real analysis
limit
probability
real analysis unsolved

Can anyone give a hint to show that the set of reals $ x$ s.t.  $ 0 \le x \le 1$ and the decimal expansion(base 10) of $ x$ doesn't contain the digit 6, has Lebesgue measure 0?

consider numbers  $ 0.a_{1}a_{2}...a_{n}$ at first let's try to calculate the number of such different  nambers where at least one $ a_{j}$ is equal to $ 6$, suppose this is $ A_{n}$ then the number all different numbers of the types  $ 0.a_{1}a_{2}...a_{n}$ will be equal to $ 10^{n}$
and if $ \lim_{n \to \infty} \frac {A_{n}}{10^{n}} \equal{} 1$ then we have already solve your problem.
at first let's try to calculate the number of such numbers as $ 0.a_{1}a_{2}...a_{i}6a_{i \plus{} 2}...a_{n}$ where no one $ a_{i}$ is not equal to $ 6$ for all $ i$ , the simple calculation show that : =$ C_{n}^{1}9^{n \minus{} 1}$
after consider all numbers where only two numbers $ a_{i}$ and $ a_{j}$ ($ i \neq j$) is equal to $ 6$ answer will be  $ C_{n}^{2} 9^{n \minus{} 2}$. let's do this operation $ n$-th times and the last number will be $ \underbrace{0,66...6}_{n}$ and the number of such numbers is $ C_{n}^{n}9^{n \minus{} n}$
. finally we have :
$ C_{n}^{1}9^{n \minus{} 1} \plus{} C_{n}^{2}9^{n \minus{} 2} \plus{} ... \plus{} C_{n}^{n}9^{0} \equal{} 10^{n} \minus{} C_{n}^{0}9^{n} \equal{} 10^{n} \minus{} 9^{n}$
hence
$ \lim_{n \to \infty} \frac {A_{n}}{10^{n}} \equal{} \lim_{n \to \infty} \frac {10^{n} \minus{} 9^{n}}{10^{n}} \equal{} 1$
done.
please note that there is no important that $ x$ does not containt digit $ 6$. solution is same for all digit.
by $ C_{n}^{k}$ I denote $ \binom{n}{k}$

[quote="Extremal"]consider numbers  $ 0.a_{1}a_{2}...a_{n}$ at first let's try to calculate the number of such different  nambers where at least one $ a_{j}$ is equal to $ 6$, suppose this is $ A_{n}$ then the number all different numbers of the types  $ 0.a_{1}a_{2}...a_{n}$ will be equal to $ 10^{n}$
and if $ \lim_{n \to \infty} \frac {A_{n}}{10^{n}} \equal{} 1$ then we have already solve your problem.[/quote]

First of all I thank you for your reply. But I am unable to get your logic. Moreover, I think, your statements have grammatical mistakes. So I unable to understand your point.

I try to explain more good.
At first I fix $ n$ which is positive integer.
Now, I consider numbers  which have only $ n$ digits in the decimal expansion (base 10).
or I mean different numbers like this :
$ 0,a_{1}a_{2},...,a_{n}$ where $ a_{j}$ take values between: $ 0 \minus{} 9$ integers.number all such numbers is equal to $ 10^{n}$.
consider all this  numbers  as the set $ A \equal{} \{\underbrace{0,0..0}_{n \plus{} 1},...,\underbrace{0.9...9}_{n \plus{} 1}\}$ and normalise Lebesgue measure on this set,
say $ \mu(A) \equal{} 1$.
My next step is to calculate number all such different  numbers, where at least one $ a_{i} \equal{} 6$
answer is $ 10^{n} \minus{} 9^{n}$.
therefore Lebesgue measure of this new set $ B$ will be $ \mu(B) \equal{} \frac {10^{n} \minus{} 9^{n}}{10^{n}}$
Note that $ A \minus{} B$ will be set of numbers which in decimal expansion does not contain digit the $ 6$ and
$ \mu(A \minus{} B) \equal{} \mu(A) \minus{} \mu(B) \equal{} \left( \frac {9}{10} \right)^{n}$.
Now let's $ n \to \infty$ then we get answer for your question.
Is there some non understable explanation  :)  ?

Can you do it by partitioning the interval $ [0,1]$ as we do for the Cantor set?

[quote="Extremal"]consider all this  numbers  as the set $ A \equal{} \{\underbrace{0,0..0}_{n \plus{} 1},...,\underbrace{0.9...9}_{n \plus{} 1}\}$ and normalise Lebesgue measure on this set,
say $ \mu(A) \equal{} 1$.  [/quote]

I don't understand this. Could you please clarify this?

ok.

let's say: each element from $ A$ has measure $ \mu$ equal to $ \frac{1}{10^{n}}$
now I'm talking about http://en.wikipedia.org/wiki/Probability_measure

Let's consider all subsets of the set $ A$ let's denote it by $ B$.
then $ \mu$ will be probability measure on the Algebra $ B$.
...
I don't understand well, your question. what are you asking in exact ? :maybe:


[quote="thegod277"]Lam the nao de co chu ky khi post bai len vay(o phia duoi), va vd mot so bai post co ghi la Argady wrote:.......Lam sao lam duoc nhu vay the :maybe: ? mong moi nguoi giup do! :D[/quote]
Chu ki thi minh khong biet! Con de co sb wrote:... thi cau nhan vao chu Quote o phia duoi bai trich dan la duoc. O tren la 1 VD.

the con phan chu ki? nhung nguoi co kinh nghiem dau het ca roi?
Anh N.T.TUAN lam on chi bao?

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analysis, cam on cau rat nhieu :lol: !
Thank you very much!


A line $ l$ is tangent to a circle of radius $ r$ at a point $ P$. From a variable point $ R$ on the circle, a perpendicular $ RQ$ is drawn to $ l$ with $ Q \in l$. Determine the maximum area of triangle $ PQR$.

Let $ O$ be the center of the circle and $ M$ the midpoint of the chord $\overline{PR}.$ Note easily that $ \triangle PQR \sim \triangle OMP.$

$ \frac {[\triangle PQR]}{[\triangle OMP]} \equal{} \frac {PR^2}{OP^2}  \Longrightarrow \ [\triangle PQR] \equal{} [\triangle POR] \cdot \frac {PR^2}{2r^2}$

Let $ \angle POR \equal{} \theta.$ Then we have

$ [\triangle POR] \equal{} \frac {_1}{^2}r^2\sin \theta \ , \ PR^2 \equal{} 2r^2(1 \minus{} \cos \theta)$

$ [\triangle PQR] \equal{} \frac {_1}{^2}r^2( \sin \theta \minus{} \sin \theta \cos \theta)$

$ [\triangle PQR]$ is maximum $\Longleftrightarrow$ $ f(\theta) \equal{} \sin \theta \minus{} \sin \theta \cos \theta$ is maximum.

Note that $ \frac {\partial f(\theta)}{\partial \theta} \equal{} \cos \theta \minus{} 2\cos^2 \theta \plus{} 1$

Therefore, $ \frac {\partial f(\theta)}{\partial \theta} \equal{} 0$ yields $ \Longrightarrow \ \cos \theta \equal{} \minus{} \frac {_1}{^2}$ and $ \cos \theta \equal{} 1$

As a result,  $ [\triangle PQR]$ is maximum when $ \theta \equal{} 120^{\circ} \ , \ ( 180^{\circ} \ge \theta \ge 0^{\circ})$

Then, the maximum area  is given by $ [\triangle PQR] \equal{} \frac {3\sqrt {3}}{8}r^2$


algebra
polynomial
ratio
symmetry
real analysis
real analysis unsolved

suppose $P$ is a polynomial,which has been fixed first.(And $P$ is not always equal to 0)
$u,v$ are two real numbers such that $P(u)+P(v)=0$
Prove:$u+v$ is bounded.

For a solution,see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=102700[/url]

We may assume that $P$ has odd degree (otherwise both $u$ and $v$ must stay inside a bounded interval), which means that the ratio between $Q(x)=-P(-x)$ and $P$ goes to $1$ as $x\to\pm\infty$. The problem is now a particular case of:

Let $P,Q$ be monic polynomials of degree $n$. Then there is a constant $M$ such that $|x-y|\le M$ whenever $P(x)=Q(y)$.

We may as well assume we're working only with positive arguments (i.e. we're only interested in what happens for large $x$ and $y$). A moment's thought will show that if $t$ is large enough, then the coefficients of the polynomial $P_{t}(x)=P(x+t)$ are larger than the corresponding coefficients of $Q(x)$. By symmetry, $t$ can be chosen large enough so that this goes the other way around: the coefficients of $Q(x+t)$ are larger than those of $P(x)$. This means that there is a $t$ such that $|x-y|\le t$ if we're to have $P(x)=Q(y)$.

[quote="grobber"] A moment's thought will show that if $t$ is large enough, then the coefficients of the polynomial $P_{t}(x)=P(x+t)$ are larger than the corresponding coefficients of $Q(x)$.[/quote]
Thank you,the moment's thought is very nice.


Prove that the equation $ x^2\plus{}y^2\minus{}z^2\equal{}1997$ has infinitely many solutions in integers $ x,y$ and $ z$.

[quote="SimonM"][quote="Wickedestjr"]If you add z^2, then you get x^2 + y^2 = 1997 + z^2. If x^2 = 1997, then that means that y^2 = z^2, and there is an infinite amount of possible numbers that can equal z and y.

Ha ha!!!! My first proof. How do you rate this proof on a scale of 1 to 10. This is the first proof I have ever actually done successfully.

How do you rate this solution?[/quote]

0. $ x^2 \equal{} 1997$ doesn't have any solutions in [b]INTEGERS[/b][/quote]

Dang it!!! Then this wasn't my first proof.  :wallbash_red:

[quote="joh"]Prove that the equation $ x^2 \plus{} y^2 \minus{} z^2 \equal{} 1997$ has infinitely many solutions in integers $ x,y$ and $ z$.[/quote]

[hide]
Let $ y \equal{} n, z \equal{} n\plus{}1 \Longrightarrow x^2 \equal{} 1998 \plus{} 2n$, and there's an infinite number of even squares ..   :roll:
[/hide]

[quote="Wickedestjr"][quote="azjps"][quote="joh"]Prove that the equation $ x^2 \plus{} y^2 \minus{} z^2 \equal{} 1997$ has infinitely many solutions in integers $ x,y$ and $ z$.[/quote]

[hide]
Let $ y \equal{} n, z \equal{} n \plus{} 1 \Longrightarrow x^2 \equal{} 1998 \plus{} 2n$, and there's an infinite number of even squares ..   :roll:
[/hide][/quote]

Good job, but how did you know to do that?[/quote]you should practice a lot. this problem is quite easy.

First of all, you should become familiar with the difference between solving problems in [b]INTEGERS[/b] and [b]REAL NUMBERS[/b]. It is actually quite different. For example, while you can solve equations like this (in real numbers) quite straightforwardly: \[ 2x \plus{} 3y \equal{} 4 \\
6x \plus{} 8y \equal{} 99\] Solving equations like \[ 7x \plus{} 11y \equal{} 1\] in INTEGERS is very different.

...conclusion: the Pre-Olympiad section is not the best place to go around looking for problems that you can do your first proof on.  :wink:

Don't be THAT harsh... :o
At least, I learned greatly about how to write proofs by browsing this section. I also successfully attempted a few proofs myself (and fixed a few minor errors once they were pointed out).

So I say you should learn how to write proofs first. Do not rely on intuition, for example "it is obvious that two consecutive integers have exactly one of them that is divisible by 2."

Exercise: Prove the above statement. Rigorously. :)

Oh, and...

[quote="Wickedestjr"]So is the whole thing about substituting things in for the variables to make it something like;

(some even number) + 2n = x^2

?...
[/quote]

No, the whole thing is [b]not[/b] about substituting things in the exact form
\[ \text{(some even number)} \plus{} 2n \equal{} x^2
\]
Rather it's about finding creative and rigorous solutions at the same time. The above result was just one product of such creativity.

Perhaps someone could post some motivation for the solution for him?

I pm-ed him before, though I guess I stressed the wrong points, which explains the post

[quote="Wickedestjr"]So is the whole thing about substituting things in for the variables to make it something like;

(some even number) + 2n = x^2

?...[/quote]

We have $ x^2\minus{}1997\equal{}(z\plus{}y)(z\minus{}y)$, as long as $ x^2\minus{}1997$ is odd, we can let $ z\minus{}y\equal{}1$, $ z\plus{}y\equal{}x^2\minus{}1997$.

Exactly.  The reason this problem is straightforward is that $ y^2 \minus{} z^2$ can take on [i]any[/i] integer value not congruent to $ 2 \bmod 4$, so all that has to be done is to choose $ x$ such that this is not the case and one can always find $ y, z$.


integration
calculus
calculus computations

This problem comes from an examination of our Department and no one can solve it correctly.

Find the volume of the solid generated by $ x^2 \plus{} y^2 \equal{} 1$, $ y^2 \plus{} z^2 \equal{} 1$.

[hide]$ \frac {16}{3}$[/hide]

This is easy. $ V\equal{}2\int_0^1 (2\sqrt{1\minus{}t^2})^2 dt\equal{}\frac{16}{3}$

This is called Steinmetz solid, for more.

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


Find the largest integer $ n$ for which $ 12^n$ evenly divides $ 20!$.

$ 12\equal{}2^2*3$

there are $ 10\plus{}5\plus{}2\plus{}1\equal{}18$ 2's in 20, so there are 9 $ 2^2$s.

there are $ 6\plus{}2\equal{}8$ 3's.

obviously 9>8 lol

answer : 8

alternately, divide $ 20!$ by $ 12$ until it becomes a decimal( with a scientific calculator :wink:  )

[quote="FantasyLover"]$ 12 \equal{} 2^2*3$

there are $ 10 \plus{} 5 \plus{} 2 \plus{} 1 \equal{} 18$ 2's in 20, so there are 9 $ 2^2$s.

there are $ 6 \plus{} 2 \equal{} 8$ 3's.

obviously 9>8 lol

answer : 8[/quote]

Why do you only divide 20 , 10 , 5 , and 2 ?

\[ \frac{20}{2}\equal{} 10\qquad\qquad\frac{10}{2}\equal{} 5\qquad\qquad\frac{5}{2}\equal{} 2.5\qquad\qquad\frac{2}{2}\equal{} 1\]

I listed out all the factors of 2 by hand and got the correct answer, I just don't understand the faster method presented in the solution.

Thank you for your time.

You divide 20 by increasing powers of 2, 20/2, 20/4, 20/8...until the power of 2 exceeds 20, which is the same as dividing the numerator by 2.  :lol:

[hide=my sol]$20!=1\cdot2\cdot3...\cdot20.$ This means there are 8 3s(3, 6, 9, 9, 12, 15, 18, 18.) and 18 2s, so the answer is $\boxed{8}$[/hide]

[quote=Silverstorm]Could someone provide more insight? I'm not sure I understand the above solutions. First of all, why are there 18 2's in 20? And how are there any 3's at all in 20?[/quote]

Do flrfunction(20/a)+ flrfunction(20/a^2)+...

[quote=Silverstorm]Could someone provide more insight? I'm not sure I understand the above solutions. First of all, why are there 18 2's in 20? And how are there any 3's at all in 20?[/quote]

First, there are no 3's in 20, they are in $20!=20\cdot 19 \cdot 18...\cdot 1$.
There are at least floor($\frac{20}{2}$)$=10$ factors of 2 in 20!, since $20!=20\cdot 19\cdot 18...\cdot 1$, and every other term is even, and thus has a factor of 2. However, some of these factors in 20!, such as 4 or 8, have multiple factors of 2. So, we have to add the number of multiples of 4 to the previous count. This is equal to the floor of
$$\frac{20}{4} = 5.$$ Similarly, we must count the multiples of $8$ now (we counted each multiple of 8 once when counting multiples of 2, and again when counting multiples of 4, but 8 has 3 factors of 2 in it, so we must count them again). This adds floor($\frac{20}{8}$)$ = 2$ to our count.
Finally, we do the same thing with $16$ (we stop here since any power of 2 greater than this is either not a factor of 20! or has already been counted). This yields floor($\frac{20}{16}$)$=1$. Adding these up gives us:
$$10+5+2+1=18.$$ Factors of $2$. We can do the same thing with $3$, and then find the greatest value of $n$ that satisfies the inequalities in the official solution.

[quote=DevanshShekhar]Can someone say the solution in a different way? It is very confusing.  :wacko:[/quote]

it’s trivial by legendre’s (which was what dotted used in the post above me)

Observe that $12$‘s distinct prime factors are $3$ and $2$. Now you find the highest powers of $2$ and $3$ in the prime factorization of $20!$ (where legendre’s comes in)

You take the smaller number, since while you have 18 2s you can factor out, you only have 8 3’s to accompany this.

[quote=DottedCaculator]$\nu_p(n!)=\sum_{i=1}^\infty\left\lfloor\frac n{p^i}\right\rfloor$. $\nu_3(20!)=8$. $\nu_2(20!)=18$. $\boxed8$.[/quote]

Thanks so much

[quote=DevanshShekhar]Can someone say the solution in a different way? It is very confusing.  :wacko:[/quote]


EDIT: This feature has been disabled due to various technical and user issues.

We are, as of today, allowing users to change their username whenever they want to (just like you change your password.) However you can only change [b]ONCE[/b] per year. Only you can use your previous usernames as well.

If you have any questions post here.

[quote="Valentin Vornicu"]We are, as of today, allowing users to change their username whenever they want to (just like you change your password.) However you can only change [b]ONCE[/b] per year. Only you can use your previous usernames as well.

[/quote]
If we change it, can we change it back to our original username?

[quote="suma_milli"]Do we just have to send a pm to an administrator or something?[/quote]No, you can change it yourself. No PMs, no topics, just go in your profile and change it. As I said do not "test" this feature as the username change will be permanent (for 1 year :).

@myyellowducky82: yes, you can change back to your original one in 1 year! (only [b]you[/b] can change back, so nobody will ever be able to claim myyellowducky82 as his own.)

Will there be any way for other users to see how usernames are linked?

For example, suppose I go on vacation for a few months and users change their name; when I come back, will there be some way to determine, for example, that the user formerly posting as "Valentin Vornicu" is now posting as "SexxyBeast1337"?   :wink:

Yes, you will have, under your profile, the option to show the previous username of that user. Note that 1 year changes are not cummulative (ie if you don't change your username in two years you don't get to change it twice within a week.)

[quote="JBL"]... for example, that the user formerly posting as "Valentin Vornicu" is now posting as "SexxyBeast1337"?   :wink:[/quote]
You're joking, of course, but I think those of us in the upper half of the front page under the "memberlist" button are going to maintain more continuity than that with our established brands and personas.

[quote="AIME15"]Question: Under "Username" there is "Username change reason", then email/password stuff.  So, where do we put our new username?[/quote]You just edit your username in the box (it's editable now.)

Sorry, but I didn't really understand. I go to profile. OK, get the registration information tab automatically.

Fields: Username - Bugi
can't be edited
Username change reason: empty box, can be edited
E-mail
Password...

So how do I change my username?

[quote="Bugi"]Fields: Username - Bugi
can't be edited
[/quote]Oops!  :blush:
Since I can change anyone's username, I had that field editable, but I forgot to flip the switch for everyone else too :) You can change it now.

There is a problem though. I was doing FTW, and when I changed my username, I lost all my badges, game count, etc.... Also, when I played a game, the game count didn't go up or down.. Can this be fixed?  :(

Also, I'm losing rating when I'm winning??  :o

If somebody changed their username how would I be able to see who they were before they changed.
[quote="For Example $ LaTeX$"]...I changed my username...[/quote]
So if I wanted to see who $ LaTeX$ was before he changed his name how would I find out

I don't think that is possible just yet.....  :P

[quote="r15s11z55y89w21"]By badges, he means FTW. FTW has badges.[/quote]

We don't need no stinkin' badges!

^ That wasn't necessary

I know. You can always just put a list, maybe in the right corner of the signature, that says:

[code]<$LaTeX$'s post>

______________

(align-right)
Previously known as:

Utter Loser[/code]

[quote="$ LaTeX$"]Perhaps easier, would be to put the old username  next to the new username in the profile page...... or under the name of the author....[/quote]Yeah that will be the case.

[quote="Valentin Vornicu"]However you can only change [b]ONCE[/b] per year.[/quote]

What do you mean by year? Does it start jan 1, and end dec 31st(in which case, you could change your username on 2 consecutive days)? or is it you cannot change your username until 365 days after yuor previous change?

[quote="abacadaea"] is it you cannot change your username until 365 days after yuor previous change?[/quote]Correct.

I changed my username... Now my rating on FTW keeps on going baacck to 990... Is there any way to fix it?

[quote="levans"]This is a known issue. I'll be looking for a fix for it when I get a chance.[/quote]

Thanks

AoPSWiki user pages don't seem to be updated with the correct new username when a user changes their username.

I think you should maintain a (public) log of all username changes.

This has already been suggested. Read earlier posts.

I fail to see which earlier post you are talking about.

Perhaps you don't know what I mean by "log" - I want a list of all username changes that have ever occured, on one list.

Also:
Given any user whose current username I know, I want to be able to find out his past usernames (already talked about here).
I also want to be able to find a user's current username, given his past username. (For example, suppose I don't check the forums for a year, and when I come back, "Kent Merryfield" no longer exists. I would like to find what Kent Merryfield changed his name to.)

Apparently, the "username change" ability is deleted now? Or am I looking in the wrong place?

Yes, it is. You are no longer able to change your username.

Er...can we change our username once a year?
I don't know whose answer is right: Latex's, or Valentine Vornicu's.

I think you used to be able to change it once a year, but they took that out, so you cannot change your username, as LaTeX said.

No, you can not change your username. This caused far too many problems for our site and users.

I would've liked to change my username, but it's not avaliable anymore.


geometry
geometric transformation
rotation
trapezoid
circumcircle
dilation
congruent triangles

You do it by composing rotations. In general, let $R(X,\alpha)$ be the rotation of center $X$ and angle $\alpha$:

You can obtain $M$ from $N$ by the transformation $R(A,\frac \pi 2)\circ R(P,\frac \pi 2)\circ R(A,\frac \pi 2)$. The composition of rotations is, however, another rotation, its angle is the sum of the angles of the three rotations, and the center of that rotation is the fixed point of the composition of the three rotations. It's easy to see that the fixed point is $P$, and the angle is $\frac{3 \pi}2$, and from here the result follows.

What about a solution using similarity of triangles?

We complete the isosceles right triangles ACM and ABN to squares ACOM and ABQN. Then we denote by U the midpoint of the segment QO. Now I will show that both triangles BCU and MNU are isosceles right triangles, with the right angle at U. When this will be proven, it will follow that U = P (in fact, both BCU and BCP are isosceles right triangles, and they have the base BC in common, so that the points U and P must coincide), and consequently, it will follow that the triangle MNP is isosceles right, and the problem will be solved. Hence it will be enough to show that both triangles BCU and MNU are isosceles right triangles.

It will even be enough only to show that the triangle BCU is isosceles right; in fact, the proof for the triangle MNU will be analogous (in order to understand why, just look at the two triangles ABC and ANM and the squares ACOM and ABQN; you see that the triangles ABC and ANM are completely "equal in rights").

Hence I will show that the triangle BCU is isosceles right. (This is, by the way, an old result of Oene Bottema.)

Let R and S be the feet of the perpendiculars from the points Q and O on the line BC. Let H be the foot of the A-altitude of triangle ABC, and let W be the midpoint of the side BC.

The square ABQN tells us (yes, squares can speak) that < QBA = 90°. In the right triangle AHB, we have

  < BAH = 90° - < ABH = 180° - 90° - < ABH
      = 180° - < QBA - < ABH = < QBR.

This, together with AB = BQ (this was again told me by the square ABQN) and < AHB = < BRQ = 90°, shows that the triangles BAH and QBR are congruent. Thus, AH = BR and BH = QR. Similarly, AH = CS and CH = OS. Now, from AH = BR and AH = CS, we see that BR = CS, and together with WB = WC (since the point W is the midpoint of the segment BC), this gives WR = WB + BR = WC + CS = WS. Hence, the point W is the midpoint of the segment RS. Now, the quadrilateral QRSO is a trapezoid (since its sides QR and OS are parallel to each other, being both perpendicular to the line BC); since U and W are the midpoints of its legs QO and RS, respectively, it follows that the segment UW is the midparallel of this trapezoid. Thus, the segment UW is parallel to the legs QR and OS of this trapezoid, and the length of this segment is $UW = \frac{QR+OS}{2}$. Now, from the parallelism of the segment UW to the legs QR and OS, it follows that $UW \perp BC$ (since $QR \perp BC$). Thus, < BWU = 90°. Also, from $UW = \frac{QR+OS}{2}$, together with BH = QR and CH = OS, we get $UW = \frac{QR+OS}{2} = \frac{BH+CH}{2} = \frac{BC}{2} = BW$ (the last equality is because W is the midpoint of the segment BC). Altogether, we have < BWU = 90° and UW = BW; thus, the triangle BWU is an isosceles right triangle, and it follows that < UBW = 45°. In other words, < UBC = 45°. Similarly, < UCB = 45°. Thus, the triangle BCU is an isosceles right triangle, and this completes the proof.

I know that all of this can be proved much more quickly by using rotations, but for some reason I like my proof...

  Darij

Thanks for your solutions Grobber and Darij  :) , but i didn't understand the solution with rotations. :blush:
I insist on a solution with similarity. I know for sure that a such exists.

Here is, for Iris, a more detailed version of Grobber's solution.

We begin with a well-known fact:

[b]Theorem 1.[/b] The composition of n rotations with angles $\alpha_1$, $\alpha_2$, ..., $\alpha_n$ is

- a rotation with angle $\alpha_1 + \alpha_2 + ... + \alpha_n$ if the angle $\alpha_1 + \alpha_2 + ... + \alpha_n$ is not a multiple of 360°;
- a translation if the angle $\alpha_1 + \alpha_2 + ... + \alpha_n$ is a multiple of 360°.

Now, we denote the rotation with center X and angle $\alpha$ by $R\left( X,\alpha \right) $.

WLOG assume that the triangle ABC is labeled counterclockwisely, i. e. that when moving along the circumcircle of triangle ABC in a counterclockwise direction, starting at the point A, then one encounters the vertices of triangle ABC in the order A, B, C. (Equivalently, the angles < CAB, < ABC, < BCA are clockwisely oriented.)

Then we consider the composition of rotations $R\left( A,90^{\circ }\right) \circ R\left( P,90^{\circ }\right) \circ R\left( A,90^{\circ }\right) $ (this means: first rotate with center A and angle $90^{\circ }$, then rotate with center P and angle $90^{\circ }$, and then again rotate with center A and angle $90^{\circ }$). By Theorem 1, this composition is a rotation with angle $90^{\circ }+90^{\circ }+90^{\circ }=270^{\circ }$. We denote this rotation by $R\left( Z;270^{\circ }\right) $, where Z is the center of this rotation. We will now prove that Z = P.

Actually, a rotation (with an angle not being a multiple of $360^{\circ }$) has only one fixed point, namely its center. Hence, in order to prove that Z = P, it is enough to show that the point P is a fixed point of the rotation $R\left( Z;270^{\circ }\right) $; in fact, once this will be showed, it will follow that the point P is the center of this rotation $R\left( Z;270^{\circ }\right) $, so that we will immediately get P = Z.

To show that the point P is a fixed point of the rotation $R\left( Z;270^{\circ }\right) $, we remember that this rotation $R\left( Z;270^{\circ }\right) $ can be written as the composition $R\left( A,90^{\circ }\right) \circ R\left( P,90^{\circ }\right) \circ R\left( A,90^{\circ }\right) $. Let P' be the image of the point P under the rotation $R\left( A,90^{\circ }\right) $, then let P" be the image of the point P' under the rotation $R\left( P,90^{\circ }\right) $. If we succeed to show that the image of the point P" under the rotation $R\left( A,90^{\circ }\right) $ is P again, then it follows that the point P is a fixed point of the composition $R\left( A,90^{\circ }\right) \circ R\left( P,90^{\circ }\right) \circ R\left( A,90^{\circ }\right) $, and hence P is a fixed point of the rotation $R\left( Z;270^{\circ }\right) $. This will entail that Z = P as I have said before.

So we will now prove that the image of the point P" under the rotation $R\left( A,90^{\circ }\right) $ is P again. In fact, since the point P' is the image of the point P under the rotation $R\left( A,90^{\circ }\right) $, we have AP' = AP and < PAP' = 90°. Thus, the triangle PAP' is right isosceles. Hence, < AP'P = 45° and $PP^{\prime }=\sqrt{2}\cdot AP^{\prime }$. Since the point P" is the image of the point P' under the rotation $R\left( P,90^{\circ }\right) $, we have PP" = PP' and < P'PP" = 90°, so that the triangle P'PP" is isosceles right, too. Thus, < P"P'P = 45° and $P^{\prime }P^{\prime \prime }=\sqrt{2}\cdot PP^{\prime }$. From < AP'P = 45° and < P"P'P = 45°, it follows that < AP'P = < P"P'P, so that the points P', A, P" lie on one line. From $PP^{\prime }=\sqrt{2}\cdot AP^{\prime }$ and $P^{\prime }P^{\prime \prime }=\sqrt{2}\cdot PP^{\prime }$, we get $P^{\prime }P^{\prime \prime }=\sqrt{2}\cdot \left( \sqrt{2}\cdot AP^{\prime }\right) =2\cdot AP^{\prime }$, so that the point A is the midpoint of the segment P'P". Thus, AP" = AP'. Together with AP' = AP, this yields AP" = AP. We also have < P"AP = 180° - < PAP' = 180° - 90° = 90°. Hence, the image of the point P" under the rotation $R\left( A,90^{\circ }\right) $ is the point P. Hence, after what I said above, we get Z = P.

Hence, we can rewrite our rotation $R\left( Z;270^{\circ }\right) $ as $R\left( P;270^{\circ }\right) $. But we know that this rotation is the composition $R\left( A,90^{\circ }\right) \circ R\left( P,90^{\circ }\right) \circ R\left( A,90^{\circ }\right) $; hence, it maps the point N to the point M (in fact, the rotation $R\left( A,90^{\circ }\right) $ maps the point N to the point B, since the triangle NAB is isosceles right; then, the rotation $R\left( P,90^{\circ }\right) $ maps the point B to the point C, since the triangle BPC is isosceles right; and finally, the rotation $R\left( A,90^{\circ }\right) $ maps the point C to the point M, since the triangle CAM is isosceles right; together, it follows that the composition of these rotations maps the point N via B and C to the point M). Hence, we now know that the rotation $R\left( P;270^{\circ }\right) $ maps the point N to the point M; this yields PM = PN and < NPM = 90° (if I would have used directed angles, I would have written 270° instead). Thus, the triangle MPN is isosceles right. Tombstone ($\blacksquare$).

That's a good example of a quite easy proof which is very hard to write up...

[quote="Iris Aliaj"]I insist on a solution with similarity.[/quote]

Hey, congruent triangles are similar! So what do you have against my first solution?

One remark about the problem itself:

[quote="Iris Aliaj"]In the exterior of the triangle ABC with m(<B) > 45, m(<C) > 45 one constructs the right isosceles triangles ACM and ABN such that m(<CAM)=m(<BAN)=90 and in the interior of ABC the right isosceles triangle BCP with m(<P)=90. Show that the triangle MNP is aright isosceles triangle.[/quote]

The conditions m(<B) > 45 and m(<C) > 45 are not really necessary. The triangle ABC can be arbitrary; however, then one must define the triangle BCP in a slightly different way: << Let BCP be a right isosceles triangle such that < BPC = 90° and the points P and A lie in the same halfplane with respect to the line BC. >>

  Darij

Something similar with [b]Grobber[/b]'s:

Let $D$ be the midpoint of $BC$; a $45^\circ$ rotation about $B$, followed by a $\frac{1}{\sqrt 2}$ dilation will swap the triangles $NBP$ and $ABD$, the same sequence of transformations, with the rotation in the reverse direction will swap the triangles $MCP$ and $ACD$, thus $MD=ND=AD\sqrt 2$ and $\angle MPN=2\cdot 45^\circ=90^\circ$, this way finding also $MN=2AD$ , but that is a well known relation.

Best regards,
sunken rock


geometry
circumcircle
geometric transformation
reflection
rectangle
trigonometry
geometry unsolved

[hide="My Solution"][u]Construction: [/u]
Let $ N'$ be the reflection of $ B$ on $ N$.
Join $ DF,N'N,N'C,N'A$

[u]Proof[/u]
$ N,F$ are respectively midpoints of $ BN',BA\implies N'A||NF$ and $NF\perp AC\implies N'A\perp AC$
$ N'A\perp AC\text{ and }N'B\perp BC\implies CN'$ is the diameter of $ \Delta ABC$
$ N,D$ are respectively midpoints of $ BN',BC\implies$
$ ND \equal{} \frac {1}{2}N'C \equal{} \frac {1}{2}$(diameter of $ \Delta ABC$)$ \equal{}$(circumradius of $ \Delta ABC$)
[b]QED[/b]

Edit: I forgot to include the fig.[/hide]

It is easy to prove P,O,C are collinear =>
PA⊥AC   NF∥AP   => N is the midpoint of PB  =>  ON⊥PB
=> ONBD is a rectangle => DN=OB

Akashnil i am having trouble with this problem. In the diagram you drew i label the intersection of $ AB$ and $ ND$ as $ X$ and the intersection of $ AC$ and $ NF$ as $ L$. Now  $ ALF$ and $ NFX$ are similar and so angle $ NFX$ is $ 90 \minus{} A$.
Now angle $ DFB$ is $ A$ because of midpoint theorem. Implies angle $ NFD$ = angle $ NFX$ + angle $ XFD$ = $ 90 \minus{} A$ + $ A$ = $ 90$. But you proved that angle $ NBD$ is $ 90$ and i proved that angle $ NFD$ is $ 90$. This implies $ NBDF$ is concyclic. This means that angle $ XBD$ is equal to angle $ FNX$. But from the similarity i wrote earlier angle $ FNX$ is $ A$. Also angle $ XBD$ = $ B$. Which implies angle $ A$ =angle $ B$ . However this was not stated in the problem which means........

[quote="Akashnil"][quote="befuddlers"]Now  $ ALF$ and $ NFX$ are similar[/quote]
I don't think this is true. If you thought $ ND\parallel{}AC$ you are wrong. I've not read the rest of your post. I need to understand this part first.[/quote]
Dude i never assumed $ ND\parallel{}AC$ . The two triangles are similar because of $ AA$ similarity which follows from the perpendicular condition and vertically opposite angles.

I can't see the $ AA$ similarity. It is right that $ \angle ALF\equal{}90^0$, but could you pls elaborate which one is the $ 90^0$ in $ \Delta FNX$? and why?

hey can anyone tell me approx how many marks will i get if i have done the way as plane geometry has done but solved everything in my mind and by mistake forgot to proove p,o,c...........to be collinear????????????
 :rotfl:

btw......i hav e  written in beautiful handritting .......
that onbd is a rctangl.....and prooved it by congruency..............
pl tell//////////

[quote="Akashnil"]I can't see the $ AA$ similarity. It is right that $ \angle ALF \equal{} 90^0$, but could you pls elaborate which one is the $ 90^0$ in $ \Delta FNX$? and why?[/quote]
Never mind. Stupid me. Since you didn't modify the typo in the question and since i didn't read your proof properly i was confused. Now i understand my flaw.

@arnav : They won't take off a lot of points. So don't worry arnav.

\[ \begin{gathered} Let \hfill \\
NF \cap AC\_at\_X \hfill \\
\angle BFN \equal{} 90^o \minus{} \angle A \hfill \\
BNXC\_cyclic\_quadrilateral \hfill \\
\angle BNF \equal{} 180^o \minus{} \angle C \hfill \\
\frac {{BF}} {{\sin \angle BNF}} \equal{} \frac {{BN}} {{\sin (90^o \minus{} \angle A)}} \hfill \\
\Leftrightarrow \frac {{BF}} {{\sin \angle BNF}} \equal{} \frac {{BF}} {{\sin (180^o \minus{} \angle C)}} \equal{} \frac {{BF}} {{\sin \angle C}} \equal{} \frac {{BN}} {{\sin (90^o \minus{} \angle \angle A)}} \equal{} \frac {{BN}} {{\cos \angle A}} \hfill \\
\Leftrightarrow \frac {{BF}} {{\sin \angle C}} \equal{} \frac {{BN}} {{\cos \angle A}} \hfill \\
\Leftrightarrow BN \equal{} \frac {{BF\cos \angle A}} {{\sin \angle C}} \equal{} \frac {{\frac {c} {2}\cos \angle A}} {{\sin \angle C}} \equal{} \frac {{c.\cos \angle A}} {{2\sin \angle C}} \equal{} R\cos \angle A \hfill \\
\hfill \\
OD \equal{} R\cos \angle A\equal{}BN\hfill \\
\therefore \Delta OBD \cong \Delta NBD \Leftrightarrow DN \equal{} OB \equal{} R \hfill \\
\end{gathered}
\]

now i read the question correctly i got it :D
In triangle $ NFD$ it can be proved that angle $ N$ is $ B$,$ NFD$ is 90. and by midpoint formula we have FD is b/2
So by sine rule ND is $ b/2sinB$ which is $ R$ which is what we needed to prove

I think that using trigo you can make the answer much more precise and short and I have one simple and short answer here.

The answer goes something like this:

Using Midpoint Theorem on Points $F$ and $D$ we have $FD ll AC$.
which makes $\angle A= \angle BFD$

Applying Circumradius $R= \frac a{2 \sin A}$ we have
\[R \&=\frac{2 \cdot BD}{2 \cdot \sin BFD} \\
  \\  \&=\frac{BD}{ \sin BFD} \\
  \\ \& = \frac{BD}{ \sin BND}   \ \ \ \ \ \ (BNFD \text{ cyclic}) \\
  \\ \& = \frac{ND}{ \sin NBD}    \ \ \ \ \ \ (\text{sin rule in } \triangle BND) \\
  \\ \& = \frac{ND}{ \sin 90^{\circ}} \\
  \\ \& = \boxed{ND}\]

hey, i think there is a much elegant soln than trigo..
for that, just shift d triangle so that C falls on D.
then its obvious that ND is d circum radius of the triangle..

here is my detailed solution
 by shifting ABC so that C falls on D,
we get F as the midpoint of AC, bcoz of midpoint theorem.
and as B becomes the midpoint thus N becomes the circum-centre, as
perpendicular from F  to AC is the perpendicular bisector.thus ND is the circum-radius.

$NBDF$ is cyclic,so $\angle BND=\angle BED$,also $FBDO$ is cyclic ($O$ is the circumcenter),$\angle BFD=\angle BOD$,hence we get $NBDO$ is cyclic,so it is rectangle.therefore $OB=ND$.

Let $E$ be the mid point of $AB$.
If we prove that $NBOE$ is cyclic we are done($O$ is the circumcentre)
This is because we already have $\angle OEB=90$. So We'll get that $\angle BNO=90$ which means that $ODBN$ is a rectangle.(D is the mid point of $BC$).As $\angle BOC=2C$ and as $\angle BNE=180-C$..
Also note that $OAB$ is isosceles. So $\angle BOE=C$. Thus we see that $NBOE$ is cyclic..
DONE. :)

My solution:
Clearly by angle chasing $\angle{OBD}=90-\angle{A}$.Hence $BD=RsinA$ and $NB=RcosA$ as $\angle{NBD}=90$.Therefore,$ND^2=NB^2+BD^2=R^2cos^2A+R^2sin^2A=R^2$.  Hence $ND=R$.

[quote=Akashnil]Let $ ABC$ be an acute angled triangle; let $ D,F$ be the midpoints of $ BC,AB$ respectively. Let the perpendicular from $ F$ to $ AC$ and the perpendicular from $ B$ ti $ BC$ meet in $ N$: Prove that $ ND$ is the circumradius of $ ABC$.

[15 points out of 100 for the 6 problems][/quote]

Wait  aminute  ; what does it mean to say that the perpendicular from B to bc meets at N ? Yes I uunderstood that before reading your post , thanks !

[quote=kk108][quote=Akashnil]Let $ ABC$ be an acute angled triangle; let $ D,F$ be the midpoints of $ BC,AB$ respectively. Let the perpendicular from $ F$ to $ AC$ and the perpendicular from $ B$ ti $ BC$ meet in $ N$: Prove that $ ND$ is the circumradius of $ ABC$.

[15 points out of 100 for the 6 problems][/quote]

Wait  aminute  ; what does it mean to say that the perpendicular from B to bc meets at N ?[/quote]

Yes it is little confusing.The problem means to say that a line perpendicular to BC passes through B and meets at AN.

Let $FN \cap AC=G$, Clearly, $\angle CGN=\angle DFN=90^{\circ} \implies FDBN \text{ and } FODB \text{ is cyclic}$
Hence, $180^{\circ}-\angle FNB=\angle FDB=\angle FOB \implies FOBN \text{ is also cyclic} \implies \angle OFB=\angle ONB=90^{\circ}$
Now, we already know that, $\angle ODB=\angle NBD=90^{\circ} =\angle ONB \implies ODBN \text{ is a rectangle } \implies \boxed{ND=BO=R}$

This problem can be killed with homothety.
Yes it truely does take homothety with sc.factor 2 on BNFD and see how the magic does AVADA KEDAVRA to the problem and there will be a little left required to prove.

Let the perpendicular from $F$ on $AC$ meet $AC$ at $E$. By mid point theorem, we get that $DF \parallel AC$.
$\Rightarrow \angle CEF= \angle DFN$ (corresponding angles). As $\angle DFN= \angle DBN= 90^{\circ}$, $FNBD$ is cyclic.
$\angle CAB=\angle DFB= \angle A$ (corresponding angles).
$\angle DFB= \angle DNB= \angle A$.

By sine rule, $\frac{BC}{sin \angle A}=2R$, where $R$ is the circumradius of $\Delta ABC$.
In $\Delta NBD$, $sin \angle BND= sin \angle A = \frac{BD}{ND}$
$\Rightarrow ND= \frac{BD}{sin \angle A} \Rightarrow ND= \frac{2.BD}{2.sin \angle A}$
$\Rightarrow ND= \frac{BC}{2.sin \angle A} \Rightarrow ND= \frac{2R}{2} = R$.

Thus, $ND$ is the circumradius of $\Delta ABC$.

[b]Claim:[/b] $OBDN$ is a rectangle. Note that proving this claim completes our problem because the diagonals of a rectangle are equal.
[b]Proof:[/b] Join $OD$ and $OF$. Notice that points $O,D,B,F$ are concyclic.
Now $\angle FDB=\angle C$ as $FD \parallel AC$. Now use the fact that quadrilateral $CBNK$ is cyclic to get $\angle BNF = 180^{\circ}-\angle C$
Thus $ODBN$ is cyclic too. As, any three points determine a unique circle, we have that ponts $O,D,B,N,F$ are concyclic. Thus $\angle BNO=\angle BFO=90^{\circ}$. Our proof is thus complete.


topology
advanced fields
advanced fields unsolved

Let Q be the metric space of rational numbers on the real line with the euclidean metric.
LET S the subset of rational s on [0 , pi) .
Determine whether or not
1- S is open in Q
2- S is Closed in Q
3- S is Compact in Q
4- S in Connected in Q

sparky: Please don't start new topics in "Solved Topics." Nothing should start there; it's a place to move things when we're done with them.

As for the answers: The set $S$ is closed in $\mathbb{Q}$ but it is not open, compact, or connected.

It's closed because it contains all its limit points. Note that $\pi$ is not a limit point because there is no such thing as $\pi$ in the universe we inhabit here.

It's not open because it doesn't contain a neighborhood of $0.$

It's not compact because a compact subset of a metric space must be complete. We can find Cauchy sequences that do not converge.

It's disconnected. For example, $\{x\in S: x^2<2\}\cup\{x\in S: x^2>2\}$ splits $S$ into two open sets.


[quote="rem"]Is there a server in Aops that allows you to translate all pages into another language? :maybe:[/quote]

right click anywhere on the screen and press the second button from the bottom

Well, the second option from the bottom in my left-click menu is View Page Source, which iirc rem was looking for a little while ago. Maybe instead you would like, say, Babelfish (http://babelfish.altavista.com), which translates whole pages, but it's not that accurate. :) (Babelfish translates the Chinese for bug as in flaw in a computer program stinking insect.)


Let $ a,b,c \in R^ \plus{}$
Prove that
\[ \frac{a^2\plus{}b^2}{a\plus{}b}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{c^2\plus{}a^2}{c\plus{}a} \leq \frac{3(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c}\]

i think it's very easy :

$ \sum\frac{a^2\plus{}b^2}{a\plus{}b}\equal{}\sum a\plus{}b\minus{} \frac{2ab}{a\plus{}b}$

so we have to prove :

$ 2(a\plus{}b\plus{}c)\leq 2(\sum\frac{ab}{a\plus{}b})\plus{}\frac{3(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c}$

wich is trivial since $ \sum\frac{ab}{a\plus{}b}\leq \sum a$

[quote="apollo"]Let $ a,b,c \in R^ \plus{}$
Prove that
\[ \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a} \leq \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}
\]
[/quote]
SOS kills it :)
we need to prove $ \frac{1}{(a\plus{}b\plus{}c)}(\sum\frac{ac(a\minus{}c)\plus{}bc(b\minus{}c)}{a\plus{}b}) \leq 0$
equivalent to $ \frac{1}{(a\plus{}b\plus{}c)}(\sum ac(a\minus{}c)(\frac{1}{a\plus{}b}\minus{}\frac{1}{b\plus{}c}) \leq 0$
equivalent to $ \frac{1}{(a\plus{}b\plus{}c)}(\sum \frac{\minus{}ac(a\minus{}c)^2}{(a\plus{}b)(b\plus{}c)}) \leq 0$ which is true :)

[quote="apollo"][quote="mehdi cherif"]i think it's very easy :

$ \sum\frac {a^2 \plus{} b^2}{a \plus{} b} \equal{} \sum a \plus{} b \minus{} \frac {2ab}{a \plus{} b}$

so we have to prove :

$ 2(a \plus{} b \plus{} c)\leq 2(\sum\frac {ab}{a \plus{} b}) \plus{} \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}$

wich is trivial since $ \sum\frac {ab}{a \plus{} b}\leq \sum a$[/quote]
Why ?[/quote]
$ \sum\frac{ab}{a\plus{}b}\leq\sum\frac{ab}{a}\equal{}\sum a$

[quote="mehdi cherif"][quote="apollo"][quote="mehdi cherif"]i think it's very easy :

$ \sum\frac {a^2 \plus{} b^2}{a \plus{} b} \equal{} \sum a \plus{} b \minus{} \frac {2ab}{a \plus{} b}$

so we have to prove :

$ 2(a \plus{} b \plus{} c)\leq 2(\sum\frac {ab}{a \plus{} b}) \plus{} \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}$

wich is trivial since $ \sum\frac {ab}{a \plus{} b}\leq \sum a$[/quote]
Why ?[/quote]
$ \sum\frac {ab}{a \plus{} b}\leq\sum\frac {ab}{a} \equal{} \sum a$[/quote]
but from this you get $ 2\sum\frac{ab}{a\plus{}b}\leq2\sum a$ so you can you give link with what you say and the original inequality ?

[quote="apollo"]Let $ a,b,c \in R^ \plus{}$
Prove that
\[ \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a} \leq \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}
\]
[/quote]
It remains to prove that:
$ \sum\frac{c(a^2\plus{}b^2)}{a\plus{}b} \le a^2\plus{}b^2\plus{}c^2$
or:
$ (a^2\plus{}b^2)(\frac{2c}{a\plus{}b}\minus{}1)\plus{}(b^2\plus{}c^2)(\frac{2a}{b\plus{}c}\minus{}1)\plus{}(a^2\plus{}c^2)(\frac{2b}{a\plus{}c}\minus{}1) \ge 0$
we assume that $ a \le b \le c$,we have:
$ (a^2\plus{}b^2)(\frac{2c}{a\plus{}b}\minus{}1)\plus{}(b^2\plus{}c^2)(\frac{2a}{b\plus{}c}\minus{}1)\plus{}(a^2\plus{}c^2)(\frac{2b}{a\plus{}c}\minus{}1) \ge (a^2\plus{}b^2)(\sum\frac{2a}{b\plus{}c}\minus{}3) \ge 0$

[quote="mehdi cherif"][quote="apollo"][quote="mehdi cherif"]i think it's very easy :

$ \sum\frac {a^2 \plus{} b^2}{a \plus{} b} \equal{} \sum a \plus{} b \minus{} \frac {2ab}{a \plus{} b}$

so we have to prove :

$ 2(a \plus{} b \plus{} c)\leq 2(\sum\frac {ab}{a \plus{} b}) \plus{} \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}$

wich is trivial since $ \sum\frac {ab}{a \plus{} b}\leq \sum a$[/quote]
Why ?[/quote]
$ \sum\frac {ab}{a \plus{} b}\leq\sum\frac {ab}{a} \equal{} \sum a$[/quote]

It is $ \sum\frac {2ab}{a \plus{} b}\leq \sum a$

which can be proved by full expansion. After that all is obvious. :wink:

You don't need expansion; it's the cyclic sum of $ \frac{a\plus{}b}{2} \ge \frac{2ab}{a\plus{}b} \iff \frac{(a\minus{}b)^2}{2(a\plus{}b)} \ge 0$ :)

[quote="apollo"]Let $ a,b,c \in R^ \plus{}$
Prove that
\[ \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a} \leq \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}
\]
[/quote]
$ \iff (a \plus{} b \plus{} c)(\frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a}) \leq 3(a^2 \plus{} b^2 \plus{} c^2)$
$ \iff \frac {c(a^2 \plus{} b^2)}{a \plus{} b} \plus{} \frac {b(c^2 \plus{} a^2)}{c \plus{} a} \plus{} \frac {a(b^2 \plus{} c^2)}{b \plus{} c}\leq a^2 \plus{} b^2 \plus{} c^2$
$ \iff (c^2 \minus{} \frac {c(a^2 \plus{} b^2)}{a \plus{} b}) \plus{} (b^2 \minus{} \frac {b(c^2 \plus{} a^2)}{c \plus{} a}) \plus{} (c^2 \minus{} \frac {a(b^2 \plus{} c^2)}{b \plus{} c}\geq 0$
$ \iff \frac {ac(c \minus{} a)}{a \plus{} b} \plus{} \frac {bc(c \minus{} b)}{a \plus{} b} \plus{} \frac {cb(b \minus{} c)}{c \plus{} a} \plus{} \frac {ab(b \minus{} a)}{c \plus{} a} \plus{} \frac {ca(a \minus{} c)}{b \plus{} c} \plus{} \frac {ba(a \minus{} b)}{b \plus{} c} \geq 0$
$ \iff \sum \frac {ac(a \minus{} c)^2}{(a \plus{} b)(c \plus{} b)} \geq 0$
the inequality is right . :)


MATHCOUNTS search function FTW AMC AIME USA(J)MO

How helpful is the All Time Greatest Mathcounts Problems book for someone who is trying to make MC nationals next year? Also, do you think that new book, Practice Competitions for Mathcounts by Josh Frost will be good?

Yes, you should get it. It's a wonderful book packed with problems and solutions very beneficial to MathCounts.

I actually didn't like it. As long as you can get your hands on old mc tests you should be fine.

Don't get the first book...as Ihatepie said, you can (with some effort) trade with others to get old tests (getting started is the hardest part, but usually there are people who are willing to help) to get many more problems than the book will give you. The book doesn't even have that many good problems from the old tests anyway, and the solutions are not that great. As for Mr. Frost's book, it should be very good.

There are a lot of ways to get problems without paying/trading. If you go to the search function and type in, for example, "state" and select the forum "FTW Problems Discussion", bam! you just got tons of state problems. If you want, specifically, national countdown problems, type in "national countdown". Years work too. Also, lots of people post old problems or really good original problems in marathons. Go to the search and type in "marathon". There's additionally tons and tons of mock contests. Type in some keywords like "mock contest", "mock mathcounts", etc. There are so many ways to get free resources on AoPS, and it's really easy too.

Yeah...for the win apparently has 7600 mathcounts problems...that's way more than enough.

Good game!  Countdown is very similar to Mathcounts count down, if you want countdown problems.

FTW! has [b]easy[/b] problems that were [b]designed to be solved quickly[/b].  If you really want to get better at MATHCOUNTS, and succeed at the state/national level, you need to be able to solve harder problems.  FTW! doesn't provide that.

Um no? FTW has lots of state and chapter sprints and targets, in addition to national problems.

They're all problems that are designed to be solved under 45 seconds.  By the way, not all state and chapter sprints and targets are problems that are going to make or break your day :wink:.  Neither are all national problems.

Well, yeah, but aren't all state/chapter problems? It's quite possible that all or most MATHCOUNTS problems of the state and chapter level are in FTW, actually. Not sure what you mean by:

[quote="AIME15"]By the way, not all state and chapter sprints and targets are problems that are going to make or break your day :wink:.  Neither are all national problems.[/quote]

Please explain ?

AIME 15 is just saying that you should get hard problems if you want to get better at MC. Problems on FTW are designed for cd, in which the problems aren't very hard. Also, they don't have national level problems. The state problems that they have aren't very hard, (not like #30 and #29 in the sprint round) so it won't help your written score very much. Tell me if this made sense.

Also, Frost's new book should be very good, but its kinda pricey. its like $ 25$ dollars for 4 practice test! You could barter 8 old tests for free! As for the other book, I don't have it , so I can't review it.  :wink:

EDIT: 35 dollars for four practice tests. Only 190 Problems! Pricey?

Personally, I bought it and was kind of disappointed. I thought it was pricey for such a small book! It was like 3/4 a centimeter thick!!!! I was extremely disappointed that I wasted like 15 bucks for that small book.  :(

But it does have some good problems in there.

Small books can contain a wealth of information.

However, I do agree that the All-Time Greatest MATHCOUNTS Problems is a little overpriced.  What do you expect?  They need somewhere besides sponsors to get money for...let's see! Funding 200+ flights to Orlando, 100+ rooms in a good hotel, renting out a venue for the National competition, 50+ state competitions.  Hosting the website and hiring staff also costs some money.

If you guys are complaining about overpriced things, what about the Competition Sets from MATHCOUNTS?  25 dollars for one year (chapter/state/national, with sprint/target/team/countdown/answers).  Kind of a ripoff, if you ask me.  A chapter competition will contain MAYBE 1 or 2 good problems, a state from 1 to 4, and a national competition under 10.  At least with the All-Time Greatest MATHCOUNTS Problems book, they make an effort to choose some of the more challenging problems.  Some of the Target and Team round problems in that book make you think.

To connaissance: It's four high-quality tests for 35 dollars, as opposed to two low-quality and one medium-quality test for 25 dollars.  Anyway, you have to appreciate the time and effort that goes into writing a good math question.  In this case, that means writing almost 200 problems, along with detailed solutions.

If you don't want to purchase these books for 17 and 35 dollars, respectively, you can always try the AMC site.  1950-1974 AHSME's costs 20 dollars, and contains 25 old AHSME's with solutions.  Same price for 1975-1999.  Also, AJHSME's from 1985-present costs 20 dollars, also with solutions.  To me, the AHSME's are a better value, but if you are specifically preparing for MATHCOUNTS, I suppose you'd want to try one of their books?

[quote="AIME15"]To connaissance: It's four high-quality tests for 35 dollars, as opposed to two low-quality and one medium-quality test for 25 dollars.  Anyway, you have to appreciate the time and effort that goes into writing a good math question.  In this case, that means writing almost 200 problems, along with detailed solutions.[/quote] No offense to Mr. Frost, but I would assume that a Mc Nationals test would be of the same, if not better quality as his tests.

However, if you've already taken all the previous National tests, any new test would be great practice.

Also, Mock Mathcounts tests written on AoPS are usually ...bad... Of course, there are some exceptions.

[quote="AIME15"]They're all problems that are designed to be solved under 45 seconds.[/quote]

This statement is false.

You will find that some of the problems are extremely easy while others are quite difficult.
I used it before chapter and state, and i made nats this year, so i guess its helpful.

[quote="ThinkFlow"]Speaking of mock MATHCOUNTS tests, can anyone point me to pythag's 2001i mock test? I tried searching and couldn't find it.[/quote]

I haven't posted it on aops and i plan to use the problems, so I will not be posting it on aops.

Additionally, its definitely much much harder than nats level.

LOL i'm guessing it has USAMO level problems.. :o

I'm guessing that many of them are considered "trivial".  :P

I know it may be a little late to post this, but I have a question about the new Practice Competitions for Mathcounts by John Frost.

So I'll be participating in Nats, and if I assume that I order it now and I receive it one week later, do you think I have enough time (2 weeks) to work on all (or the majority) of the problems?


[quote="dina013"]Let $a, b, c$ be positive real numbers and $abc=1.$ Prove that:

        $\sum \frac{1}{a^{2}+2b^{2}+3}\le \frac{1}{2}$.[/quote]

$\frac{1}{a^{2}+2b^{2}+3}\leq \frac{1}{2ab+2b+2}$.
I think it was posted here already.


MATHCOUNTS
AMC
AIME
geometry
number theory

how do i prepare?

1. any book recommendations?

2.any website recomendations?

3.i have previously prepared by using this site to ask questions and doing old tests...but its not helping  :(  :mad:

***please help!!***

how hard are the problems? Moems level? Mathcounts level? AIME level? IMO level? :D

i would get AoPS volumes 1 and 2 found on this site.
both books help with just about any math contest
this is probably the best site for helping, as you get to see solutions of other people that may contain helpful content.
doing old tests won't help that much if you get no gaining from them.
if you have a specific area of math you want to improve, you can buy one of the AoPS subject books i.e. intro to number theory, intro to geometry, etc. or maybe take an online class on AoPS that concentrates on one area, or do many problems you fund elsewhere in that subject.

generally prepare over a period of time....

if its like next weekend for example (And the difficulty is not within ur current limit) there's no point in going...

except im preparing for an MC tourny and im not shure whatim doing half the time  :roll:

well, the difficulty of the problems varies on the tournament (or school, if you will.) some schoolds have very hard question, some not so hard..like in one of my previous posts (inf. radical) you will see i ask this question $\sqrt{6+\sqrt{6+\sqrt{6+...}}}=x$ and this question was on a sixth grade test...

so im not really sure :(  :|

[hide]Call that x. x=sqrt(6+x). x^2=6+x. x^2-x-6=0 (x-3)(x+2)=0, and since x can't be negative, x=[b]3[/b][/hide]

i did not need the answer to the question...but thanks

try Elias Saab's math workouts. ill give you the URL if you need it..? or have you heard of them?


algebra
function
domain
integration
geometry
3D geometry
sphere

Because  $ D$ is symmetric to the plane  $ z\equal{}0$ we calculate the volume of $ D_{\plus{}}$, which is  the space of $ D$ above plane $ z\equal{}0$.

The common surface of $ D_{1}$ and $ D_{2}$ is the plane $ z\equal{}\frac{4}{\sqrt{5}}$, because $ \left\{{\begin{array}{c}
x^2\plus{}y^2\equal{}4\minus{}z^2\\\noalign{\vspace{0.1cm}}
\frac{z^2}{16}\equal{}1\minus{}\left({x^2\plus{}y^2}\right)
\end{array}}\right\}$, and
the common line of the sphere $ x^2\plus{}y^2\plus{}z^2\equal{}4$ and the ellipsoid   $ x^2\plus{}y^2\plus{}\frac{z^2}{16}\equal{}1$ is the circle  $ x^2\plus{}y^2\equal{}\frac{4}{5}$.

The volume of $ D_{\plus{}}$ between of plane $ z\equal{}\frac{4}{\sqrt{5}}$ and the part of the sphere which lies above it is
$ V({\check{D}_{1}})\equal{}\displaystyle\iiint_{\check{D}_{1}}{dz\,dy\,dx}\equal{}\int_{\minus{}\frac{2}{\sqrt{5}}}^{\frac{2}{\sqrt{5}}}\int_{\minus{}\sqrt{\frac{4}{5}\minus{}x^2}}^{\sqrt{\frac{4}{5}\minus{}x^2}}\int_{\frac{4}{\sqrt{5}}}^{\sqrt{4\minus{}x^2\minus{}y^2}}{dz\,dy\,dx}\equal{}$

$ \displaystyle\int_{\minus{}\frac{2}{\sqrt{5}}}^{\frac{2}{\sqrt{5}}}\int_{\minus{}\sqrt{\frac{4}{5}\minus{}x^2}}^{\sqrt{\frac{4}{5}\minus{}x^2}}{\sqrt{4\minus{}x^2\minus{}y^2}\minus{}\frac{4}{\sqrt{5}}\,dy\,dx}\equal{}$

$ \displaystyle\int_{0}^{2\pi}\int_{0}^{\frac{2}{\sqrt{5}}}{\rho\,\sqrt{4\minus{}\rho^2\,\cos^2{\phi}\minus{}\rho^2\,\sin^2{\phi}}\minus{}\frac{4\rho}{\sqrt{5}}\,d\rho\,d\phi}\equal{}$

$ \displaystyle\int_{0}^{2\pi}\int_{0}^{\frac{2}{\sqrt{5}}}{\rho\,\sqrt{4\minus{}\rho^2}\minus{}\frac{4\rho}{\sqrt{5}}\,d\rho\,d\phi}\equal{}\frac{16\pi}{3}\minus{}\frac{176\pi\,\sqrt{5}}{75}$.

The volume of $ D_{\plus{}}$ between of plane $ z\equal{}0$ and the part of the ellipsoid which lies above it is
$ V({\check{D}_{2}})\equal{}\displaystyle\iiint_{\check{D}_{2}}{dz\,dy\,dx}\equal{}\int_{\minus{}1}^{1}\int_{\minus{}\sqrt{1\minus{}x^2}}^{\sqrt{1\minus{}x^2}}\int_{0}^{4\sqrt{1\minus{}x^2\minus{}y^2}}{dz\,dy\,dx}\equal{}$

$ \displaystyle\int_{\minus{}1}^{1}\int_{\minus{}\sqrt{1\minus{}x^2}}^{\sqrt{1\minus{}x^2}}{4\,\sqrt{1\minus{}x^2\minus{}y^2}\,dy\,dx}\equal{}\int_{0}^{2\pi}\int_{0}^{1}{4\rho\,\sqrt{1\minus{}\rho^2}\,d\rho\,d\theta}\equal{}\frac{8\pi}{3}$.

The volume of $ D_{\plus{}}$ between of plane  $ z\equal{}\frac{4}{\sqrt{5}}$ and the part of the ellipsoid which lies above it is
$ V({\check{D}_{3}})\equal{}\displaystyle\iiint_{\check{D}_{3}}{dz\,dy\,dx}\equal{}\int_{\minus{}\frac{2}{\sqrt{5}}}^{\frac{2}{\sqrt{5}}}\int_{\minus{}\sqrt{\frac{4}{5}\minus{}x^2}}^{\sqrt{\frac{4}{5}\minus{}x^2}}\int_{\frac{4}{\sqrt{5}}}^{4\sqrt{1\minus{}x^2\minus{}y^2}}{dz\,dy\,dx}\equal{}$

$ \displaystyle\int_{\minus{}\frac{2}{\sqrt{5}}}^{\frac{2}{\sqrt{5}}}\int_{\minus{}\sqrt{\frac{4}{5}\minus{}x^2}}^{\sqrt{\frac{4}{5}\minus{}x^2}}{4\,\sqrt{1\minus{}x^2\minus{}y^2}\minus{}\tfrac{4}{\sqrt{5}}\,dy\,dx}\equal{}\int_{0}^{2\pi}\int_{0}^{\frac{2}{\sqrt{5}}}{4\rho\,\sqrt{1\minus{}\rho^2}\minus{}\tfrac{4\rho}{\sqrt{5}}\,d\rho\,d\theta}\equal{}$

$ \displaystyle\frac{8\pi}{3}\minus{}\frac{56\pi\,\sqrt{5}}{75}$.

So the volume of $ \check{D}_{4}$ which is the part of the ellipsoid between the planes $ z\equal{}0$ and  $ z\equal{}\frac{4}{\sqrt{5}}$ is
$ V({\check{D}_{4}})\equal{}\displaystyle\frac{8\pi}{3}\minus{}\left[{\frac{8\pi}{3}\minus{}\frac{56\pi\,\sqrt{5}}{75}}\right]\equal{}\frac{56\pi\,\sqrt{5}}{75}$.

So the volume of $ D$ is $ V({D})\equal{}\displaystyle2\,V({D_{\plus{}}})\equal{}2\left({\frac{8\pi}{3}\minus{}\frac{176\pi\,\sqrt{5}}{75}\plus{}\frac{56\pi\,\sqrt{5}}{75}}\right) \equal{}8\pi\left({\frac{2}{3}\minus{}\frac{\sqrt{5}}{5}}\right).\quad\square$

P.S.2.  After the  above calculations it isn't difficult to calculate the volume of $ P\equal{}\left({D_{1}\backslash{D}_{2}}\right)\cup\left({D_{2}\backslash{D}_{1}}\right)$.
P.S.1.  In the former post Kunny is using a formula which gives the volume of the space created by an axis rotation of a curve, but doesn't calculate it  correctly.


Galois Theory
superior algebra
superior algebra unsolved

I need to find the Galois group of (x-1)(x+1)(x-2)=0 over the rationals Q. And then what is the Galois group of x^3-1=0 over the rationals Q.
What is the process for doing so.  Please help
 :)


$ {O,H}$ is the circumcentre and the othercentre of triangle $ {ABC}$, let $ {BH \cap CA \equal{} D}$ $ {OD}$ ⊥ $ {DE}$ Then prove that :$ {\angle BAC \equal{} \angle EHD}$.

Denote the circumcircle by $ k$ and let $ BH\cap k \equal{}F$, $ FC \cap ED \equal{}G$.
We know that $ HD\equal{}FD$, from butterfly theorem we have $ DE\equal{}DG$ and obvously, the angels $ \angle{EDH}$ and $ \angle{GHF}$ are equal, thus $ \triangle{DEH}$ and $ \triangle{DGF}$ are congruent, which implies $ \angle{EHD}\equal{}\angle{GFD}\equal{}\angle{CFB}\equal{}\angle{CAB}$, which is what we wanted to proof.  :)

[quote]from butterfly theorem we have ...........[/quote]
Using butterfly theorem ! Really nice solution! Thanks a lot Martin N.! :)


The graph below shows a portion of the curve defined by the quartic polynomial $P(x)=x^4+ax^3+bx^2+cx+d$. Which of the following is the smallest?

$\text {(A) } P(-1)$
$\text {(B) The product of the zeros of } P$
$\text {(C) The product of the non-real zeros of } P$
$\text {(D) The sum of the coefficients of } P$
$\text {(E) The sum of the real zeros of } P$

[hide="Solution"]
Choice A is approximately equal to 4.
Choice B is equal to the value of $d$ in the original polynomial, which is $P(0) \approx 5.5$ or something....
Choice C is equal to the product of all roots divided by the real roots, which is $\approx \frac{5.5}{2 \cdot 4} = \frac{11}{16}$.
Choice D is $P(1) \approx 2$.
Choice E is $\approx 6$.

The smallest is clearly $C$.
[/hide]

[b]EDIT:[/b] Nevermind, I keep screwing up. Just look at chess64's solution.

[quote="1234567890"]Where is the graph??? :rotfl:[/quote]

It's been there the whole time... :huh:

I can't see the attachment either... wierd... they always work only sometimes for me.  What's up with that?

edit: and yet it shows up after I post.  Hm.

 I didn't think of letting x=1 to find the sum of the coefficients.  smart.

[quote="daveed"][hide] I didn't think of letting x=1 to find the sum of the coefficients.  smart.  [/hide][/quote]

Neither did I :lol: Credit goes to justdudxit, who helped me with that in my independent study office hour.

My mistake was that I didn't think about the fact that the product of two complex roots can be positive... I was just thinking multiplying two imaginary to get a negative. :roll: I had the right answer at first, but for the wrong reason, then I changed it because I thought it was something different... I need to think about these problems a little more carefully. :P


A number $ m \equal{} 111...11$ consists entirely of 1's in its decimal representation, and is divisible by 13.
Which number below need not divide $ m$?

A) 3
B) 7
C) 11
D) 17
E) 37

Note that $ 1001 \equal{} 7\cdot11\cdot13$.

Multiplying this by $ 111$ gives us $ 111111 \equal{} 3\cdot7\cdot11\cdot13\cdot37\implies\boxed D$.

[hide="alternate solution"]
$ m\equal{}111....111 \longrightarrow 9m\minus{}1 \equal{} 10^n \longrightarrow m \equal{} \frac{1}{9}10^n \minus{} \frac{1}{9}$
$ \frac{1}{9}10^n \minus{} \frac{1}{9} \equiv 0 \mod 13 \Rightarrow 10^n \equiv 1 \mod 13$
$ 10^n \equiv (\minus{}3)^n \equiv (\minus{}1)^{\frac{n}{3}} \mod 13 \longrightarrow$ let $ n\equal{}6k$
$ 10^{6k} \equiv 1 \mod 13$

1) $ m$ has $ 6k$ digits so $ 1\plus{}1\plus{}1...\plus{}1$ is divisible by $ 3$ and so is $ m$

2) By fermats little theorum - $ a^{p\minus{}1} \equiv 1 \mod p$
$ 7\minus{}1$ and $ 37\minus{}1 \equal{} 6k$
So $ m$ is divisible by $ 7$ and $ 37$

3) $ 10^{6k} \equiv (\minus{}1)^{6k} \equiv 1 \mod 11$

This leaves $ 17$ as the only option that does not divide $ m$
[/hide]


vector
superior algebra
superior algebra unsolved

[color=purple] Prove that , it is impossible to construct a vector space of $ Z$ over an arbitrary field $ K$ with the normal additive properties.
[/color]

If the characteristic of $ K$ is zero, then $ \mathbb{Q} \subseteq K$. If not, it is $ p$, but then $ p \equal{} p\cdot 1 \equal{} 0$.


It is a problem in this book: Problem Solving Strategies by Arthur Engel if you have it.

I do have the book. But can't find it. Could you please furnish the Chapter and Page number? I believe it's from the chapter Number Theory. Thanks!

[quote="A K Bhattacharyya"]Prove that no prime can be expressed as the sum of two squares in two different ways.[/quote]

Let $ p\equal{}a^2\plus{}b^2\equal{}c^2\plus{}d^2$ with $ a>c$

Let $ a\plus{}c\equal{}uv$ and $ b\plus{}d\equal{}uw$ with $ u\equal{}\gcd(a\plus{}c,b\plus{}d)$ and so $ \gcd(v,w)\equal{}1$
Let $ a\minus{}c\equal{}w't$ and $ d\minus{}b\equal{}v't$ with $ t\equal{}\gcd(a\minus{}c,d\minus{}b)$ and so $ \gcd(v',w')\equal{}1$

$ a^2\plus{}b^2\equal{}c^2\plus{}d^2$ $ \implies$ $ a^2\minus{}c^2\equal{}d^2\minus{}b^2$ $ \implies$ $ uvw't\equal{}uwv't$ $ \implies$ $ vw'\equal{}v'w$ $ \implies$ $ v\equal{}v'$ and $ w\equal{}w'$ (since $ \gcd(v,w)\equal{}1$ and $ \gcd(v',w')\equal{}1$)

So $ a\plus{}c\equal{}uv$ and $ b\plus{}d\equal{}uw$ and $ a\minus{}c\equal{}wt$ and $ d\minus{}b\equal{}vt$

And then $ a^2\plus{}b^2\equal{}c^2\plus{}d^2\equal{}\frac{(u^2\plus{}t^2)(v^2\plus{}w^2)}{4}$

The case $ u\equal{}t\equal{}1$ or $ v\equal{}w\equal{}1$ $ \implies$ the two decompositions are the same.
So $ u^2\plus{}t^2>4$ and $ v^2\plus{}w^2>4$ and $ \frac{(u^2\plus{}t^2)(v^2\plus{}w^2)}{4}$ cant be a prime.

Hence the result.

Sorry about not replying. It is problem 131 in Number Theory and the solution is on p.152.

Suppose $ p \equal{} x^2\plus{}y^2 \equal{} a^2\plus{}b^2$. Then $ p \equal{} (x\minus{}yi)(x\plus{}yi) \equal{} (a\minus{}bi)(a\plus{}bi)$. Now $ x\minus{}yi,x\plus{}yi,a\minus{}bi,a\plus{}bi$ must all be primes in $ \Bbb{Z}[i]$ because they have prime norms. Then $ x\minus{}yi$ divides either $ a\minus{}bi$ or $ a\plus{}bi$; since these also are primes they must all be associates; hence $ \{|x|,|y|\} \equal{} \{|a|, |b|\}$ and we are done.


quadratics
algebra
number theory unsolved
number theory

Find  $x;y;z$ satisfies $x^2+y^2+z^2=3xyz+1$ and $x;y;z$ are integer numbers

You can use [b]descent[/b] method to prove that the equation has no solution in [b]N^3[/b],I suggest
 So if I am not wrong the only cases are at least one of x,y,z be 0,that makes the puzzle easier to be solved.

[quote="nalpaction"]You can use [b]descent[/b] method to prove that the equation has no solution in [b]N^3[/b][/quote]
I don't know this one :( ;can you post your solution?

[quote="nalpaction"]Assume that $(x,y,z)$ is a root of the equation on$N^3$
 We choose $(x,y,z)$ s,t $x+y+z$ is minimum as possible,also assume that$x=>y=>z$
  Now consider the eqution as a quadratic equation of vary $x$:
          $x^2-x(3yz)+y^2+z^2-1=0$
 Denote by $x_1$ the rest root so: $x+x_1=3yz$
                                            and $xx_1=y^2+z^2-1$
 So x_1 is positive integer and hence $x_1=>x$ , also:
  $(x-1)(x_1-1)=y^2+z^2-3yz$ thus $(x-1)^2=<y^2+z^2-3yz$
  combine with $x=>y=>z$ we have $x=y$ and then have a contradiction from the equation. Then we have no solution on $N^3$
 Looking for the signs of $x,y,z$ in the equation lead us to the fact that one of $x,y,z$ must be $zero$.The rest isnot large problem.[/quote]
Sorry, I CANT see WHY $x_1\ge x$ with your arguments, can you explain better.

Thx and Best regards.

RX TCM

x_1 ,y ,z are also the solution of the original equation .But then x_1 + y +z >= x+y+z because x+y+z min .So x_1 >=x  :D


The distance from the earth to the sun is about 150,000,000 km.
That is about 391 times the distance from the earth to the moon.
How many kilometers are in the distance from the earth to the
moon? Round your answer to the nearest whole number.

is there a faster way to answer/think through this?

not really this is most straightforward mmehtod


induction
number theory proposed
number theory

We will prove by induction that for every $m\geq2$ and $n\in\mathbb{N}$ we can find integers $a_1,a_2,\ldots,a_m$ such that $|a_j|\geq n$ and \[ \sum_{j=1}^{m}\frac{1}{a_j}=\frac{1}{a_1a_2\cdots a_m}.  \] For $m=2$ we have $\frac{1}{n+1}+\frac{1}{-n}=\frac{1}{-n(n+1)}.$ Let the problem be true for $m\in\{2,3,\ldots,k\}$. If we put $a_{k+1}=1-a_1a_2\cdots a_k$, then \[ \sum_{j=1}^{k+1}\frac{1}{a_j}=\frac{1}{a_1a_2\cdots a_k}+\frac{1}{a_{k+1}}=\frac{a_{k+1}+a_1a_2\cdots a_k}{a_1a_2\cdots a_{k+1}}=\frac{1}{a_1a_2\cdots a_{k+1}}  \] by the definition of $a_{k+1}$. Obviosly, $|1-a_1a_2\cdots a_k|\geq n^k-1\geq n$.

I hope I haven't made a mistake somewhere :blush:

I think this problem is the worst of problems of the ARMO 2006

I think choose $c=-(2a-1) and d=-(2a+1)$

  Firstly,consider the structure of 1/a+1/b+1/c+1/d=1/abcd.Think the identity 1/a-1/(a+1)=1/a(a+1).Now we can build four
numbers such that a=-n,b=n+1,c=n(n+1)+1,d=n(n+1)*(n(n+1)+1)+1.
  Multiply the four numbers.
  It satisfies the formula.

Seems too similar to skyletter's idea but slightly more motivated.

Multiply all by $abcd$ to get $abc+bcd+cda+dab=1 \Leftrightarrow  ab(c+d)+cd(a+b)=1$.This inspires us to pick $a+b=1$,thus $b=1-a$ and $ab=a-a^2$,from which we rewrite our identity by trivial manipulation as $(c-x)(d-x)=x^2+1$,where $x=a^2-a$.Obviously $c=x+1,d=x^2+x+1$ satisfy this equation.Now it is enough to pick $a>10^7+1$;it follows $b<-10^7$,$c>x>a>10^7$,$d>c>10^7$,Q.E.D.This obviously can be generalized to any absolute value.

[quote=darij grinberg]Show that there exist four integers $a$, $b$, $c$, $d$ whose absolute values are all $>1000000$ and which satisfy $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{abcd}$.[/quote]

I think my solution hasn’t been posted here.
Consider,
                  $a=n$
                  $b=1-n$
                  $c=n^2-n+1$
                  $d=t^2-t+1$
where $n,t \in \mathbb{N}$ such that $t=n^2-n$.
It is easy to verify that this works.


Here is one old problem that just came up to me, a classical little short problem great professor Klamkin that appeared in the journal Mathematics Magazine:

Show that one cannot inscribe a regular polygon of more than four sides in an ellipse of unequal axes.

For beginners please. :)

We suppose that we can inscribe a regular polygon with more then four
sides into a ellipse.
Let consider the polygon A_1A_2....A_n with regular sides.
Because A_1A_2...A_n is a regular polygon the points A_1,..A_n are the
images of the roots of a binomial equation such as x^n=a where n is
natural and a is complex.
Because A_1,..A_n are the images of the roots of a binomial eqution the
points are on a circle.
But the polygon is inscribed also into a ellipse with unequal axes so
our suppose is false so the only regular polygons that can be inscribed
into a ellipse are the triangle and the square.

a little less elegant but a lot shorter is simply to calculate the derivates in the possible touching points and see it's not possible :D

Actually, I'm not sure that's shorter, and certainly not simpler. PM's solution is very good.

A circle and an ellipse (or, in general, two ellipses) can intersect at at most 4 points. Because all of the points of the polygon have to be on the ellipse, and all of the points of the polygon have to be on the circle which can be circumscribed around it, there are at most four places that the points can be [edit: the maximum of 4 intersections]. Thus, it can only have 4 sides.

.dzhonatan

[quote="swiftstream"]Actually, I'm not sure that's shorter, and certainly not simpler. PM's solution is very good.
[/quote]

I prefer your solution more cause it's more "geometric".


Show that the rectangular form of the polar equation $ r\equal{}\frac{ab}{(1\minus{}a \cos \theta)}$ for $ a<1$ is:

$ (1\minus{}a^2)x^2\plus{}y^2\minus{}2a^2bx\minus{}a^2b^2\equal{}0$


analytic geometry
trigonometry
induction

Thomas is walking on a coordinate plane. Starting at the origin, he walks to $ (6,0)$. He then turns left $ \alpha$ degrees and then walks a fraction $ x$ of the distance he just walked to get to $ (9,4)$. He turns left $ \alpha$ degrees again and walks $ x$ of the distance he walked to get to another point. This process of turning and walking a smaller amount continues on and on. Thomas eventually begins to converge on a point. Find this point.

[hide="Hint"]
Treat the x- and y-coordinates separately.[/hide]
[hide="Further Hint"]
Use the formulas: \[ \cos (\theta) = \frac{ e^{i \theta}+e^{-i\theta}}{2}\] \[ \sin( \theta) = \frac{ e^{i \theta}-e^{-i \theta}}{2i}\] ;) [/hide]

I managed to find recurrence relations which give the next point if we know the two points before it. I won't give the working however as it took me about four sides of A4 to do! Let $ (a_{n}, b_{n})$ be the $ n$th point.

$ a_{n}= \frac{1}{2}(3a_{n-1}-a_{n-2})-\frac{2}{3}(b_{n-1}-b_{n-2})$

$ b_{n}= \frac{2}{3}(a_{n-1}-a_{n-2})+\frac{1}{2}(3b_{n-1}-b_{n-2})$

I tried to say as $ n \to \infty$ then $ a_{n}= a_{n-1}= a_{n-2}$ and $ b_{n}= b_{n-1}= b_{n-2}$ but I realised that this works for any $ a_{n}$ or $ b_{n}$ that we try. I also tried to use iteration to find a pattern that I could prove by induction but I got up to $ a_{6}= \frac{847}{432}$ and $ b_{6}= \frac{1705}{324}$ and gave up.


SOrry if this question has been asked before.

But is there any place I can download the old papers from 2006 onwards?

Thanks!

http://www.agmath.com/23806.html

This is good, and contains some MC tests also.


Find all triangles for which \[\tan(A-B) + \tan(B-C) + \tan(C-A) = 0.\]

Since $(A-B) + (B-C) + (C-A) = 0$, we have the following identity:

$\tan(A-B)+\tan(B-C)+\tan(C-A)=\tan(A-B)\tan(B-C)\tan(C-A)$

We now have $\tan(A-B)\tan(B-C)\tan(C-A) = 0$. Thus at least one of $\tan(A-B)$, $\tan(B-C)$, or $\tan(C-A)$ is equal to 0. WLOG, let it be $\tan(A-B)$. If $\tan(A-B)=0$, then we have $A-B=0$, or $A=B$. Thus the triangle is isoceles.

[quote="pkothari13"][quote="JSRosen3"]Since $(A-B) + (B-C) + (C-A) = 0$, we have the following identity:

$\tan(A-B)+\tan(B-C)+\tan(C-A)=\tan(A-B)\tan(B-C)\tan(C-A)$
[/quote]

Is that a given identity or did you derive that from something else?[/quote]

He uses the fact that if x, y, z are three angles such that x + y + z is a multiple of 180, then tan x + tan y + tan z = tan x tan y tan z. In particular, if x + y + z = 0, then tan x + tan y + tan z = tan x tan y tan z.

This fact can be proven very easily: Since x + y + z is a multiple of 180, say $x+y+z=k\cdot 180^{\circ }$ with integer k, we have $z=k\cdot 180^{\circ }-\left( x+y\right) $ and thus

$\tan z=\tan \left( k\cdot 180^{\circ }-\left( x+y\right) \right) =\tan \left( -\left( x+y\right) \right) $

(this is because the tangens function is periodical with period 180). Hence,

$\tan z=\tan \left( -\left( x+y\right) \right) =-\tan \left( x+y\right) =-\frac{\tan x+\tan y}{1-\tan x\tan y}$,

where I have used the tangens addition formula $\tan \left( x+y\right) =\frac{\tan x+\tan y}{1-\tan x\tan y}$. Now, we have

$\tan z=-\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{\tan x+\tan y}{\tan x\tan y-1}$,

and thus

$\tan z\left( \tan x\tan y-1\right) =\tan x+\tan y$,

i. e.

$\tan x\tan y\tan z-\tan z=\tan x+\tan y$

what immediately yields

$\tan x\tan y\tan z=\tan x+\tan y+\tan z$.

And this completes the proof.

  Darij


Solve
$ 1\minus{}\frac{12} {y\plus{}3x}\equal{}\frac{2}{\sqrt{x}}$
$ 1\plus{}\frac{12} {y\plus{}3x}\equal{}\frac{6}{\sqrt{y}}$

hello, adding both equations we get
$ 1\equal{}\frac{1}{\sqrt{x}}\plus{}\frac{3}{\sqrt{y}}$
solving this equation for $ y$ we have
$ y\equal{}\frac{9x}{(\sqrt{x}\minus{}1)^2}$
setting this in your first equation and factoring we get
$ 4x^{3/2}\plus{}x^{5/2}\minus{}4x^2\minus{}4\sqrt{x}\equal{}0$
with the positive solution $ x\equal{}(1\plus{}\sqrt{3})^2$ and with this we get $ y\equal{}3(1\plus{}\sqrt{3})^2$
Sonnhard.


search
inequalities unsolved
inequalities

Let $ a_1 , a_2 ... a_n$ and $ b_1 , b_2 ... b_n$ be positive reals satisfing these conditions:

*$ a_1 \ge a_2 \ge ... a_n$
*$ b_1 b_2 ... b_k \ge a_1 a_2 a_3 ... a_k$ For all $ 1 \le k \le n$

Prove:

$ b_1 \plus{} b_2 \plus{} ... b_n \ge a_1 \plus{} a_2 \plus{} ... a_n$


Millikan determined the charge on the electron by studying the static charges on oil drops falling in an electric field. A student carried out this experiment using several oil drops for her measurements and calculated the charges on the drops. She obtained the following data.

Droplet A had a charge of 1.60 x 10^-19 C
Droplet B had a charge of 3.15 x 10^-19 C
Droplet C had a charge of 4.81 x 10^-19 C
Droplet D had a charge of 6.31 x 10^-19 C

What is the significance of the fact that the droplets carried different charges? What conclusion can the student draw from these data regarding the charge of the electron. What value should she report for the electronic charge?

This question involves some physics concepts too. If anyone wants, I can post the question in the physics forum as well.

Needless to say, this should be in the Physics Forum.

One thing we can observe is that all measured charges are approximtely integer multiples of 1.6x10^{-19} C, which suggests that this is the charge of an electron, and all charges that exist are multiples of this one.


[quote="sri340"]Example:  The "sum of digits" of  1656  is  1+6+5+6 = 18 .HOW MANY whole numbers from 1 to 2000 have their "sum of digits" equal to 25?

A.10	 B.12	 C.13	   D.   15     E16[/quote]
[hide]Since the sum we're looking for is 25, the numbers must be at least 3 digits. Also the greatest possible sum of a 3 digit number is 27, so the only possible options for 3 digit numbers are 9-8-8 or 9-9-7. For each case, there are 3 options so we have 6 numbers. In 4-digit numbers we already have a 1 in the thousands digit and working similarly like the last one, we have 3 digits that must sum to 24. The possibilities are 8-8-8, 9-9-6, and 9-8-7. For the 8-8-8 there's only one number. For 9-9-6, there are 3 numbers, and for 9-8-7 there are 6 numbers.
6+3+1+6=16[/hide]


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

I is a two sided ideal in a  ring R AndM is left R-module.M/IM becomes a module over the quotient ring R/I.

How can I show that if M/IM is a projective R/I-module then M is a projective R-module.

for $ R/I$-modules $ N$, we have a natural bijection $ Hom_{R/I}(M/IM,N)\equal{}Hom_R(M,N_R)$, where $ N_R$ denotes $ N$ regarded as a $ R$-module. from this the claim follows.

will you please expalin in some more detail. thanks

how does the claim follow from this?

this is not enough because  there are other conditions, like finitely generation, nilpotent ideal
and flatness.

How can I use them to show this?? Somebody please help me.

no, there are no other assumptions needed.

$ M$ is a projective $ R$-module iff $ Hom_R(M,\minus{})$ carries epimorphisms to epimorphisms. this is just the definition, do you use another one? from this definition and my hint the proof works without any difficulties.

peteryellow:

Stop tripleposting. I already merged one of your posts into another one yesterday. Especially, be more patient!

why is $ Hom_{R/I}(M/IM,N)\equal{}Hom_R(M,N_R)$? supposing $ R\equal{}R_1\oplus R_2$, $ I\equal{}0\oplus R_2$, $ M\equal{}M_1\oplus M_2$ where $ R_1M_2\equal{}R_2M_1\equal{}{0}$ then $ M/IM\cong M_2$ and similar condition hold for $ N\equal{}N_1\oplus N_2$ than $ Hom_{R/I}(M/IM,N)\cong Hom_{R_2}(M_2,N_2)$ but give no information about homogeneos between M1 and N1.

well, maybe I've made some stupid mistake.
But if $ N$ is an R/I module, Since we have to prove for every epimorphism $ p: A\rightarrow A'$, $ Hom(M,A) \rightarrow Hom(M,A')$ induced by $ p$ is epimorphism, where A and A' are R-module but not necessarily R/I module, then how does assertion $ Hom_{R/I}(M/IM,N) \equal{} Hom_R(M,N)$ make sense

I think peteryellow's condition is not enough to conclude that M is projective R-module. For example in #7, $ IM$ is independent with $ R/I\minus{}M$.

read the definitions.

$ N$ is a $ R/I$-module, which is used for using the definition of injectivity over $ R/I$.

[quote="-oo-"]no, there are no other assumptions needed.

$ M$ is a projective $ R$-module iff $ Hom_R(M, \minus{} )$ carries epimorphisms to epimorphisms. this is just the definition, do you use another one? from this definition and my hint the proof works without any difficulties.[/quote]

Actually, other assumptions are needed indeed! For example let R be any commutative ring which is not semisimple. So there exists an R-module M which is not projective.

Now let m be any maximal ideal of R. Then M/mM is obviously projective as R/m module.


[quote="tranthanhnam"]Cho minh hoi ai la dan Back khoa Thanh pho HCM  , dong thoi la thanh vien diendan ML  nay thi llen day de : toan dan biet mat ti nhe . hohooho .
   Lo dau se co hop mat neu dong doa , huong ung tich cuc nhe[/quote]

Tai sao ko? Minh la dan back khoa, khoa hoa thuc pham ne. Chi co dieu la o BK duoc vai thang thoi. Chua biet duoc nhieu ve truong . :D

chào anh  :) Em biết anh đã lớn tuổi nhưng yêu cầu anh lần sau ko nên post các bài toàn trong VMEO, MFC, MAY lên đây nếu nó chưa hết hạn gửi bài  :wink:

[quote=zaizai_hoang]chao anh  Em biết anh đã lớn tuổi nhưng yêu cầu anh lần sau ko nên post các bài toàn trong VMEO, MFC, MAY lên đây nếu nó chưa hết hạn gửi bài hà[/quote]
hந động như vậy   la khong hay

Đúng rồi đó,theo em các anh không nên làm như vậy

that  co loi voi dien dan,  ah`  ma  muc nay toi post len de lam quen voi dan bk thanh pho ma` . Sao lai vo dong gop y kien lung tung the


ratio
limit
geometric series
calculus
calculus computations

hey, i need help on the main idea on how to solve these

-find whether the sequence is convergent, if convergent find limit, or divergent.

1.)r=(11/22)
lim              nr^n=?
n-> infinity

2.)An=50/(5^n)
{An} is convergent so find its limit.

for 2 i found it to 50/infinity so would the limit be 0?

any help would be great, thanks.

1) Use the ratio test (or the root test, if you're skilled)

2) It's a geometric series!  You really need to know how to find the sum of those.

so for 2.) i dont think the limit of that sequence is the same as the sum of its series which is 25/2, im having a problem finding the sequences limit since it is converget.

Oops, I thought you were posting about series.

Okay, for the first one, you want to find $\lim_{n\to\infty} nr^{n}$ with $|r|<1$.  You could rewrite it as $\lim_{n\to\infty} \frac{r^n}{\frac 1n}$ and apply L'hopital's rule, for example...

For the second one, it defines a convergent series, so the limit of the sequence must be zero.


The polynomial equation $ x^4\minus{}6x^2\plus{}56x\minus{}24\equal{}0$ has two real roots and two complex roots.  What is the sum of the two real roots?

Our choices are $ 2\sqrt{6}$, $ \minus{}4$, $ 0$, $ 4$, $ \minus{}2\sqrt{6}$.

We should assume that the problem works out nicely since the answers are nice. We could a) construct the polynomial with roots that are a+b where a,b are roots of the original equation, but that is a lot of work. So I would suggest taking $ (x^2\minus{}4x\plus{}a)(x^2\plus{}4x\plus{}b)$ and $ (x^2\plus{}2\sqrt{6}x\plus{}a)(x^2\minus{}2\sqrt{6}x\plus{}b)$ and matching coefficients to get a factorization.

[quote="mathgeniuse^ln(x)"][quote="Altheman"]We should assume that the problem works out nicely since the answers are nice. We could a) construct the polynomial with roots that are a+b where a,b are roots of the original equation, but that is a lot of work. So I would suggest taking $ (x^2 \minus{} 4x \plus{} a)(x^2 \plus{} 4x \plus{} b)$ and $ (x^2 \plus{} 2\sqrt {6}x \plus{} a)(x^2 \minus{} 2\sqrt {6}x \plus{} b)$ and matching coefficients to get a factorization.[/quote]

The answer is -4, by the way.[/quote]

The matching with $ (x^2 \minus{} 4x \plus{} a)(x^2 \plus{} 4x \plus{} b)$ works very nicely.

Further, we could simply use the "solve" function on our TI-84+silver or higher (the competition this is from allows it) and find the two real roots, and just add them, yielding the answer in seconds.  But of course, we don't like that do we....


[quote="mathVNpro"]Let $ ABC$ be a triangle inscirbed in circumcircle $ (O)$. Let $ d$ be an arbitry line such that it intersects $ AB,AC$ respectively at $ D,E$, and intersects $ (O)$ at $ P,Q$. Let $ BE\cap (O)\equiv M$, $ CD\cap (O)\equiv N$. Let $ I\equiv PM\cap QN$, $ K\equiv MN\cap PQ$. Prove that: $ IA \bot OK$.[/quote]

[img]http://img35.imageshack.us/img35/3336/giaibaicucvadoicuc.png[/img]

$ \triangle AMN$ and $ \triangle IQP$ are in perspective axially.

$ \equal{} > \ IA\ ;\ MQ\ ;\ NP$ are concurent .

$ K$ is pole of $ IA \ \equal{} > \ IA \bot OK$

Another solution:
Let $ AI\cap (O)\equal{}{A,F}$
Apply Pascal Theorem for $ AQMAPN$ and $ FQMFPN$, respectively, we have tangents to $ (O)$ at $ A,F$ intersect at $ K$.
$ \Rightarrow AK\equal{}FK,AO\equal{}FO \Rightarrow AF \perp OK$


I am interested in becoming a quant. Right now, I am on the path to a degree in mathematics. However for my last two years I have the option of taking CS classes. I am not positively certain I can get a CS major/minor within that time frame so my question is do employers care if your diploma doesn't actually say CS minor but you took enough relevant CS classes or you taught yourself the relevant CS material?

Actually my friend says you don't want to go overboard with CS because a lot of firms will try to place you in tech instead of quant or prop trading or whatever if they think you are really a CS person. (This isn't necessarily bad if that's what you want to do, and those positions have among the best starting salaries, but there's much less opportunity for advancement.)

Unless you actually want to be programming in your job, you only need to be vaguely competent with CS. For example, the place that hired me doesn't really care if I can program. They just want me to be proficient with computers and to be able to understand sophisticated algorithms. (It's a firm with quantitative trading strategies, not some Bridgewater type place.)

But if you do want to learn programming, (1) learn C++; (2) it really doesn't matter exactly what qualifications your university gives you. Just put it in you transcript somewhere and the hiring people will see it and ask you about it in the interview.

It really doesn't matter what kind of CS courses have you taken ( or you plan to take ), the good thing is that you're interested in CS, and I'm sure it will give you an advantage, looking for a job in the finance industry, but I agree with Xevarion that, it may push you in the wrong direction as far as job position.

I talked to a guy from RIFF and he told me that they usually look for people that are good in handling large amount of data, since most of the data they have isn't filtered, so learning to handle databases and learning some kind of database language ( maybe SQL ) would be a big plus. As he said that's the reason why hedge funds hire a lot physicists.

I was thinking of a CS minor as a plus to my math background and also a backup in case the quant field is so elitist I can't get my foot in the door (sure sounds insanely tough from what I read on here  :| ). My school has a major in Computational mathematics, which sounds like what Mr. Doe is talking about, but my interests lie more in 'pure' mathematics and I feel I will do better in classes if I enjoy and am interested in what I am doing.

So I'm inferring that they just look at whether you've taken programming classes in C++, databases, etc and not "majored in CS" or "minored in CS".

I have another question too: my school offers a masters in Financial Mathematics, which can be obtained in a year. Will it help entering the quant field for me to get the masters or is a math degree enough? It's hard for me to say what skillset is being sought after apart from strong quantitative/logical reasoning abilities.

1) You don't even have to take classes as long as you know the stuff. Basically, it doesn't matter how you get skills as long as you have them. But if you teach yourself programming you should at least do some kind of significant project so that you have something to show for your efforts.

2) It's hard to say if that would help. Probably depends on the specific job. You probably don't have to decide about the masters thing until after you've found out whether you'll have a job when you graduate. Also at the beginning of senior year if you think you're not going to be able to work immediately after graduation you could look at applying to PhD programs.

As Xeverion says, you are a some time from being a quant, and you can get C++ from other places.
>50% of the quant we see are either wholly self taught, or their course was a so dismal that they might as well have picked up Duffy or Stroustrup.

If your CS course teaches C++, then it might have some value, database and web stuff is much less useful.
Handling large data set in quant finance is often not really an SQL based activity. MS has some interesting tech in this, but it's not available to the general public yet.
Don't waste your time learning Java if you want to be a quant.

The value of masters in finance varies a lot, as does their price, and none are cheap. Even though I don't know which school you are at, I would strongly advise looking around. As a quant headhunter I hear disturbing things about the MS in QF at Fordham.

Let's say I am a student of a school called Stanford University and am considering the Financial Mathematics program there after I graduate. Of course, such a university is also renowned for its computer science program. (Does that help, Dominic Connor?)

I actually learned Java and C++ in high school (my high school had a CS major) but I forgot those after not using them for 2-3 years. I'm sure I can easily refresh my memory over the summer.

Stanford CS  is not bad.
I'm impressed if you did learn C++ at high school, however you ought to get up to STL, smart pointers and polymorphism.
Stanford FinMath is on our good list, so certainly you should consider it.
One question is whether you should do MS or PhD.

This discussion is weird because what language you learn is really fairly unimportant. Once you know one language it's easy to learn another following the same paradigm. "Don't bother to learn Java" while learning C++? They are almost the same thing. Learn C++ and you know Java. You should expose yourself to several different paradigms; C++, Haskell and Perl/python seems to be pretty good 3 languages to learn/take courses in. Of course there is much more to a CS degree, but I think that you should focus on C++ is a misplaced view.

[i]	This discussion is weird because what language you learn is really fairly unimportant.[/i]
The question is about quant work, and in that very precise context it is extremely important. People who know C++ have more chance of getting a given quant job than those who have done quiche languages.
That's a fact, I am a headhunter for quants (google on my name...)

There is now some demand for C#, but that's rarely taught in universities, and  Java barely shows up as background noise in this game, so much so that if a job mentions Java, I start to wonder if it's really quant, or actually housekeeping IT.

For other careers specific language matters a lot less of course.

It is true that learning C++ makes Java easy (if occasionally quite irritating), and I agree that anyone who considers themselves a serious programmer should be thinking in ways not limited to the syntax of one language, but instead draw from several. Haskell  and a scripting language  make sense, though I pick up a current moving towards F#, which is like OCaml, but done (surprisingly well) by Microsoft. A bit of Lisp/Prolog is also good for expanding your set of mental tools to address a problem.

But once you have decided to become some sort of quant, go for C++. It's a big thing, and sometimes quite tricky.
Be aware that "C++ helps you learn Java" is not symmetrical, Java and C++ have a common ancestor, but then again we share ancestors with dolphins (apologies to any creationists here)
That means lots of things are quite similar but the set of things you can do in C++ is much larger, but so is the number of concepts you need to understand.
A C++ developer can learn Java much faster than a Java programmer can learn C++.
Even for the roles we handle where other languages are used, they still prefer to have C++ experience because managers often feel it proves you are smarter.

[/i]

Ok, sorry for giving out misleading advice. I am of the belief that for most programming jobs, it is much easier to program than to do mathematics. Of course writing good software is really difficult (emacs and awk are the only good programs and all that  :D), but do you really learn that at university anyway?

It's flatly wrong to say that maths is harder than programming, on many levels.

First, programming is at its core maths, so thus you are in danger of saying that maths is easier than maths.

Next, no competitive activity with a large population is "easier" than any other.
Is writing a loop easy ? I've seen very smart people get this wrong, hell I've done this myself.

Is addition easy ? You try working in any organisation which handles serious money with maths and you will see scary errors on that front as well.

Is it easier to solve PDEs than work out why once a week, a program of 100,000 lines of code produces a slightly wrong number ?

I've seen very smart people who simply could not code, and conversely expert programmers who get all vague at undergrad calculus.

But even if I'm wrong in this, I am confident that it's a bad attitude. You need to be better than other people at the things that are important to the job. If you find programming easy, then I'm happy for you, but I am going on at length here because some people poison their chances through thinking they are "too good to program".

I agree with what you're saying but not your examples.

Your examples were basically,

Is it easier to notice small mistakes than to carry out prescribed methods, neither of which are the essence of the subjects.

I think you are wrong. In order to actually reach a contemporary level of an area in mathematics, you first need a long education. Computer science is a much much younger subject, and you can quickly reach a level where you can actually do something.


limit
Functional Analysis
advanced fields
advanced fields unsolved

Let $ x_1,x_2,\ldots$ be elements of the Banach space $ X$.
$ \displaystyle \sum_{n\equal{}1}^{\infty}x_n$ is [i]absolutely convergent[/i] if $ \displaystyle \sum_{n\equal{}1}^{\infty}\parallel{}x_n\parallel{}<\infty$.

Show that if $ \displaystyle \sum_{n\equal{}1}^{\infty}x_n$ is absolutely convergent then $ \displaystyle \sum_{n\equal{}1}^{\infty}x_n$
is convergent and $ \left\parallel{}\displaystyle \sum_{n\equal{}1}^{\infty}x_n \right\parallel{} \leq \displaystyle \sum_{n\equal{}1}^{\infty}\parallel{}x_n\parallel{}$.

Cauchy convergence (note, that for $ m>n: \,|\sum_{i\equal{}1}^mx_i\minus{}\sum_{i\equal{}1}^nx_i|\leq \sum_{i\equal{}n\plus{}1}^m|x_i|$) plus Banach spaces are complete

A more challenging (but still not hard) problem related to this would be to show that a normed space $ X$ is Banach [u]if and only if[/u] whenever $ (x_n)_n\subset X$ so that $ \sum_{n\ge1}\parallel{}x_n\parallel{}<\infty$ then $ \sum_{n\ge 1}x_n$ is convergent in $ X$.

Still not difficult - and a standard idea. (The details involve your favorite known convergent series. I'll use $ \sum 2^{\minus{}k}.$)

Suppose $ (x_n)$ is a Cauchy sequence. From the definition of Cauchy, we can find a subsequence $ x_{n_k}$ such that $ \|x_{n_{k\plus{}1}}\minus{}x_{n_{k}}\|<2^{\minus{}k}.$

Then $ x_{n_k}\equal{}x_{n_1}\plus{}\sum_{j\equal{}1}^{k\minus{}1}(x_{n_{j\plus{}1}}\minus{}x_{n_j})$

But the series $ \sum_{j\equal{}1}^{\infty}\|x_{n_{j\plus{}1}}\minus{}x_{n_j}\|<
\sum_{j\equal{}1}^{\infty}2^{\minus{}j}$ converges.

Hence by our hypothesis,

$ \lim_{k\to\infty}x_{n_k}\equal{}x_{n_1}\plus{}
\sum_{j\equal{}1}^{\infty}(x_{n_{j\plus{}1}}\minus{}x_{n_j})$ converges.

And if some subsequence of a Cauchy sequence converges, then so does the whole sequence, and to the same limit.