Find all positive integers n such that 1! + 2! + 3! + ... + n! is the square of a positive integer.

(Taken from the 2002 University of Tennessee FERMAT Contest)

Hmm. Very interesting. And fun!

Find all positive integers n such that 1! + 2! + 3! + ... + n! is the square of a positive integer.

[hide]

1! is the only odd factorial of a positive integer, so a^2 (integer a) is odd, so a is odd. We can easily find a couple numbers:

1! = 1  (a=1)

1+2!+3! = 9  (a=3)

All squares :equiv: 0 or 1 (mod 4)

1! :equiv: 1 mod 4

2! :equiv: 2 mod 4

3! :equiv: 2 mod 4

(3+b)! :equiv: 0 mod 4 for positive integers b

So a^2 :equiv: 1 mod 4 and a is odd. That's consistent with how it should be.

All squares :equiv: 0,1,3, or 4 (mod 6)

1! :equiv: 1 mod 6

2! :equiv: 2 mod 6

(2+b)! :equiv: 0 mod 6

So a^2 :equiv: 3 mod 6 for a>1. So a is of the form 3+6b

All squares :equiv: 0,1,2,4 (mod 7)

1!:equiv: 1 mod 7

2!:equiv: 2 mod 7

3!:equiv: 6 mod 7

4!:equiv: 3 mod 7

5!:equiv: 1 mod 7

6!:equiv: 6 mod 7

(6+b)!:equiv: 0 mod 7

So for a :ge: 6, we have 1+2+6+3+1+6 (mod 7) :equiv: 5 mod 7. No perfect squares there. For a=5, we have -1 mod 7 :equiv: 6 mod 7, so a=5 doesn't work. Thus, the ones I found before are the only ones. Yay.[/hide]

That's right, but can you think of a faster way to do that?

Hint: [hide]Look at the last digits for n > 3, or equivalently, think mod 5.[/hide]

[hide]My answer is n = 1 or 3.

Listing the first few values of f(n) = 1! + 2! + 3! + ... + n! in a table, we get:

n | f(n)

1 | 1

2 | 3

3 | 9

4 | 33

5 | 153

6 | 873

For n :ge: 4, f(n) will always end in a 3, because 5!, 6!, and so on contain a multiple of 2 and a multiple of 5, therefore they have a zero on the end of them so they do not affect the units digit.

We instantly see that the requirement is satisifed (i.e. f(n) is a perfect square) if n = 1 or 3.  We wish to prove that these are the only possibilities.

All perfect squares are equivalent to 0, 1, or 4 mod 5.  This is because

0:^2: :equiv: 0 (mod 5)

1:^2: :equiv: 1 (mod 5)

2:^2 :equiv: 4 (mod 5)

3:^2 :equiv: 4 (mod 5)

4:^2 :equiv: 1 (mod 5)

and for all x > 4, x mod 5 is equivalent to one of the five above numbers.

Since all f(n) for n > 4 end with a 3, they are equivalent to 3 mod 5, which cannot be a perfect square.  Thus the only possible values of n are 1 and 3.[/hide]

Hmmm...looks like gauss202 beat me to using the mod 5 method.  I'm surprised that it took me 15 minutes to type that solution out, though.

You could also use mod 10, I suppose.  It'd take a little more work, because you'd have to list ten modular congruences, not five.


Let $a,b,c,d>0$.
Prove that
$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+d}}+\frac{c}{\sqrt{d+a}}+\frac{d}{\sqrt{a+b}}\geq\sqrt{2(a+b+c+d)}$.

actually
$\sum_{cyc} af(b+c)\geq (a+b+c+d)f(\frac{\sum_{cyc} a(b+c)}{a+b+c+d})$
so we have to show that $(a+b+c+d)^2\geq 2(ab+ac+bc+bd+cd+ca+da+db)$
but it's equivalent to: $(a-c)^2+(b-d)^2\geq 0$ ;)


topology
function
real analysis
real analysis theorems

Unfortunately I have not yet studied topology, but I need a quick question to be answered. Suppose $(X,\rho)$ is a metric space and that $(Y,d)$ is also a metric space, and that $f: A \to Y$ is a function defined on a subset $A \subset X$. Is it true that $f$ is continuous by the usual $\epsilon-\delta$* definition iff for every open set $B \subset Y$ we have that $f^{-1}(B)$ is the intersection of an open set in $X$ with $A$? And is the same thing as saying that $f^{-1}(B)$ is open in the relative topology of $A$?

* Given an $a\in A$ and $\epsilon > 0$ there should be a $\delta > 0$ such that $z\in A \; , \; \rho(a,z) < \delta$ implies $d(f(z),f(a)) < \epsilon$

Yes. The metric topology is identical to the (relative) subspace topology on $A$.

Any open set in $Y$ can be written as a union of $\epsilon$-balls around its points; the inverse image of such a ball contains some $\delta$-ball around each point of the inverse image of the center, and the union of these is the inverse image of the set.


calculus
derivative
geometry
geometric transformation
dilation
absolute value
real analysis

Suppose f is defined on $[0,1]$ with values in the same set and also that the absolute value of the derivative of f in each point is at most 1. Then if we define $x_{n+1}=f(x_{n})$ the sequence $x_{2n}$ converges.

First a useful criterion for the convergence of such sequences:

If $f: I\to I$ ($I=[0,1]$ here) is continuous, and we define $x_{0}\in I,\ x_{n+1}=f(x_{n}),\ \forall n\ge 0$, then $(x_{n})_{n}$ converges iff $x_{n+1}-x_{n}\to 0$.

If $(x_{n})$ converges, we're done, so assume it does not. Our hypotheses imply that $|f(x)-f(y)|\le|x-y|,\ \forall x,y\in I$. This means that $|x_{n+1}-x_{n}|$ decreases to a positive limit $\ell$, and $(*)\ |x_{2n+2}-x_{2n}|,|x_{2n+1}-x_{2n-1}|$ also converge. Suppose now that we can find infinitely many $n$ such that $(\#)\ x_{n}<x_{n+1}<x_{n+2}$. Then we’ll be able to find infinitely many for which $x_{n}<x_{n+1},x_{n+3}<x_{n+2}$. For large $n$, this implies that $|x_{n+2}-x_{n}|$ will be close to $2\ell$, while $|x_{n+3}-x_{n+1}|$ will be close to $0$. This contradicts the convergence of the sequences $(*)$. The same thing happens if we assume that there are infinitely many $n$ for which $(\#)$ holds with reversed signs.

We have thus shown that for large $n$, the sequence $x_{n}$ increases and decreases alternatively. The convergence of $|x_{n+1}-x_{n}|$ now implies the fact that $(*)$ actually converge to $0$, and the criterion cited above finishes the proof.

The hypothesis tells us that $| f(a)-f(b) | = \left| f^\prime (c) \right| |a-b| \leq |a-b|$ (I don't think we really need the fact that the derivative exists in this problem; maybe the fact that it is dilation by a $\leq 1$ factor is sufficient). Thus, $\left| x_{n+1}-x_{n}\right| \leq \left| x_{n}-x_{n-1}\right|$, so $\left| x_{n+1}-x_{n}\right|$ converges to some $\ell$.

[i]Case 1:[/i] $\ell > 0$;

As $x_{n}$ is bounded, it must have a subsequence $\left\{ x_{n_{k}}\right\}_{k \geq 1}$ converging to some $a_{1}$. Hence, $\left\{ x_{n_{k}+1}\right\}_{k \geq 1}$ will converge to $a_{2}= a_{1}\pm \ell$. If $\left\{ x_{n_{k}+2}\right\}_{k \geq 1}$ doesn't converge to $a_{1}$, then it must converge to some other $a_{3}$. Next, we build the numbers $a_{4},a_{5},\ldots$, where $a_{k+1}= a_{k}\pm \ell$. If some $a_{k+1}$ would coincide with $a_{k-1}$, then we would run into a "cycle", thus never reaching $a_{1}$ again, which is a contradiction. So $a_{1},a_{2},\ldots$ are distinct ($a_{k+1}= a_{1}\pm k \ell$). But in this case we eventually reach a dead end, i.e. a contradiction. This implies what we wanted: $a_{1}= a_{3}$.

[i]Case 2:[/i] $\ell = 0$.

Again we take a subsequence converging to some $a$. Since $\ell = 0$, when $x$ gets close to $a$, $f(x)$ also gets close to $a$. But $f$ is continuous, so $f(a) = a$. This implies the fact that $\left\{ x_{n}\right\}_{n \geq 1}\to a$.

Is it OK? I don't how rigorous it is, but it checks with my intuition.

[Edit]Hhmm... I'm late. It took me some time to write the above.[/Edit]

Another a bit related nice problem :

Suppose $f$ is defined on $[0,1]$ with values in the same set and also that $f(f(x)) \leq (1+a)f(x)-ax$ for some $a > 0$ and $a \neq 1$ Then if we define $x_{n+1}=f(x_{n})$ the sequence $x_{n}$ converges.


function
linear algebra
matrix
symmetry
vector
linear algebra unsolved

I'm having a little trouble figuring out a problem posted by my prof:

Let $ V \equal{} \mathbb{R}^n$ Let $ A$ be an $ n \times n$ real matrix. Define a function $ < \cdot,\cdot > _A : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ by:

$ < v,w > _A \equal{} v^TAw$
$ v,w \in V$

Give conditions on $ A$ such that $ < \cdot,\cdot > _A$ is a valid Euclidean inner product on $ \mathbb{R}^n$ if and only if the conditions you give are satisfied by $ A$. (Hint: think eigenvalues/eigenvectors).

I think I'm somehow supposed to put this EIP to the test against its axioms of symetry, additivity, homogeneity and positivity but the only result I get from that is that $ A$ cannot be the zero matrix, there has to be more to it. Any help?

The "additivity" and "homogeneity" (I would have said "bilinearity") are guaranteed by the structure. Don't worry about those.

But do write down what the symmetry condition gives you. We must have $ v^TAw\equal{}w^TAv$ for all $ v,w\in\mathbb{R}^n.$ That puts a big constraint on $ A$ - do you see it? In particular, use $ v\equal{}e_i,w\equal{}e_j$ (standard basis vectors) in that statement.

After that, the big issue is positivity: that $ x^TAx>0$ for all $ x\ne 0.$ What if such an $ x$ happens to be an eigenvector of $ A?$ What then?

Sorry for the late reply, been very busy with other midterms, thanks for providing a hint, I did some work on it and it seems that A must be symmetric for symmetry to hold in the inner product, and for positivity the only thing i get is that if x is an eigenvector of A then $ \lambda x^{T}x > 0$, i dont think Im seeing this correctly, so A must have positive eigenvalues and x cant be zero.

Any elaboration? Thanks!

Eigenvectors are nonzero by definition, so that isn't an issue.

You have it right. One more useful detail: all eigenvalues of a real symmetric matrix are real, so we can actually find a basis of real eigenvectors.


calculus derivative LaTeX function algebra polynomial parameterization

I think using Boost.Regex would simplify it alot (to see patterns). Also use recursion to get the inner derivatives etc.

http://www.boost.org/
http://www.boost.org/libs/regex/doc/index.html

I assume you know how regular languages work since this is a math forum  :P ;)

Also, a simple thing would be to build a binary tree when doing calculations.

What about RPN? See the link
http://en.wikipedia.org/wiki/Reverse_Polish_notation

HTH!

[quote="IUnknown"][quote="Valentin Vornicu"]I was wondering if you guys have any tips for the simplest and quickest way to make a program that reads a string in $\LaTeX$ which should be a function (say only simple things allowed, like polynomials, cos, sin, fractions, exponents, logs) and then returns a string in $\LaTeX$ which is the derivative of the function above ...[/quote]

I think the main problem is how to apply rules of Differentiation, including chain rule. I think this may not a "simple" program within 100 lines, right?[/quote]
That's where regex comes in handy. Say you get something like

$\frac{e^{x^2}sin(x)x^2}{ln(3x)}+3x$

You call the function "der" (or something) with the parameters string="\frac{e^{x^2}sin(x)x^2}{ln(3x)}+3x", string="x" (with respect to x)

Split up the string around '+', '-'. You'll get $\frac{e^{x^2}sin(x)x^2}{ln(3x)}$ and $3x$. The last one will be easy: Call the function recursively with the parameter string="3x" and string="x". The regexp looking for $constant*x$ will get a match and hence it will return the derivative $constant$.

Now for the next string, it would be easy to parse it into $e^{x^2}sin(x)x^2*\frac{1}{ln(3x)}$. Now use regex to get hold on each part and use existing rules. Another problem that could arise is that you do NOT (at this state) want to match the inner ones, i.e., $3x$ and (the first) $x^2$. Say you get this far that you're able to write the current string as $e^{x^2}*sin(x)*x^2*\frac{1}{ln(3x)}$. So the derivative of this one will be

$der(e^{x^2}, x^2)*sin(x)*x^2*\frac{1}{ln(3x)} + e^{x^2}*der(sin(x), x)*x^2*\frac{1}{ln(3x)}+e^{x^2}*sin(x)*der(x^2, x)*\frac{1}{ln(3x)}+e^{x^2}*sin(x)*der(\frac{1}{ln(3x)}, 3x)$.

If you have Sin(x^3), it will not be hard to get a hold of the "x^3" if you're using regex, hence the chain rule will be pretty easy.

This one implemented on 5 min. So making one work for all types of log/sin/cos/tan/exp/poly/etc shouldn't be hard.

[code]<?php

\$var = "sin(3x^2)ln(5x^2e^{3x})";
\$arr_p = array("/sin\((.*?)\)/i", "/(lnn\((.*?)\))/i");
\$arr_r = array('sin(\1)*', 'ln(\1)*');

echo (\$var = preg_replace (\$arr_p, \$arr_r, \$var)) . "

";

\$arr = explode("*" , \$var);

echo "d(" . \$arr[0] . ")/dx = ";

preg_match ("/sin\((.*?)\)/i", \$arr[0], \$matches);

echo "d(" . \$arr[0] . ")/d(" . \$matches[1] . ") d(" . \$matches[1] . ")/dx

";

echo "d(" . \$arr[1] . ")/dx = ";

preg_match ("/ln\((.*?)\)/i", \$arr[1], \$matches);

echo "d(" . \$arr[1] . ")/d(" . \$matches[1] . ") d(" . \$matches[1] . ")/dx";

?> [/code]

Sorry for the fuzzyness in my input. But I hope this was somewhat clear(?).

Basically, the "hard" (yeah right ;) ) part will be to turn $x^3sin(3x^2)e^{7x}$ into $x^3*sin(3x^2)*e^{7x}$ and NOT into $x^3*sin(3x^2)*e^{7x}*3x^3*7x$

PS: I had to escape all my "\$" in the php-code. Else it was interpreted as latex code.

Remember also that some function has no derivative everywhere. For example $\text{ln } x$ is differentiable only if $x>0$.


geometry
circumcircle
Euler
geometric transformation
homothety
perpendicular bisector
similar triangles

Prove that the nine-point center of a given triangle is equidistant from the circumcenter and the orthocenter of the triangle formed by arbitrary three Simson lines of the given triangle .

$ H, O, N$ are orthocenter, circumcenter and 9-point circle center of a $ \triangle ABC.$ Let $ P, Q, R \in (O)$ be poles of 3 arbitrary Simson lines $ x, y, z$ intersecting at $ X \equiv y \cap z, Y \equiv z \cap x, Z \equiv x \cap y.$ Let $ H_S, O_S, N_S$ be orthocenter, circumcenter and 9-point circle center of the $ \triangle XYZ.$ Let $ K, L, M$ be midpoints of $ HP, HQ, HR.$ As $ (N), (O)$ are similar with similarity center $ H$ and coefficient $ \frac {_1}{^2}$ and $ \angle RPQ \equal{} \frac {_1}{^2} \angle ROQ \equal{} \angle (z,y) \equal{} \angle ZXY,$ etc., $ (N)$ is circumcircle of $ \triangle KLM$ and $ \triangle KLM \sim \triangle PQR \sim \triangle XYZ$ are directly similar. But $ K \in x \equiv YZ, L \in y \equiv ZX, M \in z \equiv XY,$ which means that $ \triangle KLM$ is obtained from the pedal triangle $ \triangle X'Y'Z' \sim \triangle XYZ$ of $ \triangle XYZ$ WRT $ O_S,$ i.e., from its medial triangle, by a spiral similarity with center $ O_S.$ $ O_S$ is therefore the common orthocenter of the $ \triangle X'Y'Z', \triangle KLM,$ while $ N_S, N$ are their respective circumcenters. Consequently, $ \triangle O_SN_SN \sim \triangle O_SX'K$ are spirally similar $ \Longrightarrow$ the angle $ \angle O_SN_SN \equal{} \angle O_SX'K$ is right, so that $ N$ is on the perpendicular bisector of $ O_SH_S$ through $ N_S.$

Very nice problem and solution :) There's a nice ( and easy by angle chase ) result:
Let $ M,N,P$ are arbitrary poiints in the circumcircle of $ \triangle{ABC}$. Let $ d_M$ is the line pass through $ M$ and parallel to the Simson line of $ M$ wrt $ \triangle{ABC}$. $ d_M$ intersects $ (ABC)$ at $ M'$. Define $ d_N, d_P, N', P'$ similarly. Let $ X\equal{}d_N \cap d_P, Y \equal{} d_P \cap d_M, Z \equal{} d_M \cap d_N$, and $ T,U$ are circumcenter and orthocenter of $ \triangle{XYZ}$. Then: $ T \equal{} (XNP) \cap (YPM) \cap (ZMN)$ and $ U \equal{} (XN'P') \cap (YP'M') \cap (ZM'N')$
This result can lead to the initial problem, but it isn't a nice solution.

Dear friends,

  M.T.

   Historical note:

   I suspect I. Newton  new about this property, for he knew mechanics very well.

   L. Euler knew this set-up for sure, he knew too much about his circle and even more

   about his line.

We'll use two well-known wesults

1) Directed angle between two Simson lines WRT $\triangle ABC$ with poles $ P,Q$ equals half of the arc $ PQ$ of the circumcircle.

2) Circumcenters of the set of similar triangles $\triangle A'B'C' \sim  \triangle  ABC$ with $A',B',C'$ on $BC,CA,AB$  lie on the perpendicular bisector of the segment connecting the orthocenter and circumcenter of $\triangle ABC.$

Let $ P,Q,R$ be the poles of the Simson  lines $p,q,r$ WRT $\triangle ABC.$ From (1) we deduce that $\triangle XYZ$ bounded by $p,q,r$ is similar to $ \triangle PQR.$ Sidelines of $ \triangle XYZ$ bisect the segments $OP,OR,OQ$ at $ M,N,L,$ since a Simson line is equidistant from its pole and the orthocenter of  $\triangle ABC.$ Therefore, $ \triangle MNL$ is homothetic to $ \triangle PQR$ through the homothety $ (H,\frac {_1}{^2})$ $\Longrightarrow$ 9-point center of  $\triangle ABC$ is circumcenter of $ \triangle MNL.$ Together with $ \triangle XYZ \sim \triangle MNL,$ the conclusion follows from assertion (2).


inequalities
geometry
circumcircle
geometry proposed

Given are $\triangle ABC$ with circumradius $R$ prove that
$R(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})MA.MB.MC\ge (R^{2}-OM^{2})$

Try this one (stricter, but I guarantee much easier) :)

$\frac{MA.MB.MC}{2r}\geq R^{2}-OM^{2}$

This is stricter because $\sum \frac{1}{a^{2}}\geq \sum \frac{1}{bc}= \frac{1}{2Rr}$


Each point of a circle is painted with one of three colours. Prove that there exist three points on the circle which have the same colour and are vertices of an isosceles triangle.

[quote="grobber"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22857]this[/url] discussion. It works for finitely many colors, no matter how many (in this case, we have three).[/quote]
did you complete your construction for circle? as i remember, you worked on line...

I attempted something there. Visit that thread again, and tell me what you think.


Let $ABC$ be a triangle with sides $a$, $b$, $c$ and (corresponding) angles $A$, $B$, $C$.
Prove that if $3A + 2B = 180^{\circ}$, then $a^2+bc=c^2$.

[b]Additional problem:[/b]
Prove that the converse also holds, i. e. prove the following:
Let $ABC$ be an arbitrary triangle. Then, $3A + 2B = 180^{\circ}$ if and only if $a^2+bc=c^2$.

[b]Similar problem:[/b]
Let $ABC$ be an arbitrary triangle. Then, $3A + 2B = 360^{\circ}$ if and only if $a^2-bc=c^2$.

[b]Theorem 1.[/b] Let $\triangle ABC$ be an arbitrary triangle. Then $3A + 2B = 180^\circ $ if and only if $a^2  + bc = c^2 $

[b]Proof.[/b]

Since $A + B + C = 180^\circ $ and  $a = 2R\sin \alpha \ \ ,b = 2R\sin \beta ,\ \ c = 2R\sin \gamma $ we have

$ 3A + 2B = 180^\circ  \Leftrightarrow  $

$ 3A + 2B + C = A + B + C \Leftrightarrow  $

$ 2A + B = C \Leftrightarrow  $

$ \sin \left( {2A + B} \right) = \sin C \Leftrightarrow  $

$ \sin B \cdot \sin \left( {2A + B} \right) = \sin B \cdot \sin C \Leftrightarrow  $

$ \left( {2\sin \frac{B}{2}\cos \frac{B}{2}} \right)\left( {2\sin \frac{{2A + B}}{2}\cos \frac{{2A + B}}{2}} \right) = \sin B \cdot \sin C \Leftrightarrow  $

$ \left( {2\cos \frac{{2A + B}}{2}\sin \frac{B}{2}} \right)\left( {2\sin \frac{{2A + B}}{2}\cos \frac{B}{2}} \right) = \sin B \cdot \sin C \Leftrightarrow  $

$ \left[ {\sin \left( {A + B} \right) - \sin A} \right]\left[ {\sin \left( {A + B} \right) + \sin A} \right] = \sin B \cdot \sin C \Leftrightarrow  $

$ \sin ^2 \left( {A + B} \right) - \sin ^2 A = \sin B \cdot \sin C \Leftrightarrow  $

$ \sin ^2 C = \sin ^2 A + \sin B \cdot \sin C \Leftrightarrow  $

$ 4R^2 \sin ^2 C = 4R^2 \sin ^2 A + 4R^2 \sin B \cdot \sin C \Leftrightarrow  $

$ c^2  = a^2  + bc $

and we are done. $\blacksquare$

[b]Theorem 2.[/b] Let $\triangle ABC$ be an arbitrary triangle. Then $3A + 2B = 360^\circ $ if and only if $a^2  - bc = c^2 $

[b]Proof.[/b]

Since $A + B + C = 180^\circ $ and  $a = 2R\sin \alpha \ \ ,b = 2R\sin \beta ,\ \ c = 2R\sin \gamma $ we have

$ 3A + 2B = 360^\circ  \Leftrightarrow  $

$ 3\left( {180^\circ  - B - C} \right) + 2B = 360^\circ  \Leftrightarrow  $

$ B + 3C = 180^\circ  \Leftrightarrow  $

$ B + 2C = 180^\circ  - C \Leftrightarrow  $

$ \sin \left( {B + 3C} \right) = \sin C \Leftrightarrow  $

$ \sin B \cdot \sin \left( {B + 3C} \right) = \sin B \cdot \sin C \Leftrightarrow  $

$ \left( {2\sin \frac{B}{2}\cos \frac{B}{2}} \right)\left( {2\sin \frac{{B + 3C}}{2}\cos \frac{{B + 3C}}{2}} \right) = \sin B \cdot \sin C \Leftrightarrow  $

$ \left( {2\cos \frac{{B + 3C}}{2}\sin \frac{B}{2}} \right)\left( {2\sin \frac{{B + 3C}}{2}\cos \frac{B}{2}} \right) = \sin B \cdot \sin C \Leftrightarrow  $

$ \left[ {\sin \left( {B + C} \right) - \sin C} \right]\left[ {\sin \left( {B + C} \right) + \sin C} \right] = \sin B \cdot \sin C \Leftrightarrow  $

$ \sin ^2 \left( {B + C} \right) - \sin ^2 C = \sin B \cdot \sin C \Leftrightarrow  $

$ \sin ^2 A = \sin ^2 C + \sin B \cdot \sin C \Leftrightarrow  $

$ 4R^2 \sin ^2 A = 4R^2 \sin ^2 C + 4R^2 \sin B \cdot \sin C \Leftrightarrow  $

$ a^2  = c^2  + bc \Leftrightarrow  $

$ a^2  - bc = c^2  \ \ \  \blacksquare$

[b]Remark.[/b] In the proof  of theorem 1 we have

$ 2A + B = C \Leftrightarrow   \sin \left( {2A + B} \right) = \sin C$

since, if it were $2A + B +C = 180^\circ$,  it would have $A = 0$.

Similarly, in the proof  of theorem 2 we have

$ B + 2C = 180^\circ  - C \Leftrightarrow  \sin \left( {B + 3C} \right) = \sin C $

since, if it were $ B +3C = C$  it would have $B = 0, C = 0$.

The condition $3A+2B=180^\circ$ is equivalent to $C=90^\circ +\frac A2$. Therefore $C$ is obtuse. Take $D$ on $AB$ such that $AD=AC$. Then $\angle BCD=C-\angle ACD=C-\frac {180^\circ -A}{2}=A$. It follows that triangles $ACB$ and $CDB$ are similar, hence $\frac ac=\frac {c-b}{a}$, so $a^2+bc=c^2$.

BTW, this problem was discussed it least twice on forum. ;)
Unfortunately, I can't find the links. Ask Darij :)

I remember only the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=4230 , where it was shown that if $a^2=c\left(c+b\right)$ (what is equivalent to $a^2-bc=c^2$), then A = 2C (what is equivalent to 3A + 2B = 360°). The converse was shown in http://www.mathlinks.ro/Forum/viewtopic.php?t=45481 post #7.

But indeed, the synthetic solution of the initial problem given by Enescu in post #3 can be also used for proving the converse, and the related problem can be solved absolutely similarly.

  darij

So 1 St part,if you solve a little bit,we get
C=90°+ (A/2)
sin C=sin{90°+ (A/2)}
sin C=cos(A/2)
Since, Sin C=(c/a)*sinA=2(c/a)*sin(A/2)*cos(A/2)
SO, sin (A/2)=a/2c
-cos C=a/2c
(c²-a²-b²)/2ab=a/2c
So, if we solve we get,
(c+b){c²-bc-a²}=0
Since c≠b
c²-bc-a²=0,a²+bc=c²( proved)
The converse can also be done some way ,just back calculation
Additional problem:
So if we solve we shall get,
(A/2)=C
Then,
cos(A/2)=a/2c
cos C=a/2c
So if we solve,
( c-b){a²-c²-bc}=0
c-b=0.          or.      a²-bc=c²( proved)
c=b
# KRISHIJIVI


function
inequalities proposed
inequalities

If
   $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}8$
then
$ \frac{ab}{c}\plus{}\frac{bc}{a}\plus{}\frac{ca}{b}>4\sqrt[6]{a^3\plus{}b^3\plus{}c^3}$

We have that $ \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b} = \frac{(ab)^2 + (bc)^2 + (ca)^2}{abc} \geq \frac{\sqrt{3a^2 b^2 c^2(a^2 + b^2 + c^2)}}{abc}$
$ = \sqrt{3(a^2+b^2+c^2)}$ from Newton's inequality.

So, we have to prove that $ \sqrt{3(a^2+b^2+c^2)} > 4 \sqrt[6]{a^3+b^3+c^3}=\sqrt[6]{8(a+b+c)^3(a^3+b^3+c^3)}$
$ \iff 27(a^2+b^2+c^2)^3 > 8(a+b+c)^3(a^3+b^3+c^3)$
which comes from $ 3(a^2+b^2+c^2)^3 \geq (a^3+b^3+c^3)(a+b+c)^3$
$ \iff \left(\sqrt{{\frac{a^2+b^2+c^2}{3}}\right)^2 \geq \sqrt[3]{\frac{a^3+b^3+c^3}{3}}\times\frac{a+b+c}{3}}$, which comes from the fact that the power mean function $ M(p) = \left(\frac{x_1^p + x_2^p + \dots + x_n^2}{n}\right)^{\frac{1}{p}}$ is log concave on the positive reals.


We see that the measure of the angle to the right of $ x$ is $ 180 \minus{} 108 \equal{} 72^\circ$. There are $ 180^\circ$ in a triangle, and we know that two of them are 26 and 72. Thus the measure of $ x$ is $ 180 \minus{} (72 \plus{} 26) \equal{} \boxed{82^\circ}$.

x and 26 are the remote interior angles to 108, so x+26=108. x=82

If $ a$ and $ b$ are two angles of a triangle, then:

$ a \plus{} b \equal{} \text{supplementary of } c$


Let $ x,y,z$ be positive numbers such that $ x^2\plus{}y^2\plus{}z^2\le\frac{3}{2}.$
Prove that $ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\ge2(x\plus{}y\plus{}z).$

[quote="rogue"]Let $ x,y,z$ be positive numbers such that $ x^2 \plus{} y^2 \plus{} z^2\le\frac {3}{2}.$
Prove that $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\ge2(x \plus{} y \plus{} z).$[/quote]
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\geq \frac {9}{x \plus{} y \plus{} z} \geq 2(x \plus{} y \plus{} z)$

$ \dfrac{3}{2}(xy\plus{}yz\plus{}zx) \ge (x^2\plus{}y^2\plus{}z^2)(xy\plus{}yz\plus{}zx) \ge (xy\plus{}yz\plus{}zx)^2 \ge 3xyz(x\plus{}y\plus{}z)$
so $ xy\plus{}yz\plus{}zx \ge 2xyz(x\plus{}y\plus{}z)$. QED

[quote="sqing"]The following inequality also belong Vyacheslav Yasinskyy.
Let $ a,b,c$ be positive numbers .Prove that\[ \frac{a^3+abc}{b+c}+\frac{b^3+abc}{c+a}+\frac{c^3+abc}{a+b}\geq a^2+b^2+c^2.\][/quote]
It's $\sum_{cyc}\frac{a(a-b)(a-c)}{ b+c}\geq0$ and the rest is obvious.


Euler
function
number theory proposed
number theory

Given positive integer $ n > 1$ and define
\[ S \equal{} \{1,2,\dots,n\}.
\]
Suppose
\[ T \equal{} \{t \in S: \gcd(t,n) \equal{} 1\}.
\]
Let $ A$ be arbitrary non-empty subset of $ A$ such thar for all $ x,y \in A$, we have $ (xy\mod n) \in A$. Prove that the number of elements of $ A$ divides $ \phi(n)$. ($ \phi(n)$ is Euler-Phi function)

$ A$ sous groupe of $ T$ so $ |A|$ divide $ |T|$


function
combinatorics proposed
combinatorics

Given a permutation $ w \equal{} w_1 \cdots w_n$ of length $ n$, we call the triple $ (i < j < k)$ a [i]good triple[/i] if $ w_i < w_k < w_j$, i.e., if the smallest of the three entries is first and the largest of the three is in the middle and the middle element by size is last.  Among all permutations, which ones have the largest number of good triples, and how many do they have?

Here's an idea.  Define two directed graphs with vertices $ w_1, ... w_n$ as follows:  $ D_1$ has an arc from $ w_i$ to $ w_j$ if and only if $ i < j$ and $ w_i < w_j$ and $ D_2$ has an arc from $ w_i$ to $ w_j$ if and only if $ i < j$ and $ w_i > w_j$.  Then the number of good triples is the sum of the entries of $ \mathbf{A}(D_1) \mathbf{A}(D_2) \mathbf{A}(D_1)$.  Moreover, $ \mathbf{A}(D_1) \plus{} \mathbf{A}(D_2) \equal{} \mathbf{A}(K_n)$.  Unfortunately, I can't think of anything intelligent to say about the structure of $ D_1$.

Hmm, probably I should fess up on this one that I thought I had a solution and was wrong, and it turns out that the answer isn't exactly "nice" -- it's [url=http://www.research.att.com/~njas/sequences/A061061]A061061[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  It's possible to write down a simple recursion using the "max" function, but it doesn't look like there'll be a very nice formula for it  :(


a = 238000 = 2^4 x 5^3 x 7 x 11 and b = 299880 = 2^3 x 3^2 x 5 x 7^2 x 17

is there an integer n so that a divides b^n  if so what is the smallest possibility for n

The problem should be "a=238000=2^4 x 5^3 x 7 x 17" and it would give n=3


Circles $ S_1$ and $ S_2$ with centers $ O_1,O_2$ respectively, intersect each other at points $ A$ and $ B$. Ray $ O_1B$ intersects $ S_2$ at point $ F$ and ray $ O_2B$ intersects $ S_1$ at point $ E$. The line parallel to $ EF$ and passing through $ B$ intersects $ S_1$ and $ S_2$ at points $ M$ and $ N$m respectively. Prove that $ MN\equal{}AE\plus{}AF$
Thank for your nice solution  :)

[color=darkred]We will prove that $ MB=AE$ and $ BN=AF$. Then the conclusion follows immediately.

From $ \angle{O_1BE}=\angle{O_2BF}$ and from the fact that triangles $ O_1BE$ and $ O_2BF$ are isosceles we get that $ \angle{O_1EB}=\angle{O_2FB}$. Therefore quadrilateral $ O_1O_2FE$ is cyclic so

${ \angle{MAE}=\angle{MBE}=\angle[BEF}=\angle{BO_1O_2}=\angle{AEB}$,

so $ AM \parallel BE$ and $ AMEB$ is an isosceles trapezoid. Thus $ AE=BM$. Similliarly $ AF=BN$.[/color]


A train has 12 boxcars linked together, and is packed with 4 cars of lions, 4 cars of tigers, and 4 cars of bears.  However, the bears cannot be in the first 4 cars, the tigers cannot be in the next 4 cars, and the lions cannot be in the last 4 cars.  Find (and explain) the number of ways that the cars can be packed.

Oh my. ;)

By "the bears cannot be in the first four cars", do you mean that not all the first four cars can be bears? Or that NO bears can be in any of the first four cars? That is, is the situation Bears Bears Bears Lions permissible?

No bears can be in the first 4, so even LLBL doesn't work for the first 4 cars.  Something like LLTT works, though.

[hide]maybe casework on bears, where | separates the 4 cars, and x is a spot for either L or T

xxxx|BBBB|xxxx and xxxx|xxxx|BBBB
$1*2=2$

xxxx|BBBx|Bxxx and xxxx|Bxxx|BBBx

the middle x must be a L, and the last 3 must be T's, the first 3 are a mix of LLLT, and the x in the second and third groups can be in 4 spaces, the reverse is the same for the second subcase
$2*4*4*4=128$

xxxx|BBxx|BBxx
middle 2 x's are L's, last 2 x's are T's, first is a mix of LLTT, 6 permutations for each
$6*6*6=216$

$\sum=346$?

not sure...my counting ability is bad[/hide]

here is an exercise, find a summation or a closed form for $n$ bears, lions, and tigers, with same conditions on 3n cars

[hide="original"]
Say there are $x$ tigers in the first set of 4.  Then, by noticing we can't have bears in the first set, and similar facts, we obtain
set one:
x tigers
4-x lions

set two:
x lions
4-x bears

set three:
x bears
4-x tigers

x can range from 0 to 4, and we can sort each set $\frac{4!}{x!(4-x)!} = \binom 4 x$ ways.  Therefore the answer is $\sum_{x=0}^4 \binom 4 x ^3 = 346$ ways
[/hide]

[hide="althemans exercise"]
I'm pretty sure, by the same thing I did before, that the closed form for $n$ would be $\sum_{x=0}^n \binom n x ^3$
[/hide]


find $k$ is smallest such that set ${1989,1990,....,1989+k}$ can be partitioned into two subset $A,B$ satisfly sum of element of $A$ and $B$ is similar

[quote="maxal"]$k=3: A = \{ 1989, 1992\}, B = \{ 1990, 1991 \}$[/quote]
Maxal, you can post full solution (your post seem spam).

That's sort of trivial. Just show (with a brute-force) that there are no such partitions for $k=0,1,2$.

I think  Maxal's solution is complete :lol:

but I want edit problem:
find all $k$ such that set ${1989,1990,....,1989+k}$ can be partitioned into two subset $A,B$ satisfly sum of element of $A$ and $B$ is similar

I think k = 4n+3
Put all numbers in the form 4n+1 or 4n in A, 4n+2 or 4n+3 in B


I have some fractions inside an eqnarray, and they look completely out of whack. Is there any way to change the text size within the equation array? Thanks.

eqnarray is obsolete so should not be used -see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ 5[/url].
You can change the size of fractions by replacing \frac by \tfrac to make them smaller or \dfrac to make them larger $ \tfrac{1}{2} \dfrac{1}{2}$

Hi pii,

just a remark, in order to use  \dfrac and \tfac you have to include amsmath:
[code]\usepackage{amsmath}[/code]
That package is recommendable whenever you write math formulas.
Btw. if you want to get short information about why not to use eqnarray you could have a look here: [url=http://texblog.net/latex-archive/maths/eqnarray-align-environment/]eqnarray vs. align[/url].

Stefan

--
[url=http://texblog.net]TeXblog.net[/url]


calculus
integration
function
complex analysis
real analysis
real analysis unsolved

Compare to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45127[/url]. It is the same problem up to a change of variable.

[quote="julien_santini"]Consider a function $f: [1,a] \rightarrow \mathbb{R}$ continuous and integrable such that $(\int_{1}^{a}{f(t)t^{n}}dt)_{n \in \mathbb{N}}$ is a bounded sequence. Does this imply that $f=0$ on $[1,a]$ ?[/quote]
Paley-Wiener Theorem. Extend  $f$ trivially to all $\mathbb{R}$, and let  $g(s) = \int_{\mathbb{R}}e^{ist}f(t)dt$ be the Fourier transform, extended as an entire function to all $s \in \mathbb{C}$.

Then $|g^{(k)}(0)| = |\int_1^af(t)t^k dt| \le C$ for all $k$ and therefore $|g(s)| = |\sum_{k=0}^\infty \frac{g^{(k)}(0)}{k!}s^k| \leq \sum_{k=0}^\infty \frac{C}{k!}|s|^k = Ce^{|s|}$.

Therefore $g$ is the Fourier transform of a function that is supported on $[-1,1]$. This is only possible if $f = 0$.

[quote="ma_go"]what if you consider the open interval $(1,a)$ instead of $[1,a]$?
does the result hold as well?[/quote]
The key assumption is that $\int_1^a f^2(x) dx < \infty$.

[quote="zhaobin"]is $f$ should be continuous?[/quote]
No, only square integrable. The conclusion then is that $f = 0$ almost everywhere outside $[-1,1]$.


Find the ordered triples of real numbers (x,y,z) that satuidt x+yz=6, y+xz=6, z+xy=6.

By symmetry, $ x\equal{}y\equal{}z$. Therefore $ x\plus{}x^{2} \equal{} 6$

$ x^{2} \plus{} x\minus{}6\equal{}0$

$ (x\plus{}3)(x\minus{}2)\equal{}0$

$ x\equal{}y\equal{}z \equal{} \minus{}3$ or $ x\equal{}y\equal{}z\equal{}2$.

[quote="AndrewTom"]By symmetry, $ x \equal{} y \equal{} z$.[/quote]

Wrong.  $ (x,y,z) \equal{} (1,1,5)$ is also a solution.

Obviously. Thanks. Again, I rushed without thinking.

Assuming two of the unknowns are equal, say $ x\equal{}y$, we have $ x\plus{}xz\equal{}6$ and $ z\plus{}x^{2} \equal{} 6$, so

$ x\plus{} x(6\minus{}x^{2}) \equal{} 6$

$ x^{3} \minus{}7x \plus{}6 \equal{} 0$

$ (x\minus{}1)(x\plus{}3)(x\minus{}2) \equal{} 0$

$ (x, y, z) \equal{} (1, 1, 5), (\minus{}3, \minus{}3, \minus{}3), (2, 2, 2)$

Then we consider whether it is possible that $ x, y$ and $ z$ might be distinct.


Find the next term of the following sequence:

1
11
21
1211
111221
...?

312211
:D

This isn't really calculus. But while we're talking about it, how about some other questions about this sequence?
Prove that the only stable sequence (that is, the only one that doesn't grow indefinitely) is 22.
And how about that crazy thing with the algebraic number of degree 71?  :P I'm not going to try to prove that.

[quote="Ramanujan"]What is this sequence Xevarion ???  :roll:[/quote]
[hide]
1 = 1 one => 11
11 = 2 ones => 21
21 = 1 two, 1 one => 1211
1211 = 1 one, 1 two, 2 one => 111221
111221 = 3 ones, 2 twos, one one => 312211
[/hide]

[quote="Beginner"]Another interesting question, When the number $4$ occure in the sequence?  :D[/quote]
 :rotfl:  :rotfl:


This might have been mentioned before, but can we have system where it displays what posts have been ranked for you so you can see the results.

For example, I was at 3.9 after a mysterious rank-up and then I dropped again to 3.8 recently. I have no idea what post(s) caused it.

That's a good idea!

Here's another: Display the ratings of posts in search (you wouldn't be able to vote on these, just see them).

Displaying ratings but not allowing you to rate them in search? What would that do?

Maybe allow you to see how good the post is before clicking on it?

Display ratings in search: As buzzer11 said, you would be able to select higher-rated posts if you are looking for some kind of educational material or anything similar. Also, you would be able to scan your posts and see what ratings you receive and for which posts.

Not be able to rate in search: For those people who have fun by killing others' ratings.

I'm not sure this is relevant to the topic, but can you rate what other people rated? I know this is kind of really stupid...Rating other people's ratings will help point out who's giving all of the bad ratings, so you have to separate ratings:

Post Rating:
Rating people's post:

Is this a possible thing to examine in the future?  :oops:

Yeah, basically if Person A rates Person X's post as a 4 and the Person B rates that same post as a 6, then the average will be a 5.

[quote="D3m0n Shad0w"]I'm not sure this is relevant to the topic, but can you rate what other people rated? I know this is kind of really stupid...Rating other people's ratings will help point out who's giving all of the bad ratings[/quote]
This isn't stupid. However, I believe that there is already a system in place which alerts the admins if someone is excessively rating posts as spam or doing other unruly rating-related behavior. So this probably isn't really necessary. The best thing you can do is that if you see a post that you think is overrated or underrated, rate it yourself too!


geometry
ratio
trapezoid
analytic geometry
graphing lines
slope
Pythagorean Theorem

The ratio of the radii of two concentric circles is 1:3.  If AC is a diameter of the larger circle, BC is a chord of the larger circle that is tangent to the smaller circle, and AB = 12, then the radius of the larger circle is
(A) 13
(B) 18
(C) 21
(D) 24
(E) 26

Let $ABCD$ be an isosceles trapezoid with bases $AB=92$ and $CD=19$.  Suppose $AD=BC=x$ and a circle with center on $\overline{AB}$ is tangent to segments $\overline{AD}$ and $\overline{BC}$.  If $m$ is the smallest possible value of $x$, then $m^2=$
(A) 1369
(B) 1679
(C) 1748
(D) 2109
(E) 8825

The second problem is pretty tough.  Enjoy!

Unless someone wants me to, I'm not going to go into a lot of detail explaining these since it'd probably involve drawing and all the side/angle names get confusing.

1. [hide]Let the smaller radius be r and larger 3r. BC is bisected by the perpendicular radius of the smaller circle. Call the center of the circle O. Draw a segment from O to the midpoint of AB which will be parallel to BC, forming a rectangle. The bisected length of AB is 6, so r, the opposite side, is 6 and 3r is 18. (B)[/hide]

2. This was tough.

[hide]Putting the trapezoid on a coordinate grid, point A can be fixed at the origin with AB running along the x-axis. The main restriction on the figure is that the semicircle must be tangent to the sides of the trapezoid. We can say a radius of the circle perpendicularly intersects somewhere along ray AD starting at A(0,0) and is not tangent to segment AD if that intersection occurs beyond point D.

Now we can find restrictions on the trapezoid such that the intersection will be on segment AD. We call the height of the trapezoid h and let D(36.5, h). The slope of AD is h/36.5, so the slope of a perpendicular radius would be -36.5/h. The equation for the radius through the center of AB (46,0) is y = (-36.5/h)(x-46). The equation for line AD is y = (h/36.5)x. The value of x in their intersection is 46(36.5)^2/(h^2 + (36.5)^2). x must be at most the x of D (36.5), so h^2 is at least 9.5(36.5). Using Pythagorean theorem, we find that AD^2, or m^2, is at least 1679. (B)[/hide]

Just a note: In general I won't have time to check many of the solutions in this forum, so I encourage peer review of solutions.


modular arithmetic
number theory
least common multiple

Prove that for all integers $n$, that the units digit of $n^6 - n^2$ is 0.

One way is to just test all 10 cases for any n $\pmod{10}$.

Otherwise, you can factor it into $n^2(n-1)(n+1)(n^2+1)$. For 0, 1,3, and 9 this clearly has a 0 at the end. for 2 or 5 you get a factor of 2 or 5 which is 10 and thus has a 0 in the end. For 4, you get a factor of 2 and 5 = units digit 10. For 6 you get a factor of 5 and a 2. For 7 the last term is 50 so it makes it end in 0. For 8 you get a factor of 65 in the last term and a 2 otherwise so that makes a 10. Now all cases are done.

[hide]This implies that $5 | n^6-n^2$ and $2 | n^6-n^2$.

$n^6-n^2=n^2(n^4-1)=n^2(n^2+1)(n^2-1)=(n-1)(n+1)(n^2)(n^2+1)$.
This must be divisible by 2 because either $n^2$ or $n^2+1$ is even.

If $n-1 \equiv 0 \pmod {5}$, then we are done.
If $n-1 \equiv 1 \pmod {5}$, then $n^2+1 \equiv 2^2+1 \equiv 0 \pmod {5}$.
If $n-1 \equiv 2 \pmod {5}$, then $n^2+1 \equiv 3^2+1 \equiv 0 \pmod {5}$.
If $n-1 \equiv 3 \pmod {5}$, then $n+1 \equiv 4+1 \equiv 0 \pmod {5}$.
If $n-1 \equiv 4 \pmod {5}$, then $n^2 \equiv 5^2 \equiv 0 \pmod {5}$.

Therefore, it must always be divisible by 2, 5, and $\text {lcm} (2,5)=10$, so its units digit must be $0$.[/hide]


WHAT IN THE WORLD IS PUTNAM??
I'm in 6th grade...Can i I take Put name?..(oops..typo..i meant Putnam)

Here are the eligibility rules for the Putnam:
- You don't have a bachelor's degree yet.
- Four times, maximum.

If you take it early, you're losing years of eligibility later. This makes it a bad idea in the vast majority of cases.

Yeah, the Putnam is intended for undergraduate students, hence why they restrict you to taking it four times.  The vast majority of people use their eligibility on their four years of college.  A rare few will use one their senior year of high school.


Where can you get your official score report for a SAT Subject Test? Do they mail them to you?


trigonometry
inequalities unsolved
inequalities

[quote="nrs.sowrabh"]I THINK THE ANGLES SATISFY SOME CONDITION.(ATLEAST THEY MIGHT BE THE ANGLES OF A TRIANGLE)       :blush:

             ELSE PUT (A,B,C)=(0,0,0)   :P[/quote]
of course  :) they are angles in a triangle

:idea: I THINK THIS MIGHT HELP. TRY WRITING [color=red][cos(A/2)]^2[/color] as {[1+[i]cosA[/i]]/2} AND THAT AS 1+4[color=red]sin[A/2]sin[B/2]sin[C/2][/color]  AND SO ON.[/i]

[quote]Let be given a triangle $ABC$. Prove that:
\[\cos^{2}\frac{A}{2}+\cos^{2}\frac{B}{2}+\cos^{2}\frac{C}{2}\le 2+\frac{1}{4}.\cos\frac{A-B}{2}.\cos\frac{B-C}{2}.\cos\frac{C-A}{2}\][/quote]
We have:\begin{eqnarray*}\cos^{2}\frac{A}{2}+\cos^{2}\frac{B}{2}+\cos^{2}\frac{C}{2}-2&=&\frac{3}{2}+\frac{1}{2}\left(\cos A+\cos B+\cos C\right)-2\\ &=&\frac{1}{2}\left(\cos A+\cos B+\cos C-1\right)\\ &=&\frac{1}{2}\left(2\cos\frac{A+B}{2}.\cos\frac{A-B}{2}-2\sin^{2}\frac{C}{2}\right)\\ &=&\sin\frac{C}{2}\left(\cos\frac{A-B}{2}-\cos\frac{A+B}{2}\right)\\ &=&2\sin\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}\end{eqnarray*}
Hence:\begin{eqnarray*}\cos^{2}\frac{A}{2}+\cos^{2}\frac{B}{2}+\cos^{2}\frac{C}{2}-2&\le&\frac{1}{4}.\cos\frac{A-B}{2}.\cos\frac{B-C}{2}.\cos\frac{C-A}{2}\\ \Longleftrightarrow 2\sin\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}&\le&\frac{1}{4}.\cos\frac{A-B}{2}.\cos\frac{B-C}{2}.\cos\frac{C-A}{2}\\ \Longleftrightarrow 8\prod\sin\frac{A}{2}.\cos\frac{A}{2}&\le&\prod\cos\frac{B-C}{2}.\sin\frac{B+C}{2}\\ \Longleftrightarrow\prod\sin{A}&\le&\frac{1}{8}\prod\left(\sin B+\sin C\right) \end{eqnarray*}
What is truly because:\[\prod\left(\sin B+\sin C\right)\ge 8\prod\sqrt{\sin{B}.\sin{C}}=8\prod\sin{A}\]


find all the values for $T$, in which it hold in this equality.

$sin7X.sinX=sin2X.sinT$

its an ugly problem, not an hard one.

i prefer not to waste ure time to solve it :D

i my self wont think about it atall,


Fie $a,b,c$ pozitive cu $abc=1$. Sa se demonstreze:

$a^{2}+b^{2}+c^{2}+9(ab+bc+ca) \geq  10(a+b+c)$

Daca tot bate vantul pe aici, o sa pun doua linkuri:
1. http://www.mathlinks.ro/Forum/viewtopic.php?highlight=discussed&t=80197
2. http://www.mathlinks.ro/Forum/viewtopic.php?t=36874

:blush: o pusesem intai aici dupa care cand am vazut ca nu se baga lumea paici am pus-o si la unsolved si asta e :)) dupa care am vaz mai tarziu ca a mai fost paici


geometry
3D geometry
sphere
calculus
integration
blogs

A circular hole with radius 5 is drilled all the way through the center of a sphere of clay with radius 13. The remaining clay is rolled to form a new perfectly spherical "bead", what is the radius of this bead?

There is $\frac{4\pi 13^3}{3}-\pi 25 \cdot 26=\frac{6838\pi}{3}$ cubic units of clay left. This is used to make a sphere so $\frac{6838\pi}{3}=\frac{4\pi}{3} r^3 \implies 1709.5=r^3 \implies r \approx 4.75 \text { cubic units.}$

Wait, this is wrong. :P I just realized that the part that's cut out isn't a perfect cylinder :( so the answer is a tiny bit too high :blush:

I cant figure out how to get the "additional volume" in elemantary way  :mad:  so I use calculus to get the volume as $25\frac{1}{3}\pi$  :P  This gives us the remaining volume to be $2254\pi$ which is $r^3=1690.5$ :arrow: $r\approx 11.91$

[quote="shyong"]...so I use calculus to get the volume as $25\frac{1}{3}\pi$  :P...[/quote]

Can you show the steps please? I know a little calculus (self-taught) but I don't know how to do these types of things with calculus... Thanks! :)

I dont know whether calculus is suitable in Intermediate so i guess i better "hide" it  :P
[hide="calculus method"]

Consider the equation $x^2+y^2=13^2$ . The "additional volume" is

$V=2\times\pi\int^{13}_{12}y^2dx=2\times\pi\int^{13}_{12}169-x^2dx
=2\times\pi\times|169x-\frac{1}{3}x^3|^{13}_{12}$
$=2\times\frac{38}{3}\pi=\frac{76}{3}\pi$

 :)
[/hide]

Ahah. I would expect a ninth grader of your skill level to have taken calculus, but thats okay.

Well we can integrate and find the area of sphere caps.  I guess you'll need to know a quite a bit of background first though.

Take calculus! I took it in 9th grade.  Not particularly useful for olympiad problems, but useful nonetheless.

[quote="fourierseries"]Take calculus![/quote]

Look at 1234567890's blog (I'm a contributor) :P (Or click "Help!" on my sig.)

I didn't understand the calc way, so I need to learn more calculus I guess. :D All I know is differentation...

[quote="krustyteklown"]This was from a "math team tryout" that i just took today, not quite sure how to do it:

A sphere of clay has radius 13. A circle with radius 5 cuts through the center of the sphere. Squish the "bead" so it makes a new sphere. What is the radius of this new sphere?

I think what it meant was basically to cut out a cylinder with radius 5, but also with some "caps" on the ends. I'm having trouble finding the volume of these extra regions.[/quote]

Let me guess: this was from your math team tryout, and you and krustyteklown go to the same school :P

why would you find the caps separately, it would just be easier to use integration to find the whole thing
look at this topic...i dont want to retype the solution
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52168[/url]
and i as well have learned a substantial amount of calculus on my own, technically i am in algebra 2, but i am like better at math than the teacher...lol
i first learned calc, but then i found the aops, and that has occupied my extra math time


ok, my problem is that if I'm graphing z=4cos(theta) and transferring it to a rectangular coordinate, x is 4, cuz its just cos, and y is 0? is that right? cuz that doesnt make much sense

Wow no bumping. Do you want me to lock this, Milo?  :P

actually, could someone move this into a more advanced forum cuz i think that might be why noones replying

I think you mean $ z$, not $ r$.
To rewrite in polar coordinates, substitute $ \theta\equal{}\tan^{\minus{}1}y/x$ and $ r\equal{}\sqrt{x^2\plus{}y^2}$.

Bumping is pointless because the forums are so inactive. Anyone who is interested in helping you will easily notice your thread when he or she gets on.

The polar graph $ r \equal{} 4 \cos \theta$ can be translated to rectangular coordinates by multiplying by $ r$.  We then have

$ r^2 \equal{} 4r \cos \theta \implies$
$ x^2 \plus{} y^2 \equal{} 4x$.


geometry
geometric transformation
reflection
symmetry

The circles $ C_1$ and $ C_2$ with centers $ O_1$ and $ O_2$ respectively are externally tangent. Their common tangent not intersecting the segment $ O_1O_2$ touches $ C_1$ at $ A$ and $ C_2$ at $ B$. Let $ C$ be the reflection of $ A$ in $ O_1O_2$ and $ P$ be the intersection of $ AC$ and $ O_1O_2$. Line $ BP$ meets $ C_2$ again at $ L$. Prove that line $ CL$ is tangent to the circle $ C_2$.

Let $ D$ be the reflection of $ B$ in $ O_1O_2$. Clearly, $ CD$ is a common tangent to $ C_1$ and $ C_2$.
In particular, this means that $ \angle{O_2DC} \equal{} 90^{\circ}$ and hence $ PO_2DC$ is cyclic.
Let $ \angle{ABP} \equal{} \alpha$. By symmetry we have that $ \angle{ABP} \equal{} \angle{CDP} \equal{} \alpha$.
Since $ PO_2DC$ is cyclic it follows that $ PO_2C \equal{} \alpha$ and hence $ \angle{PCO_2} \equal{} 90^{\circ} \minus{} \alpha$.
Finally, we see that $ \angle{O_2LB} \equal{} \angle{LBO_2} \equal{} 90^{\circ} \minus{} \alpha$.
In conclusion $ \angle{PCO_2} \equal{} \angle{O_2LB}$, which implies that $ CL$ is tangent to $ C_2$.


AMC
AIME
USA(J)MO
USAMO
summer program
Mathcamp
trigonometry

Hey guys!

I'm at MathCamp now, and this week we had a very special visitor, Sam Vandervelde.  He went to the IMO 1989, so he is one smart cookie.  He co-created the Mandelbrot Competition with Richard and Sandor, and he produces the contest to this day.  He is amazing at writing problems, and through his class this week, he taught his 20 or so students to do the same.

Together, we came up with a list of about 160 excellent problems.  We are in the process of solving them, refining them, and narrowing them down to about 100 awesome problems.  Tomorrow, we'll choose the best 10 to use for the MathCamp Challenge '05.  These questions cannot be released.

However, the other 90 amazing questions can be used freely.  I play to use the best 86 of them for a series of Mock AMCs.

First, we'll hold a Mock AMC 12A, followed by a Mock AMC 12B or equal difficulty.  Then, those who qualify will be invited to take the Mock AIME I or a Mock AIME II or equal difficulty.  As with the real AMC, you can take both Mock AMC 12s, but only one Mock AIME.  Then, those who qualify will be invited to take the Mock USAMO.

I promise you guys that these questions are truly excellent questions.  They were prepared by MathCampers, some of whom are USAMO honorable mentions.  Then, they were carefully edited by other MathCampers and Sam Vandervelde.  I'm sure they will be as good or better than previous Mock AMCs and Mock AIMEs.

To give me feedback, please take the poll above and/or post in this thread.

Sounds like a good idea. I suppose the problems created have quite a large range of difficulty if you plan on using them for Mock AMC, AIME, and USAMO, right?

I know this is a stupid request and all, and by no means do you have to listen to it, but, could you possibly take some of the Mock12 problems, add in some easier ones, and give a Mock10, for the not-so-trig-brilliant among us.

Yes.  Some of our problems are extremely challenging (we actually haven't solved them all, yet), some are easier, some are trivial, and some are impossible (literally).  We're in the process of deleting the impossible and trivial ones.  That's why our number will go down from 160 to 100 by the end of our refining process.

Sounds great! Do you have certain dates in mind?
Also, how will you be grading the mock USAMO?

Dates?  No.  This will be up to you guys.  Because I plan to offer a Mock AMC 12A and a Mock AMC 12B, it will highly unlikely you are unavailable for either one.  However, because I am offering two, I will expect answers within one or two days of posting the questions.

Grading the USAMO?  Undecided.  I really have no clue.  Maybe the AoPS instructors or senior members may pitch in...

AMC 10?  This is an interesting idea.  However, I'm pretty much saying no for 2 reasons:
1) I have forgotten the difficulty level of the AMC 10
2) There is virtually no trig in all 160 of our problems.  There's a ton of algebra, number theory, geometry, probability and combinatorics, and Olympiad-level inequalities.  This means that if you do not know that much trig, you'll still do fine.  Plus,  siorange, you got an 8 on the AIME!  You have nothing to worry about.

I think its a good idea to have in the summer, cuz there isn't much else going on, contest-wise, and I know there'd be tons of participants

Yeah, I would be interested, because I need the practice.  Please PM me whenever you choose the dates.

Sounds good.

I was thinking the fall might be better participant-wise because many people are away.  However, people are more busy.

Does anyone else have an opinion?

Please PM/email me when you decide to do it. I'd be happy to participate!

Billy

intriguing. PM upon further details, please.

I'll make a new thread when I decide to do it.  I can't PM/e-mail tons of people.  Sorry.

Right now we have a few options:

1) AMC 12 A released Tuesday, July 19.  AMC 12 B released Wednesday, August 3.  AIME I released Tuesday, August 23.  AIME II released Wednesday, September 7.  USAMO released Tuesday, October 4 and Wednesday, October 5.
2) One week after option 1.  (starts 7/26)
3) Two weeks after option 1.  (starts 8/2)
4) Three weeks after option 1. (starts 8/9)
5) Four weeks after option 1.  (starts 8/16)
6) Five weeks after option 1.  (starts 8/23)
7) Six weeks after option 1.  (starts 8/30)
8) Seven weeks after option 1. (starts 9/6)
9) Eight weeks after option 1.  (starts 9/13)
10) Nine weeks after option 1.  (starts 9/20)

Note that these all follow the same format of spacing dates as the AMC--I mean the exact same format.  These contests are going to be as realistic as possible.

[quote="bubala"] Plus,  siorange, you got an 8 on the AIME!  You have nothing to worry about.[/quote]

Hah, I got an 8 on one of the easiest AIMEs which invovled NO trig :P

But I'm up for anything that doesn't fall between 7/25-8/16 and 8/21-8/28. Yes, I know that's not much time at all, so don't worry one bit if I can't do it. I'm LAST priority. I'd just want to see the test afterwards.

But I'd like to thank Bubala for organizing this, I know what it's like organizing a small contest, like my Mathcounts Challenge, and this is about 50 times harder, and it's a huge burden, and thank you bubala for doing all this.

Sounds like a good idea.  I'll probably participate.

[quote="bubala"]Grading the USAMO?  Undecided.  I really have no clue.  Maybe the AoPS instructors or senior members may pitch in...[/quote]

Ha  :rotfl: don't count on the instructors voluntarily grading papers.  They still have to figure out how they are going to grade WOOT, their classes, and the IS's.

umm, I hope this isn't a stupid question, but how exactly does the mock amc and aime work? when do we know the questions are up, and how do we send in the answers?

Well, hopefully bubala will be able to finish the tests.  But you pm your answers after the test is over.

[quote="probability1.01"]Don't worry, bubala will be done soon.  As of a week ago or something he has already completed I think the 6th AMC problem.[/quote]

Lol. Orznorz. Whatev.  :P

[quote="probability1.01"]Don't worry, bubala will be done soon.  As of a week ago or something he has already completed I think the 6th AMC problem.[/quote]Only 6?  Hahah  :D

[quote="probability1.01"]Don't worry, bubala will be done soon.  As of a week ago or something he has already completed I think the 6th AMC problem.[/quote]

:rotfl: I hope he finishes before the real AMCs... :D , jk

Lol...In the last three posts every single person quoted probability.

Nothing really. It's just a phrase that probability said repeatedly at MathCamp, which he apparently learned from his roommate at MOP...

[quote="chesspro"]I'm going to say that this took more time that bubala expected and will, unfortunately, be unable to finish (that is, a fairly short amount of time). So, please do not expect these tests to come out. Perhaps they will come out in the future, at which time a new thread will be made.[/quote]

Duly noted. Writing good tests is not easy, it takes time. Recall, for instance, that the good folks at the AMC write their exams a year or more in advance.

[quote="chess64"][quote="chesspro"]Orznorz.[/quote]

What does that mean? :huh:[/quote]

Oh yes, "orz norz". It's a takeoff on "oh no" that was sort of invented at MOP and used in many games of Bughouse when the opponent's moves looked threatening. It also won the "Most Annoying Phrase" at MOP 2005. (See [url]http://test.kaseorg.com/MOP_2005_Hall_of_Fame_Awards#Randomness[/url]).

It's funny how many things that happen at MOP don't spread (like the USEMO!).

Gosh, I heard HaiTao using that like... let's say... a week before bughouse craze started. It was so annoying... funny how annoying things are contagious. ORZNORZ...

[quote=rjiangbz]how so orznorz[/quote]

18 year bump wow

[hide=Post #77 by rjiangbz][url=artofproblemsolving.com/community/user/404771][b]rjiangbz[/b][/url] · May 26, 2023 at 10:16 PM [url=artofproblemsolving.com/community/p27819086](view)[/url][color=transparent]helo[/color]
how so orznorz

-----------
[color=#5b7083][aops]x[/aops] 77[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

[tip=@rjiangbz][img]https://avatar.artofproblemsolving.com/avatar_404771.png?t=1645539839[/img]
AoPS User[/tip]

Bruh, why would you pointlessly bump a thread probably older than you are


calculus
integration
function
LaTeX
geometry
derivative
real analysis

I'm having trouble with my analysis homework, it would be great if someone can help me with it! Thanks

1)  Let digit(i,x) be the ith digit in the the decimal expansion of x in (0,1).  Then digit (i,x) will be an integer {0,1, ...9}.  Given a sequence {x sub n} of real numbers in (0,1), show there exists a subsequence x sub n sub k, such that for each i, digit (i, x sub n sub k) is the same for all k > i.

2)  Decide whether these converge? explain why?

a)  summation k from 0 to infi    1 / (2k+3)
b)  summation k from 1 to infi    k sin (1/k)

3)  is integral of e^(x^2) from 0 to 1 possible to integrate?

I'm new so i don't know the notations, sorry if it's hard to read!

Hint for #1: Try to make a sequence for which it doesn't work. Why can't you?

Hint for #2: (a) This is very similar to a pretty well-known series. What is that series? Prove that they both behave in the same way by a comparison.
(b) How does $\frac{\sin(x)}{x}$ behave for small $x$? What does that tell you about the limit of the terms of this series as $k$ gets large? And in turn, what does that tell you about the series itself?

Hint for #3: What conditions do you know of that make a function integrable? Look at the definition of integration. There's a pretty simple condition you should know; check if this function satisfies that condition.

edit: Hint for notation: You can mouse over other people's equations to see the code that made them. This code goes within dollar signs. Click the LaTeX link on the sidebar for a full guide.

[quote="chaosvn"]3) is integral of e^(x^2) from 0 to 1 possible to integrate? [/quote]
[quote="Xevarion"]Hint for #3: What conditions do you know of that make a function integrable? Look at the definition of integration. There's a pretty simple condition you should know; check if this function satisfies that condition.[/quote]
Xevarion was instinctively speaking as an analyst. The integral is defined as the limit of some process - the limit of Riemann sums as the mesh tends to zero, or something related (or something more advanced that still involves a limit), and he was reading the words "possible to integrate" as an analyst reads the word "integrable": does that limiting process converge? He could have been referring to a couple of different simple conditions, both of which presume that we're talking about a closed bounded interval $[a,b].$ One is that continuous functions are integrable, and we certainly have a continuous function. Another is that monotone (or piecewise monotone) functions are integrable, and  this is a monotone function. There's no doubt that this is an integrable function. The integral exists; the area under the curve is some finite number.

Beginning calculus students do not typically think in those terms. Beginning calculus students have usually not yet been taught to think like analysts. Beginning calculus students typically want the formula for everything, and in this case, that leads only to frustration. There is no method in the "techniques of integration" chapter of a calculus book by which you can do anything to the integral

$\int_0^1e^{x^2}\,dx$

to obtain a formula for it in terms of a finite number of familiar functions. (Relax "finite number" and there are useful possibilities, notably including using power series to express the result as a rapidly converging series.)

for #3
\int_0^1e^{x^2},dx\ I can use integrating derivative to do so that's the same as e^{x^2} 2x evaluating from 0 to 1 which becomes 2e^{1} right?

for #2 I figure out first part, second part with k sin (1/k) i still haven't figured out yet.

[quote="chaosvn"]for #3
\int_0^1e^{x^2},dx\ I can use integrating derivative to do so that's the same as e^{x^2} 2x evaluating from 0 to 1 which becomes 2e^{1} right?[/quote]
Wrong. In fact, as I said, this integral is a number than cannot be expressed in elementary terms.

As an analyst myself, I'm bound to react the same way Xevarion did: of course, there is such a number. Of course, the function is integrable. If you want me to approximate that number to within $10^{-12},$ I can and I will (fairly quickly, in this case). Elementary representation is not that interesting a question.

For #2: What is the limit of $\frac{\sin(x)}{x}$ as $x \to 0$? Since that limit exists, if you take any sequence $x_n$ that converges to 0, then the sequence $\frac{\sin(x_n)}{x_n}$ also converges to 0. Then take $x_n = \frac{1}{k}$, and look at what happens to your sum.

so let x = 1/k
lim k sin (1/k) k-> inf = lim sin (x)/x as x -> 0
lim sin (x)/x = 1 when x -> 0
so by limit test 1 is not equal 0 -> diverge

I think that's right?

Yes, that's exactly what you needed to do. :)

thanks kent
i seem to not being able to type those symbols  ;)
$\int_0^1e^{x^2}\,dx$

I still don't understand
1) Let digit(i,x) be the ith digit in the the decimal expansion of x in (0,1). Then digit (i,x) will be an integer {0,1, ...9}. Given a sequence {x sub n} of real numbers in (0,1), show there exists a subsequence x sub n sub k, such that for each i, digit (i, x sub n sub k) is the same for all k > i.

Here's what I understand the problem
we have x say x = 0.123456
{x sub n} = {0 1 2 3 4 5 6}

{x sub n sub k} = {0 1 2}
k = 3 so for all j < 3 each digit of j < 3 are the same for the 2 sets?

Enclose the $\LaTeX$ code in dollar signs. I just stuck a pair into your last post - click the edit button to see.

thank you

f(x) = $\int_0^{x^3}e^{t^2}\ dt$
evaluate
$6*\int_0^1x^2f(x)\,dx$ - $2*\int_0^1e^{x^2}\, dx$

it seems like e^(x^2) is hard to integrate, but this one has it I'm not sure how to do it!
My guess is to do integration by parts
u = f(x)
dv = x^2

is that a right approach?

Integration by parts :
$\\ 6 \int_0^1x^2f(x)\,dx = 6 \int_0^1 \left(\frac{x^3}{3}\right)' f(x)\,dx$

What   $f(1)$  is  ???.

 :)

I got it, thank you!

could anyone explain me this problem?

1) Let digit(i,x) be the ith digit in the the decimal expansion of x in (0,1). Then digit (i,x) will be an integer {0,1, ...9}. Given a sequence {x sub n} of real numbers in (0,1), show there exists a subsequence x sub n sub k, such that for each i, digit (i, x sub n sub k) is the same for all k > i.

the last sentence means all the digits after jth digit are the same? how is that possible?

[quote="chaosvn"]

1) Let digit $(i,x)$ be the $i$-th digit in the the decimal expansion of $x$ in $(0,1)$. Then digit $(i,x)$ will be an integer $\{0,1, ...9\}$. Given a sequence $x _{n}$ of real numbers in $(0,1)$, show there exists a subsequence $x_{n _{ k}}$, such that for each $i$, digit $(i, x_{n _{ k}})$ is the same for all $k > i$ .

[/quote]

The sequence  $x _{n}$ is bounded  therefore there exist a monotonic subsequence $x_{n _{ j}}$ that converge say to $\alpha$.

But a convergent sequence is a cauchy sequence and for a cauchy sequence we have

$\\ \forall \epsilon>0 , \, \exists \, n_{0} \in \mathbb{N}$ such that for $m , \, \ell \in \mathbb{N}$ with $m , \ell >n_{0}$ we have $|x_{n _{ m}}-x_{n _{ \ell}}| < \epsilon$.

For $\epsilon = \frac{1}{10^i}, \, \exists \, n_{0} \in \mathbb{N}$ such that for $m , \, \ell \in \mathbb{N}$ with $m , \ell >n_{0}$ we have $|x_{n _{ m}}-x_{n _{ \ell}}| < \frac{1}{10^i}$ .

Now let $y_{n _{ k}}$ to be the element of $x_{n _{ j}} , \, j >n_{0}$ that have the bigger distance from $\alpha$.

Then for $r> k>i$ it is $|y_{n _{ k}}-y_{n _{ r}}| <\frac{1}{10^i}$ . (that is what the problem ask for )

It seems correct...but i am not sure  :?  :D

You can't say anything about the original sequence. As Ramanujan said, you can find a convergent subsequence, and convergence lets you say much more. The Bolzano-Weierstrass theorem gives you this sequence.

[quote="chaosvn"]is that the proof?
how does that show all the jth digit are the same for the sub sequence and the sequence {xn} ? I don't understand[/quote]

The problem say that the terms of the sequence $x_{n_{k}}$ ($y_{n_{k}}$ in my post ) should be the same for $k >i$.

If you find a sequence such that   $|x_{n_{k}}-x_{n_{r}}|<\frac{1}{10^i}$ for $r>k>i$ then you're done . :)

it seems like you don't need the monotonicity right?  I understand up to when you let
$\epsilon = \frac{1}{10^i}\,$
could you explain more on what you did after that? Thanks.

I find a subsequence of $x_{n_{k}}$ that has the property asked.

Since $x_{n_{k}} \rightarrow \alpha$ every interval $( \alpha- \epsilon , \alpha + \epsilon)$ for all $\epsilon > 0$ contains all the terms  of $x_{n_{k}}$ for $n > n_{0}$ for some $n_{0} \in \mathbb{N}$ .

Now take $\epsilon = \frac{1}{10}$ . From those terms of $x_{n_{k}}$ that are inside this interval we choose as $y_{1}$ that one that
have the bigger distance from $\alpha$ .(Sorry for my english )  :D

Analogously for $\epsilon = \frac{1}{10^{i}}$ you take $y_{i}$ .

But by construction both $y_{i}$ and $y_{i+1}$ are inside $( \alpha- \frac{1}{10^{i}} , \alpha + \frac{1}{10^{i}})$
therefore $|y_{i+1}-y_{i}|<\frac{1}{10^i}$.
I dont think we need monotonicity .

@ All :Can we make this more rigorous ??  :? Prof Kent some help here please :)

 :)


[hide]We see by looking at the first few differences of squares (2^2-1^2, 4^2-3^2, 6^2-5^2, 8^2-7^2) that there is a pattern in the differences: 3, 5, 9, 15. The difference between two differences next to each other is 2 greater than the difference before it (eg. 15-9 is two greater than 9-5). This is the same as 2006 3's plus an arithmetic series which starts with 0, ends with 4010, and has 2006 terms. The product of 2006*3=10,028 and the sum of the arithmetic series equals 4022030.  10,028+4,022,030=4,032,058 [/hide]

I think you made a mistake, the answer should be [hide="answer"]8050078[/hide]

[quote="aahmadi"]We see by looking at the first few differences of squares (2^2-1^2, 4^2-3^2, 6^2-5^2, 8^2-7^2) that there is a pattern in the differences: 3, 5, 9, 15. [/quote]

That should be $3, 7, 11, 15$.

[b]inom:[/b]  Your numbers are too far apart.

[quote="t0rajir0u"]Compute the difference between the sum of the first $2006$ even squares and the sum of the first $2006$ odd squares.[/quote]

[hide]This is just the sum of the sequence

3+7+11+...8023=2006*8026/2=8050078[/hide]

[hide="Equivalently, it is also the sum of the sequence"] $(1+2)+(3+4)+...+(4011+4012) = 2006 \cdot 4013$ [/hide]


calculus
integration
function
real analysis
real analysis theorems

I was wondering how to prove the following:

$\int_{0}^1\int_{0}^1\int_{0}^1\frac{1}{1-xyz}dxdydz=\sum_{n=1}^{\infty}\frac{1}{n^3}$

and/or

$\int_{0}^1\cdots\int_{0}^1\frac{1}{1-\prod_{k=1}^mx_k}\prod_{k=1}^mdx_k=\sum_{n=1}^{\infty}\frac{1}{n^m}$

...if there is such a generalization.  Any suggestions?

Thanks for your help.

Your formula can be proven for $m=3$ by expanding $\frac{1}{1-xyz}$ as $1+xyz+x^2y^2z^2+x^3y^3z^3+...$  Then just integrate one by one.  First you get $\displaystyle \int_{0}^1\int_{0}^1\int_{0}^1\frac{1}{1-xyz}dxdydz=\int_{0}^1\int_{0}^1\ [1+\frac{yz}{2}+\frac{yz}{3}+\frac{yz}{4}+...]dydz
\displaystyle=\int_{0}^1\ [1+\frac{z}{2^2}+\frac{z}{3^2}+\frac{z}{4^2}+...]dz = 1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...$

Which gives the desired result.  The general case can be proven by simply repeating the above with whatever number of integrals.


lets say you stand at the top of the empire state building and you took an insanly fast gun and shot it downward. if the bullet starts off at a faster speed than the terminal speed, would it decelerate? this is hypothetical

Yes.  The drag force would exceed the force of gravity so it would decelerate to its natural terminal speed.  If you look at the solution to the differential equation of motion, you see that from below the terminal velocity, it rises asymptotically and from above, it lowers asymptotically.


Climbing the first flight of stairs takes Jimmy 20 seconds, and each following flight takes 5 seconds more than the preceding one. How many total seconds does it take to climb the first five flights of stairs?

[quote="mickeythedunce"][hide]
y1 = 20
y2 = 25
y3 = 30
So on and so forth.
You set up an explicit sequence for this and get.

y = 15 + 5n
Where n is the number of flights he climbs.

y = 15 + 5[5]
y = 40
It will take Jimmy 40 seconds to climb 5 flights of stairs.[/hide][/quote]
Incorrect. It says "5 seconds more than the previous one."

[hide="My Answer"]$20+25+30+35+40=\boxed{150}$[/hide]

[quote="nutz_for2.718281828"][hide="solution"]The total time it takes is 20+21+22+23+24. We can take the average of the first two and multiply it by 5: 20+24=44, 44/2=22, 22*5=110. Therefore, the sum is 110.[/hide][/quote]
I repeat: it says "5 seconds more than the previous flight," not 1 second more.

[quote="Chocolate milk"]My answer
[hide] $20+25+30+35+40=150$ [/hide][/quote]
When I looked at your answer I thought it was wrong (mine was 160 originally), but then I realized I made a mistake. That [b]is[/b] the correct answer.

[hide]Like pie said, 5 ADDITIONAL seconds.

First we have 20 seconds.

A general formula we can form is then:

y=20n + 5(n-1 + .... + 1)

y is the total time elapsed and n is the number of flights climbed.

Hence, after 5 flights,

y = 20(5) + 5(4+3+2+1) = 100 + 50 =150 seconds[/hide]

[hide]It will take Jimmy 150seconds because 20+25+30+35+40=150 seconds. :bruce:[/hide]


[quote="Victory.US"]let $ a,b,c$ be nonegative real number such that $ a \plus{} b \plus{} c \equal{} 3$. Show that inequality:

$ \frac {a}{ab \plus{} 1} \plus{} \frac {b}{bc \plus{} 1} \plus{} \frac {c}{ca \plus{} 1} \le \frac {3}{abc \plus{} 1}$

--------------------
High School for Gifted Students Vinh Universary Vietnam[/quote]

Dear friend, this is a nice inequality. And this is what I've just thought after seeing it. We see that $ \frac{a}{ab\plus{}1} \equal{}a\minus{}\frac{a^2b}{ab\plus{}1},$ so the inequality can be rewritten as
$ \sum \frac{a^2b}{ab\plus{}1} \plus{}\frac{3}{abc\plus{}1} \ge 3.$
According to the Cauchy Schwarz Inequality, we have
$ \left( \sum \frac{a^2b}{ab\plus{}1}\right) \left( \sum ca(ab\plus{}1)\right) \ge \left( \sum \sqrt{a^3bc}\right)^2 \equal{}9abc,$
Because $ \sum ca(ab\plus{}1) \equal{}3abc\plus{}ab\plus{}bc\plus{}ca \le 3abc \plus{}3,$ we get
$ \sum \frac{a^2b}{ab\plus{}1} \ge \frac{9abc}{ab\plus{}bc\plus{}ca\plus{}3abc} \ge \frac{9abc}{3\plus{}3abc} \equal{}\frac{3abc}{abc\plus{}1}.$
And this implies that
$ LHS \ge \frac{3abc}{abc\plus{}1}\plus{}\frac{3}{abc\plus{}1} \equal{}3.$

Dear anh Can, this is a nice problem . i have a very nice and easy proof :D use harnometric mean inequality.
 i'll post it latter  :roll:

who have any idea


Please help with the following problem:
How many solutions does $ \cos\bigg(x\plus{}\dfrac{\pi}{6}\bigg)\equal{}0.5, 0\le x\le1000$ have?

[hide="Hint"]That equation is equivalent to $ x \plus{} \frac {\pi}{6} \equal{} \pm \frac {\pi}{3} \plus{} k2\pi$, or $ x \equal{} \minus{} \frac {\pi}{6} \plus{} \pm \frac {\pi}{3} \plus{} k2\pi(k\in\mathbb{Z})$. Now, see the inequality
\[ 0\leq \minus{} \frac {\pi}{6} \plus{} \pm \frac {\pi}{3} \plus{} k2\pi\leq 1000.
\]
[/hide]


find all polynomial p(x) such that:
p(x^k+a)=(p(x))^k+a   for x in real and a in real

finally i have the solution .
first assume that p(0)=0 then
give the consequence {b}n that {b}0=0 and {b}(n+1)=(b{n})^k+a;
we easy to prove that p({b}n)={b}n;
such that p(x)=x for all x in real
now p(0)<>0 then
we easy to prove that p(x)=q(x^k);
so we have q((x^k+a)^k))=(q(x^k))^k+a;
let q(x)=h(x+a) then we have h((x^k+a)^k+a)=(h(x^k+a))^k+a
it equivalence h(y^k+a)=(h(y))^k+a;
continue with the same steps we have p(x)  :D   :D   :D  :D

First we put $a=0, k=0$ and we get $p(x^0)={p(x)}^0$ so $p(1)=1$. Then we put $x=1$ and we get $p(1+a)=p(1)^k+a=1+a$, and if we put $1+a=y$, we get $p(y)=y$, so it has just one solution.

i think your solution have some mistake because p(x)=x^k+a is the solution

you mean that i can't put a as i want? because we put the solution in the same time so i can't see your solution first... :)


function algebra polynomial calculus derivative integration limit

find all  functions which are analytic  everywhere have a zero of order two at $ z\equal{}0$ satisfy the condition $ |f'(z)|\le 6|z|$ for all z and are such that $ f(i)\equal{}\minus{}2$

If the size of an entire function ($ f'$ in this case) is bounded by a polynomial of degree $ d$, what can you say about it?

We have Liouville's theorem that an entire function bounded by a constant is constant. It would sure be nice if there were an analogous result for polynomials. Can you prove it?

if we apply louville theorem
$ f '(z)$ is bounded and since $ f(z)$ is analytic so $ f '(z)$ is analytic
so $ f '(z)$ must be constant
then $ f(z)\equal{}az\plus{}c$ where a and c are constants
then
$ f(i)\equal{}\minus{}2$
then what?

No, $ f'$ is not bounded by a constant. You cannot directly apply Liouville here.

it means $ f(z)$ must be constant but how can we say that $ f(z)$ is bounded.

You're obviously not understanding my posts. There is exactly one function that satisfies all of the conditions in your initial post. It is not constant. Its derivative is also not constant.

if $ |f(z)| \le M |z|^n$ then f(z) is a polynomial of degree at most n
$ \Rightarrow$  here $ |f'(z)|\le 6 |z|$
$ \Rightarrow$  $ f'(z)$ is a polynomial of degree one
$ \Rightarrow$  $ f'(z)\equal{}a_1 z$ where $ a_1$ is any constant
$ \Rightarrow$   $ f(z)\equal{}a_1z^2 \plus{}c$
since f(z) has a zero of order 2 at $ z\equal{}0$
so $ c\equal{}0$
$ \Rightarrow$ $ f(z)\equal{}a_1z^2$
now $ f(i)\equal{}\minus{}2$ $ \Rightarrow$ $ a_1\equal{}2$
so $ f(z)\equal{}2z^2$

[quote="JAS"]if $ |f(z)| \le M |z|^n$ then f(z) is a polynomial of degree at most n
[/quote]

Do you know why? More generally if there exists an $ R>0$ such that $ |f(z)|\le M|z|^n \ \forall \ z\in\mathbb{C}\setminus B_R(0)$ then $ f(z)$ is a polynomial of degree at most $ n$.

[hide="Proof"]Consider the contour $ \{\gamma\} \equal{}\{z| \ |z|\equal{}R_0\}$ for $ R_0>R$. Then for all $ m>n$ by Cauchy's Theorem we have

\[ |f^{(m)}(0)| \equal{} \frac{m!}{2\pi}\left|\int_{\gamma} \frac{f(z)}{z^{m\plus{}1}}\ dz\right| \le \frac{m!}{2\pi}\int_{\gamma}\left|\frac{f(z)}{z^n}\right|\frac{1}{|z^{m\plus{}1\minus{}n}|} \ |dz| \le \frac{m!M}{2\pi R_0^{m\plus{}1\minus{}n}} \int_{\gamma} |dz| \equal{} \frac{2\pi R_0 m! M}{2\pi R_0^{m\plus{}1\minus{}n}} \equal{} \frac{m!M}{R_0^{m\minus{}n}}.\]

Now this holds for all $ R_0>R$ and note that $ m\minus{}n>0$. Hence

\[ \lim_{R_0\to\infty} \frac{m!M}{R_0^{m\minus{}n}} \equal{} 0\]
implies $ |f^{(m)}(0)| \equal{} 0$ for all $ m>n$. Since $ f$ is entire, Taylor's theorem says

\[ f(z) \equal{} \sum_{k\equal{}0}^\infty \frac{f^{(k)}(0)}{k!} z^k \equal{} \sum_{k\equal{}0}^n\frac{f^{(k)}(0)}{k!} z^k.\]

Thus $ f$ is a polynomial of degree at most $ n$.[/hide]

thanks a lot.
is my solution in post 10 is right?


Let  ABC.Find  MAX
P=$ cosA\plus{}cos\frac{B}{2}\plus{}cos\frac{C}{3}$

hello, what is $ ABC$, a triangle?
Sonnhard.

yes .ABC is a  triangle
Let help me! thank!


One test for divisibility by 7 states that 10n+d is divisible by 7 exactly when n-2d is divisible by 7, where d is a digit and n is a positive integer. The fifteen digit number 123456789101112d is 5 bigger than a multiple of 7. Find the digit d for which the 16 digit number 123456789101112d is divisible by 7

[quote="xxm0rg07hxx"][hide]I wouldn't even care about the first part of the question.  Just divide it out until you get to the last digit and see what fits for being divisble by 7.  In this case it's [b]6[/b].[/hide][/quote]

How did you do that? Wouldn't it be too big of a number to put on the calculator (without 1.234...E15)?

he means long division.

The calculator thingy is what I tried too until I realized the way youre supposed to do the problem.

[hide]It only took me 17 seconds to divide out like 12/7 leaves a remainder of 5 and then 53 leaves 4 and so on.  Being a target, 17 seconds is way less than 3 minutes per problem.[/hide]

[hide="the way its supposed to be done"]
Call the smaller number $n$.

Then $2d=5$ and $d=2.5$??

Know that you also can subtract 12 from $n$ to get a multiple of 7.

Thus $d=6$
[/hide]

Meh...
I had tried understanding the first part.
But it was faster just to long divide it...
like morg07

[quote="rcv"]The fifteen-digit number is misquoted.  With the correction, this is the actual question:
[quote]One test for divisibility by 7 states that $10n+d$ is divisible by 7 exactly when $n-2d$ is divisible by 7 where $d$ is a digit and $n$ is a positive integer.  The fifteen digit number $123,456,789,101,112$ is 5 bigger than a multiple of 7.  Find the digit $d$ for which the sixteen digit number $1,234,567,891,011,12d$ is a multiple of 7[/quote][/quote]

>< Thanks.
But I don't see a difference...
[hide]Since $123,456,789,101,112\equiv 5 \bmod{7}$ and $5$ is odd, we can't add the last digit to take away twice of it to minus $5$. So then by subtracting $7+5=12$, we can add $\boxed{6}$ to the last digit as $d$. [/hide]

Instead of long division, you can just divide parts of it.  For example, first divide just $1,234,567$ by 7 and find the greatest multiple of seven less than it (it is $1234562$). There is a remainder of 5, so divide $5,891,011$ by 7; it is divisible by 7 exactly.

Then we have $12d\rightarrow 5d\rightarrow d=6$


hi
many times im looking for a problem which is (or it seems to that it is) commenly known and discussed on the forum at least once. but it is very difficult for me looking for e.g. an inequality or something like that. but some how, some people on this forum post links to threads very frequently. i wonder whether there are any tricks and tips in searching for previously asked problems.

thx peeta

Are you using the Search button at the top of the forums or are you just scanning posts?

I think some of the people who know where particular problems are have used their own bookmarks to keep track of important threads.  Others probably use the search feature.  Still others might simply remember where threads are that they've seen before.

Unfortunately, due to the symbolism of mathematics, it's not quite as easy to search math information.

Yeah, some people have them in their favorites, others retain who posted in that particular topic and search for that user's posts in that particular forum, or just search with a more common keyword in a specific forum until they find that actual post :)

While we are on the topic of searching I though I would bring something up.  Whenever I do a search and I try to go to click next it gives me a

"No topics or posts met your search criteria."

message.  So I can only view the first page of results.  I think this is a bug.  I am using IE 6.

Hmm, I have no problem with that with either Firefox 1.0.4, IE 6 sp2 or Opera 7.54 ...  :?


Isn't this question that was on Mathcounts Chapter few years ago?

I recognize this one. :)

ummm....i just made it up but there are a lot of questions similar to this one

Actually I'll change the question to be similar to AIME format but still using an elementary concept. :D

The problem has been moved and now is Problem (1) of my contest  :D


[quote="Treething"]The combined sum of three squares of integral lengths is 69. What is the largest possible sum of the perimeters of these three squares?[/quote]

[hide]49,16,4

$4(7+4+2)=4(13)=\boxed{52}$[/hide]

[quote="Treething"]The combined sum of three squares of integral lengths is 69. What is the largest possible sum of the perimeters of these three squares?[/quote]

[hide]
Sides of the sq: 7, 4, 2

areas: 49+16+4=69
perimeter:4*13=52
[/hide]


Can we use n sticks which have the same length=l to make n equil triangles whose side=l?
        Example: 6 sticks -> 4 equil triangles.

dont know what u mean...

can the segments overlap?


geometry
perimeter
inequalities
number theory
Diophantine equation

Find the number of different right triangles which have these conditions:
1) Each side length is positive integer;
2) Perimeter equals to $ 6^n$ ($ n\in{\math{N}}$)
Express your answer in terms of $ n$

Uh, I got $ \frac{(n(n\plus{}2))^2}{8}$ for number of distinct right triangles that satisfy above conditions where $ n$ is even.

But I will not post my solution because I highly doubt my formula is correct, since it claims that there are 8 right triangles with perimeter 36. :|

Anybody except FantasyLover tried to solve this problem?
Fantasy, try to find mistakes in your solution, and try again :)

[quote="FantasyLover"]Uh, I got $ \frac {(n(n \plus{} 2))^2}{8}$ for number of distinct right triangles that satisfy above conditions where $ n$ is even.

But I will not post my solution because I highly doubt my formula is correct, since it claims that there are 8 right triangles with perimeter 36. :|[/quote]

It also claims that there are $ 1/8$ distinct right triangles with perimeter 6... in fact there are 0.

Euclid's formula is sufficient for producing all Pythagorean triples:
\[ (a,b,c)\equal{}(2kmn,k(m^2\minus{}n^2),k(m^2\plus{}n^2))\]
And the sum is \[ k(2mn\plus{}2m^2)\equal{}2km(m\plus{}n).\]
And we want this to equal $ 6^x$ (using a different variable to prevent confusion with the $ n$ above). I don't have time to finish this, hopefully this will be useful.

Why don't you just solve this diophantine equation
$ a^2\plus{}b^2\equal{}(6^n\minus{}(a\plus{}b))^2$
instead of thinking hard solutions? :)

$ a^2\plus{}b^2\equal{}6^{2n}\minus{}2*6^n(a\plus{}b)\plus{}a^2\plus{}2ab\plus{}b^2$

$ 6^{2n}\minus{}2*6^n(a\plus{}b)\plus{}2ab\equal{}0$

$ 6^{2n}\plus{}2ab\equal{}2*6^n(a\plus{}b)$

$ 3^{2n}2^{2n\minus{}1}\plus{}ab\equal{}6^n(a\plus{}b)$

Then plug in Euclid's formula.

I shall run away now before people tell me that I should have done it.

[hide]$ (a\minus{}6^n)(b\minus{}6^n)\equal{}2^{2n\minus{}1}3^{2n}$
$ a\minus{}6^n$ and $ b\minus{}6^n$ are divisors of $ 2^{2n\minus{}1}3^{2n}$
Also remember that $ a$ and $ b$ have to satisfy these inequalities:
$ 0<a<2^{n\minus{}1}3^n$ , $ 0<b<2^{n\minus{}1}3^n$
since $ a$ and $ b$ are legs.
Now try to finish solution.[/hide]

Maybe someone can finish my idea..

Let $ a,b,c$ be the sides of the triangle, where $ c$ is hipotenuse.

If $ a\equal{}b$ then hipotenuse c would be irrational. Then a is different from b.

We have $ a\plus{}b\plus{}c \equal{} 6^n$

$ \equal{}> a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2bc\plus{}2ac \equal{} 6^{2n} \equal{}> 2(c^2\plus{}ab\plus{}bc\plus{}ac) \equal{} 6^{2n}$

$ \equal{}> (*) c^2\plus{}ab\plus{}bc\plus{}ac \equal{} 3^{2n}2^{2n\minus{}1} \equal{}> (a\plus{}c)(b\plus{}c) \equal{} 3^{2n}2^{2n\minus{}1}$

or find values of n for which:

$ (a\minus{}b)^2\plus{}36^n.2$ is a perfect square since this is the discriminant of equation (*).


let  x_n   such that :$ x_{n\plus{}2}\equal{}\frac{x_{n\plus{}1}}{x_n} .$
 where ; $ x_1\equal{}1998$ and$ x_2\equal{}1999$.
 find value of$ x_{2007}$ and formul  of x_n

$ z_n\equal{}\ln{x_n}$ then
$ z_{n\plus{}2}\equal{}z_{n\plus{}1}\minus{}z_n$
It is easy to find general explit of $ z_n$ so $ x_n$

[quote="TTsphn"]$ z_n \equal{} \ln{x_n}$ then
$ z_{n \plus{} 2} \equal{} z_{n \plus{} 1} \minus{} z_n$
It is easy to find general explit of $ z_n$ so $ x_n$[/quote]

Indeed, the relevant characteristic equation is $ x^2 \minus{} x \plus{} 1 \equal{} 0$, so the general formula is $ z_n \equal{} A \omega^n \plus{} B \omega^{5n}$ (for appropriate $ A, B$) where $ \omega, \omega^5$ are the primitive sixth roots of unity (hence the periodicity observed by ElChapin).


calculus
derivative
integration
inequalities
analytic geometry
graphing lines
slope

$f(x)\not\equiv 0$ and it is differential on $[a,b]$ satisfying $f(a) = f(b) = 0$. Prove there exists $c\in [a,b]$ such that
\[|f'(c)| > \frac{4}{(b-a)^2}\int_a^b |f(x)|dx\]

Assume that for $t>0$ we have $\displaystyle\max_{c\in[a,b]}|f'(c)|\le t$. We want to prove that $\displaystyle \int^b_a |f(x)|dx<\frac{(b-a)^2}4t$. Let $A(a,0=f(a)),B(b,0=f(b))$. Draw the points $U,V$ s.t. the slopes of $UA,VB$ are $t$, and the slopes of $UB,VA$ are $-t$. It is clear (in order to prove this just flip the triangle $AVB$ around the $Ox$ axis, and all the negative parts of $f$ will now be above $Ox$) that the area between $f$ and $AB$ can be fitted inside the triangle $AUB$, so $\displaystyle \int^b_a |f(x)|dx<\frac{(b-a)^2}4t$ because $AB=b-a$ and the height of triangle $AUB$ corresponding to $AB$ is clearly $\displaystyle \frac{b-a}2t$.

I'm sure there is a standard, technical solution using calculus language, but I like this one :D.

My solution in calculus langauge, but I don't know how to

Suppose $\max |f'(x)| = M$

\[\int_a^b |f(x)|dx = \int_a^{\frac{a+b}2} |f(x)| dx + \int_{\frac{a+b}2}^b |f(x)| dx\]

\[f(x) = f(a) + f'(\xi)(x-a) = f'(\xi)(x-a),\quad \xi\in[a,\frac{a+b}2]\]
\[f(x) = f(b) - f'(\theta)(b-x) = f'(\theta)(x-b),\quad \theta\in[\frac{a+b}2,b]\]

thus

\[\int_a^{\frac{a+b}2} |f(x)| dx \leq M\int_a^{\frac{a+b}2} (x-a)dx = M\frac{(a-b)^2}8\]
\[\int_{\frac{a+b}2}^b |f(x)| dx \leq M\int_{\frac{a+b}2}^b (b-x)dx = M\frac{(a-b)^2}8\]

So

\[\int_a^b |f(x)|dx \leq M\frac{(a-b)^2}4.\]

The equality cannot be obtained


calculate the following
99.......9 (n repeats of 9) x 99........9 ( n repeats of 9 ) + 199.......9 (1 followed by n repeats of 9 )

[hide]Since the difference of the squares of consecutive integers is the sum of the consecutive integers, and 1999...9(n repeats of 9)is the sum of the consecutive integers 999...9(n repeats of 9) and 1000...0(n repeats of 0). The answer is 1000...0 :^2: (n repeats of 0). This ends up being 1000...0(2n repeats of 0)[/hide]


probability geometry calculus AMC AIME algebra polynomial \/closed

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If you have any questions about a specific class, please contact us at classes@artofproblemsolving.com.

For Precalculus, should we have mastered Algebra 3 and Intermediate C/P?

[quote="math is never easy"]Whats the difference between precalculus and algebra 3.[/quote]

Algebra 3 has material from a typical school-curriculum Algebra II and non-trig pre-calculus. The AoPS Precalculus has all the precalc topics not covered in Algebra 3 (e.g. trigonometry) plus more complex numbers and vectors.

[quote="Caelestor"]Will the AIME stuff be mostly covered in WOOT? (I've taken the AMC 12 class before)[/quote]

I think most of the late AIME stuff will be covered in WOOT, meaning later than #5

[quote="Caelestor"]Will the AIME stuff be mostly covered in WOOT? (I've taken the AMC 12 class before)[/quote]

All of our Intermediate classes (including Precalculus) cover material in-depth that will be helpful on the AIME, and in other applications of mathematics.

Precalculus is the more comprehensive course, so I say precalc > C/P

[quote="remy1140"][quote="rrusczyk"]You should have many of the fundamentals of Intermediate Algebra down -- polynomials, functions, series.  You don't have to have Intermediate Counting before Precalculus.[/quote]

This may be a somewhat vague question: is Intermediate C&P or Precalculus harder? I know that the two cover completely different fields in math, but which one has more challenging problems?[/quote]

This probably depends on what you're personally good at.  I'd say they're of comparable difficulty.

does AoPS plan on giving any olympiad NT courses?

Does olympiad geometry cover all of the geometry in WOOT?

Hard question to answer.  There's overlap in general subject matter, but not everything that appears in one appears in the other.

Oops, sorry.  :blush:  I thought..
[quote=""Calculus description""]A course in single-variable calculus. This course covers limits, continuity, derivatives and their applications, definite and indefinite integrals, infinite sequences and series, plane curves, polar coordinates, and basic differential equations.[/quote]
I assume this not fully updated?

Also, how long will the Precalculus book be? Is it about the same length as Intermediate Counting & Prob?

A good bit longer than the Interm Counting book, though part of that is because vectors and matrices (and diagrams for the geometry chapters) chew up a lot of space.  Not sure what you're referring to about the Calculus...

I got a little confused with the Course Description for Calculus in the Online Classes section, which said "A course in single-variable calculus." :) Maybe I just misunderstood the description.

Why is Geometry the only sub-subject of the 4 Introductions that doesn't have an Intermediate course, but rather an Olympiad Course? Will there be the book Olympiad Geometry?

I think that because there are so many "olympiad geometry" books(geometry revisited for example), that AoPS isn't going to release their own.(I'm not sure about this)

We wrote the Olympiad Geometry course very early on, since there were so many people that clearly wanted it (and because it was fun for me to write).  Books take a lot of time and energy to produce, and we have a lot of other things on the todo list before we get to more geometry books.

Is the AIME on Thursday going to be offered at 4:30 or only 6:00?

Only at the later time.  There is an AIME class at the usual time on Wednesdays.

When will the books for calculus and precalculus be available to buy on their own?

Somewhere around the end of this calendar year.


How many nonempty subsets of {1,2,3,...12} have the property that the sum of the
largest element and the smallest element is 13?

set the smallest and largest terms.  Then you can choose either yes or no to accept any middle terms so its 2^10+2^8+2^6...+2^0 using summation its 1365.

umm so sorry to bring this back but is my answer right mdk?

yes, your answer is indeed correct. Good job.


inequalities
geometry
perimeter
triangle inequality

Prove that in any convex quadrilateral, sum of two diagonals is less than its perimeter but greater than half the perimeter....

[hide]
Let the sides have lengths a, b, c, d and the diagonals e, f.
By the triangle inequality, we have $ a\plus{}b > e$, $ c\plus{}d>e$, $ a\plus{}c>f$, $ b\plus{}d>f$.
Adding and dividing by 2, we have $ a\plus{}b\plus{}c\plus{}d>e\plus{}f$ for the first part.

The intersection of the diagonals divides e into g and h and f into i and j.
By the triangle inequality, we have $ g\plus{}i>a$, $ g\plus{}j>b$, $ h\plus{}j>c$, $ h\plus{}i>d$.
Adding, we have $ 2(e\plus{}f)>a\plus{}b\plus{}c\plus{}d$, so $ e\plus{}f>\frac{a\plus{}b\plus{}c\plus{}d}{2}$ for the second part of the problem.[/hide]


geometry MATHCOUNTS Stanford college ARML

Isn't "Bay Area" too small of a place to have its own forum?  Wouldn't a California forum be better? ;)

I don't think it's too small, but there doesn't seem to be a lot of posters here.  I say sub-sub forum

I say sub-sub form. And the people who are voting 'I don't care', why even bother voting? :D

[quote="i_like_pie"][quote="junggi"][quote="i_like_pie"]I think sub-sub-forum would be best because this is too small of an area for a sub-forum and there aren't many people how post here.[/quote]
I don't think you know what's going on right now with JLS and their MATHCOUNTS people. This forum will be super active now[/quote]
What does JLS stand for? :huh:[/quote]
JLS=Jane Lathrop Stanfor middle school(i think it was named after stanford's wife, but maybe not, and i dont care)=the coolest middle school in palo alto=jls math ppl pwned last year, but this year...um...they might not pwn

it is named after stanford's wife but still. i only liek this board and will only post here now. otherwise im offline

Just here to say that we actually have a so. cal. forum (hidden) tho. So No.

well...you need membership to get in, and for a membership you have to contact Kent Merrifield via pm.  It's for the So Cal ARML team so I doubt that you're interested...

this is going to be super active like the JLS one?

wow it's like only you freshman people posting......

I'm not even supposed to post but whatever.

hey bluestalk. i see that you are on. no i was testing smileies. ccy probably killed it, but yeah

o and btw, you have aim?


Two different numbers are randomly selected from the set
$ S \equal{} \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. The probability that their
sum is 12 would be greater if the number $ n$ had first been
removed from set $ S$. What is the value of $ n$?

To make the probability of the pair's of number's sum 12 greater, we try to remove the number that has the least "pairs" that it can be in that add up to 12. After some examination, we find that $ 1\plus{}11\equal{}12; \; 2\plus{}10\equal{}12; \; 3\plus{}9\equal{}12\dots$
This means that every number is "paired" with the number that it is added to to make 12, and only that number.

But, if we look in the middle, we see that the number $ 6$ is in the exact middle and has no pair ($ 6\plus{}6\equal{}12$, but we cant use $ 6$ twice). So, our answer is $ \boxed{6}$.


Hai I thought I could just check my homework ... the question is this

at a proce of $20 per 1000cups, 400 000 cups per week will be sold, but for every price rise of$1 per 1000 cups sales will drop by 2000 cups per week.

QUESTION -- Find a function which will determine the revenue from the sale of the cups per week

So I figured easy an put down

R = (20+X)*(400000-(2000*X))

Where R is revenue and X is dollars increase

I'm pretty sure thats right, but is there anyother way's of putting it?

Thanx
Telon

I think your equation is right. Also, i dont think this is supplosed to be in the olympiad section.

Check again.  The price you put is per 1000 cups, not per cup.


If $ \frac {1}{x} \minus{} \frac {1}{y} \equal{} \frac {1}{z}$, then $ z$ equals:

$ \textbf{(A)}\ y \minus{} x\qquad \textbf{(B)}\ x \minus{} y\qquad \textbf{(C)}\ \frac {y \minus{} x}{xy}\qquad \textbf{(D)}\ \frac {xy}{y \minus{} x}\qquad \textbf{(E)}\ \frac {xy}{x \minus{} y}$

Wait. Is that possible?

So if I say a=b-c and then you can solve for d?

hello, how can we solve the equation $ \frac{1}{x}\minus{}\frac{1}{y}\equal{}\frac{1}{x}$ for $ z$?
Sonnhard.

$ \frac{1}{x}\minus{}\frac{1}{y}\equal{}\frac{1}{z}$

$ \frac{y}{xy}\minus{}\frac{x}{xy}\equal{}\frac{1}{z}$

$ \frac{y\minus{}x}{xy}\equal{}\frac{1}{z}$

$ \frac{1}{\frac{y\minus{}x}{xy}}\equal{}\frac{1}{\frac{1}{z}}$

$ \frac{xy}{y\minus{}x}\equal{}z$

$ Answer\equal{}\mathbf{(D)}$


For positive a, b, c, this is http://www.mathlinks.ro/Forum/viewtopic.php?t=30764 . If some of a, b, c are negative, you can conclude it from the same inequality for |a|, |b|, |c|, which is surely stronger.

  Darij


Prove, that for every prime number $p$ there exist infinetely many natural $n$ such that $p | 2^{n}-n$.

if $k$ is order of 2 mod p, then $n$ which satisfies $k|n$ and $p|n-1$ has wanted property.

Yep, it is very similar to problem from 250 problems in number theory by Sierpinski, all the same except the formula $p|n 2^{p}+1$. The solution is almost identical :)