group theory
abstract algebra
linear algebra
matrix
algebra
polynomial
Galois Theory

I found a good one on a qualifier

$ F$ is an algebraic extension that is infinite over a base field $ K$ and $ G\equal{}Aut_K F.$  Then the intersection of all normal subgroups of finite index in $ G$ is trivial.

If you get troubles I can give hints.

(EDIT to clarify, in light of the following discussion: we assume, without loss of generality, that $ F$ is Galois over $ K$, because we can otherwise consider the fixed field of $ G$)

Let $ \sigma$ be an element of the intersection of these groups, $ \alpha$ any element of $ F$, and $ L$ the splitting field of $ \alpha$.

Let $ H \equal{} Aut_L F$, then we claim that $ H\triangleleft G$ and $ H$ has finite index.  But this is simply the correspondence theorem.

By the assumption, $ \sigma\in H$, so $ \sigma$ fixes $ K$, so in particular $ \sigma \alpha \equal{} \alpha$.  Since $ \alpha$ was arbitrary, $ \sigma$ must be the identity.

It has been pointed out that this proof is broken, as it assumes that the given extension is Galois.

Simply take the fixed field $ K_2$ of $ G$. Then $ F|K_2$ is galois with group $ G$.

Just realized it after posting.  Easy to get nailed on stuff like this in the quals, I suppose that's their purpose. :)

I am not convinced that olorin's argument of restricting to the fixed field gives Galois.  Where do you get separable? Normal? Consider the infinite algebraic extension over $ \mathbb{Q}$ formed by adjoining all the $ \sqrt [3]{p}$ for all primes $ p$.

The first reply can be fixed for correspondence is not needed (or valid).

[EDIT: Field definition]

In general, it is true that an algebraic extension is Galois iff the splitting field of the automorphism group is the base field.  I don't understand what you're getting at in your example.

Why do you say that correspondence is not needed?  It is certainly valid.

EDIT: corrected wording, and some other completely wrong things I said

An infinite algebraic extension is not necessarily normal.

As for correspondence, which do you refer to?  I assume you mean the Galois Correspondence.

Please read what I am writing more closely, as I never claimed that all algebraic extensions are normal.  However, any algebraic extension is both normal and separable if the base field is equal to the fixed field of the Galois group. (actually, I think that this is true even we drop the assumption that the extension is algebraic)

Try applying this to your example.  The fixed field of the Galois group of your extension is the field itself, and the trivial extension is obviously normal. :)

You assume correctly that I am referring to the Fundamental Theorem of Galois Theory.

A stronger version of my claim is Theorem 3 [url=http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-702Spring2003/0BFA69D0-B2F9-464A-8B22-9AC82B4AE0E2/0/galoistheorems.pdf]here[/url].

Elsewhere, I have seen this as the definition of a Galois extension.

Perhaps I am being dense about this, but why are we able to assume that the automorphism group is finite as in Theorem 3?

I guess this still needs more discussion.

[quote="LydianRain"]I think that this is true even we drop the assumption that the extension is algebraic.[/quote]
This is not true.
Look at the automorphism $ K(X)\to K(X)$ induced by $ X\mapsto \lambda X$ for any $ \lambda\in K^\times$.
Then $ K$ is the fixed field of those automorphisms (if $ K$ is not finite  :wink: )
and $ K(X)/K$ is not algebraic and therefore not galois.

However in this case $ K\subseteq K_2\subseteq F$, so $ F/K_2$ is algebraic.
Then for any $ \alpha \in F$ take $ f(X)\in K_2[X]$ the minimalpolynomial of $ \alpha$.
Then $ \sigma (\alpha)$ is also root of $ f(x)$ for all $ \sigma \in G$. ($ \sigma$ fixes the coefficients of $ f(X)$, so $ 0\equal{}\sigma(f(\alpha))\equal{}f(\sigma(\alpha))$)
Let $ \alpha_1\equal{}\alpha,\alpha_2,\ldots,\alpha_n$ be the different images of $ \alpha$ under $ G$. ($ n\le \deg f$)
Then $ G$ permutes $ \alpha_1,\ldots,\alpha_n$, which induces a group homo $ G\to S_n$ with kernel $ H\unlhd G$ and $ |G/H|\le |S_n|\equal{}n!$.
So $ H$ has finite index fixing $ \alpha$. (now look at LydianRain's final argument)

Also $ G$ permuting $ \alpha_1,\ldots, \alpha_n$ gives the coefficients of $ g(X): \equal{}(X\minus{}\alpha_1)\cdots(X\minus{}\alpha_n)$ fixed under $ G$.
So $ g(X)\in K_2[X]$ with $ g(\alpha)\equal{}0$ gives $ f(X)|g(X)|f(X)$, i.e. $ f(X)\equal{}g(X)$. (both monic)
So the minimalpolynomial $ f(X)$ splits into different linear factors for any $ \alpha\in F$, which gives $ F/K_2$ normal and separable.

Precisely, olorin.

To see that the automorphism group is not necessarily finite, simply note that there are uncountably many automorphisms of $ \mathbb{C}$ over $ \mathbb{Q}.$

Thanks for the nice proof, olorin.

Though the assumptions I was making turned out to be correct, I apologize for the sloppiness. (was really in too much of a hurry earlier, should have read the page more closely)

OK, took me a while to figure out,
but if $ K$ is a finite field with $ |K|\equal{}q$,
then $ Y: \equal{}{(X^{q^3}\minus{}X)(X^q\minus{}X)^q\over (X^{q^2}\minus{}X)^{q\plus{}1}}\in K(X)\setminus K$ is fixed by all $ K$-automorphisms of $ K(X)$
and $ K(X)/K(Y)$ is galois of degree $ q^3\minus{}q$.
(try to prove that  :lol: )

[quote="olorin"]if $ K$ is a finite field with $ |K| \equal{} q$,
then $ Y: \equal{} {(X^{q^3} \minus{} X)(X^q \minus{} X)^q\over (X^{q^2} \minus{} X)^{q \plus{} 1}}\in K(X)\setminus K$ is fixed by all $ K$-automorphisms of $ K(X)$
and $ K(X)/K(Y)$ is galois of degree $ q^3 \minus{} q$.[/quote]
Is nobody interested in this?  :maybe:
I think this is very beautiful.
Or should I give some hints?

I don't get what's the problem, but anyway here are some hints.

First show, that every $ K$-automorphism of $ K(X)$ is induced by
$ X\mapsto {aX\plus{}b\over cX\plus{}d}$ for some $ a,b,c,d\in K$ with $ ad\minus{}bc\not\equal{}0$. (over any field $ K$)
This is the hard part.

Next prove that $ GL_2(K)\to \text{Aut}_K K(X), \left(\begin{matrix}a&b\\ c&d\end{matrix}\right)\mapsto (X\mapsto {aX\plus{}b\over cX\plus{}d})$ is a group homo with kernel $ K^\times I_2$,
and follow that $ \text{Aut}_K K(X)\cong PGL_2(K)\equal{}GL_2(K)/K^\times I_2$ has order $ q^3\minus{}q$ for $ |K|\equal{}q$ finite,
and that the fixed field $ L$ of $ \text{Aut}_K K(X)$ satisfies $ K(X)/L$ galois of degree $ q^3\minus{}q$ and galois group $ \text{Aut}_K K(X)$.

Next show that $ Y\equal{}{(X^{q^3} \minus{} X)(X^q \minus{} X)^q\over (X^{q^2} \minus{} X)^{q \plus{} 1}}$ is in $ L$ by looking only at the automorphisms
$ X\mapsto aX, a\in K^\times$ and $ X\mapsto X\plus{}1$ and $ X\mapsto {1\over X}$. (show that those generate $ \text{Aut}_K K(X)$)

Finally show, that $ Z\equal{}X$ is root of the polynomial
$ {Z^{q^3}\minus{}Z\over Z^q\minus{}Z}\minus{}Y\left({Z^{q^2}\minus{}Z\over Z^q\minus{}Z}\right)^{q\plus{}1}\equal{}\sum_{k\equal{}0}^{q^2\plus{}q}Z^{k(q\minus{}1)}\minus{}Y(\sum_{k\equal{}0}^{q}Z^{k(q\minus{}1)})^{q\plus{}1}\in K(Y)[Z]$
of degree $ q^3\minus{}q$ with leading coefficient $ 1\minus{}Y\equal{}1\minus{}{(X^{q^3} \minus{} X)(X^q \minus{} X)^q\over (X^{q^2} \minus{} X)^{q \plus{} 1}}\not\equal{}0$.
(Set $ X\equal{}\alpha$ algebraic of degree $ 3$ over $ K$)

This gives $ K(X)/K(Y)$ of degree $ \le q^3\minus{}q$, which gives $ L\equal{}K(Y)$.


inequalities
function
inequalities unsolved

If $ x,y,x>0$ such that $ x\plus{}y\plus{}z\equal{}1$ prove that

\[ \sum \frac{1\minus{}x}{1\plus{}x}x^2 \leq \frac{1}{6},\] where the sum is cyclic over $ x,y,z,.$

A solution without thinking.

Consider $ f(x)\equal{}\frac{x^2(1\minus{}x)}{1\plus{}x}$. We have $ f'(x)\equal{}\frac{\minus{}2x(x^2\plus{}x\minus{}1)}{(1\plus{}x)^2}$, i.e. $ f$ is increasing on $ [0$, $ \frac{\minus{}1\plus{}\sqrt{5}}{2}]$. Moreover, $ f''(x)\equal{}\frac{\minus{}2(x^3\plus{}3x^2\plus{}3x\minus{}1)}{(1\plus{}x)^3}$. We have $ f''(1/4)>0$ and $ f''(1/3)<0$. Hence the function is convex on $ [0$, $ x_0]$ and concave on $ [x_0$, $ 1]$, where $ 1/2<x_0<1/3$. Let wlog $ x \ge y \ge z$. Then there are three cases to consider: $ x \ge y \ge z\ge x_0$, $ x$, $ y \ge x_0 \ge z$ and $ x \ge x_0 \ge y \ge z$.
In the first case Jensen does the work.
If $ x \ge y \ge x_0 \ge z$, then $ f(x)\plus{}f(y) \le 2 f \left( \frac{1\minus{}z}{2} \right)$ and it's left to prove that $ \frac{z^2(1\minus{}z)}{z\plus{}1}\plus{}\frac{(z\plus{}1)(1\minus{}z)^2}{2(3\minus{}z)}\le \frac{1}{6}$, which is equivalent to $ g(z)\ge 0$, where $ g(z)\equal{} 2z(1\minus{}3z)(3(z\minus{}7/6)^2\minus{}1/4)\ge 12z(1\minus{}3z) \ge 0$, with equality for $ z\equal{}0$.
If $ 1\minus{}x_0 \ge x \ge x_0 \ge y \ge z$, we have (Karamata) $ f(y)\plus{}f(z) \le f(x_0) \plus{}f(y\plus{}z\minus{}x_0)\equal{}f(x_0)\plus{}f(1\minus{}(x\minus{}x_0))$ and it's left to prove $ f(x_0)\plus{}f(1\minus{}x\minus{}x_0)\plus{}f(x) \le 1/6$. Jensen yields $ f(x_0)\plus{}f(x) \le 2 f\left( \frac{x\plus{}x_0}{2} \right)$. Set $ 1\minus{}(x\plus{}x_0)\equal{}a \ge 0$. It's left to prove that $ 2f\left( \frac{1\minus{}a}{2} \right)\plus{}f(a) \le 1/6$, but this is the same as before. The case $ x > 1\minus{}x_0$ is left. We have $ x_0>y\plus{}z$ and hence $ f(y)\plus{}f(z) \le f(0)\plus{}f(y\plus{}z) \le f(0)\plus{}f(x_0)$, because $ f$ is increasing on $ [0$, $ x_0]$. Therefore it's left to prove $ f(x_0)\plus{}f(x) \le f(x_0)\plus{}f(1\minus{}x_0) \le \frac{1}{6}$, because $ f$ is decreasing on $ [1\minus{}x_0$, $ 1]$. Jensen yields $ f(x_0)\plus{}f(1\minus{}x_0) \le 2f(1/2)\equal{}1$ and everything is proved.


algebra
polynomial
trigonometry
Galois Theory
algebra unsolved

Your question isn't very interesting.
There exist 2 real solutions: $ u\approx\minus{}1.11,v\approx0.139$. These real are roots of the polynomial $ p(x)\equal{}64x^4\plus{}128x^3\plus{}80x^2\plus{}15x\minus{}4$.
The $ p$-Galois group is $ S_4$; thus there doesn't exist any algebraic relation between the $ p$-roots and you must use the Ferrari's formula for the fourth degree equation (cf. Wikipedia) or Maple:
allvalues(RootOf(p(x)));

[quote="thanhnam2902"]$ 8x^2 \plus{} 8x \plus{} 1$=$ \sqrt {x \plus{} 5}$
--------------------------
Thank every body![/quote]

As Loup blanc said, this problem has very few interest except if it exists a huge hint (trig, for ex).
What is the origine of this problem ?

Here is the full solution :

$ 8x^2 \plus{} 8x \plus{} 1$=$ \sqrt {x \plus{} 5}$

Raise to power two :$ 64x^4\plus{}128x^3\plus{}80x^2\plus{}15x\minus{}4\equal{}0$
Let $ x\equal{}y\minus{}\frac{1}{2}$
$ y^4\minus{}\frac{1}{4}y^2\minus{}\frac{1}{64}y\minus{}\frac{7}{128}\equal{}0$

Let $ y^4\minus{}\frac{1}{4}y^2\minus{}\frac{1}{64}y\minus{}\frac{7}{128}\equal{}(y^2\plus{}ay\plus{}b)(y^2\minus{}ay\plus{}c)$

We have then the system :
$ \minus{}a^2\plus{}b\plus{}c\equal{}\minus{}\frac{1}{4}$ and so $ b\plus{}c\equal{}\minus{}\frac{1}{4}\plus{}a^2$
$ a(b\minus{}c)\equal{}\frac{1}{64}$ and so $ b\minus{}c\equal{}\frac{1}{64a}$
$ bc\equal{}\minus{}\frac{7}{128}$

So we have $ b\equal{}\minus{}\frac{1}{8}\plus{}\frac{a^2}{2}\plus{}\frac{1}{128a}$, $ c\equal{}\minus{}\frac{1}{8}\plus{}\frac{a^2}{2}\minus{}\frac{1}{128a}$ and so :

$ bc\equal{}(\frac{1}{8}\minus{}\frac{a^2}{2})^2\minus{}\frac{1}{16384a^2}\equal{}\minus{}\frac{7}{128}$

$ \frac{9}{128}\plus{}\frac{a^4}{4}\minus{}\frac{a^2}{8}\minus{}\frac{1}{16384a^2}\equal{}0$

$ a^6\minus{}\frac{a^4}{2}\plus{}\frac{9}{32}a^2\minus{}\frac{1}{4096}\equal{}0$

Let then $ a^2\equal{}\frac{1}{48}z\plus{}\frac{1}{6}$
$ (\frac{1}{48}z\plus{}\frac{1}{6})^3\minus{}\frac{1}{2}(\frac{1}{48}z\plus{}$ $ \frac{1}{6})^2\plus{}\frac{9}{32}(\frac{1}{48}z\plus{}\frac{1}{6})\minus{}\frac{1}{4096}\equal{}0$

$ z^3\plus{}456z\plus{}4133\equal{}0$

This equation has one real root easy to compute with Cardan's formula :
$ z\equal{}\sqrt[3]{\frac{\minus{}4133\plus{}3\sqrt{3458769}}{2}}\plus{}\sqrt[3]{\frac{\minus{}4133\minus{}3\sqrt{3458769}}{2}}$

And so $ a\equal{}\sqrt{\frac{1}{6}\plus{}\frac{1}{48}(\sqrt[3]{\frac{\minus{}4133\plus{}3\sqrt{3458769}}{2}}\plus{}\sqrt[3]{\frac{\minus{}4133\minus{}3\sqrt{3458769}}{2}})}$

And the two requested roots :

$ x_1\equal{}\frac{a\minus{}1\minus{}\sqrt{\frac{1}{2}\minus{}a^2\plus{}\frac{1}{32a}}}{2}$ $ \equal{}\minus{}1.109551274...$

$ x_2\equal{}\frac{a\minus{}1\plus{}\sqrt{\frac{1}{2}\minus{}a^2\plus{}\frac{1}{32a}}}{2}$ $ \equal{}0.139036812...$

Thank you very much. You are a good friend!


calculus
integration
calculus computations


Bottle A contains more Diet Coke than Bottle B.  Now do the following:

a. Pour from Bottle A into B as much Diet Coke as B already contains.
b. Pour from B into A as much Diet Coke as A now contains.
c. Pour from A into B as much Diet coke as B now contains.

Both bottles now have 54 ounces.  How many more ounces were in A than in B at the beginning?

[hide]According to the answer key, it's 48, but if that's true, then A had 78 and B had 30 in the beginning.  After step a, B got 30x2=60, and A had 78-30=48.  After step b, A got 48x2=96, and B had 60-48=12.  After step c, B got 12x2=24, and A had 96-12=84, which is a contradiction with the problem.  At first when I did this problem, I got 40.5 ounces using my interpretation of the problem.  Is my understanding of the question wrong?

PS: It probably is.[/hide]

Let $ A$ be the amount initially in A, and $ B$ the amount initially in B.  Then keep a running list of equations to keep track of the amount in each bottle.  So the first one would be
$ A\minus{}B$ for A and $ B\plus{}B$ for B.

Yeah, that's what I did...  You get 40.5 as an answer.  I did this problem like 4 times and still came up with the same answer...

[quote="Math Geek"]Bottle A contains more Diet Coke than Bottle B.  Now do the following:

a. Pour from Bottle A into B as much Diet Coke as B already contains.
b. Pour from B into A as much Diet Coke as A now contains.
c. Pour from A into B as much Diet coke as B now contains.

Both bottles now have 54 ounces.  How many more ounces were in A than in B at the beginning?
[/quote]

[hide]Okay. Do this backwards.
After c: 54 54
Before c and after b: 81 27
Before b and after a: 40.5 67.5
Before a: 74.25 33.75
So I got 40.5 as well. Maybe we need to factor in evaporation or something. :D[/hide]

I also got 40.5, but according to the mathcounts errata for 1993,

1992-1993 MATHCOUNTS Competition Season (last updated on September 15, 2004)
• Chapter Competition, Countdown Round #19: The ordered pair (99, 1) would be an
acceptable answer.
• Chapter Competition, Countdown Round #60: The word “largest” should be inserted
before “possible.”
• State Competition, Sprint Round #16: The correct answer is 21
2 of 212 inches.
• State Competition, Sprint Round #23: The term “numbers” must be replaced with
“integers” to avoid an infinite number of answers.  The question should be phrased, “If x and
y are positive integers less than 20 for which x + y + xy = 76, what is the value of x + y?

so no national ones are wrong. Are you sure this is the 1993 year?

My 93 answer key has a completely different answer (meaning a completely different problem.)

I remember I did that problem before, and I could have sworn that both bottles were at 64 ounces at the end. Are you sure that you didn't read it wrong?

Um, a friend on AoPS emailed me his copy, and I used text-recognition.  Yeah, I'm pretty sure it's 64, then.  There must have been a mistake in text-recognition, and I didn't look at the original copy.  So yeah, thanks andersonw!  That explains a lot.

IHatepie, yeah, I'm sure it's 1993.  Maybe your answer key's wrong? :maybe:

oh, yeah, I just checked, it was not number 30, but 24.

Oh yeah, whoops...  Sorry about that.  I was typing too fast...  :blush:


How to cut a 1 by 3 rectangle into 3 pieces such that the three pieces would fit together into square?

This one seems pretty difficult. A hint would be fine.

all i could think of is [hide]finding the length of each side[/hide]

Solution is not obvious to me.  Best I can do is a 3-cut, 4-piece dissection.  Maybe you can enlighten us with the solution.

I got the solution now(which i didn't come up myself)

So that's supposed to be a "hint"?  j/p

[hide]

I don't have the advantage of a graphics program, so bear with me for this text description:

Let's label the 1x3 rectangle as ABCD, AB=CD=3, BC=AD=1.  Cut from B to a point E on CD that is :rt3: from C (creating triangle BCE).  At a point F on AB that is  :rt3: from A, cut on a line perpendicular to AB from F to point G, which lies on BE.  You now have an additional triangle BFG, and a pentagon AFGED.  Translate triangle BCE so that it's point B is at the same place as point G.  Translate triangle BFG so that it's point B is at the same place as pentagon AFGED's point E.  You now have a square with an area of 3.

[/hide]

[quote="beta"]Correct!!! Picture looks like below[/quote]

Beta, that is the weakest picture ever.  Funniest thing I've seen all day.

Oh, and nice solution :-)


geometry
3D geometry
real analysis
real analysis unsolved

Let $A_{n}$ be the number of nonnegative integer solutions of equation
$a_{1}x_{1}+\dots+a_{l}x_{l}= n$, where $a_{1}, \dots, a_{l}$ are natural numbers with no common divisor greater than 1. Prove that:
$lim \frac{A_{n}}{n^{l-1}}= \frac{1}{a_{1}\dots a_{l}(l-1)!}$.

There is something about the solution that I don't understand:

[hide=" Spoiler here"]The volume of the space satisfying $a_{1}x_{1}+\dots+a_{l}x_{l}< n$ is equal to $\frac{1}{l!}\frac{n^{l}}{a_{1}\dots a_{n}}$. The  author of the solution states that $A_{n}$ is asymptotical to $\frac{d}{dn}(\frac{1}{l!}\frac{n^{l}}{a_{1}\dots a_{n}})$. How can we strictly prove the latter conclusion?[/hide]

Given $\epsilon>0$, choose a large integer $m$ so that the number of lattice points in the set $\left\{n\le \sum a_{i}x_{i}\le n+m-1\right\}$ is between $(1\pm\epsilon)m\frac{d}{dn}\left(\frac{1}{l!}\frac{n^{l}}{a_{1}\dots a_{n}}\right)$. This is possible because the volume of a "nice" set can be compared to the number of lattice points in it by packings and coverings with unit cubes (provided that the boundary effects are negligible -- this is what $m$ is for). The second step is to argue that the equations $\sum a_{i}x_{i}=n$, $\sum a_{i}x_{i}=n+1$, ..., $\sum a_{i}x_{i}=n+m-1$ have about the same number of solutions when $n$ is large. This follows from the existence of integers $v_{1},\dots, v_{l}$ (independent of $m$ or $n$) such that $\sum a_{i}v_{i}=1$.


search
inequalities unsolved
inequalities

Prove that if  $0 \le a \le b \le c \le d$ then we have  $a^b b^c c^d d^a  \ge b^a c^b d^c a^d $

I have solved this one before.. please search for the posts I made ..


A palindrome is a number which can be read from left to right or right to left and the results are the same. A palindrome has no leading zeros. A palindrome twin is a number which is palindrome in two different bases. A palindrome set is a set of palindrome twins. How to determine all palindrome sets with at least four members where numbers are from 1 to 1000000 in base ten and a base is between 2,...,10?

1000000 isn't that high.  Can't you just find all of the palindromes in base 10 and check from base 2-9 if at least four of them are palindromic?


Suppose that $E$ is a connected space covered by finitely many closed sets $F_{1},F_{2},...,F_{n}$ and suppose $x,y\in E$. Then one can find $N$ a positive integer and some numbers $x_{1},...,x_{N}$ in $1,2,...,n$ such that $x\in F_{x_{1}}, y\in F_{x_{N}}$, $F_{x_{i}}$ has a non void intersection with $F_{x_{i+1}}$ and finally $E$ is covered by the sets $F_{x_{i}}$.

Say that $x$ and $y$ are equivalent if there is such a chain between them. $E$ splits into equivalence classes, each of which is closed, being a finite union of some of $F_{i}$s. Thus, there is only one equivalence class.

NB: I did not prove that the relation is reflexive, which is not entirely trivial, but is not hard either.


Find the sum to infinity of the infinite series $ \frac{1}{10} \plus{} \frac{2}{10^{2}} \plus{} \frac{3}{10^{3}} \plus{} \frac{4}{10^{4}} \plus{} \frac{5}{10^{5}} \plus{} ...$

It can be shown using the methods above that, in general, $ \sum_{k\minus{}1}^{\infty}\frac{k}{10^k}\equal{}\frac{k}{(k\minus{}1)^2}$.  Just something to keep in mind since things like this, and more complicated ones, crop up pretty often.

[quote="gfour84"]$ \sum_{k \minus{} 1}^{\infty}\frac {k}{10^k} \equal{} \frac {k}{(k \minus{} 1)^2}$.[/quote]  This doesn't make any sense ;)


geometry
angle bisector
similar triangles

For an isosceles triangle $ABC$ where $AB=AC$, it is possible to construct, using only compass and straightedge, an isosceles triangle $PQR$ where $PQ=PR$ such that triangle $PQR$ is similar to triangle $ABC$, point $P$ is in the interior of line segment $AC$, point $Q$ is in the interior of line segment $AB$, and point $R$ is in the interior of line segment $BC$. Describe one method of performing such a construction. Your method should work on every isosceles triangle $ABC$, except that you may choose an upper limit or lower limit on the size of angle $BAC$.
[asy]
defaultpen(linewidth(0.7));
pair a= (79,164),b=(19,22),c=(138,22),p=(109,91),q=(38,67),r=(78,22);
pair point = ((p.x+q.x+r.x)/3,(p.y+q.y+r.y)/3);
draw(a--b--c--cycle);
draw(p--q--r--cycle);
label("$A$",a,dir(point--a));
label("$B$",b,dir(point--b));
label("$C$",c,dir(point--c));
label("$P$",p,dir(point--p));
label("$Q$",q,dir(point--q));
label("$R$",r,dir(point--r));[/asy]

[hide="Answer"]

We bring a parallel form $P$ to $AB$, which intersects the side $BC$ at $R'$

The triangle $PR'C$ is isosceles and similar to the triangles $ABC,PQR$.

We have $\widehat{QPR'}= \widehat{RPC}$. That's simply because the angle $\widehat{QPR}$ is a rotation of the angle $\widehat{R'PC}$

[hide="Indeed:"]
a) $R'$ between $B$ and $R \Rightarrow \widehat{QPR'}= \widehat{QPR}-\widehat{R'PR}= \widehat{R'PC}-\widehat{R'PR}= \widehat{RPC}$
b) $R'$ between $R$ and $C \Rightarrow \widehat{QPR'}= \widehat{QPR}+\widehat{R'PR}= \widehat{R'PC}+\widehat{R'PR}= \widehat{RPC}$
c) $R' \equiv R \Rightarrow \widehat{QPR'}= \widehat{RPC}= \widehat{QPR}= \widehat{A}$[/hide]

[img]http://i7.tinypic.com/24vtai1.gif[/img]

triangles $\triangle PCR,\triangle PR'Q:$
$PR=PQ$
$PC=PR'$
$\widehat{QPR'}= \widehat{RPC}$

So they are congruent $\Rightarrow RC = R'Q$, and also $\widehat{PR'Q}= \widehat{C}$

$\Rightarrow$ the triangle $\triangle R'QB$ has external angle bisector the line $R'P$
So it is isosceles and similar to $\triangle ABC \ (R'Q=R'B)$

from $RC=R'Q$ we get $RC=R'B$ wich means that the points $R,R'$ are symmetric through the midpoint $M$ of $BC$.

[size=150][color=blue]Construction[/color][/size]
We choose the point $R' \in BC$. We find its symmetric $R$ through $M$.
The parallel from $R'$ to $BA$ intersects the side $AC$ at $P$.
We draw a circle $(R',R'B)$ and let $Q$ the common point of this circle (different from $B$) with the ray $BA$.

Suppose that $Q$ is in the interior of the segment $AB$.

From the construction, the triangle $\triangle R'QB$ is isosceles (and similar to $\triangle ABC$), so $R'Q=R'B$.
Also, $\triangle PR'C \sim \triangle ABC$ (because $R'P \parallel BA$).

The line $R'P$ is the external angle bisector of $\widehat{R}$, so $\widehat{QR'P}=\widehat{B}$.
The triangles $\triangle PCR,\triangle PR'Q$ have $RC=R'Q, PC=PR'$ and $\widehat{PCR}= \widehat{QR'P}=\widehat{B}$ so they are congruent

Hence $PR=PQ$ ($\triangle PRQ$ is isosceles) and $\widehat{QPR'}= \widehat{RPC}\Rightarrow \widehat{QPR}= \widehat{R'PC}=\widehat{A}$

$Q$ is in the interior of the side $AB$ iff $R'$ is to close to the point $B$.
The perpendicular bisector of $AB$ intersects the side $BC$ at $F$. $R'$ can only move from $B$ to $F$.

If $\widehat{A}>60^{0}$ then $F$ is out of the segment $BC$, so we have solution for every position of $R'$.

If $\widehat{A}<60^{0}:$

From the similar triangles $\triangle FAB,\triangle ABC$ we get:

$\frac{BF}{AB}=\frac{AB}{BC}\Rightarrow BF = \frac{AB^{2}}{BC}$

So, for $\widehat{A}<60^{0}$ we have solutions for all points $R' \in BC$  with $BR' <\frac{AB^{2}}{BC}$[/hide]

[hide=Thought Process]When I was solving this problem I thought, "I seriously doubt that $\triangle{PQR}$ is arbitrary if it can be easily constructed via compass and ruler, and that's where I thought of the extremely useful(for this problem since the medial triangle is automatically similar) medial triangle. But the problem was that the medial triangle and the triangle the problem writer wanted had the wrong orientation except for $RP$, so I decided to have $R$ and $P$ be the midpoints of $BC$ and $AC$. It's well-known that the simplest construction for an isosceles triangle is to take a circle and randomly pick $2$ points on it's surface, and adding this with the center gives an isosceles triangle. Simply repeat that here except with $P$ as the center and $R$ as a point on the circle and $Q$ is the second intersection of $\triangle{ABC}$ with circle $P$. As a result we're done. If you want to construct midpoints just use the perpendicular bisector by the way and the bound on $\angle{BAC}$ is less than $90^{\circ}$ or otherwise $Q$ will fail the problem requirements.[/hide]


Please , post questions on molecular rearrangements.

You really have to be more specific than that?  Is this for some sort of test or self-study?  What aspect of molecular arrangement?


Can somebody write me links to [b]free online introductory[/b] books?

[quote="orl"]I can recommend http://kolho3.tiera.ru/#48[/quote]
Ah..a wonderful site you have posted there, thx a lot
But it seems to have most stuff (in Physics specially) in Russian....any such site with English content ?
Anyways Thanks a lot !  :lol:

Same problem... anyone help with a English content site?

Well, both could do. But also in Physics, some introductory books


AMC AIME USA(J)MO USAMO MATHCOUNTS geometry calculus

Hi,

I'm currently a 9th grader going into the 10th grade (and am currently taking Precalculus over the summer). I'm eager to qualify for the AIME next year but do not yet have the necessary talents to do so. Unfortunately, I have wasted much of my time up until this point, in which I have realized that I have to turn my life around: and this will involve commiting myself to mathematics since I have scientific aspirations. I have less than a year to prepare for the AIME right now, although the entire month of August is free to me.

I have just ordered the Art of Problem Solving, which has been highly recommended by numerous peoples.

Now, I have a few questions to ask...

1: If I want to enter the USAMO sometime during my high school career, is it too late if I never took mathematics seriously up until this point? (Lagrange took mathematics seriously at age 15 as well, but this is considered quite late for a mathematician; and he probably received a better education than me; I had to go through the Core-Plus program at my school). Do I have to develop the talent before puberty sets in? (since the brain strengthens connections between neurons that are most frequently used, during the process of puberty). I'm 15 btw (which implies that I'm in the middle of this process).

2: And what is generally the best way to improve in mathematics? Yes, I know, purchase the AOPS,go through old competition problems, ask for help. But what else? For one thing, there are barely any materials out there that explain challenging problems step-by-step that help me understand how challenging problems are solved (There are REA's Problem Solvers, which I am trying to use right now. But they do have some problems). Also, like Nash and Newton, I am generally a solitary worker who doesn't enjoy asking for help or communicating to other peoples. I like to do things myself (and besides, I also have Asperger's).

If I study to my absolute hardest during the next few months, do I have a chance in mathematics competitions?

And to give an idea of my current mathematics aptitude... I'm certainly above average, though not exceptional (and still, I do struggle with some Mathcounts problems). I scored a 640 on my Math SAT a few months ago (though I made many incredible mistakes), but I have scored around 700 on recent practice tests that I took. I can do the first 10-14 problems on the AMC 10 but am lost beyond that. In Houghton-Mifflin textbooks that I've used in my summer courses and in my self-study, I can do many of the B problems but few of the C problems (C are the hardest of them all). Sadly, textbooks are not the way to teach difficult math problems (and sometimes even skip answers to the hard problems, even if the hard problems are odd ones).

Btw, I live in Redmond,WA. So if anyone's in the Seattle area and knows of some math programs there, I'd love to know. I know about Summer Stretch but that's about it.

Thanks!

-Simfish

-------

TO introduce myself. (since this is my first post) I hope to become an active member of these forums and to become addicted to mathematics.. I also forum (albeit rarely) at collegeconfidential and the physics forums. My primary role model is Marie Curie and I have aspirations to the California Institute of Technology (of course, if I want to be serious about it, I absolutely must improve my skills in mathematics).

Hm. And I actually have fallen into the Calculus trap. Though my school does offer IB math.

Hello,

It will definitely be possible for you to make the AIME this year.  I made it for the first time my tenth grade year and all I did was go over previous contests ( I didn't know about AoPS or anything).  So that probably won't be too difficult for you.   If you stick with competition there is a good chance you can make the USAMO at least once in high school.

As for the step-by-step solutions,  you have to really try a problem before you look at the solution(this is what I find at least).  Just try to see how they get to the next step and so forth.  Taking the time to understand how a problem works pays off.

To get better at math you must put in time.  I can't give you a reference to a special book that will give you the answer to all problems that could possibly occur in a hs competition.  You just have to work. Work on your weak areas.  Make sure you understand what you did and don't just use a formula.  If you use a formula make sure that you can derive it or that you know how and why it works.

Ask if you need any more guidance.

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]Look through old math jams.[/url]

Level 6 or up are about AIME level, 4/5 is AMC 10/12, and above 8 is USAMO level.

Hi, and welcome to the Art of Problem Solving!

Here are a few suggestions:

1) Work carefully through AoPS Volume 1, solving as many problems as you can. On the problems that you cannot solve, ask for hints from other forum members who will be glad to help you.

2) Consider taking an [url=http://www.artofproblemsolving.com/Classes/AoPS_C_About.php]online class[/url]. I'm in a few of them now and it is certainly the fastest (and, needless to say, best) way I've ever learned problem solving techniques.

3) If you are intent on qualifying for AIME, work as many AMC 10 and AMC 12 problems as you can. You can purchase old contests (they're extremely cheap!) [url=http://www.unl.edu/amc/d-publication/publication.html]here[/url]. There are also some online classes that cater specifically to the AMC 10 and AMC 12.

Best of luck on your problem-solving endeavors! Feel free to ask if you have any questions.

[quote="JSRosen3"]3) If you are intent on qualifying for AIME, work as many AMC 10 and AMC 12 problems as you can. You can purchase old contests (they're extremely cheap!) [url=http://www.unl.edu/amc/d-publication/publication.html]here[/url]. There are also some online classes that cater specifically to the AMC 10 and AMC 12.[/quote]

Yes, ordering old tests is a very good idea.  But don't just look at the illegaly copied ones online.  A lot of people say that the official solutions are much better.

AMC 10 = 1 dollar for 1 exam and 1 solution book from one year.
AHSME/AMC 12 = 1 dollar for 1 exam and 1 solution book from one year.
AIME = 2 dollars for 1 exam and 1 solution book from one year.

The shipping and handling is pretty expensive, though.

Welcome to the forums!

I think you would definitely be able to make AIME this year. It's pretty amazing what a little bit of studying can do for you. Also, don't worry about never taking math too seriously for a long time. Some people just don't think about math that much until they reach high school or middle school. You can still enjoy a lot of success that way. If you need any help at all, don't hesitate to ask by making a post in the forums. You'll find that a lot of people will be willing to help you. Just try not to post in subjects that have been extinct for more than 2 months or so.

If you'd like to practice to make AIME and do well, I'd suggest looking over old AMC10s, AMC12s, and AHSME or whatever the acronym for that. Also, try some of the problems on the forums right now.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14138
That's the link for a practice AMC "11" that's going to happen. I advise you to check that out.
You'll find most of the stuff concerning that on the AMC forums, http://www.artofproblemsolving.com/Forum/viewforum.php?f=133

Good luck!

[quote="JSRosen3"][quote="nr1337"]The shipping and handling is pretty expensive, though.[/quote]

It's not [i]that[/i] expensive...unless you order $80 worth of old tests like me  :lol:[/quote]

Well, if you order ten dollars worth of tests, you have to pay 7 dollars.  With 7 dollars, I could buy seven more tests!

So you might as well just order all the tests you think you'll ever need and only pay the shipping and handling fee once. That's how I got $80 worth of 1 and 2 dollar tests  :)

By the way, I had two dollar signs in the same sentence so it showed up as [tex]\TeX[/tex]. Is there any way to avoid this?

$You can do it this way$

I clicked "disable BBCode in this post" which is below the box where you type your post.

If you need tex in the same post I am not sure.

[quote="Simfish"]1: If I want to enter the USAMO sometime during my high school career, is it too late if I never took mathematics seriously up until this point?[/quote]

No. A good, liberating book to read about mathematicians is the book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0817631917/learninfreed]Mathematical People: Profiles and Interviews[/url], which is an edited set of interviews with famous mathematicians. You'll find out by reading that book that some TOP mathematicians were still struggling at your age or older ages.

[quote="Simfish"]I had to go through the Core-Plus program at my school). Do I have to develop the talent before puberty sets in? (since the brain strengthens connections between neurons that are most frequently used, during the process of puberty).[/quote]

You have my sympathies for having had to suffer through Core Plus. Read the books mentioned in the old [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8674]Readable Books thread[/url] that Mathnovice opened to restructure your thinking about mathematics. Don't worry about neurons and puberty; you have plenty of time to learn lots of math. Desire is more powerful than hormones in building new connections in your brain.

[quote="Simfish"]2: And what is generally the best way to improve in mathematics? Yes, I know, purchase the AOPS,go through old competition problems, ask for help. But what else?[/quote]

The "what else" is to take a second look at concepts you "already know" from a fresh perspective. The AoPS books will help you do that, and so will the Gelfand series books in the readable books thread I just mentioned. The one and ONLY way to get better at mathematics is to do a lot of problems. (That's why a lot of the young people who populate this forum are better at mathematics than I am--they practice many more problems than I have in my whole life. I devoted my teenage years mostly to learning foreign languages, and anyone who spends as much time as the more dedicated participants do here doing math problems as a teenager will learn a lot of math.)

[quote="Simfish"]If I study to my absolute hardest during the next few months, do I have a chance in mathematics competitions?[/quote]

You live in a competitive state, but the AMC competitions available to you at your age have a national reach, and it's not too late to try for year-by-year improvement.

Welcome aboard. And thanks, by the way, to the other participants who have already replied to you with their encouraging words. We definitely want to make the beginning problem-solvers feel welcome here.

Another good place to start would be entering the Getting Started Contest. It's stickied in this forum and there seems to be a lot of people willing to do it. IMO, the most important thing is to get involved and do some math with other people, not just looking through textbooks.

http://www.mualphatheta.org also has tons of topic-tests that I used to help on certain areas.

Thanks for the very encouraging words, everyone! :) At least I now have some confidence in my ability. I've just received the AOPS in the mail; I'll begin working on them after I finish my Precalculus course (which is only a week from ending).


Hey guys! We're playing a really simple version of "The Incredibles" for a trip to Disney. It's a really jazzy piece, and there's a symbol there that looks a like trill, but it's different. Our band teachers call it a "fall". It goes from like an F above the staff to an F on the top line of the staff or from a c in the staff to a c just above the staff. How do you play that? Thanks

i'm guessing that you already found out but it is like playing a   scale REALLY fast. skip some notes in the middle and just play F to F or C to C really fast. that worked for me... :maybe:

I did think it was something like that, but I was hoping it was something easier, lol. For now I've just been slurring it from low to high, or high to low. It doesn't sound as cool as when the jazz band people do it though  :lol: . Thanks, anyway!


http://www.donpixel.com/video/?v=031125110016

look its me . . . sike

Dont mess with Integral0 or he will differentiate you like he does a constant [img]http://www.stopstart.freeserve.co.uk/smilie/dark2.gif[/img]

[quote=Valentin Vornicu]lol :) that means he will differentiate you quickly, doens't it? :)[/quote]

*gasp* it is the first user ever!
bump 14 year
also what person op


If the question is

-3/4y + 11/2 = x

You must change it to y = mx + b format...

-3/4y = x - 11/2

Now, I am unsure of what to do after this. would it be

-3y = 4x - 44/2
y = 4x/-3 - 22/-3
y= 4x/-3 - 7 1/3

OR...

Would it be :

y = 4x - 22 ?

------------------- Also, if you have the following equation:

x - y/3 = 0, how can you change it into y = mx + b format?

Hello, banquo. :)

$-\frac{3}{4}y=x-\frac{11}{2}.$  Dividing by $-\frac{3}{4}$ both sides, or multiply by $-\frac{4}{3}$ both sides,

we have $y=-\frac{4}{3}\left(x-\frac{11}{2}\right),$ yielding $y=-\frac{4}{3}x+\frac{22}{3}.$

For $x-\frac{y}{3}=0\Longleftrightarrow \frac{y}{3}=x\Longleftrightarrow y=3x.$


Well, the most elementary way will be just to use the $(a-b)^2\ge 0$ trick a few times.

[hide]Using it on $x^8,y^8,z^8$ we get:

$x^{16}+y^{16}+z^{16}\ge x^8y^8+x^8z^8+y^8z^8$

Using it once again on $x^4y^4,x^4z^4,y^4z^4$ we get:

$x^8y^8+x^8z^8+y^8z^8\ge x^4y^4z^4\cdot (x^4+y^4+z^4)$

Using it one more time on $x^2,y^2,z^2$ we get:

$x^4+y^4+z^4\ge x^2y^2+x^2z^2+y^2z^2$

Using it for the last time on $xy,xz,yz$ we get:

$x^2y^2+x^2z^2+y^2z^2\ge xyz\cdot (x+y+z)=\\=xyz\cdot  (\dfrac{xyz}{yz}+\dfrac{xyz}{xz}+\dfrac{xyz}{xy})=x^2y^2z^2\cdot (\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz} )$

As desired.[/hide]

Come on Silouan, this is just another boring application of Muirhead...


What are the square roots of the roots of the equation y=x^2+4?

The roots are $2i$ and $-2i$, written as $2(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$ and $2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$ in trigonometric form, which have roots $\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$, $\sqrt{2}(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})$, $\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$, and  $\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$, which equal $\boxed{1+i, 1-i, -1+i, -1-i}$.


For all positive real numbers $ a,b,c$, prove that
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \geq \frac {3}{2}.\]

Do a search, this has popped up quite a lot.

Well, Nesbitt's Inequality states that $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}$, as a result, it has been proven (kind of).  I think it was better if they actually altered the problem so it wouldn't be so obvious. :)

Masoud Zargar

Has it been proven that the minimum value of something like this always comes when $a=b=c$? Because I've found that that's the case every single time I've tried any problem like this... and if so, then it's obvious from there.

Use the fact that the arithmetic mean of 3 positive numbers is greater than or equal to its harmonic mean.

Specifically, use the values $a+b$, $a+c$, and $b+c$.

Upon simplification and expansion of the terms, the original inequality may be proven.


Let $n$ be a positive integer number. Prove that $(n!)^{2}\geq n^{n}$

$LHS=1(n)\times2(n-1)\times3(n-2)\times....\times(n-1)2\times n(1)$ (n terms)

$RHS=n\times n\times n \times ...\times n$ (n terms)

Each term in the LHS is greater than or equal to n as $a<b \Leftrightarrow ab \le (a+1)(b-1)$.


does anybody know how many points are needed for a 5 on the ap bioloogy exam?

AP tests are graded on a curve. I don't think that it is very high but it always depends, you definitely don't have to get everything correct to get a 5. You shouldn't worry about points when you take the test though, on the essays they won't mark off wrong answers so try to put down as much correct info as you can and you will be fine.

For most AP tests, getting around 70% or more is usually enough to get a 5.


Solve for $x$ ($0\le x<2\pi$):

$\csc x=\csc 2x+\csc 3x$

Does this simplify to

$ \sin{3x} \cos{x} =\ \sin{ \frac{5x}{2}  } \cos{ \frac{x}{2} } $

Using the fact that $\frac {e^{i\theta} - e^{-i\theta}}{2i} = \sin \theta$ I got it to simplify to

$\cos 5x - \cos 4x - \cos 3x + \cos 2x = 0$

$\sin 4x = \sin 3x$

$y^4 -y^{-4} = y^3 - y^{-3}$

Simplifying further yields

$(y-1)(y^7+1)=0$

Thus $y$ is a seventh root of "negative unity" =)

Since $\text {cis} \pi = -1$ all the solutions are of the form $\frac {\pi k}{7}$ where $k$ is an odd number so all of the solutions between 0 and $2 \pi$ are $\frac {\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \frac {9 \pi}{7}, \frac {11 \pi}{7}, \frac {13 \pi}{7}$.

$\pi$ is not possible since it would result in $\frac {1}{0}$

and what do you get from that???? :rotfl:

Nice solution, tarquin.

There's a way to get $\sin 4x=\sin 3x$ just using trig. Try it :)

[quote="236factorial"][quote="frt"]There's a way to get $\sin 4x=\sin 3x$ just using trig. Try it :)[/quote]

How can you get something that isn't true?!  :D[/quote]
Well, $\sin 4x=\sin 3x$ is true if $\csc x=\csc 2x+\csc 3x$. Look at tarquin's solution :)

All the answers ($\frac{\pi}{7}$, $\frac{3\pi}{7}$, $\frac{5\pi}{7}$ etc.) satisfy $\sin 4x=\sin 3x$.

What I meant was, tarquin used complex numbers to get the equation $\sin 4x=\sin 3x$, but you can just use trig to get the same thing.

[hide]
this was on Chicago ARML tryouts this year

$\displaystyle\frac{1}{\sin x} - \frac{1}{\sin 3x} = \frac{1}{\sin 2x}$

$\displaystyle\frac{\sin 3x - \sin x}{\sin 3x \sin x} = \frac{1}{\sin 2x}$

Using identity for top part:

$\displaystyle\frac{2\sin x \cos 2x}{\sin 3x \sin x} = \frac{1}{\sin 2x}$

$2\sin 2x \cos 2x = \sin 3x$

$\sin 4x = \sin 3x$

So solutions within the range are $x = \displaystyle\frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \frac{9\pi}{7}, \frac{11\pi}{7}, \frac{13\pi}{7}$

[/hide]


if m,n,p,q were natural number and                                             [img]http://www.mathlinks.ro/Forum/latexrender/pictures/1e74106e88f434ea21f87749f6415955.gif[/img] then m=p or m=n

[quote="ali666"]if m,n,p,q were natural number and                                             [img]http://www.mathlinks.ro/Forum/latexrender/pictures/1e74106e88f434ea21f87749f6415955.gif[/img] then m=p[/quote]

Note that for m and p different , but $m=n$ and $p=q$ , the result is true!


geometry
3D geometry
vector
parallelogram
AMC

A cube with side length $ 1$ is sliced by a plane that passes through two diagonally opposite vertices $ A$ and $ C$ and the midpoints $ B$ and $ D$ of two opposite edges not containing $ A$ and $ C$, ac shown. What is the area of quadrilateral $ ABCD$?
[asy]import three;
size(200);
defaultpen(fontsize(8)+linewidth(0.7));
currentprojection=obliqueX;
dotfactor=4;

draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4"));
draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1));
draw((0.5,1,0)--(0,1,0)--(0,1,1));
dot((0.5,0,0));
label("$A$",(0.5,0,0),WSW);
dot((0,1,1));
label("$C$",(0,1,1),NE);
dot((0.5,1,0.5));
label("$D$",(0.5,1,0.5),ESE);
dot((0,0,0.5));
label("$B$",(0,0,0.5),NW);[/asy]$ \textbf{(A)}\ \frac {\sqrt6}{2} \qquad \textbf{(B)}\ \frac {5}{4} \qquad \textbf{(C)}\ \sqrt2 \qquad \textbf{(D)}\ \frac {3}{2} \qquad \textbf{(E)}\ \sqrt3$

[hide]Since $ AB \equal{} BC \equal{} CD \equal{} DA \equal{} \frac {\sqrt {5}}{2}$, $ ABCD$ is a rhombus.
$ K \equal{} \frac {1}{2}d_1d_2 \equal{} \frac {1}{2}(AC)(BD) \equal{} \frac {1}{2}(\sqrt {3})(\sqrt {2}) \equal{} \frac {\sqrt {6}}{2}$.  $ \boxed{\text{A}}$[/hide]
EDIT: Wrong letter. :oops:

Post above: Right answer I think, but letter C?

[hide="Vectors"]The area of a parallelogram formed by adjacent sides $ \vec{a}$ and $ \vec{b}$ is $ | \vec{a} \times \vec{b} |$ (magnitude of cross product). So here, we want to find $ | \langle 0,1,.5 \rangle \times \langle 1,0,.5 \rangle |$. Computing the cross product, we get $ \langle .5,.5,1 \rangle$, whose magnitude is $ \sqrt{6}/2$.[/hide]

[quote=rxh4n]Sorry to reopen! But why is BD sqrt(2)? How do you know it is?[/quote]

Take a square with side lengths $1$. The diagonal is $\sqrt 2$.

[quote=rxh4n]Sorry to reopen! But why is BD sqrt(2)? How do you know it is?[/quote]

Bring B and D down such that it is the vertex of the the base. So the diagonal of the base is $\sqrt{2}$.

Yeah... it looks familiar. 2018 AMC 8 problem 24 is pretty much the same thing.

By symmetry ABCD is a rhombus, coordinates trivially give AC=sqrt3 and BD=sqrt2 so the area corresponds to A


calculus
integration
trigonometry
complex numbers
complex analysis
complex analysis unsolved

Hello everyone! Can you help me ? Can you use [u][b]residue[/b][/u] to calculate these integrals:
1) $\int_{-1}^{1}\!{\frac{1}{\sqrt [3]{ \left( 1-x \right) \left( 1+x \right)^{2}}}}{dx}$
2) $\int_{0}^{2\,\pi }\! \left( \sin \left( \phi \right)+\cos \left( \phi \right)+2 \right)^{-1}{d\phi}$

For the second: look at $z=e^{i\phi}$, and try to make your integral into the real or imaginary part of an integral on the circle.

It is very hard to solve the first problem !


The product of the digits of a positive three-digit integer $ n$ is zero, the sum of $ n$'s
digits is 8, and $ n$ is an odd number. What is the arithmetic mean of the possible values
of $ n$?

last digit cant be 0 because its an odd number.
also first digit cant be 0.

so middle digit is 0.

$ 107, 206, 305, \boxed{404}, 503, 602, 701$

answer : 404

Since $ n$ is an [b]odd[/b] number, you have only four possibilities.

$ 107$, $ 305$, $ 503$, $ 701$

The answer is still $ \boxed{404}$, however.


What is the smallest composite number such that all of its divisors are of different number of digits

[quote="Art of Owna"]What is the smallest composite number such that all of its divisors are of different digits[/quote]

is the answer...[hide]
6???
[/hide]

I mean all of its divisors are of different number of digits so 6 cant work because 1 2 and 3 all have 1 digit
sry about that

ya, i was being stupid and said well if it is like that it has to contain 2 prime factors of different amounts of digits9in order for it to be composite) so 11*101 would make the smallest, not realizing aobut 11^2


counting
distinguishability
combinatorics unsolved
combinatorics

How may ways to put ten indistinguishable bottles of milk and twenty indistinguishable cups of coffee on a round table  ( the positions on the round table are the same) , if
a) between any two bottles are at least 1 cup ?
b)no condition?

You can arrange the bottles on the table in 10! ways

because you say that between each bottle there must be at least one cup

its the same as asking "arrange 20 objects in 10 cells, so that there is at least one object in each cell"

you put 1 object in each cell

that leaves you 10 objects to arrange between 10 cells

there are 10-1 dividers between the cells

So the number of ways to arrange 10 cups between 10 bottles is
D(10,10) = (10-1+10)!/10!0!=19!/10!9! = (19*18*17*16*15*14*13*12*11*10!)/10!9!
So the number of ways to arrange 10 bottles and 20 cups so that between every 2 bottles there is at least one cup is :19*18*17*16*15*14*13*12*11/9! and multiply that by the number of ways to arrange the bottles . meaning 10! = which is 19!/9!

[quote="Lena"]You can arrange the bottles on the table in 10! ways[/quote]

The bottles are indistinguishable, so you can arrange them in only 1 way.

In that case the number of ways to arrange cups between bottles would be only D(10,10)=19*18*17*16*15*14*13*12*11/9!


[1]
Find all polynomials $P(x)$, such that$P(x^2 ) - P(y^2 ) + 2x + 1 = P(x + y)P(x - y)$ ,$x,y \in \Re$

[2]
If
$\left\{ \begin{array}{l} \frac{x}{{a + \alpha }} + \frac{y}{{b + \alpha }} + \frac{z}{{c + \alpha }} = 1 \\ \frac{x}{{a + \beta }} + \frac{y}{{b + \beta }} + \frac{z}{{c + \beta }} = 1 \\ \frac{x}{{a + \gamma }} + \frac{y}{{b + \gamma }} + \frac{z}{{c + \gamma }} = 1 \\ \end{array} \right.$ is true.Prove,without solve the system,that $x+y+z=a+\alpha+b+\beta+c+\gamma$

[3]

If $x_1$,$x_2$ are roots of $x^2-(a+d)x+ad-bc=0$,$a,b,c,d \in \Re$ then find $A$
$Q(y)=y^2-(a^3+d^3+3abc+3bcd)y+A$ such $Q(x_1^3)=Q(x_2^3)=0$

Problem 3:
[hide]$(x_1+x_2)=a+d$ $x_1x_2=ad-bc.$ Cubing a+d, $a^3+3a^2d+3ad^2+d^3.$  Using the first equation, $a^3+3a(x_1x_2+bc)+3d(x_1x_2+bc)+d^3$.  Noting the similarity between this and the b term of the $y^2$ function, the middle term of that function can be written as:
$(x_1+x_2)^3-3x_1x_2(x_1+x_2)$.  Putting y as $x_1^3$, $x_1^6-x_1^6-3x_1^5x_2-3x_1^4x_2^2-x_1^3x_2^3+3x_1^5x_2+3x_1^4x_2^3+A$.  After simplifying, the only terms that are left is $-x_1^3x_2^3+A=0$ so A is $x_1^3x_2^3$.  In fact, the middle term's cancelling terms was due to the fact that it wasn't fully simplified.  The middle term on y fully simplified is $-(x_1^3+x_2^3)$.  Showing the equation this way makes it clear that the equation is symmetric.  A is constant regardless of which root is plugged in.[/hide]

Man... Problem 1 is very slick, took me a while...

Problem 1:

[hide]
well, plug in $x = 0$ and $y = 0$ we get $1 = P(0)^2$, which implies $P(0) = \pm 1$.  Now plug in $y = x$ we get $2x+1 = P(0)P(2x)$, so $P(2x) = \pm(2x+1)$, let $u = \frac{x}{2}$ and plug $u$ in we get $P(u) = \pm (u+1)$, which is the desired polynomial.  HOWEVER, plugging back in the original one we see that $u+1$ doesn't work, so our solution is $P(x) = -x-1$.
[/hide]


linear algebra
matrix
graph theory
advanced fields
advanced fields unsolved

Let $ M$ be a $ (0, 1)$ matrix; that is, a matrix where each of whose entries is either a 0 or a 1. A line in $ M$ is either a row or a column of $ M$. Use Konig's theroem to prove that the minimum number of lines containing all the 1's of $ M$ is equal to the maximum number of 1's, no two of which are in the same line.

My attempt:

Let $ G$ be a bipartite graph with vertex bipartition $ {X, Y}$ , such that $ A$ is an adjacency matrix of graph $ G$, where $ X$ is a set of vertices corresponding to the rows of matrix $ A$, and $ Y$ is the vertex set corresponding to the columns. So the result follows from Konig's thereom. I am not sure how the adjacency matrix helps for making this bipartition. I would like some help on this one. Thanks in advance.

It doesn't. You construct a graph first, apply the theorem to that graph, and read the result off for the matrix constructed this way. You're not using any linear algebra; this is really still a purely graph-theoretic result.


Prove that there does not exist a function $f:C \rightarrow C$ such that $f(f(z)) = az^2 + bz + c$.

I believe that this problem has been posted before (and that Pierre posted a solution).  But I couldn't find the thread in any of my searches.  Can somebody please give me the link or post a solution?

~Mike

To be rigourous, I've not posted a solution for in the complex case. But, I gave a reference where you could find a solution of that one (and more).

http://www.mathlinks.ro/Forum/viewtopic.php?t=14151&highlight=function

Pierre.

Let $g(x)=ax^2+bx+c$.
The equation $g(x)=x$ has two solutions $x,y$.
The equation $g(g(x))=x$ has four solutions $x,y,u,v$ and $g(u)=v, g(v)=u$.
Now if $f(u)=w$ then easily we deduce $g(g(w))=w$ hence $w$ is one of
$u,v,x,y$. Obviously it cannot be $x,y,u$. If $w=v$ then f(v)=v$ and so $g(v)=v$ contradiction

Iura,

I came up with the identical idea, eventually.  It's very nice, isn't it?


let $0<z\leq y \leq x \leq3$ such that :
         a$\dfrac{3}{xy} +\dfrac{2}{y^2} \geq1$
         b$\dfrac{18}{x^2y}+\dfrac{4}{y^2z}+\dfrac{3}{z^2x}\geq3$
 find the maximum of expression :
  $A= \dfrac{xyz}{2} +\dfrac {80x^3}{27} +\dfrac{18y^3}{8}$


Find all pairs $ (a,b)$ of integers such that
$ a^2\plus{}7ab\plus{}12b^2\plus{}a\plus{}3b\minus{}9\equal{}0$.

[hide]We can factor this to $ (a\plus{}3b)(a\plus{}4b\plus{}1) \equal{} 9$

We now have multiple cases to consider. Since $ a$ and $ b$ must be integers, we have the following cases:

$ a\plus{}3b \equal{} 1$
$ a\plus{}4b\plus{}1 \equal{} 9$

$ a\plus{}3b \equal{} 3$
$ a\plus{}4b\plus{}1 \equal{} 3$

$ a\plus{}3b \equal{} 9$
$ a\plus{}4b\plus{}1 \equal{} 1$

$ a\plus{}3b \equal{} \minus{}1$
$ a\plus{}4b\plus{}1 \equal{} \minus{}9$

$ a\plus{}3b \equal{} \minus{}3$
$ a\plus{}4b\plus{}1 \equal{} \minus{}3$

$ a\plus{}3b \equal{} \minus{}9$
$ a\plus{}4b\plus{}1 \equal{} \minus{}1$

Each of these systems of linear equations must be solved. We therefore get the following pairs:

$ (\minus{}30, 7)$
$ (\minus{}20, 7)$
$ (0, \minus{}1)$
$ (6, \minus{}1)$
$ (26, \minus{}9)$
$ (36, \minus{}9)$
[/hide]


Make a table for this:
$ y \equal{} \frac {\sqrt {5}}{5} \left( \frac {1 \plus{} \sqrt {5}}{2}\right) ^x \minus{} \frac {\sqrt {5}}{5} \left( \frac {1 \minus{} \sqrt {5}}{2}\right) ^x$

[quote="herefishyfishy1"]Umm... I didn't actually simplify all those radical expressions... I knew that the table would look like that. So I just typed the Fibonacci numbers.[/quote]Oh, but you at least got the catch.

I have no idea why this is in the games forum. It is not a game.

:ddr:

[quote="AIME15"]Um...because you put it in the games forum?[/quote] Whoooops.

Does anyone know what special value of $ \frac {1 \pm \sqrt {5}}{2}$ is?

[hide="Hint"]
Think Greeks and their architecture.
[/hide]

Could it have anything to do with....$ \pi$ ???? xD its actually $ \phi$


inequalities
number theory proposed
number theory

$ (m,n)\in N$
$ \sqrt{7}\minus{}\frac{m}{n}>0$
Prove that
$ \sqrt{7}n\minus{}m>\frac{1}{m}$
b)If 1 is a best constance such that the inequalities hold for all $ (m,n)$

[quote="TTsphn"]1)Prove that
$ \frac{1}{2n}<\{{\sqrt7}.n\}< 1\minus{}\frac{1}{6n}$
2)Find the best c>0 such that
$ \{\sqrt2 n\}\ >\frac{c}{2\sqrt2 n}$[/quote]
You are wrong. For best rational aproximate we have $ 7n^{2}\minus{}m^{2}\equal{}3$, therefore $ \{\sqrt 7 n\}\equal{}n\sqrt 7\minus{}m\equal{}\frac{3}{m\plus{}n\sqrt 7 }>\frac{3}{2\sqrt 7 n}$. Therefore best constants are
$ \frac{3}{2\sqrt 7 n}<\{\sqrt 7 n\}<1\minus{}\frac{1}{2\sqrt 7 n}.$
In these whay $ \frac{1}{2\sqrt 2 n}<\{\sqrt 2 n\}<1\minus{}\frac{1}{2\sqrt 2 n}.$

[quote="Rust"][quote="TTsphn"]1)Prove that
$ \frac{1}{2n}<\{{\sqrt7}.n\}< 1\minus{}\frac{1}{6n}$
2)Find the best c>0 such that
$ \{\sqrt2 n\}\ >\frac{c}{2\sqrt2 n}$[/quote]
You are wrong. For best rational aproximate we have $ 7n^{2}\minus{}m^{2}\equal{} 3$, therefore $ \{\sqrt 7 n\}\equal{} n\sqrt 7\minus{}m \equal{}\frac{3}{m\plus{}n\sqrt 7 }>\frac{3}{2\sqrt 7 n}$. Therefore best constants are
$ \frac{3}{2\sqrt 7 n}<\{\sqrt 7 n\}< 1\minus{}\frac{1}{2\sqrt 7 n}.$
In these whay $ \frac{1}{2\sqrt 2 n}<\{\sqrt 2 n\}< 1\minus{}\frac{1}{2\sqrt 2 n}.$[/quote]
Which problem do you say?
The 1 It true .

Yes your statments are true, but constants are not best.

At the first problem I do not want to find $ c$ best
At the Problems 2 this c is exist
At problems 3 :I think it exist with Pell equation.

For 1 and 2 I give best constants.
Problem 3 is not true. If $ n\equal{}1, m\equal{}2$ we have $ \sqrt 7\minus{}2>\frac{1}{m}$  and best constant  is $ 2\sqrt 7\minus{}4.$
I think best question is $ \frac{m}{n}\minus{}\sqrt 7>0$, prove that $ m\minus{}n\sqrt 7 >\frac{1}{2m}$.

We have $ \sqrt{7}n > m$ then $ 7n^{2}> m^{2}$

For all integers $ m$ the number $ m^{2}\plus{}1,m^{2}\plus{}2$ not square , so $ 7n^{2}\ge m^{2}\plus{}3$.

Then :

$ 7n^{2}\ge m^{2}\plus{}3\ge m^{2}\plus{}2\plus{}\frac{1}{m^{2}}\equal{}\left(m\plus{}\frac{1}{m}\right)^{2}$ then $ \sqrt{7}n\minus{}m\ge\frac{1}{m}$ the equality not hold  :)

Prove that
$ |m\minus{}\sqrt{2005}n|>\frac{1}{90n}$ for all $ (m,n)\in N$


group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

Prove that for subgroups $ H,K\subseteq G$, $ \left|HK\right|\equal{}\frac{\left|H\right|\left|K\right|}{\left|H\cap K\right|}$.  Also, can we make the generalization

\[ \left|H_{1}H_{2}\cdots H_{n}\right|\equal{}\frac{\left|H_{1}\right|\left|H_{2}\right|\cdots\left|H_{n}\right|}{\left|H_{1}\cap H_{2}\cap\cdots\cap H_{n}\right|}?\]

the number of pairs $ (h,k)$ with $ h\in H$, $ k\in K$ is clearly $ |H\parallel{}K|$. now we need to show that for a given $ x$ in $ HK$, the number of pairs $ (h,k)$ with $ hk \equal{} x$ is $ |H\cap K|$. let $ hk\in HK$. note that $ hk \equal{} h'k'$ iff $ h'^{ \minus{} 1}h \equal{} k'k^{ \minus{} 1}$. these last two expressions represent an element in $ H\cap K$. conversely, every element $ a\in H\cap K$ gives a pair $ (h',k')$ for which $ h'k' \equal{} hk$--namely, the pair $ (h',k') \equal{} (ha^{ \minus{} 1},ak)$.

No version of the generalization can work- not even if we include all possible intersections of subsets, with various coefficients. For example, consider subgroups of $ (\mathbb{Z}/p\mathbb{Z})^3$, viewed as vector spaces over a finite field.

If I set $ H_1\equal{}\text{span}\{(1,0,0)\},H_2\equal{}\text{span}\{(0,1,0)\},$ and $ H_3\equal{}\text{span}\{(0,0,1)\}$, I get $ |H_1|\equal{}|H_2|\equal{}|H_3|\equal{}p$, and each intersection of two or more is trivial. In this case, $ |H_1\plus{}H_2\plus{}H_3|\equal{}p^3$.

If instead $ H_1\equal{}\text{span}\{(1,0,0)\},H_2\equal{}\text{span}\{(0,1,0)\},$ and $ H_3\equal{}\text{span}\{(1,1,0)\}$, the intersections are all the same sizes and $ |H_1\plus{}H_2\plus{}H_3|\equal{}p^2$.


This is what I sent in an email to classes@artofproblemsolving.com, but decided to post it here for suggestions.

"Hello.

I was considering enrolling in the AMC 12 Problem Series, but I realized that it has already started and that two classes have gone by. I was wondering if the notes for those weeks would be available if I joined late. Also, I was wondering if the class would be offered again later on during the year.

I'm also deciding between the AMC 12 Problem Series and the AIME Problem Series. The last time I took a practice AMC 12, I got a 109.5 and for AIME, I got 5 questions right. Which would you suggest that I take?

Thank you very much"

Also, would it be a good idea to take both, or do the two cover similar things? (Edit: I meant both the AMC 12 and the AIME, was kind of unclear)
Thank you.

You can get the transcripts from the past 2 weeks for the AMC 12.

The two AIME classes discuss different material under the same subject I think.

I took the AMC 10 Problem Series course over the summer.

I think the AIME classes cover roughly same level problems, topics, but obviously different problems

If those were the scores you got, it'd definitely be more beneficial to take the AIME problem series. Also, along with that you may also consider going over the Intermediate Series books and/or the classes, they are really helpful for AIME prep and easy-ish olympiad stuff.


linear algebra
matrix
calculus
derivative
inequalities
function
real analysis

Let \[n \ge 1 \] be an integer and \[\varphi : R^{n}\to R^{n}\] be a \[C^{1}\]
mapping. Suppose that there exists a real number C < 1 such that for all \[x \in R^{n}\] \[\parallel \varphi '(x)\parallel \le C \]

Prove that the mapping \[f: R^{n}\to R^{n}\] given by setting \[f(x) = x+\varphi (x) \] is then a C-1 bijection whose inverse also is C-1.

What can be said if we relax the condition on the derivative of \[\varphi \] to the statement that for all \[x \in R^{n}\] we have \[\parallel \varphi '(x)\parallel < 1 \]

Since $|\varphi(x)-\varphi(y)|\le C|x-y|$, it follows that $|f(x)-f(y)|\ge |x-y|-|\varphi(x)-\varphi(y)|\ge (1-C)|x-y|$. This proves that $f$ is injective. The double inequality $(1-C)|x-y|\le |f(x)-f(y)|\le (1+C)|x-y|$ can be used to prove that $f(\mathbb R^{n})$ is both open and closed. The inverse is $C^{1}$ by the inverse function theorem.

With the relaxed condition $f$ is still injective, because $|\varphi(x)-\varphi(y)|<|x-y|$. But it does not have to be surjective, and the inverse is not necessarily $C^{1}$. Examples: $f(x)=\arctan(x)$ is not onto, $f(x)=c(\arctan(x))^{3}$ with sufficiently small $c>0$ is not onto, and the inverse is not $C^{1}$.


Here are the problems  of the  final  round  of our national  olympiad .

Nice work, nickolas. :D

You can create a particular topic for each problem
so that discussion is done!


hello. My math teacher told me a story about the number 1 billion. he pointed us to a website about it. I dont know if he wrote that himself or if he just read it and told us. Can someone read it please and tell me if its true. i think my math teacher might be wrong. thank you

the website is http://www.tomyn.com/billion

That is a really interesting way to put a billion into context
 :huh:

i predict that your teacher is telling the truth, cuz teachers most of the time do...  :rotfl:


Define
\[\Omega(n)=\sum_{p^{e}\parallel n}e \]
for each prime $p$. Find
\[\sum_{d|n}g(d) \]
where $g(1)=1$ and $g(k)=(-1)^{\Omega(k)}$.

[hide]It is evident that $g(x)$ is multiplicative function. The same is true for $I(x) \equiv 1$ Hence the function $f(x)=(f\circ I)(x)= \sum_{d|n}g(d)I(\frac{x}{d})= \sum_{d|x}g(d)$ is also multiplicative.
Now the all we need is to calculate $f(x)$ at points $x=p^{k}$:
$\Omega (p^{k})=1+...+k=\frac{k(k+1}{2}$
$g(p^{k})=1$ for $k\equiv 0,3 (mod 4)$ and $g(p^{k})=-1$ othervice.
It is evident how to finish:
$f(x)=0$ if $x$ is not perfect square and
$f(x)=(-1)^{m}$, if $x=u^{2}$ and $u=p_{1}p_{2}... p_{m}$ - prime decomposition with $p_{1}\le p_{2}\le ...$, i.e we count all entries of primes.[/hide]

[quote="Phelpedo"]What do you mean by $I \equiv 1$?[/quote]

$\equiv$ is used to stand for an identity, or a definition; i.e. something is always true.  $=$, on the other hand, often denotes a condition we use to solve for a variable that is only true in certain cases, and we want to identify those cases.

Anyway, Digger is just using the group of multiplicative arithmetic functions under Dirichlet convolution.

[b]Definition:[/b]  A function $f : \mathbb{N}\to \mathbb{C}$ is said to be an arithmetic function.

[b]Definition:[/b]  An arithmetic function $f$ is said to be multiplicative if $gcd(a, b) \implies f(ab) = f(a) f(b)$.

[b]Definition:[/b]  The Dirichlet convolution of two arithmetic functions $f, g$ is given by

$h(n) = \sum_{d | n}f(d) g \left( \frac{n}{d}\right)$

[b]Lemma:[/b]  The Dirichlet convolution of two multiplicative arithmetic functions is also multiplicative.

[quote="t0rajir0u"][b]Definition:[/b]  An arithmetic function $f$ is said to be multiplicative if $gcd(a, b) \implies f(ab) = f(a) f(b)$.[/quote]

Just to say he means $gcd(a,b)=1$

Edit: Or indeed she means. Ive honestly no idea.


inequalities
search
inequalities proposed

Put $f(x,y,z)=\text{LHS}-\text{RHS}$, then easy see that $f(1,1,1)\geq 0,f(1,1,0)\geq 0,f(1,0,0)\geq 0$. Done! :P

[quote="kunny"]Sorry, I can't understand. Could I have expalnation in detail?

kunny[/quote]
I used an assertion in note of  Lee . And it is on internet  :wink:

Sorry  :oops: His name is Hojoo Lee. Please search!

Oh, is he? I see. Thank you, N.T.TUAN. :)

[quote="kunny"]For $x>0,\ y>0, \ z>0,$ prove the following inequality.
\[\frac{x^{3}+y^{3}+z^{3}}{3}-xyz\geq 2\left\{\left(\frac{x+y+z}{3}\right)^{3}-xyz\right\}\]
[/quote]
$\frac{x^{3}+y^{3}+z^{3}}{3}-xyz\geq 2.25\left\{\left(\frac{x+y+z}{3}\right)^{3}-xyz\right\}$ is true too.

I would like to see more basic solution, because this problem is to be solved by Japanese high school students.

[quote="arqady"]kunny, your inequality equivalent to
 $6\cdot\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)+\sum_{cyc}(x^{3}-xyz)\geq0.$ :wink:[/quote]

Thank you very much, arqady. :)

kunny


function
logarithms
real analysis
real analysis unsolved

I have that following question:

$f(x) = \ln(\sqrt{1+x^2}+x)$

The condition for a function to be odd : $f(x)=-f(-x)$

Therefore:

$f(x) = \ln(\sqrt{1+x^2}+x) = -\ln(\sqrt{1+x^2}-x)$

$= \ln(1) - \ln(\sqrt{1+x^2}-x) =$

$\ln \frac{1}{ln(\sqrt{1+x^2}-x)}$

$= \ln \frac{\sqrt{1+x^2} + x} { (\sqrt{1+x^2} +x) (\sqrt{1+x^2} -x)}$

and because the denominator equals 1.
QED.

That proof is wrong.

A: Why is that wrong?
B: Prove it correctly.

[quote="RealTime"]I have that following question:

$f(x) = \ln(\sqrt{1+x^2}+x)$

The condition for a function to be odd : $f(x)=-f(-x)$

Therefore:

$f(x) = \ln(\sqrt{1+x^2}+x) = -\ln(\sqrt{1+x^2}-x)$

$= \ln(1) - \ln(\sqrt{1+x^2}-x) =$

$\ln \frac{1} {\sqrt{1+x^2}-x}$

$= \ln \frac{\sqrt{1+x^2} + x} { (\sqrt{1+x^2} +x) (\sqrt{1+x^2} -x)}$

and because the denominator equals 1.
QED.

That proof is wrong.

A: Why is that wrong?
B: Prove it correctly.[/quote]


function
geometry
geometric transformation
rotation
analytic geometry
trigonometry
integration

How do you calculate  the average distance $\langle d \rangle$ between the points
in the region $D$ to the line $x+y=0$. The region $D$ is the quarter disk
$x^{2}+y^{2}\leq a^{2}$ and $x\geq 0$ and $y\geq 0$
(use meanvalue theorem)

[hide="my try"]
I assume one should find the function $d(x,y)$ , which gives the distance to each point to the line
and then $\langle d \rangle=\frac{1}{area(D)}\iint_{D}d(x,y)dxdy$
but I cant figure out the function $d(x,y)$
[/hide]

[hide="answer (not solution)"]
$\frac{4a\sqrt{2}}{3\pi}$
[/hide]

So, your method of solution is fine; all you are left to do is to find the distance from a point $(x_{0}, y_{0})$ to the line $x+y = 0$.  There is a well-known general formula for this, which I'm surprised you haven't seen.  One simple way (especially in this case) to calculate the desired value is to apply a rotation of 45 degrees to your coordinate plane (keeps distances fixed); after this, the distance between the (new) point and (new) line will be easy to calculate, and you can just plug it into your formula.

Let's set the whole thing up in polar coordinates. The region $D$ is described by $0\le r\le a, 0\le\theta\le\frac{\pi}2.$ The line $x=-y$ is now the line $\theta=\frac{3\pi}4$ (or $\theta=-\frac{\pi}4.$)

Consider a point $(r,\theta)$ in the first quadrant. Draw a triangle whose vertices are the origin, that point, and the nearest point on the perpendicular line $\theta=\frac{\pi}4.$ This is a right triangle, and we can see from it that the distance from the point to the line is $r\cos\left(\theta-\frac{\pi}4\right).$

The average distance is then

$\frac{4}{\pi a^{2}}\int_{0}^{a}\int_{0}^{\frac{\pi}2}r^{2}\cos\left(\theta-\frac{\pi}4\right)\,d\theta\,dr$

$=\frac{8}{\pi a^{2}}\int_{0}^{a}\int_{0}^{\frac{\pi}4}r^{2}\cos\left(\theta-\frac{\pi}4\right)\,d\theta\,dr$

$=\frac{8}{\pi a^{2}}\cdot\frac{1}{\sqrt{2}}\int_{0}^{a}r^{2}\,dr$

$=\frac{8a^{3}}{3\pi a^{2}}\cdot\frac{1}{\sqrt{2}}= \frac{4a\sqrt{2}}{3\pi}$

Plausibility check: the answer is an average distance, so it should be in the units of distance - in particular, it should be a constant mulitple of $a,$ which it is. The smallest distance is $0$ and the largest distance is $a,$ so the answer should be of the form $ca$ for some $0<c<1.$ Since $\frac{4\sqrt{2}}{3\pi}\approx .6,$ this is a plausible answer.

Thanks a lot.  :)
[b]JBL[/b]; you mean [url=http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html]this[/url] formula? (This formula would create an awful expression wouldn't it? (even if we substitute $x=-y$))

The method you mentioned of rotating is good thanks, it simplifies the whole problem to just considering one coordinate. If rotated the region $D$ 45 degrees anticlockwise and switched to the $x$-axis instead of $x+y=0$ the $d(x,y)=y$ and the integral would have become
\[\frac{4}{a^{2}\pi}\iint_{D'}ydydx \]
which is easier.

[b]Kent Merryfield[/b] Thanks a lot for your reply. Just one question, shouldn't the line be translated into polar coordinates as $x=-y$ which is $\frac{-\pi}{4}$and $\frac{3\pi}{4}$?

I cleaned up a typo or two. In fact, I had worked the whole problem originally as if the line was $x=y,$ and gotten the answer

$\frac{8a}{3\pi}\left(1-\frac1{\sqrt{2}}\right).$

The methods are essentially the same - in that case, I had the absolute value of a sine to contend with. When I realized what the problem actually stated, I went though and changed things, but I didn't quite get everything.

The formula I meant is equation 8 on that page.  Since all your points are in the first quadrant, you'll be able to drop the absolute values and the actual result won't be messy.

Also, I would have proceeded slightly differently than you did: the rotated region is less nice than the given region (unless you're going to integrate in polar coordinates, as Kent suggests), so rather than rotating and integrating, I would rotate, find $d(x, y)$, and then rotate back before integrating.  Then we have to apply the rotation to our distance function, which converts it from the very nice $y$ to the not-quite-so-nice-but-still-pretty-good $\frac{x+y}{\sqrt{2}}$, and we can integrate this over the original region $D$.  (This is, of course, the same value that one would get if one just used the formula directly.  In fact, this basic method (rotate, calculate in rotated coordinates, rotate back) is one method of deriving the formula in Mathworld.)

It is right, except that the bounds on $x$ should be $\frac{\pm a}{\sqrt{2}}$, not $\frac{\pm 1}{\sqrt{2}}$.


Moved from RT to a more appropriate forum.

[quote="elcholoperuano"][color=green]Moderator says: I've fixed your dollar signs for you -- they trigger LaTeX on the forum.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 or [[LaTeX]] for what this means.[/color]

So the question is as follows and I can't figure out what the equation would be.  Any ideas?

Four lots were sold for a total of 80,000 dollars. Lot 1 sold for 3000 dollars
more than lot 3.  Lot 2 sold for 1500 dollars more than lot 1. Lot 4 sold for 500 dollars more than lot 2.  How much was paid for lot 3?

Thanks in advance! : )

Montoya[/quote]

$ x_1 \equal{} x_3 \plus{} 3000$

$ x_2 \equal{} x_1 \plus{} 1500$

$ x_4 \equal{} x_2 \plus{} 500$

$ x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}80000$


real analysis
integration
limit
probability and stats

Give with proof necessary and sufficient condition for sequences $ a_{n}$ and $ b_{n}$ for which sequence of random variables
$ X_{n} \sim  N(a_{n},b_{n})$ is uniformly integrable.

Are you sure Alekk what if $ Var(X_n)=\sigma$ and $ E(X_n)=n$
Then don't we have
$ Limsup_{a\to+\infty}sup_{n\in\N}\int_{|x|>a}xe^{-\frac{(x-n)^n}{2\sigma^2}}=\infty$ ?

:blush:  you're right TheBridgeTwo, one needs a condition like "$ a_n$ bounded" too ..

the following extremely useful lemma gives the solution to this exercice in no time:
A sequence $ X_n$ of random variables is uniformly integrable iff $ E[|X_n|]$ is bounded and
\[ \lim_{P[A] \to 0} E[ |X_n| ; A] \equal{} 0
\]

This is the first time in my life I can point a mistake by Alekk

Alekk now leads 1875 to 1

:)


Let $ (X,\parallel{}\cdot\parallel{})$ and $ (Y,\parallel{}\cdot\parallel{})$ separable banach spaces. Prove that $ X\times Y$ is separable

This is roughly the same as showing that when $ A$ and $ B$ are countable sets, so is $ A \times B$.

Just take $ W \equal{} \{w_n\}$ and $ V \equal{} \{v_n\}$ as respective countable dense subsets. Then define $ W \times V$ as the pointwise cartesian product of these sets, i.e. all elements $ \{(w_i,v_j) \in X \times Y \, : \, i,j \in \mathbb{N}\}$.

Now $ W \times V$ is countable (I leave this for you). It is obviously a subset, so we must show it is a  dense subset.

Take arbitrary $ \epsilon > 0$ and $ (x,y) \in X \times Y$. There is some $ i$ with $ \parallel x \minus{} x_i \parallel_X < \epsilon$ and likewise $ j$ with $ \parallel y \minus{} y_j \parallel_Y < \epsilon$.

Assuming we have $ \parallel (x,y) \parallel_{X\times Y} \equal{} \sqrt{\parallel x\parallel_X \plus{} \parallel y \parallel_Y}$, we then get as $ (x,y) \minus{} (x_i, y_j) \equal{} (x\minus{}x_i, y\minus{}y_j)$ that

$ \parallel (x\minus{}x_i,y\minus{}y_j) \parallel^2_{X\times Y} \equal{} \parallel x \minus{} x_i \parallel_X \plus{} \parallel y \minus{}y_j\parallel_Y < 2 \epsilon$, and hence

$ \parallel (x,y) \minus{} (x_i, y_j) \parallel_{X\times Y} < \sqrt{2 \epsilon}$. We are done.


geometry
3D geometry
function
algebra unsolved
algebra

Find a formula for the n-th term of the sequence of all natural numbers which are not a:
a) square (2,3,5,6,7,8,10,11,..) b) cube (2,3,4,5,6,7,9,10,11,12..).

For a), take $x_n=[n+\sqrt{n}+\frac12]$ (here,$[\quad ]$ means the integer part.)  Next, examine why this works and find an answer for b) yourself :)

Enescu, please tell me how did you find that formula , I knew the answer, but i wanted to see how to get to it.

$n+[\sqrt[3]{(n+[\sqrt[3]{n}])}]$

More generally what is generated by $n+[\sqrt[k]{(n+[\sqrt[k]{n}])}]$  ;)   :?:

I don`t have Crux 1993, but I really want to know how to get to that formula.
Will someone solve it, please.
tµtµ If you are right, then $[\sqrt{n+[\sqrt{n}]}]=[\sqrt{n}+\frac{1}{2}]$.
How will you proove that?

[quote="tµtµ"]$x_n=n+[\sqrt[k]{(n+[\sqrt[k]{n}])}]$[/quote]

Beat, I'll write the solution for k=3, but it does hold for an arbitrary $k\geq 2$.
First note that in $\{1,2,...,p^3\}$ there are $p^3-p$ terms of the sequence, the last of them being $p^3-1$. Hence
    $x_{p^3-p}=p^3-1$
and after that
    $x_{p^3-p+1}=p^3+1$
    $x_{p^3-p+2}=p^3+2$
    ....................................
    $x_{(p+1)^3-(p+1)}=(p+1)^3-1$

So
(*)     $x_n=n+p$ iff $n\in\{p^3-p+1, p^3-p+2, ..., (p+1)^3-(p+1)\}$.
It remains to write $p$ as a function of $n$.

  $p^3-p+1\leq n\leq (p+1)^3-(p+1)\ \Longleftrightarrow\ p^3-p< n\leq (p+1)^3-(p+1) \ \ (1)$
For a given number $n$, there is a single $p$ which satisfies conditon (1), hence we may assume that (1) is true and just obtain a necessary form for $p$ (we don't need equivalence).
From (1) we get first
  $(p-1)^3<n<(p+1)^3$,
so $p-1<\sqrt[3]{n}<p+1$  (2)

Now from (1) and (2) we obtain
  $p^3-n\leq p-1<\sqrt[3]{n}$
and
  $(p+1)^3-n\geq p+1>\sqrt[3]{n}$,
so
  $p^3<n+\sqrt[3]{n}<(p+1)^3$,
hence $p=[\sqrt[3]{n+\sqrt[3]{n}}]$.
 Now remeber the sentence (*) above and obtain
  $x_n=n+[\sqrt[3]{n+\sqrt[3]{n}}]$

Oups...it's not exactly tµtµ's formula !!! However, it's a good one.


Given the sets $A,\ B,\ C$ which is made by the finite fixed points on a plane. Denote the number of element of each set by $a,\ b,\ c$. Select one point from each set at random and denote each point by $P,\ Q,\ R$. Assume that the probability such that $P=Q$ is $\frac{1}{6}$ and the probability such that $P,\ Q,\　R$ are colinear is $\frac{1}{abc}$. If all sets of $A,\ B,\ C$ aren't the sub set of other set, find the pair of $(a,\ b,\ c)$.


real analysis
function
trigonometry
algebra
polynomial
real analysis unsolved

What are the idempotent elements of

a) $ L^1(\mathbb{R}^d)$?
b) $ L^2(\mathbb{R}^d)$? :)

under convolution?  (that is, functions $ f$ such that $ f*f \equal{} f$)

e: edited the title name, the original was stupid/undescriptive and i apologize

Let's start with the second question: in $ L^2.$  Since the Fourier transform turns convolution into pointwise mulitplication, this is the same as asking what the idempotents are under pointwise multiplication. And for a function to be an idempotent under pointwise multiplication, its values must be either $ 0$ or $ 1.$

Hence:

$ f\in L^2(\mathbb{R}^d)$ is an idempotent under convolution if and only if $ \widehat{f}\equal{}\chi_E$ for some measurable set $ E$ of finite measure.

----

Does that answer the question for $ L^1?$ Well, no, since such functions tend not to be in $ L^1.$ In one dimension, the (inverse) Fourier transform of the characteristic function of an interval behaves like $ \frac{\sin ax}{x}$ for large $ x,$ and that's not in $ L^1.$ I suspect (but am to lazy to prove right now) that $ L^1(\mathbb{R}^1)$ contains no idempotents. I also suspect that the answer may be different for $ d>1.$

I should add this: $ L^2(\mathbb{R}^d)$ with convolution is NOT an algebra, since the convolution of two $ L^2$ functions need not be in $ L^2.$ (We can say that $ L^2*L^2\subset C_0,$ where by $ C_0$ I mean the continuous functions that tend to zero at infinity. But $ C_0\not\subset L^2.$)

$ L^1(\mathbb{R}^d)$ with convolution is a Banach algebra, albeit without an identity.

yeah, it may not really make sense to talk about idempotents for $ L^2$ (the original problem was for $ L^1$, and i added the bit about $ L^2$ because it's different and a lot easier)

Well, the same idea works for $ L^1$, doesn't it? The Fourier transform of the convolution is still the product of the Fourier transforms, so, on one hand, it should take only values $ 1$ and $ 0$; on the other hand, it should be continuous. Now there are not an awful lot of such   functions :).

D'oh! The proof is easy, and I was overlooking it. $ L^1(\mathbb{R}^d)$ has no nonzero idempotents, for any $ d.$

Suppose $ f*f \equal{} f,$ with $ f\in L^1.$ Then $ \widehat{f}\in C_0$ and $ (\widehat{f})^2 \equal{} \widehat{f}$ at every point. But that makes $ \widehat{f}$ a continuous function which takes only $ 0$ and $ 1$ as its values. Hence $ \widehat{f}$ is either constantly $ 0$ (trivial) or constantly $ 1$ (not in $ C_0.$)

Now $ L^1(\mathbb{T})$ (the set of periodic integrable functions) does have nontrivial idempotents under convolution. They are precisely the trigonometric polynomials whose nonzero coefficients are all exactly $ 1.$

[color=green][Message composed before I saw fedja's response.][/color]

If $ M(\mathbb{R}^d)$ is the set of finite signed Borel measures on $ \mathbb{R}^n$, which is a Banach algebra under convolution, then what are its idempotents? In particular, can you find any other than the Dirac delta?


The school population has increased from $ 3,200$ to $ 3,600$ this year. If this percent rate of increase remains constant, what will be the school population in the end of the year?


Show that
tan (pi/13) · tan (2pi/13) · tan (3pi/13) · tan (4pi/13) · tan (5pi/13) · tan (6pi/13) = sqrt 13


Click [url=http://mathideas.org/public/math/mock2007A.pdf]here[/url] to access the test.

There is a typo on #24 but I am about to fix that. :) (It should be $=0$ at the end.)

I cannot access it.

Says it is unavailable.

I encountered an error on the link.

Can you just put it into a word document and make it an attachment?

copy link location and put a . before pdf

[quote="Xantos C. Guin"]Click [url=http://mathideas.org/public/math/mock2007A.pdf]here[/url] to access the test.[/quote]
That still doesn't work for me.
EDIT: Never mind.  :blush:

My mistake. The link is now fixed. Everyone has five extra minutes to turn in their answers.

(Which means that they are due by exactly 8:25PM.)

[quote="SplashD"]We have to pm our answers before 8:15?!?[/quote]

I said, they are due by [b]8:25PM[/b]... :roll:

[quote="chess64"][quote="eryaman"]Uh, number 1 is worded kind of weird.  :?[/quote]

Take 2007 and add $x$ to it $y$ times...[/quote]K yeah, got it now.  Not sure what I was thinking before.

THe pdf wouldnt update for me -,-

turned in answers already, i failed.

[quote="lotrgreengrapes7926"][quote="SplashD"]There aren't errors in 20 and 24 right? I just submitted mine and I left those blank.[/quote]
If in 24 you mean it should be $da$, chess64 says it's right as it is.

I just sent mine in. My highest possible score is 118.5. :( I probably got around 90.[/quote]

I just mean errors in general. You left 7 blank?

[quote="SplashD"]I don't see anything wrong with it.[/quote]

I guess I messed up...  :oops:

[quote="Max300"]Could someone pm me the solution to #6?

I just can't get it and its bugging me.[/quote]

I'll do it if chess64 verifies that he received your answers

chess, thanks a lot for the test, there were a few hicups during the test, but it still was fun! i liked the problems a lot! especially the sum of reciprocals of odd numbers, clever! lol...i just realized that it is not to my advantage to guess C on every problem that i did not get, oh well, at least i will not do it on the real test, i forgot this time lol

I sent my answers too...this test was hard :P

I wasn't able to take it during the designated time, but I just took it and submitted my answers. Wow, I got owned  :blush: Am I the only one who didn't get an answer to Question 3?!?

Heh, I got owned too.  123 on the 12B last year, 105 on this. :P

I don't know if I got owned yet.

116 on AMC 12A, and at most 141 on this.

i thought it was 2.5 points per one left blank

[quote="chess64"]Starting 2007 it's only 1.5.[/quote]

Yeah, to encourage more guessing, (which is what I did) :D


The tournament will be 32 people. Seeding will be determined by highest rating. I will NOT participate. This will be single elimination.

Current entrants:

1. mewto55555 (1527)
2. FantasyLover (1502)
3. ernie (1451)
4. RomanianGenius (1438)
5. budi713 (1412)
6. ash-Pokemon (1370)
7. isabella2296 (1350)
8. thdanh90 (1349)

OMG, ther r so many tourneys goin on right now...

i cant even count them on my fingers  :rotfl:

but i think ill join...  :D

EDIT: wh00ps, i 4got 2 add, highest rating = 1451.

______________________________________________________________________________

THIS IS JUST LIKE THE COUNTDOWN CUP!

COPYCAT!  :D

ill join

highest - 1502

6th tourney i think  :rotfl:

I'll join too.

1350(I know, really low... :oops: )

But it's higher than mine, izzy.  :wallbash:  :wallbash:

[quote="mewto55555"]What about method of seeding? Where have I seen that before?[/quote]

hmmm... :ewpu:  :rotfl:

But imitation is the greatest form of flattery.

w00t!

The tournament will be 32 people. Seeding will be determined by highest rating. I will NOT participate. This will be single elimination.

Current entrants:

1. mewto55555 (1527)
2. FantasyLover (1502)
3. ernie (1451)
4. RomanianGenius (1438)
5. budi713 (1412)
6. tacoman206 (1399)
7. AIME_is_hard (1375)
8. ash-Pokemon (1370)
9. isabella2296 (1350)
10. thdanh90 (1349)

Games to be Played:

4 budi713 VS 13 myyellowducky82 (FR-1)
8 ash-Pokemon VS 9 AIME15 (FR-2)
5 tacoman206 VS 12 pinkpiglet (FR-3)
7 AIME_is_hard VS 1 FantasyLover (QF-3)
6 Thanatos VS 11 thdanh90 (FR-5)

People who DO NOT have to play:

2 kl2836
3 vallon22

Eliminated:

10 isabella2296

[quote="isabella2296"][quote="lotsofmath"]

Eliminated:

10 isabella2296[/quote]

*sniff*[/quote]

Awww... poor izzy.  But it's OK.  Before CTC starts, you have like 10 other tourneys to participate in. :roll:

well, if she didn't sign up for it, then she can't play in it...

I signed up for a lot. But no one ever told me who to play.  :huh:

See thats the chaos. If no one ever tells you who to play, TOURNEY MANIA ruins the tournaments. Also, the excessive amountsof spam are bad...If you look at my tourney, the first 7 or so pages are very productive and thats why my tournament will finish.

ok, fine
i will make a few changes in my tourny to make it more efficient
check out my new SUMMER tourny thread for more details

FantasyLover vs. AIME_is_hard

AIME_is_hard forfeits.

FantasyLover wins.

ONE ROUND BYE

1 FantasyLover
2 kl2836
3 vallon22

NO BYE

4 budi713
5 tacoman206
6 Thanatos
7 AIME_is_hard
8 ash-Pokemon
9 AIME15
10 isabella2296
11 thdanh90
12 pinkpiglet
13 myyellowducky82

FIRST ROUND

4 budi713 VS 13 myyellowducky82 (FR-1)
8 ash-Pokemon VS 9 AIME15 (FR-2)
5 tacoman206 VS 12 pinkpiglet (FR-3)
7 AIME_is_hard DEFEATS 10 isabella2296 (FR-4)
6 Thanatos VS 11 thdanh90 (FR-5)

QUARTER FINALS

winner of FR-1 VS 2 kl2836 (QF-1)
winner of FR-2 VS winner of FR-3 (QF-2)
1 FantasyLover DEFEATS 7 AIME_is_hard (QF-3)
winner of FR-5 VS 3 vallon22 (QF-4)

SEMI FINALS

winner of QF-1 VS winner of QF-2 (SF-1)
1 FantasyLover VS winner of QF-4 (SF-2)

FINALS

winner of SF-1 VS winner of SF-2

3RD PLACE MATCH

loser of SF-1 VS loser of SF-2

Games to be played:

4 budi713 VS 13 myyellowducky82 (FR-1)
8 ash-Pokemon VS 9 AIME15 (FR-2)
5 tacoman206 VS 12 pinkpiglet (FR-3)
6 Thanatos VS 11 thdanh90 (FR-5)

People who DO NOT have to play:

1 FantasyLover
2 kl2836
3 vallon22

Eliminated:
7 AIME_is_hard
10 isabella2296

[quote="lotsofmath"]
Games to be played:

4 budi713 VS 13 myyellowducky82 (FR-1)
8 ash-Pokemon VS 9 AIME15 (FR-2)
5 tacoman206 VS 12 pinkpiglet (FR-3)
6 Thanatos VS 11 thdanh90 (FR-5)
[/quote]

...i havent seen budi, ash, tacoman, thanatos or dan in like a month...

tacoman206 told me that he'd be gone for 6 days, I think its been 6 days though

omg
i'm getting bored waiting for this....
can u pplz just use the coin method?

Budi713 has not been on for weeks, he is eliminated by default and myyellowducky82 is the winner of the match.

I haven't seen ash-Pokemon since I played him in a cd last month, he is also eliminated by default. AIME15 wins the match.

I'll give tacoman206 a few more days before I eliminate him.

Sorry, I can't eliminate BOTH Thanatos and thedanh90. I'll prod them sometime.

FIRST ROUND

13 myyellowducky82  DEFEATS 4 budi713(FR-1)
9 AIME15 DEFEATS 8 ash-Pokemon (FR-2)
5 tacoman206 VS 12 pinkpiglet (FR-3)
7 AIME_is_hard DEFEATS 10 isabella2296 (FR-4)
6 Thanatos VS 11 thdanh90 (FR-5)

QUARTER FINALS

13 myyellowducky82 VS 2 kl2836 (QF-1)
9 AIME15 VS winner of FR-3 (QF-2)
1 FantasyLover DEFEATS 7 AIME_is_hard (QF-3)
winner of FR-5 VS 3 vallon22 (QF-4)

SEMI FINALS

winner of QF-1 VS winner of QF-2 (SF-1)
1 FantasyLover VS winner of QF-4 (SF-2)

FINALS

winner of SF-1 VS winner of SF-2

3RD PLACE MATCH

loser of SF-1 VS loser of SF-2

Games to be played:

5 tacoman206 VS 12 pinkpiglet (FR-3)
6 Thanatos VS 11 thdanh90 (FR-5)
13 myyellowducky82 VS 2 kl2836 (QF-1)

People who DO NOT have to play:

1 FantasyLover
3 vallon22
9 AIME15

Eliminated:
4 budi713
7 AIME_is_hard
8 ash-Pokemon
10 isabella2296


inequalities
geometry
circumcircle
trapezoid
trigonometry
congruent triangles
geometry unsolved

Let O be the intersection of diagonals of a convex quadrilateral ABCD and
let P and Q be the circumcenters of triangles ABO and CDO respectively.
Prove the inequality
AB + CD ≤ 4PQ.

Hi, I also think it is unusual. I found problem on the site
http://www.imocompendium.com/index.php?options=oth|Yug|Ne|Ne&ttn=Serbia%20and%20Montenegro&p=0,
year 1996

Take isosceles trapezoid ABCD with AD || BC and circumradius R of congruent triangles AOB, COD. M, N are midpoints of AD, BC. When $ B, C \longrightarrow N,$ then $ \angle AOB \equal{} \angle COD \longrightarrow 2 \angle A$ $ \Longrightarrow$ $ \angle APB \equal{} \angle CQB \longrightarrow 4 \angle A$ $ \Longrightarrow$ $ \angle MNP \equal{} \angle MNQ \longrightarrow \angle A$ $ \Longrightarrow$ $ \frac {AB \plus{} CD}{4 PQ} \longrightarrow \frac {4R \sin 2A}{8R \sin A} \equal{} \cos A.$ Then let $ \angle A \longrightarrow 0.$


a^3+b^3=-2c^3,c=ab^0,5 find the value of
a^3+2ba^2+ab^4+ab^2+b^5+2ca^2+2acb+2cb^4=?
tell ur opinion about my problem

First of all,why aren't you trying to write it in Latex? Secondly,i'm sorry to say this to you,but I think that these kind of problems don't say nothing. Of course these problems can be solved using some well-known identities or by brute calculation. ;)


vector
linear algebra
linear algebra unsolved

This is a problem from a textbook, Riley Hobson and Bence 'Mathematical Methods for Physics and Engineering'. It asks to check the validity of a vector identity. If [b]a[/b], [b]b[/b] and [b]c[/b] are general vectors satisfying [b]a[/b] x [b]c[/b] = [b]b[/b] x [b]c[/b], does this imply [b]c[/b] . [b]a[/b] - [b]c[/b] . [b]b[/b] = c|[b]a[/b]-[b]b[/b]|

According to the book the identity is satisfied, but I think that there should be a plus/minus sign before the RHS, since we don't know whether the vectors [b]a[/b]-[b]b[/b] and [b]c[/b] are in the same or opposite directions. What do you think?

you're right. $ \mathbf{a}, \mathbf{b}$ may be swapped, negating the LHS.


limit
logarithms
function
integration
calculus
derivative
calculus computations

can anyone prove this limit..

$ \lim_{x\to\infty}(1 \plus{} \frac {1}{x})^x \equal{} e$

I don't have this limit's proof in my textbook..
I was doing the proof by L'Hopital,, I think it's surely wrong.. :blush:
thanks for your help :)

Hint: the first step is to take the logarithm.

I know about the step you wrote,,professor Kent..

the proof I made like this,,

$ \lim_{\frac{1}{x}\to0}(1\plus{}\frac{1}{x})^\frac{1}{\frac{1}{x}}\equal{}p$
$ \frac{1}{x}\equal{}a$
$ \lim_{a\to0} \frac{\ln(1\plus{}a)}{a}\equal{}\ln p$
$ \equal{}\lim_{a\to0}\frac{[\ln(1\plus{}a)]'}{a'}\equal{}\ln p$
$ \equal{}\lim_{a\to0}\frac{1}{1\plus{}a}\equal{}1\equal{}\ln p$
$ p\equal{}e$

of course that's incorrect,, because $ [\ln(1\plus{}a)]'$ is come from the limit in this topic..
is that right what I said?

So now the issue comes down to this: what is your first definition? In developing the exponential and logarithmic functions, what do you start with? Do you start by defining the logarithm by $ \ln x\equal{}\int_1^x\frac{dt}{t}\,?$ Do you start by defining the exponential function as the power series? Do you start somewhere else - perhaps even by [i]defining[/i] $ e\equal{}\lim_{n\to\infty}\left(1\plus{}\frac1n\right)^n\,?$ There are many different definitions and paths. They all work, and the end product is equivalent.

I think you're right..
It depends on the point of view of each person to prove the limit,,..
but,, my teacher (sensei) told me that, before L'hopital method was created, there was another proof about the limit in this thread..
he also doesn't really know about that proof. He said that the proof I've written is not valid, because of the last line of the post I've written before.
it seems like,, there's a cycle..we can prove one limit with the differential method,, and we can prove the differentiation with limit..

so,, is the solution I've written valid? :maybe:

Here's an alternative way to get your proof (in the attachment).

The only part I find difficult is finding the motivation for taking the derivative of $ \ln(x)$ at $ x \equal{} 1$.

This limit can also be seen by applying Euler's method from numerical analysis to the differential equation $ y'\equal{}y, y(0)\equal{}1$ on the interval $ [0,1]$ and letting the stepsize approach 0. Then write down the formula for the approximation to $ y(1)$ as the stepsize approaches 0. It is the limit above.


analytic geometry
graphing lines
slope
topology
advanced fields
advanced fields unsolved

[b]
Is there anyone knows example about [u]Regular Space [/u]which is [u]not Completely regular [/u]other than Tychonoff Corckscrew ?.

I need easy one not as that which it depends on ordinals numbers.

Thx.[/b]

Ah yes, there is one that isn't so bad conceptually (but kind of hard to write down). This was in A. Mysior Proc. AMS. April 1981.

Take $ [0,\infty)\times [0,1]$ so a slice of the plane and throw in an extra point $ \{x^*\}$ where $ x^*\notin\mathbb{R}\times [0,1]$ (so the point is "below" the slice).

Your open sets are the typical ones if you are in the slice $ [0,\infty)\times (0,1]$ (open balls about those points that are small enough to stay in the slice). If you have a point on the line $ [0, \infty) \times \{0\}$ then the open set is a vertical line from your point and a line of slope $ 1$ coming from the point. So this looks like a V. Formally, $ V_x\equal{}\{(x, t)\in [0,\infty)\times [0,1] : 0\leq t \leq 1\}\cup \{(s,t)\in [0,\infty)\times [0,1] : x\leq s \leq x\plus{}1, t\equal{}s\minus{}x\}$ and we are allowed to remove finitely many points from these two "lines."

Now the only open sets we have left are the ones about $ x^*$. Take the basis elements to be $ B_*\equal{}\{\{x^*\}\cup ([a,\infty)\times [0,1]) : a\in [0,\infty)\}$. These open sets are just your point not in the slice with a portion of the slice to the right of sum number.

This may look bad, but it is really pretty simple. Just note what your three types of open sets look like. In the slice are normal open balls, on the bottom line of the slice are V shapes minus a finite number of points, and the point off the slice are chunks to the right of some point. Try drawing each type, it will seem nicer.

These spaces in general are hard to come by.

[b]
I'm so thankful HilbertThm90.

it strange space [Mysior Space].But, it looks very nice one.

Thank you again dude. :)  :D  :wink: [/b]

I'm thinking to prove that Mysior space is Completely regular on different way as follows :

I'll depend on about $ 0\minus{}dimensional$ space ( every space has a clopen base ).

Clearly , Mysior space has a Clopen Base ( Yes or Not )????

If yes, we can depend on theorem " every $ 0\minus{}dim$ space is completely regular .

Give me your opinion ....