The radius of a spherical balloon is increasing by $ 2$ cm/sec. At what rate is air being blown into the balloon at the moment when the radius is $ 10$ cm?

The rate at which air is being blown in is the same as the rate at which the volume is changing.  So we have
$ V\equal{}\frac{4}{3}\pi r^{3}\Rightarrow\frac{dV}{dr}\equal{}4\pi r^{2}\frac{dr}{dt}$.  Now you have everything you need.

[quote="JRav"]
$ \frac {dV}{dr}$.  [/quote]
 U mean $ \frac{dV}{dt }$

[hide]$ \frac{dV}{dt}\equal{}\frac{dV}{dr}*\frac{dr}{dt}$

$ \frac{dV}{dt}\equal{}4\pi r^2*\frac{dr}{dt}$

$ \frac{dV}{dt}\equal{}4\pi 10^2*2\equal{}\boxed{2513.3\frac{cm^3}{sec}}$[/hide]

Correct.  I would just leave your answer as $ 800\pi$.

I suggest you read your textbook and go through the examples more carefully-or at least seek tutoring or extra help since the concepts you ask about are are rather important-but also rather simple.


An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?

the answer is 121

anyone want to explain this thoroughly, moreso than the solution found in the resources section does?

[hide="Here is the short answer"]
The final point can be anywhere in the 11 x 11 square centered about the origin.[/hide]

[hide]We can consider the extreme points, where the object moves farther away from the origin with every step. In the first quadrant these points are (10,0), (9,1), (8,2), ..., (0,10). This is the same for the other quadrants. Thus we have a square with vertices (10,0), (0,10), (-10,0), and (0,-10) with side length 10+1=11.
All points within this square are also final points, because the object can always "move back". Thus, the number of possible final points is equal to the number of square units in the area of the square because of the one-to-one correspondence. So our answer is 11*11=121.[/hide]

how does a square with those vertices have a side length of 11, wouldn't it be 10sqrt(2)

The taxicab length of the square is $ 11$, because there are $ 11$ points along one side of the "square".

You are thinking of Euclidean geometry, which is a perfectly acceptable thought process, but note that along any line of the form $ x \plus{} y \equal{} c$ for $ \minus{} 10 \le c \le 10$ for even values of $ c$, there are $ 11$ points along any given line and there are $ 11$ valid lines ($ 11$ even numbers between $ \minus{} 10$ and $ 10$ inclusive), so $ 11 \times 11 \equal{} \boxed{121}$.

another way of thinking about it...

the sum of the absolute values of x and y is going to be an even number (0,2,4,6,8, or 10).

If you plot each of these and count the number of lattice points that lie on the graphs, you get 4*(10+8+6+4+2)=8(5+4+3+2+1)=120 plus the orgin so 121


trigonometry
geometry
trig identities
Law of Sines
Law of Cosines

[url]http://www.nspsmo.org/documents/2003TRIG-03.pdf[/url]

can someone give me a hint? heres what I have so far[hide]
I can find CE and CG (pt G is where the line CF hits the bottom line when you extend it) using law of cosines.
I can figure out what BC is using the law of sines if I know what <BCF is.  $ \frac {sin < BCF}{BF} \equal{} \frac {sin < BFC}{BC}$  I'm don't know how to find <BCF.   [/hide]

thanks![/hide]

I think the problem is unsolvable. you can only get the first part using law of cosines. By the way, what you stated is the law of sines.

This is solvable, albeit extremely annoying. I'm going to skip over the actual computations, and just go over the proof.

[hide]$ FA$ is computable by the law of cosines. $ \angle ABF$, $ \angle BFA$, and $ \angle BAF$ are also computable by the law of sines. You can also find $ \angle FCB$ and and $ \angle FBC$ by the law of sines.

Consider $ \triangle CEF$. By the laws of cosines and sines, you can find $ CE$, $ \angle CEF$ and $ \angle ECF$.

Since $ BC$ and $ AD$ are parallel, $ \angle EDA \equal{} \angle DCB$. Extend $ AB$ to a point $ G$. $ \angle CBG$ is known, since $ \angle CFB$ and $ \angle FBA$ are known. Since $ BC$ and $ AD$ are parallel, $ \angle DAB$ is known. Since $ \angle FAB$ is known and $ \angle DAB$ is known, $ \angle DAF$ is known. Also, $ \angle DFA$ is known since $ \angle BFA$ is known and $ \angle DFB$ is known. Thus, since $ \angle DFA$ and $ \angle DAF$ are known, $ \angle FDA$ is known. Since $ FA$ is known, we may apply the law of sines to find $ DF$. We may apply the law of cosines on $ \triangle FCD$ to find the length of $ CD$.

I apologize if that was a bit unclear. It helps if you label each angle/length you've found. [/hide]

Yup sorry about that. I did not see that AB was fixed. Yes its solvable.


ok i'm doing Mathcounts and i kind of understand what it's about. Can someone give me an over view on what you do?  :wink:

I am taking geometry, and too bad never did math counts because wasnt on math team. I am working my head off this year though in geometry.

Let's see math counts is a district state and national event.
How far you get into the tournament depends how good you are, or how smart you are. Only 4 make it to nationals in each state, if i am correct.

you mena 4 groups.. and i don't think it determinds how smart you aer... becuase you could be the smartest kid in your school... but still fail a test>>>>> :(

by how smart you are i meant compared to other mathcounts contestants in math in general.

we have math team yet we still don't do mathcounts. We just have these math meets every once in a while but I know its not that important because once I proved that the answer to the question was wrong.

 :D

how could we do mathcounts?

...The answer key to one of the questions on the '05 national team round was wrong. Mistakes happen.

The poor grammar and spelling makes everything very hard to read...try to make your posts more coherent.

Ihatepie,

my son's school also had a math team last year, but not Mathcounts. So I registered my son as an individual with me as his coach. (Went to the principal first and asked for permission promising to pay all the fees and do all the work too). My son  did get to State level and did there pretty well too (better then many teams in our chapter anyway).

This year my son organized a team (found a couple of boys interested in math). School does not do anything, but does not object either.
So all the students paid their $20 registration fee, I registered a team with me as a coach, we meet at least once a week at our house and practice...

Registration deadline is sometime in December, so you still have time.

PM me if you have questions and good luck!

I'm going to chapter...last week of Feb.  :D

[quote="spiderboy95"]Really? I'm going on the 14th to chapter.[/quote]

Same.  Which chapter?  I might see you there. :)

[quote="Ihatepie"]How'd you guys do?[/quote]

Second. :(

Lost to Josh Cho in both written and cd.

Aww man, Stanley Cai lost first round.  He's beast. :(

Were any of them there last year? I might've seen them.

I'm not in your chapter, but the people that made state cd last year all got pwned.

[quote="Ihatepie"]O. Now that I think of it, I think I faced Stanley in State Cd last year. Not sure about Josh Cho.

(was 1st cd and 2nd written)[/quote]

Josh Cho is from Indian Mountain.  I don't think he was participating last year.  Stanley, I believe, is home-schooled.

Here were rankings from NW Chapter:

1 Josh Cho
2 me
3 Menthing Wang (sorry if I spelled your name wrong)
4 Elizabeth Porter? <- something like that
5 Stanley Cai

Last Year, it was:

1 Jenny Park (she made Nats as a 7th grader, now a freshman)
2 Menthing Wang
3 Stanley Cai
4 some kid i beat in CD round
11 (:() me

And woah, wasn't it Top 25 make state last year?  This year, it's only Top 8...

Oh yeah, Jenny Park. That was hilarious. The guy who got 3rd guessed on the target question which put him from 5 to 3.  :lol: Jenny would've been 4 but missed nats by a single point.

ERNIE: yeah, i thought more people made it last year! WOW you are in my chapter. I got eliminated FIRST (a bit embarrassing.)

IMS was good this year and last year, Josh Cho was there last year but didn't win.

Wait, spiderboy95, who are [b]you[/b]?  Did you make state?  Gah, I got owned by Josh Cho this year. :(

[quote="spiderboy95"]That was a very intense match between you and Stanley at States!!!!!!

I was like, "who's going to win?" I really didn't know!

Good job to both of you! Good luck at nats, Stanley!!!!![/quote]

Haha yeah, it was 2-0, and if you saw me, I was praying for an easy question which I could skim. :)

[quote="spiderboy95"]Which you got.....[/quote]

Yup!  Three of 'em...


inequalities
number theory proposed
number theory

Let $ (a,b,c)$ be distinct positive integer number
$ ab\plus{}ac\plus{}bc\geq 3k^{2}\minus{}1$
Let
A=$ \frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{3}\minus{}abc$
Find least value of A
I'm sorry I post in Own problems.

If $ a>b>c$, then min A when $ a\equal{}k\plus{}1,b\equal{}k,c\equal{}k\minus{}1$ and equal 3k.


[quote="wonderful"][url=http://www.imagehosting.com][img]http://www.imagehosting.com/out.php/i1811033_TTinequality1.gif[/img][/url]

Have A Great Day![/quote]

By Holder inequality:

$ LHS^2 ( \sum ab(ab\plus{}c^2)) \ge (ab \plus{} bc \plus{} ca)^3$

So $ LHS^2 \ge \frac{(ab \plus{}bc \plus{}ca)^3}{\sum ab(ab\plus{}c^2)} \equal{} \frac{(ab\plus{}bc\plus{}ca)^2}{(ab\plus{}bc\plus{}ca)^2}\equal{}ab\plus{}bc\plus{}ca$

We are done!


Determinr whether there is subset $X$  os integers eith th following  property: for any $n$ is integer  there is exactly one solution of $a+2b=n$ with $a,b\in\X$.

it's quite well-known
Assume that we have a set S with k elements and m be the smallest integer can not be represent as a+2b with a.b in S.Assume that |a|<M for all a in S
Then add 2 news element 5M+m;-10M-m we have new set with m is attainable and  it does not coincide any existing solutions, since |2b + a|, |2a + ai|, |2b + ai|, |a + 2ai|, |ab + 2ai|  are all ≥ 3N


function
trigonometry
algebra solved
algebra

Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ satisfying $f(x+y) + f(x-y) =2 \cdot f(x) \cdot cos(y)$.

make the following substitutions for (x,y): $(0, t)$, $(\frac{\pi}{2}+t,\frac{\pi}{2})$, $(\frac{\pi}{2},\frac{\pi}{2} + t)$

to get: $f(t)+f(-t)=2f(0)\cos t$,  $f(\pi+t)+f(t)=0$,  $f(\pi+t)+f(-t)=-2f(\frac{\pi}{2})\sin (t)$

setting $f(0)=a$ and $f(\frac{\pi}{2})=b$

we get $f(t)=a\cos t + b\sin t$


[quote="Fierytycoon"]Whoa, did a forum section just get pruned or something, because I noticed that my post count just dropped by about a hundred posts. I know there is auto-pruning for the Fun & Games section, but still...[/quote]

What's the worry about post count, anyway? Why not establish a reputation for quality rather than quantity?

[quote]What's the worry about post count, anyway? Why not establish a reputation for quality rather than quantity?[/quote]

Don't I have a right to be curious about it?

[quote="MithsApprentice"]how did it just drop by 100?...ur still at 1842...i dont recall u ever being any higher...

but of course, i may be wrong. I didn't notice my post count drop...[/quote]

It was in the 1900's before (a couple days ago, at least).

Hmm...something's fishy is going on here...and with that jobro_gold oddity...

It definitely dropped. I was over 2500, and now I'm not. I don't mind, but I am also curious.

Me, too -- my posts per day dropped significantly.  I imagine someone went and deleted a bunch of stuff manually.

Somebody may have deleted a few threads.  I am investigating a couple of thread deletions now, but I don't yet know what happened or how many other threads might have been deleted.  If anyone figures this out please let me know.

[quote="probability1.01"]Do really old threads get automatically deleted, or are they stored somewhere?[/quote]

F&G is the only board that experiences auto-pruning. I believe they still "exist" but are not shown on the board.

[quote="rrusczyk"]I believe one of our moderators manually cleaned out very old spam-laden threads from the MATHCOUNTS forum.  If moderators want to do this in the future without affecting individual post counts, they can simply move the threads to the F&G forum and let the auto-pruner do the rest (unless the thread is a poll, in which case manual deleting is the way to go, and post counts will be affected no matter what).[/quote]

Hmm...except I never post in the MATHCOUNTS forum. Oh well, no big deal.

There were probably a bunch of old polls clipped from the Fun & Games thread, too.

The last time I remember cleaning out some threads in the Mathcounts forum was late January.  It may have been Mystic, though.

More than 100 of my posts have disappeared.  That seems like a lot to have been trimmed in MC and F&G.


How many quarts of cream containing 36% butterfat must be added to 100 quarts of milk containing 3% butterfat to produce 4% milk?

I got 2400 qt. of cream is this correct?

I got 2328.   :maybe:  Here's what I did.  Out of that 100 quarts of milk only 97 quarts is actually milk.  We want the milk to only be 4% of the total so we have $ 97 \equal{} \frac {1}{25}x \implies x \equal{} 2425$.  This means we need to add $ 2425 \minus{} 97 \equal{} 2328$ quarts of cream.  Note that I ignored the amt of butter fat in the cream because as long as it's not milk it doesn't matter.  I think this is right,  I'll just wait to see what other people think.

Edit:  Oh I think I know what you did.  You probably used 100 quarts as the amount of milk right?

I saw this problem on CC  :P   The wording is confusing... but I'm assuming that 4% milk is actual milk that contains 4% butterfat... but who knows!  I can't believe this was actually a question on a real SAT.

Well it's really annoying that they include the information about the butterfat if the whole thing counts as milk.  If it's an SAT question then hopefully it doesn't include both 2328 or 2400 as an answer.

Also um what's CC?

Yeah I don't see how they got away with this!  CC stands for College Confidential - [url]http://talk.collegeconfidential.com/[/url]

I thought perhaps the % butterfat was irrelevant. Even though the milk contains 36% butterfat it is still 100% milk, likewise, with the cream. So I thought if we have a 100 qt. of milk then that is a 100% milk. To get 4% milk how many qt. of cream would we have to add to the 100 qt. of milk?
So I set up the equation:
100=.04(100+x)
x=2400

So we have 2400 qt. of cream.
100 (qt. of milk)/ 2500 (qt. cream + milk)= 4% milk

I am not sure if this is correct though...

Yeah I think you are right now.  That makes more sense.  Also if it's an SAT question it probably shouldn't be very complicated. :P

lol I just googled this question to see if there was any "official" answer.  But all I found was something on yahoo answers and all the people there ended up with really low things like 4 quarts which makes no sense at all.

Yeah I saw people's responses on yahoo answers too. The only way it could be the 3.125 qt answer is if we were trying to produce 4% butterfat right?

Yeah adding 3.125 quarts of cream gives us exactly 4% butterfat.


Let $a_1;a_2;...;a_n \in R$. Prove that:
$\displaystyle \sum_{i;j=1}^n \frac {a_ia_j}{1+|i-j|} \geq 0$

Is it true that $\det A>0$, where $\displaystyle A=(a_{ij}),\ a_{ij}=\frac 1{1+|i-j|}$? I think this would solve the problem.


limit
logarithms
Euler
real analysis
real analysis unsolved

Compute
\[ \lim_{n\rightarrow \infty} \left(\sum_{k \equal{} 1}^n\frac {1}{k} \plus{} \ln\left(\frac {e}{n}\right) \minus{} c\right)^n\]
where $ c$ is the euler constant defined by $ c \equal{} \lim_{n\rightarrow\infty} \sum_{k \equal{} 1}^n\frac {1}{k} \minus{} \ln(n)$.

Denote $ H_n: \equal{} \sum_{k \equal{} 1}^{n}\frac {1}{k}$, the n-th harmonic number of order 1. (You have a mistake - you sum the term 1/n instead of 1/k ...)

Now we have
\[ \left (H_n \plus{} \ln\frac {n}{e} \minus{} c\right )^n \equal{} \left (1 \plus{} (H_n \minus{} \ln n \minus{} c)\right )^n.\]
It is very good known that
\[ \lim_{n\to\infty}(H_n \minus{} \ln n) \equal{} c,\]
where $ c \equal{} 0.577...$ is the Euler-Mascheroni constant. This implies that $ \{H_n \minus{} \ln n \minus{} c\}_{n \equal{} 1}^{\infty}$ is a null sequence. Moreover, the temrs of this sequence are positive. Hence,
\[ \lim_{n\to\infty}\left (1 \plus{} (H_n \minus{} \ln n \minus{} c)\right )^n \equal{} \exp\left (\lim_{n\to\infty}n\cdot\ln\left (1 \plus{} (H_n \minus{} \ln n \minus{} c)\right )\right ) \equal{} \exp\left (\lim_{n\to\infty}n\cdot\left (H_n \minus{} \ln n \minus{} c\right )\right ).\]
It can be deduced the asymptotic behaviour of n-th harmonic number in the form
\[ H_n \equal{} \ln n \plus{} c \plus{} \frac {1}{2n} \plus{} O\left (\frac {1}{n^2}\right ),\qquad {\text{as}}\quad n\to\infty.\]
For the proof see [url]http://www.myoops.org/twocw/nctu/upload/fourier/supplement/euler%20summation.pdf[/url].

This implies that your limit equals $ \exp\left (\frac {1}{2}\right )$.


algebra
polynomial
search
superior algebra
superior algebra unsolved

Let $ K/F$ be a field extension and let $ \alpha \in K$. Also, assume that the degree of the minimal polynomial of $ \alpha$ over $ F$ is odd. Prove that $ F(\alpha) \equal{} F(\alpha ^2)$.

Use that $ [M: K] \equal{} [M: L][L: K]$ for fields $ K \subset L \subset M$. Or use the search function.


quadratics
topology
Functional Analysis
advanced fields
advanced fields theorems

Given a Banach space $X$ over real scalars, let $X^{*}$ be the space of all continuous linear functionals from $X$ to $\mathbb R$. The space $X^{*}$ is normed by $\|f\|: =\sup\{f(x)\colon x\in X,\|x\|\le 1\}$. Each element $x\in X$ acts as a continuous linear functional on $X^{*}$ by evaluation: in a formula, $e_{x}\colon X^{*}\to\mathbb R$ is defined by $e_{x}(f)=f(x)$. The space $X$ is [b]reflexive[/b] iff there are no other continuous linear functionals on $X^{*}$. To put it another way, $X$ is reflexive iff the canonical isometric embedding $i\colon X\to X^{**}$, defined by $i(x)=e_{x}$, is onto.

  This definition is sometimes misinterpreted to mean that $X$ is reflexive iff $X^{**}$ is isomorphic to $X$. This is wrong: there is a nonreflexive space (constructed by R.C. James) that is even isometric to its second dual. A simplified form of James' space (which is isomorphic, but not isometric to its second dual) is very easy to describe: it is the space of all real sequences with bounded quadratic variation (with the norm being the square root of quadratic variation). The quadratic variation of a sequence $(x_{i})$ is $\sup\left(|x_{i_{0}}|^{2}+\sum_{k}|x_{i_{k}}-x_{i_{k-1}}|^{2}\right)$, where the supremum is taken over all finite increasing integer sequences $i_{k}$.

[b]Jacobson[/b] asked in another thread:
[quote]It is well known that finite-dimensional vectorspaces are reflexive. However infinite-dimensional ones seem to behave bad. What is 'exactly' the reason they behave like this?[/quote]
The best answer I could think of was "because they lack weak compactness", which is more like a non-answer. Does anyone have a better, more intuitive, explanation of nonreflexivity?

Nobody elebaroting it seems? I was interested about the

"because they lack weak compactness"

sentence, since I don't know what you mean with that :)

The weak topology on a Banach space $X$ is defined by declaring all half-spaces $\{x\in X\colon f(x)>a\}$ to be open, for any $a\in\mathbb R$ and any continuous linear functional $f\colon X\to\mathbb R$. (Naturally, the unions and finite intersections of half-spaces are open, too). A Banach space is reflexive if and only if its closed unit ball is compact in the weak topology. One part of this statement (reflexive->weakly compact) is a special case of the [url=http://en.wikipedia.org/wiki/Banach-Alaoglu_theorem]Banach-Alaoglu theorem[/url].

There is also a characterization of reflexivity (due to R.C. James) that is very easy to state and quite hard to prove: $X$ is reflexive iff every continuous linear functional on $X$ attains its maximum on the closed unit ball.

[quote]The best answer I could think of was "because they lack weak compactness", which is more like a non-answer. Does anyone have a better, more intuitive, explanation of nonreflexivity?[/quote]

Any recent progress made by someone?

I found: http://www.ams.org/proc/1998-126-08/S0002-9939-98-04691-7/S0002-9939-98-04691-7.pdf   as suggested by mlok, but it's too advanced for me. An intuitive answer will do fine :)


How do you add matrices in LaTeX? Because I have
[code]\[ \left( \begin{array}{cc}
a & b\\
c & d  \end{array} \right)\]
+
\[ \left( \begin{array}{cc}
e & f\\
g & h  \end{array} \right)\]
=
\[ \left( \begin{array}{cc}
a+e & b+f\\
c+g & d+h  \end{array} \right)\]
[/code]

giving

\[ \left( \begin{array}{cc} a & b \\
c & d \end{array} \right)\]
+
\[ \left( \begin{array}{cc} e & f \\
g & h \end{array} \right)\]
=
\[ \left( \begin{array}{cc} a \plus{} e & b \plus{} f \\
c \plus{} g & d \plus{} h \end{array} \right)\]

, which is of course bad.

The symbols \[ and \] begin and end lines in a certain way; they have nothing at all do do with matrices. The matrices begin and end with the \begin{array} and \end{array} commands (which you should change - see below).

Try taking all of your \[ and \] commands out of your code, except for the one at the very beginning and the one at the very end. Or take those out as well and replace them by $ and $.

And then: don't use \left(\begin{array}{cc} a&b\\c&d \end{array}\right). Rather, write it this way:

\begin{pmatrix} a&b\\c&d \end{pmatrix}

Note that you don't even have to specify the number of columns up front; the code takes care of that. Use {pmatrix} for parentheses and {bmatrix} for square brackets.

And vmatrix gives determinants:
$ \begin{vmatrix} 1 & 2 \\ 2 & 1\end{vmatrix}$


Are there any good problems that range in difficultly from the last half of the AIME to easier olympiad problems?

Zeitz's Art and Craft of Problem Solving fills that niche pretty nicely.


I cant understand what does this questions mean:
1) b^2-100c^2
2)32h^2-18
3) 9/4g^2-h^2

Find the value of the qusetions using alegbra way to solve
1) 897^2-103^2
2) 854^2-754^2

Expand $(A+B)(A-B),$ risinglegend. :)

kunny

kunny, i know dats the solution but i'm really mixed up with how to solve...mayb some solutions? plz this is important..if i flunk my maths test tomorrow, i'm gonna flunk myself of my house too[die]lol juz try to help me k..thanxx

kunny, thanxx u r a great help! i kinda understand the concept nw

Don't mention it, risinglegend. :)

Good luck!

kunny


calculus
derivative
function
calculus computations

Calculus does not belong in this forum. Try "Calculus Computations and Tutorials".

Thanks.

[quote="kAwz"]Hi, please help me with this question, thanks in advance:

Ship A is sailing south at 24 km/h while ship B, which is 48 kilometres due south of A, is sailing east at 18km/h.

i) At what rate are they approaching or seperating one hour later?

ii) When do they cease to approach one another and how far apart are they at this time?

iii) What is the significance of the distance in part b)?

Many Thanks Mr. Green[/quote]

The formula for the distance between them is $6\sqrt{25x^{2}-64x+64}$. The derivative of this is $\frac{150x-192}{\sqrt{25x^{2}-64x+64}}$.

i) The derivative at x = 1 is $-\frac{42}{5}$, meaning they are getting closer at 8.4 km/h in an hour.

ii) They start getting further away when the distance function reaches its minimum, or when the derivative of it reaches 0. It is zero when the numerator is zero, meaning they start getting further apart in $\frac{32}{25}$ hours.

iii) Their distance at the time is the closest they'll ever be to each other.

kAwz:

1. This sort of question belongs in "Calculus Computations and Tutorials."

2. Be patient in waiting for answers. Not everyone here has the time or inclination to deal with what appear to by typical student concerns - that's tutoring, not problem solving. You probably will get an answer eventually, but this is not primarily a tutoring website.

3. DO NOT POST THE SAME PROBLEM MORE THAN ONCE! At this point I think you have multiple copies of some problems all here (after they've been moved here by various moderators.) If I can figure out which ones are the duplicates, I'll delete them - I don't have time to do that right now.

4. Which brings up another point - can you please give you problems more distinctive titles? For instance "Ship distance" for this, "Lamppost and shadow" for another. That way we can tell which is which.


geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
geometry unsolved

Let $ ABC$ be a triangle with the circumcentre $ O$. Let A',B',C' be the reflections of A,B,C respectively in the opposite side $ BC,CA,AB$.
a) Prove that 6 circles $ AOA',BOB',COC',AB'C',BC'A',CA'B'$ concur at one point $ P$.
b) Let $ A_{1}\equal{}OA'\cap B'C',B_{1}\equal{}OB'\cap C'A',C_{1}\equal{}OC'\cap A'B'$. Solve that $ AA_{1},BB_{1},CC_{1}$ concur
at the isogonal conjugate of $ P$.

[quote="nktp"]Let $ ABC$ be a triangle with the circumcentre $ O$. Let A',B',C' be the reflections of A,B,C respectively in the opposite side $ BC,CA,AB$.
a) Prove that 6 circles $ AOA',BOB',COC',AB'C',BC'A',CA'B'$ concur at one point $ P$.
b) Let $ A_{1}\equal{} OA'\cap B'C',B_{1}\equal{} OB'\cap C'A',C_{1}\equal{} OC'\cap A'B'$. Solve that $ AA_{1},BB_{1},CC_{1}$ concur
at the isogonal conjugate of $ P$.[/quote]

[b]b)[/b] is an open problem (proved by computer algebra, but not synthetically) in my [url=http://forumgeom.fau.edu/FG2003volume3/FG200311index.html]Forum Geometricorum paper[/url], found by Bernard Gibert. Good luck at proving it...

  Darij

This problem I have solved part a) by classical geometry but part b) I haven't solved yet!

My solution ( the main idea is using the inversion and the symmetric )
Problem a) is Kosnita Theorem through an inversion has $ O$ to pole ( my proof for Kosnita theorem also using the inversion and the symmetric, I posted it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120538]here[/url] )
Problem b)
Let $ f$ is an inversion has $ A$ to pole and $ AB.AC$ to constant. Call the inversive image of $ X$ as $ f(X)$. $ At$ is the bisector of $ \angle BAC$. Now, we have:
$ f(B),f(C)$ are the symmetric points of $ C,B$ wrt $ At$
$ f(O)$ is the symmetric point of $ A$ wrt $ f(B)f(C)$, it means that $ f(O)$ is the symmetric point of $ A'$ wrt $ At$
$ f(A')$ is the center of the circle which passes through $ A,f(B),f(C)$, it means that $ f(A')$ is the symmetric point of $ O$ wrt $ At$
$ f(B')$ is the point lies on $ AB'$ and $ Af(B') \equal{} AC$, it means that $ f(B')$ is the symmetric point of $ C'$ wrt $ At$. Similarly, $ f(C')$ is the symmetric point of $ B'$ wrt $ At$
So:
$ A,f(O),f(A')$ are the symmetric points of $ A,A',O$ wrt $ At$
$ A,f(B'),f(C')$ are the symmetric points of $ A,C',B'$ wrt $ At$
But:
$ f(A_{1})$ is the intersection $ (\neq A)$of the circle passes through $ A,f(O),f(A')$ and the circle passes through $ A,f(B'),f(C')$
$ P$ is the intersection $ (\neq A)$ of the circle passes through $ A,A',O$ and $ A,C',B'$
Then, we have $ f(A_{1})$ is the symmetric point of $ P$ wrt $ At$. It means that $ AA_{1}$ is the isogonal line of $ AP$ wrt $ \angle BAC$, which ends the proof.

Good .But there aren't any one check his solution.
I think this solution very good.
He is my friend ànd he talk me that he use new method so can you check solution.
I think he solved the open question.

Another proof for question b, from my friend (nguyenhuukiensp)
Denote:
$ A_{2},B_{2},C_{2}$ are the symmetric points of $ O$ wrt $ A,B,C$
$ A_{3},B_{3},C_{3}$ are the centers of the circles $ (AOA'),(BOB'),(COC')$
$ A_{4},B_{4},C_{4}$ are the symmetric points of $ O$ wrt $ A_{3},B_{3},C_{3}$
$ M\in AA_{4}\cap B'C'$
$ H$ is the orthocenter of triangle $ ABC$
Easy to have that $ H$ is the nine-point center of triangle $ A_{2}B_{2}C_{2}$
We have the following well-known results:
$ A_{4}, B_{4},C_{4},P$ are collinear
$ H$ is the symmetric point of $ A'$ wrt $ B_{2}C_{2}$. (Similarly with $ B',C'$)
$ O_{2}$ is the isogonal conjugate of $ H$ wrt $ A_{2}B_{2}C_{2}$, then $ O,O_{2},P$ are collinear
Moreover $ O_{2}$ is the circumcenter of triangle $ A'B'C'$
$ OA_{4}\parallel B'C'$, $ A_{4}M$ is the perpendicular bisector of $ A_{2}O$
And, note that $ A_{2}B' \equal{} A_{2}C'$
Now, in order to prove that $ AA_{1}$ is the isogonal line of $ AP$ wrt $ \angle BAC$, we will prove that$ \angle AA_{1}A' \equal{}\angle AOP$
We have $ \angle AOP \equal{}\angle AA_{4}P$
Cause $ OA_{4}\parallel B'C'$, so $ A'AA_{1}M$ is cyclic, then $ \angle AA_{1}A' \equal{} AMA'$
So we need to prove that the line $ \overline{A_{4}B_{4}C_{4}}\parallel A'M$
Note that $ \overline{A_{4}B_{4}C_{4}}\perp OP$, or $ \overline{A_{4}B_{4}C_{4}}\perp OO_{2}$
We will prove that $ OO_{2}\perp A'M$, which is equivalent to $ MO_{2}^{2}\minus{}A'O_{2}^{2}\equal{} MO^{2}\minus{}A'O^{2}$
But we have $ MO_{2}^{2}\equal{} MB'.MC'\plus{}O_{2}B'^{2}\equal{} MA_{2}^{2}\minus{}A_{2}B'^{2}\plus{}O_{2}A'^{2}\equal{} MO^{2}\plus{}O_{2}A'^{2}\minus{}A_{2}B'^{2}$
And $ A_{2}B' \equal{} A_{2}H \equal{} OA'$
So ends the proof  :wink:


[hide]This means that x-4 and x+4 are both positive both negative, or x-4=0.  If both x-4 and x+4 are positive, then x>4.  If they are both negative, x<-4.  If x-4=0, x=4.  Thus:
[b]x>=4 or x<-4[/b][/hide]


simplify

sinx+cosx

as a single cosine

in radians
or
degrees

$sinx+cosx=\sqrt{2}(\frac{1}{\sqrt{2}}sinx+\frac{1}{\sqrt{2}}cosx)$

By composite angle formula: $sinx+cosx= \sqrt{2}cos(x-\frac{\pi}{4})$


As the title says :) I noticed a topic for the 2004 attendance list, so I thought I could carry the tradition on

If you're going to PROMYS this year, please post in here!

I can't wait to check out Boston :D

So, how old is everybody who's going?
I'm 15 (rising junior)

Hi,

I'm 14 and a rising sophmore.  I don't really know what to expect and am a little bit nervous...   :?

everyone is so young...

i'm 17 and i'm a rising senior  :blush:


Stanford
college
calculus
SuMAC
topology
group theory

SMT contest problems are usually original. But sometimes they do silly things. (See calculus 2002 #10 for an example. That was 7 free points for knowing the answer before entering the test room for me.)

As long as this thread says "Stanford" I might as well post under this...

Did anyone else here apply to SUMaC 2004?
Or if you've attended SUMaC in past years, how was it (in all aspects)?
I've already seen ComplexZeta's post about SUMaC in the other thread. How many kids do you think apply to this camp?

jimmy

I don't know how many apply, but in 2001, there were 35 students at the camp. 25 were in the group theory program, and 10 were in the algebraic topology program.


geometry
MATHCOUNTS
trigonometry
rhombus
ratio
angle bisector

[quote="pgpatel"]Does anybody know the area of a pentagon with all of its angles equaling 108 degrees in terms of the side length, x?[/quote]
pgpatel posed the above question in the MATHCOUNTS forum a few days ago.  Unfortunately all answers used trigonometry.  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=240837#p240837]The original post[/url] now lives in the High School Forums.

Now, let's try again.  ABSOLUTELY NO TRIGONOMETRY WILL BE PERMITTED in this thread!  Any response that even hints at trig will be deleted.

Starter question.  For those who say this is beyond MATHCOUNTS, consider the following question, taken from the 1997-98 School Handbook Workout #1, Problem #3:

[quote="1997-98 MATHCOUNTS School Handbook, Workout #1"]3.  Sir Roger Penrose created the [i]Penrose tiles[/i] in 1974.  These tiles consist of a [i]kite[/i] (EFGH) and a [i]dart[/i] (ABCD), which, if placed together, form a rhombus.  How many centimeters are in the length of $\overline{\hbox{AC}}$?  Express your answer as a decimal to the nearest thousandth.  (See diagram, below.)[/quote]
Once you know the length of AC, perhaps you could determine the area of triangle ACD.  Knowing the area of triangle ACD might help you determine the area of a regular pentagon.

Run with it.  ABSOLUTELY NO TRIG IS ALLOWED!

[quote="236factorial"]Hang on a second, what [i]is[/i] the length of AC?
What is the length of the diagonal of a rhombus with a side [i]s[/i]?[/quote]
Workout problems are typically multi-step problems.  So, expect the solution to require a bit more than a simple formula.
[hide="hints"]Chase all of the angles in the [i]kites[/i] and [i]darts[/i].  Look at the triangles they have already given you.  Then carefully consider where you might draw a line segment to create smaller similar triangles.[/hide]

You can't find the diagonals of a rhombus when only given its side length... you can squash it or stretch it or whatever and still maintain the same side lengths, so you would require the angle too.

Well, don't you have a golden triangle, so the length would be the golden ratio, or (sqrt5-1)/2 which would be about 0.618?

[quote="sirorange"]Well, don't you have a golden triangle?[/quote]
Yes.  But can you show why?

Beginning with the fact that EFH is isosceles, you can chase all angles in the [i]kite[/i] and [i]dart[/i], as shown in the illustration.  Then, since the two base angles of triangle ACD are equal, you can conclude that ACD is isosceles.  Then, set the length of its legs equal to [i]x[/i].

Solve for [i]x[/i].  Show your work.

Triangle ACD is similar to Triangle ADH, so $\frac{1}{x}=\frac{x+1}{1}$. X can't be $\frac{-1-\sqrt{5}}{2}$, because that is negative. So it must be $\frac{\sqrt{5}-1}{2}$.

[quote="worthawholebean"]Triangle ACD is similar to Triangle ADH, so 1/x=(x+1)/1. X can't be (1+$\sqrt{5}$)/2, because that violates the triangle ineguality postulate in triangle ADH. So it must be ($\sqrt{5}$-1)/2.[/quote]
Nice job.  Let me expand upon the algebra:

\begin{eqnarray*}
\frac{1}{x}&=&\frac{x+1}{1}  \\
1&=&x(x+1) \\
0&=&x^2+x-1 \\
x&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\
x&=&\frac{-1\pm\sqrt{1+4}}{2} \\
x&=&\frac{\sqrt{5}-1}{2}\approx 0.618 \\
\end{eqnarray*}

Now, is there enough information to derive the area of a regular unit pentagon?

This is how I did it. $\triangle ABC$ is an isoscoles triangle with a vertex angle of 108 and $\triangle BCE$ is an isoscoles triangle with a vertex angle of 36. ED is an angle bisector of $\angle BCE$ producing isoscoles triangles $\triangle BED$ and $\triangle EDC$. Because the sidelength of the regular pentagon is $x$, then $BE = ED = DC = x$.

If the length of BC is $a$ then EC is also $a$, and by the angle bisector theorem, the ratio of BD to DC is $x$ to $a$. We set up proportions

$\frac xa = \frac {a-x}{x}$

$x^2 = a^2 - ax$

$a^2 - xa -x^2 = 0$

Solving for $a$ in terms of $x$

$a = \frac {x \pm \sqrt {x^2 + 4x^2}}{2}$

$a = \frac {x(1 + \sqrt 5)}{2}$

Solving for the height of $\triangle ABC$ we have

$\sqrt {x^2 - \left(\frac {x(1 + \sqrt 5)}{4}\right)^2}$

$\sqrt {\frac {16x^2 - 6x^2 - 2x^2\sqrt 5}{16}}$

$\sqrt {\frac {x^2(10 - \sqrt 5)}{16}}$

$\frac {x\sqrt {10 - 2 \sqrt 5}}{4}$

Solving for the height of $\triangle BEC$ we have

$\sqrt {\left(\frac {x(1 + \sqrt 5)}{2}\right)^2 - \frac {x^2}{4}}$

$\frac {x\sqrt {5 + 2 \sqrt 5}}{2}$

The area of the entire pentagon is then

$x \cdot \frac {x\sqrt {5 + 2 \sqrt 5}}{2} \cdot \frac 12 + \left(\frac {x(1 + \sqrt 5)}{2}\right)\left(\frac {x \sqrt {10 - 2 \sqrt 5}}{4}\right)$

$\frac {x^2(2 \sqrt {5 + 2 \sqrt 5} + \sqrt {10 - 2 \sqrt 5} + \sqrt {50 - 10 \sqrt 5})}{8}$

Squaring and square rooting the expression above reveals the formula

$\frac {x^2 \sqrt {25 + 10 \sqrt 5}}{4}$

If we construct the other diagonal of the rhombus, we see that the rhombus is two 54-54-72 triangles. A pentagon is made up of five of these. So the area of a pentagon is $\frac{5}{2}$ times the area of the rhombus, given the just-drawn-in diagonal a value of $s$.

similar triangles, how can I forget??? I just had a test on that, and I got 104! Looks that didn't pay off here...

tarquin:  You have a couple of typographical errors, but the final answer is correct.  Nice job!

its weird that there are so many attachments here...ive never seen any other attachments at this board and now i see a bunch...weird


modular arithmetic
number theory proposed
number theory

Prove that $\forall a \geq 4$ there exist infinitely many free-square numbers like $n$ that $n | a^n-1$

Great problem, I saw it last year given in a bulgarian olympiad. I will not spoil the topic and not post a solution, but I could not say that it is a very nice problem.

Yes,This problem is very beautiful.I solved it in a beautiful way in the exam.

I'm not sure this is Ok.

Assume we have found $p_1,\ldots,p_k$ s.t., if we denote $S_t=p_1\cdot\ldots\cdot p_t$, then $S_{k-1}|a^{S_{k-1}}-1,S_k|a^{S_k}-1$. We have $(\frac{a^{S_k}-1}{a^{S_{k-1}}-1},a^{S_{k-1}}-1)=(a^{S_{k-1}}-1,p_k)=p_k$.
This means that we can find $p_{k+1}\not|S_k$ s.t. $p_{k+1}|a^{S_k}-1|a^{S_{k+1}}-1$, so $S_{k+1}|a^{S_{k+1}}-1$, unless $\frac{a^{S_k}-1}{a^{S_{k-1}}-1}$ is a power of $p^k$.

Let $b=a^{S_{k-1}}$. Assume $b^{p-1}+\ldots+b+1=p^t$ for some prime $p$ (we can take $p=p_k$, to treat the case above). It's easy to see that this implies $p|b-1$, so let $b=kp+1$. We find $p^t=M_{p^2}+(p-1)kp+p$, so we get $p|k-1$. This means that $b=up^2+p+1$. Now, it's easy to see that unless $p=2$, this is impossible (because we have $b^{p-1}+\ldots+b+1\equiv p\pmod{p^2}$).

In our situation, we can only run into problems when $p_1=2$, and we can't find any $p_2\ne 2,\ p_2|a^2-1$. This means that $a^2-1$ is a power of $2$, and we easily take care of this case.

We can easily create $n_1$, ie the smallest prime divisor of $a-1$.
We attempt to create the recurrence $n_{k+1} = p_k \cdot n_k$, $\gcd(p_k,n_k)=1$ where $n_{k+1}|a^{n_k}-1$

Consider: $n_k|a^{n_{k-1}p_{k-1}}-1 = (a^{n_{k-1}}-1)\sum_{i=0}^{p_{k-1}-1}a^{in_{k-1}}$
But $\gcd (a^{n_{k-1}}-1,\sum_{i=0}^{p_k-1}a^{in_{k-1}})|p_{k-1}$
However, $\sum_{i=0}^{p_k-1}a^{in_{k-1}}$ cannot be purely a power of $p_{k-1}$, and thus we can choose another prime $p_k$ factor.

EDIT: Argh 4 minutes late, and the above proof is more complete  ;)

[quote="namanhams"]Anyone can explain for me : what is square free ?[/quote]

all prime factors only happen once. for example $30 = 2 \cdot 3\cdot5$ is square-free, $24=2^3\cdot3$ and $60=2^2\cdot3\cdot5$ are not.

Indybar, use binomial expansion and a few substitutions to simlify the algebra, then it's obvious.

Bomb

Omid Otami you said that you solved this in a beautiful way. can you post your solution ?Thanks

My solution is something like the grobber's solution


geometry
geometric transformation
reflection

The positive integer $ k$ has the property that for all natural numbers $ m$, $ k|m$ implies $ k|m_r$ where $ m_r$ and $ m$ are mutual reflections like $ 1234$ and $ 4321$.Show that $ k|99$

Is $ k \equal{} 1$? If $ k|m$, where m is any natural, $ k|1$ which means that $ k \equal{} 1$, and so $ k|99$ because $ 1|99$

But... it sounds strange...
What did I get wrong?

[quote="Gabuntu94"]Is $ k \equal{} 1$? If $ k|m$, where m is any natural, $ k|1$ which means that $ k \equal{} 1$, and so $ k|99$ because $ 1|99$

But... it sounds strange...
What did I got wrong?[/quote]

The problem does not ask to find an example of $ k$, it asks you to find all k that have that property and to show that these k divide 99

Ok... I meant: "In my opinion the only $ k$ which satisfies that is $ k\equal{}1$, so this is why $ k|99$. Is it right?"
I know I should find all k with that property, but if $ k | m$, where $ m$ is the whole set of natural numbers, $ k$ must be equal to 1.  :blush:

So... have I misunderstood the question?  :maybe:

it is saying if k|m where m is a natural number, not if m is a natural number and k|m...

so k|1 may be false


linear algebra
matrix
group theory
abstract algebra
topology
calculus
derivative

Suppose that $A$ is a $3\times 3$ matrix such that \[\|A\mathbf u\times A\mathbf v\|=\|\mathbf u\times \mathbf v\|\] for all $\mathbf u, \mathbf v \in\mathbb{R}^{3}$.

Prove that \[\|A\mathbf u\|=\|\mathbf u\|\] for all $\mathbf u\in\mathbb{R}^{3}$.

I suggest the following proof:

Suppose that A is a matrix that has the desired property, i.e. that

$| Au \times Av | = | u \times v |$ for all $u,v \in \mathbb{R}^{3}$.

First notice that A must be an invertible matrix, because if $ker A \neq 0$ and $u \neq 0$ such that $Au = 0$ then $Au \times Av = 0$ for all $v \in \mathbb{R}^{3}$. But it is easy to find a v such that $u \times v \neq 0$.

Now let G = all matrices with the desired property. We have just proved that $G \subset GL_{3}$.

If $A,B \in G$ then

$| ABu \times ABv | = | Bu \times Bv | = | u \times v |$  and $| A^{-1}u \times A^{-1}v | = | AA^{-1}u \times AA^{-1}v | = | u \times v |$ so G is in fact a group, a subgroup of $GL_{3}$.

Next, notice that if A is an orthogonal matrix then

$| Au \times Av | = |Au\parallel Av|sin\phi$, where $\phi$ is the angle between Au and Av, $= |u\parallel v|sin\phi = |u \times v|$, because A preserves angles and lengths. So $A \in G$.

I have shown that $G \supset O_{3}$, the orthogonal group.

The final step is to prove equality. I shall set up a smooth map $\phi : GL_{3}\to \mathbb{R}^{6}$ such that $G = \phi^{-1}(0)$. Put $\phi_{i}(A) = |Ae_{j}\times Ae_{k}|^{2}-1$ for i =1,2,3 and i,j,k distinct with j < k.

To set up the other 3 equations, recall that $2<x,y> = |x+y|^{2}-|x|^{2}-|y|^{2}$. Let $B : u \mapsto Au \times Av$ and $C : u \mapsto u \times v$, they are linear maps. Then we easily show that $<Bx,By> = <Cx,Cy>$ for all $x,y \in \mathbb{R}^{3}$ if $A \in G$.

We put $\phi_{3+i}(A) = <Ae_{j}\times Ae_{i}, Ae_{k}\times Ae_{i}>$ for i =1,2,3 and i,j,k distinct and j < k.

It should be fairly evident from what has been said that $G = \phi^{-1}(0)$, an algebraic group. For general manifolds, to deduce a smooth submanifold, we must now prove that 0 is a regular value of $\phi$, but since G is a group it is enough to show that $1 \in GL_{3}$ is a regular point for $\phi$. Take all 9 partial derivatives of $\phi$ at $1 \in GL_{3}$ and show that they span $\mathbb{R}^{6}$. This calculation can safely be left to the reader.  :)

We have found that G is a Lie subgroup of dimension $dim G = dim GL_{3}-dim \mathbb{R}^{6}= 9-6 = 3$. But $dim O_{3}= 3$ and this compact real Lie group, a closed 3-dimensional manifold, cannot be extended without increasing the dimension. Therefore we arrive at last at the conclusion that $G = O_{3}$.

If you find this proof ridiculously involved I'm inclined to agree.  :) I was looking for a more direct and elementary proof but didn't find any.


algebra
polynomial
Euler
logarithms
function
number theory unsolved
number theory

Let    $\Phi_n(x)$ be   the  cyclotomic polynomial of degree Euler Phi(n), i.e. $\deg (\Phi_n(x))=\varphi(n)$, prove that

1) The $\Lambda(n)=\log(\Phi_n(1))$  where $\Lambda$ is the function related with the Zeta function,

2) Find a closed relation between Mobious Function and \Phi_n(1), \Phi_n(-1).

3) Is well known that   $\quad X^n-1=\prod_{t|n}\Phi_t(x)$ , but I can't find the following identity

Let $m$ and $n$ positive integers, then   $\Phi_m(x^n)=\prod_{t|N_2}\Phi_{N_1\cdot m \cdot t}(x)$, where $n=N_1\cdot N_2$ and the election of $N_2$ is maximum in the way that $gcd(N_2,m)=1$.

good luck.

[quote="RobertuX"]Let    $\Phi_n(x)$ be   the  cyclotomic polynomial of degree Euler Phi(n), i.e. $\deg (\Phi_n(x))=\varphi(n)$, prove that

1) The $\Lambda(n)=\log(\Phi_n(1))$  where $\Lambda$ is the function related with the Zeta function,

2) Find a closed relation between Mobius Function and $\Phi_n(1)$, $\Phi_n(-1)$.

3) Is well known that   $\quad X^n-1=\prod_{t|n}\Phi_t(x)$ , but I can't find the following identity

Let $m$ and $n$ positive integers, then   $\Phi_m(x^n)=\prod_{t|N_2}\Phi_{N_1\cdot m \cdot t}(x)$, where $n=N_1\cdot N_2$ and the election of $N_2$ is maximum in the way that $gcd(N_2,m)=1$.

good luck.[/quote]

Well, it's OBVIOUS!!!!!!!!!  :D

More seriously:

1. Note first that $\Lambda(p^n)= \log (p)$ for prime powers $p^n$ and $0$ otherwise by definition.
Let's state it as:
$\Phi_{p^n}(1)=p$ for primes $p$ and $\Phi_m(1)=1$ otherwise (except $m=1$).
Now using $n=1+1^2+1^3+...+1^{n-1}= \prod_{1 \neq d|n } \Phi_d(1)$, the problem gets trivial by induction.

2. Well, $\Phi_{2\cdot p^n}(-1)=p$ for primes $p$ and additionally $\Phi_m(-1)=1$ for all other numbers except $m=1,2$ (can be done as above).
No idea indeed what the relation should be.

3. Could probably be done by looking at the degrees and zeros of both sides...


Let $ABC$ and $A'B'C'$ be two non-congruent triangles whose sides are respectively parallel. Prove that the three lines $AA'$, $BB'$, and $CC'$ (extended) are concurrent.

Also, does Stewart's Theorem come up often enough for me to know it by heart?

It's not hard to know it,

man+dad=bmb+cnc

A man and his dad put a bomb in the sink.

However, don't gloss over the labeling. Triangle ABC with sides a, b, c. The cevian is drawn to side a. m and c are adjacent; n and b are adjacent. d is the length of the cevian.

Thanks for the caveat  :wink:

Can anyone prove the statement about the homothetic triangles?


limit
trigonometry
real analysis
real analysis unsolved


A particle $P$ is in the $x-y$ plane and its coordinates satisfy the equations:

$\frac{d^{2}x}{dt^{2}}=-n^{2}x$ and $\frac{d^{2}y}{dt^{2}}=-n^{2}y$.

When $P$ is at $(5,0)$, $t=0$, $\frac{dx}{dt}= 0$, $\frac{dy}{dt}= 4n$.

Find the locus of $P$.

This is quite a nice problem from my Mathematics Text Book: Mechanics Section.

[hide="Hint"]

Prove that $\frac{d}{dx}(\frac{v^{2}}{2})= a$[/hide]

i found x=x(t)=5 cos nt+4 sin nt and y=y(t)=0 from the differential eq. p''+n^2*p=0 when p is x or y and establishing the conditions.
 So the locus(i understood  it's the moovement in  x-y plane ) is a line.

It's not a line. Find a cartesian form of the locus.

$(x, y) = (5 \cos (n t), 4 \sin (n t) )$.

Thus, it is an ellipse.

i agree.i have mistaken the initial conditions


For any AoPS folk who have been either
(a) waiting until the Interesting Test became available before applying to the Hampshire College Summer Studies in Mathematics,
or
(b) wondering why it hadn't been emailed to them yet, as they applied a while ago,

IT's ready!

To apply, go to http://www.hcssim.org
and if you have already applied, but don't receive IT via email by Wednesday 29th evening, let me know (either by PM or by posting here).

Y'Peeeeeeeeee!
*dances around shrieking, figuratively*

My mom was urging me to bug you about it, just in case you had published it but had forgotten to send it to everyone in Colorado...

Well, O Excited One, you and your mom may also wish to come to our Math Jam (held next week, April 6th) to lend your expertises.  That goes for all HCSSiM alumns, of course; most of you will be more eloquent than me in describing the program!

See, I knew there'd be a reason to stay home Spring Break!

Other than USAMO practice of course.

As a former HCSSiMer, I can say in full honesty that this is the most fun and most educational program you will attend.  Everyone should apply, especially given that this is year 34 = 2*17.

Is it possible to get the IT even if I'm not planning on applying???

JB

[quote="jackb1115"]As a former HCSSiMer...
Is it possible to get the IT even if I'm not planning on applying???
[/quote]

Yes, we're willing to share IT with alumns.  However, since you're still eligible to attend, you'll have to wait until the application/selection/invitation process is over.  Email me in late June, and I'll send you a copy.

definitely a great program to attend.

I will try to attend the math jam if time permits, I can offer the unique perspective of a lost immature kid who is now ... gosh who am I kidding ;)


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ S$={$ (x,y): x\plus{}y \in \mathbb{Z}$} $ \subset \mathbb{A}^2(\mathbb{Q})$.
Compute the Zariski closure $ \bar S \subset \mathbb{A}^2(\mathbb{Q})$.

We can prove that $ I(S)\equal{}(0)$ using the following lemma:

Let $ f(x,y) \in k[x,y]$ with term the degree of $ x$,$ y$ is not greater then $ n$,let $ X_1,X_2$ be two sets of $ k$ with $ n\plus{}1$ element.

If $ f(x,y)\equal{}0;\forall x \in X_1,y \in X_2$.

Then $ f(x,y)\equal{}0$(k is a field)

For proving the lemma,if $ f\neq 0$ you should first write $ f$ as
$ f(x,y)\equal{}f_k(y)x^k\plus{}f_{k\minus{}1}(y)x^{k\minus{}1} \cdots\plus{}f_{0}(y)$
$ k$ be the largest integer such that $ f_k(y) \neq 0$,then $ k \le n$.
by $ deg(f_k(y)) \le n$ we can find there is some $ y_0 \in X_2$ such that $ f_k(y_0) \neq 0$.
Then consider the polynomial
$ f(x)\equal{}f_k(y_0)x^k\plus{}f_{k\minus{}1}(y_0)x^{k\minus{}1} \cdots\plus{}f_{0}(y_0)$,we can find an some $ x_0 \in X_1$ such that $ f(x_0) \neq 0$
Thus  $ f(x_0,y_0) \neq 0$ contradiction.

In fact you can generalize the lemma for $ n$ variables.

Then $ \bar S\equal{}V(I(S))\equal{}\mathbb{A}^2(\mathbb{Q})$


God Tag!

Congratulations on your opening Communities! :)

kunny

God dag kunny  :)
Thank you!!
Thank you for the help too. It would be difficult without your help :)

Oh! Thanks. You had a good job, borz.

kunny


$ 20\%$ of the retail price of an article is profit for the shopkeeper. What percent of the shopkeeper's original cost is this profit?

I'm not sure, but the answer is supposed to be 25.
So...
$ 100\minus{}20\equal{}80$
$ 20/80\equal{}25\%$
But I'm really not sure what the problem means.


Are there any possibilities of our team competing in contests other than ARML? Seems like lots of other states have competitions to go to, such as HMMT. It would be fun experience, although I can imagine the enormity of the cost.

Although any decision wouldn't really affect me, I have to say that the reason a lot of other teams travel to other states to go to competitions is because states are a lot smaller on the other coast. You can drive for a couple of hours on the East coast and go through a couple of states, while in Washington, you drive for a couple of hours and you're still in Washington. And in general, I think that there are quite a few math competitions already in Washington. Off the top of my head, I can think of Math is Cool, Fall Classic, Blaine, MAO state, Eastlake, SIGMA, BATH, GAME... The schedule's pretty packed as it is.

what you could do is to take the WA ARML team and compete in local competitions...so you wouldn't be part of your school but rather part of the WA Team who is going against other schools (in the same state  :P )

I lol'ed at the post above.

But I don't think so since 3-4 Bellevue people belong in the arml team.

[quote="Uh oh!"]And if you did that, Bellevue would still dominate.  :D[/quote].

yeah edward is way too optmistic about Bellevue's math club!


algebra
polynomial
function
calculus
derivative
inequalities
linear algebra

Write
$ |\sum_{k\equal{}0}^{N} C_{k} \plus{} \sum_{k\equal{}0}^{N} X_{k}|^{2}$

as a sum of terms where each term  either depends on $ C_{k}$ only or  $ X_{k}$ only e-g  the solution can be like $ |C_{k}|^{2}\plus{}|X_{k}|^{2}\plus{}2Real(|C_{k}|)$ (note: this is not the solution, just wanted to explain wht it means that the solution should be the sum of termswhere each term is either $ C_{k}$ dependent or $ X_{k}$).
$ C_{k}$ and $ X_{k}$ are complex variables.
Solution can have inequalities.

I have the solution in the form
[hide]$ |\sum_{k\equal{}0}^{N} C_{k}|^{2} \plus{}|\sum_{k\equal{}0}^{N} X_{k}|^{2} \plus{} 2 Real (\sum_{k\equal{}0}^{N} C_{k}\sum_{k\equal{}0}^{N} X_{k}^{'}  )$

So the problem here is the third term $ Real (\sum_{k\equal{}0}^{N} C_{k} \sum_{k\equal{}0}^{N} X_{k}^{'} )$ which is dependant upon both $ C_{k}$ and $ X_{k}$.
Any idea to separate them?[/hide]

This is clearly not possible (without inequalities).  Even in the case $ N \equal{} 0$, it's easy to prove that the polynomial $ (c \plus{} x)^2 \equal{} c^2 \plus{} 2cx \plus{} x^2$ cannot be written in the form $ f(c) \plus{} g(x)$ for any functions $ f, g$.  The problem is that as one variable is held constant the $ cx$ guarantees that the growth rate of the function in terms of the other depends on the value of the first; in other words, the partial derivatives can't match up ($ f, g$ must be differentiable since the above is).

This is very similar to your other question (and has the same issue as far as solvability); are you thinking about these questions independently or do they come from somewhere?

hi,
in the last post it is said that the solution is impossible WITHOUT inequalities. OK then wht is the solution WITH inequalities?

I have my solution with inequalities:

[hide]$ |\sum_{k\equal{}0}^{N} C_{k} \plus{} \sum_{k\equal{}0}^{N} X_{k}|^{2} \leq |\sum_{k\equal{}0}^{N} C_{k}|^{2} \plus{} |\sum_{k\equal{}0}^{N} X_{k}|^{2}\plus{}2|\sum_{k\equal{}0}^{N} C_{k}| |\sum_{k\equal{}0}^{N} X_{k}|$

but here again problem is the same that the third term contains both $ C_{k}$ and $ X_{k}$
[/hide]

PS: @t0rajir0u  :These questions regard a certian research problem I am facing.

Are you expecting a unique solution or are you looking for the tightest possible bound?  I must give the same answer as the answer to the previous problem:  [b]only very loose bounds are possible.[/b]

I hope the following argument will give you an intuitive understanding why:  the desired expression depends only on the sum of all of the values of $ C_k$ and $ X_k$, which means it does not change value when we permute any of these values - in particular, we can switch around values of $ C_k$ and values of $ X_k$ without changing the value of the sum and we can, for example, set all of the values of $ X_k$ to zero and add them to values of $ C_k$.

Could you specify what this research problem is?  I don't see a motivation for trying to separate out a function of many variables into inequalities bounded by functions of less variables; again, do you understand what I'm trying to say conceptually when I say the partial derivatives don't match?


logarithms
real analysis
real analysis unsolved

This is part of solving ODE by Frobenius method (not the same as before), and I'm having little recursion issue :D

the general formula is:
$ a_n \equal{} \minus{} \frac {a_{n \minus{} 2}}{(n \plus{} \lambda \minus{} 1)(n \plus{} \lambda \minus{} 1)}$

For $ \lambda \equal{} 1$

So the first few terms are: $ a_2 \equal{} \minus{} \frac {a_0}{2\cdot4},\ a_4 \equal{} \frac {a_0}{2\cdot4\cdot4\cdot6},\ a_6 \equal{} \minus{}\frac {a_0}{2\cdot4\cdot4\cdot6\cdot6\cdot8},\ \ldots$

So general recursion formula is:

$ a_{2n} \equal{} ( \minus{} 1)^n\frac {a_0\cdot2}{(2n)!!(2n)!!}$

Since I have product of even numbers (I have added $ \frac {2}{2}$). How can I simplify it?

$ (2n)!!$ should be $ 2\cdot4\cdot6\cdot\ldots\cdot(2n \minus{} 2)\cdot2n$ for $ n\in\mathbb{N}$, and also $ (2n)!! \equal{} 2^nn!$, and since this was from my class, my assistent wrote that the general recursion is:

$ a_{2n} \equal{} \frac {( \minus{} 1)^n}{2^{2n}}\cdot\frac {a_0}{n!(n \plus{} 1)!}$

So can someone explain how he got that? I tried to find some formula that could explain this, but I'm out of luck...

Thanx!

I may be missing something subtle, but your various equations don't appear to be consistent with one another.

Your first expression:
$ a_n \equal{} \minus{} \frac {a_{n \minus{} 2}}{(n \plus{} \lambda \minus{} 1)(n \plus{} \lambda \minus{} 1)}$
For $ \lambda \equal{} 1$

leads to $ a_{n} \equal{} \minus{}\frac{a_{n\minus{}2}}{n^{2}}$ and not what you have written, which was:
$ a_2 \equal{} \minus{} \frac {a_0}{2\cdot4},\ a_4 \equal{} \frac {a_0}{2\cdot4\cdot4\cdot6},\ a_6 \equal{} \minus{} \frac {a_0}{2\cdot4\cdot4\cdot6\cdot6\cdot8},\ \ldots$

And the above does not, as far as I can see, simplify to
$ a_{2n} \equal{} ( \minus{} 1)^n\frac {a_0\cdot2}{(2n)!!(2n)!!}$

So... I think that we should start from your middle expression:
$ a_2 \equal{} \minus{} \frac {a_0}{2\cdot4},\ a_4 \equal{} \frac {a_0}{2\cdot4\cdot4\cdot6},\ a_6 \equal{} \minus{} \frac {a_0}{2\cdot4\cdot4\cdot6\cdot6\cdot8},\ \ldots$
and see where we can go from there.

By doing what you did before, we put a 2 in the numerator and one in the denominator, giving two 2s on the bottom. But there is only one of the highest number on the bottom, and this leads to:
$ a_{2n} \equal{} \frac{(\minus{}1)^{n} 2 a_{0}}{(2 \cdot 4 \cdot 6 \cdots 2n)(2 \cdot 4 \cdot 6 \cdots 2n\plus{}2)} \equal{} \frac{(\minus{}1)^{n} 2 a_{0}}{(2n)!! (2n\plus{}2)!!}$
which is equal to:
$ a_{2n} \equal{} \frac{(\minus{}1)^{n} 2 a{0}}{2^{n} n! 2^{n\plus{}1} (n\plus{}1)!}$
so the 2 on the top cancels with one in the $ 2^{n\plus{}1}$ leading us to:
$ a_{2n} \equal{} \frac{(\minus{}1)^{n} a_{0}}{2^{2n} n! (n\plus{}1)!}$

Ooops!  :blush:
It should be

$ a_{n}\equal{}\minus{}\frac{a_{n\minus{}2}}{(n\plus{}\lambda\minus{}1)(n\plus{}\lambda\plus{}1)}$

So after $ \lambda\equal{}1$ the expression is:

$ a_n\equal{}\minus{}\frac{a_{n\minus{}2}}{n(n\plus{}2)}$

And I see where I got wrong... Thnx!

I have another problem, while solving another ODE with frobenius, but I don't want to open new topic.

ODE is $ xy''\minus{}y\equal{}0$

Indicial equation is:

$ a_0(\lambda(\lambda\minus{}1))\equal{}0,\
\lambda_1\equal{}1, \lambda_2\equal{}0$

Recursion:

$ a_n\equal{}\frac{a_{n\minus{}1}}{(n\plus{}\lambda)(n\plus{}\lambda\minus{}1)}$

First solution is:

$ a_n\equal{}\frac{a_0}{n!(n\plus{}1!)},\
y_1(x)\equal{}a_0x\sum_{n\equal{}0}^{\infty}\frac{1}{n!(n\plus{}1)!}x^n$

The second solution is of the form:

$ y_2(x)\equal{}d_{\minus{}1}y_1(x)\cdot\ln(x)\plus{}\sum_{n\equal{}0}^{\infty}d_n(\lambda_2)x^{n\plus{}\lambda_2}$

Since putting $ \lambda\equal{}0$ gives no solvable recursion I have to find the solution with:

$ y_2(x)\equal{}\frac{\partial}{\partial\lambda}\Big[(\lambda\minus{}\lambda_2)y(\lambda,x)\Big]\Big|_{\lambda\equal{}\lambda_2}$

The first several terms are:

$ a_1\equal{}\frac{a_0}{\lambda(\lambda\plus{}1)},\ a_2\equal{}\frac{a_0}{\lambda(\lambda\plus{}1)^2(\lambda\plus{}2)},\ a_3\equal{}\frac{a_0}{\lambda(\lambda\plus{}1)^2(\lambda\plus{}2)^2(\lambda\plus{}3)}$

So to derive the 'general' solution $ y(\lambda,x)$ I would need some kind of closed recursion.

So what would it be?

Hi,

from what I can see, your
$ a_{1}\equal{}\frac{a_{0}}{\lambda(\lambda\plus{}1)},\ a_{2}\equal{}\frac{a_{0}}{\lambda(\lambda\plus{}1)^{2}(\lambda\plus{}2)},\ a_{3}\equal{}\frac{a_{0}}{\lambda(\lambda\plus{}1)^{2}(\lambda\plus{}2)^{2}(\lambda\plus{}3)}$

is equivalent to your
$ a_{n}\equal{}\frac{a_{n\minus{}1}}{(n\plus{}\lambda)(n\plus{}\lambda\minus{}1)}$
at the start of your post.

I may have misunderstood your question. Apologies if so.

Hm, I don't see how it's the same  :huh:

For second solution (if I put $ \lambda\equal{}0$) my recursion fails:

$ a_1\equal{}\frac{a_0}{1\cdot0}$

So instead of putting $ \lambda$ in it, I'm just writing few terms:

$ y(\lambda,x)\equal{}a_0 x^{\lambda}\left[\frac{x}{\lambda(\lambda\plus{}1)}\plus{}\frac{x^2}{\lambda(\lambda\plus{}1)^2(\lambda\plus{}2)}\plus{}\frac{x^3}{\lambda(\lambda\plus{}1)^2(\lambda\plus{}2)^2(\lambda\plus{}3)}\plus{}\ldots\right]$

That is then multiplied by $ (\lambda\minus{}\lambda_2)\equal{}\lambda$ and derived with the respect to $ \lambda$ as in:

$ y_{2}(x)\equal{}\frac{\partial}{\partial\lambda}\Big[(\lambda\minus{}\lambda_{2})y(\lambda,x)\Big]\Big|_{\lambda\equal{}\lambda_{2}}$

So to derive that, I would need some kind of recursion. There are infinite numbers of terms, so how to derive it? I'm guessing by the help of logarithm derivation ($ y\equal{}x^x\rightarrow \ln(y)\equal{}x\ln(x)$ and so on...)


analytic geometry
geometry
search
pigeonhole principle
area of a triangle

Prove that any convex pentagon whose vertices (no three of which are collinear)
have integer coordinates must have area > 5/2

Back to your old ways, I see?  At least, posting problems [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1717259558&t=159358]you've posted before[/url] and that have been [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1717259558&t=175008]solved on the forum[/url].

We claim that the minimum possible area is $ \boxed{\frac{5}{2}}$.  This can be formed by having points at $ (0,0) (1,0) (0,1) (2,1) (1,2)$.
Consider the parity of the coordinates of each of the points.  Then by pigeonhole, at least two points, $ A$ and $ B$ will have the same parities.  Then the midpoint of $ AB$ will be a lattice point.
Case 1: $ A$ and $ B$ are not consecutive vertices on the pentagon.
If this is the case, the midpoint of $ AB$ will lie in the interior of the pentagon.  Draw lines from this midpoint to each of the vertices, and we see 5 triangles are formed.  It is easy to show that the smallest possible area of a triangle with lattice coordinates is $ \frac{1}{2}$ so the minimum possible area of the pentagon in this case is $ \frac{5}{2}$.
Case 2: $ A$ and $ B$ are consecutive vertices on the pentagon.
Then the $ AB$ must be at least two.  We see that there must be a triangle with base $ AB$ that has area at least 2.  But we still have two triangles left over so the total area must be at least $ 2\plus{}2\frac{1}{2}\equal{}3$.


linear algebra
matrix
group theory
abstract algebra
superior algebra
superior algebra solved

let H =

{  | 1    n | , such  that n is a member of Z }.
    | 0    1 |

prove that H is a cyclic subgroup of GL(2, R)

Since $\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}^{k}=\begin{pmatrix}1 & k \\ 0 & 1\end{pmatrix}$ for all $k\in\mathbb{Z}$ this follows immediatly.


If Sam can buy five notebooks for $ \$6$ and then sell any two notebooks for $ \$3$, how many notebooks must she sell to make a total profit of $ \$12$?

From the information given, we see that Sam can buy one notebook for $ \frac {\$6}{5} \equal{} \$1.20$, and sell one for $ \frac {\$3}{2} \equal{} \$1.50$.  So for every notebook Sam sells, she makes a profit of $ \$1.50 \minus{} \$1.20 \equal{} \$0.30$.  Thus, she must sell $ \frac {12}{.3} \equal{} \boxed{40}$ notebooks to make a total profit of $ \$12$.


trigonometry
geometry
circumcircle
power of a point
cyclic quadrilateral

Hi, i've been struggling with this problem and am not sure how to solve it.

I dont think its any good if i describe it, so i'll link it.

http://www.mandelbrot.org/resources/sample_test.pdf

#7 on there.

Any help would be appreciated, thanks.

[color=cyan]You might try trigonometry -- you'll have four equations to solve with three angles and h as variables to solve for.  I don't know if it's really doable, but you might give it a shot.[/color]

i dont know, that will be really messy (if doable) i think, but right now its all i have, so i'll try it.  But im hoping theres a somewhat cleaner way to do this.

I can't even access the website, but I'm pretty sure Mr. Rusczyk can help the most.

Click here for the picture

Solution below.

[hide]Since BFE and BAE are subtended by same segment BE, B, E, F and A must be on the same circle. Therefore chord EF subtends congruent angles <CBF and <CAE. Therefore triangles CBF and CAE are similar. We obtain 40/h = (9+h)/10; solving for h and discarding the negative solution, we get h = 16.[/hide]

Not quite, here we use BBCode, not the usual html code, mostly because BBCode's are much easier.

nice solution, by the way.

[color=cyan]It works, I belive.

Also, the method I suggested will work as well, although with a fair bit more work.[/color]

[color=green]It is an important property of circles.  You should learn it in any high school geometry course.[/color]

[quote="cats..."]actually, im not sure how you know that they're all on the same circle.  I dont understand that argument, can you please clarify?[/quote]

im not too sure about this, but if two angles subtend the same arc, then those angles are congruent. furthermore, the 4 points that form the two angles form a cyclic quad. now if you draw a circ through those 4 pts, you will get two secant emanating from an exterior point. power of a point: h(9+h)=10(40). solve, then done!

btw, can anyone explain number 3? i keep getting 1/6..

Well, since angles BAE and BFE are equal as they subtend the same arc, and since angles ADB and FDE are equal as they are vertical angles, angles ABD and FED are also the same. By that last equality, BAEF is a cyclic quadrilateral.

(spoiler)

[hide]Consider the previous diagram. To prove that B, E, F and A are concyclic, we need to prove that the opposite angles of this quadrilateral are complementary.

We have <BAE = <BFE = a; also <BDA = <EDF. Therefore triangles BDA and EDF are similar with BD / DA = ED / DF, <ABF = <AEF = b. Rearranging yields BD / DE = AD / DF. Since we have <BDE = FDA, triangles BDE and FDA are similar; therefore <EBF = <EAF = c, <BED = <AFD = d. Note that a + b + c + d = 180 degrees. Then we have <BEA + <FEA + <BAE + <FAE = d + b + a + c = 180 degrees.

[/hide]

Edit: Uhh, I mushed up the labelling. Relabel and apply...

Thanks everybody!

Edit 2: NICE, s0mp! I forgot to use power of a point and went for the messier stuff instead.

Power of a point: From a point P outside a circle, draw two secants intersecting the circle at A, B and C, D respectively; then PA * PB = PC * PD.

If you drew a secant and a tangent, with the secant intersecting the circle at A, B and the tangent point being C, then PA * PB = PC^2.

Joel, I think you had a better high school geometry class than most if you learned it there!  I sure didn't learn about cyclic quads until MOP.

Specifically, if you have 4 points A,B,C,D, and angle BAC = angle BDC, then A, B, C, D all lie on a circle.  (See if you can prove it - start with the circumcircle of ABC, then think about what happens to angle BDC if D is inside or outside the circle.)

Finding a cyclic quadrilateral is often the key step to unwinding olympiad level Euclidean geometry problems, so this is definitely a geometric principle you want to master.

[quote="cats..."]sqrt(3)/2>sqrt(rand)>sqrt2/2

3/4>rand>2/4

(3/4-2/4)/1 = 1/4.[/quote]

oh...oops..i see my error now. thanks.

thanks everyone, i understand now, i didnt know thats what we were going for though.


I know its got nothing to do with math, but

Just curious, does any body know ususally how many Canadians do the US ivy league schools admit each year for undergrad?

If you are a sufficiently good student, then they will accept you.  It's really about as simple as that.  I imagine that some Ivies that are more 'Canadian-friendly', like Princeton, could accept 20-25 Canadians, or even more, per year.

Speaking of Princeton, they have an amazing scholarship programme for Canadian students.  If you're thinking of applying to an ivy leage school, but in need of financial support, you should seriously consider Princeton..

They don't give out scholarships.  They just treat internationals the same as domestic students for purposes of financial aid.

Do the ivy universities mainly look at grade 11 and 12 marks or like all the marks from grade 9 and up?


geometry
perimeter
inequalities unsolved
inequalities

See http://www.mathlinks.ro/Forum/viewtopic.php?t=18742 .

  darij

[quote="fabycv"]Let $a,b,c$ are sides of a triangle, where perimeter is 1,
Show that:

$a^2 + b^2 + c^2 + 4abc < \frac{1}{2}.$
and why couldn't be equal??

Thanks in advanced!!

Faby   :rotfl:[/quote]
to prove
$a^2 + b^2 + c^2 + 4abc < \frac{1}{2}.$
we just need to prove
$(a+b-c)(a+c-b)(b+c-a)>0$

[quote="davidlam"]
to prove
$a^2 + b^2 + c^2 + 4abc < \frac{1}{2}.$
we just need to prove
$(a+b-c)(a+c-b)(b+c-a)>0$[/quote]

mmmmm ....  well .

davidlam ..
could you explain the way: From this to the original ineq??
it's not direct .... i guess!!!

Thanks!!

Faby ;)

[quote="davidlam"]
to prove
$a^2 + b^2 + c^2 + 4abc < \frac{1}{2}.$
we just need to prove
$(a+b-c)(a+c-b)(b+c-a)>0$[/quote]

$a^2 + b^2 + c^2 + 4abc < \frac{1}{2}.$
since $a+b+c=1$ we should prove
$(a^2 + b^2 + c^2)(a+b+c) + 4abc < \frac{(a+b+c)^3}{2}.$
and then expand and factorize it...


inequalities
limit
inequalities proposed

[quote="quykhtn-qa1"][quote="TRAN THAI HUNG"]Let a,b,c>0.abc=1Prove that:
$ 1 < \frac {1}{{1 \plus{} a^2 }} \plus{} \frac {1}{{1 \plus{} b^2 }} \plus{} \frac {1}{{1 \plus{} c^2 }} \le \frac {3}{2}$ :)[/quote]
the inequaliti 1 is true
the inequaliti 2 is false[/quote]What do you mean 1 and 2? :)

I think inequality 2 should be $ \frac{1}{{1 \plus{} a^2 }} \plus{} \frac{1}{{1 \plus{} b^2 }} \plus{} \frac{1}{{1 \plus{} c^2 }} \le \frac{3}{2}$ as b, c can be made arbitrarily small, it's wrong.

The first estimation is easy to prove:
Let $ a\equal{}min\{a,b,c\}$
Then:
$ \frac{1}{1\plus{}a^{2}}\plus{}\frac{1}{1\plus{}b^{2}}\plus{}\frac{1}{1\plus{}c^{2}}\geq \frac{1}{1\plus{}a^{2}}\plus{}\frac{a^{2}}{1\plus{}a^{2}}\plus{}\frac{1}{1\plus{}c^{2}}>\frac{1\plus{}a^{2}}{1\plus{}a^{2}}\equal{}1$ :wink:
As you previously said the second inequality does not hold...

[quote="polskimisiek"]The first estimation is easy to prove:
Let $ a \equal{} min\{a,b,c\}$
Then:
$ \frac {1}{1 \plus{} a^{2}} \plus{} \frac {1}{1 \plus{} b^{2}} \plus{} \frac {1}{1 \plus{} c^{2}}\geq \frac {1}{1 \plus{} a^{2}} \plus{} \frac {a^{2}}{1 \plus{} a^{2}} \plus{} \frac {1}{1 \plus{} c^{2}} > \frac {1 \plus{} a^{2}}{1 \plus{} a^{2}} \equal{} 1$ :wink:
As you previously said the second inequality does not hold...[/quote]

Are you sure? I think $ a \equal{} min\{a,b,c\}$ will be modified $ c\equal{}max\{a,b,c\}$.

Sorry to my bad English.

Well, I was not clear enough. You are right, we can assume even $ a\leq b \leq c$
Then $ (\frac{1}{1\plus{}b^{2}}\geq \frac{a^{2}}{1\plus{}a^{2}})\iff (c\geq ab)$
and because $ a\leq 1$ then it is obvious. :wink:

The problem might be:

 :arrow: [i]Find the best constant $ k$ such that:

$ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2} < k$

holds for all positive real numbers $ a,b,c$ with $ abc \equal{} 1.$[/i]

We'll show that $ \min k \equal{} 2.$ For $ k \equal{} 2$ we have:

$ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2} < 2\Longleftrightarrow$

$ \Longleftrightarrow\displaystyle\sum_{cyc} a^2 b^2 \plus{} 1 > 0,$ which is obviously true.

Now for $ a \equal{} b \equal{} m\in (0,\infty)\Longrightarrow c \equal{} \frac {1}{m^2}.$

So: $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2} \equal{} \frac {1}{1 \plus{} m^2} \plus{} \frac {1}{1 \plus{} m^2} \plus{} \frac {m^4}{1 \plus{} m^4}$ $ \equal{} \frac{2}{1\plus{}m^2}\plus{}\frac{m^4}{1\plus{}m^4}\equal{}\frac{m^4 (3\plus{}m^2)}{(1\plus{}m^2)(1\plus{}m^4)}\plus{}\frac{2}{(1\plus{}m^2)(1\plus{}m^4)}.$

We have: $ \displaystyle\lim_{m\to 0}\frac{m^4 (3\plus{}m^2)}{(1\plus{}m^2)(1\plus{}m^4)}\plus{}\frac{2}{(1\plus{}m^2)(1\plus{}m^4)}$ $ \equal{}\displaystyle\lim_{m\to 0}\frac{1}{1\plus{}\frac{1}{m^4}}\cdot \left (1\plus{}\frac{2}{1\plus{}m^2}\right )\plus{}\displaystyle\lim_{m\to 0}\frac{2}{(1\plus{}m^2)(1\plus{}m^4)}\equal{}0\plus{}2\equal{}2.$ Thus the best constant is $ 2.$

[quote="Maxim Bogdan"]The problem might be:

 :arrow: [i]Find the best constant $ k$ such that:

$ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2} < k$

holds for all positive real numbers $ a,b,c$ with $ abc \equal{} 1.$[/i]

We'll show that $ \min k \equal{} 2.$ For $ k \equal{} 2$ we have:

$ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2} < 2\Longleftrightarrow$

$ \Longleftrightarrow\displaystyle\sum_{cyc} a^2 b^2 \plus{} 1 > 0,$ which is obviously true.

Now for $ a \equal{} b \equal{} k\in (0,\infty)\Longrightarrow c \equal{} \frac {1}{k^2}.$

So: $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2} \equal{} \frac {1}{1 \plus{} k^2} \plus{} \frac {1}{1 \plus{} k^2} \plus{} \frac {k^2}{1 \plus{} k^2} \equal{} \frac {2 \plus{} k^2}{1 \plus{} k^2} \equal{} 1 \plus{} \frac {1}{1 \plus{} k^2}.$ For $ k\to 0\Longrightarrow 1 \plus{} \frac {1}{1 \plus{} k^2}\to 2.$ Thus the best constant is $ 2.$[/quote]
Very nice then.You've made it be true but I wonder why you put a=b=k. :).

The following inequality is also nice:
Let $ a,b,c$ be positive real numbers satisfying $ abc\equal{}1$. Find the best constant $ k$ such that
$ \sum_{cyc} \frac{1}{\sqrt{1\plus{}a^2}} \le k$    :)

[quote="TRAN THAI HUNG"]
Very nice then.You've made it be true but I wonder why you put a=b=k. :).[/quote]

I've done a little mistake. I've edited my post. :)


geometry
circumcircle
geometric transformation
reflection
cyclic quadrilateral
Pythagorean Theorem
geometry unsolved

A point $ H$ belongs to a diagonal $ AC$ of convex quadrilateral $ ABCD$ such that $ BH$ is perpendicular to $ AC$. Let $ AB$ be perpendicular to $ BC$, and $ AO$ perpendicular to $ DH$, where $ O$ is the circumcenter of triangle $ ACD$. Prove that $ AB\equal{}BC$.

Sorry... it should be "prove that $ AB\equal{}AD$"  :blush:

[quote="mstoenescu"]I think something is wrong :huh:[/quote]

Not a horrible mistake!!!

Just a typo :D

In fact we should prove: $ AB\equal{}AD$

Ok,lets solve the problem,$ AK$ is the diameter of the circumcircle of $ \triangle ACD$,then:

$ \angle AHM\equal{}\angle ACK\equal{}\frac{\pi}{2}$,and $ MHCK$,is cyclic quadrilateral.

as the power of point $ A$,to the circle of cyclic quadrilateral,we have:

$ AM.AK\equal{}AH.AC  \Longleftrightarrow AM.(AM\plus{}MK)\equal{}AH.(AH\plus{}HC)$

$ AM^2\plus{}AM.MK\equal{}AH^2\plus{}AH.HC \Longleftrightarrow$

Accorcing to Pythagorean theorem $ AH^2\minus{}AM^2\equal{}MH^2$

Again,according to power of points$ H$and$ M$,to the circles with diameters$ AC$ and $ AK$,respectively,

We can conclude: $ AH.HC\equal{}HB^2$ and $ AM.MK\equal{}MD^2$

Then: $ AM^2\plus{}AM.MK\equal{}AH^2\plus{}AH.HC \Longleftrightarrow MD^2\equal{}MH^2\plus{}HB^2$

Accorcing to Pythagorean theorem $ AD^2\minus{}AM^2\equal{}MD^2$

\Longleftrightarrow  $ AD^2\minus{}AM^2\equal{} MH^2\plus{}HB^2 \Longleftrightarrow  AD^2\equal{}AM^2\plus{} MH^2\plus{}HB^2$

Accorcing to Pythagorean theorem $ MH^2\plus{}AM^2\equal{}AH^2  \Longleftrightarrow AD^2\equal{}AH^2\plus{}HB^2$

Accorcing to Pythagorean theorem $ AB^2\equal{}AH^2\plus{}HB^2$

$ AD^2\equal{}AB^2  \Longleftrightarrow AD\equal{}AB$

[quote="massnet"] [color=darkred]A point $ H$ belongs to a diagonal $ AC$ of convex quadrilateral $ ABCD$ such that $ BH$ is perpendicular to $ AC$. Let $ AB$ be perpendicular to $ BC$, and $ AO$ perpendicular to $ DH$, where $ O$ is the circumcenter of triangle $ ACD$. Prove that $ AD \equal{} AB$.[/color] [/quote]
[color=darkblue][b][u]Proof[/u][/b] (similarly to [b][u]Ashegh[/u][/b]'s).

Denote the circumcircle $ w$ of the triangle $ ACD$ and the power $ p_w(X)$ of the point $ X$ w.r.t. the circle $ w$ . Thus,

 $ \left\|\begin{array}{ccc} HA\perp HB & \Longrightarrow & HA^2 \plus{} HB^2 \equal{} BA^2 \\
 \\
BA\perp BC & \implies & HA\cdot HC \equal{} HB^2 \\
 \\
DH\perp AO & \implies & DA^2 \minus{} DO^2 \equal{} HA^2 \minus{} HO^2 \\
 \\
p_w(H) & \implies & OD^2 \minus{} HO^2 \equal{} HA\cdot HC\end{array}\right\|\ \ \bigoplus$ $ \implies$ $ AD^2 \equal{} AB^2$ $ \implies$ $ AD \equal{} AB$ .[/color]

notice that $ \angle ADH\equal{}\angle ACD$ hence $ AD^2\equal{}AH.AC\equal{}AB^2\Longrightarrow AB\equal{}AD$

[quote="Amir.S"]notice that $ \angle ADH \equal{} \angle ACD$ hence $ AD^2 \equal{} AH.AC \equal{} AB^2\Longrightarrow AB \equal{} AD$[/quote]
[color=darkred] [size=117][b]Very nice the your proof, [u]Amir S.[/u] ![/b][/size][/color] [color=darkblue] Denote the reflection $ E$ of the point $ A$ w.r.t. the point $ O$ .

Then $ DA\perp DE$ and $ \widehat {ADH}\equiv\widehat {AED}\equiv\widehat {ACD}$ , i.e. $ \widehat {ADH}\equiv\widehat {ACD}$ a.s.o. [b]Thank you ![/b] [/color]

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension.[/u][/b] Let $ ABCD$ be a convex quadrilateral.  Denote the circumcircle $ (O)$ of the triangle $ ACD$
and the point $ H\in AC$ for which $ DH\perp AO$ .  Prove that $ AD \equal{} AB\Longleftrightarrow \widehat {ABH}\equiv\widehat {ACB}$ . [/color][/quote]


function
probability
expected value
probability and stats

Loosely inspired by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302600]this topic[/url], but then it's not really the same thing.

The game show "Deal or No Deal" proceeds as follows:

The contestant is shown a collection of people - 20 or 25 - each holding a closed briefcase. Each briefcase contains a label inside (hidden from sight) indicating that it is worth a certain amount of money. These amounts range from \$0.01 to \$1,000,000, in a crude approximation of a geometric progression.

The game the proceeds. The contestant chooses one briefcase which is set down beside him or her, but remains closed. Hypothetically, the unknown amount of money in that case will be the contestant's prize. However, he or she must first chose a batch of several other briefcases that will be opened, showing the enclosed amounts.

After each such round of openings, the "banker," communicating through the show's host, offers to "buy" the contestant's briefcase for a certain amount of money. To a first approximation, the offer is often fairly close to the expected value of the amounts in the still-unseen briefcases. The contestant may than either accept this deal, ending the game, or refuse the deal. After refusing the deal, the contestant must then have another batch of other briefcases opened. (In practice, the size of the batches gets smaller as the number of remaining briefcases gets smaller; eventually it comes down to one at a time.)

Since the banker's offer is often approximately the expected value of the unseen briefcases, it goes up if a batch uncovers mostly small values and down if that batch uncovers some big values.

Does there exist some kind of optimal strategy for the contestant? (It's possible that to answer that, you'd need more details than I've given - the exact list of values, the amounts in each batch, deviations of the banker's behavior from strictly offering the expected value. And there are some utility of money issues as well.)

something i think about when i watch this show is the variance of the remaining briefcases.
when the variance is high near the end and theres a couple small values and a couple large values, the banker seems to offer lower than the expected value..

perhaps the optimal stopping time would be when the bankers offer is within a certain fraction of the current standard deviation? it seems like each game would have its own "best strategy" as your first picks progressed...?

I've found that the banker's offers are substantially discounted from the expected value of the remaining cases, especially the earlier offers. That being the case, the ideal strategy would be to collude with other contestants and agree to keep opening cases as long as banker offers an amount less than the expected value - and then to evenly split the winnings among all participating contestants. I suspect this would be disallowed by the producers for obvious reasons - and that enforcing the even split of money would be legally difficult.

But if you're playing on your own, you've hit on the key point - which is what matters most in the game of Deal or No Deal is not the amount of money, but the utility of the money to the individual playing. You also cannot ignore the psychological factor of "feeling of loss" - when someone has a large offer early and loses it later on. So I think the optimal strategy is to decide on a minimum amount that you would be "happy" winning, and keep opening cases as long as there is less than 10% chance that all cases greater than or equal to that value could be eliminated by opening the next set of cases.

of course, if the banker always offered you the expected value, then like the thread that inspired you, you gain not edge by waiting..
 assuming that there are n remaining cases of value $ v_1,...,v_n$ and you have an equal chance of picking one

the bankers offer on step n would be $ X_n\equal{}\frac{1}{n}\sum_{k\equal{}1}^n v_k$
then the next step having n-1 cases would have

$ E[X_{n\minus{}1}]\equal{}\sum_{k\equal{}1}^n \frac{1}{n} \frac{nX_n\minus{}v_k}{n\minus{}1}\equal{}\frac{n(n\minus{}1)X_n}{n(n\minus{}1)}\equal{}X_n$

but if the banker considers the variance, then waiting may have an effect..

since the variance on step in is a r.v. $ s_n\equal{}\sum_{k\equal{}1}^n \frac{1}{n}(X_n\minus{}v_k)^2\equal{}\frac{1}{n}\sum X_n^2\minus{}2X_nv_k\plus{}v_k^2$

so $ E[s_{n\minus{}1}]\equal{} \sum_{j\equal{}1}^n \frac{1}{n} \sum_{1\le k \le n, k\ne j} \frac{1}{n\minus{}1}(\frac{nX_n\minus{}v_j}{n\minus{}1}\minus{}v_k)^2$
$ \equal{}\frac{1}{n(n\minus{}1)}\sum_{j\equal{}1}^n\sum_{1\le k \le n, k\ne j}[\frac{(nX_n\minus{}v_j)^2}{(n\minus{}1)^2}\minus{}2v_k\frac{nX_n\minus{}v_j}{n\minus{}1}\plus{}v_k^2]$
$ \equal{}\frac{1}{n(n\minus{}1)^3}\sum_{j\equal{}1}^n\sum_{k\ne j}[(nX_n\minus{}v_j)^2\minus{}2v_k(nX_n\minus{}v_j)(n\minus{}1)\plus{}v_k^2(n\minus{}1)^2]$
$ \equal{}\frac{1}{n(n\minus{}1)^3}\sum_{j\equal{}1}^n\sum_{k\ne j}[(n^2X_n^2\minus{}2nX_nv_j\plus{}v_j^2\minus{}2(n\minus{}1)(nX_nv_k\minus{}v_jv_k)\plus{}v_k^2(n\minus{}1)^2]$
$ \equal{}\frac{1}{n(n\minus{}1)^2}\sum_{j\equal{}1}^n v_j^2(n^2\minus{}2n\plus{}2)\plus{}n^2X_n^2\minus{}2n^2X_nv_j\plus{}2\sum_{k\ne j}v_jv_k$
$ \equal{}\frac{1}{n(n\minus{}1)^2}\left(n^3s_n\minus{}2\sum_{j\equal{}1}^n nv_j^2\minus{}\sum_{k\equal{}1}^n v_jv_k\right)$
$ \equal{}\frac{1}{n(n\minus{}1)^2}\left(n^2(n\minus{}1)s_n\minus{}n\sum_{j\equal{}1}^n (v_j^2\minus{}X_n^2)\right)\equal{}\frac{n}{n\minus{}1}s_n\minus{} \sum_{j\equal{}1}^n \frac{v_j^2\minus{}X_n^2}{(n\minus{}1)^2}$

since $ \sum_{j\equal{}1}^n \frac{v_j^2\minus{}X_n^2}{(n\minus{}1)^2} >0$  we get $ E[s_{n\minus{}1}\minus{}s_n]>0$ iff $ (n\minus{}1)s_n > \sum_{j\equal{}1}^n v_j(v_j\minus{}X_n)$
and $ E[s_{n\minus{}1}\minus{}s_n]<0$ when $ (n\minus{}1)s_n < \sum_{j\equal{}1}^n v_j(v_j\minus{}X_n)$

so if the banker offers you some linear form like $ Y_n\equal{}aX_n\plus{}bs_n$ then
$ E[Y_{n\minus{}1}\minus{}Y_n]\equal{}\frac{b}{n\minus{}1}\left[(n\minus{}1)s_n \minus{} \sum_{j\equal{}1}^n v_j(v_j\minus{}X_n)\right]$

so i guess this formula is more useful near the the end of the show when theres $ \le 8$ cases,
of course i did a lot of assumptions on the bankers part

or you could say that you are trying to maximize a utility function $ \mathcal U$ (typically a concave function, like $ \frac{x^{1\minus{}\lambda}}{1 \minus{} \lambda}$). If the banker offers you an amount $ X_n$, you have to compare $ \mathcal U(X_n)$ with your expected utility, which is $ \frac{\sum_{k} \mathcal U(v_k)}{n}$ (where the $ v_k$ are the remaining boxes). The concavity of $ \mathcal U$ simply means that there is a big difference (happiness) between 100£ and 200£, while the difference is not that big between 100100£ and 100200£. The concavity immediately implies that if the banker offers you the expected sum $ \frac{\sum v_k}{n}$, you should accept. The critical value is, under the assumption that your utility function is $ \mathcal U(x) \equal{} x^{\alpha}$ (with $ 0<\alpha<1$), the $ \alpha$-mean:
\[ X_* \equal{} (\frac{ \sum v_k^{\alpha}}{n})^{\frac{1}{\alpha}} < \frac{\sum v_k}{n}\]


Basically, if $ x^{100} < y^{100}$, then $ x<y$. So when your teacher took off the zeroes, he/she was basically doing that.

I know, but it can't be that simple.

So with the information you've given me, you are basically saying that in any case you can just take off the zeroes to determine the numbers?

Granted, I know that 2^100<3^100...but the ratio of the real numbers to the numbers without the zeroes are not the same?

You can if it's an inequality, if the exponents are positive, and the base numbers are positive.

You're not trying to prove the ratio, you're just proving which ones are bigger.

So what you are saying is that no matter what, if everything is positive, you could take any numbers and simplifly their exponents, and the answer will be the same?

It is in general true; you divide through by a common positive factor of both exponents without changing the sense of the inequality.

Here is a complete, general proof for positive exponents using logarithms and the fact that $ \log_b a^c=c\log_b a$

For $ n,c,d$ positive integers (the use of log base two is arbitrary; it doesn't matter which base you choose)

\begin{align*}
a^{nc}& &b^{nd}\\
\log_{2} a^{nc} & & \log_{2} b^{nd}\\
nc\log_2 a &  &nd\log_2 b \\
c\log_2 a & & d \log_2 b \\
2^{c\log_2 a}& & 2^{d\log_2 b}\\
2^{\log_2 a^c}& & 2^{\log_2 b^d}\\
a^c& &b^d\\
\end{align*}

Whichever way the inequality goes, the sense is preserved throughout (we performed only "legal" and identical operations to each side).  The restriction to positive exponents is important because dividing through by negative numbers changes the sense of the inequality.

I understand everything!

Except when the 2 is brought in front of the 'log.'

Is it because it's base 2?
I understand all of the steps before and after that!

Yeah, he's basically exponentiating the logs to get rid of them and he uses 2 since it is base 2. Since log basically means "the power you raise 2 to in order to get whatever number", so if you raise 2 to this power you get that number.


I still don't know how to read the names of some mathematicians. Because the pronuncations change when they are translated into Chinese.

for example,

Cauchy
Poisson
Poincare
Lagrange

....
it seems that "oi" in French sounds [wa], isn't it?

As far as I knoiw, 'oi' is [wa], but I have always pronounced Poincare with [oi] instead of [wa], because it sounds nicer to me :D

but what about the other vowels?
like 'au' in Cauchy, 'a' and 'e' in Poicare?

[quote="liyi"]Cauchy Poisson Poincare Lagrange
it seems that "oi" in French sounds [wa], isn't it?[/quote]

Indeed "oi" is [wa].

.Cauchy = [ko][\inti], with a closed "o"  (or cowtchee in the future Langua universalis announced by Grobber :D). chi=chy=[\inti] (like in Chirac - not sure it's te right exemple in fact :?).
.Poisson = p[wa][s]
.Poincar = p[w~e]kare (hard to write phonetic letters, put '~' over the 'e' ....)
.Lagrange = lagr:z

How do you pronounce (and spell correctly!) Buniakovski ? Chebyshev ? Brahmagupta ?

Again, as far as I know, those last 3 names you wrote there are pronounced exactly as they are written. I'm not sure abt the spelling (there might be a 'y' in Buniakovski, but i'm not sure, and I've never really known if 'Chebyshev' is correct).

the common writing is "Bunyakovsky" or "Bunyakovski"

[quote="grobber"]pronounced exactly as they are written. [/quote]

Mmmm, I'm not sure it helps  :D  :D

Just to take "u" :

Is-it [u] like in german, [y] like in french, "you" like in english, ......

Sorry... What country are you from? I was blinded by my Romanian pronounciation  :D I think it's [u] (German type of u).

[quote="grobber"]What country are you from? [/quote]

From the country of Cauchy, Poisson, Poincare and Lagrange  :)

hmm... Many famous mathematicians are from France.

Goethe said, "Mathematicians are like Frenchmen: Whatever you say to them, they translate it into their own language, and forthwith it means something entirely different."

[quote="liyi"]Goethe said, "Mathematicians are like Frenchmen: Whatever you say to them, they translate it into their own language, and forthwith it means something entirely different."[/quote]

 :D  :D

Not entirely false !

more (how to read)

Darboux
Cesro
Stolz
Galois
Bernoulli
Dedekind
Pascal (is this pars-gol or pars-gar?)
Schwarz

Darboux = [dar][bu]
Galois = [ga][lwa]
in Stolz S is  \int and z is pronounced like in teh word tzar (or tsar = Russian emperor). The same goes for Schwarz, and w is [v].
 In fact:[url]

http://www.nsm.iup.edu/ma/gsstoudt/pronounce.html

[/url]

Thank you grobber for this nice url (even if the pronouciation of Monge, Gaspard (1746-1818) (MNZH)
is a bit strange !)

Still don't know however how to say Buniakovski or Chebyshev absent from the list (a cold war side effect ?) !

[quote="grobber"]in Stolz S is  \int and z is pronounced like in the word tzar (or tsar = Russian emperor)
[/quote]
what about 'o' in Stolz? it is [o](like 'o' in 'dog'), [u] or anything else?

thanks for grobber's url. very nice, it is.

[quote="grobber"]Darboux = [dar][bu]
[/quote]
is the accent on [dar] or [bu]?

i find that many French names have their accents on the second syllable.

Every time I've heard someone say Darboux, the accent fell on the second syllable.

[quote=cool972]i think this may be the longest bump ever
EDIT: Yes, it is. The record is now 5200 days
I searched it up.[/quote]

Congrats

[quote=cool972]i think this may be the longest bump ever
EDIT: Yes, it is. The record is now 5200 days
I searched it up.[/quote]

[url=https://artofproblemsolving.com/community/q7h6601p23214]This[/url] bump is longer.

[quote=math_is_life_2004][quote=ythomashu]mine is direct[/quote]

Your creates spam and nobody cares. KiwiYum is awesome.[/quote]

no u

How about "Euler"?
Is it "yoo-ler" or "oiler"???

I think it's oiler.
And Euclid is yoo-clid.

Remember when all of these people were actually active?

this is old umm Pneumonoultramicroscopicsilicovolcanoconiosis


Would you like to explain for me how it could be that the cartesian product of AxB is not the same as BxA.
If A = {1,2} and B = {a, b} then the carteisan product for AxB = {(1,a), (1,b),(2,a), (2,b)}. Why is not for example (1,a) the same as (a,1)? What way is there to proof that AxB is not the same as BxA?

Thank you for any help.

Read the definition of Cartesian product carefully.
$A\times B$ are sets of ordered pairs $(a,b)$ where $a\in A$ and $b\in B$.

By definition, two ordered pairs $(a,b)$ and $(a',b')$ are equal if and only if $a=a'$ and $b=b'$. So $(1,a)$ is not equal to $(a,1)$ since $1\neq a$ and $a\neq 1$.

What do you think about the fact that {a, 1} is not considered to be the same thing as {1, a}?  Where is the diffrence, why should it be like this?

$\{a,1\}$ is the same thing as $\{1,a\}$ (since those are sets). But $(a,1)$ and $(1,a)$ are different things. You can think of $(a,b)=\{\{a\},\{a,b\}\}$ (note that this is not symmetric).


$t_1,t_2,t_3,\ldots,t_n$ are positive reals.

$(t_1+t_2+t_3+\ldots+t_n)(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\ldots+\frac{1}{t_n})<n^2+1$

Prove that for all $i,j,k$ that $1\leq i<j<k\leq n$ $t_i,t_j,t_k$ are lengths of sides of triangle.

The best constant 's $(n-3+\sqrt{10})^2$ (replece $n^2+1$)


Hello everybody, I am having trouble understanding the difference between incomplete dominance and codominance.  :blush:  To me they seem to be essentially the same thing at the molecular level. Here is my (probably flawed) reasoning:

The common example of incomplete dominance is the pink snapdragon which is a hybrid of a red flower and white flower; it has the genotype 'RW'. I understand that it gets its pink color due to the fact that a single 'R' gene does not make enough red pigment to keep the flower distinctly red. The 'W' gene does not make any pigment, but if it coded for another color - for instnace blue - then the flower would be purple becuase each cell would make some red pigment and some blue pigment.

A common example of codominance is blood type. The A and B alleles are both codominant over the O allele. So, when the genotype of an individual is 'AB' then the A allele and the B allele are both expressed, leaving the blood having both A antigens and B antigens. [u]But isn't this exactly the same as a 'RW' flower producing both red pigments and "white pigments" in the example of incomplete dominance above?[/u]

Another question that i have about codominance is with the example of shorthorn cattle. A hybrid roan possesses both red hair and white hair, [u]but what determines whether a given strand of hair is red or white?[/u]

I am very confused about all this and if somebody understands where my logic is wrong PLEASE explain!!

I'm not sure that a flower with a red and a blue gene would be purple; it might be codominant too.

Further to your question, often the line between different types of dominance blurs when you look at the molecular level.  Tay-sachs disease, for instance, is completely recessive at the organism level (i.e., a person with only one copy of the allele won't have it), but at the molecular level, its incompletely dominant (cells produce fewer gangliosidases, but not few enough to cause any real problems).

Blood type isn't the same as the RW flower, simply because "both" phenotypes aren't expressed discretely.  The flower isn't "both" red and white, its "only" pink, whereas blood cells are "both" A and B.  Its mainly an issue of semantics; flower color is a continuum, antigen expression isn't.

Equine coat color is actually a much more complex topic than simple codominance; there are actually genes that code for the color of specific patches of hair.  Its worth looking up in wikipedia.  An easier example might be the tortoiseshell cat, which has patches of red and black fur.  This is due to the fact that the cat inherits its coat color genes on its x-chromosomes;  a heterozygous female cat has both genes, and due to random x-inactivation, the effect is codominance.


Let $ f(x) \in \mathbb{Z}/p\mathbb{Z}[x]$ such that $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \in \mathbb{Z}/p\mathbb{Z}[x,y]$. Then $ f(x)$ must be of the form $ a_{0}x \plus{} a_{1}x^{p} \plus{} a_{2}x^{p^2} \plus{} ... \plus{} a_{m}x^{p^m}$.

If you check equality in k[x,y] be looking at the "homogenous" terms, then you see that f(x+y)=f(x)+f(y) if and only if b*(x+y)^n = b*x^n + b*y^n where b*t^n is a monomial of f.  In other words, notice that b*(x+y)^n is a sum of monomials all of total degree n.


linear algebra
matrix
complex analysis
complex analysis unsolved

Is there a way to solve a set of linear equations analytically?

A system of n linear equations in n variables (in $ \mathbb{C}$).
I can imagine that it can be done (theoreticaly) using complex analysis.

There's no need for complex analysis; this is the province of [url=http://en.wikipedia.org/wiki/Linear_algebra]linear algebra[/url].

That's because nobody can figure out what exactly you are asking. State your question in an understandable way, and you'll get an understandable answer in return :roll:

Assuming you mean something like:
$ ax_{1}\plus{}bx_{2}\plus{}cx_{3}\equal{}d$
$ ex_{1}\plus{}fx_{2}\plus{}gx_{3}\equal{}h$
$ \vdots$
where $ x_{1}$, $ x_{2}$, and $ x_{3}$ are your variables, I would say kunny gave you a pretty helpful answer.  Just turn your system into a matrix and do some [url=http://en.wikipedia.org/wiki/Gaussian_elimination]Gaussian elimination[/url].


number theory
greatest common divisor
prime numbers
number theory proposed

Prove that $240$ divides all numbers of the form $p^4-q^4$, where $p$ and $q$ are prime numbers strictly greater than $5$.
Prove also that $240$ is the greatest common divisor of all numbers of the form $p^4-q^4$, with $p$ and $q$ are prime numbers strictly greater than $5$

Let $p,q$ be odd primes greater than $5$.
Then $p^4-q^4 = (p^2-q^2)(p^2+q^2)$ where $p^2 = q^2 = 1 \mod [8].$ Thus $p^4 - q^4 = 0 \mod [16].$
Usinf Fermat'slittle theorem, we also have $p^2 = q^2  = 1 \mod [3]$ and $p^4 = q^4 = 1 \mod[5].$
Thus $p^4 - q^4 = 0 \mod [5]$ and $p^4 - q^4 = 0 \mod [3].$
And then $p^4-q^4 = 0 \mod [3\cdot 4 \cdot 16] = 0 \mod [240].$

It follows that the desired $\gcd$ is divisible by $240.$
But $11^4 - 7^4 = 3 \cdot 17 \cdot 240.$ And $17^4 - 7^4 = 2\cdot 13^2 \cdot 240.$
It easily follows that the desired $\gcd$ is $240.$

Pierre.


algebra
polynomial
inequalities
inequalities proposed

Prove that if $x$ and $y$ are positive reals and $x^{2}+y^{3}\ge x^{3}+y^{4}$, than:

$x^{3}+y^{3}\le 2$

Prove that if $x$ and $y$ are positive reals and $x^{2}+y^{3}\ge x^{3}+y^{4}$, then $x^{5}+y^{5}\le 2.$
By the way, $x^{6}+y^{6}\leq2$ it is already wrong: $x=0.941$ and $y=1.0456.$

I proved by Cauchy -Swartz and AM-GM that
$x^{3}+y^{3}\leq x^{2}+y^{2}\leq x+y\leq 2$

Dear Silouan,

can you, please, explain your proof

[quote="silouan"]I proved by Cauchy -Swartz and AM-GM that
$ x^{3} \plus{} y^{3}\leq x^{2} \plus{} y^{2}\leq x \plus{} y\leq 2$[/quote]

I don't understand why

[quote="silouan"] $ x^{2} \plus{} y^{2}\leq x \plus{} y\leq 2$[/quote]

Thank you very much.

Manlio this is for you  :)

We have $ x^3\plus{}y^3\leq\sqrt{x^3\plus{}y^4)(x^3\plus{}y^2)}\leq\sqrt{(x^2\plus{}y^3)(x^3\plus{}y^2)}\leq\frac{x^2\plus{}y^3\plus{}x^3\plus{}y^2}{2}$

so $ x^3\plus{}y^3\leq x^2\plus{}y^2$

Also $ x^2\plus{}y^2\leq\sqrt{x^3\plus{}y^3)(x\plus{}y)}\leq\sqrt{(x^2\plus{}y^2)(x\plus{}y)}\leq\frac{x^2\plus{}y^2\plus{}x\plus{}y}{2}$

so $ x^2\plus{}y^2\leq x\plus{}y$

but  $ x\plus{}y\leq\sqrt{2(x^2\plus{}y^2)}$ so $ x\plus{}y\leq 2$ and we are done  :)

we have ;
 $ x\plus{}x^3 \geq 2x^2$  and $ y^2\plus{}y^4 \geq 2y^3.$
  but : $ x^2\plus{}y^3 \geq x^3\plus{}y^4.$  hence : $ x\plus{}y^2 \geq x^2\plus{}y^3$
complete prove :

OH! This problem is very old! And yes, using AM-GM is the most beautifl way to do this problem. However, finding that x+y>=x^2+y^2 is the most thoughtful step! :) . I think we should try expanding by proving he case with 3-variable x,y,z!

[quote="Hong Quy"]we have ;
 $ x \plus{} x^3 \geq 2x^2$  and $ y^2 \plus{} y^4 \geq 2y^3.$
  but : $ x^2 \plus{} y^3 \geq x^3 \plus{} y^4.$  hence : $ x \plus{} y^2 \geq x^2 \plus{} y^3$
complete prove :[/quote]
   others  : we have :
 $ 1\plus{}x^2\plus{}1\plus{}y^4 \geq 2x\plus{}2y^2 \geq x^2\plus{}x^3\plus{}y^3\plus{}y^4$
 hence : $ 2 \geq x^3\plus{}y^3$

[quote="j555"]Prove that if $ x$ and $ y$ are positive reals and $ x^{2} \plus{} y^{3}\ge x^{3} \plus{} y^{4}$, then $ x^{3} \plus{} y^{3}\le 2.$[/quote]$ x^3\plus{}y^3\equal{}2\minus{}3\left(x^2\plus{}y^3\minus{}x^3\minus{}y^4\right)\minus{}(x\minus{}1)^2(2x\plus{}1)\minus{}(y\minus{}1)^2\left(3y^2\plus{}2y\plus{}1\right)\leq 2.$

This problem is from Hungary Olympics(2002) :lol:
You can find it.


The coefficient of $ x^3$ in $ (1\minus{}x)^5 (1\plus{}x)^{25}$ is:

$ \textbf{(A)}\ 1040 \qquad \textbf{(B)}\ 1140 \qquad \textbf{(C)}\ 1240 \qquad \textbf{(D)}\ 1340 \qquad \textbf{(E)}\ 1440$

[hide="Solution"]
You can simplify $ (1 \minus{} x)^5(1 \plus{} x)^{25}$ to $ (1 \minus{} x^2)^5(1 \plus{} x)^{20}$.

The coefficient of $ x^3$ will be the coefficient of $ x^3$ in $ (1 \plus{} x)^{20}$ minus the coefficient of $ x^2$ in $ (1 \minus{} x^2)$ times the coefficient of $ x$ in $ (1 \plus{} x)^{20}$.

To put this into numbers we have: $ \binom{20}{3} \minus{} \binom{5}{1}\cdot\binom{20}{1} \equal{} \boxed{1040}$.

So the answer is $ \mathbf{(A)}$.
[/hide]

If $ r_i$ is a root of $ p(x)\equal{}(1\minus{}x)^5(1\plus{}x)^{25}$ for $ 1\leq i\leq 30$ and $ i$ is an integer, then compute

$ \sum_{a\neq b\neq c}\frac{1}{r_ar_br_c}$.

[quote="mathwizarddude"][quote="1=2"]If $ r_i$ is a root of $ p(x) \equal{} (1 \minus{} x)^5(1 \plus{} x)^{25}$ for $ 1\leq i\leq 30$ and $ i$ is an integer, then compute

$ \sum_{a\neq b\neq c}\frac {1}{r_ar_br_c}$.[/quote]
What theorem is this?[/quote]It's an application of Vieta's formulas on reciprocal roots.

Using the above argument, the answer is still $ 25\binom{5}{2} \plus{} \binom{25}{3} \minus{} \binom{5}{3} \minus{} 5\binom{25}{2}$
although I derived this expression using a different argument in my solution (considering the possible combinations of $ x^3$ from the two expressions). :idea:  :D