Hmm, at MIT this year I wasn't a huge fan of the facilities for the team and guts rounds. For the team round, we were sharing one of the classrooms with another team, and it was annoying because there was no convenient way to sit so that we could be working together. Also, we were forced to talk quieter than the other teams that were in the rooms by themselves around tables.

On guts, it was kinda chaotic. Working down a row is frustrating, and I think everyone would do better around a table, and not passing messages down.

I don't know if it was better at Harvard, as this year was my first year, but I think there could be an improvement if the team and guts rounds were in like a big gym, with cafeteria tables or something for everyone. It would be much more relaxed, you could spread out your stuff (calcs, compasses, problems), and communication would be easier.

This is just a thought. Feel free to comment. Otherwise, I thought the contest was very well run. The MIT students were very friendly, very helpful, and the problems were fun.

[quote="Elemennop"]1) Why did you have to be quiet just because another team was in the room?

2) You didn't all have to be in a row... you could take multiple rows up (which most teams did).[/quote]

Exactly. We fish shared a room, too, and didn't have a problem at all. For the guts round, we  :ninja: 'd into two rows of 3 and one of 2, which made it at least fairly easy to get the problems and see each other's work.

it was fine...stop complaining...how else do you have 400 people in a room?

guts was awesome

i guess my team had our own room

PWND


for $ a,b,c\in \mathbb{R}_{\plus{}}$ prove ineq:
$ \sum_{cyc}\frac{1}{(2a\plus{}b)^{2}}\geq \frac{1}{\sum_{cyc}ab}$

[quote="Dumel"]for $ a,b,c\in \mathbb{R}_{ \plus{} }$ prove ineq:
$ \sum_{cyc}\frac {1}{(2a \plus{} b)^{2}}\geq \frac {1}{\sum_{cyc}ab}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141336
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218503


Find all perfect squares such that the division of them by 11 yields a partial quotient that is a prime number and the remainder is 4.

I know Gyan already posted the correct answer, but since I'm a "budding" problem-solver I'd thought I'd post my method to see if it's correct or get comments on how to improve it.

[hide]
p2 = 11q + 4
p2 = 4 (mod 11)
Consder all squares modulo 11. You'll get:
22 = 4  (mod 11) and 92  = 4 (mod 11), so p=2 or p=9.
Substituting back into the original equation, we see that p2=81 is the only perfect square that matches what we were given in the question.
[/hide]

Correct? :blush:

Unfortunately, not quite.
[hide]
What about p = 13 or 20 or 24 or 31 or ...
They are all 2 or 9 mod 11 as well!
[/hide]

D. You are almost there.
[hide]as you say  x= + 2 or -2  (mod 11)
This gives square number as   (11k  + 2)^2 or (11k-2)^2

or 11*11k^2  +-  44 k +4

which is    11k(11k +-  4)  + 4
When you divid by 11  the remainder is 4 but the quotient is

 k(11k +-4)
which can be prime only if k=1  and even there, you can use only the  minus sign;  as 11 +4 is not prime, hence the only solution is when k=1

and  we get  11(11-4) +4 = 81. [/hide]

Of course nothing special about 11 here (or, for that matter 4)  answer to similar problem , if one  had the number 19 instead of 11 the answer would be  (19-2)^2   and  (19+2)^2    and  only one (or none) may be valid...


Let ABC be an acute-angled triangle. K is a point in it's plane. If x,y,z are the distances from K to the sides of the triangle then:

4(x+y+z) 2  + (AB+BC+CA) 2   \geq  4(KA+KB+KC) 2

[i]unknown source[/i]

I think K must be an interior point of triagle. Must'nt it?

Namdung

Useless post, I know :D : isn't this from a Hungarian team selection test?

This problem seems to be wrong (even when K lies in interior part of triangle). Let take triangle ABC with AB=AC=2, BC=2sqrt(2). K is midpoint of BC then x=0, y=z=1, KA=KB=KC=sqrt(2) and the ineq
4(x+y+z)^2+(AB+BC+CA)^2>=4(KA+KB+KC)^2 does not hold. Ofcourse, we can varry this triangle to get an counter-example with ABC is acute-triangle and K lies in interior of triangle.

Let check again, friends!

Namdung

I looked it up among my files and the problem appears in exactly this form in a file containing the 1'st of the Hungarian selection tests for the 43'rd IMO. The file can even be downloaded from this site.


calculus
integration
limit
logarithms
trigonometry
function
real analysis

Evaluate

\[\lim_{n\to\infty}\int^\frac{\pi}{3}_\frac{\pi}{4}n\ln\left[{{\left(1+\frac{\sin \theta\sec^2\theta}{n}\right)}^{\cos \theta}\cdot{\left(1+\frac{\cos \theta}{n}\right)}^\cot \theta}\cdot{\left(1+\frac{\cot \theta}{n}\right)}^\sin \theta\sec^2\theta}\right] d\theta\]

Requesting a clarification on the notation:

Is that $\left[\ln\left(1+\frac{\sin\theta\sec^2\theta}n\right)\right]^{\cos\theta}$ ?

Or is it  $\ln\left[\left(1+\frac{\sin\theta\sec^2\theta}n\right)^{\cos\theta}\right]$ ?

Or are all three factors inside the logarithm?

To Kent Merryfield:

Oh! I'm sorry, I will correct.

Regards.

kunny

No wonder I couldn't do it. It really blasted me. Now with the appropriate correction, the problem is way easier.

$\lim_{n\to\infty}\,n\ln\left[\left(1+\frac{f(x)}n\right)^{g(x)}\right]=f(x)g(x).$

We're well away from any singularities of any of the functions, and the convergence will be uniform, so we can interchange integral and limit. We get

$\int_{\frac{\pi}4}^{\frac{\pi}3} \tan\theta+\frac{\cos^2\theta}{\sin\theta} + \sec\theta\,d\theta.$

We should be able to finish by using standard integration techniques.


Let $S$ be the set consisting of integers in the form of $x=p^{i}q^{j}$ where $p$ and $q$ are primes and $i,j$ are non-negative integers. Compute $\sum_{x \in S}\frac{1}{x}$


The number a,b,c are the digits of a three digit number which satisfy 49a+7b+c=286. What is the three digit number
(100a+10b+c)?

[hide]If $a=5$, that leaves 41 left.

So, $b=5, \ c=6$

So the number is $556$ [/hide]

[hide]
I used guess and check.
First let's start with $a$.
$49a$ has to be really close to 286, without going over (the price is right)
The closest it can be is if $a=5$.
So $49a=245$.
That leaves us with $7b+c=41$.
Same thing for $b$, the closest without going over it can be is if $b=5$.
That leaves $c=6$.
The number is thus
$\boxed{556}$[/hide]


induction
pigeonhole principle
graph theory
combinatorics solved
combinatorics

There are 51 senators in a senate. The senate needs to be divided into n committees, where each senator is on one committee. Each senator hates exactly three other senators. However, if senator A hates senator B, then senator B does NOT necessarily hate senator A. Find the smallest value of n so that it is always possible to arrange the committees so that no senator hates another senator on his committee.

We will prove that n=7.
By induction on number of senators.
Suppose we can organize 7 committees if we have m senators where each of them hates at most 3 other senators.
Consider senate with m+1 senatrors. Due to Pigeonhole principle there exists senators which is hated by at most 3 others senators. We exclude this senator from senate and remaining senators we organize to 7 committees. Now return excluded senator. He hate 3 senators and he is hated by at most 3 senators - totally 6 collisions  ==> since we have 7 committees there is committee s.t. there are no collisions for returned senator within this committee!
That's all.

You can easely construct example s.t. it is impossible organize 6 committees.
(hint: 7 senators, and for any two senators one of them hates other).

with turan theorem prove in Complementary graph of this problem there exist one
45-complete graph.
now we  all 45 senator cast in one committee and other 6 senator cast  in 6 difference comitee.
You can easely construct example  it is impossible organize 6 committees.

Let  be a graph with  vertices and without a  ( is a complete graph with n vertices).
Then the maximum number of edges in G is equal to

m(n,k)=k-2/k-1.(n-r)(n+r)/2  + r(r-1)/2

if

n=(k-1)t+r   0=<r=<k-2.

now we have not less than n^2 - 3n/2 edges in Complementary graph. then now change n=51 and k=45 in turan theorem we say
Complementary graph have a 45-regular graph .it is trivial any
45 senator not hate from other in 45. these 45 senator can cast in
one comitee and other 6 senator cast in 6 different comitee.

i cant speak and i cant live.  :(

Hmm... It is wrong application of Turan theorem. It is easy to construct example s.t. there is no 45-regular (!!!) subgraph.


hav a total of 5 problems in inqualities concerning induction.i tried to solve them pls tell me whether they r correct and in some problems i got stuck..pls help me proceed in those problems and tell whether the problems i solved r correct and if not correct me and tell me how to go about it.

--------------------------------------------------------------------------------

1.use induction to show that n^2>n+2 for n>2.
well the basic step is true.
for n=k,k>k+2is true .i hav to prove for n=k+1
adding 2k+1 to both sides i get k+2k+1>k+2+2k+1
then (k+1)>(k+1)+2+2k
since 2k is +ve (k+1)>(k+1)+2
hence true for n.

--------------------------------------------------------------------------------

2.by induction show that 2^n>2n+3 for each n>=4.
for n=k,2^k>2k+3 is true.for n=k+1
taking 22^k>2(2k+3)
2^k+1>2k+2+2k+4
then 2^k+1>2(k+1)+3+2k+1 since 2k+1 is +ve hence 2^k+1>2(k+1)+3.
hence true for n.

--------------------------------------------------------------------------------

3.by induction prove n^2>=6n+7 for n>=7.
for n=k,k6k+7
addind 2k+1 to both sides
k+2k+16k+2k+7+1
(k+1)6(k+1)+2k+2
pls pls...tell me whether this is correct or how to proceed after this.if it is wrong tell me how to do it.

--------------------------------------------------------------------------------

4.by induction prove (n+1)<2nfor n3
5.by induction prove 2^n>nfor n5
i hav tried these questions too but nothin has come out of it...no proper result.

--------------------------------------------------------------------------------

I HAV TRIED ALL THESE PROBLEMS MANY TIMES BUT STILL I AM NOT ABLE TO SOLVE THEM PROPERLY ....PLEASE HELP ME...ANY HELP WOULD BE GREATLY APPRECIATED

[quote="madhuri"]hav a total of 5 problems in inqualities concerning induction.i tried to solve them pls tell me whether they r correct and in some problems i got stuck..pls help me proceed in those problems and tell whether the problems i solved r correct and if not correct me and tell me how to go about it.[/quote]
It would be dishonorable for us to do your homework for you, so instead I will give you a lesson in Proof by Induction.

[quote="madhuri"]
1.use induction to show that n^2>n+2 for n>2.
well the basic step is true.
for n=k,k>k+2is true .i hav to prove for n=k+1
adding 2k+1 to both sides i get k+2k+1>k+2+2k+1
then (k+1)>(k+1)+2+2k
since 2k is +ve (k+1)>(k+1)+2
hence true for n.[/quote]
You seem to have a typing problem: you are losing letters. For example, you said, "for n=k, k>k+2 is true," when what you meant was "for n=k, k^2 > k+2 is true." No matter how correct your mathematics is, your teacher will mark it wrong if you do not write that mathematics correctly.

The biggest mathematical flaw with your above solution is that you did not write down the first step. You merely say, "well the basic step is true." If you did the first step, write it down! Proof by induction is like a piston pump. First you fill the cylinder with water and seal it tight (this is called "priming the pump"). Then as you pump the piston, the forward stroke forces water out of the piston into your bucket and the back stroke creates a vacuum that pulls water from the well into the cylinder. But if you don't prime the pump, the forward stroke merely compresses the air in the cylinder and the back stroke lets the air expand again. No water moves. You cannot skip the first step.

The first step for this problem is the smallest possible value for n, which is n=3. Is the statement true for n=3? Plug the number in and prove it numerically!

Once you prove the n=3 case, you proceed from that starting point, but you must do so carefully, because the proof shifts to abstraction. Your statement (corrected version), "for n=k, k^2 > k+2 is true," is too bold. The variable k represents all values for which you know the hypothesis, k^2 > k+2, is true. We know that k=3 gives a true statement, and suspect that it works for other values of k too. We want to use the n=3 case to prove the n=4 case, use the n=4 case to prove the n=5 case, use the n=5 case to prove the n=6 case, etc. However, so that we can write this up as a single step, we say k instead of 3, 4, 5, 6, etc. One step is much easier to write up than infinitely many steps.

So instead of saying, "for n=k, k^2 > k+2 is true," the proper statement is "Let k be an integer with k>2 such that k^2 > k+2." Mathematically, you know that at least one such k exists, the k=3 case that you just proved numerically; therefore, you can make that statement. The next statement is either, "Let n = k+1," or "Consider the k+1 case," depending on whether you want to use the variable n in this step or save it for the conclusion. Using n or not here is purely a matter of style. The finish this step by using the n=k case to prove the n=k+1 case. This second step is essentially creating the lemma, "If the hypothesis is true for n=k, then it is also true for n=k+1."

The third step of a proof by induction is the conculsion where you combine steps 1 and 2. It is nothing except the statement, "Therefore, by induction n^2 > n+2 for all integers n with n>2." That two-word phrase, "by induction," summarizes a great quantity of mathematical derivations. If you leave off those two words, your proof is not valid. What "by induction" summarizes is, "Since we know n^2 > n+2 is true for n=3, then by the lemma, it is true for n=4. Since we know n^2 > n+2 is true for n=4, then by the lemma, it is true for n=5. Since we know n^2 > n+2 is true for n=5, then by the lemma, it is true for n=6. Since we know n^2 > n+2 is true for n=6, then by the lemma, it is true for n=7. ... (Repeat this for all values of n)," which is much too long to actually write down.

By the way, your textbook has a rotten set of problems. All five problems are more easily proven directly than by induction. For example, the direct proof for problem 1 is:
(n-2)^2 > 0 for all n > 2 because the square of a positive number is positive.
n^2 - 4n + 4 > 0 for all n > 2
n^2 > 4n - 4 for all n > 2
n^2 > n + (3n-4) for all n>2
n > 2 means 3n > 6 and 3n - 4 > 2, so n^2 > n + 2 for all n > 2

For a classic problem by induction, prove by induction that the sum of the first n positive odd integers is n^2. For example the sum of the first 4 positive odd numbers is 1+3+5+7 = 16 = 4^2.

[quote="madhuri"]
4.by induction prove (n+1)<2nfor n3
5.by induction prove 2^n>nfor n5
i hav tried these questions too but nothin has come out of it...no proper result. [/quote]

If these are prove n+1 < 2n for n>3 and prove 2^n > n for n>5, then they are easier than the ones you have done already.

Erin Schram
Mathematician

Please madhuri, avoid putting titles in all caps. I'd like them used only for acronyms, like for contests and the like.


Assume that $x+\frac{1}{x}=1,xy+\frac{1}{xy}=\sqrt2$.  Find the value of $y+\frac{1}{y}$.

That's right, mathfanatic.
Excuse me,Can you use complex number?If so,you would solve this problem by that.

kunny


If a fair, six-sided die is tossed 10 times and the numbers are recorded each time. What is the number of 3's that will appear within this sequence of 10 numbers with the greatest probability?

[hide]The probability of a $3$ being rolled is $\frac17$, so the number of $3$s that will appear with the greatest probability will be $\frac{10}7$ rounded, or $1$.

The second most likely would be $2$.[/hide]

[hide="Sol"]The probability of exactly one three is given by
$\frac{10 * 5^{9}}{6^{10}}$
We choose which of the ten dice has the three, and for the other 9 dice there are five options each.
P(2 threes)
Choose two dice, assign 5 values to other 8:
$\frac{\binom{10}{2}\cdot5^{8}}{6^{10}}$

In general, for $N \le 10$
P(N threes) = $\frac{\binom{10}{N}\cdot5^{10-N}}{6^{10}}$
A small chart of that shows:
P(0) = $\frac{5^{10}}{6^{10}}$ = 0.1615
P(1) = 0.323
P(2) = 0.290
P(3) = 0.155
P(4) = 0.054

From this it becomes clear that we expect to get exactly ONE three per ten rolls.
The expected value can be computed fairly easily as well.
[/hide]

EDIT:
Darn it!
Beaten again.
Nikita, you got the right answer, but with the wrong probability.


vector
invariant
geometry
geometric transformation
rotation
induction
linear algebra

[quote="capriccioso"]Let V be a finite dimensional vector space over C, with hermitian
inner product (,) - think of C^n with (x,y)=x*y if you wish.

The key lemma for any linear map A: V -> V which lets us show that it
can be diagonalized with respect to an orthonormal (or unitary) basis
when A is symmetric (or hermitian) is this:

	If a subspace W of V is A-invariant, then its orthogonal
	complement W^\perp is A*-invariant, where A*:V->V is defined
	by the property (A*x,y) = (x,Ay).

PROBLEM B.2.  Prove this lemma.

In the following, we will apply this lemma to show we can diagonalize
not only symmetric (or hermitian) maps, but also skew-symmetric (or
skew-hermitian), unitary, or, more generally, normal (A*A=AA*) maps in
the same way.

PROBLEM B.3.  Check that hermitian (A*=A), skew-hermitian (A*=-A), and
unitary (A*A=I=AA*) maps are all normal.

To apply this do this you will need another lemma (cf. the first part
of Curtis 32.13, but this is more general):

     Suppose L, M are commuting linear maps V -> V, that is LM = ML.
     Then L and M have a common 1-dimensional invariant subspace W,
     i.e. a common eigenvector w with span({w}) = W.

PROBLEM B.4.  Prove this lemma.  [Compare to the lemma from 4/20, and
contrast with the following: commuting linear maps need not have the
same invariant subspaces, since obviously a scalar multiple of the
identity (for which any subspace of V is invariant) commutes with any
L (which may have only particular invariant subspaces).  You might also
think about whether the following is true for a vector space V over an
arbitrary field K:

     Suppose L, M are commuting linear maps V -> V, that is LM = ML.
     Then L and M have a common irreducible invariant subspace W.

In particular, when K is algebraically closed, W is 1-dimensional,
and when K is a real field, 1- or 2-dimensional; for example any
pair of real 2-by-2 rotation matrices commute, but generally R^2
is the only invariant subspace.]

Now, to show that a normal linear map can be diagonalized, let L=A and
M=A*, and observe that A**=A.  Thus the n-1 dimensional subspace
W^\perp is also invariant under under both A and A*, so we can use
the dimension reduction argument as before (induction on n if you like).

PROBLEM B.5.  Carry out the details of the above paragraph.[/quote]

I am quoting a copy of this post for posterity; it appears to be part of a take-home final for this http://www.gang.umass.edu/~kusner/class/545hw course.

yes, this is part b of a takehome final from my advanced linear algebra class. I am done with part B now, and struggling with part C.

To compute eigenvectors and eigenvalues of hermitian matrix A, we find the maximum of the associated quadratic form Q(x)=x*Ax among the unit vector x. which being a closed bounded set. Show that the maximum pooint v is an eigenvector of A belonging to the largest eigenvalue, the maximum value Q(v).

any ideas?


How many digits are in the decimal representation of: $10^{100}-9^{100}$?

[hide="guess"]I am guessing 100 becuase 10^2 and 9^2 has 2 integers and 3 for 3 etc. so.....believe that it is 100[/hide]

I think it is [hide]
99. 1 second :D[/hide]

i think its[hide]100 one second also because it is exactly 101 with 10^100 but 9 ^100 is considerably smaller that it wont get rid of a digit in the number so that there will only be one digit lost in the subtraction of 9^100. This is kind of confusing but I dont know a better way to explain it.[/hide]

[quote="biffanddoc"]yes but 10^100 has 100 digits right? then all you would need is to subtract 1 to get rid of a digit[/quote]

doesn't $10^{100}$ have $101$ digits?

[quote="biffanddoc"]o yeah duh...
then 10 is the right answer[/quote]

You mean $100$??  :P

aaaaahhhhhhh
im losing my brain (not like i ever had any)
 :fool:  :wallbash_red:

[hide="nevermind"]can we use a calculator? I got
[hide]5[/hide]

way that my dad showed me (still uses calculator though)

[hide]$\log_{10}n+1$ rounded down to nearest whole number gives the number of digits for n in decimal representation

$10^{100}$ has 101 digits

$9^{100}$ has $\log_{10}9^{100}+1=100*\log_{10}9+1=96.424...$ (used calculator)

which rounds down to 96

$101-96=5$[/hide][/hide]

[quote="Klebian"][quote="t0rajir0u"][quote="pianoforte"]can we use a calculator? I got
[hide]5[/hide]

way that my dad showed me (still uses calculator though)

[hide]$\log_{10}n+1$ rounded down to nearest whole number gives the number of digits for n in decimal representation

$10^{100}$ has 101 digits

$9^{100}$ has $\log_{10}9^{100}+1=100*\log_{10}9+1=96.424...$ (used calculator)

which rounds down to 96

$101-96=5$[/hide][/quote]

Er, that would be the approximate number of digits in the [i]quotient[/i], not the difference :D  If you subtracted a $1$-digit number from a $6$-digit number would you expect to get a $5$-digit number?[/quote]

I would. 100,000 - 1 = 99,999[/quote]

By the same token, $100,000 - 50,000 = 50,000$ but we are subtracting a $5$-digit number from a $6$-digit number.  :D

UUHhhhhh so is the answer 100 digits or 99 digits.........hmmmm i guess it is 100......... What is the REAL answer to this problem?

The real answer is:[hide]Are you sure you want to know?[hide]100[/hide][/hide]


For three points $A,B,C$ in the plane, we define $m(ABC)$ to be the smallest length of the three heights of the triangle $ABC$, where in the case $A$, $B$, $C$ are collinear, we set $m(ABC) = 0$. Let $A$, $B$, $C$ be given points in the plane. Prove that for any point $X$ in the plane,

\[ m(ABC) \leq m(ABX) + m(AXC) + m(XBC).  \]

Hmmm I'm surprised no one has posted a solution to this yet. Anyways here goes:

We first prove a couple lemmas. Let $M(ABC)$ be the length of the longest side of triangle $ABC$.

Lemma 1. Given a point P either inside or on a side of triangle ABC, $M(ABP) \le M(ABC)$.

Suppose on the contrary that this is false; then WLOG $AP > M(ABC)$. But extend AP to hit BC at Q, and suppose R is on AB such that RQ || AC. We have

$AP \le AQ \le AR+RQ = \frac{CQ \cdot AB}{BC}+\frac{BQ \cdot AC}{BC}\le \max(AC, BC) \le M(ABC)$,

a contradiction. Hence lemma 1 is proven.

Lemma 2. Given a point P on segment AB of any triangle ABC, we have $m(PAC) \le m(ABC)$.

Let Q be the point on AC s.t. BC || PQ. By lemma 1, $M(PAC) \ge M(AQP)$. Then

$m(PAC) = \frac{2[PAC]}{M(PAC)}\le \frac{2[PAC]}{M(AQP)}= \frac{2[ABC]}{M(ABC)}= m(ABC)$,

where the second-to-last equality comes from $[PAC] = \frac{AP}{AB}\cdot [ABC]$ and $M(AQP) = \frac{AP}{AB}\cdot M(ABC)$. This proves lemma 2.

Now consider the original problem.

Case I: X lies in the interior of or on a side of ABC.

$m(XAB)+m(XBC)+m(XCA) = \frac{2[XAB]}{M(XAB)}+\frac{2[XBC]}{M(XBC)}+\frac{2[XCA]}{M(XCA)}$
$\ge \frac{2[XAB]+2[XBC]+2[XCA]}{M(ABC)}= \frac{2[ABC]}{M(ABC)}= m(ABC)$,

with the inequality resulting from lemma 1.

Case II: One of the vertices of ABC lies in the interior of or on a side of the triangle formed by X and the other two vertices (WLOG A lies in XBC).

Extend BA to meet XC at D. We have

$m(XAB)+m(XBC)+m(XCA) \ge m(XBC) \ge m(BDC) \ge m(BAC)$

where the last two inequalities follow from lemma 2.

Case III: A, B, C, and X form a convex quadrilateral (WLOG in that order).

Let AC and BX meet at Y. Using lemma 2, we have

$m(XAB)+m(XBC)+m(XCA) \ge m(AYB)+m(CYB)+m(XCA)$
$\ge m(AYB)+m(CYB) = m(AYB)+m(CYB)+m(AYC)$.

However, as was shown in Case I, $m(AYB)+m(CYB)+m(AYC) \ge m(ABC)$.

Therefore, the inequality holds in all three cases, and these are all the cases, so we are done.


algebra
polynomial
trigonometry
AMC
AIME
number theory
relatively prime

The polynomial $ P(x)=\left(1+x+x^2+x^3+\ldots+x^{17}\right)^2-x^{17}$ has $ 34$ complex roots of the form $ z_k=r_k[\cos\left(2\pi a_k\right)+i \sin \left(2 \pi a_k\right)]$, $ k=1,2,3, \ldots , 34$, with $ 0 < a_1 \le a_2 \le a_3 \le \ldots \le a_{34}<1$ and $ r_k >0$. Given that $ a_1+a_2+a_3+a_4+a_5=\frac{m}{n}$, where $ m$ and $ n$ are relatively prime positive integers, find $ m+n$.

I'm going to post my personal solution, because I want to know if there's a better way to do it, especially the part where I use some rules about cosine:[hide="Solution"]Since $ 1+x+x^2+x^3+\ldots+x^{17}$ is a finite geometric series with $ a=1$ and $ r=x$, we can rewrite $ P(x)=0$ as:
\begin{align*}&\left(\frac{1-x^{18}}{1-x}\right)^2-x^{17}=0 \\
&\implies 1-2x^{18}+x^{36}=x^{17}-2x^{18}+x^{19} \\
&\implies x^{18}+\frac{1}{x^{18}}=x+\frac{1}{x}\end{align*}If we let $ x=r \cdot\left( \cos\theta + i \sin \theta\right)$ Then, the equation becomes $ \cos 18 \theta = \cos \theta$, so either $ 18\theta=2\pi k + \theta$ or $ 18\theta=2\pi k - \theta$. The equations imply $ \theta=\frac{k\left(2\pi\right)}{19}$ or $ \theta=\frac{k\left(2\pi\right)}{17}$.
Thus, the five smallest values of $ a_k$ = $ \frac{1}{19}$, $ \frac{2}{19}$, $ \frac{3}{19}$, $ \frac{1}{17}$, $ \frac{2}{17}$, so $ a_1+a_2+a_3+a_4+a_5=\frac{159}{323}$, so $ m+n=\boxed{482}$.[/hide]

[hide="Note that"]
\begin{align*} \left(\frac {x^{18} - 1}{x - 1}\right)^2 - x^{17} & = \frac {x^{36} - 2x^{18} + 1 - x^{17}(x - 1)^2}{(x - 1)^2} \\
& = \frac {x^{36} - x^{19} - x^{17} + 1}{(x - 1)^2} \\
& = \frac {(x^{17} - 1)(x^{19} - 1)}{(x - 1)^2} \end{align*}
The rest is the same.

Edit: Wait actually, thats what you did. But in this form it directly follows the solutions are $ \text{cis}\,\left(\frac {2\pi k}{17\, \text{or}\, 19}\right)$, with the exception of $ x = 1 \Longleftrightarrow k = 0$.
[/hide]

Edit: This problem is from an AIME; this is the method I used when taking that AIME as a practice test, so I already had the solution in mind.  :P

@azjps: that's a much better way to solve it! How long did this problem take you?


function
induction
algebra
polynomial
algebra proposed

I created this problem, I hope it is a good one  :lol:
Find all the functions $f: R \rightarrow R$ satisfiying :
$(f(x))^3-(f(y))^3=f(x-y)(f(x^2)+xy+f(y^2))$ for all real numbers $x,y$

[quote="senouy"][quote="erdos"]ehhh..... do you find it too easy or too difficult ? :?[/quote]

Iam still trying to solve it.I reached easily the following facts wich help me conjecture that your function is f(x)=x:
f(0) is 0
f(1)is 1
f is an odd function
for every n f(n)=n
f(half)=half
Any hint?
Thanks.[/quote]

By using the fact that f is odd,we can show that f(1/n)=1/n:comparing, using the caracteristic of the funtion,and the fact that f is odd
cubic f(2/n)-cubicf(1/n)
cubicf(1/n)-cubicf(-1/n)
This has been the most difficult fact to show for me.
After this ,using the indution we can show that f(p/n)=p/n
And here i used the fact that every real number is a limit of a rationel sequence supposing f continuos.
It lead to f(x) = x.

Setting $x=y$ yields $f(x)^3=f(x)f(x^2)$. Therefore $f(0)=0$ and $f(x^2)=f(x)^2$ for all real $x$ [EDIT: if $f(x) \ne 0$, but that does not interfere with the rest of the proof].

Now let a be an arbitrary real number sufficing $f(a) \ne 0$.
Set $x=2a$ an $y=a$ to obtain $f(2a)^3-f(a)^3=f(a)(f((2a)^2)+2a^2+f(a^2)) \Leftrightarrow f(2a)^3=f(a)(f(2a)^2$ $+2f(a)^2+2a^2)$ (1).
Set $x=a$ and $y=-a$ to obtain $f(a)^3-f(-a)^3=f(a-(-a))(f(a^2)+a(-a)+f((-a)^2)) \Leftrightarrow f(2a)=\frac{2f(a)^3}{2f(a)^2-a^2}$ (2).
Plugging (2) into (1) results $2(f(a)^2-a^2)(8f(a)^6+2f(a)^4a^2-4f(a)^2a^4+a^6)=0$ but since the second factor, regarded as a polynomial in $\frac{a^2}{f(a)^2}$, has no positive roots one gets $f(a)^2=a^2$.

The result so far is: For every real x $f(x)=x$, $f(x)=0$ or $f(x)=-x$.

Furthermore, I have a proof for $f$ being the zero function or the identity on the rational numbers, based on some inductive arguments like $f(\frac{1}{2^k})=\frac{1}{2^k} \Rightarrow f(\frac{1}{2^{k+1}})=\frac{1}{2^{k+1}}$ and $f(\frac{n}{2^k})=\frac{n}{2^k} \Rightarrow f(\frac{n+1}{2^k})=\frac{n+1}{2^k}$ but I do not really see how to show this for real numbers without using another condition like continuity or monotony.

[quote="NoName"]Setting $x=y$ yields $f(x)^3=f(x)f(x^2)$. Therefore $f(0)=0$ and $f(x^2)=f(x)^2$ for all real $x$.[/quote]

Why? If $f(x) = 0$ how can you derive from it that $f(x^2)=0$?

Pierre.

Very nice exercice Badr!

f=0 satisfie.
So sup f#0
1) x=y=0  then f(0)=0
2) x#0 , y=0  then f(x)^3=f(x).f(x^2)
                     then f(x)=0 or f(x)^2=f(x^2)
                     As f"# 0      then f(x)^2=f(x^2)
     It follows that if x>=0   then f(x) >=0
3) f(x)^2=f(x^2)=f((-x)^2)=f(-x)^2  then f(x)=f(-x) or f(x)=-f(-x)
   Sup f(x)=f(-x) then f(x)^3-f(-x)^3=0=f(2x).(2f(x^2)-x^2)
                                        then f(2x)=0 or f(x^2)=(x^2)/2
                                        or f#0 eject
        sup f(x^2)=(x^2)/2 we have f(1)=1/2 contradiction with (2) f(1)=f(1)^2
        So f is odd and only x who verifie f(x)=0 is x=0 otherwise f=0.
4)Using now what Noname did f(x)^2=x^2 so f(x)=x or f(x)=-x
  As f(x)>=0 if x>=0 we eject f(x)=-x
And we done! ;)

[quote="AYMANE"] then f(x)=0 or f(x)^2=f(x^2)
As f"# 0 then f(x)^2=f(x^2)[/quote]
I think there is a mistake here  :?
you have $\forall x \in \mathbb R : f(x)=0$ or $f(x)^2=f(x^2)$
you should have $\forall x \in \mathbb R: f(x)=0$ or $\forall x \in \mathbb R : f(x)^2=f(x^2)$
then you could say : since $f$ is not the zero function, we have $\forall x \in \mathbb R : f(x)^2=f(x^2)$

Sorry i wrote in French :blush:
we supposed thet f#0.
f is odd and the only value which verifies f(x)=0 is x=0 :D

nice problem Badr ! i knew u r a good problem solver, but this is the first time i see your own problem, & its very good  :D

When i had lunch i saw my mistake.
5) sup x>y>0   then f(x)>=0 and f(y)>=0
 f(x)^3-f(y)^3=f(x-y).(f(x^2)+f(y^2)+xy)
                     =f((x-y)^1/2))^2.(f(x)^2+f(y)^2+xy)
                     >=0
So f(x)^3-f(y)^3=(f(x)-f(y))(f(x)^2+f(y)^2+f(x)f(y))>=0
then  f(x)>=f(y) f is increasing!

So f must be in the form :
f is odd
f(x)=0 if x is in [0,a[
f(x)=x if x is in [a,&[   where a real >=0

Try to prove a=0

Now sup that a#0 , 0<y<a and x>a then f(y)=0 and f(x)=x
f(x)^3-f(y)^3=f(x)^3=x^3=f(x-y)(f(x)^2+xy)=f(x-y)(x^2+xy)
and choose x and y so that  x-y<a then f(x-y)=0 then x^3=0 contradiction
So a=0

and only fonctions are f(x)=0 or f(x) =x

Correct me if not

What is your solution Badr :?

[quote="AYMANE"]So f must be in the form :
f is odd
f(x)=0 if x is in [0,a[
f(x)=x if x is in [a,&[ where a real 0 [/quote]
Well, I didn't understand how you got that  :?
I will post my solution next week  :lol:

Cause NONAME proved
for x real f(x)(f(x^2)-x^2)=0
and not f(x^2)=x^2 :D

Pourquoi une semaine :D

I think AYMANE is right. I cannot explain why I have not seen $f(x) \ge 0$ for $x \ge 0$ but the result follows immediatly.

many problems can be created in the same way, pick an identity and replace some of it terms by the function! :)

Really ?  :?
So why didn't you create this problem before me ?
I was the first who thought of that, you see ?  :lol:

Well you were the first one to use this identity! and to solve what you created using this identity! I am not against you, I was just saying that most functional equations are created this way.... ;)


algebra
polynomial
number theory proposed
number theory

Obviously from Prime Number Theorem !

[quote="nhat"]Prove that there do not exist polynomials $P$ and $Q$ such that :
             $t(x)=\frac {P(x)} {Q(x)}$ for all $x \in N$   ($t(x)$:the number of primes $p$ with $p \leq x$)[/quote]


[quote="gb2124"]Suppose that I is the incenter of triangle ABC and l' is a line tangent to the incircle. Let l be another line intersecting AB,AC,and BC at
C',B' and A',respectively.Draw a tangent other than BC from A' to the incircle, and let this tangent inersect l' at A1.Define B1 and C1 similarly.Prove that AA1,BB1,CC1 are concurrent.[/quote]
Forget about $ A', A_1$ for a moment, consider just $ B' \in CA, C' \in AB.$ Incircle tangents form $ B', C'$ other then $ CA, AB$ cut its arbitrary tangent $ l'$ at $ B_1, C_1.$ Hexagon $ BCB'B_1C_1C'$ is tangential to the incircle of $ \triangle ABC$ $ \Longrightarrow$ $ BB_1, CC_1, B'C'$ are concurrent at a point $ P$ by Brianchon theorem. Now add $ A' \equiv BC \cap B'C'$ and tangent form $ A'$ other than $ BC$ cuts $ l'$ at $ A_1.$  In exactly the same way, $ CC_1, AA_1, C'A'$ are concurrent. But $ CC_1$ cuts the line $ l \equiv A'B'C'$ at exactly one point $ P$ $ \Longrightarrow$ $ AA_1$ also goes through $ P.$

This problem had already been posted. Iran TST 2008-2

http://www.mathlinks.ro/viewtopic.php?p=1132694#p1132694


limit
integration
calculus
real analysis
real analysis unsolved

Show that $ \lim_{n \to \infty} \int_{0}^{1} \frac{nx}{1 \plus{} n^2x^2} dx \equal{} 0$.

[quote]General Integral(1)
Show that $ \lim_{n \to \infty} \int_{0}^{1} \frac{nx}{1 \plus{} n^2x^2} dx \equal{} 0.$ [/quote]

$ \lim_{n \to \infty} \int_{0}^{1} \frac{nx}{1 \plus{} n^2x^2} dx$ = $ \lim_{n \to \infty}\frac{\int_{0}^{1} \frac{2n^{2}x}{1 \plus{} n^2x^2} dx}{2n}$

on evaluating the integral we finally get this as

 $ \lim_{n \to \infty} \frac{\ln(1\plus{} n^{2})}{2n}  \equal{}  0$

Alternatively: One could bound $ f_{n}(x) \equal{} \frac{nx}{1\plus{}n^{2}x^{2}}$ by $ g(x) \equal{} x^{\minus{}\frac{1}{2}}$.
Since $ f_{n} \to 0$, the result follows by the Dominated Convergence Theorem.


[quote="Frogger012"]Ok, as probably many of you probably know the compound interest formula is A=P(1+(R/N))^(NT)
A=Final Amount
P=Principle
R=Interest Rate
N=Compound Rate(times per yr.)
T=Time(yrs)

I wanted to solve this equation for all variables and got hung up on N and R.

Any help solving for N and R would be appreciated.[/quote]

May be because it is a formula, not an equation. Of course it may be seen as an equation, but then, it does not have too much sence to solve it.

Best,

Thanks miraculouspostmaster

The reason I want this solved for all the different variable is because I'm writing a program for my calculator, and I just wanted it to be able to solve for all variables, whether I use them or not.


[hide="Solution"]CRT guarantees uniqueness $\bmod 180$.  Moreover, $x \equiv 2 \bmod 36$.  From here it is not hard to see $\boxed{x \equiv 38 \bmod 180}$.[/hide]

[quote="parrot"]find all numbers x that satisfies the congruences: x is congruent to 2 mod4, x is congruent to 3 mod5 and x is congruent to 2 mod9[/quote]

[hide="Chinese Remainder Theorem"]
\begin{eqnarray*} x&\equiv &2 \pmod {4}\\ x&\equiv & 3 \pmod {5}\\ x&\equiv & 2 \pmod {9} \end{eqnarray*}

$M_1=45,M_2=18,M_3=10$

$45b_1\equiv 1\pmod {4}\implies b_1\equiv 1 \pmod {4}$
$18b_2\equiv 1 \pmod {5}\implies b_2\equiv 2 \pmod {5}$
$10b_3\equiv 1 \pmod {9}\implies b_3\equiv 1 \pmod {9}$

$\therefore x\equiv (2)(1)(45)+(3)(2)(18)+2(1)(10)\equiv \boxed {38} \pmod {180}$
[/hide]


Harold, Tanya, and Ulysses paint a very long picket fence.
Harold starts with the first picket and paints every $h$th picket;
Tanya starts with the second picket and paints everth $t$th picket; and
Ulysses starts with the third picket and paints every $u$th picket.
Call the positive integer $100h+10t+u$ $\textit{paintable}$ when the triple $(h,t,u)$ of positive integers results in every picket being painted exaclty once.  Find the sum of all the paintable integers.

[hide]
Note that none of the numbers can be pairwise relatively prime or they will paint the same picket at some point. Look at the $4$th picket.

CASE 1: $h$ paints the $4$th picket. Then $h = 3$.

Sub-case 1: $t$ paints the $5$th picket. Then $t = 3$ and $u = 3$. This gives $333$.
Sub-case 2: $u$ paints the $5$th picket. Then $u = 2$. But that means $(h,u) = 1$. So it doesn't work.

CASE 2: $t$ paints the $4$th picket. Then $t = 2$.

Sub-case 1: $h$ paints the $5$th picket. Then $h = 4$ and $u = 2$. This gives $424$.
Sub-case 2: $u$ paints the $5$th picket. Then $u = 2$. But that means $h = 0$. So it doesn't work.

CASE 3: $u$ paints the $4$th picket. Then $u = 1$. But that would mean $u$ paints everything. So it doesn't work.

Our answer is $333+424 = 757$.[/hide]

[quote="joml88"]Call the positive integer $ 100h \plus{} 10t \plus{} u$ $ \textit{paintable}$ when the triple $ (h,t,u)$ of positive integers results in every picket being painted exaclty once.[/quote]

I don't get this part of the question.  Aren't there an infinite amount of $ 100h \plus{} 10t \plus{} u$?  What are $ h$, $ t$, and $ u$?  Thanks!

[bump]**My question is still unanswered.  Could someone please help?  Thanks!**[/bump]

I agree that that's badly worded. As it turns out, the only possible h, t, and u are between 0 and 9, so $ 100h \plus{} 10t \plus{} u$ is just the 3-digit integer [i]htu[/i].

Basically, the question's asking you to sum together $ 100h \plus{} 10t \plus{} u$ for every valid (h,t,u).

An alternate first step...

[hide="Solution"]Harold paints, $ 1/h$ of the pickets, etc. Thus, $ 1/h \plus{} 1/t \plus{} 1/u \equal{} 1$.

Working within $ \{ 0 .. 9 \}$, the only solutions are $ (3,3,3)$, $ (2,4,4)$, $ (2,3,6)$, and their permutations.

$ \blacktriangleright$ $ (3,3,3)$ obviously works.

$ \blacktriangleright$ For $ (2,4,4)$, we need $ t \equal{} 2$, otherwise Harold and Ulysses will conflict.

$ \blacktriangleright$ For $ (2,3,6)$, note that whoever gets the $ 2$ will paint all pickets of a particular parity, and since $ 3$ is odd, $ m \plus{} 3n$ must take both even and odd values for any $ m$, so this can't work.

$ \textbf{Answer: } \; \; 333 \plus{} 424 \equal{} 757$[/hide]


What is the sum of the three distinct prime factors of 47,432?

$ 2 \plus{}7\plus{}11\equal{}\boxed{20}$.  Basically we divide all factors of 2, then all factors of 3 (none), then 5 (none) then 7(2) etc.


Given a circle S with center O, radius r. The inverse of a point P (P != O) to be the point P' on tne OP whose distance from O satisfies the equation OP.OP' = r*r.

1) Assume that OP > r. Let l1, l2 be two tangent lines to the circle S of radius r passing through the point P and let Q1, Q2 be the points of contact. Prove that the point P' is an intersection of the segments OP and Q1Q2.

2) Let O1 be the middle point at the segment OP. Take circle S1 of radius r1 = O1P = O1O.  Prove that the points Q1 and Q2(problem 1) are the points of intersection of the circles S and S1.

Anyone here familiar with Inversion Geometry? The top paragraph is the basic description.

$ Q_1Q_2 \cap OP\equal{}P_1$,and$ PQ_1 \perp OQ_1$,and$ PQ_2 \perp OQ_2$
$ 1)$it is obvious that $ Q_1Q_2 \perp OP$,the similarity between $ \triangle OQ_1P_1$and$ OQ_1P$,shows us:$ OP_1.OP\equal{}r^2$,thus$ P_1\equal{}P'$
$ 2)$we know that $ \angle OQ_1P\equal{}\frac{\pi}{2} \Longleftrightarrow OO_1\equal{}PO_1\equal{}Q_1O_1$,and it's done!


Euler
algebra
function
domain
inequalities
Cauchy Inequality

:D  :D  :D

$ \left\{ {\begin{array}{*{20}c} {x \plus{} y \minus{} \sqrt {xy} \equal{} 3} \\
{\sqrt {x \plus{} 1} \plus{} \sqrt {y \plus{} 1} \equal{} 4} \\
\end{array}} \right.$

There is a great solution! Can you find?  :D

Really?
I think $ x\plus{}y\minus{}xy\equal{}\color{red}{\minus{}}\color{black}{3}$

[quote="Kouichi Nakagawa"]Really?
I think $ x \plus{} y \minus{} xy \equal{} \color{red}{ \minus{} }\color{black}{3}$[/quote]

I'm sorry, the first equation is $ x \plus{} y \minus{} \sqrt {xy}  \equal{} 3$

[hide]
The first equation shows that both $ x,y \ge 0$, since $ xy \ge 0$, but if $ x,y \le 0$ then $ x\plus{}y\minus{}\sqrt{xy} \le 0 < 3$.

By AM-GM, we have $ 3 \equal{} x\plus{}y\minus{}\sqrt{xy} \ge x\plus{}y \minus{} \frac{x\plus{}y}{2} \equal{} \frac{x\plus{}y}{2} \Longrightarrow x\plus{}y \le 6$, and by AM-QM, we have $ 4 \equal{} \sqrt{x\plus{}1} \plus{} \sqrt{y\plus{}1} \le 2\sqrt{\frac{x\plus{}y\plus{}2}{2}} \Longrightarrow x\plus{}y \ge 6$. Thus $ x\plus{}y \equal{} 6,\ xy \equal{} 9$, and $ \boxed{x\equal{}y\equal{}3}$.

[/hide]

Interchanging $ x$ and $ y$ does not alter the system, hence $ x\equal{}y$, then its simple

[quote="|nSan|ty"]Interchanging $ x$ and $ y$ does not alter the system, hence $ x \equal{} y$, then its simple[/quote]
$ \left\{\begin{array}{l}
x\plus{}y\minus{}\sqrt{xy}\equal{}39 \\
\sqrt{x\plus{}1}\plus{}\sqrt{y\plus{}1}\equal{}9
\end{array}\right.
\iff
(x, y) \equal{} (3, 48), (48, 3)$

$ \left\{\begin{array}{l}
x\plus{}y\minus{}\sqrt{xy}\equal{}633 \\
\sqrt{x\plus{}1}\plus{}\sqrt{y\plus{}1}\equal{}28
\end{array}\right.
\iff
(x, y) \equal{} (3, 675), (675, 3)$

$ \left\{\begin{array}{l}
x\plus{}y\minus{}\sqrt{xy}\equal{}855 \\
\sqrt{x\plus{}1}\plus{}\sqrt{y\plus{}1}\equal{}35
\end{array}\right.
\iff
(x, y) \equal{} (15, 960), (960, 15)$

[quote="Bugi"]L_Euler, you are posting many problems, and you never say what is the domain for the solutions(integer,real,natural).[/quote]

You solve all my equation and equation system in $ R$!.  :)

[quote="azjps"][hide]
The first equation shows that both $ x,y \ge 0$, since $ xy \ge 0$, but if $ x,y \le 0$ then $ x \plus{} y \minus{} \sqrt {xy} \le 0 < 3$.

By AM-GM, we have $ 3 \equal{} x \plus{} y \minus{} \sqrt {xy} \ge x \plus{} y \minus{} \frac {x \plus{} y}{2} \equal{} \frac {x \plus{} y}{2} \Longrightarrow x \plus{} y \le 6$, and by AM-QM, we have $ 4 \equal{} \sqrt {x \plus{} 1} \plus{} \sqrt {y \plus{} 1} \le 2\sqrt {\frac {x \plus{} y \plus{} 2}{2}} \Longrightarrow x \plus{} y \ge 6$. Thus $ x \plus{} y \equal{} 6,\ xy \equal{} 9$, and $ \boxed{x \equal{} y \equal{} 3}$.

[/hide][/quote]

That's intelligent solution!! :)  :)  This is my solution:

Condition: $ (1) \leftrightarrow xy \ge 0$ and $ (2) \leftrightarrow \left\{\begin{array}{l} x \ge \minus{} 1 \\
y \ge \minus{} 1 \\
\end{array} \right.$

if:  $ \left\{\begin{array}{l} 0 > x \ge \minus{} 1 \\
0 > y \ge \minus{} 1 \\
\end{array} \right.$. we have $ x \plus{} y \minus{} \sqrt {xy} < 0 \ne 3$ (discard)

if: $ \left\{\begin{array}{l} x \ge 0 \\
y \ge 0 \\
\end{array} \right.$

Use Cauchy Inequality to (1): $ 3 \equal{} x \plus{} y \minus{} \sqrt {xy} \ge x \plus{} y \minus{} \frac {{x \plus{} y}}{2} \leftrightarrow x \plus{} y \le 6$

Use Cauchy-Schwarz to (2): $ 4^2 \equal{} (\sqrt {x \plus{} 1} \plus{} \sqrt {y \plus{} 1} )^2 \le 2.(x \plus{} y \plus{} 2) \leftrightarrow x \plus{} y \ge 6$

So $ x \plus{} y \equal{} 6$. Happend "=" in 2 inequalities so $ x \equal{} y \equal{} 3$

Enjoy!!! :)


limit
calculus
integration
calculus computations

I am sorry for the previous post. I am a newbie here and just pasted the problem from a MS word document :(  and din't care to see the post after it was submitted. Anyway what I intended is same as the one given by kent

That's completely unreadable. If you're going to write formulas in plain text, at least include "^" to indicate exponentiation, if appropriate. And parentheses.

This is what I think inexorable said - mind you, I'm only guessing in a couple of places:

Let $S(n,p)=\sum_{k=1}^nk^p$

Show that $\frac{(p+1)S(n,p)-n^{p+1}}{n^p}\to\frac{p+1}2$ as $n\to\infty.$

You should break some of those long expressions up onto multiple lines- it doesn't look very good when it splits automatically in the middle of a $p+1$.

Some naive calculus-based ideas:

$S(n,p)=\sum_{k=1}^nk^p=\int_0^nx^p\,dx+\sum_{k=1}^n\int_{k-1}^k(k^p-x^p)\,dx.$

Right away we see that $S(n,p)<\frac{n^{p+1}}{p+1}.$

Now we work on estimating that second term By the mean value theorem:

$k^p-x^p=(k-x)(p\xi^{p-1}$ for some $k-1<\xi<k.$ That means that

$(k-x)p(k-1)^{p-1}<k^p-x^p<(k-x)pk^{p-1}$

$\frac p2(k-1)^{p-1}<\int_{k-1}^k(k^p-x^p)\,dx<\frac p2k^{p-1}$

$\frac p2\sum_{k=0}^{n-1}k^{p-1}< \sum_{k=1}^n\int_{k-1}^k(k^p-x^p)\,dx <\frac p2\sum_{k=1}^nk^{p-1}$

$\frac p2\left(S(n,p-1)-n^{p-1}\right)<\sum_{k=1}^n\int_{k-1}^k(k^p-x^p)\,dx < \frac p2S(n,p-1).$

Now, this sum is clearly $O(n^p)$, so we have that $S(n,p)=\frac{n^{p+1}}{p+1}+O(n^p).$

This also applies to $S(n,p-1).$ By this estimate, $S(n,p-1)=\frac{n^p}p+O(n^{p-1}).$

So we now have that

$S(n,p)=\frac{n^{p+1}}{p+1}-\frac p2\cdot\frac{n^p}p+O(n^{p-1})$

$(p+1)S(n,p)=n^{p+1}-\frac{p+1}2n^p+O(n^{p-1}).$

That is exactly what we've been asked to prove.

If I'm not mistaken, we can apply Cesaro-Stolz in the obvious way to get a solution somewhat shorter than xirti's.

For $p>0$ we have that

\begin{eqnarray*} \lim_{n \to \infty} \frac{(p+1) S(n,p) - n^{p+1}}{n^p} &\stackrel{\textrm{CS}}{=}& \lim_{n \to \infty} \frac{(p+1) \left[ S(n+1,p) - S(n,p) \right] - (n+1)^{p+1} + n^{p+1}}{(n+1)^p - n^p} \\ \ &=& \lim_{n \to \infty} \frac{(p+1) (n+1)^p - (n+1)^{p+1} + n^{p+1}}{(n+1)^p - n^p} \\ \ &=& \lim_{n \to \infty} \frac{(p+1) \left( 1 + \frac1{n} \right)^p - \left( 1 + \frac1{n} \right)^p (n+1) + n}{\left( 1 + \frac1{n} \right)^p - 1} \\ \ &=& \lim_{n \to \infty} \frac{(p+1) \left( 1 + \frac{p}{n} + O \left( n^{-2} \right) \right) - \left( 1 + \frac{p}{n} + \frac{p(p-1)}{2n^2} + O \left( n^{-3} \right) \right) (n+1) + n}{\left( 1 + \frac{p}{n} + O \left( n^{-2} \right) \right) - 1} \\ \ &=& \lim_{n \to \infty} \frac{p+1 + \frac{p(p+1)}{n} - (n+1) - \frac{p(n+1)}{n} - \frac{p(p-1)(n+1)}{2n^2} + n + O \left( n^{-2} \right)}{\frac{p}{n} + O \left( n^{-2} \right)} \\ \ &=& \lim_{n \to \infty} \frac{\frac{p^2}{n} - \frac{p(p-1)(n+1)}{2n^2} + O \left( n^{-2} \right)}{\frac{p}{n} + O \left( n^{-2} \right)} \\ \ &=& \lim_{n \to \infty} \frac{\frac{p^2}{n} - \frac{p(p-1)}{2n} + O \left( n^{-2} \right)}{\frac{p}{n} + O \left( n^{-2} \right)} \\ \ &=& \lim_{n \to \infty} \frac{\frac{p+1}{2n} + O \left( n^{-2} \right)}{\frac1{n} + O \left( n^{-2} \right)} \\ \ &=& \lim_{n \to \infty} \frac{\frac{p+1}2 + O \left( n^{-1} \right)}{1 + O \left( n^{-1} \right)} \\ \ &=& \frac{p+1}2 .\end{eqnarray*}


group theory
superior algebra
superior algebra solved

How to show that the number of conjugacy classes of the symmetric group $\mathcal{S}_n$ is equal to the number of partitions of $n$?

To each partition corresponds a unique cycle structure for the decomposition of a permutation into disjoint cycles. The cycle structure decomposition determines the conjugacy class.

Oh right :D That's neat. I forgot about the "cycle structure" thing :)


calculus
integration
calculus computations

Suppose that $x=(x_{1},x_{2},\ldots,x_{n})$ is in $R^{n}$ ($R$ is the real numbers).  Please help show that there exists a constant $C>0$ such that for a fixed real number $y$ we have
$\int_{\{x: |x|^{n}\ge y\}}|x|^{-2n}~dx \le C y^{-1}$


calculus
derivative
inequalities proposed
inequalities

Let $a,b,c,d$ be nonegative real numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}=1,\] disprove that \[\frac{a^{2}}{a+b+c}+\frac{b^{2}}{b+c+d}+\frac{c^{2}}{c+d+a}+\frac{d^{2}}{d+a+b}\geq\frac23.\]

easy with derivatives :blush: : Call the LHS $f(a,b,c,d)$. Now $\frac{\partial f}{\partial a}(\frac12,\frac12,\frac12,\frac12)=\frac13\ne 0$.
[hide]To be more detailed, this shows that $(\frac12,\frac12,\frac12,\frac12)$ cannot be a local minimum, thus the ineq cannot hold in a whole neighborhood of $(\frac12,\frac12,\frac12,\frac12)$. As a counterexample we would have either (didn't check) $(\frac12+\epsilon,\frac12-\delta,\frac12-\delta,\frac12-\delta)$ or $(\frac12-\epsilon,\frac12+\delta,\frac12+\delta,\frac12+\delta)$ for small $\epsilon>0$ and appropriate $\delta>0$. [/hide]:)


AMC
AIME
MIT
college
Putnam
geometry
geometric transformation

Hi I was wondering. If one is applying to a college that has an optional section or whatever for ones AMC and or AIME scores, such as MIT, and they are not very good. Is it better to just leave it blank? However if you do that and you do make note of the fact that you were on math team will they just assume that you did really badly and so didn’t mention it so would you have to just not mention that either? In fact if that is the case it’s probably not safe to mention any involvement in AOPS activities either?  I mean I know they say a lot of people don’t even know what the AMC/AIME is, but in this case will they be like you knew what it was you tried it and did badly. This is probably something that very few if any here need to worry about but  I thought maybe you could shed some light on the matter.

P.S. Ive never posted anything before so please let me know if I made any errors in terms of where I posted or how I did it hopefully though this has ended up where it was intended to.

What are they? They might actually be worth including.

In any case, there are plenty of high schools that don't do the AMC's, so don't worry about that. And admissions officers won't read that much between the lines. [b]The key thing[/b] is that you should include things that illustrates your personality, interests, goals, priorities, and perspective.

Feel free to also follow up with any other information that might help us give such a suggestion, or any other suggestions. Information may or may not include the theme of your application, projected major/career path, domestic/international student, current grade, etc.

A girl who received early action at my school deliberately left the AMC section blank. I don't believe it will make a large difference, and chances are you'll probably be better off writing them down anyway.

Girls are judged by different standards than boys in MIT's admissions process, so your example does not necessarily lead to relevant advice for the OP.

But I do agree that, most likely, your scores are worth putting up. Feel free to post them, or any of the other things I mentioned earlier, if you want more reliable advice.

Personality/Story/ Projected Major or Career Path

Alright so here is the thing...I am a junior in high school and I absolutely love math and science (specifically physics) I mean what can I say it is amazing, so beautiful,so elegant and its everywhere and everything. There is nothing quite like working on a really interesting and engrossing math problem. In school though I always found math class rather easy still I pressed forward accelerating through classes and what not thinking that once I was taking an advanced enough class the math would become more interesting. It didn't it was just pretty much the same I mean I learned stuff I didn't know before but it wasn't all that deep if you know what I mean, just regular old school math. Then I happened across math contests! Here were problems that were actually interesting and fun, problems that dealt not only with stuff like algebra but with lots of other really cool math, stuff like number theory and combinatorics etc. There was just one problem I sucked at contest problems. Even though I had always found math class easy, these new problems I encountered were unlike anything I had ever seen before. Had I not liked math I suppose I probably wouldn't have cared much, but as it was this really sucked for me being terrible at something that I enjoyed so much. So I decided to work on it and try and get better. So ya that's how I started doing the AMC and joined AOPS, the math team and what not. Ive learned a lot of cool new math from AOPS and gotten better. However, I am still bad at contests (one of my biggest problems being I always have trouble figuring how I should begin to approach a problem, of course now that I think about I am actually getting better at that). So anyway the other day I was thinking about math and stuff and all of these things just kinda popped into my head and I I suddenly realized mmm... after this year even if I do good on the AMC the contest is after college applications and stuff are to be sent in so it can't help me and this of course prompted the concerns I presented in my original post. It's a problem cause I like math a lot it really is a very big part of who I am and how I see the world around me. But like I said I suddenly began to wonder whether talking about this would lead to the concerns I stated in my original post. A very very small part of me very rarely but sometimes wonders whether I would be better off not liking math then I wouldn't care or need to worry, but then again I can't imagine myself actually disliking something so beautiful.

As for what I wanna do in the future I know people change and everything so of course nothing is set in stone but I really want to study theoretical physics and mathematics and become a professor/researcher. I am also hoping to get good at math so that unlike the AMC and other my other high school contest experiences I can do well on the Putnam and other contests when I go to college.

Thanks To You All

P.S I hope this was along the lines of the additional information that TZF mentioned would be helpful. Also I apologize if this was a little jumbled and long.

If you got good grades in math but didn't get superb AMC scores, put them down anyway! I'd say they're probably worth recording if you got at least an 82.5 on the 12 or a 96 on the AMC 10. If you made AIME's, then you should [i]definitely[/i] put them down. But here's the trick -- you have to bring out your passion/interest/love for the subject in your essays; otherwise even a good AMC score won't help you much.

Having said that, try to do well on the AMC's this year. After you take the test, if you think you did well (write down your answers and unofficially score yourself against the answer sheet after the test), consider writing a letter to the admissions office telling them why you are so interested in the school, and including your projected new scores. You can even say it along the lines of, hey, I really love this stuff, and I'm getting a lot better, and I'm really excited to go to your school to learn more about it, as well to (other things).

What if I don't record them though what about the stuff I originally mentioned which if any of those things will be an issue?

If you're going to talk about how you care about math contests, leaving the scores off reflects poorly on you. You definitely want the scores to go with the story, even if they're nothing special.

That's kinda what I was thinking however, the emphasis is not that I care about math contests it's that I care about math and the contests were just a way for me to get better at math.

Exactly: the point is that, though your performance wasn't the best (as demonstrated by the AMC/AIME scores you wrote down), you still enjoyed participating, because it was an opportunity to learn and share with others.

[i]That[/i] perspective is certainly more appealing than "I have no particular interest in math, yet I managed to solve the Riemann hypothesis and the USAMO problems in about an hour."

I would send them in, no mater how badly you do.  It probably won't matter too much unless you make USAMO, or do really well on AIME, but I really doubt that putting a score down would hurt you.  They would probably just look at it if they are that bad, and then move on to the rest of your application and concentrate on that.

So if I understand this correctly everyone's consensus here seems to be that I should not worry about what they will think if I don't send them in and just send them in regardless of what they may be?

Hello? Was the previous post a correct statement? If not what are the ramifications of not recording as I asked before can I still talk about math team participation and AOPS?? TZF commented that this really shouldn't be a concern since admissions probably won't make that big of an effort to read between the lines, do others agree?

I was going to say what simo14 said but he beat me to it:

[quote="simo14"]just say what you said in your previous post. you rly show your love for the subject, which is all what really matters.[/quote]

Here's the thing, you can't just brute force your way into MIT by beasting a bunch of tests/competitions (well, if you beast the IMO or something, then maybe).  Along the same lines, you wont get immediately turned down because of your scores.  You've got to show them you're passionate.  While having a good array of achievements (e.g. research, projects, etc.) are good indications of this, your essay is a key place to bring attention to your passions your pursuit of them.

I am not sure whether is is appropriate to string this to this thread if not plz let me know. Anyway here goes:

Ive recently heard about what sound like lots of really cool summer math programs like mathcamp, Ross, HCSSim, etc. These look really cool and so I was wondering knowing what you do from reading the previous posts on this thread would you say any of these programs are worth my applying too? That is to say based on where I am right now in terms of math could I get into any of these programs?

If you are very interested in going to one of these camps, why don't you apply, and take the admissions quiz/test/whatever? If you crash and burn on the admissions test, don't count on getting in. If you do reasonably well, you can have more confidence in getting in.

Meanwhile, you can consider applying to some interesting science summer programs, applying for a job, starting a personal science project, doing research, and a whole host of other things that, in my opinion, are probably more relevant to your goals (I believe you said you like theoretical physics). Start looking around and talking to people and getting a feel of what's out there in terms of opportunities -- and there are lots of good ones. You should understand what it means to proactively search for opportunities early on in life; that's a skill that will take you a long way. Frankly, I think that's more worthwhile than going to some math camp.

But hey, it never hurts to apply to a math camp, and word is, they can be very fun  :)

Ross is definitely worth applying to.

http://www.math.ohio-state.edu/ross/

I feel compelled to add that Mathcamp (http://www.mathcamp.org/) is also very much worth applying to.

I am a student from India and REALLY passionate about science especially physics. In fact I am aspiring to do research in nuclear physics to help in he development of nuclear fusion as an energy source. Currently in class 11, I have secured 90%+ in my 10th exams. Besides I have won some good competitions like -
[hide]i. Selected for final award of scholarship in the National Talent Search Scheme
ii. Qualified for Indian National Maths Olympiad through the Regional Maths Olympiad
iii. Qualified for the Indian National Astronomy Olympiad Jr. conducted by Homi Bhabha Centre for Science and Education through the National Standard Examination in Physics and was placed among top 10% of 35 students enrolled at the centre.
iv. Qualified for 2nd Round of National Science Olympiad and attained all India rank of 126
v. Achieved all India ranks 60, 67 and 163 in different years in National Level Science Talent Search Examination conducted by Unified Council.
vi. Qualified for 2nd round of National Cyber Olympiad and attained All-India Rank of 172
vii. Won 1st prizes in Inter-Branch Olympiads for Physics(twice) and mock NTSE conducted by Inspection Department, City Montessori School
viii. 1st prize in Elocution and Quiz competitions conducted by Regional Science City.
ix. 1st position in Science Talent Search Exam in West Bengal
x. Came in top 3 in the written round of Birla Scholarship exam and won the interview round.
xi. Came third in the regional round Discovery Kids Quiz
xii. Cleared the written round for KVPY fellowship and awaiting interview results(Can be called India's RSI)[/hide]
Besides [hide]I'm a good poet, writer(both stories and plays), amateur singer, lyricist, like making models at school..anything from trebuchets to Van de Graaff generators :p
I am also a good orator and have won a few prizes in essay writing/debates etc.
A few summer camps like an electronics one at the Regional Science City have also been attended by me.
I have given my first SAT and am awaiting my results on 12th. I am pretty confident about crossing 2200 in one or two attempts.[/hide]
Considering these what are my chances of getting into MIT, seeing as I have never qualified for any International Olympiad :(
I will try my best to get at least to the TC for at least one IO next year though :D
Thanks!

TZF I am curious as to who/what  you were referring too when you said talk to people,are you suggesting something like asking a professor if I can work in their lab or something of that nature? (I have actually thought about this but am unsure as to the best way to find a suitable professor who will actually allow me to do some kind of work that I may enjoy)

Also I think any endeavor that would help me in improving my math skills is certainly very relevant to my goal, since to study mathematics/theoretical physics one must have a very strong command of mathematics. I suppose even beyond the concerns expressed in my original post as far as long term is concerned this is what truly scares me about how I have done on the AMC.

Also what good camps are available for international students in USA?  :huh:

Hello? Is this thread still being read or do I like need to post my most recent comment(s) in a new one or something. I sincerely apologize if I am just being impatient but like I said I have never posted anything before and I have no idea what the typical response time is or even if its just that my queries have been deemed to no longer be worth responding too. I just thought I might ask since all previous responses were rather quick in coming.

P.S. Thanks to all of you for all the helpful advice.

[quote="metafor"][i]That[/i] perspective is certainly more appealing than "I have no particular interest in math, yet I managed to solve the Riemann hypothesis and the USAMO problems in about an hour."[/quote]

Actually, if you prove the Riemann hypothesis as a high school student, any school in the world will beg you to choose them.  And you'll get a million dollars from the Clay Institute.

Greetings, and welcome to the internet. Here on the internet there are websites called "search engines." They are very useful for finding a wide variety of information. I suggest you choose your favorite search engine and type in the words "Clay institute" to see what it gives you. (On the first page of results, my favorite search engine [url=http://www.claymath.org/millennium/]gives[/url] [url=http://www.claymath.org/]several[/url] [url=http://en.wikipedia.org/wiki/Clay_Mathematics_Institute]links[/url] that look very promising.)

Thank you! The internet does seem to be a useful place :)


function
limit
calculus
calculus computations

Let $f: \mathbb{R}\rightarrow \mathbb{R}$ be a continuous function and $\lim_{h\to 0^{+}}\frac{f(x+2h)-f(x+h)}{h}=0$. Prove that $f$ is constant on its domain.

Using continuity we get $\lim_{h\to+0}\frac{f(x+h)-f(x)}h=0$ for any $x$.
Let's suppose that $f(a)$ and $f(b)$ are different, say, $f(a)>f(b),a<b$.
Consider function $g(x)=f(x)+c(x-a),\ c=\frac{f(a)-f(b)}{2(b-a)}>0$ Then $g(b)<g(a)$. There exists $d\in[a;b)$ such that $g(d)=\max_{x\in[a,b]}g(x)$ But this is impossible because $g'_{+}(d)=c>0$, where $g'_{+}(x)=\lim_{h\to+0}\frac{f(x+h)-f(x)}h$

[quote="RIP"]Using continuity we get $\lim_{h\to+0}\frac{f(x+h)-f(x)}h=0$[/quote]
Could you please elaborate on this?

\[|f(x+h)-f(x+\frac h{2^{n}})|\leqslant\sum_{k=1}^{n}|f(x+\frac h{2^{k-1}})-f(x+\frac h{2^{k}})|\leqslant\]
\[\leqslant\varepsilon(\frac{h}{2}+\frac{h}{4}+\ldots+\frac h{2^{n}})<\varepsilon h\]
for any $\varepsilon>0$ and all small $h$. Then take $n\to\infty$


For a positive integer $n$, let $\omega(n)$ denote the number of positive prime divisors of $n$. Find the smallest positive tinteger $k$ such that $2^{\omega(n)}\leq k\sqrt[4]{n}\forall n\in\mathbb{N}$.

$\pi(n)$  is the number of  prime $\leq  n$

My solution:
Let $ p_0=2<p_1<p_2<\cdots $ be the sequence of all prime number. Then, there exist an index $ i $ such that \[ p_0p_1p_2\cdots p_{i-1}\leq n<p_0p_1p_2\cdots p_i \] We have $ \omega(n)\leq i $ and \[ \frac{2^{\omega(n)}}{\sqrt[4]{n}} \leq \frac{2^i}{\sqrt[4]{p_0p_1\cdots p_{i-1}}} \] Hence \[ k=\text{max} \{0,\lfloor \frac{2^2}{\sqrt[4]{2\cdot 3}} \rfloor\,\lfloor \frac{2^2}{\sqrt[4]{2\cdot 3\cdot 5}} \rfloor, \lfloor \frac{2^2}{\sqrt[4]{2\cdot 3\cdot 5\cdot 7}} \rfloor, \cdots \}=\lfloor \frac{2^2}{\sqrt[4]{2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13}} \rfloor=5 \] satisfied the given condition.
Easy to see that $ k=4 $ doesn't satisfy with $ n=2\cdot 3\cdot 5\cdot 7=210 $


[quote="chien than"]Prove that with $x;y;z;a;b;c \in R$ we have:
$ax+by+cz+\sqrt{(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})}\geq \frac{2}{3}(a+b+c)(x+y+z)$[/quote]
It's Vasc's inequality ( from 1989 year ).


Putnam
AMC
USA(J)MO
USAMO
articles
probability
expected value

According to http://www.washingtonpost.com/wp-dyn/content/article/2006/09/19/AR2006091901779_2.html

"He wants to be a patent attorney. Growing up with parents who arrived in the United States with very little and now work at the post office and in real estate, money was not a problem exactly, but . . . he'd like to have a career that ensures he doesn't have to think about money anymore."

Now why become a patent attorney after all that? Jeez...

Also, why wouldn't have have to worry about money if he was a patent attorney? It doesn't make sense to me...

And is it really better to take less classes to attempt more AP classes on your own instead of just taking more AP classes?

[quote="aufha"]I'm still curious about the motivation for his choices despite the ranting nature of this post.[/quote]

who cares?  first of all, for whatever reasons, otherwise intelligent people make decisions all the time which are not in their best interests (at least as far as maximizing the expected value of their future earnings is concerned).  maybe his parents or other authority figures in his life have exaggerated the value of a law degree or maybe he underestimates how much he could possibly make in other fields.  maybe he suspects that he isn't smart enough to get a quant type job.

the biggest thing though, is that i don't really see how this kid is all that smart.  he's certainly bright but passing a lot of AP tests in high school, or taking a bunch of not particularly advanced courses in university are not exactly good indicators of high ability in my book.  if he found the time to get a putnam HM in there or was taking all graduate courses i might be more impressed but there are probably hundreds of math or physics majors each year who are smarter in the 'raw intelligence' category.  to your average washington post reader i'm sure it's pretty impressive, but as someone who finished their undergrad in three years and could have certainly done it in two if getting done was my number one priority, i just don't think it's that big of a deal.

A lot of AoPSers could do it if they'd wanted to..

But taking such a heavy workload doesn't leave much time for anything else, does it? School should be fun.

They never mentioned him making the USAMO either :D

[quote="undefined117"]A lot of AoPSers could do it if they'd wanted to[/quote]

I doubt it. Seriously. 37/15 ~ 2.5 times the normal workload in his second semester. Imagine doing high school 2.5 times simultaneously. You can ignore PE and health, if that helps. Now, add in that UVA is a great college, and the classes are more advanced than at most public schools. And he placed out of 72 credits of easy classes, so it's not like he's in single variable calculus.

Also, if you've ever registered full time at college, you'll realize that it can be hard to optimally schedule even the normal load of classes if you're pursuing a double major. He must've done his homework on this one (along with a bit of luck).

Finally, and most importantly, blitzing through college in 1 year takes balls. It's easy to look back and say, theoretically, anyone as smart as him should be able to do that, because he did this this and this, and none of those steps were particularly mindblowing. The fact is that he did it. Props.

I'm sure he had reasons for his priorities. Imagine having a hellish schedule, and then have a masters at age 19! Not bad at all.

Finally, don't judge him based on what he said about patent law. First of all, most freshmen, including many brilliant ones, don't yet know what they want to do with their lives... many don't even know what they want to major in! Figure that one of the down sides about finishing college in 1 year is that he hasn't had the rich exposure or the time to help him figure out what he wants to do with his life. No matter, I'm sure he'll make a good choice, if not the "right" choice.

Also, even if he goes into patent law, he definitely has the time management skills and organization needed to do very well in his first couple years on the job (when you start, you are accountable for your time in 6-minute intervals). I am more concerned (if at all) that there is no evidence in the article that he is inclined towards law. Although, there is probably some background interest tucked in those 72 AP credits. He also stated he liked math because it helped his ability to construct a good argument.

Finally, what's wrong with law? I admit, there are more [i]colorful[/i] aspects of law than patent law, but seriously... yeah, I know, most lawyers are pretty bad and don't make much money. So? Most math teachers, and many math professors, are also pretty horrendous. Right now, I have about a 50% chance of going into law. I don't think its fair for someone to conclude that I am not smart enough to do quantitative analysis just based on that. Now, if someone says I'm not smart enough to be a quant because I never made MOP, they're welcome to  :wink:


geometry
geometric transformation
rotation

Six distinct charms are placed on a bracelet that does not have a clasp.  In how many ways can these charms be arranged?

[hide]
There are $ 6!$ such ways, but you must divide by 2 and 6 because you can flip the bracelet and different positions are the same.This gives $ \boxed{60}$
[/hide]

Yes, thank you, however, I do not remember why you divide by 6; I understood this on my own a couple years ago, however, now all I see to do is divide by two, but there's only to endpoints so I don't see why to divide by 6.

Divide by 6 because the 6 rotations of the same arrangement are all the same

[quote="aahmadi"]Yeah, I guess the right side of my brain is now broken, for when I first saw a problem like this a long time ago I quickly found out what to divide by and why.  Now, I can think about this bracelet for three days and get nowhere.  So could someone please tell me a formula for how many arrangements there are for n distinct charms on a bracelet that does not have a clasp?[/quote]

[hide]$ \frac{n!}{2n}$ I think[/hide]

Thanks mathnerd, I just posted a question asking for a proof of this.  And sorry for saying this through adding another post, but I wasn't able to edit my last post :(


If $0º < x < 90^{\circ}$, and $cosx=3/sqrt{10}$, compute $log10(sin x) + log10(cos x) + log10(tan x)$.

Source:  MATC Mathematics Club

[hide]This expression simplifies to log(sinxcosxtanx), which simplifies further to log[(sinx)^2].

This simplifies even further to log[1-(cosx)^2], or [b]-1.[/b][/hide]

[hide]　The answer is $2\log \sin x =2\log \frac{1}{\sqrt{10}}=2\cdot \left(-\frac{1}{2}\right)=\boxed{-1}.$[/hide]

Sorry, I forgot to include the exact value of cosx in the original problem.

In that case,

[hide]The answer is $\log_{10}\frac{1}{10}=-1$.[/hide]


[quote="erbed"][b]Let  $ABC$ be an acute angled triangle. Squares $ACKL$ and $BCNM$ are constructed outside the triangle. Let $T$ be the intersection point of $AM$ and  $BL$ lines. Prove that $CT$ is perpendicular to $AB$.
[/b]
I am told that this problem has an elementary solution but i cold not find it.[/quote]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=39101 for a generalization (obviously, squares are similar rectangles).

  darij

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme vol. 5  p. 13
Sincerely
Jean-Louis


This is very very simple, but it is nice! :)

Prove that in any triangle the following inequality holds:

\[( a+b+c)^{2} \left( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \right) \geq \left( \sqrt{ \frac{h_{a}}{r^{2}}}+\sqrt{ \frac{h_{b}}{r^{2}}} + \sqrt{ \frac{h_{c}}{r^{2}}} \right)^{2}.\]

cheers! :D


What is the probability that the roots of a perfect fourth power equation written down at random are all real ?

I'm not sure that the problem is well-defined...
Anyway, in an old issue of the College mathemùatics magazine a related problem has appeared :

Let $A,B,C$ be three independent random variables uniformly distributed on $[0,1]$. The probability that all the roots of the cubic equation $x^3+Ax^2+Bx+C=0$ are real is $\frac 1 {2880} .$

Pierre.


A particle of mass $2m$ is held at a height $H$ above a horizontal plane.It is released from rest and allowed to fall vertically under gravity.At the same instant, a particle of mass $m$ is projected vertically upwards from the plane with speed $v$ . The particles collide and coalesce . Given that $v^2>3gH$ ,show that the combined particle moves vertically upwards immediately after the collision .
Given that $v^2=4gH$ , show that the energy lost in the collision is $\frac{1}{3}mv^2$ , and find , in terms of $v$ , the speed of the combined particle just before it strikes the plane .

I seem to be getting something wrong again and again.
I think i'll post my solution later.

[quote]A particle of mass  is held at a height  above a horizontal plane.It is released from rest and allowed to fall vertically under gravity.At the same instant, a particle of mass  is projected vertically upwards from the plane with speed  . The particles collide and coalesce . Given that  ,show that the combined particle moves vertically upwards immediately after the collision .
Given that  , show that the energy lost in the collision is  , and find , in terms of  , the speed of the combined particle just before it strikes the plane .[/quote]

er?
okay. lets see, then, what to do?
well because they're both accelerating down from gravity, you know that the time t it takes them to collide is h/v

and then you figure out the momenta each particle has when they do collide, yes?
so for the particle going up that would be m(v-g*(h/v)), which is mv-mgh/v
and for the particle falling it would be 2mgh/v

so then you see the vectors and add them up and, if your positive direction is up, then you have sum of momentum p is equal to mv-3mgh/v
so you have an upward momentum, and thus upward movement when v^2>3gh
yay!
i dont know how to use latex yet, never bothered.  sorry.  but i do believe thats the right answer to your first question.  it's getting quite late over here(past my bedtime,    :ninja: so i'll try to get back to you on the second one).

yay!

That was exactly my solution but it seems that the inequalitites don't staight away give the required answer

I mmean that after the eqns are writtn and it is required to prove that the net momentum is upward, i am not getting that so easily and fool proof

I guess i need a step by step guide here cause I cant see why the time taken to collide is h/v .... :blush: .Also how we know the momentum of the falling particle is 2mgh/v ? :oops:

What I was trying to say was that when i wrote down the eqns, the result which the question required wasn't clear.

I'll post my work tomorrow.Maybe it'll be allright now.

[quote="daveed"]i dont exactly know what rushil means by the hang-up, but this i'll hope he explains it and everything works out

but, anyway, as to your question, well think about it.
if you jump off a plane(with a parachute, yes) with two balls and drop them at the same time you go off, well, the distance between them stays the same.  But if you throw one of them down and drop one of them, from your point of view, the thrown ball will be going away from the dropped ball at whatever velocity you threw it at.
think about the kinematics equation, displacement=Vi * t + 1/2*a*t^2
if one of them has a velocity at your start time, the displacement due to gravity is 1/2*a*t^2 for both of them.  If you're looking at the distance between them, the displacement due to gravity is null, and you only have a changing distance  between the objects due to the initial velocities.

so, from that, you see that the distance between the two balls is H
the velocity at which the upwards-moving ball is v

so, because gravity is pulling them both down, it doesnt really matter where they are when they hit.  it just matters how long it takes

how long will it take for the balls to cross that distance of h, if traveling at v?
its just h/v, then.
that's the time it takes from the beginning to the impact.
therefor, you can use that time to calculate the momentums:
the velocity of the ball on top is going to be g*time, which is gh/v
but then it's momentum is 2mgh/v because it's mass is 2m

err, that answer your question?[/quote]

Thanks for your help . Im now trying to understand it :P

no problemo- are you still having trouble with the rest of it?


calculus derivative function Euler logarithms

Can anyone show me how to diferentiate the Hadamard product representation for the Riemann Zeta function, and thereby find the derivative of the zeta funcion?
All my attempts give an extra Euler constant when I tried s=o

This approach seems better than the Wallis product approach

Thanx  :D

(Read that in  Hardy and Wright's [u]An introduction to the Theory of Numbers[/u]) :
If we differentiate $\zeta(s) = \sum_{n=1}^{\infty} \frac 1{n^s}$ term by term with respect to $s$, we get :
\[ \zeta'(s) = - \sum_1^{\infty} \frac{\log n}{n^s} \] ($s>1$).

No I can easily get it as a series, but I want it simply as the derivative of the Hadamard Product

mathworld link-
[url]http://mathworld.wolfram.com/HadamardProduct.html[/url]


topology
complex numbers
advanced fields
advanced fields unsolved

Let $ K$ be a Kahler manifold (that is, a complex manifold equipped with a Hermitian metric whose imaginary part $ \omega$ is closed, differential 2-form). Let $ N$ be a complex submanifold of $ K$. Then the restriction of $ \omega$ to $ N$ gives a symplectic form on $ N$.

It is quite clear that the restriction is a non-degenerate, differential 2-form. I'm having trouble showing that it's closed, though. Any help is appreciated.

it is closed because it's the restriction (i.e. the pull-back through the inclusion map) of a closed form: more generally, whenever you have a differentiable map $ f: X\to Y$ between differentiable manifolds, pulling back (differential forms) commutes with taking differentials, that is $ f^*(d\eta) \equal{} d(f^*\eta)$.

in our case, you take $ \eta\equal{}\omega$ to be your symplectic form on $ K$ and $ f\equal{}i: N\to K$ the inclusion. this assures that $ \omega|_N \equal{} i^*\omega$ is closed: $ d\omega|_N \equal{} i^*(d\omega) \equal{} 0$.

Thought it would be something simple like that. Thanks. :)


calculus
derivative
trigonometry
calculus computations

I need help with worked solutions for these problems

find derivative of
f(x)=sec^3(2x)tan2x

f(x)=(1+sin^3x)^(1/2)

write equation of the tangent line to the curve
y= 1/(1+sq.root(x))  at the point (4, 1/3)

1) $f'(x)=3\sec^{2}(2x)\sec(2x)\tan(2x)2\tan(2x)+\sec^{3}(2x)\sec^{2}(2x)2$. So I tried to keep it in the order I performed the operations as my work. The whole thing is a product rule $\sec^{3}(2x)\cdot\tan(2x)$, then chain rule twice on first term, the second term is just the double chain rule too. So simplified answer would be $f'(x)=6\sec^{3}(2x)\tan^{2}(2x)+2\sec^{5}(2x)$

2) Try this one the same way.


wow!  way to much information!! :P

seriously, you need to tell us more.

and i can dowload it


I guess you meant this: http://www.mathlinks.ro/Forum/viewtopic.php?t=18908

  darij


$ M$ is a point inside rectangular $ ABCD$ such that $ \angle AMB \plus{} \angle CMD \equal{} 180$.prove that $ S_{ABCD} \equal{} |AM|.|CM| \plus{} |BM|.|DM|$

Apply a sliding motion such that $ D$ will map $ A$ and $ C$ will map $ B$; the same movement will carry $ M$ to $ M'$ such that $ MM' \parallel_{\equal{}} AD$. Now $ AM'BM$ is cyclic and, from Ptolemy, $ AB\cdot MM'\equal{}AM'\cdot MB \plus{} AM \cdot M'B$ ( 1 ), but $ MM' \perp AB$ , $ AM' \equal{} MD$ and $ BM' \equal{} CM$. As $ S_{ABCD} \equal{} AB \cdot AD$, we are done.

Best regards,
sunken rock


Bekah has four brass house number digits: 2, 3, 5 and 7. How
many distinct numbers can she form using one or more of the
digits?

[quote="ernie"][i]Case 1: She uses one digit.[/i]

She can form $ 4 \text{ P } 1$ or $ 4$ numbers.

[i]Case 2: She uses two digits.[/i]

She can form $ 4 \text{ P } 2$ or $ 12$ numbers.

[i]Case 3: She uses three digits.[/i]

She can form $ 4 \text{ P } 3$ or $ 24$ numbers.

[i]Case 4: She uses all her digits.[/i]

She can form $ 4 \text{ P } 4$ or $ 24$ numbers.

Adding, we get $ \boxed{64}$.[/quote][quote="thkim1011"][hide="A different solution"]Did anyone notice that $64=4\left(\binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}\right)$. I think this is true because the number of combinatons with no order is $\binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}=2^4=16$. So if we multiply by 4 it would give us the correct answer. Can someone correct me if I'm wrong?[/hide][/quote]
Yeah, I don't understand why we don't use combinations?

[quote=LiBoy][quote="Scientest"]
Yeah, I don't understand why we don't use combinations?[/quote]

The order of the digits matter, as 23 is a different house number from 32.[/quote]

Also, there are single digit numbers like $3$ and $7$.


topology
function
real analysis
algebra
polynomial
Functional Analysis
advanced fields

Let $X$ be a compact Hausdorff topological space. Prove that the following are equivalent:

(a) The space $C(X,\mathbb R)$ of continuous functions from $X$ to the reals (with the usual topology, given by the $\sup$ norm) is separable;

(b) $X$ is metrizable.

if $X$ is metrizable then $C(X, \mathbb{R})$ is separable:

To do this, it suffices to show that there exists a subset $F$ of functions, these functions
not necessary continuous, dense in $C(X, \mathbb{R})$ for the $sup$ norm. To this end, one
will construct, for every $n > 1$, a partition $P_n$ of $X$ in disjoint subsets of diameter at most $\frac{1}{n}$:
just take a finite covering of $X$ by balls of diameter $\frac{1}{n}$ and then make them disjoint.
Now consider the countable set $F_n$ of functions that are constant on each subset of $P_n$ and take
only rational values. Then $F$, the union of the $F_n$, is a dense subset in $C(X, \mathbb{R})$. To see
this, take a continuous function $f$ and an $\epsilon > 0$. Extract a finite covering of $X$ from the covering
$\cup_{x \in \mathbb{R}} f^{-1}(]x;x+ \epsilon[)$, and take $n$ large enough such that $\frac{1}{n}$ is a Lebesgue
number of that covering. The conclusion is clear.

if $C(X, \mathbb{R})$ is separable, $X$ is metrizable:

I will use the well-knwon fact that if a Banach space $E$ is separable, then there is a distance $D$
on the unit ball of the dual space $E'$ that induces the weak* topology. It is well adapted here because
$(C(X, \mathbb{R}), |.|_{\infty}$ is precisely a separable Banach space. So if one finds a homeomorphism
$\Phi: (X,T) \to  (E',w*)$ that maps $X$ insid the unit ball of $E'$, the problem will be solved.
(Here $T$ is the topology on $X$, and $w*$ is the weak* topology on $E'$).
A natural candidate for this homeomorphism is: $\Phi(x)=\Phi_{x}$, where $\Phi_x(f)=f(x)$. This is clearly
an injective function, it maps $X$ inside the unit ball of $E'$, and it remains to show that it
is continuous (since $X$ is compact, it will be then an homeomorphism).
Because of the definition of the weak* topology, it suffices to show that for any fixed $h_0 \in (C(X, \mathbb{R})$,
$x \to  \Phi_x(h_0)=h_0(x)$ is continous, which is obvious. So one can define (I think ..) a distance on $X$
by $d(x,y)=D(\Phi_x, \Phi_y)$.

I think that alekk's solution of first the implication is not correct.I mean, I don't have slightest idea why it would be enough to construct discontinuous functions which are dense in $ C(X, \mathbb{R})$. Note that most of the functions that alekk has given can't be approximated uniformly by continuous functions.

But I think that another aproach works. Let $ d$ be the metric on $ X$. Take the sequence $ a_1, a_2, \ldots$ which is dense in $ X$. Such sequence exists since, like in alekk's solution, we can make covering by balls of the radius $ \frac{1}{n}$ and since $ X$ is compact we can find such finite covering. And for $ a_i$ just take the center's of these balls.
Now take algebra of functions generated by the functions $ \{d(x, a_i)\}_{i \in \mathbb{N}} \cup \{1_x\}$ (where $ \cup\{1_x\}$ means function which is constantly $ 1$) . Such algebra consists of all polynomials in $ d(x, a_i)$ and with coefficients from $ \mathbb{R}$ (of course these polynomials are in several variables).
It is clear that this algebra seperates points, since if $ a \neq b$, we can take $ a_k$ very close to $ a$ and then $ d(a, a_k) \neq d(b, a_k)$ so the function $ d(x, a_k)$ seperates $ a$ and $ b$. So, from the Stone's-Weierestrass Theorem this algebra is dense in $ C(X, \mathbb{R})$.
So now it sufficies to find a countable family of functions in this algebra which are dense in it. Take the set of all functions which are polynomials in $ d(a_i, x)$ (in several variables as before) but with the rational coefficients. It is clear that this set is countable, so we just have to show that it is dense. Take an arbitrary polynomial from our algebra, it is of the form $ f \equal{} c_m (...) \plus{} c_{m\minus{}1} (...) \plus{} \ldots \plus{} c_1 (...) \plus{} c_0$ where $ (...)$ represents some monomials and $ c_i \in \mathbb{R}$. Take rationals $ p_1, \ldots, p_m$ such that $ |p_i \minus{} c_i| < \epsilon$ and then take the same function $ g$ but the coefficient $ c_i$ is replaced by $ p_i$. Then $ |f\minus{}g| < \epsilon( |(...)| \plus{} |(...)| \plus{} \ldots \plus{} |(...)|)$ from the triangle inequality. But since $ X$ is compact it has finite diameter, let's say $ C$. And then $ |f\minus{}g| < \epsilon m \cdot C^l$ where $ l$ is the degree of the polynomial. So clearly we can approximate this polynomial uniformly by previously mentioned functions and this ends the proof.
Forgive me if I have written too many obvious things.


Let $ f(n)$ denote the sum of the digits of $ n$.  It's not too hard to see that the sequence
\[ f(n), f(f(n)), f(f(f(n))), ...\]
eventually becomes constant as a single digit, and call this constant value the digital sum of $ n$.  Prove that the digital sum of the product of any two twin primes, besides 3 and 5, is 8.

$ f(n)\equiv n\pmod{6}$. All primes except 2 and 3 take the form $ 6k\pm1$ for some positive integer $ k$. Since we are dealing with twin primes, our two primes are $ 6k\minus{}1$ and $ 6k\plus{}1$.

$ (6k\minus{}1)(6k\plus{}1)\equiv36k^{1}\minus{}1\equiv8\pmod{9}$, so the digital sum is 8 as long as the neither 2 or 3.

2 is not part of a pair of twin primes, so (3,5) is the only other case.


geometry
combinatorics proposed
combinatorics

For any set $S$ of five points in the plane, no three of which are collinear, let $M(S)$ and $m(S)$ denote the greatest, and smallest areas, respectively, of triangles determined by three points from $S$. What is the minimum possible value of $M(S)/m(S)$?


polygon
geometry
counting
Obtuse triangle
IMO Longlist
IMO Shortlist

Given a regular $n$-gon $A_{1}A_{2}...A_{n}$ (with $n\geq 3$) in a plane. How many triangles of the kind $A_{i}A_{j}A_{k}$ are obtuse ?

- If $n = 2k+1$ is odd :
The triangle $A_1A_iA_j$ is obtuse at $A_1$, where $i < j$, if and only if $i \in \{2, \ldots, k+1 \}$ and $j \in \{ k+1+i, \ldots, n \}$.
Thus, for each $i \in \{2, \ldots, k+1 \}$ there are $k+1-i$ possibles values for $j$. Summing over $i$, it leads to $\displaystyle \frac {k(k-1)} 2$ obtuse triangles at $A_1$.
Similarly, there are as many obtuse triangles at $A_i$ for each $i$, and non is counted twice, since a triangle cannot have more than one obtuse angle...

Thus, there are a total of $\displaystyle \frac {(2k+1)k(k-1)} 2$ obtuse triangles in that case.

- If $n = 2k$ is even :
The triangle $A_1A_iA_j$ is obtuse at $A_1$, where $i < j$, if and only if $i \in \{2, \ldots, k \}$ and $j \in \{ k+1+i, \ldots, n \}$.
Thus, for each $i \in \{2, \ldots, k+1 \}$ there are $k-i$ possibles values for $j$. Summing over $i$, it leads to $\displaystyle \frac {(k-2)(k-1)} 2$ obtuse triangles at $A_1$.
Similarly as above, it leads to a total of $k(k-2)(k-1)$ obtuse triangles in that case.

Pierre.


The measures of the three angles of a triangle form an arithmetic sequence. If the smallest angle measures $ 45^\circ$, what is the number of degrees in measure of the largest angle?

The sequence would be $ x,x\plus{}d,x\plus{}2d$. So,

$ x\plus{}x\plus{}d\plus{}x\plus{}2d \equal{} 180$
$ 3x\plus{}3d\equal{}180$
$ x\plus{}d\equal{}60$
$ 45\plus{}d\equal{}60$
$ d\equal{}15$

$ x\plus{}2d\equal{}45\plus{}2*15\equal{}\boxed{75^{\circ}}$

Since this is a cd prob, the fast way is to know that the middle term is $ 60$ no matter what, so $ 60\minus{}45\plus{}60\equal{} 75$


complex analysis
vector
calculus
derivative
geometry
geometric transformation
rotation

Suppose $ f: U \rightarrow \mathbb{C}$ is $ C^1$ on an open set $ U \subseteq \mathbb{C}$. Let $ P \in U$. Assume that $ f$  "stretched equally in all directions at P'', i.e.
[quote]if $ |f'(P)|\neq 0$, then the directed angle from $ w_1$ to $ w_2$ equals the directed angle from $ D_{w_1} f(P)$ to $ D_{w_2} f(P)$, where $ w_1$ and $ w_2$ are arbitrary directions.[/quote]
Prove that either $ f'(P)$ or $ (\overline{f})'(P)$ exists.

[hide]I'm searching for a proof without using the concept of linearity of $ f'$, when looking at it as a real vector valued function. BTW I have some problems of proving it in that way, too. I'm sure that I have seen a typical proof for this problem, but I cannot figure it out right now.

Here by $ C^1$ I mean that the partial derivatives of $ f$ exist and are continuous, and $ D_{w_i} f(P)$ denotes the directed derivative of $ f$ in the $ w_i$ direction at the point $ P$.[/hide]

Choose appropriate $ w_1$ and $ w_2$ to show that the $ f$ satisfies the homogenous C-R equation. That is, the total derivative $ f'$ has the form of a constant times a rotation matrix.

Well, by choosing $ w_1 \equal{} x$ and $ w_2 \equal{} y$ I can show that it satisfy the C-R eq's. How can I conclude that $ f'$ is of that form?

That's just what the C-R equations mean. So if you've shown the C-R equations are satisfied, you are done.

Oops  :|
It seems that the problem is with the C-R eq's.
Would you give me a clue?

Well, I have no idea, how to show that.
I've just got the $ \frac{\partial u}{\partial x}\frac{\partial u}{\partial y}\plus{}\frac{\partial v}{\partial x}\frac{\partial v}{\partial y} \equal{} 0$
But I cannot find any other vectors to get better results :!:


Fie ABC un triunghi oarecare si punctele D,E exterioare lui, astfel incat triunghiurile ABD si ACE sa fie dreptungice isoscele in B respectiv C .
Demonstrati ca CD si BE se intersecteaza pe inaltimea din A a triunghiului ABC.

Construim patratele $ABDF$ si $ACEG$ in exterior, precum si patratul $BCHI$ pe latura $BC$. Rezulta ca, rotind $\Delta\ ABI$ cu $90^{0}$ in sens trigonometric, acesta se suprapune peste $\Delta\ DBC$ acestea fiind congruente. Rezulta ca $AI\perp\ CD$. Asemanator demonstram $AH\perp\ BE$ rotind pe $\Delta\ ACH$ peste $\Delta\ ECB$
Translatam pe $BE$ paralel cu el insusi pana ajunge $B$ in $I$, iar pe $CD$ paralel cu el insusi pana ajunge $C$ in $H$. Inaltimea din $A$ in $\Delta\ ABC$ este inaltime in $\Delta\ AIH$ si trece prin punctul de intersectie al $BE$ si $CD$ $\Rightarrow$ trece si prin punctul de intersectie al $BE$ si $CD$.
Simpatica problema...am mai facut-o prin tineretile mele  :D

Nu stiu daca se poate face analitic(am incercat si a mers pana intr-un punct), dar sigur merge cu Teorema Sinusurilor si Ceva in forma trigonometrica.
[hide="Indicatie(solutia decurge imediat)"]Din triunghiul $\ BCF$ il aflam pe $\sin CBF=\frac{b\cos C}{BF}$.Analog il aflam si pe $\sin ABF=\frac{b\sqrt 2\sin (A+45)}{BF}$.Ca sa aplicam Ceva trebuie sa gasim raportul sinusurilor unghiurilor $\ ABF,FBC$, care este $\frac{\cos C}{\sqrt 2\sin\ (A+45)}$.
Ne mai trebuie sinusurile din $\ A$.Le gasim si pe ele mai usor de data asta si apoi iese usor.[/hide]

Pentru [i]Stancioiu[/i]:

Fie $T$ piciorul perpendicularei din $A$ pe $BC$. Luam $Ox = BC$ si $Oy=TA$.

Avem $T(0,0)$, $A(0,a)$, $B(b,0)$, $C(c,0)$, unde $a>0, \, b<c$.

De asemenea, $D(b-a,-b), \, E(a+c,c)$.

$CD$: $\frac{x-c}{y-0}= \frac{b-a-c}{-b-0}\, \, \Longleftrightarrow \, \, bx+(b-a-c)y-bc = 0$.

$BE$: $\frac{x-b}{y-0}= \frac{a+c-b}{c-0}\, \, \Longleftrightarrow \, \, cx+(b-c-a)y-bc =0$.

$CD \cap BE$: Scadem cele doua ecuatii si obtinem $(b-c)x = 0$, de unde rezulta $x=0$ pentru ca $b \neq c$. Dar cum ecuatia lui $AT$ este chiar $x=0$, problema e gata.

Da, multumesc Tiberiu.Nu stiam smecheria aia cu intersectia.

[color=red]Generalizare 1[/color]
Fie triunghiul $ABC$ si punctele $D, E$ exterioare lui astfel incat $\angle ABD=\angle ACE=90^{0}$ si $\angle DAB=\angle CAE$.
Sa se arate ca dreptele $CD$ si $BE$ se intersecteaza intr-un punct situat pe inaltimea din $A$ in triunghiul $ABC$.

[color=red]Generalizare 2[/color]
Fie triunghiul $ABC$ si punctele $D, E$ exterioare lui astfel incat $\angle ABD+\angle ACE=180^{0}$ si $\angle DAB=\angle CAE$.
Daca $CD\cap BE=\{P\}$ si $AP\cap BC=\{Q\}$, sa se arate ca $\sin \angle AQB=\sin\angle ABD$.


Solve fo x:
$ 2008x^2  \minus{} 4x \plus{} 3 \equal{} 2007\sqrt {4x \minus{} 3}$
I found the solution by squaring the two side but it's really long and rude.Does anyone have another idea? :cleaning:

Let y´2=4x-3, then you have the equation 2008x´2- y´2=2007y

hello, after squaring and factoring i have got
$ 4(x\minus{}1)(\minus{}3021039\plus{}1007016x\plus{}1004000x^2\plus{}1008016x^3)\equal{}0$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, after squaring and factoring i have got
$ 4(x \minus{} 1)( \minus{} 3021039 \plus{} 1007016x \plus{} 1004000x^2 \plus{} 1008016x^3) \equal{} 0$
Sonnhard.[/quote]
Yeah, I've also got that but how we solve  $ \minus{} 3021039 \plus{} 1007016x \plus{} 1004000x^2 \plus{} 1008016x^3 \equal{} 0$ :wallbash_red:


inequalities
geometry
function
inequalities proposed

For any three nonnegative reals a, b, c, prove the inequality

$\left(a-2\left|b-c\right|\right)\left(\left|a-b\right|-\left|b-c\right|\right)+\left(b-2\left|c-a\right|\right)\left(\left|b-c\right|-\left|c-a\right|\right)+\left(c-2\left|a-b\right|\right)\left(\left|c-a\right|-\left|a-b\right|\right)$
$\geq 0$.

Merry Christmas MathLinks!

  Darij

The normal trick seems to work:

1) $a=c+x+y$, $b=c+y$: we get $3x^2+y^2-3xy \geq \left( y-\frac{3}{2}x \right)^2 \geq 0$

2) $a=b+x+y$, $c=b+y$: we get $3x^2+y^2-xy \geq \left( y-\frac{1}{2}x \right)^2 \geq 0$

Equality holds iff $x=y=0$ iff $a=b=c$.

I think you have a nicer solution. Am I right?

And BTW, what does "abstruse" mean?

http://wordnet.princeton.edu/perl/webwn?s=abstruse

Darij, you are always complaining about your style, but.. : http://www.poestories.com/wordlist.php
;)

@ Mathmanman: No, that's not the meaning of "style" I am usually referring to when I say I have a bad style. Try to read one of my geometry solutions and you will see what I mean...

@ Perfect_Radio: Alas :oops: I have to disappoint you, as my solution is not much better.

First, I should have said that a, b, c can be arbitrary reals, not necessarily nonnegative. And here is my [i]solution[/i] of the problem:

We will denote cyclic sums by the sign $\sum$. We have to prove that $\sum\left(a-2\left|b-c\right|\right)\left(\left|a-b\right|-\left|b-c\right|\right)\geq 0$ for any three reals a, b, c.

Define the function

$f\left(a;\;b;\;c\right)=\sum\left(a-2\left|b-c\right|\right)\left(\left|a-b\right|-\left|b-c\right|\right)$.

Then, we want to show that $f\left(a;\;b;\;c\right)\geq 0$ holds for any three reals a, b, c.

Let's WLOG assume that a is the smallest among the reals a, b, c (we can assume this, since our inequality is cyclic). Then, $b\geq a$ and $c\geq a$, so that $b-a\geq 0$ and $c-a\geq 0$. Denote x = b - a and y = c - a. Then, $x\geq 0$ and $y\geq 0$.

Now, for any real d, we have

$f\left(a+d;\;b+d;\;c+d\right)=\sum\left(\left(a+d\right)-2\left|\left(b+d\right)-\left(c+d\right)\right|\right)\left(\left|\left(a+d\right)-\left(b+d\right)\right|-\left|\left(b+d\right)-\left(c+d\right)\right|\right)$
$=\sum\left(a+d-2\left|b-c\right|\right)\left(\left|a-b\right|-\left|b-c\right|\right)=\sum\left(\left(a-2\left|b-c\right|\right)\left(\left|a-b\right|-\left|b-c\right|\right)+d\left(\left|a-b\right|-\left|b-c\right|\right)\right)$
$=\underbrace{\sum\left(a-2\left|b-c\right|\right)\left(\left|a-b\right|-\left|b-c\right|\right)}_{=f\left(a;\;b;\;c\right)}+\underbrace{\sum d\left(\left|a-b\right|-\left|b-c\right|\right)}_{=0}=f\left(a;\;b;\;c\right)$.

Particularly, we can set d = -a, and obtain $f\left(a+\left(-a\right);\;b+\left(-a\right);\;c+\left(-a\right)\right)=f\left(a;\;b;\;c\right)$. Since a + (-a) = 0, b + (-a) = b - a = x and c + (-a) = c - a = y, this becomes $f\left(0;\;x;\;y\right)=f\left(a;\;b;\;c\right)$. Hence, instead of proving that $f\left(a;\;b;\;c\right)\geq 0$, it is enough to show that $f\left(0;\;x;\;y\right)\geq 0$.

Now,

$f\left(0;\;x;\;y\right)$
$=\left(0-2\left|x-y\right|\right)\left(\left|0-x\right|-\left|x-y\right|\right)+\left(x-2\left|y-0\right|\right)\left(\left|x-y\right|-\left|y-0\right|\right)+\left(y-2\left|0-x\right|\right)\left(\left|y-0\right|-\left|0-x\right|\right)$
$=-2\left|x-y\right|\left(x-\left|x-y\right|\right)+\left(x-2y\right)\left(\left|x-y\right|-y\right)+\left(y-2x\right)\left(y-x\right)$
$=y^2+2\left(x-y\right)^2-\left|x-y\right|\left(x+2y\right)+2\left|x-y\right|^2$.

Now, denote |x - y| = z. Then, $\left(x-y\right)^2=\left|x-y\right|^2=z^2$. Hence, this becomes

$f\left(0;\;x;\;y\right)=y^2+2z^2-z\left(x+2y\right)+2z^2=y^2+4z^2-z\left(x+2y\right)$.

Since $x-y\leq\left|x-y\right|=z$, we have $x+2y=\left(x-y\right)+3y\leq z+3y$; this, upon multiplication with the nonnegative number z (nonnegative since z = |x - y|), becomes $z\left(x+2y\right)\leq z\left(z+3y\right)$, and thus

$f\left(0;\;x;\;y\right)=y^2+4z^2-z\left(x+2y\right)\geq y^2+4z^2-z\left(z+3y\right)$
$=\left(y-\frac32z\right)^2+\frac34z^2\geq 0$,

completing the proof of the inequality.

  Darij

[quote="darij grinberg"]@ Mathmanman: No, that's not the meaning of "style" I am usually referring to when I say I have a bad style. Try to read one of my geometry solutions and you will see what I mean...
[/quote]
Yes, I know, that was a joke :)
Anyway, I find your solutions extremely interesting because you detail them and explain pretty well, so I always learn new stuff!

Merry Xmas ;)


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