abstract algebra
linear algebra
linear algebra unsolved

Suppose that $ f: V\rightarrow V$ is a linear transformation and $ f\circ f\equal{}f$ (where the dimension of $ V$ is finite).  Prove that

\[ V\equal{}\textrm{Image}\left(f\right)\plus{}\textrm{ker}\left(f\right).\]
Clearly, the RHS is a subset of the LHS.  Also, I've shown that the intersection of the image of $ f$ and the kernel of $ f$ is the zero vector.  How do we show that if $ v\in V$, then $ v\in\textrm{Image}\left(f\right)\plus{}\textrm{ker}\left(f\right)$?

[quote="amcavoy"]How do we show that if $ v\in V$, then $ v\in\textrm{Image}\left(f\right) \plus{} \textrm{ker}\left(f\right)$?[/quote]
What about $ v\equal{}f(v) \plus{} v \minus{} f(v)$

Alternatively compare the dimensions of LHS and RHS using the dimension formula applied to the endomorphism f.


I don't get the grammar of this problem no matter how many times I read it. Could some one explain it to me?
The length of a median to the hypotenuse of an isosceles, right triangle is 10 units. What is the length of a leg of the triangle , in units? Express your answer in simplest radical form.
Thanks!

isosceles right triangle: a right triangle, meaning it has one right angle. since it's isosceles, the angles are 45-45-90, and the two legs are congruent.

median: the line segment that connects one of the vertices to the middle of the opposite side

hope this helps

Could you draw it for me please?  :) I'm still confused about the median part...

The median is $ 10$, so half of the hypotenuse is $ 10$ also.  The whole hypotenuse is $ 20$.  The length of a side is then $ \sqrt{10 \times 20} \equal{} \sqrt{200} \equal{} \boxed{10\sqrt2}$


calculus
derivative
function
real analysis
real analysis unsolved

Let $ f$ be a function which is differentiable on $ [a.b]$. Is it always true that $ f'$ is integrable on $ [a,b]$?


[quote="Bictor717"]For each subset with sum greater than 232, there is a corresponding subset with sum less than 232: its complement.[/quote]
It should be "For each subset with sum greater than 232, there is a corresponding subset with sum less [b]or equal[/b] than 232: its complement".
Anyway, my full solution would be:
The cardinal of the set $S\equiv\{1,2,\dots,30\}$ is $|S|=30$, so it has $2^{30}$ possible subsets (each element can appear or not in a subset, so there are two choices for every element).
Let's call $s(A)$ the sum of the elements of a set $A$.
$s(S)=1+2+\dots+30=30\times 31/2=465$.
If $S'$ is a subset of $S$ and $S''$ is the complementary subset $S''\equiv S\setminus S'$ , then $s(S')+s(S'')=s(S)=465$.
So, for every subset $S'$ with $s(S')\leq 232$ the complementary subset $S''$ has $s(S'')\geq 233$ and vice versa.
Therefore, there are $2^{29}$ subsets with sum greater than $232$.

[quote="Leao"]Just to avoid starting a new thread, here is another problem that I didn't quite understand, it also came from ACoPS:
Ten children order ice-cream cones at a store featuring 31 flavors. How many orders are possible in which at least two children get the same flavor?
Thanks in advance for your help!
[hide="my attempt"]
The book's solution involved looking at the complement, and I understood that solution, but why doesn't mine work:
You choose 2 of the ten children at the start to get the same flavor, and this can be done in ${10\choose 2}*31$ ways. Then, the remaining 8 children can get their ice creams in $31^{8}$ ways, so the total is $45*31^{9}$ which is totally wrong...[/hide][/quote]

You overcount.

Say you choose two children at the beginning to get vanilla, and then two other children get vanilla, and the others get some other flavors.  But you could have chosen the two other children to get vanilla at the beginning, and then the original two could have been assigned vanilla, etc.

If you don't want to count the complement, you need to use PIE.  First suppose two children get the same flavor.  This can happen in $\binom{10}{2}31^{9}$ ways.  As blahblahblah explained, we overcounted for situations in which more than two children got the same flavor.

Suppose three children get the same flavor.  This can happen in $\binom{10}{3}31^{8}$ ways.  Now we need to know how many times we counted each one of these possibilities in the previous case.  This is just $\binom{3}{2}=3$, because we counted each possibility once for each pair of people.  Therefore, we need to subtract each possibility for this case twice.

You can eventually get the answer this way after considering 4 people, 5 people, etc., but you have to do a lot of calculations.  The method using the complement is much easier.

[quote="Leao"]Just to avoid starting a new thread, here is another problem that I didn't quite understand, it also came from ACoPS:
Ten children order ice-cream cones at a store featuring 31 flavors. How many orders are possible in which at least two children get the same flavor?
Thanks in advance for your help!
[hide="my attempt"]
The book's solution involved looking at the complement, and I understood that solution, but why doesn't mine work:
You choose 2 of the ten children at the start to get the same flavor, and this can be done in ${10\choose 2}*31$ ways. Then, the remaining 8 children can get their ice creams in $31^{8}$ ways, so the total is $45*31^{9}$ which is totally wrong...[/hide][/quote]

To avoid overcount, you have to separate the problem into two scenarios:
1. more than 2 children got the same flavor
2. all the children get the DIFFERENT flavor

so the number of ways for more than 2 children get the same flavor is equal the number of ways for 10 children to select icecream minus the number of ways that all the children get the different flavor.

number of ways for 10 children to select icecream=${31}^{10}$
number of ways all the children get different flavor=$_{31}P_{10}$ or $31*30*29*28*27*26*25*24*23*22$

Therefore, the answer should be $31^{10}-_{31}P_{10}$

what you think?

[quote="ch1n353ch3s54a1l"][quote="Leao"]How many subsets of the set $[1,2,3...29,30]$ have the sum of its elements greater than $232$ ?
[hide="ACoPS hint"]
What is special about the number 232?
[/hide][/quote]
[hide]
The max sum of the elements in a subset is 465. $\lfloor 465/2\rfloor=232$  :)

find the number of subsets and divide by two...[/hide][/quote]
you are right, however, you need to show that every number under 465 can be construct with the elements 1...30.
or formally, prove that every number under 1+2+3+...+n can be construct with the number 1...n, using every number only once. (it isn't difficult as it seems...)

Going along with what Allan Z said, but I think this solves it for if it is asking for unique orders, as in one kids takes all the orders in a list and goes to place the order, aka assignment of flavors to child does not matter.

Then the number of unique orders is $\binom{10+31-1}{10}$

So the total number is $\binom{10+31-1}{10}-\binom{31}{10}$

[quote="srulikbd"]you are right, however, you need to show that every number under 465 can be construct with the elements 1...30.
or formally, prove that every number under 1+2+3+...+n can be construct with the number 1...n, using every number only once. (it isn't difficult as it seems...)[/quote]

[hide]Use induction.  The subset $\{1\}$ has sum 1.  If some subset $A$ has sum $n<465$, then a subset $B$ with sum $n+1$ can be constructed as follows:

If $A$ does not contain 1, then take $B=A\cup\{1\}$.  If the subset $A$ does contain 1, then consider the least element of $\{1,2,\cdots ,30\}$ that $A$ doesn't contain, where $1\le k\le 30$.  (If $A$ contained all elements, its sum would be 465, a contradiction.)  We get $B$ by replacing $k-1$ with $k$.[/hide]

[quote="Leao"][quote="breez"]What's the solution?

[hide]I think it is $31^{9}$, because you just need the 10th kid to have the same flavor as any of the other kids.[/hide][/quote]

Solution (from ACoPS): We assume the children are distinguishable. First, we count all possible orders without restrictions, which is just $31{}^{1}{}^{0}$. Now we count orders where there is no duplication of flavor; this is just $31*30*29...*22 = P(31,10)$. The answer is then the difference $31{}^{1}{}^{0}-P(31,10).$[/quote]

Oh, so my solution is correct  :P


inequalities
quadratics
inequalities proposed

hello, it is aequivalent to
$ \left(x\plus{}1\plus{}\frac{y}{2}\right)^2\plus{}\frac{3}{4}\left(y\plus{}\frac{2}{3}\right) \ge 0$.
Sonnhard.

Consider the inequality as quadratic trinomial,and then just prove that the discriminant $ D\leq 0$. :wink:


[quote="pvthuan"]The following is true and beautiful: \[(a+b+c)^{6}\geq32(a^{2}+b^{2}+c^{2})(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}).\] [/quote]
But SOS kills it  (for $a,b,c$ non-negative) enough easy. :wink:

[quote="SpongeBob"]What is SOS?[/quote]
Sum Of Squares. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127

[quote="arqady"]
But SOS kills it  (for $a,b,c$ non-negative) enough easy. :wink:[/quote]

Yes, arqady. But I solved it in one line without any expansion. And an even harder is

Let $x,y,z$ be real nonegative numbers such that $x+y+z=2$, prove that \[(x^{4}+y^{4}+z^{4})(xy+yz+zx)\leq2.\]

That is right, Eagle. And the following is also true and solved \[(x+y+z)^{6}\geq 32(x^{4}+y^{4}+z^{4})(xy+yz+zx).\] Another question is to find the greatest possible number $a$ such that the following inequality holds for all nonegative real numbers $x,y,z$ \[(x+y+z)^{8}\geq a(x^{5}+y^{5}+z^{5})(x^{2}y+y^{2}z+z^{2}x+xy^{2}+yz^{2}+zx^{2}).\]

[quote="pvthuan"] And the following is also true and solved \[(x+y+z)^{6}\geq 32(x^{4}+y^{4}+z^{4})(xy+yz+zx).\] [/quote]
It's wrong: $x=3$ and $y=z=1.$ :wink:


trigonometry algebra polynomial quadratics logarithms symmetry induction

[hide="Answer"]$ x \plus{} \frac {1}{x} \equal{} \sqrt {2} \implies x^2 \minus{} \sqrt {2} x \plus{} 1 \equal{} 0 \implies \left( x \minus{} \frac {1}{\sqrt {2}} \right)^2 \equal{} \minus{} \frac {1}{2}$

$ \therefore x \equal{} \frac {1}{\sqrt {2}} \pm \frac {i}{\sqrt {2}} \equal{} \text{cis} \left( \pm \frac {\pi}{4} \right)$

$ \therefore x^{2004} \plus{} x^{ \minus{} 2004} \equal{} \text{cis} (501 \pi) \plus{} \text{cis} ( \minus{} 501 \pi) \equal{} ( \minus{} 1) \plus{} ( \minus{} 1) \equal{} \boxed{\minus{} 2}$[/hide]

[hide]
Squaring and rearranging, we know
$ x^2\equal{}\minus{}\frac{1}{x^2} \implies x^{2004}\equal{}\frac{1}{x^{2004}}$

So we only need to find $ 2x^{2004}$

$ x^2\equal{}\minus{}\frac{1}{x^2} \implies x^4\equal{}\minus{}1$ if $ x\not\equal{}0$
$ x^{2004}\equal{}x^{4(501)}\equal{}\minus{}1$

So the answer is $ \boxed{\minus{}2}$

[/hide]

[hide="Wow, these problems have tons of solutions"]
Define a sequence $ a_n = x^n + \frac {1}{x^n}$. Obviously, $ a_1 = \sqrt2$ and $ a_0 = 2$. To come up with a recurrence relation, we have:
\begin{eqnarray*} \sqrt2a_n & = & a_1a_n \\
& = & \left(x + \frac {1}{x}\right)\left(x^n + \frac {1}{x^n}\right) \\
& = & a_{n + 1} + a_{n - 1} \\
a_{n + 1} & = & \sqrt2a_n - a_{n - 1} \end{eqnarray*}

$ a_{n + 2} - \sqrt2a_{n + 1} + a_n = 0$

This recurrence relation's characteristic polynomial is $ x^2 - \sqrt2x + 1$, whose roots are $ \cos\frac {\pi}{4} + i\sin\pm\frac {\pi}{4}$. Plugging in, we have that the two coefficients of the closed form are both $ 1$, so $ a_n = 2\cos\frac {\pi}{4}n$ by de Moivre. Therefore, $ a_n = 2\cos(501\pi) = \boxed{ - 2}$.
[/hide]

[quote="chess64"][hide]Let $ x \equal{} \cos \theta \plus{} i\sin \theta$. Then $ x \plus{} \frac {1}{x} \equal{} 2\cos \theta \equal{} \sqrt {2}$, so say $ \theta \equal{} \frac {\pi}{4}$. By [url=http://mathworld.wolfram.com/deMoivresIdentity.html]de Moivre's identity[/url],
\[ x^{2004} \plus{} \frac {1}{x^{2004}} \equal{} 2 \cos 2004\theta \equal{} \minus{} 2.
\]
[/hide][/quote]You can apply de Moivre's identity to the whole expression $ x^{2004} \plus{} \frac {1}{x^{2004}}$?

[quote="t0rajir0u"]chess64 is skipping a few steps.

[b]Lemma:[/b]  If $ x \plus{} \frac {1}{x} \equal{} 2 \cos \theta$ then $ x^n \plus{} \frac {1}{x^n} \equal{} 2 \cos n \theta$.

Proof:  Let $ x \equal{} e^{i \theta}$ and use de Moivre's.  Note that $ \theta$ can be complex, although in this case it isn't.[/quote]

It isn't that big a point, but I think that as long as we are trying to be rigorous: we don't just "Let $ x \equal{} e^{i \theta}$"...we have to show this (by using the quadratic formula). I also believe that this is how you would go about trying to solve the problem if you didn't already know the result.

[quote="Altheman"]we don't just "Let $ x \equal{} e^{i \theta}$"...we have to show this (by using the quadratic formula). [/quote]

We do if we define $ \cos \theta \equal{} \frac{ e^{i \theta} \plus{} e^{\minus{}i \theta} }{2}$.  :)  Like I said, $ \theta$ can be complex and the identity is still valid, so in that context it makes more sense to adopt this definition of the cosine.

I know that any real number can be written in that form. But you cannot start with one relation between r and theta, and assume another. Of course you can say that $ x \equal{} e^{i\gamma}$, but how do you know $ \theta \equal{} \gamma$ or for that matter that only one such $ \theta$ exists (well thats not true anyway).

[quote="t0rajir0u"]chess64 is skipping a few steps.

[b]Lemma:[/b]  If $ x \plus{} \frac {1}{x} \equal{} 2 \cos \theta$ then $ x^n \plus{} \frac {1}{x^n} \equal{} 2 \cos n \theta$.

Proof:  Let $ x \equal{} e^{i \theta}$ and use de Moivre's.  Note that $ \theta$ can be complex, although in this case it isn't.[/quote]

I have to double check on this but this looks like an old AHSME problem (back in long time ago...).

[quote="Altheman"]Of course you can say that $ x \equal{} e^{i\gamma}$, but how do you know $ \theta \equal{} \gamma$ or for that matter that only one such $ \theta$ exists (well thats not true anyway).[/quote]

True.  The complex logarithm is multivalued.  The quadratic method is fine as far as rigor, then.

Again, it isn't that big a point and it is probably spam to post all these posts.

But another thing to note is that there are two principal values of $ \theta$. If we were say, trying to find the value of $ x^{10}$, we would not get a unique answer. But the symmetry of $ x^{10} \plus{} x^{ \minus{} 10}$ in this case makes it so there is only one answer.

Alternately, a fairly elaborate proof would be to show by induction that the [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev polynomials of the first kind[/url] describe the behavior of both functions, but that would only prove for positive integral $ n$.

[quote="t0rajir0u"]chess64 is skipping a few steps.

[b]Lemma:[/b]  If $ x \plus{} \frac {1}{x} \equal{} 2 \cos \theta$ then $ x^n \plus{} \frac {1}{x^n} \equal{} 2 \cos n \theta$.

Proof:  Let $ x \equal{} e^{i \theta}$ and use de Moivre's.  Note that $ \theta$ can be complex, although in this case it isn't.[/quote]Could you explain this thoroughly? Sorry I'm not understanding.

Recall the identity $ e^{i \theta} \equal{} \cos \theta \plus{} i \sin \theta$. Negating $ \theta$ gives $ e^{ \minus{} i \theta} \equal{} \cos \theta \minus{} i \sin \theta$. Adding the two yields $ e^{i \theta} \plus{} e^{ \minus{} i \theta} \equal{} 2 \cos \theta$

If $ x \plus{} \frac {1}{x} \equal{} 2 \cos \theta$ for some $ \theta$, then $ x \equal{} e^{\pm i \theta}$ by the identity above. Since this provides two solutions and the original equation was a quadratic, these are the only values of $ x$. Let's just use $ x \equal{} e^{i \theta}$, since the answer will be the same.

$ x^n \equal{} (e^{i \theta})^n \equal{} e^{i(n \theta)}$, and similarly $ x^{ \minus{} n} \equal{} e^{ \minus{} i (n \theta)}$

Add to get $ x^n \plus{} x^{ \minus{} n} \equal{} e^{i(n \theta)} \plus{} e^{ \minus{} i (n \theta)}$

Using the cosine formula we derived, $ x^n \plus{} x^{ \minus{} n} \equal{} 2 \cos n \theta$

[hide="very very sweet solution"]
let $ a_n \equal{} x^n\plus{}\frac 1{x^n}$
notice that $ a_2\equal{}0$ and that $ a_2a_m\equal{}a_{m\plus{}2}\plus{}a_{m\minus{}2}$
this implies that $ a_n\equal{}a_{n\plus{}8}$, so $ a_{2004}\equal{}a_4\equal{}\minus{}2$
[/hide]

Yeah, you just used recurrence, but I like Zuton's solution the best. :D

I liked RisingStar's solution because I thought it was the most elementary


calculus
integration
induction
calculus computations

Please excuse me of my naivety (if you will) as I am in only my first quarter of calculus.  I would be so grateful if you could help me on all (any) of the following problems; for I am preparing for a final exam that will occur this comming Thursday.  Anyway, the "questions" are...

1.)  evaluate the SUM {n=0->infinity} of 1/[n! (n+2)], I evaluated SUM {n=o->5} which indicates that the sum is ~1, but I'm not sure how to irrefutably demonstrate this via "Power Series" (or what have you).

2.) Prove: for all n an element of positive integers, that the integral {o->infinity} of (x^n)*(e^-x)dx = n!

3.) Please confirm that the "Power Series Representation" for xe^x is the SUM {n=0->infinity} of [x^(n+1)]/n!.  If it isn't then please help me to discover it's true PSR (...over all real numbers).

Thnx,
Chris

#1: Write the term as $\frac{n+1}{(n+2)!}$. Turn it into a telescoping series.
The sum is exactly 1.

#2: Induction. The induction step is integration by parts.

#3: Yes, that's it.

Thx for your hasty response jmerry, that helps a lot.

One more question; I don't necessarily want you to solve this for me (but at least give me a few hints):

Problem:

 Let B be the set of all binary sequences.  Prove or disprove: there is a sequence whose image (range) is B.

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

note:  Î denotes “an element of…”

My Attempt to Prove (disprove):

Assume a sequence (call it “g”) such that the “Image of g” equals B: I g = B,
Then for each n Î Z+ (Positive Integers), g(n) Î B.

Therefore, for each k Î Z+, g(k) = 0 or g(k) = 1

Let N be a subset of Z+, then

For any N, let g(k) = 1, if k is in the subset
                    g(k) = 0, if k is not in the subset

Therefore g(k) has image B, and is an infinite “binary sequence”.

Therefore, B (set of all binary sequences) has one-to-one correspondence with the set of all subsets of the positive integers (N)… or natural numbers.

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

I don't think that this satisfies the conditions of the proof.  I'm pretty ignorant when it comes to sequences and proofs for that matter (if you haven't already deduced that); but I think I actually established (in this proof) that there's a binary sequence that has image B, where B = the set {0,1}, not the "set of all binary sequences" like it is supposed to be.  Can someone plz help me out here?  I'm in desperate need for help...

The 1-1 correspondence between infinite binary sequences and subsets of the natural numbers is easy- we map a sequence to the set of indices at which the sequence is 1.

This isn't the question. We want to know whether there is a surjective map from the natural numbers to the infinite binary sequences.

[hide="Hint"]The answer is no. Given a list of sequences, construct a sequence that isn't in the list. This is known as Cantor's diagonal argument.[/hide]

thnx yet again jmerry,

now you are positive about this right?

Don't worry, I'm joking...no...wait, no I'm not.  I don't understand how my teacher could expect me to use a method in which he never even took the time to mention during his lectures.  Can this be interpreted another way (or is my teacher just evil)? :?

My intention isn't to mock you or anything like that, it's just that I want to be 100% sure that I'm being guided in the right direction.  I cannot think of a worse scenario than putting this much time into a proof, having had the correct answer, then erasing it, only to replace it with a proof that I (quite frankly) have to decipher like a secret code and then try to put into the correct context (I'm about to go insane).

Again, I'm sry about my level ignorance relative to the average person that posts here.

It's a famous theorem and a famous proof. The proof is worth knowing. It doesn't belong in a calculus class.

The problem doesn't come out clearly in the language given, but I'm pretty sure it is what I said.

[quote="Calc_quandary"]Again, I'm sry about my level ignorance relative to the average person that posts here.[/quote]
There's nothing wrong with ignorance. You're trying to learn.

My responses in this thread have been short largely because I'd rather guide you toward the right idea than simply give the answers.


That is correct.
(after realizing I can't divide by 6)

How about this one:
Still using the equation $f_n = f_{n-1} + f_{n+1}$
Find the sum of the first $10^n - 4$ terms in terms of $f_0, f_1, f_2,$ and $n$.
(No terms defined)

Let $f_0 = a, f_1 = b, f_2 = c$.  Then the series alternates $a, b, c, -a, -b, -c, \cdots$ as shown before.  Therefore, the sum of every next 6 is 0.  So we merely have to find $10^n - 4 \pmod 6$.
We have $4^n - 4 \pmod 6$.  Clearly this is $0 \pmod 2$, and we see it is equal to $1^n - 1 \equiv 0 \pmod 3$, which implies that it is always $0 \pmod 6$.  Therefore, the sum is always $0$.


geometry
circumcircle
inequalities unsolved
inequalities

Let $ ABC$ be a triangle such that $ \frac{a}{c\minus{}a}\equal{}\frac{c\plus{}a}{b}$. Find $ \angle A: \angle C$.

[quote="johan.gunardi"]Let $ ABC$ be a triangle such that $ \frac {a}{c \minus{} a} \equal{} \frac {c \plus{} a}{b}$. Find $ \angle A: \angle C$.[/quote]

$ ab\equal{}c^{2}\minus{}a^{2}$ --> $ 4R^{2}sinAsinB\equal{}4R^{2}(sin^{2}C\minus{}sin^{2}A)$ ----> $ sinAsinB\equal{}\frac{1\minus{}cos2C}{2}\minus{}\frac{1\minus{}cos2A}{2}\equal{}\frac{1}{2}(cos2C\minus{}cos2A)\equal{}\frac{1}{2}*2sin(C\plus{}A)sin(C\minus{}A)\equal{}sin(\pi\minus{}B)sin(C\minus{}A)$ ---> $ sinAsinB\equal{}sinBsin(C\minus{}A)$---> $ sinB(sinA\minus{}sin(C\minus{}A))\equal{}2sinBsin(\frac{C}{2})sin(\frac{2A\minus{}C}{2})\equal{}0$

$ sinB\equal{}0$ -----> $ B\equal{}\pi k$
$ sin(\frac{C}{2})\equal{}0$ ---> $ \frac{C}{2}\equal{}\pi n$ ---> $ C\equal{}2\pi n$
$ sin(\frac{2A\minus{}C}{2})\equal{}0$ ----> $ \frac{2A\minus{}C}{2}\equal{}\pi m$ ----> $ 2A\minus{}C\equal{}2\pi m$ ---> $ 2A\minus{}C\equal{}0$ ----> $ \angle A: \angle C\equal{}2$

The condition is equivalent to $ c^2 \equal{} a(a \plus{} b)$.

Define a point D on the ray $ \vec{BC}$ such that CD = b, and B and D are on the opposite side to C.

Then, from the condition we obtain that the circumcircle of $ \triangle ACD$ is tangent to $ AB$.
This implies that $ \angle A \equal{} \angle CBA \equal{} \frac {\angle C}{2}$.

[quote="johan.gunardi"]Let $ ABC$ be a triangle such that $ \frac {a}{c \minus{} a} \equal{} \frac {c \plus{} a}{b}$. Find $ \angle A: \angle C$.[/quote]
$ \frac {a}{c \minus{} a} \equal{} \frac {c \plus{} a}{b}$ deduces $ \frac {sinA}{sinC \minus{} sinA} \equal{} \frac {sinC \plus{} sinA}{sin(C\plus{}A)}$. Thus: $ \frac {sinA}{2cos\frac{C\plus{}A}{2}sin\frac{C\minus{}A}{2}} \equal{} \frac {2sin\frac{C\plus{}A}{2}cos\frac{C\minus{}A}{2}}{2sin\frac{C\plus{}A}{2}cos\frac{C\plus{}A}{2}}$
Therefore: $ sinA \equal{} 2sin\frac{C\minus{}A}{2}cos\frac{C\minus{}A}{2} \equal{} sin(C\minus{}A)$. Since $ 0 < C < 180$ we deduce $ {A \equal{} C\minus{}A}$ --> $ \frac{C}{A} \equal{} 2$


If $x,y,z \in R$ then:

$ min ( (x-y)^2, (y-z)^2, (z-x)^2 ) \leq \frac{ \sum x^2 }{2} $. Generalize it!

cheers! :D :D

[quote="Lagrangia"]If x,y,z are real numbers then:

min { (x-y)^2, (y-z)^2, (x-z)^2 } <=(sum(x^2))/2. Generalize it!

cheers! :D :D[/quote]

thus x²+y²-2xy or y²+z²-2yz or z²+x²-2xz
to be compared to x²/2+y²/2+z²/2

suppose the first one smallest (parallel for the others)
then x+z \leq 2y and y+z \leq 2x thus
x²/2+y²/2-2xy \leq  z²/2 ?
z²/2 \geq x²/2+y²/2+xy so yes this is true;
xy is larger than -2xy, if x and y have the same sign.

In alike ways, one can prove the inequality starting from the fact that x should not have the same sign as y.

and it seems the pows aren't working...

² is supposed to be ^2


calculus
integration
calculus computations

Help me to solve the integral, thanks

$ \int\limits_{\minus{}\frac{b}{2a}}^{\plus{}\infty }{\frac{{{x}^{m}}.dx}{{{\left( a{{x}^{2}}\plus{}bx\plus{}c \right)}^{n}}}}$

hello, what do we know about the numbers $ m$ and $ n$?
Sonnhard.

Please help me, I have solved this integral, but I want to check my result.


symmetry
linear algebra
matrix
probability
expected value

There are three points on a circle: $A, B, \text{ and }C$.  A bug starts on $A$.  At any moment, he may move to either point that he is not on.  If he ever goes $ABCA$, he is squashed where he stands.  What is the expected waiting time for his death?

So that the thread doesn't die...[hide="Hint"]Let $A$, $B$, $C$, and $DEAD$ be states.  [hide="Hint 2"]Use Markov chains.  [hide="Hint 3"]Add the states $B'$ and $C'$ [/hide][/hide][/hide]

a solution that doesn't involve Markov chains

[hide]

Let $E_{A}$, $E_{B}$, $E_{C}$ be the expected values if the bug starts at $A$, $B$ or $C$, respectively.  Let $E'_{B}$ be the expected value after the bug travels $AB$.  After 3 moves there is a 1/8 chance the bug is dead, 1/8 chance the bug is back at $A$, a 1/8 chance the bug is at $B$ without previously visiting $A$, a 1/8 chance the bug is at $C$ without previously visiting $A$ a 1/4 chance the bug ends visiting $A$ then $B$, and a 1/4 chance the bug ends visiting $A$ then $C$.  Hence, $E_{A}= \frac18*3+\frac18*(E_{A}+3)+\frac18*(E_{B}+3)+\frac38*(E_{C}+3)+\frac14*(E'_{B}+3) = \frac18(E_{A}+E_{B}+3E_{C}+2E'_{B}+24)$.

Now if the bug starts at $B$ it either moves to $A$ or $C$, so it's expected value at $B$ is $E_{B}= \frac12(E_{A}+1)+\frac12(E_{C}+1)$, and similarly $E_{C}= \frac12(E_{B}+1)+\frac12(E_{A}+1)$.  Adding these last two equations together yields $E_{B}+E_{C}= \frac12(E_{B}+E_{C})+E_{A}+2$.  Therefore $E_{B}+E_{C}= 2E_{A}+4$.

By the symmetry of the solution, we can conclude that $E_{B}= E_{C}= E_{A}+2$.

If the bug travels $AB$ then it has a 1/4 chance of dieing by traveling $BCA$, a 1/2 chance of ending at $B$ otherwise by traveling $BCB$ or $BAB$, and a 1/4 chance of ending at $C$ with $BAC$, so $E'_{B}= \frac14*2+\frac14(E_{C}+2)+\frac14(E_{B}+2)+\frac14(E'_{B}+2)$.  Rearranging $E'B = \frac23+\frac13(E_{C}+E_{B}) = 4+\frac23E_{A}$.

Plugging these expressions into the expression for $E_{A}$ and solving for $E_{A}$ gives us $E_{A}= 24$.  Hence it should take 24 moments for the bug to be squashed.

[/hide]

Edit: Fixed some major flaws.

[quote="calculuslover800"][hide="Answer"]$24$[/hide][/quote]I got a different answer.[hide="My answer"]$17$[/hide]Which is correct?

[hide="I got the same answer as calculuslover"]If the bug is both not on $A$ and has deviated from the $ABCA$ sequence, it will take him an expected two moves to get back to $A$ by a simple geometric sum.

Now suppose the bug is on $A$ beginning a possible $ABCA$. He has a 1/2 chance of moving to $C$, and that adds 3 expected moves: one to move to $C$, expected two to get back to $A$.

The alternative is that he moved to $B$ with 1/2 chance. Now he has a 1/2 (1/4 cumulative) chance of moving back to $A$, adding 2 moves ($ABA$). Otherwise, he's still going.

Now he's on $C$. There's a 1/2 (1/8 cumulative) chance of moving to $B$ and adding 5 moves ($ABCB$ plus the two moves to return to $A$). Else, he moves to $A$, gets squished, and finishes with 3 moves ($ABCA$).

So we have the following possibilities:
$AC$: +3 moves (1/2 = 4/8 chance)
$ABA$: +2 moves (1/4 = 2/8 chance)
$ABCB$: +5 moves (1/8 chance)
$ABCA$: +3 moves and finish (1/8 chance)

So the expected value is $3 \cdot 4+2 \cdot 2+5+3 = \fbox{24}$[/hide]

[hide="My method"]
Let there be $6$ states: $A, B, C, D, B', C'$.  $D$ represents dead, $B'$ represents $B$ coming from $A$, $C'$ represents $C$ coming from $B'$.  $A$ can go to $B'$ or $C$, $B$ to $A$ or $C$, $C$ to $A$ or $B$, $B'$ to $A$ or $C'$, $C'$ to $D$ or $B$, and $D$ to $D$.  Make a stochastic matrix with these states and multiply it by itself repeatedly and see how many steps it takes for $AD$ to be $> \frac{1}{2}$.  The answer I got using a calculator was $17$.
[/hide]

miyomiyo, the thing you've cacluated isn't the expected value.  If we have a random variable $X$ from which we take repeated samples and a possible outcome $x_{0}$, the expected number of samples necessary to get $x_{0}$ is not necessarily the same as the number of samples for which the probability of getting $x_{0}$ exceeds $\frac{1}{2}$.

24 is correct, incidentally.


1) Prove that for $x,y,z>0$ then
\[ 2\leq\dfrac{x^2+y^2}{x^2+yz+xy}+\dfrac{y^2+z^2}{y^2+xz+yz}+\dfrac{z^2+x^2}{z^2+yx+zx}. \]

2) Prove that if $x,y,z\geq0$ then
\[ \dfrac{x^2+yz}{x^2+y^2}+\dfrac{y^2+zx}{y^2+z^2}+\dfrac{z^2+xy}{z^2+x^2}\geq\dfrac52. \]

your second ineq is not true
take z=0 , x=0.5,y=0.25


[quote="lingomaniac88"]In right $\triangle ABC$ with right $\angle C$, point $D$ is on hypotenuse $\overline{AB}$ such that $\angle DAC \cong \angle DCA$.  Given that $AC=7$ and $BC=24$, find $CD$.[/quote]

[hide]$CD$ is a median drawn to the hypotenuse of a right triangle and is therefore $\frac{\sqrt{7^{2}+24^{2}}}{2}=\boxed{12.5}$[/hide]

[quote="lingomaniac88"]In right $\triangle ABC$ with right $\angle C$, point $D$ is on hypotenuse $\overline{AB}$ such that $\angle DAC \cong \angle DCA$.  Given that $AC=7$ and $BC=24$, find $CD$.[/quote]
[hide]${\angle A = \tan^{-1}(\frac{24}{7}}) = 73.7^\circ$

$\frac{CD}{\sin(A)}= \frac{AC}{\sin(180-2A)}$

$CD = \boxed{12.5}$[/hide]

[hide]This is just a median of the right triangle.

So since it is to the hypotenuse it is 12.5

$12.5$[/hide]


A variable rectangle [b]PQRS [/b]has its sides parallel to fixed directions. [b]Q[/b] and [b]S [/b]lie respectively on the line [b]x=b,x=-a [/b]and [b]P [/b]lies on the[b] x-axis.[/b] Then the locus of R is
[b](a) a straight line
(b) a circle
(c) a parabola
(d) none of these.
[/b]
The answer is (a). Please explain the problem [u]if possible with figure[/u].Thanks for your help a lot.

[quote="AMC 14"]you pretty much have a rectangle only defined by three lines, where the points touch on every line except for one. We want to find the possible value for the fourth point. Let's consider the rectangle as a square, so P is between a and b. D is right above point P. As the square is warped, the rectangle becomes elongated and point D is the opposite corner of the rectangle. When P is infinitely close to either a or b, then D will be exactly at the other side. Our second condition assures us that the height and width are held by the three lines, which only one possible height from D to the x-axis. Thus, the locus of D is a line.

I hope my solution is of help, and I sincerely apologize for not being able to provide a diagram.[/quote]

AMC14 I can't understand your proof. Could you be more clear .  :blush:


Let $ \sigma \in S_n$ and $ \alpha <2$. Evaluate$ \displaystyle\lim_{n\to\infty} \displaystyle\sum_{k\equal{}1}^{n}\frac{\sigma (k)}{k^{\alpha}}$.

$ A(n)\equal{}f(1)\plus{}...\plus{}f(n)\geq (n^ 2)/2$

where f is in $ S_n$

denote u(n) the general term of the serie
with Abel tranformation

$ u(n)\geq \sum_{k\equal{}1}^{n\minus{}1}\frac{k^2}{2}(\frac{1}{k^a}\minus{}\frac{1}{(k\plus{}1)^a})$

the general  in RHS is equivalent to $ \frac{C}{k^{a\minus{}1}}$ but a<2 it diverge

the require limit is +oo


[quote="BanishedTraitor"]Can someone explain to me how Carnot's Engine and the Second Law of Thermodynamics are linked.

Thanks.[/quote]

Carnot's engine is the most efficient engine possible, even in theory. On a S vs. T graph the Carnot cycle is a rectangle whose sides are two isothermal lines, and two adiabatic (constant entropy) lines. The net change in entropy for a Carnot cycle is zero. The efficiency of a Carnot engine is given by: $ \eta \equal{} 1 \minus{} \frac{T_C}{T_H}$, where $ T_C$ is the absolute temperature of the low temperature isotherm, and likewise $ T_H$ is the absolute temperature of the high temperature isotherm. Note: Even this efficiency is less than 100%, and it only occurs for an engine with the minimal increase in entropy, i.e. zero.

The Second Law of Thermodynamics implies that for a closed system, one where particles can neither enter nor leave, of which an engine is a primary example, entropy cannot decrease. Entropy is a measure of disorder. Increased entropy represents energy that has gone into making the system more chaotic rather than doing work. Thus increased entropy means decreased efficiency.

[quote="BanishedTraitor"]Can someone explain to me how Carnot's Engine and the Second Law of Thermodynamics are linked.[/quote]

Start by answering the following two questions:

1) What is a Carnot's cycle?

2) What statement of the second law are you considering?

And don't forget that we are discussing Thermodynamics, so microscopic concepts like "molecule" or "particle" are not needed.


probability
combinatorics unsolved
combinatorics

On a small island there are 25 inhabitants. One of these inhabitants starts a rumor which spreads around the isle. Any person who hears the rumor continues spreading it until he or she meets someone who has heard the story before. At that point, the person stops spreading it. What is the probability that all 25 inhabitants will eventually hear the rumor?

Generalize to any number of inhabitants.

There needs to be some model of how the people interact -- what's the probability space we're working over?

I guess, the most natural assumption is that people only meet in pairs, and at any point in time, the next
pair to meet is uniformly chosen. However, calculations with small numbers of inhabitants suggest
that in this case the answer will not be very nice.


Let ABC be a triangle and AD be the bisector of ABC. Denote I, J be the projections of D on AB, AC.  Let U, V be respectively intersections of DI, DJ and (O) such that they lie on the small arcs AB and AC. Let K be the intersection of UJ and VI. Prove or disprove AK is perpendicular to BC.  :maybe:

I drew a dynamic sketch, and it clearly appears not to be true.


search
number theory unsolved
number theory

Prove that if $4^n+2^n+1$ is a prime then $n$ is a power of $3$.

I'm sure it was posted at least two times before.

OK I found them:
http://www.mathlinks.ro/Forum/viewtopic.php?t=69143
http://www.mathlinks.ro/Forum/viewtopic.php?t=69141 (this one contains a lot of other links)
http://www.mathlinks.ro/Forum/viewtopic.php?t=33850
http://www.mathlinks.ro/Forum/viewtopic.php?t=27620
http://www.mathlinks.ro/Forum/viewtopic.php?t=1489
http://www.mathlinks.ro/Forum/viewtopic.php?t=291

http://www.mathlinks.ro/Forum/viewtopic.php?t=291
http://www.mathlinks.ro/Forum/viewtopic.php?t=1489
http://www.mathlinks.ro/Forum/viewtopic.php?t=27620
http://www.mathlinks.ro/Forum/viewtopic.php?t=33850
http://www.mathlinks.ro/Forum/viewtopic.php?t=47701
http://www.mathlinks.ro/Forum/viewtopic.php?t=69143

 :D

 :( Once more, I'm too slow...  :(   ;)

Thanks - I should have used search button in a first place...


find all  $ a$ and $ b$ in $ N$ such
$ a^2\plus{}a\minus{}7\equal{}b^2$

[quote="greatestmaths"]find all  $ a$ and $ b$ in $ N$ such
$ a^2 \plus{} a \minus{} 7 \equal{} b^2$[/quote]

Hint

[hide]Multiply by 4 and find some way of gettin squares.[/hide]

[quote="TZF"]I am. Do you disagree with my solution?

Please share your thoughts :)[/quote]
$ \{a,b\}\subset\mathbb N.$ :wink:

Hmm well, I think that it might be a factoring problem. A^2 + A - 7 = B^2, but we can subtract B^2, subtract A and add 7 to get the equation; A^2 - B^2 = 7 - A, so we can factor the side on the left to get;
(A + B)(A - B) = 7 - A. Hmm well I can't really do anything else from their, so that's all I can do. Well, tell me if it helped any.

It's already solved and if you don't have whole solution than don't post.

I suck at these kind of problems. How do you guys know how to solve them?

It's really easy to observe that if a is larger than 7, $ a^2<a^2\plus{}a\minus{}7<(a\plus{}1)^2$ and can hence not be a perfect square. The only solution  is then (7,7)

Hmm, that helped a little bit, but I still don't understand 100%, only about 50%.

there is a fairly standard method for these types of problems.
See the post n(n+16) in Intermediate topics if you are confused.

[quote="TZF"][hide]$ 4a^2 \plus{} 4a \plus{} 1 \minus{} 29 \equal{} 4b^2$
$ (2a \plus{} 1)^2 \minus{} (2b)^2 \equal{} 29$
$ (2a \plus{} 2b \plus{} 1)(2a \minus{} 2b \plus{} 1) \equal{} 29$

$ ( \minus{} 8,7); \; ( \minus{} 8, \minus{} 7) ; \; (7,7); \; (7, \minus{} 7)$[/hide][/quote]
After arriving at $ (2a \plus{} 2b \plus{} 1)(2a \minus{} 2b \plus{} 1) \equal{} 29$, is there a quick way to determine the right system of two equations (out of four) to get $ (a,b) \in \mathbb{N}^2$ instead of solving all four systems and then determine the right solution?

Well, if $ 2a \plus{} 2b \plus{} 1$ is negative, then at least one of $ a$ and $ b$ will be negative, which we don't want.

Also, we want $ 2a \plus{} 2b \plus{} 1 > 2a \minus{} 2b \plus{} 1$, since we want $ b$ to be positive.

So, $ 2a \plus{} 2b \plus{} 1 \equal{} 29$ and $ 2a \minus{} 2b \plus{} 1 \equal{} 1$ is the only system we care about.

Multiply both sides by 4:
4a^2 + 4a - 28 = 4b^2
(2a+1-2b)(2a+1+2b)=29
Consider all divisors of 29 (2a+1+2b>2a+1-2b)
Therefore a=7 and b=7 :lol:

$ a^2  \plus{} a \minus{} 7 \equal{} b^2  \leftrightarrow \left( {a \plus{} \frac{1}{2}} \right)^2  \minus{} 7\frac{1}{4} \equal{} b^2  \leftrightarrow \left( {a \plus{} \frac{1}{2}} \right)^2  \minus{} b^2  \equal{} \frac{{29}}{4}$

$ \leftrightarrow \left( {a \plus{} \frac{1}{2} \plus{} b} \right)\left( {a \plus{} \frac{1}{2} \minus{} b} \right) \equal{} \frac{{29}}{4} \leftrightarrow \left( {2a \plus{} 1 \plus{} 2b} \right)\left( {2a \plus{} 1 \minus{} 2b} \right) \equal{} 29$

We feel 29 is a prime number. SO:
$ \left[ \begin{array}{l}
 \left\{ \begin{array}{l}
 2a \plus{} 1 \plus{} 2b \equal{} 29 \\
 2a \plus{} 1 \minus{} 2b \equal{} 1 \\
 \end{array} \right. \\
 \left\{ \begin{array}{l}
 2a \plus{} 1 \plus{} 2b \equal{} 1 \\
 2a \plus{} 1 \minus{} 2b \equal{} 29 \\
 \end{array} \right. \\
 \left\{ \begin{array}{l}
 2a \plus{} 1 \plus{} 2b \equal{}  \minus{} 1 \\
 2a \plus{} 1 \minus{} 2b \equal{}  \minus{} 29 \\
 \end{array} \right. \\
 \left\{ \begin{array}{l}
 2a \plus{} 1 \plus{} 2b \equal{}  \minus{} 29 \\
 2a \plus{} 1 \minus{} 2b \equal{}  \minus{} 1 \\
 \end{array} \right. \\
 \end{array} \right.$

I think that's right@@@ :wink:

Nice solution. Even one that I can understand. How do you guys get so good at problems like these?


Alyssa has four marbles of different colors, red, blue, green and yellow. In how many
different ways can she form a set consisting of some of the marbles if she must choose at least one marble?

[hide="I don't think this is correct because"] {2,3} is different from {3,2}, so the answer should be 64, but they get 15[/hide]

[hide]what's 2,3 and 3,2 if you mean like blue red and read blue, then yeah, that's the same. So each color can either be in or out. So 2^4=16, but it can't be all out so -1=15[/hide]

[quote="penny13"][hide="I am curious, wouldn't this be"] 19 because we can make a chart

1 marble chosen----- 4 choices
2 marbles chosen---  6 choices (4x3 divide by 2 because we don't care about the order)
3 marbles chosen---  8 choices  (same as above but divide by our 3 ways we can choose)
4 marbles chosen---  1 choice   (there is only one way for a marble to be chosen)[/hide]

[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Hide your solutions.[/url][/color][/quote]
That was my answer at first. The answer said 15 though.

[quote="penny13"]
3 marbles chosen---  [color=#BF0000]8[/color] choices  (same as above but divide by our 3 ways we can choose)[/quote]

That should be $4$, and hence the answer is $15$.

[hide = "Official Solution"]Each of the four marbles can either be in Alyssa's set, or not. That gives two possibilities for each marble, for a total of $2^4=16$ possible sets. However, we are told the set must have at least one marble. We need to subtract one to eliminate the case of an empty set, which we counted. Our final answer is $16-1=\boxed{15}$ sets.[/hide]

I agree with AoPS and Radioactive. Either way, its just 4 choose 1 plus 4 choose 2 plus 4 choose 3 plus 4 choose 4

Doesn't tell you enough information, I didn't know that you could only choose 2 marbles per set, so yeah.

this was my first post :)
i am posting 8 or 9 months later :)
now i have 208 posts

this wasn't my first post... my first one was in my C&P class... but the thread is locked so i can't do anything

[hide=process]Though you are supposed to do this with complimentary counting, with casework counting you could easily do it by saying one set marbles there are 4 two set marbles there are 6 3 set marbles there are 4 and 4 set marbles there are 1 and add to 15[/hide]

[hide=Hint]
You can do complementary counting (subtracting from total) or use combinatorics (adding to get total) to solve this. Also the color of the marbles has nothing to do with the solution.
[/hide]
[hide=Process of Combinations]
We are told that Alyssa must choose at least $1$ in $4$ marbles. So that means that she can choose $1$, $2$, $3$, or $4$ marbles from $4$ marbles. That's $\binom{4}{4}+\binom{4}{3}+\binom{4}{2}+\binom{4}{1}=1+4+6+4=\boxed{15}$ ways she can form a set with the specifications.
[/hide]


topology
advanced fields
advanced fields unsolved

Hi all,

I need to show that for a space $ X$ the following three conditions are equivalent:
(a)  Every map $ \phi: S^1 \rightarrow X$ is homotopic to a constant map, with image a point.
(b)  Every map $ \phi: S^1 \rightarrow X$ can be extended to a map $ \phi^*: D^2 \rightarrow X$.
(c) $ \pi_1(X,x_0)\equal{}0$ for every $ x_0 \in X$.
Deduce that a space $ X$ is simply-connected iff all maps $ S^1 \rightarrow X$ are homotopic.

For $ (a) \Rightarrow (b)$, am I not able to do the following?
If $ \phi(S^1)\equal{}x_0$, some $ x_0 \in X$, then define $ \phi^*(x)\equal{}x_0$ for every $ x \in D^2$.  Then, $ \phi^*$ agrees with $ \phi$ on the subspace $ S^1 \subset D^2$.  Is there more to the definition of a map "extended" to $ D^2$?

For the rest, I'm pretty stuck.

Any help or hints would be very much appreciated.

A map "extended" to $ D^2$ means a continuous map. The map you defined is not necessarily continuous.

Another way to think about extension is that if $ i: S^1 \to D^2$ is the inclusion map, then $ \phi \equal{} \phi^* \circ i$.

Think about the conceptual relationship between (1) and (2). Essentially, if a map can be extended to a map from $ D^2$, then the image of $ D^2$ can be thought of as the region enclosed by the image of $ S^1$. In other words, $ S^1$ actually bounds a region in $ X$. This means that $ S^1$ can be squished to a point through this region. This should help you with (2) to (1).

To get (1) to (2), notice that we try to deduce the existence of a map from a two-dimensional space $ D^2$ based on properties about a map from a one-dimensional space $ S^1$. This property, specifically, is the existence of a certain homotopy. How does this property help us create a map from a higher dimensional object? ;)

To understand why (1) and (3) are related, think about what an element of the fundamental group actually is.


modular arithmetic
number theory proposed
number theory

Let $k$ be a positive integer. A natural number $m$ is called [i]$k$-typical[/i] if each divisor of $m$ leaves the remainder $1$ when being divided by $k$.

Prove:

[b]a)[/b] If the number of all divisors of a positive integer $n$ (including the divisors $1$ and $n$) is $k$-typical, then $n$ is the $k$-th power of an integer.

[b]b)[/b] If $k > 2$, then the converse of the assertion [b]a)[/b] is not true.

If $n=p_1^{a_1}\ldots p_t^{a_t}$, then the number of divisors is $(a_1+1)\ldots(a_t+1)$. If this number is $k$-typical, then $a_i+1=M_k+1$, meaning that all $a_i$s are multiples of $k$, and this proves a). In order to prove b), just take numbers of the form $p^{k(k-2)}$. This is a $k$'th power, and the number of divisors is $(k-1)^2$, which is divisible by $k-1$, which is not $1\pmod k$.


This is from USAMO, but is appropriate for this level.

Suppose $P(x)$ is a polynomial of degree $n\geq1$ such that $P(k)=\frac{k}{k+1}$ for $k=0,\ 1,\ 2,\ ...,\ n$. Find $P(n+1)$

good hints, but are you going to solve the problem?

Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=289576&highlight=#289576]HERE[/url]. That problem's almost the same. You only need to do a little more algebra to solve it completely.

[hide="Here you go"]So you build a polynomial [tex]Q(x)=(x+1)\cdot P(x)-x[/tex] of degree [tex]n+1[/tex] with roots [tex]{0,1,...,n}[/tex].
The polynomial has the form of:
[tex]Q(x)=A\cdot (x)\cdot (x-1)\cdot ...\cdot (x-n)[/tex].
You also know:
[tex]Q(-1)=(-1+1)\cdot P(-1)-(-1)=1[/tex] and hence:
[tex]Q(-1)=A\cdot (-1)\cdot (-1-1)\cdot ...\cdot (-1-n)=A\cdot (n+1)!\cdot (-1)^{n+1}=1\ \Longrightarrow\\ \Longrightarrow\ A=\frac{(-1)^{n+1}}{(n+1)!}[/tex].
Now you can compute:
[tex]Q(n+1)=\frac{(-1)^{n+1}}{(n+1)!}\cdot (n+1)\cdot (n+1-1)\cdot ...\cdot (n+1-n)=\\=\frac{(-1)^{n+1}}{(n+1)!}\cdot (n+1)!=(-1)^{n+1}[/tex].
Transforming the first expression you get:
[tex]P(n+1)=\frac{Q(n+1)+n+1}{n+2}=\frac{n+1+(-1)^{n+1}}{n+2}[/tex].
It means [tex]P(n+1)=1[/tex] for odd [tex]n[/tex] or [tex]P(n+1)=\frac{n}{n+2}[/tex] for even [tex]n[/tex].[/hide]

good job, very nicely done, and CORRECT!


so i am thinking about buying The Art and Craft of Problem Solving.  i am wondering if this book has answers and what kinds of problems it provides the reader.

The problems are from hard AMC 12 up to Olympiad level.  Most are around AIME to the easier side of the olympiad spectrum though.

There is a hint section at the end.  But there is no solution manual available to the public.  The best alternative is to post problems that you cannot figure out on these boards.

are there any books like this that have solutions manuals (other than AoPS), or are the hints very useful?

I think Problem Solving Strategies and Problem Solving Problems would work.

I've looked at the books fouier series mentioned, but they seem too dry. For Olympiadproblem solving materials,I am getting Math Oly Challenges and Treasures, which seem more interesting. I'll probably get PSS next year

does problem solving through problems come with solutions?

No, but you shouldn't really need a solution guide. If you aren't able to solve the problems, you're probably giving up too quickly. Post questions on the board if you really need help.

ACOPS is really good BECAUSE it doesn't have a solutions manual.  It makes you think about problems, and keep thinking about them, too.  I liked ACOPS as much as AoPS, and I am trying to buy (not borrow) a copy, but it's pretty expensive...


A dart board has two concentric circles. Landing a dart in the outer ring will net you seven  points. The number inside the inner circle is faded away and thus impossible to read, however, Albatross remembers that the highest score impossible to get is $5^3$ . What is the point value gained by landing a dart in the inner circle?

[hide]Use formula to get 7x - 7 - x = 125. Solve to get $\boxed{22}$.[/hide]

[hide]Apply Chicken McNugget, $7n-7-n=125$
$6n=132$
$n=22$.[/hide]

[hide]Chicken McNugget will give you 22.  Checking that it is relatively prime to 7, which it is, then that is your answer.

22.0 to the nearest tenth

$22$[/hide]


search
Pythagorean Theorem
geometry
geometry open

Hi!
Please help me to find problems which use the following well knownlemma

 $ AC\perp BD$$ \Longleftrightarrow$ $ AB^2 \minus{} AD^2 \equal{} CB^2 \minus{} CD^2$.

I spent 2 hours on search but I've found only a few problems.

Please give some links, if somebody remembers.

Thanks a lot!

hello, what do you know about the point $ D$?
Sonnhard.

hello, if $ \overline{AC}\perp\overline{BD}$, then let $ S$ the intersection point. Then we get after the Pythagorean theorem
$ \overline{AB}^2\equal{}\overline{BS}^2\plus{}\overline{AS}^2$ (1)
$ \overline{AD}^2\equal{}\overline{AS}^2\plus{}\overline{DS}^2$ (2)
so we get from (1) and (2):
$ \overline{AB}^2\minus{}\overline{AD}^2\equal{}\overline{BS}^2\minus{}\overline{DS}^2$.
Analogously we get $ \overline{CB}^2\minus{}\overline{CD}^2\equal{}\overline{BS}^2\minus{}\overline{DS}^2$.
From here follows
$ \overline{AB}^2\minus{}\overline{AD}^2\equal{}\overline{CB}^2\minus{}\overline{CD}^2$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, if $ \overline{AC}\perp\overline{BD}$, then let $ S$ the intersection point. Then we get after the Pythagorean theorem
$ \overline{AB}^2 \equal{} \overline{BS}^2 \plus{} \overline{AS}^2$ (1)
$ \overline{AD}^2 \equal{} \overline{AS}^2 \plus{} \overline{DS}^2$ (2)
so we get from (1) and (2):
$ \overline{AB}^2 \minus{} \overline{AD}^2 \equal{} \overline{BS}^2 \minus{} \overline{DS}^2$.
Analogously we get $ \overline{CB}^2 \minus{} \overline{CD}^2 \equal{} \overline{BS}^2 \minus{} \overline{DS}^2$.
From here follows
$ \overline{AB}^2 \minus{} \overline{AD}^2 \equal{} \overline{CB}^2 \minus{} \overline{CD}^2$.
Sonnhard.[/quote]

Of course.. But I asked about problems.. And not easy ones, if it's possible.

usually you can't solve hard problems with this lemma
but with this lemma you can prove the karno principle that has many hard problems


Find the smallest integer $ n>0$ such that $ n^2\minus{}n\plus{}11$ is the product of four primes (not necessarily distinct).

[hide="Impossible Prime divisors of n^2 - n + 11"]Notice that neither prime from $ \{2,3,5,7\}$ can divide $ n^2 \minus{} n \plus{} 11$[/hide]
[hide="First candidate"]Checking $ 11^4$ that no integer solution to $ 11^4 \equal{} n^2 \minus{} n \plus{} 11$ by evaluating $ n \equal{} 11^2$ and $ n \equal{} 11^2 \plus{} 1$[/hide]
[hide="Second Candidate (Lucky one)"]Checking $ 11^3 \times 13$ and noticing $ \boxed{11^3 \times 13 \equal{} (11\times 12)^2 \minus{} (11\times 12) \plus{} 11}$[/hide]


[quote="seamusoboyle"]Or we can use fermats last theorem![/quote]

That's not the first thing I would have attempted. But keep in mind that it applies to natural numbers, not the reals. I guess it doesn't matter in this case.

Ah. Didnt see the real part. My mistake.

Well , the best way is this - standard method for this type of problems !

We write the equation in the form

     $(x^2 +x-2)^3 + (2x^2 -x-1)^3 +(3-3x^2)^3 = 0$

 Since the sum of the bases is $0$ , the equation becomes

          $3 (x^2 +x-2) (2x^2 -x-1) (3-3x^2) = 0$

 But this can be solved very easy and the solutions are  $x = 1 , x = -1 , x = -2 , x= -\frac{1}{2}$ ,

[b]Remember that if    $a +b+c = 0$ , then $a^3 + b^3 + c^3 = 3abc$ [/b]

[u]Babis[/u]

If we simply factorize the two sides we will come to an equation : $9(x+2)(2x+1)(x+1)=0$..Earlier we got a solution $x=1$...Thus the other solutions are clearly $ -1,-2,\frac{-1}{2}$.


The billionaire Dagobert Grinberg arranges his $m$ bag with gold coins. Surprised he realizes that the bag can be arranged in a triangular and quadratic form without residue. How many bags does Dagobert have, if $m \in [1000,2000].$ There was even bigger surprise when Dagobert realized that he can arrange the gold coins in a triangular and quadratic way without residue too. How many coins does Dagobert have at least ?  :)

Hmm, if you say me how much euros a gold coin is worth, I will see whether I will be happy with my $1225\cdot 1$ (or $1225\cdot 36$?) gold coins...

  Darij

Hey Darij,

you have 1225 bags with gold coins not 1225 gold coins !!!  :D  But you have not yet determined the number of coins.  Maybe one gold coin for one pound sterling ?  :)

I thought the number of gold coins was the number of gold coins in each bag... in any case, the number of gold coins can be only 1, 36, 1225, 41616 or etc. ([url=http://mathworld.wolfram.com/SquareTriangularNumber.html]square triangular numbers[/url]), but I don't know how little you assume your minimum to be...

  Darij

[quote="orl"]The [b]billionaire[/b] Dagobert Grinberg arranges his $m$ bag with gold coins. [/quote]

Ah well, this is really a clue, but is it a German billionaire or an English one? I. e., how much is a billion?

  Isotomion (= the alter ego of Darij Grinberg, replacing the main ego when the latter has troubles with posting on ML during a firewall idiosyncracy of the internet room where he is posting from)

Well, in German it is a "Milliardr". So please find the smallest number $\geq 10^9.$


calculus
integration
arithmetic sequence

The sum of all numbers of the form $ 2k + 1 $, where $ k $ takes on integral values from 1 to $ n $ is:
\[ \text{(A)}\ n^2 \qquad \text{(B)}\ n(n+1) \qquad \text{(C)}\ n(n+2) \qquad \text{(D)}\ (n+1)^2 \qquad \text{(E)}\ (n+1)(n+2) \]

[hide="Sol"]
$ 2k+1 $
1 will repeat n times so Sum of $ 1 $ is $ n $,
Now, $ 2k $,
$ 2[1+2+3+.....+n] $
Arithmetic sequence,
$ 2n+n(n-1) $
$ n^2+n $
then,
$ n^2+2n $
$ n(n+2) $
Answer=$ \boxed{\text{(C)}} $
[/hide]

[hide]we see an arithmetic sequence 3, 7, 11, 15, etc., and it goes to $ 2n+1 $, and there are n terms.

$ \frac{(3+2n+1)(n)}{2}=n(n+2) $. [/hide]


Is there a way to get this size?

[img]http://img116.imageshack.us/img116/5891/holabz6.png[/img]

I actually use 11pt but is it possible to set 10.5pt or the size of the image?

You can try[code]\usepackage{fix-cm}[/code] then use [code]\fontsize{10.5pt}{10pt}\selectfont[/code]for where you want 10.5pt font. Change the second parameter 10pt to whatever line spacing looks best.

Oops... I have a doubt...

In the preamble I have

\documentclass{article}
\usepackage{fix-cm}
\fontsize{10.5pt}{10pt}\selectfont

What do I have to put after \selectfont and \documentclass[]{article} to get de 10.5pt size?

Here's a sample document
[code]\documentclass[12pt]{article}
\usepackage{fix-cm}
\begin{document}
This is in the size given by documentclass
\fontsize{10.5pt}{10pt}\selectfont This is in 10.5pt
\end{document}[/code]


function
vector
real analysis
real analysis unsolved

$f: V \to \mathbb{R}$ a $C^{1}$ function where $V$ is a Banach space. Suppose that $m = \inf_{x \in V}f(x) >-\infty$. Prove that for every $\epsilon>0$ there exist $v \in V$ such that:
$c \leq f(v) \leq c+\epsilon$ and $\| Df(v) \|_{V'}\leq \epsilon$

I assume that $c=m$ (otherwise the problem doesn't make much sense). Assume the contrary. Start with any point $x_{0}\in V$ such that $f(x_{0})<m+\varepsilon$. Then we must have $\|Df(x_{0})\|\ge \varepsilon$ and, therefore, there exists a vector $v_{0}\ne 0$ such that
\[f(x_{0}+v_{0})\le f(x_{0})-\frac{\varepsilon}{2}|v_{0}|\,.\]
Note that for any such vector $|v_{0}|\le 2$. So, we may choose the "almost longest" $v_{0}$ (the exact meaning being that $|v_{0}|$ is at least half of the supremum of lengths of such vectors). Now let $x_{1}=x_{0}+v_{0}$. Choose $v_{1}$ for $x_{1}$ in the same way and so on. We shall have $\sum_{j}|v_{j}|\le \frac{2}{\varepsilon}(f(x_{0})-m)\le 2$, so $x_{j}$ is a Cauchy sequence. Let $x$ be its limit. Choose the vector $v\ne 0$ such that
\[f(x+v)\le f(x)-\frac{2\varepsilon}{3}|v|\,.\]
Then, because of continuity of $f$, we will have
\[f(x_{j}+v)\le f(x_{j})-\frac{\varepsilon}{2}|v|\,.\]
for all sufficiently large $j$. But $|v_{j}|\to 0$, so $|v_{j}|<\frac{1}{2}|v|$ for large $j$, which contradicts the choice of $v_{j}$.

Note that the continuity of $Df$ has not been used in this proof: pointwise differentiability is enough.


Three circles of radii $R_{1}$, $R_{2}$, $R_{3}$ touch each other externally. Show that there exist exactly two circles of radii $r$ and $R$ respectively (with $r<R$) which touch all three these circles, and we have $\frac{1}{r}\pm\frac{1}{R}=\frac{2}{R_{1}}+\frac{2}{R_{2}}+\frac{2}{R_{3}}$.
(Plus or minus depends on some condition.)

Namdung

This is a very well-known result 'the Kiss precise' from Soddy, but it goes back at least to Descartes that, if 3 circles ar externally tangents, with radii a,b,c respectively, then the circles which are tangents to each of these 3 circles (internally or externally) have radius d such that :
2( 1/a 2  + 1/b 2  + 1/c 2  + 1/d 2 ) = (1/a + 1/b + 1/c + 1/d) 2 .

Pierre.

This was solved b4 on this forum:

[url]http://www.mathlinks.ro/viewtopic.php?t=352[/url]

I hope it's Ok. I don't know if ne1 has really checked it thoroughly  :D


modular arithmetic
algebra
polynomial
number theory unsolved
number theory

prove [i]Lucas' Theorem[/i]:
   if   $ m \equal{} a_{0} \plus{} a_{1}p \plus{} ... \plus{} a_{t}p^{t}$ with $ 0\leq a_i<p$
      $ n \equal{} b_{0} \plus{} b_{1}p \plus{} ... \plus{} b_{t}p^{t}$ with $ 0\leq b_i<p$
    then $ C_{m}^{n}$ $ \equiv$ $ \prod_{i \equal{} 1}^{t}C_{a_{i}}^{b_{i}}$ ($ mod p$)
  where $ p$ is a prime number .
         we suppose that $ C_{a}^{b}\equal{}0$ if $ b<a$

Use congruence polynomial.
This problem has some application
1)Find all  such that $ C_{n}^{k}\equal{} 0(mod2)$ for all $ 1 < k < n$
2)--------------------$ C_{n}^{k}$odd for all $ 1 < k < n$
You don't need suppose that because we can caculas $ C_{n}^{k}$ for all  $ (n,k)$ by series express.

It's equivalent to
\[ \binom{pa\plus{}x}{pb\plus{}y}\equiv\binom{a}{b}\binom{x}{y}\pmod{p}\]
where $ x,y<p$ and $ \binom{m}{n}$ is the binomial coefficient, equal to $ C_{n}^{m}$ apparently.
[hide]
The left hand side is the coefficient of the $ t^{pb\plus{}y}$ term in the expansion of $ (1\plus{}t)^{pa\plus{}x}$. But $ (1\plus{}t)^{p}\equiv 1\plus{}t^{p}\pmod{p}$, as all other coefficients in the binomial expansion ($ \binom{p}{1},\binom{p}{2},\cdots\binom{p}{p\minus{}1}$) are all clearly divisible by $ p$.

So $ (1\plus{}t)^{pa\plus{}x}\equiv (1\plus{}t)^{x}(1\plus{}t^{p})^{a}\pmod{p}$. In this new polynomial, the only way to get a $ t^{pb\plus{}y}$ term is to multiply a $ t^{y}$ from $ (1\plus{}t)^{x}$ with a $ t^{pb}$ from $ (1\plus{}t^{p})^{a}$. Why is this true? The exponent we need from $ (1\plus{}t)^{x}$ is uniquely determined to be $ y$ because it must be congruent to $ y$ modulo $ p$, but is also at most $ x$ and therefore less than $ p$. Then the exponent we need from the expansion of $ (1\plus{}t^{p})^{a}$ must be $ pb$ in order to end up with a $ t^{pb\plus{}y}$ term.

Therefore, the coefficient of $ t^{pb\plus{}y}$ in $ (1\plus{}t)^{x}(1\plus{}t^{p})^{a}$ is $ \binom{x}{y}\binom{a}{b}$. Since $ (1\plus{}t)^{x}(1\plus{}t^{p})^{a}\equiv (1\plus{}t)^{pa\plus{}x}\pmod{p}$, we have
\[ \binom{pa\plus{}x}{pb\plus{}y}\equiv\binom{a}{b}\binom{x}{y}\pmod{p}\]
[/hide]

[quote="hoangclub"]prove $ lucas theorem$
   if   $ m \equal{} a_{0}\plus{}a_{1}p\plus{}...\plus{}a_{t}p^{t}$
      $ n \equal{} b_{0}\plus{}b_{1}p\plus{}...\plus{}b_{t}p^{t}$
    then $ C_{m}^{n}$ $ \equiv$ $ \prod_{i \equal{} 1}^{t}C_{a_{i}}^{b_{i}}$ ($ mod p$)
  where p is a prime number .
         we suppose that $ C_{a}^{b}$=1 if b<a[/quote]
well-known //http://ecademy.agnesscott.edu/~lriddle/ifs/siertri/LucasProof.htm  :wink:


Show that the sequence $\{a_n\}_{n\geq1}$ defined by $a_n = [n \sqrt 2]$ contains an infinite number of integer powers of $2$. ($[x]$ is the integer part of $x$.)

[quote="Rust"] then $n_{k+1}=2n_{k}-1, \theta_{k+1}=2\theta_{k}.$ [/quote]
i don not understand


The [i]Simson theorem[/i] states that, for any point $P$ on the circumcircle of a triangle $ABC$, the orthogonal projections of the point $P$ on the lines $BC$, $CA$, $AB$ lie on one line. This line is called the [i]Simson line[/i] of the point $P$ with respect to triangle $ABC$.

[b]a)[/b] Let $P$ and $Q$ be two points on the circumcircle of triangle $ABC$, and let $s$ and $t$ be the Simson lines of these points $P$ and $Q$ with respect to triangle $ABC$. Let $R$ be an arbitrary point on the circumcircle of triangle $ABC$. Prove that $\measuredangle\left(s;\;t\right)=\measuredangle QRP$, where we use directed angles modulo $180^{\circ}$.

[Feel free to work with non-directed angles in the proof if you are more familiar with them.]

[b]b)[/b] For any line $d$ in the plane, prove that there exists exactly one point on the circumcircle of triangle $ABC$ whose Simson line with respect to triangle $ABC$ is parallel to the line $d$. Construct this point.

  darij

[hide="a)"]Let $Q_{1}, Q_{2}, Q_{3}$ be the feet of perpendiculars from $Q$ to $BC, AB, AC$ respectively.

Lemma 1
Let $S$ be the intersection between the perpendicular from $Q$ to $BC$ and the circumcircle, then $AS \| t$.

We have $\angle QQ_{1}Q_{2}= 90^{\circ}-\angle Q_{2}Q_{1}B = \angle B-\angle QQ_{2}Q_{1}$.
Since $QQ_{1}BQ_{2}$ is cyclic we have $\angle QQ_{1}Q_{2}= \angle B-QQ_{2}Q_{1}= \angle QBA = \angle QSA$.
Also, in the same way, $AR \| s$.

Back to the problem.
$s$ intersect the circumcircle in $M, H$ and $t$ in $L, K$. Then $\angle (HM, KL) = \frac{\smile HK+\smile ML}{2}$. Let $S, R$ be the intersections between $QQ_{1}, PP_{1}$ and the circumcircle, respectively, where $P_{1}$ is the orthogonal projection of $P$ on the line $BC$. Then $\smile PQ = \smile SR$ and for Lemma 1, $\smile AK = \smile SL$ and $\smile HA = \smile MR$. Hence $\angle (HM, KL) = \angle (s, t) = \frac{\smile PQ}{2}$.[/hide]

First, sorry for a mistake in the problem. I have just fixed it (replaced < PRQ by < QRP).

Andreas - your proof is correct, but hard to understand... e. g. you should have defined the point $P_{1}$ (as the orthogonal projection of the point P on the line BC) and the point R (as the point where the perpendicular to BC through P intersects the circumcircle of triangle ABC, apart from P).

[hide="Anyway, here is my solution."][i]Solution.[/i] [b]a)[/b] Let X and Y be the orthogonal projections of the point P on the lines BC and CA. Then, the Simson line s of the point P with respect to triangle ABC is the line XY. Thus, < (s; CA) = < XYC. Now, since < CXP = 90° and < CYP = 90°, the points X and Y lie on the circle with diameter CP; hence, < XYC = < XPC. Now, in the right-angled triangle PXC, we have < XPC = 90° - < PCX. Thus, < (s; CA) = < XYC = < XPC = 90° - < PCX = 90° - < PCB. Similarly, < (t; CA) = 90° - < QCB. Hence,

$\measuredangle\left(s;\;t\right)=\measuredangle\left(s;\;CA\right)-\measuredangle\left(t;\;CA\right)=\left(90^{\circ}-\measuredangle PCB\right)-\left(90^{\circ}-\measuredangle QCB\right)$
$=\measuredangle QCB-\measuredangle PCB=\measuredangle QCP$.

Since the points C, P, Q, R lie on one circle, < QCP = < QRP, so that < (s; t) = < QRP, and problem [b]a)[/b] is solved.

[b]b)[/b] Fix an arbitrary point P on the circumcircle of triangle ABC, and let s be the Simson line of this point P with respect to triangle ABC. Consider a point Q on the circumcircle of triangle ABC, and the Simson line t of this point Q with respect to triangle ABC. We want to show that $t\parallel d$ holds for exactly one position of the point Q on the circumcircle of triangle ABC, and construct this point Q.

According to part [b]a)[/b], we have < (s; t) = < QCP (since C is a point on the circumcircle of triangle ABC). But $t\parallel d$ holds if and only if < (s; t) = < (s; d). Since < (s; t) = < QCP, this is equivalent to < QCP = < (s; d). Now, C and P are fixed points and < (s; d) is a fixed angle; the locus of all points T in the plane such that < TCP = < (s; d) is a line through the point C. Hence, the equation < QCP = < (s; d) holds if and only if the point Q is the point of intersection of this line with the circumcircle of triangle ABC different from C; clearly, there is exactly one such point, and its construction is obvious. Thus, part [b]b)[/b] of the problem is solved.[/hide]

  darij


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ p$ be a prime and let $ G$ be an abelian group of order $ p^{n}$. Prove that the nilradical of the group ring $ \mathbb{F}_pG$ is the augmentation ideal.

Use : the augmentation ideal in the group ring $ RG$ is generated by {$ g$ - 1|$ g\in G$}.

the augmentation ideal is per definition the kernel of a homomorphism $ \mathbb{F}_p G \to \mathbb{F}_p$, hence a prime ideal. thus every nilpotent element lies in this ideal. conversely, $ (g \minus{} 1)^{p^n} \equal{} 0$ shows the other inclusion. $ G$ abelian just guarantees that $ \mathbb{F}_p G$ is a commutative ring, so that the nilradical is, in fact, an ideal.


Set $ A$ contains all the two-digit integers that equal the product of their tens and units digits divided by the quotient of their tens and units digits. What is the square root of the product of all integers in set $ A$?

Let the number be $ 10a\plus{}b$.

Thus, $ \frac{ab}{\frac{a}{b}}\equal{}10a\plus{}b \implies b^2\equal{}10a\plus{}b$

The solutions are $ (2,5), (3,6)$.

Thus, $ \sqrt{25 \cdot 36}\equal{}\sqrt{5^2 \cdot 6^2}\equal{}5 \cdot 6\equal{}30$.


Thank you rjyanco for posting the State meet links  :!:

Does anyone know which schools quallified for the New England meet  :?:

There are different number of schools qualifying in each
section apparently based on the category (Large/Medium/Small)

This year I heard more schools attended in certain sections
hence more qualifying spots...  :maybe:

From my league (in Connecticut) Hopkins and Choate are going.

Worcester County, MA is sending $\boxed{\text{Shrewsbury HS}}$ into large division yay

We're gonna fail  :(

In the Small Schools Division the two teams that will go from the Connecticut State Match will be:

Hopkins
Brunswick School


Let $x$ and $y$ be positive integers such that $xy$ divides $x^{2}+y^{2}+1$. Show that
$\frac{x^{2}+y^{2}+1}{xy}=3$.

Set$x^{2}+y^{2}+1 =kxy.$
Choose root$(x_{0},y_{0})$ of this equation so that $x_{0}+y_{0}$ min.
Use Viet theorem , we have $k=3$(Easy to prove that)

Sorry thachpbc, do you can post the complete solution??
I dont know Viet theorem...  :oops:

[quote="Yimin Ge"]This problem has been posted soooo many times :D

http://www.mathlinks.ro/Forum/viewtopic.php?t=97705
http://www.mathlinks.ro/Forum/viewtopic.php?t=71760
http://www.mathlinks.ro/Forum/viewtopic.php?t=40207
http://www.mathlinks.ro/Forum/viewtopic.php?t=91590

 :P[/quote]
How do you find so many links on it?
 :)


probability
calculus
geometry
trigonometry
number theory
\/closed

Hi I'm kinda new to AOPS, but I'm interested in the courses that it has to offer. I bought the Intro to Number Theory and Intro to Counting and Probability books, but I also want to take the class. Should I read them over first or wait until the class? Moreover, I'm taking pre-calculus next year. Does anyone have a good suggestion on the class I should take here that would prepare me for that?

Furthermore, I've taken Algebra I, II and Geometry before, but I don't think the teacher covered it in depth. Should I also take the class and/or buy the book?

Thanks  :lol:

Reading them before the class should help you with the harder problems presented in class and as homework, even just skimming through sections.  If you've taken Algebra I, II, and Geometry in school, and understood the stuff they taught there, school pre-calculus should be fine.

As for your last question, buying AoPS Volume 1 and/or 2 might be the best (and more cost-effective) choice.  Basically, they both cover a wide range of topics in the "AoPS-style", which is more in depth than what is taught at school

Thanks, do you think I should take Algebra III?

That would probably be a good idea, because the AoPS Alg 3 course covers not just Algebra 2 but also some pre-calculus(I think?) I'm taking the course this June, and going into pre-calc this fall.

K, cool. How many classes do you think I should take or can handle if I"m also taking Biology over the summer?

Read ahead in the book.  I always did this in college, and it made college *very* easy for me.

As for classes to take, Algebra 3 will get you well ahead of your class in non-trig precalc.  We'll have a precalc class in the fall that will go with our precalc book.  We haven't set the dates, but it will likely start just after the Algebra 3 class ends.

You can probably manage two classes during the summer, particularly if you get ahead early on.  It's usually much tougher to carry two subject classes (as opposed to the problem series, which are less demanding time-wise) during the school year.


function
inequalities
algebra solved
algebra

Prove that there does not exist any function f : [0, +∞) → (0, +∞) such that for all real x>= 0 we have f'(x) >=f(f(x)).

Pretty nice and easy. f is of course increasing. Thus, when we integrate from o to x the inequality we will have $ f(x)\geq\frac{x}{2}\cdot f(f(\frac{x}{2}))$. Thus, for all $ x>2$ we have $ f(\frac{x}{2})<x$ and thus $ f(x)<2x$ for all $ x>1 $. Thus, using again the found inequality we obtain $ f(f(\frac{x}{2}))<4 $ for $ x>2 $ and thus $ f(f(x))<4 $ for all $ x>1 $. But $ f(x)>\frac{x}{2}\cdot f(f(\frac{x}{2})>\frac{x}{2}\cdot f(f(1))$ for all $ x>2$  and thus also $ 4> f(f(x))>\frac{x}{2}\cdot f(f(1)) $ for all $ x>2$ , false.


number theory
relatively prime
prime factorization
prime numbers
number theory open

Let's try to find the number of primes less than 30=2*3*5. Since 30 < 49 = 7*7, all numbers less than 30 and relatively prime to it are prime, so the number of primes less than 30 is phi(30)-1+3 (accounting for 1,2,3,5) = 4*2+2 = 10.

Now apply this method to 210=2*3*5*7. The only nontrivial numbers less than 210 that are relatively prime to it and composite are 11*11,11*13,11*17,11*19,13*13 (which is easy to figure out using a calculator with integer multiplication). Then 1 is relatively prime to 210 but composite, and 2,3,5,7 are not relatively prime to 210 but are prime. Then pi(210) = phi(210) - 1 - 5 + 4 = phi(210) - 2. Then phi(210) is easy to calculate using its prime factorization, for it is (2-1)(3-1)(5-1)(7-1) = 2*4*6 = 48. So pi(210) = 48-2=46 which is correct :)

Notice that in that calculation, all we had to know was information about relatively small prime numbers (i.e. those < 20) and perform multiplication. This is a fairly simple way to find the number of primes less than 210. I'm not sure how well this works for other numbers, but are there any interesting implications of this method?

Computing $ \pi(p_1\cdot p_2\cdot\dots\cdot p_n)$ by your method will be harder and harder as $ n$ grows. In particular, one would need to account all semi-primes with prime factors from the interval $ [p_n \plus{} 1,\sqrt {p_1\cdot p_2\cdot\dots\cdot p_n}]$ (of exponentially growing length).

If you are interested in efficient way to compute $ \pi(x)$, take a look at this paper:
M Deleglise, J Rivat [url=http://cr.yp.to/bib/1996/deleglise.pdf]Computing π (x): the Meissel, Lehmer, Lagarias, Miller, Odlyzko method[/url]

Although this thread's a little old, [url=http://terrytao.wordpress.com/2007/06/05/open-question-the-parity-problem-in-sieve-theory/#more-89]Terence Tao[/url] has an interesting post related to this subject.


The answers are now posted on AOPS. How did people do?

I found this AIME to be slightly easier than last year's AIME 2, and much easier than last year's AIME 1.

 What do you think the cutoff will be for USAMO?

I did all right, got 11 so my index is 236 which should be enough I think.
I believe it was easier, because there was not as many counting problems and ugly calculations.
I think cutoff will be around 215-220.

I think I got 11 as well... this makes my index 227. I did pretty badly on AMC lol

I should get a 10 for an index of 236.5.  Hopefully the USAMO will be kind to me.

Congratulations, guys. I messed up on AIME very badly, so I won't be writing USAMO, but our school do have two Juniors who will make it through the floor value.  :)

How did your schools do?

Foundation


function geometry ratio trigonometry probability limit search

In honor of today, pi day, let's make a list of problems whose answer is $\pi$ and do [i]not[/i] mention $\pi$ in them.  Starting with:

1.  For the function $y = x[sin\left(\frac{180}{x}\right)][cos\left(\frac{180}{x}\right)]$, as $x$ goes to infinity, what does $y$ approach?

EDIT: Also, feel free to do any solutions to people's problems.

Okay...

Compute $\sqrt{6\sum_{i=1}^{\infty}\frac{1}{i^{2}}}$.

Let $a$ be the area of the region in the first quadrant bounded by the graphs $y=0$, $x=0$, and $y=\sqrt{1-x^{2}}$.  Compute $4a$.

It's a nice idea. Here's one (though not intermediate level)

Find the limit of $a_{n}$, as $n\to \infty$, where \[a_{n}= \sqrt{ 6 \left( 1+\frac 1{2^{2}}+\frac 1{3^{2}}+\cdots+\frac 1{n^{2}}\right) }.\] :)

[quote="Valentin Vornicu"]What's the ratio of a circle's length to its diameter? :D[/quote]

suprisingly or not, this question was on a local contest...cough bad contest cough:

a circle passes through (0,0) and (3,1), what is the ratio of the circumference to the diameter of that circle

Haha Altheman NSML contest 5 last year right? I remember that, the sad thing is that only me and one other person from my school got it right.

Evaluate $\frac{2}{\cos \frac{\pi}{4}\cos \frac{\pi}{8}\cos \frac{\pi}{16}...}$

Edit:  Let $p$ be the probability that two randomly selected positive integers are relatively coprime.  Evaluate $\sqrt{ \frac{6}{p}}$.

Edit 2:  Evaluate $\frac{1}{2}\cdot \left( \lim_{n \to \infty}\frac{n!}{ n^{n+\frac{1}{2}}e^{-n}}\right)^{2}$

$\frac{2}{p}$ where p is the probability that if you have two parallel cracks on the floor, and a stick that's the length of the distance between them, when you drop the stick it lands on a crack.

the area of a circle over the radius squared.

Does anybody want to show any proofs of these?

[quote="diophantient"]$\sqrt[4]{90 \sum^{\infty}_{x=1}x^{-4}}$[/quote]

The message contains only quote and code tags. Please add some text of your own.

As long as we're still doing this, how about

$\frac{1}{2}\cdot \left( \int_{-\infty}^{\infty}e^{ \frac{-t^{2}}{2}}\, dt \right)^{2}$

Or $\sqrt{ 8 \sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}}$, which seems to have been neglected, the poor thing.

Or how about the beautiful

$2 \cdot \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot ... }{1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot ... }$

That's not pi!!!!! (read the topic...) :wink:

Sorry  :oops: It is $e$ and $e\neq \pi$.

I'm going to do one that's not a formula...

$n$ random numbers between 0 and 100 are chosen.  We split them into $\frac{n}{2}$ pairs of $2$.  If the sum of the squares of a given pair is $< 10000$, we add $1$ to a given counter $c$.  Otherwise, we add $1$ to a given counter $d$.  Find $\lim_{n\to\infty}\frac{4c}{c+d}$.

The surface area of a sphere divided by the square of the diameter.

The volume of a sphere divided by 4/3 of the cube of the radius.

approx. 22/7

To Mr. Vornicu's first expression:
I wanted to attempt a proof.

It is well known that the sum of the inverses of the squares of the natural numbers is $\frac{\pi^{2}}{6}$ (the Basel Problem)

Thus, we have $a_{n}= \sqrt{6 \cdot \frac{\pi^{2}}{6}}$

from which we have $a_{n}= \sqrt{\pi^{2}}$

and finally, $a_{n}= \pi$.

That hardly constitutes a proof.  The content of the problem is the solution to the Basel Problem.

[quote="1=2"]approx. 22/7[/quote]

Incidentally, there's a fascinating integral related to this.  Calculate

$\frac{22}{7}-\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}}\, dx$

[quote="cincodemayo5590"]I think we have to whip out [hide="a whip"] $\arcsin$ of a Taylor series. [/hide][/quote]

But being in radians implies pi.

Area of circle with diameter 2? ^^ Hopefully it hasn't be posted yet.

[quote="Karth"]Or you could just be like:

$\frac{\ln{(-1)}}{i}$[/quote]

Even though that expression might equal $\pi$, could it really be evaluated? Is $\ln{(-1)}$ actually defined?

The problem with taking logs of negative numbers is that they're not uniquely defined.  It's more correct to say that

$e^{i(\pi+2 \pi k)}=-1$

Which of the methods listed so far converges fastest?

Bailey or Ramanujan are extremely fast convergers.

Ramanujan developed certain series which converge to pi that can generalize, then mathematicians with computers were later able to come up with formulas with huge numbers that converged super quick - 50 correct digits per iteration!!

[b]miyomiyo:[/b]  http://www.artofproblemsolving.com/Forum/viewtopic.php?p=831263#831263

[b]AstroPhys:[/b]  http://mathworld.wolfram.com/EulerProduct.html

all of them - how else would they be known?  although i would like to see a few more shown, as all of the ones i know have been linked to or stated.

but yeah some of the wallis formulas/etc. are really nasty.

Happy Pi Day!

I feel a strange need to post as my school for some weird reason celebrated pi day today May 18 instead of March 14.

This year's wisconsin Mathcounts pin - wisconsin - pizza :)


One I see alot:

Find the sum of all positive factors of 504.

My question is, is there a better way of solving this problem besides finding all the factors then adding them?

There's this really cool trick:  Say you want to find the sum of the factors of 72.  First, you prime factorize to get $ 2^3*3^2$.  Then you add all the powers of 2 and powers of 3, and multiply: $ (1\plus{}2\plus{}4\plus{}8)(1\plus{}3\plus{}9)\equal{}195$.

Similarly, the prime factorization for 504 is $ 2^3*3^2*7^1$.  $ (1\plus{}2\plus{}4\plus{}8)(1\plus{}3\plus{}9)(1\plus{}7)\equal{}\boxed{1560}$.  I hope this helps!  :)

If you don't understand how this product represents the sum of all the factors, distribute and you'll see this covers all possible combinations of the availible prime factors.

Where can I find a proof for that trick?

You can prove it for yourself, just multiply it out (in Algebra of course) and you can see that all possible additions are covered.


function
algebra
floor function
calculus
calculus computations

1). Prove that the limit (-1)^[x] as x approaches infinity does not exist. Bracket indicates floor function

2).Suppose that the limit of f(x)=L as x approaches a and L is posotive. Show that there exists an interval I containing a such that f(x)>0 and x E I

Thank you  :)

1) The possible values of $ (\minus{}1)^{[x]}$ is only $ 1$ and $ \minus{}1$, because $ [x]$ is always an integer regardless of $ x\in\mathbb R^\plus{}$. So, if there exists the limit, we can find $ M>0$ such that $ x>M\Rightarrow f(x)\equal{}1$(without loss of generality, we assumed the value of the limit is $ 1$).By completeness property of $ \mathbb R$, it is well known that the set of natural number $ \mathbb N$ is not bounded above . So, we can find an odd natural number $ c$ larger than $ M$, this is contradiction because $ (\minus{}1)^{[c]}\equal{}\minus{}1$.

2)For all $ \varepsilon >0$, there exists $ \delta >0$ such that $ 0<|x\minus{}a|<\delta\Rightarrow |f(x)\minus{}L|<\varepsilon$($ L>0$). The contradictive method also gives the way in this problem. Sorry, I have to go now!

@ J.Y. Choi, thank you so much for explaining it to me. I quite understand it now.  :)


The exterior wall of a building forms a right angle with the ground at its base.  A 25 foot ladder is placed against the wall so that the foot of the ladder is 7 feet from the base of the wall.  The ladder slips and its upper end slides 4 feet down from the wall.  How many feet did the foot of the ladder slide along the ground?

[hide]
my answer was 8.  Is that right?[/hide]

yes. at the beginning it is a 7-24-25 right triangle. when it slides down 4 feet it becomes a x-20-25 right triangle, x, of course, being 15. so 15-7=8. your correct! hooray! give yourself a cookie!  :)


[url=http://www.lunchlady.org/files/seizure.htm]i am recovering in the hospital[/url]

I had a seizure! OMG then I passed out!
it reminds me of a site which you were supposed to look realy closely to find differences in pictures and then a beast thing appeared and screamed real loudly after like a minute!
BTW does anyone know which site that was? I redid my emails so i have no clue....

I laugh at your predicament.

Anybody that posts a link to eBaumsWorld in this topic will be disemboweled.

Contrarily, anybody who posts a link to lunchlady will be reemboweled.

[quote="funcia"]AND ANYONE who has had someone have a seizure right before their eyes and is joking like this after crying like a girl before is crazy for caring about their bowels.
really my head started to hurt after 10 min. of staring at it![/quote]
me and my buddy made something like that during web design class at school and spent the whole period staring at it. good stuff.


I've been writing this class, and I was wondering if I could simplify this using the switch statement. The problem is that I don't know how. Anyone want to help me?

[code]	private String message;

	public String getLetterGrade(int gradePercent){
		KeyboardReader reader = new KeyboardReader();

		if (gradePercent >= 96 && gradePercent <= 100){
			message = "Grade is A+";
		}
		else if (gradePercent >= 92 && gradePercent <= 95){
			message = "Grade is A";
		}
		else if (gradePercent >= 90 && gradePercent <= 91){
			message = "Grade is A-";
		}
		else if (gradePercent >= 86 && gradePercent <= 89){
			message = "Grade is B+";
		}

		else if (gradePercent >= 82 && gradePercent <= 85){
			message = "Grade is B";
		}

		else if (gradePercent >= 80 && gradePercent <= 81){
			message = "Grade is B-";
		}

		else if (gradePercent >= 76 && gradePercent <= 79){
			message = "Grade is C+";
		}

		else if (gradePercent >= 72 && gradePercent <= 75){
			message = "Grade is C";
		}

		else if (gradePercent >= 70 && gradePercent <= 71){
			message = "Grade is C-";
		}

		else if (gradePercent >= 66 && gradePercent <= 69){
			message = "Grade is D+";
		}

		else if (gradePercent >= 62 && gradePercent <= 65){
			message = "Grade is D";
		}

		else if (gradePercent >= 60 && gradePercent <= 61){
			message = "Grade is D-";
		}

		else if (gradePercent >= 0 && gradePercent <= 59){
			message = "Grade is F";
		}
		return message;
	}[/code]

Switch only works where each case is a single value - so here, where you are testing ranges, you can't really use it.

There are a couple of things you could do to simplify it - firstly, assuming gradePercent is always going to be something between 0 and 100, a lot of those conditions are unnecessary - you can get rid of all the <= conditions (other than <= 100), since they are dealt with by the 'else if' part.

But the best way to simplify it would be to use an array/loop - I get the feeling that this code is something you'd write before you started using these at all though. (You could store all the grade cutoffs and grade letters in arrays; then just loop through the array, find the highest grade cutoff that is less than your mark, and use the respective letter).

You could let the case statements fall through:
[code]
switch (gradePercent) {
    case 100:
    case 99:
    case 98:
    case 97:
    case 96:
        message = "Grade is A+";
        break;
    case 95:
    case 94:
    case 93:
    case 92:
        message = "Grade is A";
        break;
    ...
}
[/code]
I don't know if it's simpler than what you already have though :D.


AMC USA(J)MO USAMO algebra polynomial geometry ratio

Just received this information from Maria Monks, one of the coaches along with Allison Miller, at the China Girls Math Olympiad:

Carolyn Kim with a score of 45 got a bronze
Jenny Jin with a score of 39 got a bronze
In Young Cho with a score of 54 got a bronze
Colleen Lee with a score of 48 got a bronze
Joy Zheng with a score of 48 got a bronze
Wendy Mu with a score of 63 got a silver!
Jenny Iglesias with a score of 81 got a GOLD!!
Lynnelle Ye with a score of 87 got a GOLD!!!

That's a medal for each one on our team!
Congratulations to all the team members (and their coaches!), great work!

I am posting this in the AMC forum since the team members were selected on the basis of their USAMO scores and their training was at MOSP.  Many regular readers will know them.

Steve Dunbar
MAA Director, American Mathematics Competitions

Congratulations to all! Just curious, how exactly is the CGMO scored (save the speculative answers...they're generally not too helpful)? Also, are the problems anywhere on the internet?

There are a couple problems in the olympiad section. Look in the Combinatorics unsolved.

Hopefully this isn't too speculative for you.

http://www.msri.org/specials/gmo
[quote]Sherry got first place with 114/120 points[/quote]

20 points per problem, I suppose. This would make a score of $ 87$ quite impressive, as well as all of the others.

Aren't there 8 problems?

When will the problems be officially released?

EDIT: I have found 3 problems from CGMO 2008.
7.	[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221517[/url]
5.	[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221516[/url]
1.	[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1228708#1228708[/url]

Hmm I'm going to be a total hypocrite now. But it seems that those are scores are all multiples of 3. I find it unlikely that problems are scored by one person, out of 15 points, that seems rather inconvenient. I suspect 3 people grade the paper from 0 to 5. But this may be complete BS.

[quote="inyoung"]There are 8 problems, each worth 5 points each.
They just multiply your score by 3 to... I dunno make you feel good.
Uhh, I have the problems with me, but am I allowed to post them?[/quote]

3 have already been posted, so it should be okay. Everyone starts talking about the IMO problems like immediately after the test ends, too.

Congrats to the teams! Upward trend-- this is a good thing! :-)

[quote="inyoung"]There are 8 problems, each worth 5 points each.
They just multiply your score by 3 to... I dunno make you feel good.
Uhh, I have the problems with me, but am I allowed to post them?[/quote]
Last year, Alison said that there were 5 steps, if you will, that you get to in the process of solving the problem. If you get the first step, you get 3 points; the second, 6, and so on, and if you solve the problem, you get 15. So it's easier to get partials than on, say, the USAMO.

And the obligatory disclaimer that must come with everything one says on a public forum: If I've mangled her explanation and the scoring isn't actually done that way, I take full responsibility.

[quote="minsoens"]Inyoung, what's going on in [url=http://web.mit.edu/monks/www/ZhongShan1/images/p7310405.jpg]this[/url] picture? :huh: .... lol hmm[/quote]
Haha oh man, they made you do the aerobic dancing thing again! How'd you guys do in the competition? (the REAL competition you went to China for-- not the math thing, silly!)

[quote="mihail911"][quote="rrusczyk"]Lot of AoPSers on that list, indeed, including one who, I believe, took the first two AoPS classes ever (over 5 years ago!)[/quote]
Woah. This site is really old. :D[/quote]
woah, I think I feel really old, haha. ):

[quote="Piggypi314"]Haha oh man, they made you do the aerobic dancing thing again! How'd you guys do in the competition? (the REAL competition you went to China for-- not the math thing, silly!)[/quote]
We got "most energetic." :lol:
And yeah, I might have spent more time doing aerobics than I spent doing math.

[quote="minsoens"]Inyoung, what's going on in [url=http://web.mit.edu/monks/www/ZhongShan1/images/p7310405.jpg]this[/url] picture? :huh: .... lol hmm[/quote]

Dude, it's a girl's olympiad. They like those things.  :D

[quote="orl"]Dude, it's a girl's olympiad. They like those things.  :D[/quote]
I, for one, didn't. I'm surprised I didn't singlehandedly disqualify our dance team from "most energetic".

Wait...dance? Is this actually part of the main event?

[quote="Piggypi314"]
And the obligatory disclaimer that must come with everything one says on a public forum: If I've mangled her explanation and the scoring isn't actually done that way, I take full responsibility.

[end quote]

What Piggypi314 explained is what I was told about scoring before I had ever gone to the competition.  When I went to the competition, I found out that scoring works on a partial credit system similar to the USAMO/IMO (although not entirely the same: on one problem points were taken off for minor arithmetic errors on solutions that would have received full credit at the IMO).

And yes, as said before: 4 hours for 4 problems each day, 2 days, problems graded on a score of 0-15 in multiples of 3.  I've been told this is because the Chinese are used to tests graded out of 120...

[quote="Alison"]
I've been told this is because the Chinese are used to tests graded out of 120...[/quote]
Whoa whoa whoa where did you hear that? It's out of 100, just like anywhere else...

[quote="bookaholic"]Hmm...my parents generally took tests out of 120 back when they were high-schoolers in China...

[quote]on one problem points were taken off for minor arithmetic errors on solutions that would have received full credit at the IMO[/quote]
Like 4!=12 *sigh*

I think the dance lessons/competition were just to keep us busy in place of having to organize more touring...[/quote]
Wait you serious? My parents said they took tests out of 100. I've never heard of the 120 thing until now.

[quote="serialk11r"]
Wait you serious? My parents said they took tests out of 100. I've never heard of the 120 thing until now.[/quote]
Yeah, i went to ask my parents, and they said 100 as well...

Well it probably differed from school to school then. Like, in every French book I've seen, it says tests in France are all out of 20, but I've heard of schools in France which use 100.

[quote="minsoens"][url]http://www.msri.org/specials/gmo/2008[/url]
[url]http://web.mit.edu/monks/www/Pictures.html[/url]
Inyoung, what's going on in [url=http://web.mit.edu/monks/www/ZhongShan1/images/p7310405.jpg]this[/url] picture? :huh: .... lol hmm[/quote]

Is there a web-page which has the problems, possibly with solutions, and complete results?

Hello, these problems are from Inyoung:

1. (a) Determine if the set $ \{1,2,\cdots,96\}$ can be partitioned into 32 sets of equal size and equal sum.
(b) Determine if the set $ \{1,2,\cdots,99\}$ can be partitioned into 33 sets of equal size and equal sum.

2. Let $ \varphi(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ be a polynomial with real coefficients. Given that $ \varphi(x)$ has three positive real roots and that $ \varphi(0) < 0$, prove that
\[ 2b^3 \plus{} 9a^2d \minus{} 7abc \leq 0.
\]
3. Determine the least real number $ a$ greater than $ 1$ such that for any point $ P$ in the interior of the square $ ABCD$, the area ratio between two of the triangles $ PAB$, $ PBC$, $ PCD$, $ PDA$ lies in the interval $ \left[\frac {1}{a},a\right]$.

4. Equilateral triangles $ ABQ$, $ BCR$, $ CDS$, $ DAP$ are erected outside of the convex quadrilateral $ ABCD$. Let $ X$, $ Y$, $ Z$, $ W$ be the midpoints of the segments $ PQ$, $ QR$, $ RS$, $ SP$, respectively. Determine the maximum value of
\[ \frac {XY \plus{} YW}{AC \plus{} BD}.
\]
5. In convex quadrilateral $ ABCD$, $ AB \equal{} BC$ and $ AD \equal{} DC$. Point $ E$ lies on segment $ AB$ and point $ F$ lies on segment $ AD$ such that $ B$, $ E$, $ F$, $ D$ lie on a circle. Point $ P$ is such that triangles $ DPE$ and $ ADC$ are similar and the corresponding vertices are in the same orientation (clockwise or counterclockwise). Point $ Q$ is such that triangles $ BQF$ and $ ABC$ are similar and the corresponding vertices are in the same orientation. Prove that points $ A$, $ P$, $ Q$ are collinear.

6. Let $ (x_1,x_2,\cdots)$ be a sequence of positive numbers such that $ (8x_2 \minus{} 7x_1)x_1^7 \equal{} 8$ and
\[ x_{k \plus{} 1}x_{k \minus{} 1} \minus{} x_k^2 \equal{} \frac {x_{k \minus{} 1}^8 \minus{} x_k^8}{x_k^7x_{k \minus{} 1}^7} \text{ for }k \equal{} 2,3,\cdots
\]
Determine real number $ a$ such that if $ x_1 > a$, then the sequence is monotonically decreasing, and if $ 0 < x_1 < a$, then the sequence is not monotonic.

7. On a given $ 2008 \times 2008$ chessboard, each unit square is colored in a different color. Every unit square is filled with one of the letters C, G, M, O. The resulting board is called [i]harmonic[/i] if every $ 2 \times 2$ subsquare contains all four different letters. How many harmonic boards are there?

8. For positive integers $ n$, $ f_n \equal{} \lfloor2^n\sqrt {2008}\rfloor \plus{} \lfloor2^n\sqrt {2009}\rfloor$. Prove there are infinitely many odd numbers and infinitely many even numbers in the sequence $ f_1,f_2,\cdots$.

Just to be sure, for #4, are you sure it is XY+YW?  I just want to be sure that the Z does not matter.

I'm fairly sure that it should involve Z, since I remember the numerator being the sum of the diagonals of XYZW (although I could be wrong here)

[quote]4. Equilateral triangles $ ABQ$, $ BCR$, $ CDS$, $ DAP$ are erected outside of the convex quadrilateral $ ABCD$. Let $ X$, $ Y$, $ Z$, $ W$ be the midpoints of the segments $ PQ$, $ QR$, $ RS$, $ SP$, respectively. Determine the maximum value of
\[ \frac {XY \plus{} ZW}{AC \plus{} BD}.
\]
[/quote]yeah I think this makes sense. I can't edit my post though  :(

Darn, Minseon, you suck!

I already solved the problem as it was originally stated: XY+YW.

(The answer is $ \frac {1 \plus{} \sqrt {3}}{2}$.)

Darn darn darn.  Well is anyone certain about the correct statement of the problem?

(And anyway, XY is not a diagonal?  So perhaps you mean XZ+YW?)

[quote="gighiuhui"]Darn, Minseon, you suck! I already solved the problem as it was originally stated: XY+YW. (The answer is $ \frac {1 \plus{} \sqrt {3}}{2}$.)[/quote]good job im so proud of you[quote="gighiuhui"]Darn darn darn.  Well is anyone certain about the correct statement of the problem? (And anyway, XY is not a diagonal?  So perhaps you mean XZ+YW?)[/quote]I don't know why I didn't think about actually drawing the diagram before I posted. XZ+YW makes sense, but if you want reassurance from someone who has more credibility than I now do, PM or IM Inyoung or Lynnelle or Joy.

Yeah, it's XZ+YW; I checked my problem sheet

[quote="minsoens"][quote="gighiuhui"]Darn, Minseon, you suck! I already solved the problem as it was originally stated: XY+YW. (The answer is $ \frac {1 \plus{} \sqrt {3}}{2}$.)[/quote]good job im so proud of you[/quote]

Hey, hey no need for all that sarcasm.

Anyway, it turns out the answer is the same either way, but whatever the equality case of XY+YW is a subset of that for XZ+YW.

Hrmm..

[quote="orl"]So does anyone know the complete results? We are looking for the relative performance of the US team comparably to other nations, don't we?  :)[/quote]

We had to leave before the closing ceremony, so we didn't get much information at all about the results (all we know is our own scores/medals, and the gold medal cutoff). Zuming might have the team results by now, but I'm not sure about that.

So are these problems appreciably easier compared to last year's?

I recall struggling with a few of them last time, especially that ridiculous un-symmetric inequality that I never solved (and which Sherry said she Lagrange murderplied).

This year, I was just kind of like blah blah blah.  Although perhaps this is a just a consequence of (most of) the problems being appreciably nicer than last year's if not easier.


$(a)$There are two sequences of numbers, with $2003$ consecutive integers each, and a table of $2$ rows and $2003$ columns
$\begin{array}{|c|c|c|c|c|c|} \hline\ \ &\ &\ &\cdots\cdots&\ &\ \\ \hline \ &\ &\ &\cdots\cdots&\ &\ \\ \hline \end{array}$
Is it always possible to arrange the numbers in the first sequence in the first row and the second sequence in the second row, such that the sequence obtained of the $2003$ column-wise sums form a new sequence of $2003$ consecutive integers?
$(b)$ What if $2003$ is replaced with $2004$?

a) since a sequence of 2003 consecutive integers may be written as $x+1, x+2, ..., x+2003$, it's enough to consider the case where both sequences are $1, 2, ..., 2003$. the following arrangement shows that it's possible:

$\left(\begin{array} {ccccccccc} 1 & 1003 & 2 & 1004 & 3 & 1005 & {...} & 2003 & 1002 \\ 1002 & 1 & 1003 & 2 & 1004 & 3 & {...} &1001&2003 \\ \end{array} \right)$

b) consider more generally the case of two sequences each of length $2k$. as in (a) we may assume both sequences are $1, 2, ..., 2k$. the sum of all the elements of both these sequences is $2\cdot \frac{2k(2k+1)}{2} = 2k(2k+1)$. on the other hand, assume they produced a set of $2k$ consecutive integers $m+1, m+2, ..., m+2k$. the sum of these numbers is $2km+k(2k+1) = k(2m+2k+1)$, and should equal the sum $2k(2k+1)$. but this is impossible because the latter has one more factor of 2 than the former.


function
calculus
derivative
trigonometry
topology
real analysis
real analysis unsolved

Let $f: \mathbb{R}\to \mathbb{R}$ be a derivable function. Is the graph of $f'$ connected ?

No? One can construct a function with a derivative which has a jump discontinuity.

I'm talking about the set $G=(x; f'(x))$ as a subset of $\mathbb{R}^{2}$. For example, the graph of $x^{2}\sin(\frac{1}{x})$ is connected..

Nevermind, I realized I am wrong here. Sorry to spam up your topic.

I believe the graph is connected.

Let $f: \mathbb{R}\to\mathbb{R}$ be an everywhere differentiable function. Note that $f'$ is the limit of continuous functions. Assume the graph $G$ of $f'$ is not connected and let $A_{1}$,$A_{2}$ be a disconnection. Let $K_{j}$ be the projection of $G \cap A_{j}$ onto the real line. $K_{j}$ are $F_\sigma$ sets that fill up $\mathbb{R}$. Hence they are $G_\delta$ sets. Write $K_{1}= \cap E_{n}$, $K_{2}= \cap F_{n}$

Let $x \in \partial K_{1}\cap K_{1}$ and let $M \subset \mathbb{R}$ be an open set containing $x$. I claim that $M$ also intersects $\partial K_{2}\cap K_{2}$. The key to this is that $f'$ has the intermediate value property, so that if we have a ball $B(y,r)$ contained in $K_{2}$ then actually $\overline{B(y,r)}$ is contained in $K_{2}$. I will not prove the claim unless you don't believe me.

Now let $H = \overline{(\partial K_{1}\cap K_{1}) \cup (\partial K_{2}\cap K_{2})}$. By what has just been said, for any $n$ $E_{n}\cap F_{n}\cap H$ is an open and dense subset of $H$. By the Baire Category Theorem $K_{1}\cap K_{2}\cap H = \emptyset$ is dense in $H$, so actually $K_{1}$ and $K_{2}$ are both open. However, using again that $B(y,r) \subset K_{j}\Rightarrow \overline{B(y,r)}\subset K_{j}$ we find that $K_{j}$ cannot be open unless it is all of $\mathbb{R}$, a contradiction.

I would just like to add that the last contradiction follows easier by just observing that $\mathbb{R}$ is connected. (I am not allowed to edit my post).


inequalities
geometry
circumcircle
geometry unsolved

A frieng of mine asked me a hard problem, which I can do nothing on it.
Given a triangle ABC, let S denote the area of the triangle, and let L be the length of a chord of the circumcircle of ABC that divides ABC  into two parts with the same area. Prove that $L_{\text{min}}\leq\sqrt{\frac{2S}{\sqrt3}}$.
Hope I didn't make any grammar mistakes. :)

the problem is attractive, I have spent a lot time on it,but it seems that it is beyond my knowledge. :?