I'm starting this thread for people to discuss the C of C math meet, if they wish to do so.  I thought it was an excellent meet as usual.  There was some really good competition there.  Congrats to everyone.

Also, I wanted to share my math marathon solutions with everyone.  They usually don't have official solutions for the math meet, so I thought that posting my solutions might be good for future reference.

lol, I now have three copies of Count Down....hmm...

about the written test... , um..yeah, that's all I have to say :mad: The test was a lot harder than the past two and I don't understand how you can possibly finish all the problems in 60 minutes! *keeps on making excuses and hitting his head on the wall :wallbash_red: but realizing that there is still Furman and Coastal Carolina*

Oh and btw Joe, congradulations on winning two or three awards..

And to Asif, congrats on the written test

hmm i never decided to consider the case where b = 0 bc i got an answer without it, and assumed the answer would be distinct. whoops.  ;)

also joe, i assumed that the k could be factored out of the solution for number 3, since i assumed that f(x) was reduced already - i suppose that would be a valid assumption.

i did basically the same thing as you for number five, but it was mostly bs, even though it pretty much said the same thing as yours. im not entirely sure if it was right..oh well.  :D

and yeah. wow, the level 3 test was pretty hard. even though i was third i think the score was pretty bad  ;)

[quote="solafidefarms"]You guys going to Furman too? See ya'll there![/quote]

Yeah, I'll be going.  Asif, are you going to register and go as well?

[quote="solafidefarms"]You guys going to Furman too? See ya'll there![/quote]

solafidefarms, what is your name, because I"ll be going too..

yeah, we should be registering within the next one or two days.

Billy Dorminy; I'm from GA. (Otherwise known as the stupidest state in the South :D)

haha i dont know what you're talking about, SC is known as the stupidest state in the south  :P

when i went to national mathcounts, the team motto we said during the team roll call was, "South Carolina: Where the average IQ is lower than the average temperature."  :lol:

$\text{GA 5 USAMO Qualfiers}$ and $\text{SC 1 USAMO Qualifiers}$ in $2005$ $\Longrightarrow GA>SC$ :rotfl:

yea, the written test was definately hard and too short. heck, i didn't even have time to guess on the ones i didn't know.  :P
we're happy to get 3rd in lvl 3 with "average/decent" scores from joe, brian, and I and having to take a fill-in who probably scored around a 25 :?
what suprises us is that we made the finals in varsity elimation. 21 points just doesn't seem like it's the best score. i know its supposed to be difficult but with all the competition you'd think others would be better. o well - i'm not complaining.
i'm looking forward to the remaining competitions

how did you guys do on the first round, like 40 or something, because the guy said that Summerville was the clearcut winner and we came in second with only 29 :? (and two guys got 27 on our team so I'm dissapointed because we could've pulled off an upset against you guys if ALL four of members in our team did well... :( )

Yeah, we got 40.  (Brian-15, David-10, William-5, Me-10)

In round 2 in retrospect we could have done much better.  I mean, I put -4/10 as an answer...that's 5 points right there.  Brian only got 2 points but David pulled through for us with a 10 :) (I think Brian could have used a red bull before round 2 :P )

you think the $\frac{-4}{10}$ is a mistake, try:

I think the question was this/or similar to this:

$(x+\frac{1}{4})^2+(x+\frac{1}{4})^2+(x+\frac{1}{4})^2+(x+\frac{1}{4})^2=0$, find the x value of the minimum.

I put $-2$ :fool: and somehow completely forgot about dividing by $4$ :read:

That could've given me $10$ points on round 2 and given our team a chance with $17$...I'm sure we could've gotten close to $20$ if we worked faster too...., but oh well..

Why did the written test get progressively easier??? That serves pretty well to confuse people! I really want to see what i got though; I figured I got over 100, but I must have only gotten 80s or below.

It was a close elimination--you looked like the clear winners for a while. Besides, Hoggard's never won anything. Great place to start, I guess.

[quote="jb05"]It was a close elimination--you looked like the clear winners for a while. Besides, Hoggard's never won anything. Great place to start, I guess.[/quote]

They started getting into wordy problems and...meh.  Anyway, nice job.  You were your team's captain (the one sitting closest to me), right?  It was cool seeing you and, heh, it wasn't until I saw your name on the projection at the award ceremony that I realized I had seen you around AoPS.  It was a fun elimination and I was really glad just to be up on the stage :)

The captain was asian, so I was the other one, but I think you probably just mixed up who was captain.

I still can't figure out why my school chose to go to CofC... I'm from Glenbrook High, the random school from Illinois. I thought that CofC was really quite well run -- the variety of the sprints and such was very cool. And the weather was so.... warm! (haha, it gets cold up here in Illinois.)


geometry
circumcircle
perpendicular bisector
geometry solved

How can you construct a triangle with given three element of same vertical? I mean can you construct a triangle with h_a,m_a,n_a(bisector)?

Make a circle $\omega$ with center in the midpoint of m_a. Chose the a diameter AM and construct the segment AH=h_a, with $H \in \omega$. Construct AD=n_a, with $D \in \bar{HM}$. Let $P=AD%Error. "capr" is a bad command.
$ where r is the perpendicular bisector of $\bar{BC}$. So the circuncenter O is $O=r%Error. "caps" is a bad command.
$ where s is the perpendicular bisector of $\bar{AP}$. Now contruct the circuncircle of ABC and find B and C.


I'm guessing the three beggars are unique so they can be chosen in $ \binom{3}{2}$

Assumption: Each beggar gets atleast 1 blanket

Ans is $ 20 \binom{3}{2} \equal{} 60$

no  that's not the right answer also some of them can be left with no blankets and yes each of them is unique...
also use the fact that some two of them must get same number of blankets

Yeah I used that fact that da.

Its like this.. Choose any number from 0-10. Give that many blankets to the two beggars each resp. And automatically remaining blankets go to the third beggar. Except (7,7,7).

$ 10 \binom{3}{2}$ = 30.

it is same as the no. of non-negative integral solutions to $ X_1\plus{}X_2\equal{}21$ subject to the condition that $ \frac {X_1}{2}\ne X_2$ wich is 21; this is is becoz 21 is an odd number and hence every solution $ (X_1,X_2)\equiv (0,1) (mod 2)$ and hence on of the 2 is always an even number.

It should be 30 .

We can get the answer by simply listing down the ways .
$ (0 , 0 , 21)$
$ ( 1,1,19)$
 $ ....$
$ (10 , 10 , 1)$

Here $ (7,7,7)$  is not included. Each of these can be permuted in 3 ways.

In the solution given by hell_ever, if $ X_1 \equal{} 5 \plus{} 11$ and $ X_2 \equal{} 5$ , it is still a solution but it is not counted .( i'm not sure if i understood that method rightly)

(0,0,21) (1,1,19) (2,2,17) (3,3,15) (4,4,13) (5,5,11) (6,6,9) , (8,8,5) (9,9,3) (10,10,1)

are the only tuples possible. But since ordering becomes important here. Each can be arranged in 3 ways which make it 10x3 = 30.

What is wrong with my logic?

Edit: I didn't copy bruteforce.. He posted exactly 1 minute ahead of me while I was typing.

u r rite shreyas and bruteforce. My mistake.
i actually tried listing out the solutions using my method and the pattern was something like
(0,0,21), (0,21,0), (1,10,10), (10,1,10),....
basically i got all tuples but only 2 out of 3 possible permutaions were appearing for each tuple. hence i must add the remaining 9 distinct solutions so that the answer is 30

[quote="pardesi"]i had the same 'mistake' as hel_ever but the answer given is 21 :)[/quote]

What book da? YG File nu solladha?

RSM(2 yr version ) Part 3  Combination and Permutations pg 18 exercise 5 q)2


Let $ABC$ be a triangle . $M$ is inside $\triangle ABC$ . $x,y,z$ are three positive reals . Prove that :
 $x\cdot MB\cdot MC+y\cdot MA\cdot MC+z\cdot MA\cdot MB \geq \min\left\{xbc , yac, zab\right\}$
  Have you seen it before ?

Let $m = \frac{x}{a}, n = \frac{y}{b}, p = \frac{z}{c}$. Then the problem is equivalent to $m\cdot \frac{MB\cdot MC}{bc}+n\cdot \frac{MC\cdot MA}{ca}+p\cdot \frac{MA\cdot MB}{ab}\geq min(m, n, p)$
 This is true since $\sum \frac{MB\cdot MC}{bc}\geq 1$.
 Or am I mistaken somewhere?


AMC
AIME
trigonometry
number theory
relatively prime
algebra
difference of squares

In triangle $ABC$, angle $C$ is a right angle and the altitude from $C$ meets $\overline{AB}$ at $D$. The lengths of the sides of $\triangle ABC$ are integers, $BD = 29^3$, and $\cos B = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Denote the triangle sides a = BC, b = CA, c = AB, $a^2 + b^2 = c^2$. From the similar right angle triangles $\triangle ABC \sim \triangle CBD$, $BD = \frac{a^2}{c},\ c = \frac{a^2}{BD} = \frac{a^2}{29^3}$. Since c is an integer, $a = k \cdot 29^2$, where k is an integer. Then $c = k^2 \cdot 29$ and

$b^2 = c^2 - a^2 = k^4 \cdot 29^2 - k^2 \cdot 29^4 = k^2 \cdot 29^2 (k^2 - 29^2)$

Therefore, $k^2 - 29^2$ must be a full square. In other words, we are looking for a Pythagorean triple (i, j, k) with one leg equal to i = 29. This Pythagorean triple is generated by integers p > q, such that

$i = p^2 - q^2,\ \ j = 2pq,\ \ k = p^2 + q^2$

One of the legs must be even (2pq), that's why we can put $i = p^2 - q^2 = 29$. Every odd number 2q + 1 can be expressed as a difference of squares of 2 consecutive integers

$(q + 1)^2 - q^2 = 2q + 1$

Therefore, 2q + 1 = 29, q = 14, p = q + 1 = 15. Now we can generate the Pythagorean triple (i, j, k):

$(i, j, k) = (p^2 - q^2,\ 2pq,\ p^2 + q^2) = (29, 420, 421)$

Since $\cos \widehat B = \frac{a}{c} = \frac{k \cdot 29^2}{k^2 \cdot 29} = \frac{29}{k}$, knowing k, we get $\cos \widehat B = \frac{29}{421} = \frac m n$. So the answer is m + n = 29 + 421 = 450.

This is pretty easy for a problem on the 1994 AIME... especially #10. Anyway:

Use similar right triangles to get $BC = 29^2k$, $AC = 29k\sqrt{k^2-29^2}$, and $AB = 29k^2$. We know $29^2|BC$ because we can relate $AD$ in terms of $BC$, like so - $AD = \frac{BC^2}{29^3}-29^3 \Rightarrow 29^2|BC$.

So then $k^2-29^2$ is a square, so $k = \frac{29^2+1}{2} = 421$ (generating a Pythagorean triple with smallest leg $29$).

Then $\cos{\angle B} = \frac{BC}{BA} = \frac{29}{k} = \frac{29}{421} \Rightarrow 29+421 = 450$.


floor function
advanced fields
advanced fields unsolved

prove that the set $ \mathbb{D}\equal{} \frac{m}{10^p}$ is dense in $ \mathbb{R}$ where $ m$ is an integer and $ p$ is positive integer .

physical set definition ...

mathematical: $ \mathbb{D} \equal{} \{m/10^p : m \text{~integer~}, p \text{~positive integer~}\}$

this is dense since every real number $ x$ is the limit of $ \lfloor 10^p x \rfloor / 10^p$.


What follows may sound silly because I am not a native English speaker.

I was considering in order a book from an online publisher (Springer) and one of the methods of payment is "Bill me". What exactly is this? Thanks in advance.

I think that it basically is a way to pay with your credit card.
Correct me if I'm wrong; I'm only 13.

Without context it's hard to be certain, but my first guess is that it means "send me a bill" (which would be useful if, for example, you had to pay by cheque rather than credit card).

Is it just "Bill me"? I know that many online stores are adding an option called [url=https://www.billmelater.com/index.xhtml]Bill Me Later[/url], though to me it sounds like the type of thing a clothing store would use (or some other store that wants people to overspend).

Yes, it is just bill me. You can see on the Springer internet site, where there are three options: Bill Me, Credit card, and Bank account.


calculus
integration
real analysis
function
real analysis unsolved

Let $f: [a,b]\rightarrow R$ be an integrable function. Suppose that $g: [a,b]\rightarrow R$ and there is a finite set $S\subset[a.b]$ such that $f(x)=g(x)$ if $x\notin S$

Show that $g$ is integrable and that its integral equals to integral of $f$

what do u mean by "Depends what integral you are talking about and how much you know for each of them. In both cases the result holds anyhow."??

Riemann integral? Is that what you tried to say? Or something else?

If you are considering the Riemann integral, you can say that $f$ is continuous a.e because Riemann-integrable, and since $S$ is finite, $g$ will be continuous a.e as well, whence Riemann-integrable, and with same integral (because Riemann and Lebesgue integrals coincide on the segment, and the Lebesgue-integral of $g$ is clearly that of $f$).
You may find this answer unsatisfying if you are not aware of a certain number of facts, in which case you need to go with the brute-force work on the subdivisions.

Errr.... I still dont get it. Need more explanation, please.
Thx.

This is an interesting question, I'd like to know if it's deal to the Riemann Integrable function.
Can anyone show it how? Thx in advance.

Quoting the theorem about a.e. continuity is overkill. You don't need that.

Choose a partition that isolates each of the exceptional points in a very thin interval - say the sum of the lengths of all $n$ of those subintervals (where we have $n$ points at which $f$ differs from $g$) is $\delta.$ (We'll see what we need $\delta$ to be.)

Now find a refinement of this partition for which the upper and lower sums for the integral of $f$ differ by less than $\frac{\epsilon}2.$

$f$ and $g$ are both bounded functions. $f$ is bounded because it's Riemann integrable. As bounds for $g$ we can use whichever is larger of the bounds for $f$ and of the finitely many values of $g$ at the points at which it is different. Suppose $|f(x)|\le M$ and $|g(x)\le M$ for all $x.$ Then the contribution of the $n$ subintervals to the upper sum for $g$ is no more than $\delta M$ and the contribution of the same subintervals to the lower sum for $g$ is no less than $-\delta M.$ These differ by at most $2\delta M.$ Go back to the beginning of the proof and choose $\delta=\frac{\epsilon}{4M.}.$ But this all together and we have a partition on which the upper and lower sum for $g$ differ by less than $\epsilon.$ This makes $g$ integrable. Furthermore, if we examine the proof, we'll see that the upper sum for $g$ differs from the upper sum for $f$ by no more than $\frac{\epsilon}2$, and similarly for the lower sums. Thus the integral of $g$ is the same of the integral of $f.$

This proof did very much use that the exceptional set was finite. For functions that differ on an infinite set, it matters how those points are arranged. For instance, the set $\left\{\frac1n: n\in\mathbb{N}\right\}$ and the set of rational numbers in $[0,1]$ are both countably infinite sets. However, the function which is $1$ at $\frac1n$ and zero otherwise is Riemann integrable (with integral zero), while the function which is $1$ at every rational and zero otherwise is not Riemann integrable.


trigonometry
LaTeX
geometry
geometric transformation
reflection
power of a point
radical axis

These are the first and second problems of RMO India 2005.(in the attachment)
solve them.

#2:  [hide]If we call the 6 integers a,b,c,d,e,f, then without a loss of generatlity, assume that a is opposite e, b is opposite f, and c is opposite d.  $abc+abd+bec+bed+efc+efd+fac+fad=2004.$  This can be factored into $(a+e)(b+f)(c+d)$.  2004 has prime factors of $2^2*3*167$.  The only possibilities other than this one is where the extra 2 is multiplied somewhere else.  This yields 2,6,167, 2,3,334 and 2,2,501.  So the T can be 174, 339, 175, 505.[/hide]

solution for #1

[hide]

[b]Notation[/b]: let the perpendicular touch line $L$ at $H$, and the point of tangency $T$

[b]Lemma 1[/b]: $PH<PT$

the way that i proved it was make it on a cooridinate grid, the circle is $x^2+y^2=1$, and $T =(\cos b, \sin b)$, (where $b$ is the angle that $OT$ makes with the postive $x-axis$) and $-90<b<90$ because we only need to have those angles because if $b=\pm 90$, then the tangent from $T$ and line $L$ are parallel, and if $b>90$, then the tangent goes on the other side and we can just flip the figure because it is symmetric

and line $L$ is $y=a$ with $a>1$

so we find the tangent to be $y-\sinh=-\frac{\cos b(x-\cos b)}{\sin b}$, and it intersects line $L$ at
$x=-\frac{\sin b (a-\sin b)}{\cos b }+ cos b$, and $y=a$

so $PH=-\frac{\sin b (a-\sin b)}{\cos b} + \cos b$

and we find $PT = \frac{a-\sin b}{\cos b}$

so we need to show that $PT>PH$, so after a bit of simplification, we need to show

$\cos b + \frac{(1+\sin b)(a-\sin b)}{(\cos b)} + \cos b > 0$

this is true because everything term is positve

we are only considering the range $-90<b<90$, so $\cos b$ is positive in that range, so only consider the first 2 terms

$-1<\sin b<1$ in the interval, so $0<\sin b +1<2$, and since $a>1$, $a-\sin b>0$
so all of the steps are reversable, and we have $PT>PH$

[b]Lemma 2[/b]:  $PT<PO$

because $\angle OTP$ is a right triangle, the hypotenuse $PO$ is longer than one of the legs of that triangle $PT$

[b]The Proof[/b]

consider the locus of points $K$ such that $PT=PK$, clearly this is a circle (call this circle $K$), and if this intersects $OH$, we are done because we have a desired point on $OH$

consider when $K$ intersects line $L$, say it intersects at $A$ where $PA=PT$, since $PA=PH+HA$, and $PA>PH$, then $AH>0$, so $A$ lies beyond line $OH$ in respect to $P$

draw $OP$, when $K$ intersects $OP$, call this intersection $B$, so $OP=OB+PB$ (with $B$ in the middle), so $OB>0$, therefore $OB$ is positive when it is $B$ is between $OP$, so it must be positive, so $B$ lies on the same side of like $OH$ as $P$ does

the locus $K$ has a point on both sides of line $OH$, the arc that connects $AB$ must cross $OH$ because it is a circle $K$ is continuous (this is the same idea as the intermediate value theorem)

$\mathbb{QED}$

[/hide]

EDIT: I added Latex so it is more readable.

Thanks for the solutions.

Altheman, what do you mean by " Lemma ". :?:

A lemma is an auxiliary theorem that you prove first in order to use in your solution to the original problem

[b]The first method.[/b] The line $l$ is the radical axis between the circle $\Gamma=C(O,R)$ and the point $Q$ (a circle with the null radius). I note $S\in l$, $OS\perp l$.We construct a tangent $ST$ to the circle $\Gamma$ ($T\in \Gamma$). Then $SQ=ST$, $S\in (OQ)$.
[b]The second method.[/b] I note $\{M,N\}\subset OS\cap \Gamma$ and the harmonic conjugate point $R$ of the point $S$ w.r.t. the points $M$, $N$. Then the reflection of the point $R$ w.r.t. the point $S$ is the in demand point $Q$.


analytic geometry
combinatorics unsolved
combinatorics

Only know is just this name.
Please tell me more about it
thanks

Lexicographic ordering essentially means sorting by the first coordinate, with the next coordinate used as 1st tiebreaker, then the coordinate after that as second tiebreaker, and so on.

In other words $(a_1, a_2, \dots a_n) <_{lex} (b_1, b_2, \dots b_n)$ iff there is some $k$ such that $a_j=b_j$ for $j<k$ and $a_k<b_k$.

For example, the 24 permutations of $\{1, 2, 3, 4 \}$ are ordered lexicographically as

\[ 1234<1243<1324<1342<1423<1432<2134<2143<2314< \dots <4231<4312<4321 \]

Ah,thanks
but if i want to get a given one in the Lexicographic ordering
such as the 15th one in the permutations of {1,2,3,4}
how?


Hi there,
I have another problem which I can't solve! Thanks to whoever solves it (and to whoever tries!)

[b]Let A and B be n*p matrices. Show that ther is a p*p matrix D with B = AD exactly when C(B) is contained in C(A)[/b] note: C(B) referes to Columb Space of B, and C(A) refers to Column Space of A.

It is somehow trivial. Just note that the columns of $C$ are the linear combinations of columns of $A$,where$C=AB$.


Bobby Fischer invented chessX, where X is a positive integer. He called it this because it was a variation of chess in which there were X starting positions. His objective was to create a game in which one would have to think about his or her opening, as opposed to memorizing common attacks and defenses (starting to sound like problem solving?). Setup was as follows: both white and black have symmetric first file-setups and all 8 pawns are in the second file. The only conditions for the first file are that the bishops must be on opposite squares, and the king must be in between the two rooks.

Find X.

im confused. would it be $ 8!/(2!^{3})$? which would be 7! which is 5040?


If charlies utility function is (x^4)(y), the price of apples is Pa, and the price of bananas is Pb, and his income is m, then charlies demand for apples is....

So I got....

U(x,y)= (x^4)(y)

MUx= 4x^3
MUy= 1

MRS= MUx/MUy
MRS= 4x^3

MRS= Pa/Pb

Pa/Pb= 4x^3

Plug into budget line

m= PaX+PbY

to solve for X

Im pretty sure I took the MRS wrong, becuase you end up with having to take the third root of something and thats not right. I thought MUx and MUy were just partial derivs of the utility function?

Any advice?


Let a,b,c are positive intergers such that $ a\plus{}b\plus{}c\equal{}3$.Prove the inequality:
$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}3abc \geq \sum \frac{ b(3c\minus{}a^2)(a\plus{}b)}{2}$

I want to see nice solutions from MathLinkers. :P

The inequality is equivalent to (All sums are cyclic)
\[ \qquad 2\sum a^2b^2 + 6abc\ge \sum b(3c - a^2)(a + b) = 9abc - \sum a^3b + 3\sum b^2c - \sum a^2b^2 \\
\iff 3\sum a^2b^2 + \sum a^3b\ge (a + b + c)(a^2b + b^2c + c^2a) + abc(a + b + c) \\
\iff 2\sum a^2b^2\ge \sum 2abc(a + b + c) \\
\iff \sum a^2(b - c)^2\ge 0
\]
which is obviously true.

EDIT: Another solution:
\begin{align*}2\sum a^2b^2 + 6abc & \ge \sum b(3c - a^2)(a + b) \\
\iff 2\sum a^2b^2 + 2abc(a + b + c) & \ge \sum b(a + b)(c(a + b + c) - a^2) \\
& = \sum b(a + b)(c(a + b) + c^2 - a^2) \\
& = \sum bc(a + b)^2 + \sum (a + b)(bc^2 - a^2b) \\
& = \sum bc(a + b)^2 + abc(a + b + c) - \sum a^3b + \sum b^2c^2 - \sum a^2b^2 \\
& = \sum bc(a + b)^2 + abc(a + b + c) - \sum a^3b \end{align*}
Hence it remains to show that
\begin{align*} & 2\sum a^2b^2 + abc(a + b + c) + \sum a^3b\ge \sum bc(a + b)^2 \\
\iff & 2\sum a^2b^2 + \sum a^3b + abc(a + b + c)\ge \sum a^2bc + 2\sum ab^2c + \sum b^3c \\
\iff & 2\sum a^2b^2\ge \sum 2abc(a + b + c) \\
\iff & \sum a^2(b - c)^2\ge 0\end{align*}
obviously true.

Oh I wasn't think about that way.To made the inequality I use Schur. :huh: May be it's not hard because I don't have experience in making inequalities.(I'm just in grade 9)


Let the vertex of each curve be $\pm a$ and the two foci be $\pm c$

How can one show that the directrix, $d$, is equal to $\frac{a^2}{c}$ and the eccentricity ($pf: pd$, where p is a point on the conic section) is equal to the $\frac{c}{a}$?

[hide="bad approach?"]This is how far I got:

$e = \frac{c-a}{a-d} = \frac{b^2}{ac}$

I got the second part of the equality by using the equation for y of a hyperbola centered at origin.

Since $y = \pm b\sqrt{x^2 - a^2}$, then the distance of point $(c, 0)$ to the conic section must be $\pm b\sqrt{c^2 - a^2} = \pm b^2$, or just $b^2$ as I'll stick to the right side directrix.

However, I am sort of stuck at this point and have no idea how to isolate the d and simplify it to equal $\frac{a^2}{c}$[/hide]

[hide="**EDIT: simple proof**"] Here's a simple derivation I found:

$\frac{a-c}{d-a} = e$
$a-c = de-ae$

$\frac{a+c}{d+a} = e$
$a+c = de + ae$

Combining equations gives:
$2c = 2ae$
$2a = 2de$

$\frac{c}{a} = e$
$\frac{a}{e} = d = \frac{a^2}{c}$

$QED$

This derivation proves also that ellipses and hyperbolas share the same relationship of variables to the value of directrix and eccentricity.

A corollary of this is that since $a>c$ in an ellipses, $e<1$. Similarly, since $a<c$ in hyperbolas, $e>1$
Circle's directrix is infinitely far from the conic section, and so $e$ is $0$. Now it is very clear that eccentricity signifies a conic's deviation from a circle.[/hide]

Does a hyperbola have a directrix/directrices?

Yes, both hyperbolas and ellipses have 2 directrices, one for each foci.

Wait...how?
What properties does that directrix have?

A directix is a line perpendicular to the axis containing the foci of a conic section. The distance PF is proportional to the distance PD, where P is a point on the conic, D is the directrix, and F is the foci. As you can see there are 2 foci in ellipses and hyperbolas, thus two directrices exist.


Adam buys oranges at the rate of three oranges for \$1. He sells them at the rate of five oranges for \$2. In order to make a profit of at least \$10, what is the minimum number of oranges he must sell?

150. 150 oranges will cost him 50 dollars, and selling them will earn him 60 dollars, or 10 dollars profit.

3 oranges for 1 dollar = 15 oranges for 5 dollars
5 oranges for 2 dollars = 15 oranges for 6 dollars

so, you earn 1 dollar of profit every moment you sell 15 oranges.

therefore, if you want to earn 10 dollar of profit, you must sell 15*10=150 oranges.

Answer : 150

Every time you sell 15 oranges you make 1 dollar profit.


Chunky Nuggets dry dog food is 6 parts meat and 3 parts filler. How many pounds of filler are in 50 pounds of the food? Express your answer rounded to the nearest pound.

There are $ 3 \plus{} 6 \equal{} 9$ total parts, for the denominator of our filler ratio. $ \frac {3}{9} \equal{} \frac {1}{3}$.
$ \frac {1}{3}(50)$ rounded to the nearest pound is $ \fbox{17}$ pounds.


Given four positive reals $a,x,y,z$ satisfying $\frac{x}{a}=\frac{a+z}{a+y}$, $\frac{y}{a}=\frac{a+x}{a+z}$ and $\frac{z}{a}=\frac{a+y}{a+x}$, show that $x=y=z=a$.

This was part of a solution to an Irish MO problem from this year which I constucted.  My own proof is a bit messy, and I'd like to see a neater approach if someone has one.

[hide="My Solution"]
Multiplying the three equations together gives $xyz=a^3$.

Cross multiplying and then adding gives $xy+yz+zx=3a^2$.

Finally, we have $\frac{a}{x}+\frac{x}{a}+\frac{a}{y}+\frac{y}{a}+\frac{a}{z}+\frac{z}{a}=\frac{a+z}{a+y}+\frac{a+y}{a+z}+\frac{a+x}{a+z}+\frac{a+z}{a+x}+\frac{a+y}{a+x}+\frac{a+x}{a+y}$.

If we expand this, make the substitutions $xyz=a^3$, $xy+yz+zx=3a^2$ and $x+y+z=pa$ and simplify, we get $p+3=\frac{2p^2+9p+3}{p+5}\Rightarrow p=3$ (since $p>0$).

So $x,y,z$ are some permutation of the roots of the equation $(t-a)^3=0$, and the result follows.
[/hide]

We consider the equivalent set of equations (by dividing by $a$)

\[x = \frac{1+z}{1+y}, y = \frac{1+x}{1+z}, z = \frac{1+y}{1+x}\]

Then we have $xyz = 1$.
If any 2 are equal, for example $x = y$, then $z = 1$ and we have $x=y=z=1$.

So now we consider if all 3 are distinct. WLOG, assume $x$ is greatest, so $x>y,  z, x>1$.

From the second equation, $y > 1$, so $z < 1$, otherwise $xyz > 1$.
But then from the first equation, we have $x < 1$, contradiction.


trigonometry
inequalities unsolved
inequalities

a)$\forall x\in(0,1]$ we have
$\dfrac{\sin\pi x}{\pi x}\ge\dfrac{1-x^2}{1+x^2}\cdot[1+(x-x^2)^3]$
b)$\forall x\in[1,2]$ we have
$(\sin\dfrac{\pi}{x})^{x}\ge\sin\dfrac{\pi (x-1)}{2}\cdot(1+\dfrac{(x-2)^2}{2})$
c)$\forall x\in(0.\dfrac{\pi}{2})$we have
$(\dfrac{\sin x}{x})^2+\dfrac{\tan x}{x}\ge2+\dfrac{37}{180} x^3\tan x$

I tried solving them but I didn't suceed.

It's American Mathematical Monthly magazine. ;)


[quote="orl"]Problems with solutions and results for the 2nd Romanian Master of Mathematics are appended.[/quote]

The 2010 Romanian Master of Mathematics took place during the last week.
Can anybody post the problems?
(It seems that they moved from a 1day/4problems to a 2days/6problems IMO-style format.)


calculus
derivative
calculus computations

Find a formula for the nth derivative:

y = 1 / [1 + x]

Just calculate a few of them, and look for a pattern.

Generally,
$ \frac{d^{n}}{dx^{n}}\frac{1}{(ax\plus{}b)^m} \equal{} \frac{(\minus{}1)^n (m\plus{}n\minus{}1)!\cdot a^{n}}{(ax\plus{}b)^{m\plus{}n}(m\minus{}1)!}.$


function
ratio
limit
floor function
probability
real analysis
real analysis unsolved

Find a function $ f: [0,1] \to [0,1]$ such that
for each $ I\subseteq[0,1], f(I) \equal{} [0.1].$

I think I have a solution, but not a rigorous proof (yet).

Well, if the problem does not require a 'constructive' one, then we may use AC to find one. (Actually, I wonder if such function can be constructed without AC.)

For $ x, y, \in [0, 1]$, define $ x \sim y \Leftrightarrow x \minus{} y \in \mathbb{Q}$. Then $ \sim$ is an equivalence relation. By Axiom of Choice, one can choose a set of representatives $ R \equal{} \{ q_\alpha \}_{\alpha \in A} \subset [0, 1]$ for the set of equivalence classes $ [0, 1] / \sim$.

Clearly, $ R \times \mathbb{N}$ is equipotent to $ [0, 1]$, so $ |R| \equal{} \left|[0, 1]\right|$. This means that there is a bijection $ \varphi$ from $ R$ to $ [0, 1]$.

Now define $ f : [0, 1] \to [0, 1]$ by $ f(x) \equal{} \varphi(q_\alpha)$ if $ x \sim q_\alpha$. Clearly $ f$ is well defined.

Suppose $ \varnothing \neq (a, b) \subset [0, 1]$. If $ r \in [0, 1]$, then $ f^{ \minus{} 1}(r) \equal{} [\varphi^{ \minus{} 1}(r)] \equal{} \{ x \in [0, 1] \, | \, x \sim \varphi^{ \minus{} 1}(r) \}$ is dense in $ [0, 1]$.  So $ (a, b) \cap f^{ \minus{} 1}(r)$ is nonempty and $ f((a, b))$ contains $ r$. But $ r$ is arbitrary, so $ f((a, b)) \equal{} [0, 1]$.

The previous post nicely proved that such a function exists, as counter-intuitive as it might be.  Now for an example of just such a crazy function.

Let $ r(x)$ be the ratio of the number of twos to the number of ones in the decimal expansion of $ x$.  More formally, we could say $ T_n(x)$ = the number of twos in the first n digits of x.
$ O_n(x)$ the number of ones in the first n digits of x.
And, if there are no ones in the first n digits of x, define $ O_n(x)$ to be 1.
Then  $ r(x) \equal{} \lim_{n\to\infty}\frac{T_n(x)}{O_n(x)}$
It is certainly possible for this ratio to increase without bound, so $ r(x)$ is not defined for all x in [0,1].
So we define $ f(x) \equal{} r(x)$ if $ r(x)$ exists and $ r(x)\le1$ and 0 otherwise.

Now, given any interval, I = (a,b) contained in [0,1], and given any $ y$ in [0,1] we need to show that we can always find an $ x, a<x<b$ with $ f(x) \equal{} y$
Say a and b agree in the first n digits of their decimal expansion and
Let $ x_0$ be (b-a)/2, rounded to n+1 digits.
Define $ q$ by $ q\equal{}d_0d_1d_2 \cdots d_n$ where $ d_k$ is the kth digit.
And $ d_k \equal{} \lfloor ny \rfloor \minus{} \lfloor (n\minus{}1)y \rfloor \plus{} 1$
Now, $ r(q) \equal{} y$ (proof?) and since the first n digits will not affect the limit, we can say that $ x \equal{} x_0 \plus{} 10^{\minus{}n\minus{}1}q$ is a desired $ x, a<x<b$, with $ f(x) \equal{} y$

Details: We ask for a function with range $ [0,1]$, not the whole positive axis. The ratio $ \frac{O(n)}{n}$ would work better.
Also, increasing without bound isn't the problem; it's about erratic oscillations. The limit frequently doesn't exist.
With probability $ 1$, the limit exists and is equal to the "average" value: $ 1$ for your original formulation, $ \frac1b$ where $ b$ is the base in mine.


Two numbers are successively selected at random and with replacement from the set {1,2,....,100}. What is the probability that the first one is greater than the second?

We can examine case-wise:

let n be the first and m be the second number chosen

if $n=1$, $p(m>n)=99/99$
if $n=2$, $p(m>n)=98/99$
if $n=3$, $p(m>n)=97/99$
$\vdots$
if $n=x$, $p(m>n)=\frac{100-x}{99}$

The probability of each of these 100 cases occuring is 1/100, so the probability is $\frac{1}{100} \cdot \sum_{n=1}^{100} \frac{100-n}{99}=\frac 1 2$
(incidentally, the probability is 1/2 regardless of how many numbers you choose from)

(Note/Edit: this is actually the solution to the problem without replacement)

Oops.  Saw: "with replacement"  Read: "without replacement"

[quote="towbomb"]
(incidentally, the probability is 1/2 regardless of how many numbers you choose from)[/quote]

What if we have 1 number?  What if we have 2 numbers?

[hide="solution"]if the second # is one, we can have 99 cases where the first one is greater.
if the second # is two, we can have 98 cases of such..

and if we keep going up by one, then we can find all the cases such that fist term is greater than second term.

total possibilities of that: $1+2+...+99 = 4950$

total possibilities: $100 \cdot 100 = 10000$

$\displaystyle \frac{4950}{10000} = \frac{99}{200}$[/hide]

[quote="blahblahblah"]What if we have 1 number? What if we have 2 numbers?[/quote]

If there are 2 numbers, the probability is still 1/2, because there is a 1/2 chance that you pick the smaller of the two.  If there is 1 number, the problem is trivial because you cannot select a second number.

[quote="Bictor717"]WITH replacement.
[hide]There are three cases: the first is greater; the second is greater; the numbers are the same. The first two cases have the same probability of occurring because each pair in the first case has a corresponding pair in the second case, i.e., its reverse. Since these are all possible cases, the sum of their probabilities is 1. To find the probability that the first is greater, subtract the probability of getting the same number from 1 and divide by 2.

$P\left(1^{\text{st}}\text{ number }> 2^{\text{nd}} \text{ number}\right)=\frac{1-\frac1{100}}{2}=\frac{99}{200}$

Also by this reasoning, you can quickly arrive at 1/2 for problems without replacement, as towbomb demonstrated above.[/hide][/quote]

i understand that 99/200 is the right answer (i did it $\frac{100C2}{10000}$), but why do you subtract $\frac1{100}$ for the prob that both are the same? ...and not 1/10000?

The first number you pick doesn't matter, so that's 1. Multiply by the probability that the second number you pick is the same as the first. There is only one such number out of 100, so 1/100.
Alternatively, 1/10000 is the probability that you pick a specific pair of numbers. There are 100 pairs of numbers such that the two number are the same. 100*1/10000=1/100


I'm new at LaTeX(that's not on the forum), so I was wondering how to get a space between two blocks of text, i.e. how to get

1. blah blah blah blah here's my solution blah blah blah blah blah blah blah blah blah blah blah blah

2. blah blah blah blah blah blah blah blah blah here's my solution blah blah blah blah blah blah blah

rather than

1. blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah
2. blah blah blah [...]

Thanks in advance!

You could always just throw in a \newline command.  So
[code]
This is on the first line.\\
\\
This is on the second line.[/code]
or
[code]
This is on the first line.
\newline
\newline
This is on the second line.[/code]
I tend to use \\ and \newline interchangeably.

Or \vskip 3mm or \vskip 5mm or vskip .33in or whatever - you need a number and a unit of measurement, and you can tweak the number until it looks like exactly what you want it to look like.

Hi,

I'm preferring the LaTeX-macro \vspace instead of \vskip - that's used internally by \vspace.

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


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $ A$ be a symmetric entries, all entries being polynomials $ f_{ij}(X_{1},X_{2},...,X_{m})$ with real coefficients in $ m$ variables. Suppose that when we choose real numbers $ x_{i}$, the matrix with coefficients $ f_{ij}(x_{1},x_{2},...,x_{m})$ is positive semidefinite. Show that $ A$ is a sum of squares of symmetric matrices with entries in $ \mathbb{R}[X_{1},X_{2},...,X_{m}]$.

Harazi, this problem has been already solved by Cristopher Hillar in a short paper from Prooc. of AMS. Some days ago, I have discussed a non-commutative version of Hilbert 17-th with prof. Florin Radulescu. :)
See here http://www.math.tamu.edu/~chillar/files/matrixhilbert17thfinal.pdf


Find $k$ such that $3^{k}\equiv 8 \bmod 65537$.

I immediately spot that
$65537 = F_{4}= 2^{2^{4}}+1$
I think that helps.

I spotted that too, of course.

I tried to solve it using the standard procedure, but, since $2$ is the only prime factor of $2^{16}$, it doesn't seem to work.


In a quadrilateral $ ABCD$,where $ AD\equal{}BC$;$ AC$ and $ BD$ meets $ (O)$ at $ O$ ,and the angle bisector of the angles $ DAB$ and $ CBA$ meets at $ I$.Prove that the midpoints of $ AB;CD$ and $ OI$ are collinear

To Mods: Please lock this topic, this is a problem from M&Y, issue 9, and we can't discuss it now.

To mods: please DELETE this topic. The problem was reposted in http://www.mathlinks.ro/Forum/viewtopic.php?t=167953 .

  darij


Hi,

I took biology as my freshman year science and this year, I'm taking chemistry. Since I'm in gifted student program, I can take any science as I want in this school now. People say that I should take Adv. Bio and Adv. Chem because in my school, physics is for seniors ONLY (of course, I'm an exception :D ). But I saw that many colleges DO look for physics and it looks like that many people in AoPS are taking physics in their junior or sophomore year. I'm pretty determined to take physics but I wonder what will be the best option (Physics vs. Adv. Bio and Adv. Chem).

Thanks!

It's kind of silly to reserve physics for seniors only. How is a few months' difference in age going to affect your understanding? Pretty much everyone at my high school took physics before senior year.
If you understand the reasoning and the teacher is decent enough to explain the concepts behind the formulas, no one should have a problem learning physics.
AP Physics C covers the material more deeply than high school physics, but not as many topics. You need to have the basics down solid to solve the problems, which are more than just substituting into formulas. The curve on the test is ridiculous, though. My class didn't finish the curriculum, so I left half of the free response blank, but I still managed a 5.

What makes Adv. Bio and Adv. Chem. different from their "regular" counterparts? Are they continuations or do they start at the beginning and do everything over in greater detail? I'm just curious.

Actually, over here in fairfax county it's a required 11th grade course.

I think I have all things needed to physics. I'm not sure how good the teacher is though. Everyone said that it's really boring (the class).

Adv. Chem and Adv. Bio.. Well.. I didn't take the class yet so I don't really know much but I think it's just biology and chem beyond. I heard Adv. Bio isn't a very good course (many students tell me that he doesn't teach well; he often times expect you to know by yourself but again, my school isn't really smart anyway). On the other hand, Adv. Chem I heard is an excellent course that goes beyond regular chem and helps you prepare to college level. In fact, some students came back after first semester and told the teacher that their chemistry in their college was mostly review.

I agree that taking physics ONLY in senior is bit stupid. I don't see why you shouldn't learn it early.

depending on what you might be looking for your college to see in you, you have a couple of choices..
you can take physics as a junior then adv chem as a senior- to show that you are familiar with all the sciences and did well in them.
or you can take adv chem junior year, especially if you take the AP test and do well on it, as that shows you're adept with college-level work.
also, is there a possibility to take more than one science class- that migh be somethign you can look into...

[quote="LordoftheMorons"]By advanced, do you mean AP? If so, take them, otherwise take physics.[/quote]

No. I don't think it's AP Chem although I'll ask my teachers if I can take AP exam by paying for it (according to what I hear, the school won't pay for me on science as they did for AMC for me  :D ).

Yes, I can take two science classes as I stated in the first post (Adv. Chem and Adv. Bio). I can't take two next year though.. I won't have any spaces (or maybe I can?? but I rather go with physics).

Thanks for advices. I'll go with physics. Anyone know a good physics book that I can look during summer? It has to be basic.  :roll:

I had to pay for all my APs, so it's not out of the ordinary. Try to get a fee waiver or some reduction.
In some places, students get paid to take APs. There was an article on the Times a few weeks ago about businesses that sponsor the tests and have lotteries for people who take them...
You can PM for the link, but you have to pay to read all of it, I think.

I would really look into what those particular programs are like at your school.  For example, if it happens that your school's chemistry teacher is excellent whereas the physics teacher isn't quite as good, I'd go with chem.  See which classes other people enjoyed more Be wary of who you ask though and ask them specific questions.  I know that many people at my school will say that a class is great if it's a piece of cake.  Of course, that's not my ideal class.  So continue to ask around.  Also consider which class [i]you[/i] think you will like the most.  I know plenty of people who loved European History at my school.  Nonetheless, history isn't my thing so I didn't take that class.

I took/will have taken AP Chem and AP Physics B in high school.  I chose these over biology.  Those two courses would still be my preference if I had to do high school again :D

Yes, my school has those similar views about class being "good." In my freshman year, I took biology over earth science (which was what most people took as freshman science but it's completely useless; everyone pass it because teacher basically tell you what to do etc..). I don't think there's any programs as far as science go. My school is in very rural area and it was me last year who "introduced" AMC to my high school.  :D  I'd like to compete in some science olympiads too but that's not my major concern. For Adv. Bio, people say that the class will be good for ME since I do extra studying every night ( :roll: ). But as for classroom teaching, I heard it's not very good so I'm not putting much idea on there. I heard Adv. Chem is good but hmm, I have two reasons for not taking this next year (I want to hear what you guys think about this). In college, I'm very positive that my first year science will be biology or chemistry so taking physics as senior year won't be such a good idea. Second, I'm going for early admission and I heard Adv. Chem is pretty tough to get 100 (that's how it is in Gen. Chem; I can get 99 but not 100! grr).

By the way, how useful are the AP credits? My school obviously offer only ONE AP class and that's AP Calc AB.

Physics is pretty cool,sorry i don't know the exact educ. system out there but physics is much better than bio and pretty much better than chemistry,at least fer me.

[quote="LordoftheMorons"]Physics isn't supposed to be that hard. AP physics, however, is supposed to be insane. Of course, this is based on information from upperclassmen at TJHSST (Best highschool in the nation.... at least according to wikipedia =p)[/quote]

I don't know how they do it over at TJ (presumably pretty well, looking at the number of USAPhO people last year-- but maybe that was because of the team meets), but physics isn't too hard, and AP definitely isn't impossible.  All you need to do is be comfortable with your math(which you seem very capable of), and read everything until you "get" it.  Quite a few people are taking Physics(first year) in Sophomore year, and some places do it for even freshmen.  Def. don't worry about the difficulty, and I would reccomend you take it  :P
Actually, why don't you try Physics+Adv. Bio?  (I'm biased against chem... haha, hate doing labs)  That might be fun.


OH THERE IS A CONF I MEANT $OG=OK'$[/quote]


What is the value of $ x$ in the equation $ (17^6\minus{}17^5)\div16\equal{}17^x$?

Can someone solve this and include an explanation?

[quote="Mewto55555"]Sure!

We factor out $ 17^5$ from the first part.

$ \frac{17^6\minus{}17^5}{16}\equal{}\frac{17^5(17\minus{}1)}{16}\equal{}\frac{17^5\cdot16}{16}\equal{}17^5$ so $ x\equal{}5$[/quote]
Mewto, can you tell me why $17^6-17^5\equal{} 17^5(17-1)$?
Thanks!
~Jigglypuff

Sure Jigglypuff!
$17^6-17^5= 17^5(17-1)$ because you can use distributive property and distribute the $17^5$. You can check that by factoring $17^5(17-1)$.


What is the greatest number of whole 4-inch by 6-inch notecards that can be cut from a 25-inch by 27-inch sheet of poster board if every cut must be made parallel to a side of the poster board? No taping or gluing is allowed.

The maximum number that can be made is the larger area divided by the smaller area, then flooring the expression.

\[ \left\lfloor\frac{27\cdot25}{6\cdot4}\right\rfloor\equal{}\boxed{28}
\]

[quote="hunter34"]The maximum number that can be made is the larger area divided by the smaller area, then flooring the expression.
\[ \left\lfloor\frac {27\cdot25}{6\cdot4}\right\rfloor \equal{} \boxed{28}
\]
[/quote]

This formula simply isn't true.  If I give you a strip of paper 1 inch wide and 24,000,000 inches long, the formula says:
$ \left\lfloor\frac {24,000,000}{24}\right\rfloor \equal{} 1,000,000$ notecards are available, which is nonsense. You can't get any notecards out of it.

Certainly, 28 is an [i]upper bound[/i] to the maximum possible, but it's not necessarily physically possible.

It looks like you just have to pack them in efficiently as possible. All the arrangements I tried yielded $ \boxed{24}$ with fairly little leftover.  Let me know if I missed a clever packing!

EDIT:  I would like to add that
$ \left\lfloor\frac {\text{long side of paper}}{\text{long side of card}}\right\rfloor\cdot\left\lfloor\frac {\text{short side of paper}}{\text{short side of card}}\right\rfloor$ is also a faulty formula.  For instance, a sheet of dimensions 22 by 12 can be evenly divided up into 11 cards, but that formula says 9.  This new formula gives a [i]lower bound[/i] to the number of cards available.  (Here the original formula "gave the right answer", but that is only a coincidence, since the cards fill up the sheet completely.)

Not everything has a formula you can mindlessly apply, so when answering this type of question, ask yourself if the physical shape of things is going to work!  Or else, you risk a lot of embarassing screwups in front of your boss!

I just did this problem, and got it wrong. The answer says it's $ \boxed{26}$

That's because you can take cards and put them sideways, they don't just have to be facing one way or the other.  Yes, the answer is $ \boxed{26}$.


If $\frac{\sin x+\sin y+\sin z}{\sin(x+y+z)} = \frac{\cos x+\cos y+\cos z}{\cos(x+y+z)} = 2$, what is the value of $\sin x\sin y+\sin y\sin z+\sin z\sin x$?

hello, i have not a complete proof yet but i think the searched sum has the value $ \minus{}\frac{3}{4}$.
Sonnhard.


modular arithmetic
function
algebra
polynomial
algebra unsolved

For a fixed $ d\in \mathbb{N}$, define a sequence $ \{ a_k\} _{k\in \mathbb{N}}$
\[ a_k \equal{} \sum_{j \equal{} 1}^d{a_{k \minus{} j}}\pmod{10} \ \ \ \ \ \text{for } k > d
\]
i.e. each term is the units digit of the sum of the previous $ d$ terms.  It is easy to see that the sequence must be periodic, i.e. there exists $ \tau \in \mathbb{N}$ so that $ a_k \equal{} a_{k \plus{} \tau }$ for all $ k\in \mathbb{N}$.  In general, we should expect that $ \tau$ depends on the initial values $ \{a_k\}_{1\leq k\leq d}$.

- What is a good upper bound for $ \tau$ as a function of $ d$? ($ 10^d$ is an obvious upper bound, but not very tight)
- What can be said about the number of different values $ \tau$ can take on?

$ \bmod 2$, the characteristic polynomial $ x^d \equal{} x^{d\minus{}1} \plus{} ... \plus{} 1$ is congruent to $ \frac{x^{d\plus{}1} \minus{} 1}{x \minus{} 1}$ and so $ \tau_2 | d \plus{} 1$.  When $ d \equal{} 2$, this is the familiar case of the Fibonacci numbers, which, $ \bmod 2$, do indeed repeat with a period of $ 3$.

$ \bmod 5$, the characteristic polynomial (henceforth referred to as $ P_d(x)$) has more complicated behavior.  For example, when $ d \equal{} 2$ we have $ P_2(x) \equiv x^2 \plus{} 4x \plus{} 4$, so the sequence itself has closed form $ a_n \equal{} (An \plus{} B) (\minus{}2)^n$.  The linear part repeats with a period of $ 5$ while the exponential part repeats with a period of $ 4$, so we expect the sequence to generically repeat with $ \tau_5 | 20$.  Unfortunately, the same situation cannot occur if, say, $ d > 20$, and so we should expect the roots of the characteristic equation to behave in more complciated ways.

Generically, I don't really know what happens.  The algebraic approach may not be best for establishing bounds in the regime where $ d$ becomes large and the modulus $ m$ is fixed; it's really much better suited to the case where $ d$ is fixed and $ m$ becomes large.


calculus
integration
conics
parabola
analytic geometry
geometry

number of integral points inside the parabola ,$ y^{2}\equal{}4ax$ and $ x^{2}\plus{}y^{2}\equal{}16$
by the way where is the coordinate geometry section

okay so, y^2=4ax and x^2+y^2=16

x^2+y^2=16 which is also y^2=16-x^2

x^2+y^2=16 is basically a circle with a center of (0,0) and radius of 4
Integral points are (0,4) (0,-4) (4,0) and (-4,0)

BUT: x^2+y^2=16 equals y^2=16-x^2
SO: 4ax=16-x^2

Which means:
if x=0, there is no soultion ("a" can't be anything)
if x=4, "a"=0, which means y^2=4ax is not a parabola
if x=-4, "a"=0, which means y^2=4ax is not a parabola

Thus I doubt if there is any integral points


probability
inequalities
geometric sequence
geometric series

A tournament sequence ${t_i}$ is a sequence defined recursively by $t_n\leq{2t_{n-1}}$ for $n\geq{1}$. Suppose that $t_1=1$. Find, with proof, the probability that this tournament sequence will be a geometric sequnce.

Define a tournament sequence of order $k$, ${t^k_i}$ as a sequence defined recursively by $t_n\leq{kt_{n-1}}$ for $n\geq{1}$. Suppose ${t^k_1}=1$ and $k=r!$, for some integer $r$. Find, with proof, the probability that this tournament sequence is a geometric sequence.

are there any other restrictions than that inequality? It seems like there are any infinite number of possible choices for T(n) so the probabilty of it being a geometric sequence is 0.

I see what ur saying. Since there are finitely many geometric series, yet infinitely many series, the probability is 0. Sorry about that  :oops:  (If there were also infinitely many geometric series, then the problem would be more interesting)
Modify the problems then:
For the first one, find the probability of it being a finite geometric sequence within the first 10 terms.
On the second one, assume that the sequence has length c and find the probability.


There appears to be a bug in the "Allowed Extensions and Sizes" page--see attached screenshot.

I'm pretty sure that means the size is unlimited.

My point wasn't that several of them were 0 bytes, but that everything was on top of each other, leading to a decidedly ugly-looking image.  The menu and proof-without-words should be in the margin, NOT running down the left side of the image.


induction
inequalities
combinatorics proposed
combinatorics

Grobbers problem made me remember this one:

Let $S$ be a set with $n$ elements and let $A_1,A_2...,A_n$ be subsets of $S$. for every pair of elements of $S$, there is exactly one $i$ such that $A_i$ contain this pair.  Prove that for every $i,j$ (different!) we have that $A_i$ and $A_j$ have exactly one element in common.

$|A_i|>1$!!!!!!!!!!!!!!!!!!!!!!!!!!!

edited.

I must be misunderstanding the problem...

Suppose the elements of $S$ are $1, 2, ..., n$. Then take $A_1={1, 2, ..., n}$, $A_i={i}$ for $i$ between $2$ and $n$. Then the first condition is satisfied (for any pair it is in $A_1$ and no other subset) but the second isn't... Is some additional condition missing?

I'm pretty sure Pascual meant to say that no set is the full set (i.e. no $A_i$ is $\{1,2,\ldots,n\}$). Another way to put it is that all the sets have at least two elements.

let $S=\{d_1,d_2,...d_n\}$.suppose that there exist $A_i,A_j$ such that $|A_i \cap A_j| \neq 1$, there cannot be $|A_i \cap A_j| > 1$, so $|A_i \cap A_j|=0$
let $x_i$ be the number of subsets containing $d_i$, let $a_i=|A_i|$, we have $\sum a_i=\sum x_i$, $\sum {a_i \choose 2}={n \choose 2}$, and $\sum {x_i \choose 2}<{n \choose 2}$. we have also $a_i \leq x_s$, where $d_s$ doesn't belong to $A_i$. we can show by induction that we can create $n$ pairs $(d_{i_1},A_{i_1})$ such that every $d,A$ will be in only one pair, and $d_{i_j} \notin A_{i_j}$. so we can obtain inequalities $x_{i_1} \geq a_{i_1}$, we add those $n$ inequalities, and we have $\sum x_i \geq \sum a_i$ so there must be $x_{i_1} = a_{i_1}$ and $x_{i_1}^2 = a_{i_1}^2$, so ${n \choose 2}> \frac{1}{2}\sum x_i^2-x_i = \frac{1}{2} \sum a_i^2-a_i ={n \choose 2}$, we got a contradiction

[quote="Zorro"]... we can show by induction that we can create $n$ pairs $(d_{i_1},A_{i_1})$ such that every $d,A$ will be in only one pair, and $d_{i_j} \notin A_{i_j}$ ...[/quote]
Is there easy reasons for induction? At least, I only see how to prove it with Hall's theorem.

Nice solution!

thanks
induction is easy if we will prove that for sets such that $|A_i|=1$ too, the situation with all sets $A_i : \ |A_i|>1$ is a particular case


inequalities
algebra
polynomial
three variable inequality

let$ x\equal{}1,y\equal{}\frac{1}{2}$ and$ z\equal{}0$ . then $ C \leq  \frac{5}{2}$
  ineq  becomes :
$ 2(x^3\plus{}y^3\plus{}z^3) \plus{}  5(xy^2\plus{}yz^2\plus{}zx^2)  \geq  7(x^2y\plus{}y^2z\plus{}z^2x)$

Applying CID (Cyclic inequality oF degree 3) theorem, we [i]can let[/i] $ c \equal{} 0$ in the inequality. It becomes
\[ x^3 \plus{} y^3 \plus{} cx^2y\ge (c \plus{} 1) xy^2.
\]
Thus, we have to find the minimal value of
\[ f(y) \equal{} \frac {y^3 \minus{} y^2 \plus{} 1}{y^2 \minus{} y} \equal{} y \plus{} \frac {1}{y(y \minus{} 1)}
\]
when $ y > 1$. It is easy to find that
\[ f'(y) \equal{} 0 \Leftrightarrow 2y \minus{} 1 \equal{} (y(y \minus{} 1))^2 \Leftrightarrow y^4 \minus{} 2y^3 \plus{} y^2 \minus{} 2y \plus{} 1 \equal{} 0.
\]
Solving this symmetric equation gives us:
\[ y \plus{} \frac {1}{y} \equal{} 1 \plus{} \sqrt {2} \Rightarrow y \equal{} \frac{1 \plus{} \sqrt {2} \plus{} \sqrt {2\sqrt {2} \minus{} 1}}{2}
\]
Thus we found the best value of $ C$ is
\[ y \plus{} \frac {1}{y(y \minus{} 1)} \equal{} \frac {1 \plus{} \sqrt {2} \plus{} \sqrt {2\sqrt {2} \minus{} 1}}{2} \plus{} \frac {1}{\sqrt {\sqrt {2} \plus{} \sqrt {2\sqrt {2} \minus{} 1}}}\approx 2.4844
\]

[quote="t0rajir0u"][quote="hungkhtn"]Applying CID (Cyclic inequality oF degree 3) theorem, we [i]can let[/i] $ c \equal{} 0$ in the inequality.[/quote]

You mean $ z \equal{} 0$?  Can you give a statement and proof?[/quote]

Here you are:

http://www.mathlinks.ro/viewtopic.php?p=1130901#1130901

A very strange computational problem to give in an olympiad...
anyways to see why we only need to check when $ z\equal{}0$ is a simple argument.
note that the inequality can be rewritten as
$ \sum (x\plus{}y)(x\minus{}y)^2\geq (2c\plus{}1)(x\minus{}y)(y\minus{}z)(x\minus{}z)$ so decreasing each number by $ k$ would only decrease the RHS and leave LHS fixed.  :wink:

[quote="Albanian Eagle"]A very strange computational problem to give in an olympiad...
anyways to see why we only need to check when $ z \equal{} 0$ is a simple argument.
note that the inequality can be rewritten as
$ \sum (x \plus{} y)(x \minus{} y)^2\geq (2c \plus{} 1)(x \minus{} y)(y \minus{} z)(x \minus{} z)$ so decreasing each number by $ k$ would only decrease the RHS and leave LHS fixed.  :wink:[/quote]

Yes, that is simpler, although the idea is exactly the same as the classical entirely-mixing-variable method.

[quote="hungkhtn"] $ \frac {1 \plus{} \sqrt {2} \plus{} \sqrt {2\sqrt {2} \minus{} 1}}{2} \plus{} \frac {1}{\sqrt {\sqrt {2} \plus{} \sqrt {2\sqrt {2} \minus{} 1}}}\approx 2.4844$
[/quote]

This answer is correct. Just in case, it can be rewritten as $ \frac{1}{2}\left(\sqrt {13\plus{}16\,\sqrt {2}}\minus{}1\right)$. It's the root of the polynomial $ k^4\plus{}2k^3\minus{}5k^2\minus{}6k\minus{}23$.

Sergey Markelov

hi markelov how did you get the equation $ k^4\plus{}2k^3\minus{}5k^2\minus{}6k\minus{}23\equal{}0$

I found nice solution with Vornicu Schur Inequality
\[ C \equal{} \minus{} \frac12 \plus{}\frac {7}{2\sqrt{28\sqrt {2} \minus{} 35}} \minus{} \frac{1}{ \minus{}\frac12 \plus{} \frac {1}{14}\sqrt {28\sqrt {2} \minus{} 35}} \minus{}\frac {1}{\frac12 \plus{}\frac {1}{14}\sqrt {28\sqrt{2} \minus{}35}}\approx 2.48443533
\]

$ x^3\plus{}y^3\plus{}z^3\plus{}\frac{\sqrt{13\plus{}16\sqrt{2}}\minus{}1}{2}\left(xy^2\plus{}yz^2\plus{}zx^2\right)$

$ \minus{}\frac{\sqrt{13\plus{}16\sqrt{2}}\plus{}1}{2}\left(x^2y\plus{}y^2z\plus{}z^2x\right)$

$ \equal{}\left(\frac{1\plus{}\sqrt{8\sqrt{2}\minus{}11}}{2}y\plus{}\frac{1\minus{}\sqrt{8\sqrt{2}\minus{}11}}{2}z\minus{}x\right)\left(x\plus{}\frac{\sqrt{5\plus{}4\sqrt{2}}\minus{}1}{2}y\minus{}\frac{\sqrt{5\plus{}4\sqrt{2}}\plus{}1}{2}z\right)^2$

$ \plus{}2x\left(x^2\plus{}y^2\plus{}z^2\minus{}yz\minus{}zx\minus{}xy\right)\geq0,$

which is clearly true for $ x \equal{} \min\{x,y,z\}.$

[quote=Litlle 1000t]Find the maximum number $ C$ such that for any nonnegative $ x,y,z$ the inequality
$ x^3 \plus{} y^3 \plus{} z^3 \plus{} C(xy^2 \plus{} yz^2 \plus{} zx^2) \ge (C \plus{} 1)(x^2 y \plus{} y^2 z \plus{} z^2 x)$ holds.[/quote]
Prove that for all nonnegative numbers $a,b,c$ the following inequality holds
$$ x^3\plus{}y^3\plus{}z^3\plus{}\frac{\sqrt{13\plus{}16\sqrt{2}}\minus{}1}{2}\left(xy^2\plus{}yz^2\plus{}zx^2\right)\geq\frac{\sqrt{13\plus{}16\sqrt{2}}\plus{}1}{2}\left(x^2y\plus{}y^2z\plus{}z^2x\right).$$
Proof of dragonheart6:


Let $ n \in \mathbb{N}$. Let us consider all natural numbers $ 0 \le m \le n\minus{}1$. Which of these numbers satisfy the following property: $ n \mid m^q$ for some $ q \in \mathbb{N}$ ?

the numbers whose set of prime divisors contains that of $ n$. i don't know of an easy characterization of those numbers  :|

Let P be the product of positive prime divisor of n.
no. of m values = [n/P]; where [] is the step function.

The value $ P$ that praneeth described is known as the [i]radical[/i] of $ n$, denoted $ \text{rad}(n)$.  It figures in the [url=http://en.wikipedia.org/wiki/Abc_conjecture]abc conjecture[/url].


The difference of the squares of two odd numbers is always divisible by $ 8$.  If $ a>b$, and $ 2a\plus{}1$ and $ 2b\plus{}1$ are the odd numbers, to prove the given statement we put the difference of the squares in the form:

$ \textbf{(A)}\ (2a\plus{}1)^2\minus{}(2b\plus{}1)^2 \\
\textbf{(B)}\ 4a^2\minus{}4b^2\plus{}4a\minus{}4b \\
\textbf{(C)}\ 4[a(a\plus{}1)\minus{}b(b\plus{}1)] \\
\textbf{(D)}\ 4(a\minus{}b)(a\plus{}b\plus{}1) \\
\textbf{(E)}\ 4(a^2\plus{}a\minus{}b^2\minus{}b)$

[hide="Click for solution"]
$ (2a\plus{}1)^2\minus{}(2b\plus{}1)^2\equal{}4a^2\plus{}4a\plus{}1\minus{}4b^2\minus{}4b\minus{}1$.

Combining like terms and factoring gives:

$ 4[a(a\plus{}1)\minus{}b(b\plus{}1)]$.

Since the product of two consecutive integers is divisible by $ 2$ the last expression is divisible by $ 8$.  Thus the answer is $ \boxed{\textbf{(C)}}$.
[/hide]


[quote="behdad.math.math"]if xy+xz+yz+2xyz+1 prove that x+y+z>=3/2[/quote]
Maybe you mean $ xy \plus{} xz \plus{} yz \plus{} 2xyz \equal{} 1$ and $ \{x,y,z\}\subset(0, \plus{} \infty)$ $ ?$ :wink:
If so, it's Nesbitt's inequality.
See here:
http://en.wikipedia.org/wiki/Nesbitt%27s_inequality
 :wink:

[quote="behdad.math.math"]I'm sorry.that is xy +xz+ zy+2xyz=1[/quote]
Try $ x\equal{}y\equal{}\minus{}2$ and $ z\equal{}\minus{}\frac{3}{4}.$ :wink:


There are 2 rows with 3 slots each.  A randomly chosen integer from 1 to 9 is placed in each slot (with repetition).  Find the probability that the sum of the the numbers in the first row equals the sum of the numbers in the second row.

I got a weird answer [hide]$\frac{3629}{59049}$[/hide] so I wanted to verify.

What's the fastest way to find out how to make the sum of 3 numbers equal?  I can do it, but I don't feel like doing so many calculations.

i used generating functions...but that process was rather tedious


Compute the sum of all of the roots of $(2x+3)(x-4)+(2x+3)(x-6)=0$.

[hide]
simplify(or unsimplify, however you interpret it) to get :
$(2x-10)(2x+3)=0$
thus either $2x-10=0$ or $2x+3=0$
so x=5,-3/2
the sum is 7/2
[/hide]


Let $ A$, $ M$, and $ C$ be nonnegative integers such that $ A \plus{} M \plus{} C \equal{} 12$. What is the maximum value of $ A \cdot M \cdot C \plus{} A\cdot M \plus{} M \cdot C \plus{} C\cdot A$?

$ \textbf{(A)}\ 62 \qquad \textbf{(B)}\ 72 \qquad \textbf{(C)}\ 92 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 112$

[hide]$ AMC \plus{} AM \plus{} MC \plus{} CA \equal{} (A\plus{}1)(M\plus{}1)(C\plus{}1) \minus{} (A \plus{} M \plus{} C \plus{}1) \equal{} (A\plus{}1)(M\plus{}1)(C\plus{}1) \minus{} 13$. Maximized at 112.[/hide]

[hide="Solution"]
The format $ amc \plus{} am \plus{} mc \plus{} ca$ reminds us of the expansion of $ (a \plus{} 1)(m \plus{} 1)(c \plus{} 1) \equal{} amc \plus{} am \plus{} mc \plus{} ca \plus{} a \plus{} m \plus{} c \plus{} 1$.

Note that $ a \plus{} m \plus{} c \plus{} 1 \equal{} 13$. So:
$ (a \plus{} 1)(m \plus{} 1)(c \plus{} 1) \equal{} (amc \plus{} am \plus{} mc \plus{} ca) \plus{} 13$
$ amc \plus{} am \plus{} mc \plus{} ca \equal{} (a \plus{} 1)(m \plus{} 1)(c \plus{} 1) \minus{} 13 \equal{} pqr \minus{} 13$, where $ p \equal{} a \plus{} 1$, etc.

We want to maximize the product $ pqr$. We are only given $ a \plus{} m \plus{} c \equal{} 12$, but we can modify this to give us helpful information:
$ a \plus{} m \plus{} c \equal{} 12 \implies (a \plus{} 1) \plus{} (m \plus{} 1) \plus{} (c \plus{} 1) \equal{} 15 \implies p \plus{} q \plus{} r \equal{} 15$

Recall the AM-GM inequality: for positive reals, the arithmetic mean is always at least the geometric mean. This applies to p,q,r as well:
$ \frac {p \plus{} q \plus{} r}{3} \geq \sqrt [3]{pqr}$

We know $ p \plus{} q \plus{} r \equal{} 15$, so $ \sqrt [3]{pqr} \leq \frac {15}{3} \equal{} 5 \implies pqr \leq 5^3 \equal{} 125$

The maximum value of $ pqr \minus{} 13$ is $ 125 \minus{} 13 \equal{} \boxed{E. \; 112}$.

This value happens when $ p \equal{} q \equal{} r \equal{} 5 \implies a \equal{} b \equal{} c \equal{} 4$
[/hide]

[hide="Another possible solution"]
For sake of typing convenience, let $ a \equal{} A$, $ b \equal{} M$, $ c \equal{} C$. We notice that the latter part of the expression appears in the well-known inequality

$ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca \implies (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca) \ge ab \plus{} bc \plus{} ca \implies ab \plus{} bc \plus{} ca \le 12^2/3 \equal{} 48.$

Now we can maximize $ abc$ by using AM-GM.

$ abc \le \left(\frac{12}{3}\right)^3 \equal{} 64$

Adding the two inequalities maximizes the desired expression to $ \boxed{112}$.
[/hide]

sunehra, this is exactly my solution, but I think it needs to be said that this is correct only because both maxima occur by meeting the same condition: $ a\equal{}b\equal{}c\equal{}4$.

Oops, I forgot that, and I'm one who usually remembers to check the equality condition... :blush: Sorry.


Compute the smallest positive integer $ x$ greater than 9 such that

$ \lfloor{x \rfloor} \minus{} 19 \cdot \lfloor { \frac{x}{19} \rfloor} \equal{} 9 \equal{} \lfloor {x \rfloor} \minus{} 89 \cdot \lfloor { \frac{x}{89} \rfloor}$

That is just saying that $ x$ is 9 mod 19 and 89...so x=9+lcm(19,89)=1611

Uh, I didn't get that.

[hide="Perhaps your error on calculation.."]
LCM(19,89) = 1691 so answer is 1700. I verified with calculator that LCM(19,89) is indeed 1691. I don't know how you got 1611 as your answer. [/hide]


geometry
trapezoid
angle bisector
geometry unsolved

X and Y are points on the sides BC and AC of a triangle ABC respectively such that

Angle AXC = Angle BYC and
BX = XY

Prove that AX bisects the angle BAC.

Let's use parallel lines and a circumscribed circle well.

How do you show those four points are concyclic?

I think it's easier to prove that A, X, Y, and B are concyclic.

[quote="Bictor717"]How do you show those four points are concyclic?[/quote]
$ XY$ is a tangent line of the circumscribed circle.

C,X,Z,Y are concyclic , where  Z is the intersection of BY and AX
X,Y,A,B are concyclic
∠BAX=∠BYX=∠XBY=∠XAY
AX bisects the angle BAC

From <AXC = <BYC we get triangles AXC and BYC similar, so <CAX = <CBY hence ABXY is cyclic. Adding BX = AY we get ABXY isosceles trapezoid only.
In order to get AX angle bisector we need BX = XY, i.o. BX = AY!

Best regards,
sunken rock


Show that $ n|m \Rightarrow F_n|F_m$

Show that: [tex]7|x^3+y^3+z^3 \Rightarrow  7|xyz.

[hide="Problem 2"]Consider cubes modulo $ 7$. $ 0^3 \equiv 0$. $ 1^3 \equiv 1$. $ 2^3 \equiv 1$, $ 3^3 \equiv 6$. $ 4^3 \equiv 1$. $ 5^3 \equiv 6$. $ 6^3 \equiv 6$. Assume for the sake of contradiction that $ 7$ does not divide $ xyz$. This implies that there exist three numbers $ x\plus{}y\plus{}z$ where each of $ x$, $ y$, and $ z$ is either $ 1$ or $ 6$, and they sum to a number divisible by $ 7$. By observation, this is impossible. Therefore, the assumption is false, so $ 7$ must divide $ xyz$.

[/hide]


i think noone can solve this pro :D  me too :P
but, it maybe very easy???????


complex numbers
complex analysis
complex analysis unsolved

Let D be a closed disc in the complex plane. Prove that for all positive integers n, and for all complex numbers  z1, z2, ..., zn in D, there exists a z in D such that z^n=z1.z2.(...).zn

What is the official solution for this problem??


search
number theory unsolved
number theory

Prove that there are infinitely many primes p such that for every p there exists a natural number n for which n! is divisible by p, and p-s is not divisible by n.

If I understand ther problem correctly the problem is pretty easy.

$ p|n!$ for every $ n\geq p$ (only then there is a factor $ p$ in $ n!$)
and $ n\geq p$ then n will not divide $ p \minus{} s$ because $ p \minus{} s < n$ for every $ s > 0$

sorry, folks, put instead of "n!" "n!+1" and instead of "p-s" put "p-1". Next time I'll be more attentiv

He meant:

[quote="Inequalities Master"]Prove that there are infinitely many primes $ p$ such that there exists a natural number $ n$ for which $ n! \plus{} 1$ is divisible by $ p$, and $ p \minus{} 1$ is not divisible by $ n$.[/quote]

This was posted before [url=http://www.mathlinks.ro/viewtopic.php?t=140463]here[/url]. These primes are called Pillai primes. For a start, check [url=http://en.wikipedia.org/wiki/Pillai_prime]this[/url].

Btw.. No offense, but you could firstly search and use a better title/description :)
Moderators, please lock this topic.


Umm.. the title is pretty self explanatory. I need help with changing the background of my blog and replacing it with an image, i used to do this easily in html but this is slightly different. Can someone give a step-by-step process?
Thanks in advance

Blog Options -> Blog Layout -> Background image.

You will need to put it in syntax:

[code]url('URL GOES HERE')[/code]

You can also do it by directly editing the CSS code.  :) The method mentioned is more "user-friendly", however.


A car is moving on a surface in which the coefficient of friction increases with a linear distance X. The car is moving with maximum possible acceleration. How is the kinetic energy of the car related to X?

I hope no info is missin....

OK - now what is "maximum possible acceleration" in mathematical terms?

Yeahhh...max possible acceleration...does it mean its constannt?I think it would keep changin

hey , if i aint wrong this question is in physics MCQ s by Mukherjee , right?

hmm i 'll try to solve  it , though i know the answer , its at the back......

$\text{K.E}$inversely proportional to $X$
I mean as the distance increases the $\text{K.E}$ would not exactly decrease in velocity but it wouldnt increase as much either.

what i cant get is ,  the accelereation is constant ,  but the backward force is increasing (friction) hence the body should start , accelerate in the beggining , slowly its acceleration would come down to 0 , then it would start retarding and finally stop , what next???

also is it that u have to use

$v^2$ Proportional to $as$ but ke = 1/2 $mv^2$ hence ke proportional to as, but s is increasing linearly.......

ne way i ';ll think further....

See,If we take the engine power to be finite,then as the friction incerases,the engine wont be able to maintain the constant acceleration...so what i say is that first $K.E PROP TP DISTANCE$ and then as distance ibncreases a lot further then it becomes inversely proportional.....I guess :blush:

By the way,what does it mean by the line "the coefficient of friction increases"???

You need to specify the force of engine on car to solve this problem precisely. It will involve somee good diff. eqns. but the soln. will be perfect rather than speculation.!!! Otherwise, with assumptions., other sons. are correct

If engine power is constantly $P$, then the force it creates is $F_a=\frac{P}{v}$. The problem statement reads how coefficient of friction increases but this is not a very good way to slow the car down - friction between the road and tires is actually the only force that accelerates the car! :) So let's say that the car is being slowed down by a force which incerases linearly with $x$:

$F_b=kx$.

Acceleration of the car is then

$a=\frac{F_a-F_b}{m}=\frac{\frac{P}{v}-kx}{m}$

or

$mva=P-kxv$.

This can be approximated numerically but I'd like to see an exact solution - anybody know how to do this?

This picture shows my simulation for

k = 20 N/m
mass = 1200 kg
power = 75 kW

It shows kinetic energy on the y axis (max = 1 MJ) and distance on the x axis (max = 2 km). (Click the picture to see the lines properly.) Notice that the car will never stop - that's because of the constant power assumption.

Same thing i was sayin in my post,i agree with u bus.

[quote="eminem"]hey bus  :?: how come the car wont stop , what if the coefficient of friction goes (theoretically) off to INFINITY? i guess it would stop then ?[/quote]The problem is that we assume constant power. With constant power, the slower the car goes, the greater the force the engine creates to drive it forward: $F=\frac{P}{v}$. So as the resistance increases, the car slows down to make engine provide more force and this force can be theoretically infinitely big as well, the car only needs to go infinitely slow. Of course, this is not how it works in real life.

hmmmm bus i think i am interpreting power differently. wont constant power mean that constant forwaRD  force  but the backward force(friction) is increasing constantly so the car will ultimately stop or am i wrong?

Power and force are different. Power is how much energy does the engine produce in one second. If we assume that all of this energy is converted to work wich pulls the car forward, the car can't ever stop - the work done by the engine would be zero since the car wouldn't be moving.
On the other hand, you could specify the source of the resistant force - such as wind - and then you could say that the engine actually does some work by slowing the wind down. It's just a really problematic approach. We had a lot of problems with power, work and efficiency of a wind car in the last year's IYPT...

bus,iypt?

Anyway,doesnt power depend on force,since work =F *s *costheta?Or am i wrong :huh:  :huh:
I think the car would after all become still,even though the engines on,no displacement,i'm not sayin ur wrong at all bus,u are right i guess...i have to think on waht u said :D

[quote="shadysaysurspammed"]bus,iypt?[/quote]It's International Young Physicists' Tournament. Kubus mentioned it some time ago in the "Experiments in Home" topic, it's a really interesting competition. Rushil was very sad that India doesn't participiate though...

[quote="shadysaysurspammed"]Anyway,doesnt power depend on force,since work =F *s *costheta?Or am i wrong :huh:  :huh: [/quote]It does. But if we consider power to be a constant, you should say force depends on power. And "s" changes with speed - or rather:

$\mathrm{power=\frac{work}{time}=\frac{force\times distance}{time}=force\times velocity}$

(the cos theta part is equal to 1)

If power is constant, force changes with velocity. If velocity is 0, force has to be infinity since power is constant > 0. Now you can think about it :) .

[quote="eminem"]hey bus , can you then tell me the exact change in velocity so then wee can judge the force thingy using the formula p = f*v you gave ,cos i aint understandingyour graph properly... :blush:  :blush:  :blush:[/quote]And I probably don't understand your question, I'm sorry :blush: . But if you're asking for an exact solution, as I said, I can't solve the equations. The graph I gave comes from a simulation. If you don't understand the graph: it shows distance $x$ on the horizontal axis and kinetic energy on the vertical axis. (There's no time in the graph if this is what confuses you.) As you know from Newton's second law, acceleration of the car is $a=\frac{F}{m}$. Force $F$ is the sum of resistance force $F_r=-kx$ (it is negative because it slows the car down and it grows linearly with distance $x$ as the problem states - more or less) and the force created by engine $F_e=\frac{P}{v}$ (which you're not sure about, right?). So $a=\frac{\frac{P}{v}-kx}{m}$. This is the equation that I don't know how to solve but I do know how to simulate on computer and that's where the graph comes from.


LaTeX
inequalities
geometry
circumcircle
inequalities proposed

[quote="quykhtn-qa1"]This inequaliti easy but nice:
Let $ a,b,c$ be positive numbers such that: $ abc = a + b + c + 2$ .Prove that:
$ (a^2 - 1)(b^2 - 1)(c^2 - 1)\le27$.
Good luck. :lol:[/quote]

[b]quykhtn-qa1[/b], just write $ \$\dots\$$   between the math formula to $ \text{\LaTeX}$ it.  :wink:

About the problem:

The condition $ abc = a + b + c + 2$ can be written in the following equivalent form: $ \frac {1}{1 + a} + \frac {1}{1 + b} + \frac {1}{1 + c} = 1$.
Let $ x = \frac {1}{1 + a}$ and so on. $ \Rightarrow a = \frac {1 - x}{x} = \frac {y + z}{x}$, since $ x + y + z = 1$.

The inequality will turn into the form: $ 27x^2y^2z^2\ge(1 - 2x)(1 - 2y)(1 - 2z) = (x + y + z)(x + y - z)(y + z - x)(z + x - y)$.
The last inequality has signification only if $ x,y,z$ are sides of triangle $ \Bigg($when all of the variables are less than $ \frac {1}{2}\Bigg)$.
Let $ S$ is the area of the triangle with sides $ x,y,z$, and $ R$ be the radius of circumcircle of this triangle.
Then the inequality will become $ 27S^2.16R^2\ge 16S^2$ or $ 27R^2\ge 1$ which is obvious by Blundon - $ p\le 2R + (3\sqrt {3} - 4)r\le \frac {3\sqrt {3}}{2}R$, where $ p=\frac{1}{2}$ donates the semiperimeter of the triangle.

See the stronger here:  :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=256282

[quote="quykhtn-qa1"]This inequaliti easy but nice:
Let a,b,c be positive numbers such that:abc=a+b+c+2.Prove that:
(a^2-1)(b^2-1)(c^2-1)≤27.
Good luck. :lol:[/quote]
It is easy,we just need to let x=$ \frac{b\plus{}c}{a}$,y=$ \frac{c\plus{}a}{b}$,z=$ \frac{a\plus{}b}{c}$
the ineq<=> $ (b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)(a\plus{}b\plus{}c)^3$≤$ 27(abc)^2$
We have a stronger one :
$ 9(abc)^2$≥$ (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a^2\plus{}b^2\plus{}c^2)$
(this is easy to prove by using $ 9R^2$≥$ a^2\plus{}b^2\plus{}c^2$


geometry
geometric transformation
reflection
number theory unsolved
number theory

Let $ F_n$ be the n-th Fibonacci number.Prove that for all $ n$≥$ 4$,$ F_n\plus{}1$ is not a prime.

[b]Dorin Andrica[/b], Problem O64, Mathematical Reflections, Issue 5, 2007.

Thank you very much for your help.Could you give me the link for downloading Mathematical Reflections of the other year(2005,2006,2008...) please ? Thanks.


function
inequalities
topology
triangle inequality
real analysis
real analysis unsolved

X is a compact metric space, and f:X --> X is a shrinking function (ie d(x,y) > d(f(x),f(y)) for all distinct x and y in X. Define F: X --> R as F(x) = d(x,f(x)). Show that F is continuous.

$ f \in C(X)$
let $ g(x) \equal{} (x,f(x))$
$ \Rightarrow g\in C(X,X^2)$
of course $ d \in C(X^2)$
so,$ F \equal{} dg \in C(X)$
$ X$ is a common metric space.

By the triangle inequality, $ d(x,f(x))\leq d(x,y)\plus{}d(y,f(y))\plus{}d(x,f(x))$.  Therefore, $ d(x,f(x))\minus{}d(y,f(y))\leq d(x,y)\plus{}d(f(x),f(y))<2d(x,y)$.  Repeating the same argument with the roles of $ x$ and $ y$ interchanged shows that $ |F(x)\minus{}F(y)|<2d(x,y)$, so $ F$ is continuous (in particular, it attains a minimum on $ X$, which is forced to be a fixed point).


Determine the sum of all single-digit replacements for $ n$ such
that the number $ 42,789,n37$ is divisible by 3.

If the number is divisible by 3, the sum of the digits must too. The sum is 40+n, so n is 2, 5, or 8. The answer is 2+5+8=[b]15[/b].


calculus integration geometry probability conics hyperbola logarithms

Let R and S be independent and uniformly distributed on the interval (0,1).  Let X = R+S and Y=RS.

Compute the density of X and Y.

I'm not really sure where to begin.  I'm guessing an integral needs to be taken, but I'm not sure what exactly I'm integrating here...

There are several ways to do this. I'll start with the slowest, because it's also the way in which each step makes the most sense and is most easily explained. This slowest method works with the cumulative distribution function.

Let's start with finding the joint density of $ (R,S).$ Since they are independent and uniform, that part is straightforward: the joint density is $ 1$ on the square $ 0\le r \le 1,0\le s\le 1$ and zero otherwise. That also means that the probabilities we compute will simple be areas of regions within that square. Yes there are integrals to compute, but I may not actually write them all out as integrals.

Let's start with $ X\equal{}R\plus{}S.$ The possible values of $ X$ lie in $ [0,2],$ so that's what we'll concern ourselves with.

Suppose $ x\in[0,2].$ Compute $ F(x)\equal{}P(X\le x)\equal{}P(R\plus{}S\le x).$

The condition $ r\plus{}s\le x$ places us below and to the left of the diagonal line $ r\plus{}s\equal{}x.$ We observe how that intersects the square, and observe two cases.

Case 1: $ 0\le x\le 1.$ Then $ P(R\plus{}S\le x)$ is the area of a right triangle with legs $ x$ and $ x$. That area is $ \frac12x^2,$ so in this case $ F(x)\equal{}\frac12x^2.$

Case 2: $ 1\le x\le 2.$ Then it's easier to compute the complementary probability. The probability that $ R\plus{}S>x$ is the area of a right triangle with legs $ 2\minus{}x$ and $ 2\minus{}x.$ (You really should be drawing pictures on a piece of scratch paper as you read this). Hence $ P(R\plus{}S>x)\equal{}\frac12(1\minus{}x)^2$ or $ F(X)\equal{}1\minus{}\frac12(1\minus{}x)^2.$

Summarizing: $ F(x)\equal{}\begin{cases}0,&x<0\\ \frac12x^2,&0\le x\le 1\\ 1\minus{}\frac12(1\minus{}x)^2,&1\le x\le 2\\ 1,&x>2\end{cases}.$

Now we can find the density by differentiating this:

$ f(x)\equal{}\begin{cases}x,&0<x<1\\ 1\minus{}x,&1<x<2\\ 0,&\text{otherwise}\end{cases}.$

Now look at $ Y\equal{}RS.$ The possible value of $ Y$ are between 0 and 1, so that's what we'll concentrate on. Let $ y\in(0,1).$ We want to compute $ P(RS\le y).$ The boundary for this is the hyperbola $ rs\equal{}y$ or $ s\equal{}\frac{y}{r}.$ Looking at the picture (still got that scratch paper out?), it looks like it would be easer to work with the complement.

$ P(RS>y)\equal{}\int_y^1\int_{\frac{y}{r}}^11\,ds\,dr\equal{}\int_y^1\left(1\minus{}\frac{y}{r}\right)\,dr$

$ \equal{}\left.\left(r\minus{}y\ln r\right)\right|_y^1\equal{}(1\minus{}0)\minus{}(y\minus{}y\ln y)\equal{}1\minus{}(y\minus{}y\ln y).$

This is the complement of what we want; we subtract from $ 1$ to get $ F(y)\equal{}P(RS\le y).$

$ F(y)\equal{}\begin{cases}0,&y\le 0\\ y\minus{}y\ln y,&0<y<1\\ 1,&y\ge 1\end{cases}.$

We differentiate this to get the density:

$ f(y)\equal{}\begin{cases}\minus{}\ln y,&0<y<1\\0,&\text{otherwise}\end{cases}.$


for every positive a, b, c
can we prove that
$ \frac{a}{\sqrt{a^2\plus{}b^2}}\plus{}\frac{b}{\sqrt{b^2\plus{}c^2}}\plus{}\frac{c}{\sqrt{c^2\plus{}a^2}}\equal{}\frac{a}{\sqrt{a^2\plus{}c^2}}\plus{}\frac{b}{\sqrt{b^2\plus{}a^2}}\frac{c}{\sqrt{c^2\plus{}b^2}}$

It is not true for all $ x,y,z$ . You can file an example with it.


[quote="gabriel ponce"]We know that if $n$ is a positive intenger then there exist intengers $x$ and $y$ such that $n=x^{2}+y^{2}$ if each prime factor of $n$ of the form $4k+3$ appears an even number of times.
So if $n \in S$ we have that $n-1$,$n$ and $n+1$ satisfies the propertie above.So if each prime factor of $n-1$ of the form $4k+3$ appears an even number of times, and each prime factor of $n+1$ of the form $4k+3$ appears an even number of times we have that each prime factor of $(n+1)(n-1)=n^{2}-1$ of the form $4k+3$ appears an even number of times,thus $n^{2}-1$can be expressed as a sum of two squares.Of course the same is true for $n^{2}$ since each prime factor of it of the form $4k+3$ will appears an even number of times too!.So we just have to prove now that $n^{2}+1$ can be expressed as a sum of two squares.But that is obviously since $n$ is intenger and $1=1^{2}$!!!.
So we solved the problem! :lol:

...........................

Gabriel Ponce[/quote]

actualy.....we know that $n$ could be written as a sum of squares  if each prime factor of the form "4k+3"appears an even number of times but it is not true that if every prime factor of that form appears an even number of times then the number can really be written like a sum of squares.see for example $7^{2}$.
$7=4*1+3$ and it appears an even number of times but $7^{2}$ can not be written that way.but i think rust's solution is the shorteest way to kill the problem which is pretty easy in fact.

[hide=Solution]
[quote=China Western MO 2006 P5]Let $S=\{n|n-1,n,n+1$ can be expressed as the sum of the square of two positive integers.$\}$. Prove that if $n$ in $S$, $n^{2}$ is also in $S$.[/quote]
[b][color=#000]Solution:[/color][/b] Suppose, $n \in S$, then we must have $$\begin{cases} a^2+b^2=n-1 \\ k^2+\ell^2=n \\ p^2+q^2=n+1 \end{cases}$$
Now for any $x \in \mathbb{Z}$ $\implies$ $x^2 \equiv 0,1 \pmod{4}$ $\implies$ $n-1,n,n+1 \equiv 0,1,2 \pmod{4}$ respectively! Now $n^2+1$ $=$ $(k^2+$ $\ell^2)^2$ $ +1$. Also, Using the Identity
$$(a^2+~b^2)(c^2+~d^2)=(ac-~bd)^2+~(ad+~bc)^2 \implies n^2-~1=(a^2+~b^2)(p^2+~q^2)=(ap-~bq)^2+~(aq+~bp)^2$$
Also, using the same Identity,
$$n^2=(k^2+\ell^2)^2=(k^2+\ell^2)(k^2+\ell^2)=(k^2-\ell^2)^2+(2k\ell)^2$$
Also, $k$ is not equal $\ell$, hence, $n^2 \in S$ $\qquad \blacksquare$
[/hide]


Let $ f$ is a function on $ (X,d)$ - metric space. Where $ X \equal{} \bigcup\limits_{t\in T}G_{t}$ ($ G_{t}$ open),$ \forall t\in T f|_{G_{t}}$  continuous . Show that $ f$ continuous.

Let $ U$ be an open subset of the co-domain of $ f$. we need to show that $ f^{-1}(U)$ is open. since the union of the $ G_t$ is $ X$, we have $ f^{-1}(U) = \bigcup_{t\in T} f^{-1}(U)\cap G_t$. now, $ f^{-1}(U)\cap G_t$ is open in $ G_t$ because $ f|G_t$ is continuous. since $ G_t$ is open in $ X$, it follows that $ f^{-1}(U)\cap G_t$ is open in $ X$ too. (every open set in a subspace is the intersection of an open set in the super-space with the subspace). thus $ f^{-1}(U)$ is a union of open sets, hence open.


induction
real analysis
irrational number
real analysis unsolved

Does there exist a dense, uncountable set $ S$ of real numbers such that any two elements of $ S$ differ by an irrational number? what if irrational is replaced by rational?

The answer to the second question is no because clearly any such set is countable.  I'll think about the first question.

[b]Edit:[/b]  We'll build $ S$ up iteratively.  We first take countably many steps as follows.  At any step in which $ S$ is at most countable, there are at most countably many $ x \in \mathbb{R}$ such that at least one element $ x \minus{} s, s \in S$ is rational so in particular in every interval there is at least one $ x$ such that all elements $ x \minus{} s, s \in S$ are irrational.  Pick an enumeration of the intervals of the form $ [a, b], a, b \in \mathbb{Q}$ and add one such $ x$ for each such interval; the resulting set $ S$ is dense, countable, and satisfies the desired property.  Now, since $ S$ is still countable, by transfinite induction we can continue adding elements until $ S$ is uncountable.  (I am not totally sure this is rigorous, by the way.)

Declare two real numbers equivalent if their difference is rational. Each equivalence class is countable, therefore we have uncountably many equivalence classes. Pick one element from each (Axiom of Choice at its finest).

Incidentally, this is the standard example of a Lebesgue-nonmeasurable set.

Of course it does not have to be dense. Missed that condition.  :blush:

Start the set with the rational multiples of $ \sqrt{2}$.  That guarantees denseness right away.  Then apply mlok's procedure for all the numbers not equivalent to those.

Unfortunately, the rational multiples of $ \sqrt {2}$ are countable.

Edit: I think I found a way to do it. Is my reasoning correct?

First divide the real numbers into equivalence classes that differ by a rational.
Since every class is dense everywhere, given any interval with positive measure, their intersection is nonempty, and so you can pick a member of the equivalence class which is contained in this interval.

So first, select a countable subset of the equivalence classes.
Also, take the set of all intervals of positive measure and rational endpoints. Obviously, there are countably many of these.
Now pair them up and take a member from each one. The resulting set has a member in every rational interval, so it must have a member in every interval as well, and hence be dense everywhere. However, it is not yet uncountable.

Now take all of the equivalence classes that were not chosen before and use the axiom of choice to choose a member from each one.

Now you're left with an uncountable dense set with no two members a rational distance apart.

That's right. Having a countable base saves the day again.

[quote="0714446459923"]Unfortunately, the rational multiples of $ \sqrt {2}$ are countable.[/quote]
I agree, but I don't see why that prevents my idea from working.  Maybe I should say it with more details, and you can tell me if there is a problem.  Let $ A$ be the set of rational multiples of $ \sqrt{2}$.  Let $ B$ be the set of real numbers not equivalent to some number in $ A$.  Following mlok, let $ C$ be formed by choosing one representative from each equivalence class of $ B$.  Then the set $ A \cup C$ is a dense, uncountable set such that every pair of numbers differs by an irrational number.

[quote="Ravi B"][quote="0714446459923"]Unfortunately, the rational multiples of $ \sqrt {2}$ are countable.[/quote]
I agree, but I don't see why that prevents my idea from working.  Maybe I should say it with more details, and you can tell me if there is a problem.  Let $ A$ be the set of rational multiples of $ \sqrt {2}$.  Let $ B$ be the set of real numbers not equivalent to some number in $ A$.  Following mlok, let $ C$ be formed by choosing one representative from each equivalence class of $ B$.  Then the set $ A \cup C$ is a dense, uncountable set such that every pair of numbers differs by an irrational number.[/quote]

Wow that's even simpler then what I had. Great idea.

Although my method has the advantage of working not only for the rationals, but for any countable set that is closed under addition and subtraction. For example, you could construct a dense uncountable set where no two members are a computable distance apart.

Edit: Any countable set that is closed and dense.

My method works for any countable set, but I guess people don't like transfinite induction when they can write down something slightly more "explicit" (although what that means when we're all using the axiom of choice I can't fathom).

Here is an explicit example. Start with continuum of subsets of positive integers such that every two subsets have infinite symmetric difference (for instance, $ E_\alpha\equal{}\{[\alpha m]: m\in\mathbb N\}$ with $ 2<\alpha<3$ will work just fine). Consider the set of numbers of the form $ \sum_{j\in E_\alpha} \frac{1}{j!}$. It is of cardinality continuum, the differences are irrational (the classical proof of irrationality of $ e$ works here), so the only problem is that it is not dense yet. Now, observe that we can shift each element of this set by any rational number and nothing will change in terms of rationality of differences. But, obviously, we can constructively choose countably many elements in our set and choose shifts to approximate any rational with any precision.

Yet another proof of existence of such a set, inspired by mlok's idea, but slightly modified and developed.

On $ \mathbb{R \setminus Q}$ define a relation "$ \sim$" as follows: for $ a, b \in\mathbb{R \setminus Q}$ let $ a \sim b \iff (\exists p, q \in \mathbb{Q} \setminus \{0\})\ ap - bq \in \mathbb{Q}$.
It is easy to check that "$ \sim$" is equivalence relation on $ \mathbb{R \setminus Q}$. Let $ \mathcal{S}$ be system of all equivalence classes of relation "$ \sim$" and $ T$ be transversal for $ \mathcal{S}$, i. e. a set, where $ (\forall C \in \mathcal{S})\ |C \cap T| = 1$.
Now for each $ t \in T$ denote $ \mathbb{Q}_t = \{a \in \mathbb{R}: (\exists q \in \mathbb{Q})\ a = pt\}$. It is not difficult to verify that ${{ (\forall t_1, t_2 \in T)\ t_1 \neq t_2 \Rightarrow \mathbb{Q}_{t_1}} \cap \mathbb{Q}_{t_2} = \emptyset}$.
Finally, denote $ A = \bigcup_{t \in T}\mathbb{Q}_t$. Each of $ \mathbb{Q}_t$'s is dense, thus $ A$ is dense, a fortiori. $ \mathbb{Q}_t$'s are pairwise disjoint and countable, $ T$ is uncountable and this implies that $ A$ is uncountable. Are the differences of distinct elements of $ A$ irrational?
If $ a, b \in A, a \neq b$ then either:
1) $ (\exists t \in T)\ a \in \mathbb{Q}_t \wedge b \in \mathbb{Q}_t$
In this case $ a = tp$ and $ b = tq$ for some distinct $ p, q \in \mathbb{Q}$. So we have $ b - a = tq - tp = t(p - q)$. $ t(p -q)$ must be irrational, otherwise $ t$ would be rational.
or
2) $ (\exists t_1, t_2 \in T)\ t_1 \neq t_2\wedge a \in \mathbb{Q}_{t_1} \wedge b \in \mathbb{Q}_{t_2}$
In this case $ a = t_1p$, $ b = t_2q$ for some $ p, q \in \mathbb{Q}$. The difference $ b - a =  t_2q - t_1p$ must be irrational, otherwise $ t_1$ and $ t_2$ would be equivalent and that is a contradiction.

Lack of this construction is that the Axiom of Choice has been used. :( I am so sad...


function
calculus
derivative
complex analysis
limit
analytic geometry
complex numbers

i was asked to find out if the function : f: R 2 -> R 2    defind by:
(x/(x 2 +y 2 )^ :beta: , y/(x 2 +y 2 )^ :beta:)

is continuous and if so , to explore if it differentiable.

i know that for continuos we first can replace y with mx and to find out if those lines are continuos.  but i know that some functions are continuos in the lines but not in some curves.

here the critic point is (0,0) and the differentiable is depend on the continuous of the partial derivatives (as far as i know ,  i do not know how to find this out other way )

i found that the function continuos where  :beta: < 1/2 but not sure about the curves i mentioned.

this is common problem i asked some times before ,  but never knew  clearly how to solve it properly

Consider complex numbers : $z = x + iy$ , $ f(z) = \frac{z}{|z|^{2\beta}}$
  :cool:  Do you know the holomorphic functions ? :cool:

it is not should involve complex numbers...

not in this one

Continuity for a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ is the same as for a function from $\mathbb{C}$ to $\mathbb{C}$; the distance function is even the same.

$x^2+y^2$ is the nice quantity here; it is the square of the distance from the origin.

On the other hand, we should stay in $\mathbb{R}^2$ for differentiability; that is different in the complex numbers.

We can express $|f(x)|$ cleanly in terms of $|x|$; that should clear up continuity immediately.

The other way to deal with differentiability is to go straight to the definition: $f$ is differentiable at 0 if there exists a linear mapping $D$ such that $\lim_{x\to 0} \frac{f(x)-Dx}{|x|}=0$. If such a mapping exists, the partial derivatives are its coefficients. Here, the partial derivatives indicate that $D$ is a constant multiple of the identity if it exists.

for  :beta:  >= 1/2 this function is not continuous ,since the first coordinate (also the second) is not continuous (we consider f as f=(f1,f2)) to clear why f1 is not continuous consider the line (0,y) on this line f1 is constantly 0 but considering the line (t,t) it turns out that the value at (0,0) should be 1/ :sqrt: 2 if  :beta: =1/2 or (0,infinity) if  :beta: >1/2. but you are not right when you say that existence and continuity of partial derivatives yields differentiability. a counterexaple is RxR->R   f(x,y)=xy/(x*x +y*y) which is even real-analytic in either variable (more than C-infinity function) but is not even continuous in both the variables

Uhh... it [i]is[/i] true that existence and continuity of the partial derivatives implies differentiability. qw's example does not have continuous partial derivatives. I'd rather leave finding the answers in this exercise to the original asker.


What integer is closest to $ \sqrt{500}$?

Well, $ \sqrt{500}$ roughly equals $ 22.36$ so you round down to $ 22$

Of course, when you're in countdown, you unfortunately do not have the aid of a calculator. The closest square to 500 is $ 22^2 \equal{} 484$, so the answer is $ \boxed{22}$.


Prove by the definition of the limit of a squence:  $\lim_{n\to \infty}{\frac{n+1}{n(n-1)}}=0$.

For any $\epsilon>0$, let $N = [\frac{1}{\epsilon}]+1$.
Then for any $n>N$,
$\left|\frac{n+1}{n(n-1)}\right| = \frac{n+1}{n(n-1)} < \frac{2n}{n(n-1)} = \frac{1}{n-1} < \epsilon$.


induction
inequalities
calculus
derivative
inequalities unsolved

Use induction to prove
$(1+a_1)(1+a_2) \cdots (1+a_n) \ge \frac{2^n}{n+1}(1+a_1+ \cdots + a_n)$

[quote="zhaobin"]it is certainly wrong when $a_i \to 0$
what about $a_i \geq 1$?[/quote]

It's true for $a_i \ge 1$.

Yes, the inequality should read $x_i\geq 1$.  Here is a plodding solution by induction (there are much faster ones without induction)

It's enough to show that

$(1+a_1+\ldots+a_n+a_{n+1})+a_{n+1}(a_1+\ldots+a_n)\geq \frac {2(n+1)}{n+2}(1+a_1+\ldots+a_n+a_{n+1})$

or equivalently

$a_{n+1}(a_1+\ldots+a_n)\geq \frac {n}{n+2}(1+a_1+\ldots+a_n+a_{n+1})$

So we need to show that

$(a_1+\ldots+a_n)(a_{n+1}-\frac {n}{n+2})-\frac {n}{n+2}(1+a_{n+1})\geq 0$

Clearly this expression is increasing in $a_{n+1}$ (take the derivative or whatever), so it's enough to consider the case $a_{n+1}=1$.  But it's trivial from here, we need only consider the inequality

$\frac {2S-2n}{n+2}\geq 0$ which is obvious from $x_i\geq 1$ (I set $S=\sum x_i$ to save my fingers in the last step)

It's clear from all this that inequality holds iff the $x_i$'s are all $1$

Let  [tex]x=a_1+a_2+...+a_n , y=a_{n+1} [/tex] The  induction  step  becomes  [tex] n(x-1)(y-1)+2(xy-n)\geq0 [/tex]  which is  true  since  [tex] x\geq1 , y\geq1 , xy\geq n [/tex]

Source : Loren Larson , Problem Solving Through Problems , page 244. There is  a  different  solution  in the  book .


Let $ n>3$ be an integer and $ a_1, a_2,..., a_n \geq 0$ whose sum of squares is 1. Prove that:

$ \frac{a_1}{a_2^{2}\plus{}1}\plus{}\frac{a_2}{a_3^{2}\plus{}1}\plus{}...\plus{}\frac{a_n}{a_1^{2}\plus{}1} \geq \frac{4}{5}(a_1\sqrt{a_1}\plus{}a_2\sqrt{a_2}\plus{}...\plus{}a_n\sqrt{a_n})^2$


Let $x,y,z$ be nonegative real numbers such that $x+y+z=1$, prove that
\[ (xy+yz+zx)\Bigl(\frac x{y^2+z}+\frac y{z^2+x}+\frac z{x^2+y}\Bigl)\geq\frac34.  \]
When does equality hold?

[quote="pvthuan"]Let $x,y,z$ be nonegative real numbers such that $x+y+z=1$, prove that
\[ (xy+yz+zx)\Bigl(\frac x{y^2+z}+\frac y{z^2+x}+\frac z{x^2+y}\Bigl)\geq\frac34.  \][/quote]

Sorry but counterexample: $x=\frac23$, y = 0, $z=\frac13$.

  Darij

[quote="pvthuan"]And is the following true?

For $x,y,z\geq0$ and $x+y+z=1$, prove that
\[ \frac x{y^2+z}+\frac y{z^2+x}+\frac z{x^2+y}\geq\frac94.  \][/quote]

I used Cauchy Swarz inequality and killed the problem...

What about you?

Davron

The last inequality was given at last Federal competition (2006. 1.st problem for third grades) in Serbia and Montenegro (see http://www.imo.yu.org-problems from olympiads-Serbia and Montenegro-2006 MO Final)
Davron,can you please write your solutin?

After Cauchy, it remains to prove that
\[ xy^2+yz^2+zx^2+xy+yz+zx\leq\frac49. \]
But ....

After Cauchy we have to prove that
$4(x+y+z)^2\geq 9(xy^2+yz^2+zx^2+xy+yz+zx)$  or
$4x^2+4y^2+4z^2\geq 9(xy^2+yz^2+zx^2)+xy+yz+zx$  or
$(4x^2+4y^2+4z^2)(x+y+z)\geq 9(xy^2+yz^2+zx^2)+(xy+yz+zx)(x+y+z)$
$4x^3+4y^3+4z^3+3x^2y+3y^2z+3z^2x\geq 6(xy^2+yz^2+zx^2)+3xyz$

But $x^3+y^3+z^3\geq 3xyz$  so it suffices to prove that
$x^3+y^3+z^3+x^2y+y^2z+z^2x\geq 2(xy^2+yz^2+zx^2)$

But by AM-GM we have that $x^3+z^2x\geq 2x^2z$   ;)


calculus
real analysis
real analysis theorems

Why are the steps you follow to maximize and minimize a quantity the same?

If you can fully explain why, your awsome.

Minimizing $Q$ is the same thing as maximizing $-Q$.


Which of these elements has the smallest ionic radius for its most commonly found ion?

(A) O
(B) La
(C) Rb
(D) Mg
(E) N

My answer: D because $Mg^{2+}$ has the most protons for its electrons among the smallest atoms, whose ions are isoelectronic: O, N, Mg

Use the following answer choices for the following questions:

(A) Heisenberg uncertainty principle
(B) Pauli exclusion principle
(C) Hund's rule (principle of maximum multiplicity)
(D) Shielding effect
(E) Wave nature of matter

Can be used to predict that a gaseous carbon atom in its ground state is paramagnetic

My answer: I don't know.  What exactly does "paramagnetic" mean?

Explains the experimental phenomenon of electron diffraction

My answer: I don't know.  What exactly is "electron diffraction"?

The elements in which of the following have most nearly the same atomic radius?

(A) Be, B, C, N
(B) Ne, Ar, Kr, Xe
(C) Mg, Ca, Sr, Ba
(D) C, P, Se, I
(E) Cr, Mn, Fe, Co

My answer: E

Which of the following represents the ground state electron configuration for the $Mn^{3+}$ ion? (Atomic number Mn = 25)

(a) $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{4}$
(b) $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{2}4s^{2}$
(c) $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}4s^{1}$

Any help at all would be greatly appreciated.  Thanks!!!

[quote="Zakary"]Can be used to predict that a gaseous carbon atom in its ground state is paramagnetic[/quote]
Answer: B

[quote="Zakary"]Explains the experimental phenomenon of electron diffraction[/quote]
Answer: E