ are inside the same circle.
But $f '(y)=0 \Longleftrightarrow (y=0 \mbox { or } y=\frac{1}{n}-1)$ and $\mid \frac{1}{n}-1\mid =1-\frac{1}{n}\geq \frac 12$, so this root is not inside the circle above, contradiction.

*See http://mathworld.wolfram.com/LucassRootTheorem.html for Lucas's theorem.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ A$ be a finite abelian group, and let $ \varphi : A \to \mathbb{C}^\times$ be a homomorphism which is not the trivial homomorphism ($ \varphi(x) \equal{}1$ for all $ x$). Prove that $ \sum_{a\in A} \varphi(a)\equal{}0$.

[i]Wlog[/i] we can assume that $ A$ is cyclic. Let $ n$ be the order of $ A$. If $ g$ is a generator of $ A$, it follows that $ \phi(g)$ is an $ n$-th root of unity. Hence, $ [\phi(g)]^{n} \equal{} 1$ and whence $ (\phi(g) \minus{} 1)([\phi(g)]^{n \minus{} 1} \plus{} [\phi(g)]^{n \minus{} 2} \plus{} \ldots \plus{} 1) \equal{} 1.$ Since $ \phi$ is not identically equal to $ 1$, we derive from the previous equation that $ \phi(g^{n \minus{} 1}) \plus{} \phi(g^{n \minus{} 2}) \plus{} \ldots \plus{} 1 \equal{} 0$. Clearly, as the exponents of the generator vary between $ 0$ and $ n \minus{} 1$, we [i]recover[/i] every single element of $ A$. DONE.

Okay, now do it without the "Abelian" condition...

@Omegatheo: $ \mathbb{C}^{\times}$ is the multiplicative group of the nonzero complex numbers. It doesn't matter how we denote the operation in $ A$.

  darij

$ \phi(A)$ is finite cyclic group

$ \sum \phi(a)\equal{}|ker\phi|*0\equal{}0$

$ \phi(b) \sum_{a \in A} \phi(a) \equal{} \sum_{a \in A} \phi(a)$.

(1/card(G))sum(phi(g)) is the number of times phi contains the unit representation.
Patrick


Imagine you sell those metallic digits used to number houses, locker doors, hotel rooms, etc. You need to find how many of each digit to ship when your customer needs to number doors/houses:

 - 1 to 100
 - 51 to 300
 - 1 to 2,000 with zeros to the left

Is there a formula where you can plug in the starting and ending numbers, and the digit number, and it will return a count of occurrences for that digit number?

I'm glad to see you followed my suggestion!

Let $ T_i(n)$ denote the number of times the digit $ i$ occurs when writing the numbers from $ 0$ to $ n$, with all the necessary zeroes on the left.  Note that it's enough to know how to compute this number, since to get an arbitrary range all you have to do is subtract.  It will be easiest to describe $ T_i(n)$ recursively instead of giving an explicit formula, as follows:

- If $ n \equal{} 10^k \minus{} 1$, then $ T_i(n)$ counts the number of times digit $ i$ is used when writing all $ k$-digit numbers down.  In total, $ k \cdot 10^k$ digits are written down, and each digit is used the same number of times, so $ T_i(10^k \minus{} 1) \equal{} k \cdot 10^{k\minus{}1}$.

- If $ n \equal{} a \cdot 10^k \minus{} 1, 1 \le a \le 9$, then the answer is almost $ ak \cdot 10^{k\minus{}1}$ except that we have to add the contribution of the first digit.  If $ i < a$, then the digit $ i$ is the first digit $ 10^k$ times and $ T_i(a \cdot 10^k \minus{} 1) \equal{} ak \cdot 10^{k\minus{}1} \plus{} 10^k$; otherwise, it's not used at all and $ T_i(a \cdot 10^k \minus{} 1) \equal{} ak \cdot 10^{k\minus{}1}$.  (This is the kind of thing that makes it annoying to write down an explicit formula.)

- If $ n \equal{} a \cdot 10^k \plus{} m, m < 10^k$, then the answer is $ T_i(a \cdot 10^k \minus{} 1) \plus{} T_i(m) \plus{} m \plus{} 1$.  The second and third terms describe the digits you get going from $ a \cdot 10^k$ to $ m$.

[quote="t0rajir0u"]
- If $ n \equal{} a \cdot 10^k \plus{} m, m < 10^k$, then the answer is $ T_i(a \cdot 10^k \minus{} 1) \plus{} T_i(m) \plus{} m \plus{} 1$.  The second and third terms describe the digits you get going from $ a \cdot 10^k$ to $ m$.[/quote]

Seems this is not quit true. For instance, n = 23, i = 1
$ T_i(n)$ = $ T_1(2 * 10^1 + 3)$ = 12 + 1 + 3 + 1 = 17

If $ n \equal{} a \cdot 10^k \plus{} m, m < 10^k$ then
  If i >= a then $ T_i(a \cdot 10^k \minus{} 1) \plus{} T_i(m) \plus{} m \plus{} 1$
  else               $ T_i(a \cdot 10^k \minus{} 1) \plus{} T_i(m)$

Actually it should be 13.
I think we should check for whether i < a at this case as well.
Please advise.


[quote="john lai"]If x*x + 2x - 35 = 0,
then x = ?[/quote]

[hide]
$x^{2}+2x-35=0$
By factoring it, you get $(x+7)(x-5)=0$

So then, you just have to make it:

$x+7=0$, so $x=-7$.

And for the other one, $x-5=0$, so $x=5$. Therefore, the two solutions for it are $\{-7, 5\}$.[/hide]


Well,we know that when the force is acted on tho object,it causes m*a in the opposite direction.So there also exists the force of gravity,but m*g is in the direction of this force.How can you explain this?

It looks like you're suffering from a common misconception about Newton's Third Law.  Say it with me now:

[b]Equal and opposite forces act on different objects.[/b]

That is to say, when you have an object interacting gravitationally with the Earth, the Earth exerts a force of $mg$ on the object, but the [i]object also exerts a gravitational force on the Earth.[/i]

Because the Earth is so large, this force is negligible, and so we usually do not consider it in calculations.

I just wanted to check your knowledge.I also know the answer.So if you want to answer the question,just answer it,OK?

Oh, sorry.  I thought you were asking a question, not posing a problem.   :oops:


calculus
integration
inequalities
geometry
real analysis
real analysis solved

Let $f \in {C^2([0,+\infty])}$ and $\int_0^{+\infty}|f(x)|dx<+\infty$. Prove the following inequality
$|f(0)|^2\leq2\int_0^{+\infty}|f(x)|dx\int_0^{+\infty}|f''(x)|dx$

I think I have somewhat of a solution, but it's rather intuitive and geometric :).

First of all, let's assume $f(0)>0$, for simplicity. We have to show that $\frac 12\cdot |f(0)|\cdot\frac{|f(0)|}{\int^{\infty}_0|f''(x)|dx}\le\int^{\infty}_0|f(x)|dx$.

We can look at the last inequality as being an inequality of areas. The RHS is the area between the graph of $f$ and $(Ox$, while the LHS is the area of a right triangle with legs $f(0)$ (we assumed $f(0)>0$) and $x_0=\frac{|f(0)|}{\int^{\infty}_0|f''(x)|dx}$. If we show that the segment with endpoints $A=(0,f(0)),B=(x_0,0)$ is below the graph of $f$, we are done.

In order to show that, it suffices to show that $\int^{\infty}_0|f''(x)|dx\ge\sup_x |f'(x)|$, or, in other words, if we fix $a$, we have to show that $\int^{\infty}_0|f''(x)|dx\ge |f'(a)|\ (*)$. There exists a sequence $a_n\to\infty$ s.t. $|f'(a_n)|<\frac 1n$ (because of the initial condition on $f$ of being $L^1$). Now take $n$ large enough so that $a_n>a$. We have $\int^{\infty}_0|f''(x)|dx\ge\int^{a_n}_a|f''(x)|dx\ge\left|\int^{a_n}_a f''(x)dx\right|=|f'(a)-f'(a_n)|$, which is $\ge||f'(a)|-|f'(a_n)||$. Making $n\to\infty$, we get $(*)$.

Is it even remotely Ok? :)

Nice solution, grobber. I think it is ok.Thanks


LaTeX
Asymptote
analytic geometry
graph theory

Does anyone know any freeware math rendering programs? I don't think a Google trek will be very useful, and asking is probably better...

Sorry, forgot to mention that I'm looking for a graph rendering program. I'm already using LaTeX.

What exactly do you mean by "graph?"  Do you mean Euclidean plane graphing with coordinates or graph theory with vertices and edges?  Asymptote is probably the best math drawing program, but I prefer to use Canvas to draw anything if I need to.

Do you mean something like this?

http://www.padowan.dk/graph/


geometry
circumcircle
incenter
inradius
trigonometry
quadratics
geometric transformation

Given a triangle $ABC$($BC$ not equal to $AB$) and its circumcircle. If $BD$ is the altitude of the triangle and two more circles ,touching the circumcircle internally, are drawn, one tangent to $BD,AD$ and other tangent to $BD,CD$.Prove that the other transverse common tangent(other than $AD$) of these two circles bisect $AC$ iff $a,b,c$ are in $A.P$. ($a,b,c$ are sides of the triangle $ABC$ )

The problem contains a typo and the conditions are not stated clearly. The last sentence should read:

Prove that the other common internal tangent (other than $BD$) of these two circles bisects $CA,$ iff the sides $a, b, c$ of the triangle are in an aritmetic sequence (A.P. = arithmetic procession).

The sides a, b, c being in an arithmetic sequence means that $b = \frac{a+c}{2}.$ Let O, I be the triangle circumcenter and incenter, R, r the circumradius and inradius, and let $(I_{1}),\ (I_{2})$ be the circles tangent to (O), BD, and to AD, CD, respectively. These circles are easy to construct. Let $T_{1},\ T_{2}$ be the tangency points of $(I_{1}),\ (I_{2})$ with (O). Let the circumcircle tangents perpendicular to CA meet the circumcircle tangent parallel to CA, on the opposite side of CA than the vertex B, at points $D_{1},\ D_{2},$ where $D_{1}$ is on the opposite side of BD than A and $D_{2}$ is on the opposite side of BD than C. The tangency points $T_{1},\ T_{2}$ are the external similarity centers of the circle pairs $(O),\ (I_{1})$ and $(O),\ (I_{2})$ and consequently, $D_{1}D$ passes through $T_{1}$ and $D_{2}D$ passes through $T_{2}.$ The centers $I_{1},\ I_{2}$ are then intersections of the circumcircle radii $OT_{1},\ OT_{2}$ with the bisectors of the right angles $\angle BDA,\ \angle BDC.$ In order to prove the problem proposition, we have to prove the following theorem:

[color=darkblue][b]Theorem:[/b] In any triangle $\triangle ABC$ with $\angle A,\ \angle C$ acute, the incenter I is the midpoint of the segment $I_{1}I_{2}$ and the inradius is equal to $r = \frac{r_{1}+r_{2}}{2},$ where $r_{1},\ r_{2}$ are radii of the circles $(I_{2}),\ (I_{2}).$[/color]

This theorem was proved using inversion for the special case, when the angle $\angle ABC = 90^\circ$ is right, in the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=60053]Common tangents[/url] (compare with the computational solution on [url=http://www.kalva.demon.co.uk/imo/imo69.html]Kalva homepage[/url]). In the general case, the inversion approach fails, so we resort to calculations. Let $S_{1},\ S_{2}$ be the tangency points of $(I_{1}),\ (I_{2})$ with CA and WLOG, assume $a < c,\ \angle A < \angle C.$ The larger $\angle C$ is acute, otherwise the circle $(I_{2})$ would not be inside of the circumcircle (O). Let B' be the midpoint of CA. Let a line through O parallel to CA meet the radii $I_{1}S_{1},\ I_{2}S_{2}$ at $Q_{1},\ Q_{2}.$ Using Pythagorean theorem for the right triangle $\triangle I_{1}Q_{1}O,$

$OI_{1}^{2}= I_{1}Q_{1}^{2}+Q_{1}O^{2}$

$(R-r_{1})^{2}= (r_{1}-OB')^{2}+(r_{1}-AD+AB')^{2}$

$(R-r_{1})^{2}= (r_{1}-R \cos B)^{2}+(r_{1}-c\ \cos A+b/2)^{2}$

This quadratic equation for the radius $r_{1}$ looks complicated, but it is actually quite easy to solve. Multiplying things out and substituting $b = 2R \sin B,\ c = 2R \sin C,$ the quadratic equation becomes

$r_{1}^{2}+2Rp\ r_{1}+4R^{2}q = 0$

$p = 1+\sin B-\cos B-2 \sin C \cos A$

$q = \sin^{2}C \cos^{2}A-\sin B \sin C \cos A$

First, we have to square-root the discriminant $D = 4R^{2}(p^{2}-4q^{2})$ (if we cannot do this, the approach is hopeless).

$p^{2}-4q^{2}= 1+\sin^{2}B+\cos^{2}B+4 \sin^{2}C \cos^{2}A$ $+$

$+\ 2 \sin B-2 \cos B-4 \sin C \cos A-2 \sin B \cos B-4 \sin B \sin C \cos A+4 \cos B \cos C \cos A$ $-$

$-\ 4\sin^{2}C \cos^{2}A+4 \sin B \sin C \cos A =$

$= 2(1-\cos B)+2(1-\cos B) \sin B-4(1-\cos B) \sin C \cos A =$

$= 4 \sin^{2}\frac{B}{2}\ (1+\sin (C+A)-2 \sin C \cos A) =$

$= 4 \sin^{2}\frac{B}{2}\ (1-\sin (C-A)) =$

$= 4 \sin^{2}\frac{B}{2}\ \left(\cos^{2}\frac{C-A}{2}-2 \cos \frac{C-A}{2}\sin \frac{C-A}{2}+\sin^{2}\frac{C-A}{2}\right) =$

$= \left[2 \sin \frac{B}{2}\ \left(\cos \frac{C-A}{2}-\sin \frac{C-A}{2}\right)\right]^{2}$

Thus we succesfully transformed the discriminant to a full square. Using $\sin \frac{B}{2}= \cos \frac{C+A}{2},$ etc,

$\sqrt{p^{2}-4q^{2}}= \pm\ 2 \left(\cos \frac{C+A}{2}\cos \frac{C-A}{2}-\cos \frac{C+A}{2}\sin \frac{C-A}{2}\right) =$

$= \pm\ (\cos C+\cos A-\sin C+\sin A)$

Using the root $r_{1}$ with the positive sign in front of $\sqrt{D}$ (the other root comes out negative),

$r_{1}= \frac{-2Rp+2R\sqrt{p^{2}-4q^{2}}}{2}=$

$= R(-1-\sin B+\cos B+2 \sin C \cos A+\cos C+\cos A-\sin C+\sin A)$

Substituting $r = R(\cos A+\cos B+\cos C-1),$ $s-a = \frac{b+c-a}{2}= R(\sin B+\sin C-\sin A),$ and $c = 2R \sin C,$

$r_{1}= r-(s-a)+c \cos A = r-AS+AD$

where S is the tangency point of the incircle (I) with CA and D the foot of the B-altitude. In exactly the same way, we can get

$r_{2}= r-(s-c)+a \cos C = r-CS+CD$

$r_{1}+r_{2}= 2r-(AS+CS)+(AD+CD) = 2r$

Since all 3 circles $(I),\ (I_{1}),\ (I_{2})$ are tangent to CA, it follows that I is the midpoint of $I_{1}I_{2}$ and the tangency point S of the incircle (I) with CA is the midpoint of $S_{1}S_{2},$ where $S_{1},\ S_{2}$ are tangency points of the circles $(I_{1}),\ (I_{2})$ with CA. Thus the above theorem is completely proved.

Let $P \in BD$ be the internal homothety center of $(I_{1}),\ (I_{2})$ (the intersection of their common internal tangents) and let their other common internal tangent different from BD meet CA at E. $DI_{1},\ DI_{2}$ bisect the right angles $\angle BDA \equiv \angle PDA,\ \angle BDC \equiv \angle PDC,$ while $EI_{1},\ EI_{2}$ bisect the angles $\angle PEA,\ \angle PEC.$ Therefore, both the angles $\angle I_{1}DI_{2}= \angle I_{1}EI_{2}= 90^\circ$ are right, the quadrilateral $I_{1}I_{2}DE$ is cyclic and its circumcenter is the midpoint I of $I_{1}I_{2}.$ Thus S (the incircle tangency point with CA) is the midpoint of ED and the condition $B' \equiv E$ (i.e., E is the midpoint of CA) is equivalent to S being the midpoint of B'D, which in turn is equivalent to

$AS = \frac{AB'+AD}{2}\ \Longleftrightarrow\ s-a = \frac{b+2c \cos A}{4}= \frac{b^{2}+b^{2}+c^{2}-a^{2}}{4b}\ \Longleftrightarrow$

$2b^{2}+2bc-2ab = 2b^{2}+c^{2}-a^{2}\ \Longleftrightarrow 2b(c-a) = (c+a)(c-a)\ \Longleftrightarrow b = \frac{a+c}{2}$ (because by the problem conditions, $a = BC \neq AB = c$).

[quote="yetti"]In order to prove the problem proposition, we have to prove the following theorem:

[color=darkblue][b]Theorem:[/b] In any triangle $\triangle ABC$ with $\angle A,\ \angle C$ acute, the incenter I is the midpoint of the segment $I_{1}I_{2}$ and the inradius is equal to $r = \frac{r_{1}+r_{2}}{2},$ where $r_{1},\ r_{2}$ are radii of the circles $(I_{2}),\ (I_{2}).$[/color].[/quote]

Let $Q_{1},\ S_{1}$ be the tangency points of the circle $(I_{1})$ with the lines BD, AC and let $Q_{2},\ S_{2}$ be the tangency points of the circle $(I_{2})$ with  the lines BD, AC. Then the lines $Q_{1}S_{1},\ Q_{2}S_{2}$ meet at the incenter I of the $\triangle ABC$ (see below). Since $\angle S_{1}S_{2}I \equiv \angle DS_{2}Q_{2}= 45^\circ = \angle DS_{1}Q_{1}\equiv \angle S_{2}S_{1}I,$ the triangle $\triangle IS_{1}S_{2}$ is isosceles right. Let T be the midpoint of its base $S_{1}S_{2}.$ Then $r_{1}+r_{2}= DS_{1}+DS_{2}= 2 IT = 2r.$

It is sufficient to show that $Q_{1}S_{1}$ cuts the internal bisector of the angle $\angle ABC$ at the triangle incenter I. The proof is similar to the proof of the lemma in the thread [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=728293]Concyclic points with triangle incenter[/url], except the circles $(O),\ (I_{1})$ are tangent internally, so that we use an internal bisector of the $\angle ABC,$ let $K \in CA$ be its foot. Let $Q_{1}S_{1}$ meet BK at J (we do not know that this is the incenter) and let  $Q_{1}S_{1}$ meet AB at $P_{1}.$ By Menelaus theorem for the right triangle $\triangle ABD$ cut by the transversal $P_{1}Q_{1}S_{1},$

$\frac{P_{1}B}{P_{1}A}= \frac{Q_{1}B}{Q_{1}D}\cdot \frac{S_{1}D}{S_{1}A}= \frac{Q_{1}B}{S_{1}A}$

because $Q_{1}D = S_{1}D.$ By Menelaus theorem for the triangle $\triangle ABK$ cut by the same transversal $P_{1}JS_{1}\equiv P_{1}Q_{1}S_{1},$

$\frac{JB}{JK}= \frac{S_{1}C}{S_{1}K}\cdot \frac{P_{1}B}{P_{1}A}= \frac{S_{1}A}{S_{1}K}\cdot \frac{Q_{1}B}{S_{1}A}= \frac{Q_{1}B}{S_{1}K}= \frac{Q_{1}B}{AK-AS_{1}}$

If a = BC, b = CA, c = AB are sides of the $\triangle ABC,$ then $AK = \frac{bc}{c+a}.$ By [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=67398]Purser theorem[/url] for the $\triangle ABC$ with the circumcircle (O) and with the circle $(I_{1})$ internally tangent to (O),

$c \cdot (b-AS_{1}) = c \cdot CS_{1}= a \cdot AS_{1}+b \cdot BQ_{1}$

$bc-(c+a) \cdot AS_{1}= b \cdot BQ_{1}$

$\frac{JB}{JK}= \frac{(c+a) \cdot BQ_{1}}{bc-(c+a) \cdot AS_{1}}= \frac{c+a}{b}$

Thus $Q_{1}S_{1}$ cuts BK in precisely the same ratio as the internal bisectors of the angles $\angle BCA,\ \angle CAB,$ which means that $J \equiv Q_{1}S_{1}\cap BK \equiv I$ is the incenter of the $\triangle ABC.$


algebra polynomial articles AMC AIME LaTeX

[code]
%%blah blah blah
\begin{document}

%% \begin{titlepagehack}
 \pagenumbering{roman}
 \thispagestyle{empty}
 \begin{center}
  \begin{tabular}{||c||}
   \hline\hline \\
   \LARGE American Invitational Mathematics Examination\\\\
   \Large (Mock AIME 2007IV)\\\\
   \textsc{By Adeel Khan}\\
   \textsc{\today}\\\\
   \hline\hline
  \end{tabular}
 \end{center}
%% \end{titlepagehack}
[/code]

That's from something else but it's the same thing. (I can't find the TeX version of the polynomials paper right now :()

But stevem's way looks like it's probably better. ;)

[quote="rnwang2"]I'm trying to make a title page with just a title, my name, and a date all in a huge box, like the "A Few Elementary Properties of Polynomials" article in the Student Papers section. How do you do it?[/quote]

or you can just use MSWord...

Um, let me put it this way: LaTeX is for typesetting. Word is for word processing. The two do not overlap. :)

Anyway, here's my hack which permits infinite customization of the title page. You can see an example on my paper immediately after chess's paper:
[code]
%Title Page Hack
\pagenumbering{roman}\thispagestyle{plain}
\begin{center}\vspace{-.2in}
\begin{tabular}{|c|}
\hline \\
\Huge Predicting Improper Fractional Base\\
\Huge Integer Characteristics\\\\
\Large Billy Dorminy\\\\
\Large Sola Fide School\\\\
\Large Mathematics\\\\
\Large Senior Division\\\\
\hline
                   \end{tabular}                                \end{center}
\pagebreak[/code]

I'm sure it could be cleaned up, but you get the idea.

[quote="stevem"]Maybe the author of the document will say how they did it, but meanwhile you could replace \maketitle with
\fbox{\parbox{10cm}{\maketitle}}
replacing 10cm with whatever is suitable.

For example:[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\begin{document}
\title{My title goes here}
\author{My name goes here}
\fbox{\parbox{10cm}{\maketitle}}
\clearpage
Text starting on a new page
\end{document}[/code][/quote]

Oh this looks great! But how do you change the date? I want it to be "Dec 12, 2006".

Under \title{...}, put \date{Dec Whatever, Whenever}.


algebra
polynomial
calculus
integration
complex analysis

I believe the following is true although considering my solution to the one with $ n\equal{}2$ has holes, I'd prefer to put this one in a new thread.  Hope that's ok.

For a suitably chosen determination of $ \sqrt[n]{z}$ with $ C(r)$ encompassing $ a$ and $ b$, show:

$ \oint_{C(r)}\sqrt[n]{(z\minus{}a)(z\minus{}b)^{n\minus{}1}}dz\equal{}\minus{}\frac{\pi i(n\minus{}1)(a\minus{}b)^2}{n^2}$

I think this can be expressed in full generality:

For any polynomial of degree $ n$:

$ P_n(z) = c(z - z_0)(z - z_1)...(z - z_n)$

I believe the following expression is true:

$ \oint_{D_k}\sqrt [n]{P_n(z)}dz = \frac { - \pi i e^{2k\pi i/n}\sqrt [n]{c}}{n}\left\{(z_0^2 + z_1^2 + ... + z_n^2) - \frac {1}{n}(z_0 + z_1 + ... + z_n)^2\right\}$

where $ D_k$ is any closed contour around disjoint branch $ k$ of the n'th root  of  $ P_n(z)$ such that $ D_k$ encloses all the roots of $ P_n(z)$.  Note the factor of $ e^{2k\pi i/n}$ and $ \sqrt[n]{c}$ is the principal n'th root of c.

Here's an example:  The first plot below is the imaginary multi-sheet for $ \sqrt [3]{(z + 1)(z - 1)(z - 2)}$.  The three "disjoint" branches are color coded.  I choose the principal branch (red), and form the yellow contour encircling all the zeros.  Then:

\begin{align*} \oint_{Yellow}\sqrt [3]{(z + 1)(z - 1)(z - 2)}dz & = - \frac {\pi i}{3}\left\{(1 + 1 + 4) - 1/3( - 1 + 1 - 2)^2\right\} \\
& = - 14 \pi i/9\approx - 4.887i \end{align*}

Which agrees with a numerical integration over the yellow contour (real and imaginary sheets).

[edit]  Added factor of $ e^{2k\pi i/n}$.

[img]http://img205.imageshack.us/img205/9073/cubicsheetsuc4.jpg[/img]

I think this is so cool. :)


inequalities
geometry
circumcircle
complex numbers
rearrangement inequality

Let $a,b,c$ be complex numbers such that $|a|=|b|=|c|=r$.

Prove that
\[|a-b|\cdot |b-c|+|c-a|\cdot |a-b|+|b-c|\cdot |c-a|\le 9r^{2}\]

[hide="Solution"]
Interpret it geometrically, with $z_{a},z_{b},z_{c}$ the complex numbers representing vertices of a triangle with sidelengths $a,b,c$ denoted in the usual way, and with circumradius $R$ (instead of the lower-case $r$ in the problem). We can use the relation:
\[9 OG^{2}= 9 R^{2}-(a^{2}+b^{2}+c^{2}) \]and the rearrangement inequality (or equivalently, $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\geq 0$) , giving:

\[ab+bc+ac \leq a^{2}+b^{2}+c^{2}=9R^{2}-OG^{2}\leq 9R^{2}\]
with equality in both cases when the triangle is equilateral, i.e.
\[\arg( z_{c})= \arg(z_{b})+\frac{2 \pi}{3}= \arg(z_{a})+\frac{4 \pi}{3}\]
[/hide]


Find the largest positive integer n such that 1 + 2 + 3 + · · · + n^2 is divisible by
1 + 2 + 3 + · · · + n.

[hide="hint "]use the closed form for those sums[/hide]

$\frac{ \frac{n(n+1)(2n+1)}{6} }{ \frac{(n)(n+1)}{2} }=\frac{2n+1}{3}$, so it is true for all positive $n \equiv 1 \mod 3$...are you sure that the problem is correct, and elemennop, are you sure that the $\sum k^2$ is on the bottom, as it stands, that should be on the top, perhaps the problem is written wrong, and that is the intended problem...at least it seems that way

I think you may have the wrong summation. It is $\sum_{i=1}^{n^2}i$, not $\sum_{i=1}^n i^2$.  I also got Elemennop's answer, but it looks...small.[hide]\[ 1+2+\ldots+n^2=\frac{(n^2)(n^2+1)}{2}, \] \[ 1+2+\ldots+n=\frac{(n)(n+1)}{2}. \] Since $\frac{n}{2}$ divides both terms, we only need $(n+1)$ to divide $n(n^2+1)$. From division, we have \[ n^3+n=(n+1)(n^2-n+2)-\frac{2}{n+1}. \] So $n+1$ needs to divide $2$, and the largest such $n$ is $1$.[/hide]

my solution is quite similar to towerfreak2006's:
[hide]
as he pointed out $n+1$ needs to devide $n(n^2+1)$. since $\gcd(n,n+1) = 1$, $u: = n+1$ needs to devide $n^2+1 = (u-1)^2 + 1= u^2-2u+2$. so $u$ must devide $2$. therefore the greatest $n$ is $1$
[/hide]

haha yea nice solution, i have a similar one but here it is

[hide]

using sum of finite arithmetic sequence, we have:

\[ \frac{{\frac{(n^2+1)(n^2)}{2}}{\frac{(n+1)(n)}{2}} = \frac{n^3+n}{n+1} \]

here we note that:

\[ n^2 - n + 1 \le \frac{n^3+n}{n+1} < n^2 - n + 2 \]

since $\frac{n^3+n}{n+1}$ must be an integer, and since $n^2 - n + 1$ and $n^2 - n + 2$ differ by exactly 1, this implies that:

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

$(n^2 - n + 1)(n+1) = n^3 + 1 = n^3 + n$

hence $n=1$

[/hide]

[hide]Let $p=\frac{1}{2} \cdot n^2(n^2+1)$ and $q=\frac{1}{2}n(n+1)$. Then $\frac{p}{q}=\frac{n^2(n^2+1)}{n(n+1)}=\frac{n^3+n}{n+1}=n^2-n+1-\frac{2}{n+1}$ by synthetic division. The largest integer $n$ such that $(n+1)|2$ is $n=\boxed {1}$.[/hide]

EDIT: Yeah, it's basically the same as towersfreak2006's solution.

Yea, my solution is similar to towerfreaks. I guess the tricky part is noticing it's the sum of the first $n^2$ integers, not the sum of the squares of the first $n$ integers.

By the way, which year/test was this?

Simply apply the rule of the remainder of a polynomial divited by a first degreeded polinomial (the remainder will be $-1((-1)^2+1)$


probability
search
probability and stats

1. Suppose 2n customers stand in line at a box office, n with $ \$$5 bills and n with $ \$$10 bills. Suppose each ticket costs $ \$$5, and the box office has no money initially. What's the probability that none of the cutomers has to wait for change?

Ans: $ ( \binom{2n}{n}  \minus{}  \binom{2n}{n\plus{}1} ) / \binom{2n}{n}$

But why is that?

2. Prove that

$ \sum_{k\equal{}0}^{n}{\binom{n}{k}}^{2} \equal{} \binom{2n}{n}$

2) is more evident when written as $ \sum_{k\equal{}0}^{n}\binom{n}{k}\binom{n}{n\minus{}k}\equal{}\binom{2n}{n}$. Both sides are equal to the coefficient of $ x^n$ in $ (1\plus{}x)^n (1\plus{}x)^n\equal{}(1\plus{}x)^{2n}$.

[quote="mlok"]2) is more evident when written as $ \sum_{k \equal{} 0}^{n}\binom{n}{k}\binom{n}{n \minus{} k} \equal{} \binom{2n}{n}$. Both sides are equal to the coefficient of $ x^n$ in $ (1 \plus{} x)^n (1 \plus{} x)^n \equal{} (1 \plus{} x)^{2n}$.[/quote]

I'm still confused...

I understand how $ \binom{2n}{n}$ is the coefficient for $ x^{n}$ in $ (1\plus{}x)^{2n}$, but how is $ \sum_{k \equal{} 0}^{n}\binom{n}{k}\binom{n}{n \minus{} k}$ the coefficient for $ x^{n}$ in $ (1\plus{}x)^{n}(1\plus{}x)^{n}$?

What's the coefficient of $ x^n$ in the product $ \left(\sum_{i \geq 0} a_i x^i \right) \cdot \left(\sum_{j \geq 0} b_j x^j \right)$?  Also, this question has appeared numerous times on the forum (including once very recently) -- a search for "Vandermonde" will bring up some examples.

For the first question, this is just one combinatorial interpretation of the Catalan numbers.  Numerous proofs are available either on the forum or on the internet.


Ερχεται Θαλης-Ευκλειδης-........
Δινoνται τα τριγωνα $ ABC$ των οποιων οι διάμεσοι $ BZ$ και $ CE$ τέμνονται κάθετα στο $ H$ .Oι διάμεσοι $ BD$ και $ CU$ του τριγωνου $ BHC$ τέμνονται στο $ K$ και εστω $ a$ η οξεία γωνια τους.
Nα δειχθει οτι τα συνημιτονα των γωνιων $ A$ και $ a$ εχουν κοινό ελαχιστο το οποιο να βρεθει

Ωραία άσκηση !  :)
Έστω ότι η $ CU$ τέμνει την $ AC$ στο $ U_1$ και η $ BD$ τέμνει την $ AB$ στο $ D_1$ . Τότε αφού οι ευθείες $ CU_1,BD_1,AM$ συντρέχουν (όπου $ M$ το μέσο της $ BC$ ) από το θεώρημα Ceva έχουμε ότι $ U_1D_1//BC$ και άρα $ \angle{AU_1D_1}\equal{}\angle{B}$ και $ \angle{AD_1U_1}\equal{}\angle{C}$ και άρα $ \angle{DKC}\equal{}\angle{A}$ άρα αφού είναι ίσες θα έχουν κοινό συνημίτονο .

Τώρα έχουμε ότι $ BH^2\plus{}HC^2\equal{}BC^2$ και ;ara  $ \frac{4BZ^2\plus{}4CE^2}{9}\equal{}BC^2$ και από το θεώρημα διαμέσων και μετα΄τις απαραίττες απλοποιήσεις
$ AB^2\plus{}AC^2\equal{}5BC^2$ . Τώρα από το νόμο συνημιτόνων έχουμε : $ \cos A\equal{}\frac{AB^2\plus{}AC^2\minus{}BC^2}{2AB\cdot AC}\equal{}\frac{2BC^2}{AB\cdot AC}$

Τώρα από την ανισότητα ΑΜ-ΓΜ έχουμε $ AB\cdot AC\leq\frac{AB^2\plus{}AC^2}{2}\equal{}\frac{5BC^2}{2}$  άρα το ελάχιστο είναι $ \min\left\{\cos A\right\}\equal{}\frac{4}{5}$

Πολύ ωραία λυση Sil .Εγω την εχω με Ν.συνημιτονων η με διανυσματα .Προεκυψε παρατηρώντας τυχαια δυο γνωστες ασκησεις.


probability counting derangement factorial

[quote="sirorange"]Suppose that 9 people checked their hats at a coat room at a banquet hall. each received a receipt, but the person in the coat room lost the receipts. if each hat is given back randomly to each owner, what is the probability that no one will receive their own hat?[/quote]
[hide]
$\frac{!9}{9!}=\frac{133496}{362880}=\frac{16687}{45360}$[/hide]

By looking at smaller cases, if there are 2 people, then the probability that they both get the wrong hats is $\frac12$. For 3 people, it's $\frac13$. For 4, $\frac14$. This leads me to believe that it's $\frac{1}{10}$ probability for 10 people, although this is not proof.
Edit: looks like I might be wrong.

What you are asking for is the [url=http://mathworld.wolfram.com/Derangement.html]number of derangements[/url]. :)

so you'd just find the number of sets of 1,2,3,4,5,6,7,8,9 that don't have any numbers in order and that is the numerator. Is the numerator 362,880 (nine factorial)?

 :clap:  :flex:

[quote="DanK"]Actually, I was thinking the fraction would be $\frac{8!}{9!}$, which is $\frac{1}{9}$

Because for the 1st person, they have a $\frac{8}{9}$ chance of not getting their hat, then $\frac{7}{8}$ for the next person, and so on, down to $\frac{1}{2}$ for the 8th person, leaving the last person with a probability of 1 for not getting their hat, which doesnt change the answer at all.  Then it's all those fractions multiplied together, giving $\frac{8!}{9!}$.  Everything in that cancels out except the 9 in the denominator, so you get $\frac{1}{9}$[/quote]
I don't think the probabilities continue like that. Becuase for the first person, if he takes the second person's, then the second person has a 100% chance of not getting his own.

Good point.  So, we need to include the fact that someone could have their hat chosen before they get one...


let $ p$ and $ q$ be odd prime numbers such that $ q |2^p\minus{}1$ prove that $ 2p |q\minus{}1.$

[quote="tchebytchev"]let $ p$ and $ q$ be odd prime numbers such that $ q |2^p \minus{} 1$ prove that $ 2p |q \minus{} 1.$[/quote]
Since $ q\minus{}1$ is even and $ p$ is odd,it is sufficiently to prove that $ p| q\minus{}1$.Now let $ d$ be the order of $ 2$ mod $ q$.Then $ q| p$,hence either $ d\equal{}p$, or $ d\equal{}1$.
Second case is impossible.So $ d\equal{}p$.But from Fermat's Little Theorem we know that $ 2^{q\minus{}1}\equiv 1(mod q)$,hence $ p\equal{}d|q\minus{}1$,and we are done.


Bon, je pensais que ce serait une bonne idée de créer ce "topic" pour ceux qui aiment discuter en français...

Je suis resté sidéré Arne quand vous  disiez que le Net en Afrique du sud est très lent !
L'image que j'ai de ce pays est un pays developpé dans un continent à 95% sous developpé.
Est ce que vous pouvez nous dire plus sur ce pays?

Oui, c'est vrai que l'Afrique du Sud est plus developpé que le (la?) plupart des pays Africains, mais cela ne signifie pas qu'il y a une bonne connection internet dans le pays entier... J'ai eu des problèmes avec ça moi-même... Et je sais qu'il y a des élèves dans le programme d'entrainement qui ont beaucoup de difficultés avec le net.

premierement je te remercie pour cette criation d'espace de discusion en français surtout que l'anglais n'est que notre 3 langues aprés l'arabe et le français . :)  et deuxiemement je veux savoir si tu es africain (sud) ou du belgique ;)

Ben, je suis Belge, mais j'ai vécu dans l'Afrique du Sud pendant l'année passée. Maintenant je fais mes études en Belgique :)

Et je dois vraiment travailler à mon français... Mon français était beaucoup plus mieux il y a un an, mais on ne parle pas vraiment le français dans l'Afrique du Sud...

Bonjour, moi aussi je suis Belge
Pas comme Arne, je n'ai rien à faire avec l'Afrique, et c'est justement pour ça que je veux être mieux informé.
J'espère que je ne vous dérange pas d'écrire ici.

On dit parfois que de tous les pays Africains, la Tanzanie et la Botswana sont les plus calme au niveau politique.  ça m'étonnait, parce que je pensais que ce serait l'Afrique du sud?  Evidemment, il y a une difference entre être le plus
riche et en paix, mais est-ce que ceci veut dire que l'Afrique du sud est encore  instabile?

Comme Arne aussi, je parle néerlandais, et français est ma deuxième langue, mais je écris beacoup plus en anglais.  Donc,  tout le monde peut me corriger si on veut.

[quote="akhrif"]merci d avoir ecrire en francais
 j'ai recement vu ce site et je veux savoir comment je peux avoir plu d'étoiles??[/quote]
ça depend du nombres de tes messages  ;)

ça fait vraiment plaisir de voir des gens qui participe dans cette discussion en francais


Prove that there exists a  triangle which can be  devided  into 2050  equal  triangles.

http://www.mathlinks.ro/Forum/viewtopic.php?t=30953

the topic proves that if a number can be written as a sum of squares, its possible.

but 2050 = 41*50, where 41 and 50 are both possible, so by theory 2050 is possible.


calculus
integration
derivative
logarithms
function
calculus computations

Can someone please help me solve (or give me a hint)

$ \frac{d^2D}{dt^2} \equal{} \frac{1}{D^2}$?

It comes from a mechanics question about falling bodies

thanks

$ \frac{d^2D}{dt^2}\equal{}\frac{1}{D^2}$

$ \Leftrightarrow\frac{dD}{dt}\frac{d}{dD}\left(\frac{dD}{dt}\right)\equal{}\frac{1}{D^2}$. Let's denote $ \frac{dD}{dt}\equal{}\dot{D}$, then this can be rewritten by
$ \dot{D}d(\dot{D})\equal{}\frac{1}{D^2}dD$ as formly.

I'll call that variable $ y$; any version of "$ d$" is a bad name to be doing calculus with. This is one-dimensional motion in an outward-pointing inverse square field. It's not gravity, but two similarly charged particles repelling each other works as a model.

Think physically. Energy is conserved: $ \frac12(y')^2 \plus{} P(y) \equal{} E$, where $ E$ is the total energy and $ P(y) \equal{} \int \minus{} \frac1{y^2}\,dy \equal{} \frac1y$ (we get to choose a constant of integration, and putting the zero-energy point at $ \infty$ is nicest).
Now $ (y')^2 \equal{} 2E \minus{} \frac {2}{y}$, so $ y' \equal{} \pm\sqrt {2E \minus{} \frac2y}$. That's separable, so $ \int \frac {dy}{\sqrt {2E \minus{} \frac2y}} \equal{} \pm\int dt$.
Now, solving that integral? It's ugly. Expect the answer to involve an inverse hyperbolic function. Then, of course, we invert it to get $ y$ in terms of $ t$.

As a side note, it's pretty easy to prove that energy formula with calculus manipulations. Differentiating $ \frac12(y')^2\plus{}P(y)$ with respect to $ t$, we get
$ \frac{dy}{dt}\cdot\frac{d^2 y}{dt^2}\plus{}\frac{dP}{dy}\cdot\frac{dy}{dt}$
$ \equal{}\frac{dy}{dt}\cdot\left(\frac{d^2 y}{dt^2}\minus{}\frac1{y^2}\right)$
That inner part is zero by the equation, so the whole derivative is zero and $ \frac12(y')^2\plus{}P(y)$ is constant.

hello, Mathematica
gives the following answer
$ \text{Solve}\left[\left(\frac{y(t)
   \sqrt{c_1\minus{}\frac{2}{y(t)}}}{c_1}\plus{}\frac{\log \left(2 y(t)
   \left(\sqrt{c_1}
   \sqrt{c_1\minus{}\frac{2}{y(t)}}\plus{}c_1\right)\minus{}2\right)}{c_1^{3/2}}\right){}^2
   \equal{}\left(c_2\plus{}t\right){}^2,y(t)\right]$
Sonnhard.

Interpreting that: we can write $ t$ as a function of $ y$. The inverse of that is not an elementary function.


Hi everyone,
Im currently the president of my school's math club, and I believe many of you here used to be / are also MC presidents. So do you guys have any recommendations about running money for Math Club, besides Candy Sales? I dont know what happens at other schools, but in my school, the majority of students ... dont like Math, so making them "give out money" for Math Club is kinda hard for me :-( I really look forward to getting replies from you guys. Thanks in advance.  :)

My math club didn't really do fund raisers much. Mostly, members of the club paid for stuff, or our coach did. But if you need money for a particular thing, you can try to get local businesses to sponsor you. Make shirts and and offer to put your sponsors on your shirts, perhaps.

Getting sponsors is a good idea. You could also try raffles, making & selling physical puzzles (math people can generally come up with/ already know good ones), or any kind of gambling where the odds are stacked in your favour (provided this is permissible).

Try asking your school's parent's association or alumni association for money. They must do some fundraising or collect donations, and you can make an argument that math club is a worthy investment.

Thanks for your replies. I'll probably try out with some brainstorms.

I have one...you know swim-a-thon! Do a math-a-thon....where you offer to give say $20 to winner but you have to pay$2 per person and it could be like a spelling bee and you could sell seats!(problem is I don't see how anyone could be able to want to come...just hopeful.)

[quote="mustafa"]Offer to do math homework for people for a small fee. :D[/quote]

=D

Do a math a thon with some of YOUR good officers in there. So if they win, which is most likely, it goes to your club.

yeah but honestly who would sign up for a mathathon

i know i wouldn't unless the prizes involved large sums of money and large amounts of caffeine


logarithms
number theory proposed
number theory

find in  $ \mathbb{N}^*$
$ x \plus{} y^x \equal{} x^y \plus{} y$

Suppose that:
x=1 so the equation holds for every y nonzero natural number.
x=2 so \[ 2 \plus{} {y^2} \equal{} {2^y} \plus{} y\] bu it's easy to see that \[ \forall y \ge 4\,\,\,\,\,{2^y} \plus{} y > 2 \plus{} {y^2}\]
so \[ y \in \left\{ {1,2,3} \right\}\]
the same method if y=1 or y=2.
Now if x>2 and y>2.
Suppose that y>x
Let , \[ f \equal{} \left[ {x \mapsto {x \over {\ln x}}} \right]\]
f is strictly increasing for \[ x \ge 3\] and so \[ {y \over {\ln y}} > {x \over {\ln x}}\] so \[ {x^y} > {y^x}\] and so
\[ {x^y} \plus{} y > x \plus{} {y^x}\] so the same thing if x>y and so for every \[ x,y \ge 3\] the equality holds if x=y.


Let $a(z),b(z),c(z)$ be prime polynomials comparing to each other such that $a+b=c$.
Prove:
\[max deg\{ a,b,c\}\leq n_{0}(abc)-1 \]
where $n_{0}(abc)$ is the number of distinct roots of $abc$.

As a result you can become Andrew Wiles for polynomials:

For a interger $n\geq 3$, show that if polynomials $p(z),q(z),r(z)$ that are prime to each other satisfy
\[p^{n}+q^{n}=r^{n}\]
then one of them is constant

I remember someone saying that the second result was easy with Mason's Theorem.

The first [u]is[/u] Mason's theorem.
By the way, take a search.

http://www.mathlinks.ro/Forum/viewtopic.php?t=1126&highlight=mason
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=mason&t=15566

God its [i]mason[/i], not meysen, miesen, ....
  :blush:


limit
algebra
system of equations
pigeonhole principle
algebra proposed

Let $n$ be a positive integer greater than $2$. Solve in nonnegative real numbers the following system of equations \[x_{k}+x_{k+1}=x_{k+2}^{2}\quad , \quad k=1,2,\cdots,n\] where $x_{n+1}=x_{1}$ and $x_{n+2}=x_{2}$.

I think your problem is: Let $x_{1},x_{2}>2, \ x_{n+2}=\sqrt{x_{n}+x_{n+1}}$. Find $\lim_{n\to \infty }x_{n}$.
(lim =2).

the answer is x1,x2,x3......xn = 2
any other is not possible as all cannot be either more than 2 or all less than 2 (as sum of the square of these numbers = 2 * sum of the numbers)
so only possibility remaining is these numbers are alternatingly more than 2 and less than 2
but then the numbers greater than 2 form a strictly decreasing sequence so they wont get back to their initial values as the series progresses

Summing all equations we get:
$ \sum{x_i^2} = \sum{2x_i}$
$ \sum{(x_i-1)^2} = n$

First case: there exists $a$ such that $x_a \leqslant 1$
$\Rightarrow x_a^2 = x_{a-1} + x_{a-2} \leqslant 1$
$\Rightarrow x_{a-1} \leqslant 1$
$\Rightarrow (x_i - 1)^2 \leqslant 1$
$\Rightarrow x_i = 0$

Second case: all numbers are greater than one
Let $a$ be such that $(x_a - 1)^2$ is maximised $\Rightarrow x_a$ is maximised
$\Rightarrow 2x_a \geqslant x_{a-1} + x_{a-2}$
By Dirichlet Principle: $(x_a-1)^2 \geqslant 1 \Rightarrow x_a \geqslant 2$
$x_{a-2} + x_{a-1} = x_a^2 \geqslant 2x_a$
$\Rightarrow x_a^2 = 2x_a \Rightarrow x_a = 2$
Obviously follows that all $x$ must be $2$, which is indeed a solution

Let $\textstyle\max_{i\le n}x_i = x_{k+2}$. Using $x_{k+2}\ge x_{k+3}$, we get $x_k\ge x_{k+2}$. From maximality, we obtain $x_k=x_{k+2}$. Now using $x_k\ge x_{k+1}$ we get $x_{k-2}=x_k$. Iterating, we find $x_k=x_{k+2}=x_{k+4}=\cdots$. Now, $x_{k+2}^2=x_k+x_{k+1} = x_{k+2}+x_{k+1}=x_{k+3}^3$. As $x_i\ge 0$, we obtain $x_{k+3}=x_{k+2}$. Repeating now the same argument we find $x_{k+1}=x_{k+3}=\cdots$ so all $x_i$ are equal. From here, we easily find $(x_1,\dots,x_n)=(2,\dots,2)$ is the only solution.


Express as an improper fraction: \[ \frac{\frac{1}{4}\plus{}\frac{1}{5}\plus{}\frac{1}{6}}{\frac{1}{5}\plus{}\frac{1}{6}\plus{}\frac{1}{8}}\]


greatest common divisor
least common multiple

What is the sum of the greatest common divisor and the least common mulitple of 50 and 75?

Because 50=25*2 and 75=25*3, their greatest common divisor is 25. Their lowest common multiple will be 25*2*3=150, so the answer is 25+150=175.

The greatest common factor of $ 50$ and $ 75$ is $ 25$, and the least common multiple is $ 150$. Thus our desired sum is $ \boxed{175}$.


counting
distinguishability
combinatorics unsolved
combinatorics

Give a combinatorial proof that:
$k^n=\binom{k}{1}1!S(n,1)+\binom{k}{2}2!S(n,2)+...+\binom{k}{n}n!S(n,n)$
The left side counts the number of ways to distribute n distinguishable items into k labeled bins with no restrictions. Also note:
$S(n,k)=$ The number of ways to distribute n distinguishable items into k unlabeled bins, at least 1 item per bin

another instance of simple local-global counting principles.
 the LHS of course is the number of ways of $k$-tuples of sets $(S_1,S_2,\ldots,S_k)$ where $\cup S_i =\{1,2,\ldots,n\}$ and $S_i\cap S_j =\emptyset$ for $i\neq j$.
for the RHS suppose that exactly $r$ of these sets are non-empty; thus we have to partition $\{1,2,\ldots,n\}$ into $r$ sets with all these nonempty and that is done is $S(n,r)$ ways.but since we are interested in ordered tuples there are $r!S(n,r)$ ways of obtaining an $r$-tuple of nonempty sets partitioning $\{1,2,\ldots,n\}$.
finally since these $r$ sets can themselves be in any positions among the $k$ the number of  such $k$-tuples is $\binom{k}{r} r!S(n,r)$.summing over $r$ does the job.


Which do you think is better, Opera, Firefox, or Internet Explorer?

obviously, IE is in last place.
i think firefox is a little better than opera but both are similar.

[quote="joml88"]Let's see what PCWorld.com has to say about it: http://www.pcworld.com/reviews/article/0,aid,120763,pg,12,00.asp :P

Firefox is ranked #1 versus Opera's #88. ;)[/quote]

Woot, but you have to realize that was before Opera was free.

I voted for Firefox.

I used to like Opera more than Firefox. (There's no point in even discussing IE.)

However, more sites actually work in Firefox than in Opera. Sure, the world should comply to web standards, but that won't happen anytime too soon. Plus, Firefox extensions are much nicer and easier to use.

Now my vote goes with Firefox.

Im using Opera right now. :D
Ive never really tried firefox though.

arrgh why isn't netscape in the poll? I still use good ol' netscape

meh IMO netscape has nothing to offer except resource hoggage.

Anyway, Firefox definitely wins because there is NOTHING opera can do that firefox cant, what with extensions. In fact, firefox can do much more.

Opera is definitley by far the best out of all of them.  Firefox brought in viruses to me computer.

math92, opera can do it too. Truth is, you can get a virus with any browser. Then again, u could use IE and get no viruses. It all depends on you. If you, knowingly or unknowingly, go to a shady website or download something, it doesnt matter what browser you use. Firefox on the other hand, is just as safe as Opera, if not better. By safe, it doesnt mean u cant get a virus. It means that it has fewer security flaws than the others - its better against SPYWARE and the like. Viruses no browser can protect you against completely, its up to you. Firefox tho has the nice feature (opera might have it too) of not allowing u to set a .exe to run right away.

Interesting... I'm looking at one AOPS page in Firefox (1.5!) and another in Opera.  Hmmmm.
I voted for Opera.

Clearly the fact that firefox has extensions, for who uses them (like me :D) gives it a great advantage over the other two browsers. I also use IE from time to time, as many sites won't load properly in Firefox. Also a very impressive feature of Firefox is AdBlock (which is the main reason for using Firefox on my side).

Opera is faster, consumes less memory, so if you are running a 256 MB box I suggest you use it, rather than Firefox.

But again, as specified above, do not think that Firefox is safer than IE or Opera. Just for beliving that, it makes it more dangerous (many people think that- "hey I'm using the Firefox, I can't get hacked/virused".


algebra
binomial theorem
superior algebra
superior algebra unsolved

Let $ \mathbb{F}_{q}$ be a finite field with $ q$ elements, $ \mathbb{F}_{q}^a$ its algebraic closure, and $ n\geq 1$ fixed integer. Define the mapping $ f: \mathbb{F}_{q}^a\to\mathbb{F}_{q}^a$ by $ a\mapsto a^{q^n}$. Prove that $ f$ is an automorphism.

we may replace $ q^n$ by $ p\equal{}char(\mathbb{F}_q)$. but then $ (x\plus{}y)^p \equal{} x^p  \plus{} y^p$ follows from the binomial theorem and $ (xy)^p \equal{} x^p y^p, 1^p \equal{} 1$ is also clear. the surjectivity follows directly from the fact that $ \mathbb{F}_q^a$ is algebraically closed.

for generalizations, look for frobenius and perfect fields.

EDIT: old explanation too complicated. :)

Injectivity also follows from the fact that $ a^q\minus{}b^q\equal{}(a\minus{}b)^q$.


Define a sequence $ \{u_n, n\geq0\}$ by $ u_0 \equal{} u_1 \equal{} u_2 \equal{} 1$ and there after by the condition $ u_nu_{n\plus{}3}\minus{}u_{n\plus{}1}u_{n\plus{}2} \equal{} n!,n\geq0$. Show that $ u_n$ is an integer for all $ n$ (By condition, $ 0!\equal{}1$).

[hide="My idea"]Prove that $ u_{n \plus{} 1} \equal{} nu_{n \minus{} 1}$[/hide]


geometry
trapezoid
parallelogram
circumcircle
angle bisector
geometry unsolved

In triangle $ABC$ with $AB>BC$, $BM$ is a median and $BL$ is an angle bisector.  The line through $M$ and parallel to $AB$ intersects $BL$ at point $D$, and the line through $L$ and parallel to $BC$ intersects $BM$ at point $E$.  Prove that $ED$ is perpendicular to $BL$.

CE meets BL at D'. Denote N be the midpoint of BC, since ELBC is trapezoid, D',M,N are collinear, so D' and D are coincide.
Apply Menelaus's theorem to triangle DML and the transversal (BNC), show that ND=NB=NC, so angle BDC is right, hence result.

Alternatively, after the collinearity of the points $ C,\ D,\ E,$ we can say that $ NB \equal{} ND,$ from $ \angle BDN \equal{} \angle DBA \equal{} \angle DBN$ and so, we conclude that $ \angle BDC \equal{} 90^{o},$ because of $ N$ is the midpoint of the segment $ BC.$

[b][size=100][color=DarkBlue]CONSEQUENT  PROBLEM. - A triangle $ \bigtriangleup ABC$ is given with $ AB > BC$ and let $ BM,\ BL$ be its $ B$-median and $ B$-angle bisector, respectively. The lines through the point $ L$ and parallel to $ BC,\ AB,$ intersect $ BM,\ BC$ at points $ E,\ K,$ rerspectively. Prove that $ KL \equal{} (BC \plus{} EL) / 2.$[/color][/size][/b]

Kostas Vittas.

Let F be the midpoint of CL, and G the intersection point of EF and BC. ECGL is a parrelogram, and from the previous problem EC is perpendicular to BL. Hence BL is perpendicular to GL, and BLG is a right triangle. From KB=KL (KLB=ABL=LBK) K is a midpoint of BG. Hence $ KL\equal{}\frac{1}{2}BG\equal{}\frac{1}{2}(BC\plus{}GC)\equal{}\frac{BC\plus{}EL}{2}$

Let ABA’C be a parallelogram, and let the perpendicular from C to the angle bisector AL intersects AL, AM, AB and A’B at D’, E’, P and Q respectively, and let  a, b and c be the side lengths of BC, CA and AB;  it’s obvious that D’ is the middle of PC and the triangle PQB is isosceles, with BP=BQ=c-b.( 1 )
It follows that MD’ is medial line in the triangle CPB, hence MD’ || AB and D’≡D.
From (1) and BA’ = b we get A’Q = c, hence   and LE ||QA’, so LE’ || AC and E’≡E, thence DE and AL are perpendicular.

Best regards,
sunken rock

[hide="Solution"]
Define $X=MD\cap BC$, $Y=LE\cap AB$, and $Z=MD\cap LE$.  Note that since $LY$ is parallel to $BC$, $\dfrac{BE}{EM}=\dfrac{CL}{LM}$.  Similarly, since $MX$ is parallel to $AB$, it may be concluded that $\dfrac{LD}{DB}=\dfrac{LM}{MA}$.  These facts, combined with the fact that $MC=MA$, give that \[\dfrac{BE}{EM}\cdot\dfrac{MC}{CL}\cdot\dfrac{LD}{DB}=\dfrac{CL}{LM}\cdot\dfrac{MC}{CL}\cdot\dfrac{LM}{MA}=\dfrac{MC}{MA}=1,\] so by Menalaus points $C$, $D$, and $E$ are collinear.  Therefore it suffices to look at line $CD$.

By angle chasing, it can be seen that $\angle XBD=\angle DBY=\angle BDX$, so $BX=XD$.  Since $XM$ is a midline of $\triangle ABC$, $BX=XC$, so $XC=XD$ and $\angle XDC=\angle XCD$.  Finally, note that \[\angle BDX+\angle XDC=\dfrac{\angle XBD+\angle XDB+\angle XDC+\angle XCD}2=\dfrac{180^\circ}2=90^\circ,\] so $CD\perp BL$, which means that $DE\perp BL$ as desired.[/hide]

Suppose $MD \cap BC=X$,$LE \cap AB=Y$ and $MD \cap EL=Z$,clearly $YBXZ$ is a parallelogram, so ${\angle}YBX={\angle}YZX$,but ${\angle}YLB={\angle}LBC$,so ${\angle}YZX=2{\angle}YLB$,so $Z$ is the circumcenter of $EDL$, hence ${\angle}EDL=\pi/2$.

Dear Mathlinkers,
see also
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=283777
Sincerely
Jean-Louis

[quote="vanu1996"]${\angle}YBX={\angle}YZX$,but ${\angle}YLB={\angle}LBC$,so ${\angle}YZX=2{\angle}YLB$,so $Z$ is the circumcenter of $EDL$[/quote]

This is not true, my dear. I mean, the argument is not leading to the circumcenter, only says to you that $D$, $E$, $Z$ and the circumcenter of $\Delta LDE$ are concyclic.

Let $X=CD\cap BM,Y=CD\cap AB,N=MD\cap BC$
\[
MD//AB \Rightarrow \angle BDN=\angle ABM=\angle DBN \Rightarrow BN=DN\\
MD//AB \Rightarrow BN:CN=CM:BM=1:1 \Rightarrow BN=CN
\]
So,
\[
BN=CN=DN \Rightarrow \angle BDC=\angle R
\]
So, $\triangle BDC\equiv \triangle BDY \Rightarrow BY=BC$
By, Menelaus' theorem
\[
\frac{BY}{YA}\cdot\frac{AC}{CM}\cdot\frac{MX}{XB}=1\\
\Rightarrow \cdot\frac{MX}{XB}=\frac{YA}{2BY}=\frac{AB-BC}{2BC}
\]
By, $\angle CBL=\angle ABL$
\[
CL:AL=BC:AB \Rightarrow CL:ML=BC:AB-\frac{BC+AB}{2}=2BC:AB-BC\\
\Rightarrow \frac{MX}{XB}=\frac{ML}{CL}\\
\Rightarrow BC//LX
\]
So, $X=E\Rightarrow DE\bot BL$
$(Q.E.D.)$

By easy angle chasing, $\angle BLD=\angle LDM$.  But since $MD$ bisects $BC$ and $EL$ is parallel to $BC$, then $MD$ bisects $EL$ as well.  The conclusion follows easily. :D

[hide=bash outline]
Set  $B=(0,0),A=(1,-b),C=(a,ab)$. We can find $L=(\frac{2a}{a-1},0)$ and $M=(\frac{a+1}{2},\frac{b(a-1)}{2})$. Then, we only need to prove that the $x$ coordinate of $D$ and $E$ is equal, which is very doable
[/hide]

Very short.
Let $F = \overline{MD}\cap\overline{LE}$.
\[\angle FLD = \angle ELB = \angle CBL = \angle LBA = \angle LDM = \angle LDF.\]
Since $\overline{MD}$ bisects $\overline{BC}$ and $\overline{LE}\parallel\overline{BC}$, $\overline{MD}$ bisects $\overline{LE}$.

So, $FE=FD=FL$, therefore, $E, D, L$ lie on a circle with diameter $\overline{EL}$, thus $\angle EDL = 90^{\circ}$.

Let Q be the intersection of EL and MD and N be the midpoint of BC . as BC || EL and BN = CN we have EQ = QL . as BC || EL, ∠EQD = ∠DNC and ∠ELD = ∠DBC . It's obvious that ∠DNC = 2∠DBC so we have ∠QDL = ∠QLD and QL = QD . EQ = QL = QD so ∠EDL is 90 .

[hide=Shortage]
Let $MD \cap BC = G, EL \cap AB = F$ and $EL \cap MD = X$.

$\angle FBL = \angle ABL = \dfrac{\angle ABC}{2}$, $\angle BFL = 180^{\circ} - \angle AFL = 180^{\circ} - \angle ABC \implies \angle FLB = 180^{\circ} - ( \dfrac{\angle ABC}{2} + 180^{\circ} - \angle ABC) = \dfrac{\angle ABC}{2} \implies \angle XLD = \dfrac{\angle ABC}{2}$. $\angle XDL = \angle GDB = \angle DBA = \dfrac{\angle ABC}{2} \implies \overline{XL} = \overline{XD}$.
 Now, $MG \parallel AB \implies G$ is also mid pt. of $BC$, $EL \parallel BC \implies$ there exists a homothety centred at $M$ which maps $\triangle MBC$ to $\triangle MEL$, thus it also maps the mid pt of $\overline{BC}$ to the mid pt. of $\overline{EL} \implies MG \cap EL = X$ is the mid pt. of $\overline{EL} \implies \overline{XE} = \overline{XL} = \overline{XD} \implies \angle EDL = 90^{\circ}$ and we are done :D :P.
[/hide]


Ah, I am really sorry, there was typo. Below right version:

Let $ \{f_n\}_{n \equal{} 1}^{\infty}$ be a sequence in $ L^{\infty}(\Omega)$ with the following property: For each functional $ L\in L^{\infty}(\Omega)^{\star}$(the set of continuous linear functionals on $ L^{\infty}(\Omega)$) the sequence of numbers $ L(f_1), L(f_2),\cdots$ is bounded. Then show that the norms $ \|f_n\|_{\infty}$ are bounded.

$ L\in L^{\infty}$ makes no sense. Do you mean $ L\in L^1$?

Isn't it Bahach-Steinhaus theorem?
[hide="Hint"]Yes, it is :)[/hide]


compare the difficulty of USAMTS to aime and USAMO.

The USAMTS and the AIME are not really comparable: the AIME is shorter problems that are designed to be solved in a few minutes and have a definite integer answer; the USAMTS has longer problems that usually require proofs.

The USAMTS is generally easier than the USAMO.

so you would say many of the USAMTS are harder than an AIME 1-10?

Well, assuming that you did the USAMTS, I think you can tell for yourself. :wink:

yea im just not sr, maybe ive improved since I took my last AIME

They really aren't comparable. Even between USAMO and USAMTS, the timeframe difference makes it very difficult. Some people are much better at the long format.

Well, you can qualify for AIME using you USAMTS score on the first 3 rounds

I guess you could say USAMTS score on 1 round*6=AMC score

[quote="mathemonster"]i mean in terms of just the questions not the time.
If you got a USAMTS #5  on an AIME, supposing it was a numerical answer, where would you place it? 1-5,5-10-15 etc.[/quote]You can't really do this. You just can't; the contests are written completely differently.


Which is more acidic and why: acetic acid in water or acetic acid in THF?

Acetic acid in THF. :)
It is due to leveling effect :)

Like dissolves like? So maybe it is more acidic in whichever its more soluble  :maybe:

[quote="Chemrock"]Which is wrong answer or reason?[/quote]

Both are wrong. But you can explain to me what is the "leveling effect".

The term leveling effect refers to a solvent's ability to level the effect of a strong acid or base dissolved in it.
Sorry was trying for a different term to describe the answer :oops:
Probably water because of Hydration :maybe:

acetic acid in water is more acidic, as h2o is a stronger base, thf is aromatic and the lone pair on oxygen is delocalized which makes it a weaker base

sorry, :noo:  got confused with furan :wallbash_red:  :wallbash: . then probably it is again bcoz of the polarity of water.

I think hydration causes acetic acid to be a better acid in water than in THF :)

can u elucidate on this answer please?????

[hide="Answer"]Acetic acid is a better acid in water, because water can solvate (and thus stabilize) better the conjugate base.[/hide]


HCSSiM
AwesomeMath
summer program
Mathcamp
PROMYS
geometry
Mafia

I went to Hampshire (i.e., HCSSiM), and I know people who have been to all of the others.  From what I can tell, all of the ones you have mentioned are quite different, except Hampshire and MathCamp, which might be kind of similar.  I don't really know what real applied math is, or if I do, I don't know how you would "learn" it, especially at a high-school program.  Most of what I've seen I would call "physics" or "computer science" or "engineering" or "economics".  So everything is pretty much pure math.

I'm pretty sure that AwesomeMath is competition-oriented (to quote the brochure, "AwesomeMath is designed for bright students with a special interest in mathematics competitions").  This makes it different from all the others I'm discussing (except MOP).  This could be for you; it is probably not for everyone.  It has a reputation, maybe deserved, maybe not, of being "MOP lite" :  I've heard that in past years, there have been MOP-style tests, and I think that in the first year, there were different ability groups that had color names, almost exactly like MOP, although I don't think they had this afterward.  I don't know, though; I've never been, and the organizers don't seem to shy away from this description in their literature.  It is run by the coach and the former coach of the IMO team, and they're both very cool people.  As you can see from the lists, most of the faculty and staff have been at one time olympiad coaches, problem-writers, or successful participants.  Note that this camp is only three weeks long (MathCamp is five, and the other two are six), although they do have some follow-up programs.  I don't know how many people there are; I'm guessing it's kind of large.

HCSSiM is very different.  Competitions are barely mentioned in the entire six weeks.  You will have three weeks of "workshop", in which you learn about miscellaneous stuff, and then three weeks of one "maxi" course, in which you study one area (e.g., topological graph theory, geometry/origami, ) in depth, and two "mini" courses, in which you study an area in not as much depth, but still a lot (in my year, some of them were information theory, knot theory, hypergraphs, and projective geometry (extremely popular)).  Also, one of the best parts of the program (in my opinion) is the Prime Time Theorem, a self-contained, one-hour talk on an interesting topic that occurs every weekday at 17:00.  Another really nice part of the program is the Program Journal, a weekly student publication that covers what's been happening in all of the classes and Prime Times, as well as movie reviews, weekly problems, and anything else people want to throw in.  Then there are "quasis", little impromptu classes that can cover anything from math to sports.  In my year, there were also at least two real-time mafia games, many more regular mafia games, bridge games, Risk games, Diplomacy games, two cheese quasis, three complex analysis quasis, rugby quasis, and other things I don't remember at the moment.  You'll also experience math movies that will forever change the way you look at four-line conics.  There are also regular movies (such as [i]Dr. Strangelove[/i], which was the only one I saw), field trips to faculty members' houses, and lots of other stuff.  (Can you tell that I had a good time?)  I haven't even mentioned the distinctive culture.  It is six weeks long, usually has 30-40 people, and is cheaper, I think, than any other major math summer program (except for MOP and RSI, which are free).

Of all the math summer programs I know, MathCamp receives the most consistantly glowing reviews.  At MathCamp (I am told), nothing is mandatory.  It is five weeks long, and I believe the courses offered are whatever the staff want to offer.  You can get a better idea of what they are like by looking at the website.  I know some really good people (IMO participants, etc.) who have gone, but my understanding is that you don't have to be as brilliant as that to go.  I think that many people who go return again and again.  (By contrast, HCSSiM is supposed to be a once-in-a-lifetime experience for most people; you have to give a very convincing argument that you would really benefit from going a second time if you want to go again.  That said, I know a few people who have, and one person who even went three times.)  Everybody I know who went loved it and tells me I should go.

PROMYS is different again.  (Disclaimer :  I am an alumnus of RSI, and therefore slightly biased agains PROMYS.  But I will try to be objective.)  From what I hear, it is very structured, which makes it different from HCSSiM and MathCamp.  Its emphasis seems to be number theory, not competitions.  It seems to be based on problem sets.  It also seems to be based heavily on [url=http://www.math.ohio-state.edu/ross/]the Ross Program[/url]---as I understand it, the founder was an alum of Ross.  It is six weeks long, and not as cool as RSI.  (Okay, biases aside, I know a few people who've gone, though not very many, and if I remember correctly, they enjoyed it a lot.)

You might also want to look at the aforementioned Ross program (against which I am not biased).  Surprise!  It looks sort of like PROMYS.  (Hm, coincidence?)  The first-year course seems based on number theory.  I know some really good people who have gone, but, like MathCamp, you don't have to be a MOPper to get in.  It is the longest of the major summer programs for math :  eight weeks.  It is also one of the cheaper ones.  I've been given a few problems sets that are very cool (they're about Dirichlet's theorem).  I don't know that many people who have gone, but the ones I do know really liked it.

Anyway, all of these are venerable programs that have been around for a while (except for Awesomemath, but the people running it have been around for a while).  You probably can't go too far wrong with any of them, and whichever one seems right for you is probably great for you.  If you go to any one of them, you will probably be with really good people, but you don't have to be number one in the country to get into any of them.

Wow thanks a lot for your input! It is very helpful.

The reason I am asking about how many people attend is because if the camp is relatively small (say, HCSSiM, you said, had 30-40 people). Does that necessarily mean that everyone has to be at about the same mathematical level?

And I am also a little concerned (for a lack of a better word) about competition vs. pure mathematics. I am interested in learning pure mathematics, but I also want to improve my competition scores... from your description, it seems a little like the camps are black and white on this issue...

Thanks again for your long reply!

No problem!  But 30-40 people is not as small as it seems.  I think that in my year the number was more like 42 or 43, anyway.  There was a range of mathematical "levels"; there were people who had not qualified for the USAMO before, and people who came there fresh from MOP.

Learning "real" math will probably help you with competitions, because you will be using pretty much the same skills that you are using in ideal math competitions, except without the time pressure.  Competition math (done wholesomely) will also help you with "real" math, though more indirectly.  If you do a lot of competitions (in a wholesome manner, as I said), you will probably not learn very much "real" math, but you will be better at learning it than you were before you did all the competition stuff.

Put another way :  at a place like Hampshire, you will receive no explicit olympiad training.  But if you took a test before going and then took an equally difficult test afterward, you will probably do better, because you've been doing lots of math, even though it wasn't "competition" math (most of it wasn't, anyway).  Also, if you walked into an undegraduate algebra class, perhaps you would already know a lot of the content.  Now, if you spent that time at AwesomeMath, well, you should do better on the olympiad because that's why the camp's there.  If you walked into the algebra class, you would know almost none of the content, but you would be better at learning it than you would have been before AwesomeMath, because of doing a lot of math, as before.  On the other hand, this would also be true if you went to Hampshire.

Anecdote :  A good friend of mine one year didn't make the USAMO.  That summer, he went to Hampshire with me.  The next year, he got a 10 on the AIME, and then went to RSI with me the following summer.  Correlation does not imply causation, and there isn't even really correlation with a sample size of 1, etc., but I think that HCSSiM had something to do with the improvement.

I've also realized that my earlier post may be kind of biased.  You should probably learn more about the non-Hampshire programs from people who actually went to them.

[quote="buzzer11"]I'm deciding where to go this summer, and I was wondering what the pros and cons of each camp were. I know there are already similar threads, but I was yearning for updated information. :)
How many people go to the camp? What type of math do we learn (theoretical, applied, contest)? How high of a level is the math we learn? etc..

Thank you![/quote]

Mathcamp is regarded as either awesome or disappointing. Nobody complains of the math; some people complain it wasn't as fun as they wanted or recalled or dreamed. The math is whatever staff wants: MC'08 had:
theoretical classes, like planar algebra
applied classes, like dynamical systems
contest classes, like Intro and Advanced Problem Solving (and there was a combinatorics class for a week taught by Dr. Patrick, but I don't remember the specifics)

It has 120 people, probably 90% enjoy it and 50-70% would like to go again. It ranges from not-quite-AIMErs to IMOers, but has a wide selection of people who are good at math but better at something else. We had a number of USAChemO camp folks, a member of the US' ILO team, 3 of the 4 members of the USACompO team, and probably some PhysicsO people too, I don't remember. While it is [i]Math[/i]camp, not everyone there likes math most.

Mathcamp is great fun, and you should go.

[quote="Boy Soprano II"]Learning "real" math will probably help you with competitions, because you will be using pretty much the same skills that you are using in ideal math competitions, except without the time pressure.  Competition math (done wholesomely) will also help you with "real" math, though more indirectly.  If you do a lot of competitions (in a wholesome manner, as I said), you will probably not learn very much "real" math, but you will be better at learning it than you were before you did all the competition stuff.
[/quote]

Yeah, I guess that's true. So both "real" and "contest" math help each other, but "real" math helps "contest" math more than the other way around? Something like that? :)

[quote="solafidefarms"]Mathcamp is regarded as either awesome or disappointing. Nobody complains of the math; some people complain it wasn't as fun as they wanted or recalled or dreamed.[/quote]

What exactly disappoints the disappointed? Is it not enough flexibility?

(Note: I didn't mean for this to turn out so long. Oops :oops: . So, to summarize my experience of Mathcamp in case you don't want to read all of this: extremely varied, well-taught, interesting math; helpful, funny, and approachable staff; a collaborative, wacky, and accepting atmosphere; enjoyable field trips; countless exciting/silly/serious/new/goofy/complicated/short games and activities; awesome, interesting, friendly people)

To sorta echo what Boy Soprano said (I was about to reply when I refreshed the page and saw that he'd already commented the same thing)... don't worry that if you go to one of the non-competition-based programs you won't improve at math contests. I am an alumna of Mathcamp (and MathPath, but that's a middle school camp), and I only took a week of a problem solving class this summer (well, er, 4/5 of a week; I did yearbook work instead of going to the last class.) Yet in the AMC-12 class I'm taking here on AoPS the problems are suspiciously easier than AMC-12 problems have been in the past. Basically, I think going to any program would help one improve; AwesomeMath the most, but just doing math of whatever sort for five or six weeks works.

Just to note, unlike HCSSiM which apparently has no specific competition class, Mathcamp does offer problem solving classes (usually an Intro series and an Advanced series, both of which usually go through the whole camp, and sometimes Problem Solving with Calculus or other variants.) I took all five weeks of Intro classes in '07, and think I did learn a good bit from them. These courses don't, however, have practice tests in them or anything, just... solving problems :-).

Now for some Mathcamp propoganda :-P.

Mathcamp, of course, has awesome math. What I like best is the range of math; for one thing, the level of classes range from easy(ish) to insanely hard. I always seem to come across new interesting fields of math at camp. Last summer I learned Game Theory (2 weeks long) and basic Group Theory (1 week long); this summer I learned Knot Theory (4 awesome weeks long) and Hyperbolic Geometry (1 week, but it was a special "super" class and met for 3 or 4 hours each day instead of the usual hour long classes.) Not that that's only what I took--some of my favorite classes were just one week long (unfortunately), like Probabilistic Method, How to Count, Enumeration Celebration (Guess what one of my favorite mathematical fields is?), and The Mathematics of Voting. Or one day long, such as Tropical Geometry or Fractal Dimensions. The hour-long colloquium each day (I think this is practically the same as HCSSiM's Prime Time Theorem) is also fun. Over my two years at MC, I've been to ones about cutting cake fairly, playing billiard, infinity, the mathematics of love, and more.

Outside of classes, there is also TAU, or "Time: Academic, Unsecheduled." It's basically homework time. No one is forced to go and work; some weeks a camper might not even have classes with homework (or might want to do something else if taking classes with non-required homework.) It's also a great time to work on projects (in about the second week of camp, the mentors put up lots of ideas for projects, and campers may pick one and present whatever results they have at the project fair in the last week of camp. Projects have included original research, studying a topic/paper, teaching a class in the last week of camp, helping make an algorithm for the camp's class schedule, and making earrings in the forms of equivalent knots [my project this year :-). I was yearbook editor-in-chief, so I didn't think I would have time for a more involved project.]) Another special thing about TAU is that even if one is not working on math during it, one can't schedule something else (soccer, board game, etc.) in this time period. That way people tend to do any math work they have then, while the mentors and teachers are all in the same general area and available to answer questions (this summer we held TAU in the Library; in '07 it was just in an academic building. Both had large lobbies where most campers worked.)

As a note, the mentors (grad students; they teach classes and organize academics), JCs (junior counselors that have to be MC alumni and organize field trips and non-academic stuff), teachers, and directors of Mathcamp are all super approachable, whether one has a math question or needs to know if a water bottle can be rescued if it wasn't dried and started growing mold (Whoops :oops: )

On the topic of scheduling: Mathcamp has this awesome thing called, not too creatively, the Schedule Board; campers or staff can post events/announcements on it. (Some examples: yearbook meetings, Xtreme [put about any game you want in here], Blindfolded Pictionary, Write a Novel in One Hour, Spinning Around Off the Ground, Mao 101, Swing Dancing, Salsa Dancing, Contra Dancing, Waltzing, Knitting Non-Orientable Surfaces, The Annual Staff vs. Campers Ultimate Frisbee game...)

The Mathcamp yearbook is amazing. Especially the one for '08. I could be a tad biased :wink:.

Also, there is "Mathcamp Standard Time"; the "weekdays" run from Tuesday to Saturday. Each day has five class periods, TAU, and Colloquium, with a nice lunch break in the middle (and it's not at all required to go to five classes each week. Or to go to TAU. Or Colloquium.) There's Team Problem Solving on Wed evening and Relays on Sat afternoon. (the two most contest-ish things at camp, but you can choose between a "Mellow" and "Hardcore" team in each. For TPS, you have the same team through all of camp, and it's assigned [although you're not forced to show up each week]; Relay teams vary from week to week. Also, TPS requires full proofs for answers, while for Relays you just need answers. Usually. Sometimes you might have to support seven people with three feet to submit an answer in Relays. Also, when a winning team from a previous week reforms for relays, they have to take a "handicap" like using no paper or being blindfolded. It's extremely fun.) The evenings have plenty of time for events and games and hanging out and doing extra math. The "weekends" usually have one or two big field trips (again, none required), plus some hikes, staff organized outings, or trips to the movie theater (campers can go off campus by themselves any day, as long as they stay in a group of at least three, give the staff a plan before they leave, check in and out, and come back before 10 PM. Once I missed my afternoon classes to go grocery shopping :-).) The lounges usually close at 10 or 12 PM on weeknights (you can choose before camp whether you want to be on a hall that closes early or one that closes late), although one can stay up in rooms as late as one wants, but on Saturday and Sunday nights the lounges don't close. For example, you could start playing board games at 1 AM and finally go to bed at 6 AM Monday and then sleep until 3 PM (not suggested on a regular basis, but quite fun to do once at camp.)

The best part of Mathcamp, though, is the people. Everyone, staff and campers, is very accepting. I can't really describe how great everyone is; they just are. Most of my closest friends are from my two summers of Mathcamp (plus one summer at MathPath.)

Although I think Mathcamp is the awesomest program, I have heard good things about Ross and HCSSiM; this summer I knew people who had been to one of those camps the previous summers and really liked them. I didn't really ask about Ross, but apparently one difference between HCSSiM and Mathcamp is that the classes are much more interactive at HCSSiM. At Mathcamp, the classes aren't just the teacher lecturing at the blackboard, but they tend to be structured. There are, though, "Moore Method" classes each year at Mathcamp, which consist only of students teaching themselves about a subject through problems. I haven't taken one; someone who has could give a better description. In most of the classes I took, the teachers asked questions regularly to make sure everyone was following (and probably to check that people were awake; sleep deprivation is pretty common at MC :-P.) The most interactive classes I took were Knot Theory and Hyperbolic Geometry. Knot Theory because by the beginning of the fourth week we had five people and at the end we had two, so if one of us asked a question the teacher could give really long explanations or random explorations in response and not worry about boring a lot of people who already got the answer (although I've never seen one camper complain about another being confused in class; most everyone is perfectly willing to help.) Hyperbolic Geometry was the extra long class (although we had a roughly three hour break in the middle), so we did problems in class while the teachers went around helping students and such.

Whoops, I meant to mention other camps and instead I went off on a Mathcamp tangent again. Heh, that tends to happen.... anyway, HCSSiM seems to have as fun a culture as Mathcamp. Maybe I'll be able to get the person I know who went to both to post about them...

Note: Ack, now Billy has also posted before I could. Hopefully I'm not repeating too much.

[quote="buzzer11"][quote="solafidefarms"]Mathcamp is regarded as either awesome or disappointing. Nobody complains of the math; some people complain it wasn't as fun as they wanted or recalled or dreamed.[/quote]

What exactly disappoints the disappointed? Is it not enough flexibility?[/quote]

Some are disappointed because there is not exactly the class they hoped for; that's not too common, though. I haven't had to, but the staff always says that if one isn't enjoying camp as much as one hoped, he/she should just talk to the staff and they'll try as hard as they can to make it more enjoyable for that person. It's not usually a problem with other campers or a lack of activities; if there is an activity one wants to do and it hasn't shown up at camp, just put up a poster on the schedule board and hold it! There have been roommate changes more than once; the camp is fine with this if it's needed and no people are forced to interact (the camp has a decently big size; 110 to 120 people. Too big for me, cause I can't get to know all the people I want to [which is basically everyone at camp.])

Flexibility is practically the definition of Mathcamp. There are four rules (No Stupid Stuff, No Fire, Be Excellent to Each Other, and No Dividing by Zero), plus a few things like "Don't disobey the rules of the college campus we're on," "Leave the lounges when they close,"  and "Check in every night so we know you aren't eaten by a bear," and random suggestions at camp meetings like "Don't play the really loud piano late at night" and "Take showers."  :P

There is no lack of flexibility. In fact, you're generally recommended [i] not[/i] to take classes in every time slot; 4 hours of consecutive math classes isn't good for continuing to like math.

Disappointment occurs generally from not fitting into 120 complete strangers easily. It's normally easy to find people to hang out with, but if you're psychologically overwhelmed by that many new folks, you might not have as full free time as other people. [Shy people might be this way; I'm not shy.]Alternately, there could be no graph theory classes the first week of camp and you could love group theory a lot and be sad.

@scheduling your own activities: your activity will be attended, unless it involves country music. ;)

Thank you, everyone, for your input!  I'm definitely beginning to get a feel for what the camps are like now. :)

I am an avid violinist (maybe more than I am a mathematician) and I was wondering if there was sufficient time to practice at these camps? I hope [i]at least[/i] and hour or two every day.

I probably won't have trouble fitting in, so if that's the only form of dissapointment (that and the possible lack of group theory  :lol: ) I should be okay.

I would think most math camps have plenty of musicians and would allow you to practice; at most Mathcamps the directors get some room on campus for people to practice playing instruments (and, in case any piano players read this thread, there is always at least one lounge with a piano; this summer there were other pianos in the practice room the college let us use.) Serious musicians at Mathcamp seem to be able to find time to practice, as far as I know. The piano in the main lounge was played almost constantly this summer, sometimes along with other instruments and singers (Mathcamp has two a capella groups :-).)  Anyway, many mathematicians play some sort of instrument, so whatever camp you go to will probably have accommodations for serious players to practice.

If you are decently social, there's a good chance the appeal of practicing music will be less than the appeal of hanging out some more, at least at Mathcamp. I'd expect you to do half as much practicing of music than you expect. (As in, you may want to do 1-2 hours before camp, but you'll only get done .5-1 hour.)

Yeah... you would certainly be [i]able[/i] to practice at Mathcamp. Whether you would want to by about the third week is another matter :-P (as a slightly related example, I try to keep a journal about daily, but each year in camp I've given up around the end of the second week because I'm up too late doing whatever to write at night.)

That's true, haha. I never live up to my goals, though, so I'll set them high!

wow i read maybe like a paragraph about AwesomeMath.... :P

I think which camp you want to attend over the summer heavily depends on what kind of experience yo uwant to get out of the camps that you listed.

If you want to improve your competition math skills then AMSP is definitely the place to go! [btw i went to amsp '08]. As the other person already mentioned above, it is ran by former USA IMO team coach [Dr. Titu Andreescu] and current IMO coach [Dr. Zuming Feng]. they are AWESOME teachers and both of them do teach some classes there.
AwesomeMath offeres classes in NT, Combinatorics, Algebra, and Geometry. They have 3 levels for each of those fields and you can choose whichever one you want to take. The format is  you have a morning and an afternoon class.
each class consists of 90 minutes of lecture followed by 90 minutes of problem session lead by the advisors.
Not only you get lectures and problem sessions, some advisors hold math forums after dinner, and we have team contests every week [it's like a practice olympiad-> you can look at the questions from the awesomemath forum] and you get a chance to practice your presentation skills and everything.

The camp CAN be very intense if you choose to work a lot. You basically have free time after dinner if you choose not to attend the forums and etc. Many people go outside and play sports, indoor games, go to the pool, study, etc. That is totally up to the participant himself/herself.

In my case...I wasn't motivated to study so I would cram like two nights before the tests and etc[which of course...my attempt failed miserably as you all konw very well that you can't cram competition style math into your head in like 4 hours x.X]. Looking back. I wish i was a little bit more self motivated to study because then I wouldn't have to review all my notes consistently now xP

Oh and I forgot to mention about the size of the camp. It was around 140 people last year... ummm like...20 girls and 120 guys with an advisor for every dorm, and 10 coaches or so? [btw, all the professors here are highly recognized throughout the world. but just watch out for their accents if you decide to go ;) ] It's so much fun living with your dorm mates and all that. I would say it is definitely a life changing experience for me :D  Good thing about this program is that no one forces you to study [well except the fact that you have to go to your classes and the team contest rounds], and you never have to do math 24/7. There's plenty of time to socialize and just have fun being kids! :PP

If you want to improve your competition math skills, this is def the camp to go. it's PACKED with helpful resources!

(Oh, and did i mention? There are some international students too [from like Romania, korea, etc] so you get to meet all sorts of people!)

when will promys 2009 apps be up? its been a while since they say "applications will be up soon for 09"

I don't know, I've been waiting, too. HCSSiM comes out in March, right?

yea son. did u do the "friendly letter" thingy yet?

Is there anyone out there who has gone to both mathcamp and Ross?  My daughter has gone to mathcamp for the past 2 summers, but maybe it's time for a change.  I know what the websites say, but does anyone have any first-hand experience with both?

no i haven't done the FL. When do we have to do it  by buzzer11?

[quote="HCSSiM"]Here's the 2009 HCSSiM Online Application Form; we would prefer to receive it by May 15th (but that's a preference, not a deadline):[/quote]

That's what they say on their website. The FL goes in with the online application, so it should be the same.

[quote="DiscreetFourierTransform"]Perhaps I can shed some light on the HCSSiM experience.

I went to HCSSiM last summer, and took the basic three-week workshop course, as well as a three-week Polytopes course and two smaller courses in Generating Functions and Markov Modelling.

As has been said so far, HCSSiM tries to focus less on the contrived "competition problems" aspect of math, and delves thoroughly into the realms of topics like hardcore group theory, linear algebra, and other sorts of college math topics.

While technically, calculus isn't required to participate, (and they do a very good job structuring their classes such that this is true) I found that having taken AP Calc BC was a VERY useful piece of background knowledge to have. They don't assume that everyone comes in at a high level, though, and will work with individuals to help them fully grasp the concepts that are being taught.

Plus, the two main instructors (sarah-marie and Tom Hull) are amazing people and will make your life significantly more interesting. Emphasis on amazing. Not having been to other camps, I can't really make a good comparison, but I certainly CAN corroborate (at least anecdotally) that the people at HCSSiM tend to be very quirky, intelligent people who share a common passion and are willing to stay up into the wee hours of the night working out a minor glitch in a graph theory problem so that they have a writeup to present for the next day.

Anecdote: my room-mate (Gabe Karpman) and I once stayed up until around 4 am working on a solution to an incredibly evil graph theory problem by modeling a 72-node graph on the computer, and literally working out every single possible coloring in order to show that a certain property had to exist. We literally dedicated 6 hours to this problem, and neither he nor I showed any sign of stopping. It was only when we realized that we had eaten two entire boxes of granola bars that we decided it would probably behoove us to get some sleep.

That shows how committed and passionate the students are. How about the staff?
I had a Ph. D student named Adam Marcus co-leading my workshop course for the first three weeks. He always prefaced a problem with a story, and brought a smile to the faces of everyone in the workshop when he told us of his exploits in the lands of college math.

The visitors to the program (prime-time theorem) were probably my favorite part of the entire experience.
During my summer, Stephen Wolfram himself showed up and gave a talk about Mathematica and his book, [i]A New Kind Of Science[/i]. Aaron Archer (works at AT&T) gave a talk about network flow analysis. And of course, the famous professor (and leader of the program) David Kelly gave his famous talk on the mystical properties of that number which is the number of cuts needed to unfold a hypercube into a cross-shaped net of three-cubes.

HCSSiM was a focal point in my mathematical career, and I'm exceedingly glad that I was given the opportunity to learn so much math in such a short time with so many wonderfully affable people.

Plus I learned how to a throw a Hammer. How cool is that? :P

Pictures from my year:
http://hcssim.org/pix08.php

I'm the furthest back on the right in the first picture.
I think the pictures help to get a sense for the vibe.[/quote]
The tall white kid, or the indian with long hair

[quote="shelly32494"]Kay,
so for promys/mathcamp/hcssim,
how hard it is to get it?
I'm trying all the problems, but I'm a little nervous about how many I'm able to solve.
Oh, and how big is promys?[/quote]
How many have you got so far? PROMYS ones are harder than Ross I feel.


Show that the following series are:

(a) $(a_{n})_{n\in\mathbb{N^{*}}}$ with $a_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}-\ln({n+\dfrac{2}{3}}))$, strictly increasing

(b) $(b_{n})_{n\in\mathbb{N}}$ with $b_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}-\ln({n+\dfrac{1}{2}})$, strictly decreasing.

Study their limits.

I don't know  :? . what do you mean?

//was this subject moved? If yes sorry about that :)

$\gamma$ is a less well know number than $\pi$ or $e$.

While it was known that $\lim_{n \rightarrow \infty} \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$ diverges, Eular proved that $\lim_{n \rightarrow \infty} \dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{n}-\ln{n}$ converges. Eular called the number it converged to $\gamma$

I didn't heard about it.  There has to be another way to proove the monotony of this series.


This is science question so it may need to belong in the science forum but I'm not sure which one to put.

Q: The sun is located _________ of the orbits of the planets.

A. Exactly in the center
B. At the edge
C. Above and to the right
D. Slightly away from the center

Isn't this obviously:

[hide="answer"]
A?[/hide]

[quote="sky9073"]Try looking up Kepler's laws.
If I remember correctly, planet orbits are ellipses with the sun at one focus.. so I'd say it's B.[/quote]

You are correct about the fact that the sun is at a focus of the elliptical path that a celestial body takes around then sun. But why did you choose B? If I am thinking correctly, the focus of an ellipse is never on the ellipse itself... so I think that the answer might be D. Slighly off from the center of the planet's orbit.

Yup, I'd say it's D. Although I'm not too sure what's meant by C. But with Kepler's law, D looks right.

[quote="Silverfalcon"]This is science question so it may need to belong in the science forum but I'm not sure which one to put.

Q: The sun is located _________ of the orbits of the planets.

A. Exactly in the center
B. At the edge
C. Above and to the right
D. Slightly away from the center
[/quote]

Given that this is in the Round Table, a comment a little out of the problem itself

[quote="Silverfalcon"]C. Above and to the right[/quote]

Above and to the right of what? of the orbit of the planets? that does not make any sence. So many times this kind of question that drive skillful kids away from science. The definition of orbit that I remember is the path of a celestial body or an artificial satellite as it revolves around another body (do you agree?) therefore there is no center of the orbit, there is not above or below, or left or right of the orbit, hence, the only one that makes partial sence according to the definition is to the edge, if you identify path with edge, what, logically does not have any sence either, because there is more than a plannet, not all of them on one single orbit (actually, not two of them in the same orbit!), it just does not make sence.

Multiple-choice questions tend to have one or two (or more) answer choices that are totally ridiculous, depending on how guessable the question-writer wants to make them. I suppose that's what choice C was. A refers to the center of the ellipse that the orbit traces out.

[quote="sky9073"][quote="dangvh"]But why did you choose B? [/quote]

DARN, not again! :(
I always do this... choosing the wrong letter after I get the answer.[/quote]
lol

Don't say "always"! Then, if you do, you will only be fulfilling your own prophecy.

The planets travel in an elliptical path with the sun at one of the focal points.  If you think about an ellipse the focus is a little bit away from the center (the center being the intersection of the major and minor axes).  So I agree with dangvh and say it's D.

D all the way.

However, it can't be slightly away from the center in any direction.  The sun is at one focus, so it is slightly awway from the center in a direction parallel to the major axis.

Technically the orbit can be circular (focus = center).  But of course in reality this can not be perfectly achieved by planets and other such bodies.  Since the planets orbit in nearly circular paths, it would be accurate to say that the sun is located at the focus, which is near the center.  "Above and to the right" seems to refer to off the plane of orbit and slightly off center.  But as djimenez mentioned, there really is no above or below, and there certainly is no left/right.  Where did this question come from by the way?

It's never located where you think it is due to refraction

D.  Simply because although the orbits of the planets are elliptical, some of the orbits are very close to being perfect circles.

I think that the Sun lies in the same plane as most of the planets' orbits, so that choice is out of the question.

That question illustrates how lousily science is taught in most schools.