Alright, I really don't understand how to solve this problem, as I keep getting the wrong answer:

Find the H+ ion concentration in 50.0 mL of 0.10 M NaH2AsO4
(K1= 6.0*1-^-3    K2= 1.1*10^-7      K3= 3.0 * 10^-12)
a. 2.4 * 10^-2
b. 1.6*10^-3
c. 1.0 * 10^-4
d. 2.5 * 10^-5

[hide]Here's what I thought: Kb of H2AsO4- <<< dissociation constant of H2AsO4-, so we ignore the hydrolysis. After that, I got the answer as C, but the answer key (2004 no. 32 USNCO) says it's D[/hide]

Help much appreciated!

Assuming $K_w=1.01\times 10^{-14}$, these are the equillibrium conditions (courtesy of Mathematica):

$[\text{H}_3\text{AsO}_4]=4.122\times 10^{-4} M$
$[\text{H}_2\text{AsO}_4^-]=9.915\times 10^{-2} M$
$[\text{H}_1\text{AsO}_4^{2-}]=4.372\times 10^{-4} M$
$[\text{AsO}_4^{3-}]=7.988\times 10^{-11} M$
$[\text{H}^+]=2.4947\times 10^{-5}$

The most notable concentration is the first one.  Note that the completely protonated acid is almost as concentrated as the monoprotonated acid.  This is because, although $K_b$ is small, $[\text{OH}^-]$ is also very small, so the base must be fairly concentrated.  This can be done in contest conditions by noting this, and then noting that only one choice is smaller than one would get by disregarding this change.

Thank you very much for the explanation!  :)

Theres actually a general method for these problems.  In general, the pH of a solution of a salt of a polyprotic acid (i.e. a salt that can both accept and donate protons) has

pH = 1/2 (pK1 + pK2) (independent of concentration of salt, it's counter-intuitive but true)

where K1 is the acid dissociation constant of the conjugate acid and K2 is the acid dissociation constant of the salt.  I remember it by thinking that the pH is halfway between the two acids.  You can also write this as

[H+] = Sqrt(K1 * K2).

So in the problem you mention, [H+] = Sqrt(K1 K2) = Sqrt( 6.0*10^-3 * 1.1*10^-7) = 2.5 * 10^-5, or choice D.  This formula pops up often enough that its useful to know (it was in 2005 free response too).

So _that's_ how you did that free response... it's no longer the most evil question in existence.  Thanks!


Two circles and three straight lines lie in the same plane.  If neither the circles nor the lines are coincident, what is the maximum possible number of points of intersection?


linear algebra
matrix
algebra
polynomial
vector
linear algebra unsolved

We're working over an arbitrary field here.

Let $P(x)=x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ be a monic polynomial. Define the corresponding companion matrix:

$M=\begin{pmatrix}0&1&&&&\\&0&1&&&\\ \vdots&&&\ddots&\\ &&&0&1\\-a_{0}&-a_{1}&\cdots&-a_{n-2}&-a_{n-1}\end{pmatrix}$

Prove, in an elementary fashion, that $P(x)$ is the minimal polynomial for $M.$

"In an elementary fashion" means not using Cayley-Hamilton or anything equivalent to Cayley-Hamilton. It also means not constructing any splitting fields or assuming that the polynomial splits.

Let $Q(x)=x^{p}+b_{p-1}x^{p-1}+...+b_{0}$ be the M-minimal polynomial.
For all vectors $u$,  $\{u,Mu,M^{2}u,...,M^{p}u\}$ isn't a free system.
Yet if $\{e_{1},..e_{n}\}$ is the canonic basis then $\{e_{n},Me_{n},..,M^{n-1}e_{n}\}$ is a basis  of the whole space.
($M$ is a cyclic application)
Conclusion: $p>n-1$.   QED

Yep, I thought that too, but here you implicitly use Cayley Hamilton  :lol:  (and a classic proof of Cayley Hamilton uses companion matrices ..)

Hi alekk, I don't think so.
 The first assertion comes from $Q(M)=0$ therefore $Q(M)(u)=0$  for all $u$..
The second assertion comes from the fact that the matrix of the system ${e_{n},Me_{n},...,M^{n-1}e_{n}}$ in the canonic basis is triangular with diagonal 1.
   I don't use the fact that P(M)=0.

I'll post my colleague's proof later today or maybe tomorrow. One key idea: instead of right-multiplying by standard basis column vectors as loup blanc has done, try left-multiplying by standard basis row vectors.

OK alekk;
I see the trouble; I have proved only that $d^{\circ}Q\ge{n}$.

Let $\chi_{k}$ be the row vector with $1$ in the $k$th place and zeros elsewhere (the standard basis in row vectors).

$\chi_{1}M=\chi_{2}$

$\chi_{2}M=\chi_{3}=\chi_{1}M^{2}$

and so on, through

$\chi_{n-1}M=\chi_{n}=\chi_{1}M^{n-1}$

$\chi_{n}M=-a_{0}\chi_{1}-a_{1}\chi_{2}-\cdots-a_{n-1}\chi_{n}$

$\chi_{n}M+a_{n-1}\chi_{n}+\cdots a_{1}\chi_{2}+a_{0}\chi_{1}=0.$

$\chi_{1}M^{n}+a_{n-1}\chi_{1}M^{n-1}+\cdots+a_{1}\chi_{1}M+a_{0}\chi_{1}=0$

$\chi_{1}P(M)=0.$

Now for $2\le k\le n,$ we have

$\chi_{k}P(M)=\chi_{1}M^{k-1}P(M)= \left(\chi_{1}P(M)\right)M^{k-1}=0.$

Since $P(M)$ annihilates every element of a basis, it follows that $P(M)=0.$

The only thing left to prove is that no lower degree polynomial in $M$ is zero. But if $Q(x)=b_{0}+b_{1}x+\cdots b_{n-1}x^{n}$ then

$\chi_{1}Q(M)=b_{0}\chi_{1}+b_{1}\chi_{1}M+\cdots b_{n-1}\chi_{1}M^{n-1}=b_{0}\chi_{1}+b_{1}\chi_{2}+\cdots+b_{n-1}\chi_{n}.$

Since the $\{\chi_{k}\}$ are a basis, if the coefficients of $Q$ are not all zero, then $\chi_{1}Q(M)\ne0,$ which means that $Q(M)\ne0.$

[color=green][Not my proof - it's due to a colleague at my university.][/color]


induction
strong induction
combinatorics unsolved
combinatorics

G is a connected graph  when at least a vertex is connected with an odd number of other vertexes.show that the arcs of G can be colored in red and in blue in such way that for every vertex V of G  we have:
|r(V)-b(v)|<= 1
when r(V) e b(V)  are the number of leaving arcs from V in red and in blue.

[hide="Solution"]
Call $ G$ balanced if it has a red-blue coloring satisfying $ |r(v) - b(v)| \le 1$ for all verticies $ v\in G$ and say that this coloring is a good coloring of $ G$.  Say that we "flip" a graph if we turn all of the red edges to blue edges and blue edges to red edges in the graph.  Note that the flipped version of a balanced graph is also balanced.

We proceed by strong induction on the number of verticies to show that any connected graph that has at least one vertex of odd degree is balanced.  A graph of one vertex is balanced since $ r(v) - b(v) = 0$ for this vertex.  Assume that the proposition is true for $ n - 1$ verticies.  Assume that there exists a graph with $ n$ verticies for which the problem statement does not hold.  Consider the graph with the fewest edges, $ G$, and let the verticies of this graph be $ v_1, v_2, \cdots , v_n$.

[b]Lemma 1: [/b]There is no path of verticies, $ v_{a_1}v_{a_2}\cdots v_{a_{2k}}$, so that $ v_{a_i} \ne v_{a_{i + 1}}$, so that no edge is repeated, and so that $ v_{a_1}v_{a_{2k}}$ is an edge in $ G$.

[i]Proof:[/i] Assume to the contrary that there exists such a path.  Color $ v_{a_i}v_{a_{i + 1}}$ red for odd $ i$ and $ v_{a_i}v_{a_{i + 1}}$ blue for even $ i$ (indicies taken $ \bmod 2k$).  Each vertex of this cycle now has one blue and one red edge coming from it.  Delete these colored edges to form $ G'$ (with $ v_j$ in $ G$ turning into $ v_j'$ in $ G'$).  Since the number of edges of $ G'$ is less than that of $ G$, $ G'$ is balanced.  Now add the edges we deleted back in.  Note that $ r(v_j') = r(v_j) - 1$ and $ b(v_j') = b(v_j) - 1$ for any $ v_j \in \{ v_{a_1}, v_{a_2}, \cdots , v_{a_{2k}} \}$, so $ G$ is balanced, which is a contradiction.

[b]Lemma 2: [/b]No two cycles in $ G$ share a vertex.
[i]
Proof:[/i] If these graphs share no edges, but a vertex, then start at the shared vertex, traverse along the edges of one cycle, return back to the shared point, and traverse along the edges of the second cycle, returning back to the shared point.  An even number of edges have been traversed (since the sum of two odds is even), so the path mentioned in Lemma 1 exists, which is a contradiction.  If these cycles share a set of edges, then let the cycles be $ C_1$ and $ C_2$ and traverse along the edges of $ C_1\cup C_2 - C_1\cap C_2$, which has cardinality $ |C_1| + |C_2| - 2|C_1\cup C_2|\equiv 0\bmod 2$.  This is again a path mentioned in Lemma 2, which is a contradiction.

Let's continue to the problem.
[b]
Case 1: [/b]$ G$ is a tree.

Then take a leaf, say $ v_1$ (which is connected to another vertex, $ v_2$), of $ G$ and color it red.  Let $ G - v_1 = H$ and note that $ H$ is a tree as well, so there is a leaf of $ H$, and the inductive hypothesis says that $ H$ is balanced.  If $ r(v_2) > b(v_2)$ in $ H$, then flip $ H$.  This will let $ H$ remain balanced and also force $ G$ to be balanced.

[b]Case 2: [/b]

Now, say that there exists a cycle $ v_1v_2v_3\cdots v_kv_1$, with $ k$ odd.  Since there is a vertex of odd degree, one of the verticies, say $ v_j$ has a degree greater than $ 2$.  Now color ${ v_jv_{j + 1}, v_{j + 2}v_{j + 3}, \cdots , v_{j - 1}v_j}$ red and the others of this cycle blue (indicies taken $ \bmod k$).  Take an arbitrary  vertex, $ v_i$ and let it be connected to $ v_{k + 1}, v_{k + 2}, \cdots , v_{l}$ that are not in the cycle (the set of neighbors of $ v_i$ not included in the cycle does not have to have positive cardinality).  By Lemma 2, the is no path from $ v_x$ to $ v_y$, when $ k + 1\le x < y \le j$ in $ G - \{ v_1 + v_2 + \cdots + v_k\}$.

[b]Lemma 3:[/b]  If $ k + 1\le z\le j$, the connected component of $ G - \{v_1 + v_2 + \cdots + v_j - v_i - v_z\}$ containing $ v_i$ contains a vertex of odd degree.

[i]Proof:[/i] We proceed by induction on the number of cycles (note by Lemma 2 that they are disjoint) that this connected component has.  For no cycles, the component is a tree, so it has a leaf, which has degree one.  Assume that it works for $ n - 1$ cycles and less.  If the component has $ n$ cycles, then delete the edges of the cycle closest to $ v_1$.  The new connected component containing $ v_1$ (recall that no cycle other than $ v_1v_2v_3\cdots v_k$ passes through $ v_1$ by Lemma 2) has $ n - 1$  cycles or less, so by the inductive hypothesis, it contains a vertex of odd degree; adding the cycle in adds a degree of two to each of the members of this cycle.  This does not change the parity of the degree of any memeber in the component, so there exists a member in the component with odd degree.

To finish the problem, find good colorings for each of the subgraphs mentioned in Lemma 3 (range $ z$ from $ k + 1$ to $ l$ and $ i$).  Flip the colors of each connected component so that $ v_{k + 1}v_i, v_{k + 3}v_i, \cdots$ are blue and $ v_{k + 2}v_i, v_{k + 4}v_i, \cdots$ are red.  Do this for $ 1\le i\le k$, but $ i\ne j$.  If $ i = j$, then note that $ v_j$ begins with two red edges, so color $ v_{k + 1}v_i, v_{k + 2}v_i, v_{k + 4}v_i, \cdots$ blue and $ v_{k + 3}v_i, v_{k + 5}v_i, \cdots$ red.  These restraints ensure that $ |b(v) - r(v)| \le 1$ for all $ v\in G$, which is a contradiction, so any graph having a vertex of odd degree is balanced. [/hide]


number theory
relatively prime
number theory open

I wanna find number of relatively prime pairs ([b]m[/b],[b]n[/b]) for a, b
1 <= [b]m[/b] <= a
1 <= [b]n[/b] <= b
I know how to count that pairs when a=b from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30027
but how should I do this when a != b

But it's only approximate result. I need precise result for any
1 <= a,b <= 50000

I heard about fastest way to calculate that. Something like [b]O[/b](square root([b]a[/b]+[b]b[/b]))

Sorry, [b]RIP[/b] and [b]maxal[/b] I had to remove your reply as this problem comes from the current Informatics Olympiad in Poland - deadline is 20 November. It's supposed to be independent work.


Could someone show me a few different ways to prove the following?:

For $a,b\in\mathbb{N}$, $ax+by=n$ has integer solutions $\left(x,y\right)$ iff $\gcd\left(a,b\right)\,\big|\,n$.

If we have a solution $(x,y)$ we have $ax+by = n$. Clearly $gcd(a,b)$ divides the left hand side, so it must also divide the right hand side.

If $gcd(a,b) | n$ we can use Euclides' algorithm to find numbers m_1 and m_2 such that $am_{1}+bm_{2}= gcd(a,b)$. Since $n = gcd(a,b)p$ for some $p$ we can let $x= m_{1}p$ and $y=m_{2}p$.

[hide="The PROMYS proof"] Let $S = \{ k | k \in \mathbb{N}, \exists x, y \in \mathbb{Z}: k = ax+by \}$.  The set is a non-empty subset of $\mathbb{N}$ since $a^{2}+b^{2}\in S$.  Hence it has a least element; call it $d$.

Consider the remainder when $a$ and $b$ are divided by $d$.  We have

$a = dq+r, 0 \le r < d$
$b = ds+t, 0 \le t < d$

But then

$r = a-dq = a-(ax+by)q$
$t = b-ds = b-(ax+by)s$

So then $r$ and $t$ are also linear combinations of $a, b$.  But since $d$ is minimal and $r, t < d$, it follows that $r, s \not \in S$.  Therefore $r = t = 0$.

$d | a, d | b \Leftrightarrow d | gcd(a, b)$

But $gcd(a, b) | a, gcd(a, b) | b \Rightarrow gcd(a, b) | (ax+by) = d$

$\Rightarrow d = gcd(a, b)$.  QED.[/hide]

[quote="computer_nuke"]For iff proofs, don't you also need to prove that the inverse/converse is also true?[/quote]
I did. First the implication from left to right and then from right to left.

[quote="t0rajir0u"][hide="The PROMYS proof"] Let $S = \{ s | s \in \mathbb{N}, \exists x, y \in \mathbb{Z}: s = ax+by \}$.  The set is a non-empty subset of $\mathbb{N}$ since $a^{2}+b^{2}\in S$.  Hence it has a least element; call it $d$.

Consider the remainder when $a$ and $b$ are divided by $d$.  We have

$a = dq+r, 0 \le r < d$
$b = ds+t, 0 \le t < d$

But then

$r = a-dq = a-(ax+by)q$
$s = b-ds = b-(ax+by)s$

So then $r$ and $s$ are also linear combinations of $a, b$.  But since $d$ is minimal and $r, s < d$, it follows that $r, s \not \in S$.  Therefore $r = s = 0$.

$d | a, d | b \Leftrightarrow d | gcd(a, b)$

But $gcd(a, b) | a, gcd(a, b) | b \Rightarrow gcd(a, b) | (ax+by) = d$

$\Rightarrow d = gcd(a, b)$.  QED.[/hide][/quote]

I have a few questions. First, why is it a non-empty subset of $\mathbb{N}$ because $a^{2}+b^{2}\in S$? Why did you choose the least element $d$ to divide out? Also, you had:

$r = a-dq = a-(ax+by)q$
$s = b-ds = b-(ax+by)s$

But I thought $s = ax+by$? (did you have $s$ as the same variable here by mistake?)

And how did you get to the step:

$r = a-dq = a-(ax+by)q$
$s = b-ds = b-(ax+by)s$ (did you meant $t$ here on the left hand side?)

Also, why does d subsititute for $ax+by$?

Why is $r,s < d$?

This looks to be a great rigorous proof, I'm just not that good and it would be good if you or someone could clarify these questions for me =) Thanks.

[quote="computer_nuke"]I have a few questions. First, why is it a non-empty subset of $\mathbb{N}$ because $a^{2}+b^{2}\in S$? Why did you choose the least element $d$ to divide out? Also, you had:

$r = a-dq = a-(ax+by)q$
$s = b-ds = b-(ax+by)s$

But I thought $s = ax+by$? (did you have $s$ as the same variable here by mistake?)

And how did you get to the step:

$r = a-dq = a-(ax+by)q$
$s = b-ds = b-(ax+by)s$ (did you meant $t$ here on the left hand side?)

Also, why does d subsititute for $ax+by$?

Why is $r,s < d$?

This looks to be a great rigorous proof, I'm just not that good and it would be good if you or someone could clarify these questions for me =) Thanks.[/quote]
It's not empty because $a^{2}+b^{2}$ is in it and it's a subset of $\mathbb{N}$ because we defined it that way ($s\in\mathbb{N}$).

$d$ is a member of $S$ and therefore it can also be written as $s=ax+by$.

$r,s < d$ because $r$ and $s$ are remainders after dividing by $d$. I think he did mean $t$ instead of $s$.

Why is it important that $r$ and $t$ are linear combinations of $a$ and $b$? And in the step of substituting $(ax+by)$ for $d$, are the original $a$ and $b$ (The ones used first and the ones that $d$ divided) the same as the $a$ and $b$ in $(ax+by=d)$ here? I'm kind of confused on the concepts.

[quote="computer_nuke"]Why is it important that $r$ and $t$ are linear combinations of $a$ and $b$? [/quote]

1.  I chose $d$ to be the smallest positive linear combination of $a$ and $b$.

2.  I generated $r$ and $t$, which are also linear combinations of $a$ and $b$, but which are [b]smaller than $d$[/b] (since they are remainders upon [b]division[/b] by $d$).

3.  This implies that $r$ and $t$ are not positive.

4.  $r$ and $t$ are remainders, which are always non-negative.

5.  Therefore, $r = t = 0$ and $d | a, b$.

But whoops, that should be $t = b-ds$, my mistake  :blush:

In any case, this is a beautifully simple proof, one of my favorites so far :)

Edit:

[quote="computer_nuke"]And in the step of substituting $(ax+by)$ for $d$, are the original $a$ and $b$ (The ones used first and the ones that $d$ divided) the same as the $a$ and $b$ in $(ax+by=d)$ here? [/quote]

$a$ and $b$ are the same.  Since I defined $d$ to be a member of $S$, there exist $x_{0}, y_{0}$ such that $ax_{0}+by_{0}= d$.  Perhaps I should have made that distinction more clear.

Thanks for taking the time to carity, it is indeed a beautiful proof  :) . I see what you mean now, the a and b and constants and x and y are the variables. Sorry, my mistake. Thanks again though.


What is the positive difference between the sum of the first 20 positive multiples of 5 and the sum of the first 20 positive, even integers?

[hide]The difference between the $ n$th term in the first $ 20$ multiples of $ 5$ and in the first $ 20$ multiples of $ 2$ is $ 3n$. Now, we are trying to find the sum of the first $ 20$ multiples of $ 3$. Thus, our answer is $ 20\cdot63\equal{}\boxed{1260}$.[/hide]

i got a different answer

first i found the sum of the first 20 positive multiples of 5

5, 10..., 95, 100

if you add the first and last, 2nd and 2nd to last, etc. you get the same number
in this case, that number is 105, and there are 10 such pairs, so the sum is 1050

the sum of the first 20 positive even numbers is simply the sum of the first 40 numbers minus the sum of the first 20 odd numbers

the formula to find the sum of the first $ n$ numbers is $ \frac{n(n\plus{}1)}{2}$

so the sum of the first 40 numbers is $ \frac{40(41)}{2}\equal{}20\times41\equal{}820$

the sum of the first $ n$ odd numbers is $ n^2$

so the sum of the first 20 odd numbers is $ 20^2\equal{}400$

we subtract 400 from 820 to get 420

to find our answer, we subtract 420 from 1050 to get $ \boxed{630}$


pbornsztein, aren't there are two b's in the equation

$%Error. "displaymode" is a bad command.
a^2 + b^3 = b^c$

[quote="jelly119"]$a,b,c\in\mathbb{N}$, show that $a^2+b^3=b^c$ has no solution.[/quote]
$b^c-b^3=a^2$ $so$ $b^3(b^{c-3}-1)=a^2$.
We can easily show that $b^3 and b^{c-3}-1$ share no factors because $b^{c-3}-1$ has no factors of b in it.
Therefore, this means both $b^3$ and $b^{c-3}-1$ must be squares. Since b^3 must be a square, b must be in the form
$x^2$. so that means that
$x^{2^{c-3}}-1$ is a square. Which is untrue, since that +1 is a square, and x isnt equal to 0 by the very nature of the problem.
Therefore  a^2+b^3=b^c has no solution in natural numbers


Let $AB$ be a segment, with any circle (diameter $MN$) throught midpoint $I$ of $AB$ prove that:
$MA\cdot NA+MB\cdot NB \ge \frac{AB^2}{2}$

Here is my proof hope without mistake
I use scalar product
$MA.NA+MB.NB \ge \vec{MA}.\vec{NA}+\vec{MB}.\vec{NB}=(\vec{MI}+\vec{IA}).(\vec{NI}+\vec{IA})+(\vec{MI}+\vec{IB}).(\vec{NI}+\vec{IB})=2.\vec{MI}.\vec{NI}+\vec{MI}(\vec{IA}+\vec{IB})+\vec{NI}(\vec{IA}+\vec{IB})+IA^2+IB^2=\frac{AB^2}{2}$
note circle diameter $MN$ throught $I$ and $I$ is midpoint of $AB$ then $\vec{MI}.\vec{NI}=0$ and $\vec{IA}+\vec{IB}=\vec{0}$


number theory
greatest common divisor
least common multiple

Let $ a,b,c$ be natural numbers. Prove that
\[ \frac{[a,b,c]}{(a,b,c)}\equal{}\frac{[ab,bc,ca]}{(ab,bc,ca)}\]
Where $ [x,y,z]$ and $ (x,y,z)$ denote the LCM and GCD, respectively, of $ x,y,z$.

After considering the valuations $ v_p$ (number of factors $ p$) for all primes, this gets:
$ \mathrm{max}(x,y,z) \minus{} \mathrm{min}(x,y,z) \equal{} \mathrm{max}(x \plus{} y,y \plus{} z,z \plus{} x) \minus{} \mathrm{min}(x \plus{} y,y \plus{} z,z \plus{} x)$.
Now that's easy to check as we may assume $ x \leq y \leq z$, then it gets $ z \minus{} x \equal{} (y \plus{} z) \minus{} (x \plus{} y)$, being trivial.

Yes $ \zeta x$, that is one of my three solutions. I hope you'll find the other two as well. :wink:

I don't quite follow........can someone expand....?

A treatment of the general identity in the spirit of ZetaX's solution is [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316]here[/url].

[b]nayel:[/b]  Does one of your other two solutions involve considering $ (a, b), (a, c)$, and so forth?  I think ZetaX's is the most straightforward.

Yes t0rajir0u you're right. One of my solutions uses consideration of pairwise GCDs and I like that solution the most. :lol:

Sorry I completely forgot about this. Here is my solution:

First we prove that $ abc = (ab,bc,ca)[a,b,c]$. This is easy enough to prove in the same way as ZetaX's, but I'll prove it in a different way which seems nicer to me.

We have
\begin{align*}(ab,bc,ca) & = (b(a,c),a(b,c)) \\
& = (b,c)(a,c)\left(\frac {b}{(b,c)},\frac {a}{(a,c)}\right) \\
& = (b,c)(a,c)\left(\frac {[b,c]}{c},\frac {[a,c]}{c}\right) \\
& = \frac {(b,c)(a,c)}{c}\left([b,c],[a,c]\right) \\
& = \frac {(b,c)(a,c)}{c}\cdot\frac {[b,c]\cdot[a,c]}{[a,b,c]} \\
& = \frac {bc\cdot ac}{c[a,b,c]} \\
& = \frac {abc}{[a,b,c]} \end{align*}
Thus we conclude that $ abc = (ab,bc,ca)[a,b,c]\quad(1)$

In this identity we put $ ab,bc,ca$ instead of $ a,b,c$ to find that $ ab\cdot bc\cdot ca = (ab\cdot bc,bc\cdot ca,ca\cdot ab)[ab,bc,ca]$

Or $ a^2b^2c^2 = abc(a,b,c)[ab,bc,ca]$

Or $ abc = (a,b,c)[ab,bc,ca]\quad(2)$

From $ (1)$ and $ (2)$ we conclude that $ (a,b,c)[ab,bc,ca] = [a,b,c](ab,bc,ca)$, which is what we wanted to prove.


Find the # of points common to the graphs of

(x-y+2)(3x+y-4)=0
(x+y-2)(2x-5y+7)=0

(x-y+2)(3x+y-4)=0
(x+y-2)(2x-5y+7)=0

From the first equation, we can say that (x-y+2)=0 or (3x+y-4)=0.
From the second equation, we can say that (x+y-2)=0 or (2x-5y+7)=0

Thus, we can pair one of the equations that we got from the first equation with one of the equations that we got from the second equation, and solve them simultaneously. There are 4 such pairs, so there are 4 points that are solutions to the original equations. (Note that it is a good idea to check each of the 4 pairs to see whether a few of the pairs yield the same solution. However, in this case, that does not happen.)


This is my first year doing MC, since I am starting the program at my school.  It would helpful to know about what score you need to advance to state, and from state to nats.

Top students in a chapter advance to state, and top 4 in the state make nationals. The qualifying score mostly depends on your region and state. In some chapters 1st place is in the low 20s; in others, a 44 might not make states.

yeah for example in MD i believe u need to be in the top 24 to advance to states (top 16 qualify for countdown round)

And here you need a 45/46 to guarantee making it to states.

No, actually a 46 since there could be >10 dudes who get perfect scores.


Does there exist a natural number $ n$ such that $ n>2$ and the sum of squares of
some $ n$ consecutive integers is a perfect square?

$ S\equal{}(m\plus{}1)\plus{}\cdots\plus{}(m\plus{}n)\equal{}\frac{(m\plus{}n)(m\plus{}n\plus{}1)}{2}\minus{}\frac{m(m\plus{}1)}{2}\equal{}$
$ \equal{}\frac{m^2\plus{}n^2\plus{}2mn\plus{}m\plus{}n\minus{}m^2\minus{}m}{2}\equal{}\frac{n^2\plus{}2mn\plus{}n}{2}\equal{}$
$ \equal{}\frac{n(n\plus{}2m\plus{}1)}{2}$

if $ n\equal{}2k\plus{}1$ then $ S\equal{}(2k\plus{}1)(k\plus{}m\plus{}1)$
take $ m\equal{}k$.

then for every $ k$ there are $ 2k\plus{}1$ consecutive integers such as:
$ (k\plus{}1)\plus{}(k\plus{}2)\plus{}\cdots\plus{}(k\plus{}2k\plus{}1)\equal{}(2k\plus{}1)^2$

Example for $ k\equal{}2$, $ n\equal{}5$
$ 3\plus{}4\plus{}5\plus{}6\plus{}7\equal{}5^2$


linear algebra
matrix
factorial
combinatorics open
combinatorics

The problem is as follows: how many ways are there to arrange integers 1 through N^2 in a square NXN matrix such that (a) the diagonal matrix elements increase from left to right and (b) each diagonal matrix element is the smallest in its respective row and its respective column? I suspect that the problem is either very simple, or well-known, or both, but I have not been able to find the answer in a useful form, either on my own or in the literature. I can write the answer as a finite sum over products of rising factorials, but the number of terms in the sum is very large, probably of the order of ((N^2-N)^N)/N! .

Let's describe a strategy to construct such a square. (1,1) = 1, (2,2) = 2. (1,2), (2,1) can be anything else from {1...n^2} other than 1 or 2.
(3,3) has to be the min ({1...n^2}-{(1,1),(2,2)(1,2),(2,1)}, and then any (i,3) or (3,i) can be any other number from {1...n^2} other than previous entries where i != 3.
Essentially, any (i,i) has to be the min ({1...n^2} - the entries of the (i-1) by (i-1) square before it), but (k,i) or (i,k) can be anything else in {1...n^2} other than pervious entries.

Thus the solution should be (n^2)!/[(n^2)(n^2-1)(n^2-4)(n^2-9)...(n^2-(n-1)^2)].


Vieta
quadratics
algebra
polynomial
analytic geometry
conics
parabola

In an equation $ x^n \plus{} a_1x^{n \minus{} 1} \plus{} a_2x^{n \minus{} 2} ... \plus{} a_{n \minus{} 1}x \plus{} a_n$, the sum of the roots is $ \minus{} a_1$.

The only reason I wasn't sure is because I've heard Vieta's only for a quadratic equation, but not extended further.

No, it applies to all polynomials. If you let $ \sum a_nx^n\equal{}\prod(x\minus{}r_n)$ and just sort of expand stuff out, you'll see how it works.

In general, if $ r_1,r_2,\ldots,r_n$ are the roots of $ a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$, then:
\begin{align}
r_1+r_2+\cdots+r_n&=-\frac{a_{n-1}}{a_n}\\
r_1r_2+r_1r_3+\cdots+r_1r_n+r_2r_3+r_2r_4+\cdots+r_{n-1}r_n&=\frac{a_{n-2}}{a_n}\\
r_1r_2r_3+r_1r_2r_4+\cdots+r_{n-2}r_{n-1}r_n&=-\frac{a_{n-3}}{a_n}\\
&\vdots\nonumber\\
r_1r_2\cdots r_n&=(-1)^n\frac{a_0}{a_n}\nonumber.
\end{align}
The left-hand side in equation $ (k)$ is the sum of the roots taken $ k$ at a time, that is, the sum of every possible product of $ k$ of the roots. The last equation is the formula for the product of the roots.

For example, for a cubic polynomial $ a_3x^3+a_2x^2+a_1x+a_0=0$, we have:
\begin{align*}
r_1+r_2+r_3&=-\frac{a_2}{a_3}\\
r_1r_2+r_1r_3+r_2r_3&=\frac{a_1}{a_3}\\
r_1r_2r_3&=-\frac{a_0}{a_3}.
\end{align*}

[quote="ArtofMath"]Yeah, that's Vieta's. Sum of the roots = -b/2a[/quote]
sum of the roots is -b/a not -b/2a :O
-b/2a is the x-coordinate of the parabola of the vertex

[quote="AdithyaGanesh"]At school, they usually teach it for quadratics, because there usually is no need to go any further.  However, this [i]can[/i] be applied to any polynomial.

http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_Formulas[/quote]

At least your school mentioned it... My school taught me absolutely none of this stuff. I learned pretty much all the math I know out of AoPS vol 1 and 2 over the course of like 6 months last year  :|


[quote="t0rajir0u"]By the latest figures, the Democrats have $51$ seats in the Senate and the Republicans have $49$.  Each of the $50$ states sends two senators to the Senate.

1.  Show that at least one state sent one Democratic senator and one Republican senator.

2.  Show that an odd number of states sent one Democratic senator and one Republican senator.[/quote]
[hide="1"] Proof by contradiction.
Assume that all states sent both democrats or both republicans. Then there would be two times the number of states is the number of repuclicans/democrats. But since there is an odd number of democrats then there has to be one state that sent one Democratic senator and one Republican senator.[/hide]

[hide="2"]

Well, assume the opposite: that an even number of states, $2a$, sent one senator from each party.

Then, the total number of (say, Democrat) senators = $2k+2a$, for some integer $k$.

So, the number of Democrat senators is divisible by 2 -- CONTRADICTION!!!!

[/hide]

[hide="click me!"]For the first one, we can do proof by contradiction.  Assuming that only both republicans and only both democrats were chosen, it is impossible because we have an odd number of each party in their seats.[/hide]
[hide] for the second, we must have an odd number of 1 dem+ 1 rep because it is the only way to produce an odd number in the total at the end.[/hide]


[quote="Chaos"]Extra Credit Problem on math test

Given:
$ x \equal{} 3^6 \times 2^{12}$

$ y \equal{} 3^8 \times 2^8$

$ x^x \times y^y \equal{} z^z$ for some integer $ z$

find the value of $ z$

thats z to the z power, not $ z^2$
 :huh:  :huh:  :huh:[/quote]
[hide="Hint"]Take the log of both sides. [/hide]


The girls have left for China, and will be competing in ZhongShan on August 14th and 15th. The format is 4 olympiad problems in 4 hours, two days in a row.

They plan to keep a blog of the trip here: [url]http://www.msri.org/specials/gmo/2008[/url]

Good luck to all the girls!

Good luck! I expect people to be scoring a lot higher than Delong's favorite score :wink:

-0

It's the first quotation of "Quotes 1" here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1177559#1177559

Good luck to the CGMO team!

[quote="pianoforte"]good luck guys.[/quote]

Correction, good luck girls. :D  I'm just being a prick...

Anyways, girls, do better than we did in Spain.  (Not that the IMO went badly, it went quite well, but I'm still telling you to beat that)

I'm a girl so i hope i can make it next year or two :D

good luck everyone!

no argument there.... lol, jk.... ya girls good luck.


I might go... Is this tournament open to 10th graders and below?

Yes, it's open to all sophmores, freshman, and middle school students.

Well, it's on the same day as state science olympiad...well i'm still going. Though it mite cause some scheduling problems. Gosh i hope a lot of ppl don't not come. Haha double negative. My grammar teacher smited (smote?) me.

I cannot go. Science olympiad has higher precedence this time. Hopefully all of Northview can place in the top 15...

OMG, it's tomorrow!!!!!!  I just did the 2005-2006 test and found it pretty cheasy (my word for a test that is both challenging but in the end relatively easy).  I did it well below the time limit, but the bad part's that I don't have the answer key, so I don't know how well I did. :(

So it's today?
I'm sitting in a hotel in Connecticut. (I heart my laptop. And complimentary stuff. :D)
How did everyone do?

Third Place Ind. :(
128/135
Missed the free response question about the singular matrix.

Hey iniquitus,
If you don't mind saying, who are you?
And do you know who got 2nd, 4th, 5th, etc.?

I got fifth, and Claus Zheng got 4th.  So sad am I...  I made two really stupid mistakes, one on the multiple choice and one on the free response.  Dood, Tianqi is uber-beast.  He got a perfect score.  ZOMG!

Oh, sorry.
I'm Oliver H from Pace Academy.

And.. I don't know who got 2nd.. I just moved here this year so I don't know anyone yet  :P

[quote="Math Geek"]I think the 2nd place person was Krishna Desai, who also got the Love Award.[/quote]

The Love Award... suggestive much?  :rotfl: Just kidding  :P

If possible and legal, can someone put up the problems (TeX, PDF, etc.) on the site?

I have an electronic copy of the scanned JV State individual test, but I'm not sure if it's legal...

EDIT: ok, I will not post it until further notice.

[quote="Karth"]It's legal as long as it doesn't belong to the maa which I'm sure it doesn't. I think the arml coaches made it, so it's probably legal unless they put a copyright on it. If you could put it here though, I'd really appreciate it.  :)[/quote]

What does the MAA have to do with it? They produce the AMC/AIME/USAMO/etc., but they certainly don't control the copyright on all math competitions everywhere in the world! The test belongs to Barry Shildneck (or David Ferretti the author) and probably Cobb County Schools. I'm almost positive that they won't mind it being available, but please remove it until I hear back from them.

Tom Fulton

Let me expand on when you can and when you cannot post a test. You can post a test if:

1. You wrote it and retain "ownership". For instance, I would say that the authors of the Fulton County test can not post it because it now belongs to Fulton County.
2. You receive explicit permission from the write/owner of the exam.

After having spoken to the folks at Kell, they were pretty upset that the test had been posted without anyone even asking them. They plan to post the test on the website before too long.

What's the difference? Well, this is a fund raising activity and people going to their web site is advertisement for the tournament. You won't find tests from other tournament posted on other sites. Check out Ga. Tech, Furman, Lassiter, Rockdale, etc. If (and that's a big if) the old tests are posted, they are posted ONLY on the tournament website.


Find
$\mathop {\lim }\limits_{x \to 1 } \left( {\frac{{x^x - x}}{{\ln x - x + 1}}} \right)$
 :D

[hide=" Read here only if you did not find the limit after at least 15 minutes."]Denote $f(x)=\frac{x^x-x}{\ln x-x+1}=x\cdot \frac{e^{(x-1)\ln x}-1}{(x-1)\ln x}\cdot \frac{\ln x}{x-1}\cdot \frac{(x-1)^2}{\ln x-x+1}\ .$
$\boxed {\lim_{x\to 1}\frac{\ln x}{x-1}=1}\ ;\ \boxed {\lim_{x\to 0} \frac{e^x-1}{x}=1}\ .$
$t=(x-1)\ln x,\ \lim_{x\to 1} t(x) =0\Longrightarrow \lim_{x\to 1}\frac{e^{(x-1)\ln x}-1}{(x-1)\ln x}=\lim_{t\to 0} \frac{e^t-1}{t}=1\ .$
$\lim_{x\to 1} \frac{(x-1)^2}{\ln x-x+1}=\lim_{t\to 0}\frac{t^2}{\ln (t+1)-t}\ \underline {\underline {l'H}}\ \lim_{t\to 0}\frac{2t}{\frac{1}{t+1}-1}=\lim_{t\to 0} -2(t+1)=-2\ .$

Therefore, $\lim_{x\to 1} f(x)=1\cdot 1\cdot 1\cdot (-2)\Longrightarrow \boxed {\lim_{x\to 1}f(x)=-2}\ .$[/hide]

Just use L'Hopital's rule twice, the bottom turns to a nice -1 and then you can evaluate the top. Note that $\frac{d}{dx} x^x = x^x(\ln(x)+1)$

shouldn't you be able to solve intermediate problems without calculus?


Let convex quarilateral ABCD with O is intersection of two diagonals. the line [i]d[/i] across O and intersect with circles (OAB), (OBC), (OCD), (ODA) are M, N, P, Q different with O, respectively. Prove that we always have MN = PQ

Perform an inversion with pole $O$, and in general, for a point $X$, denote its image through this inversion by $X'$. Checking a few cases, we see that the statement we must prove amounts to showing that one of $OM,ON,OP,OQ$ is the sum of the other three, which, in turn, is equivalent to showing that one of the reciprocals of $OM',ON',OP',OQ'\ (*)$ is the sum of the other three. Now, $O,O,M',N',P',Q'$ are precisely the intersections of $d'=d$ with the pairs of opposite sides of the complete quadrilateral $A'B'C'D'$, which means that they are three pairs of points in involution on $d$. Write what this means explicitly in terms of cross ratios (for instance, write $(O,M';N',P')=(O,P';Q',M')$), and we'll get a relation of the type we're looking for, expressing one of the reciprocals of $(*)$ as a sum of the other three, which is which depending on the position of $d$.

Dear Grobber!
Invertion is useful tool for solving some geometry problems. Do you have a another solutions for this problem? A classcical solution!


The question is to prove that  rho n is multiplicative.
I think it is quite easy to prove this through the definition of multiplicative function .For any value of n rho n is not zero as d(n) is a multiplicative function.
Also we can write that rho n = rho n * rho 1

i think i have mistaken here calling sigma as rho.

This doesn't belong to High School Basics forum.

In regards to your question, $ \sigma(p^\alpha)\equal{}\frac{p^{\alpha\plus{}1}\minus{}1}{p\minus{}1}$. Proving $ \sigma(n)$ is multiplicative is a standard result in elementary number theory. See [url]http://www.artofproblemsolving.com/Wiki/index.php/Multiplicative_function[/url]


If 4 daps = 7 yaps and 5 yaps = 3 baps, how many daps equal 42 baps?

Heh, if I remember right, Bobby Shen got this right in like a second...

4 daps = 7 yaps ------> 20 daps = 35 yaps

5 yaps = 3 baps ------> 35 yaps = 21 baps

so answer is 40.

Or you could multiply equations together, to get:

20*daps*yaps=21*yaps*baps
divide yaps from  both sides
20 daps = 21 baps, [b]40[/b] daps = 42 baps.


Suppose that $a_{1},a_{2},...,a_{n}$ are non-negative real numbers not equal to $1$ at once and $a_{1}+a_{2}+...+a_{n}=n$. For each positive number $k$, we denote \[S_{k}=a_{1}^{k}+a_{2}^{k}+...+a_{n}^{k}-n.\] Find the maximum and minimum value of \[k=\frac{S_{4}+S_{2}}{S_{3}}.\] (the general problem for $\frac{S_{4}+rS_{2}}{S_{3}}$ is not harder).

I think the maximum value is attained for $a_{1}=a_{2}=...=a_{n-1}=0$ and $a_{n}=n$.
For $n=2$ the minimum value (unattained  :P ) is $7/3$.
For $n>2$ this value is smaller (see $a_{1}=a_{2}=...=a_{n-2}=1$ ). By RCF-Theorem, it suffices to consider the case $a_{1}=a_{2}=...=a_{n-1}$ .
For $n=3$, the minimum value is 2.248077... :oops:

I don't check if RCF theorem is valid. But it is unable for the general problem

[i]Find the minimum and maximum value of [/i] \[\frac{S_{4}+rS_{2}}{S_{3}}.\] Now, returning to the first poblem, could anyone completely solve the problem of finding maximum value? Only consider $r=1$.

I think we can apply RCF-Theorem for $r<6$ and LCF-Theorem for $r >6$.
For $r=6$, the minimum value (unattained) is 4.


Hey everyone. Here's just a couple of guidelines to follow when posting in this forum.

1. No spamming. It is worth nothing, and clogs up subjects and drives them off topic.

2. No flaming. Don't call people names or give insults.

3. Put answers in spoiler unless the author is asking for help solving a problem.  (When posting a new topic, use a message icon (? or *) to indicate "question" vs "challenge".)

4. Please put your solutions to the problems, not just an answer.

5. The Mathcounts board is not a place to brag about how you finished. Remember to not be arrogant. Modesty is good.

6. The forum is a place to discuss Mathcounts and Mathcounts problems, and problems that may be Mathcounts level. Do not post problems that are too easy or too hard. If they're too easy or too hard, a moderator will most likely move it, but try not to do this just to stress the moderators out :P .

[b]7. Do not post messages telling others not to revive old topics.[/b]
Hope you enjoy!

To add two very important guidelines to what yifman12 has said:

- Please keep posts to a level not exceeding MATHCOUNTS!

- Please keep subject material to topics that your principal would find acceptable for discussion at your middle school.

Students of all ages are encouraged to submit challenge questions.  If no one has answered your challenge within a few days, please consider providing a hint to get the problem moving.

Older students should leave the challenge questions for the younger students - answer the challenge problems that are appropriate for your age and experience level only! You are free to help out students who have questions.

Most challenge questions in this forum should be at approximately the level of Chapter or State MATHCOUNTS competition material.  However, challenge questions at the level of National MATHCOUNTS competitions are also acceptable.

Neither questions nor solutions beyond the level of MATHCOUNTS should be posted to this forum.

Questions and solutions not meeting these guidelines will be moved or deleted.

Another guideline:

 When posting, don't put as the title '...hard problem...' when it is really easy. Then the people who are thinking about joining us and are looking at what we have to offer think that really is hard for us.

Considering that we are between states and nationals right now and there are a lot of nationals competitors here (including yours truly), is it okay if there is an abundance of national level problems right now?

I think so, but I can't be sure, 'cause I'm not a moderator.

You can post mostly national problems right now, but after Nats, back to easy!

Actually, there has never been an issue with posting National MATHCOUNTS problems.  As stated above, "challenge questions at the level of National MATHCOUNTS competitions are also acceptable".  At most times of the year, we do expect the majority of challenge problems will be at the Chapter or State level.  Of course, we understand that as the competition season progresses, there will be a tendency to the harder MATHCOUNTS problems.

To be sure the moderators don't mistake your posts for harder problems, it doesn't hurt to cite the "Source" when you post the problem.  [Use the "Source" field, right under the "Subject" field when you make a new post.]

There is an issue with posting a "Mock MATHCOUNTS Competition", and filling it with the hardest of the hard National questions.  In a case where a collection of problems are posted, the administrators or moderators may bump it if they feel the overall difficulty of the problem set exceeds MATHCOUNTS level.

Also, note that seemingly simple changes such as changing a problem from "solve this" to "prove this" often turn a MATHCOUNTS-level problem into a pre-Olympiad problem.

So, please feel free to post [b]any[/b] MATHCOUNTS-level problem in this forum.  If you compose your own questions, the moderators and/or administrators may reclassify your problems to a different forum.

[quote="rcv"]To be sure the moderators don't mistake your posts for harder problems, it doesn't hurt to cite the "Source" when you post the problem.  [Use the "Source" field, right under the "Subject" field when you make a new post.]
[/quote]

What source field? I don't seem to see it.

[quote="236factorial"]What source field? I don't seem to see it.[/quote]
Hmmm.  I just learned something.  It seems you may have to make a special request to an administrator to see the "Source" field.  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23778]this post[/url] for more information.  Your question certainly explains why nobody complied with my request.  :roll:

To answerers of problems: Please don't just post the answer; explain the solution...If you get it wrong, that way we can point out where you went wrong. Also, unless otherwise specified, put your solution in [hide] tags.

To posters of problems: Wait a few days before posting the right answer or giving any indication of who is correct. People then post more solutions. Also, unless it's a problem you've been having trouble with, include in your first post "Hide your answers!".

I don't always follow my own rules either, but try to follow them. It makes for a better forum.

This is a reminder to read the first couple of posts of this thread.  Recently, many topics have been too high level for this forum.  Topics beyond the level of MATHCOUNTS do not belong in this Forum.  Also, an increasing number of topics contain subject matter that does not meet the requirement that "your principal would find acceptable for discussion at your middle school".  A couple of topics involving murder have been removed.

Again people have forgotten the rule against spam. Spam is deleted on sight. Repeat offenders will be notified and if nothing is changed the admins will be PMed with evidence. There has been much spam lately, a good deal of which has now been deleted, but  I'm sure you can find more.

I don't want to start a new topic for this, but who's the new mathcounts forum mod?

I'm assuming he/she works at AoPS, because their email address host is artofproblemsolving.com

[quote="kyyuanmathcount"][quote="yif man12"]*clip*
4. Please put your solutions to the problems, not just an answer.
*clip*[/quote]

Everyone, solutions to problems please! Lately, I've seen lots of:

[quote="Somebody"]The answer is [hide]bla[/hide][/quote]

where bla is some answer that may or may not be right.  Thanks!

BUMP!

New thing:  We don't encourage the

[quote="Random person"]Is the answer [hide]*insert some answer*[/hide]?  I'll post the solution if it's right.[/quote]

The point of posting a solution is not only to let readers know the solution, it's also so others can help you if it's not right.

Thanks[/quote]
Is it OK if we do the latter once in a while, because I am not a very fast typer, and not proficient at LateX, so solutions can take a while? I will try to keep it to a minimum  :maybe:  ... Also, if we need further explanations, perhaps we could post a request later on?

I would like to ask everyone who posts here to just read this thread once more...thanks!

(I don't want to look like a hypocrite, but if you aren't in middle school, maybe it's time to move on - please try not to give full solutions to problems unless everyone else is struggling :) )

[quote="kyyuanmathcount"]
(I don't want to look like a hypocrite, but if you aren't in middle school, maybe it's time to move on - please try not to give full solutions to problems unless everyone else is struggling :) )[/quote]
Yeah  :blush:
We can still make problems, right?

That's fine - just try to help the people actually doing the competition.

I'm going to remind people to READ THE FORUM GUIDELINES above. Please. I don't like deleting 15 posts in one night. And it doesn't help your rating to have posts deleted either.

sorry
pls delete this
i cant figure out how.

[code][hide]content[/hide][/code]

and it looks like
[hide]content[/hide]
i dont know if this is the right place to be asking this btw.


If we draw 5 cards from a standard deck of 52, what is the probability that we get 2 cards of the same value and 3 other cards with different values? The other three cards have to be different in value than the 2 identical cards.

Plz help us with this problem. We are deeply stuck and would appreciate any advice.  :)

[hide]

There are 13 different values.  There are 13 ways to choose a value to take two from, and 6 ways to determine their suits.  For the next three there are 12,11,and 10 ways to choose, and each has 4 ways of choosing the suits.  Since there is a total of 52*51*50*49*48 ways of picking four cards, the answer is

$\frac{13*6*12*4*11*4*10*4}{52*51*50*49*48}=\frac{88}{833}$

[/hide]


calculus
integration
trigonometry
calculus computations

Solution:

Let $ I \equal{} \int \frac{x^2}{(x\sin{x} \plus{} \cos{x})^2} dx$

$ \equal{} \int \frac{x}{\cos{x}} \frac{x\cos{x}}{(x\sin{x} \plus{} \cos{x})^2} dx$

Let $ u \equal{} \frac{x}{\cos{x}}, dv \equal{} \frac{x\cos{x}}{(x\sin{x} \plus{} \cos{x})^2} dx$

Then $ du \equal{} \frac{\cos{x} \plus{} x\sin{x}}{\cos^{2}{x}} dx$

$ v \equal{} \int \frac{x\cos{x}}{(x\sin{x} \plus{} \cos{x})^2} dx$

Make the substitution $ w \equal{} x\sin{x} \plus{} \cos{x}$. Then $ dw \equal{} x\cos{x} dx$.

Thus $ v \equal{} \int \frac{1}{w^2} dw \equal{} \minus{}\frac{1}{w} \equal{} \minus{}\frac{1}{x\sin{x} \plus{} \cos{x}}$

Applying integration by parts:

$ I \equal{} uv \minus{} \int vdu \equal{} \minus{}\frac{x}{\cos{x}(x\sin{x} \plus{} \cos{x})} \plus{} \int \frac{1}{\cos^{2}{x}} dx$

$ I \equal{} \minus{}\frac{x}{\cos{x}(x\sin{x} \plus{} \cos{x})} \plus{} \tan{x} \plus{} C$

$ \int \frac{x^2}{(x \sin x \plus{} \cos x)^2}\,dx \equal{} \int \frac{x \cos x}{(x \sin x \plus{} \cos x)^2} \cdot x \sec x \,dx$

$ \equal{} \minus{} \frac{x \sec x}{x \sin x \plus{} \cos x} \plus{} \int \frac{1}{x \sin x \plus{} \cos x} \cdot (\sec x \plus{} x \sec x \tan x)\,dx$

$ \equal{} \minus{} \frac{x \sec x}{x \sin x \plus{} \cos x} \plus{} \int \sec^2 x\,dx \equal{} \minus{} \frac{x \sec x}{x \sin x \plus{} \cos x} \plus{} \tan x \plus{} C$.

Hmmm...that is precisely the same method as I used...why bother repeating it in condensed form with less working?

Observe that $ x\sin x\plus{}\cos x\equal{}\sqrt{1\plus{}x^{2}}\cos (x\minus{}\tan ^{\minus{}1}x)$

$ \int \frac{x^{2}}{(x\sin x\plus{}\cos x)^{2}}dx$

$ \equal{}\int \frac{x^{2}}{1\plus{}x^{2}}\sec ^{2}(x\minus{}\tan ^{\minus{}1}x)dx$

$ \equal{}\tan (x\minus{}\tan ^{\minus{}1}x)\plus{}C$

[quote="IndoChina"]Observe that $ x\sin x \plus{} \cos x \equal{} \sqrt {1 \plus{} x^{2}}\cos (x \minus{} \tan ^{ \minus{} 1}x)$

$ \int \frac {x^{2}}{(x\sin x \plus{} \cos x)^{2}}dx$

$ \equal{} \int \frac {x^{2}}{1 \plus{} x^{2}}\sec ^{2}(x \minus{} \tan ^{ \minus{} 1}x)dx$

$ \equal{} \tan (x \minus{} \tan ^{ \minus{} 1}x) \plus{} C$[/quote]

Awesome! I haven't seen that one before.

Thanks. Most of such problems can be solved in this manner. :)


[quote="anirudh"][quote="Phelpedo"]Find $\tan 20+4\sin20$.[/quote]

[hide]

That is equal to $\frac{\sin 20+4\sin 20\cos20}{\cos20}$. When you simplify you get $\frac{2 \sin 60 \cos 20}{\cos 20}$ which is $2 \sin 60$, which after using a calculator is $\sqrt{3}$[/hide][/quote]
why do you need a calculator for that? :huh:

[quote="PenguinIntegral"][quote="anirudh"][quote="Phelpedo"]Find $\tan 20+4\sin20$.[/quote]

[hide]

That is equal to $\frac{\sin 20+4\sin 20\cos20}{\cos20}$. When you simplify you get $\frac{2 \sin 60 \cos 20}{\cos 20}$ which is $2 \sin 60$, which after using a calculator is $\sqrt{3}$[/hide][/quote]
I'd reccomend you memorize basic values of trig functions, for sine and cosine. (Tan can be expressed in sine and cosine)

You don't get a calculator on the AIME :P[/quote]

I already know that, but just to give some reason :D

[quote="anirudh"]just to give some reason :D[/quote]

Calculators are [b]never[/b] a justification for anything.  For example, I could use a calculator to "find" counterexamples to Fermat's Last Theorem.

You don't need to justify $\sin 60 = \frac{ \sqrt{3}}{2}$, since it's both well-known and geometrically simple.


Stronger affirmative/negative?
Weaker affirmative/negative?

I was taught in school that double negative mean stronger affirmative, but in the movie "Pirates of the Caribbean", I heard things like "Not unlike you...", "I ain't stealing no ship", which I am sure they meant negation.

[quote="Xevarion"]It's not very common, especially among educated people.
[/quote]

um

what?

i hear everyone using it all the time, regardless of their education

"not very common" is completely false

I think when Lord Becket said "Not unlike you", he really meant "Unlike you", not "Like you", and such a character is supposed to be well educated, maybe it IS correct in the OLd English.

Some says that double negative in Russian still means negative, are there any other languages with the same feature?

[quote="Xevarion"]"I ain't stealing no ship" is intended to be negative. But this sort of thing mainly happens when an erroneous "no" is used in place of an article. It's not very common, especially among educated people.

"Not unlike you" is a double negative and is intended to be positive. It probably is a little stronger than stating the same thing directly. Generally, simpler and more common expressions are weaker and more obscure or complicated expressions are stronger.

By the way: I think using double negatives (especially in subtle ways and for strong effect) is more British in style and less common in American English. For example, people often say something is "not bad" -- in American English, this is a pretty basic compliment, but (at least not too many years ago) in British colloquial English this was a very high compliment.[/quote]

You'd have to give us the full sentence to be really sure, I suppose -- it's possible that the screenwriters (or the actor) screwed up and no one (except you) noticed.  Without any context, though, "Not unlike you" definitely means "like you."  "Aint no" is one exception to the general rule that in English, a double negative makes a positive, and it's not "proper English."  Wikipedia has some other examples, e.g. "We don't need no education // we don't need no thought control."

There are languages in which double negatives enhance the negation -- I think it is a linguistic question that has been studied.

[quote="K81o7"][quote="Elemennop"]

i hear everyone using it all the time, regardless of their education
[/quote]

People I know use it only when jokingly imitating someone who doesn't understand double negatives.  Nobody I know uses this in normal speech.[/quote]

thats' weird, i didn't know peopel didn't use it.

Double negatives can make things harder to understand, so I maybe just look at the person's expression and what the person is doing and interpret based on that.

In colloquial English, such as country English, double-negative = massive negative. "Never wanted nothing more" = "didn't want anything else"

Double negatives are indeed common in Russian, and they are often the only way to construct a negative statement. For instance, I can't think of a way to translate "I don't want any of this" into Russian without using a double negative. However, single negatives come naturally to people who are used to playing with $ \forall$ and $ \exists$. When I learned English, I did not have no problems with negative statements. :)

Another example: in the song "Blood on the Risers", "he ain't gonna jump no more", means "he won't jump any more"

Does not bad really mean "extremely good" in British English? In my mind, "not bad" is kind of polite way of saying "neither good nor bad".

I don't need no negatives!
I use double negatives just to create a generic southern hick character.
I have never used it seriously. :roll:

There are huge differences in English between colloquial speech and edited, published writing. In general, yes, a double negative IN WRITING, for an audience of educated readers, would be interpreted as a positive statement. But I'm sure most native speakers of English whose ancestors also spoke English can be heard from time to time making statements with double (or triple, or quadruple) negations that are to be interpreted as stronger negative statements.

Double negatives are also very common in Portuguese. For example, "I don't want nothing" or "I don't know anyone" are common ways of expressing the corresponding ideas, in all levels of education.

In Greek, it is the very same thing saying "i don't want anything" (δεν θέλω κάτι) and "i don't want nothing" (δεν θέλω τίποτα), both expressions meaning there's nothing i want, used by people regardless of their educational or social background.

By the way, I rembered a Pink Floyd song entitled "We Don't Need No Education" from The Wall album.

I think double negatives are universally accepted in pretty much all languages except English.

In fact, the only languages I know of where double negative=positive are classical Latin (but not modern Romance languages) and modern English. Which is to say that Old English used double negatives all the time as negatives.

And as for Middle English, Chaucer once used the triple negative "Ther nas no man no wher so vertuous"="there wasn't no man nowhere so virtuous".

I'm in high school, and not very accomplished within the fields of language, but I do feel I'm quite skilled in all applications of my mother tongue, English, and I'm currently taking french 5.

"I ain't stealing no ship"

'Ain't' is of course, slang with the meaning of "I will not" = "I won't". What seems to be the problem here, is that instead of using a regular negation, such as 'any', he implemented the service of 'no', which in its current use, falsely generated the look and feel of a double negative.

Translation : "I'm not not stealing any ships"

(I'm going to neglect future tense here, as the pirate seems to be highly uneducated, and probably intended it to be a present-tense statement. If you desire, re-conjugation in the future-tense is acceptable.)

[quote="tokenadult"][quote]I think double negatives are universally accepted in pretty much all languages except English.[/quote]

I hope you are not taking all Indo-European languages as evidence for all human languages in general. The original questioner, I think, is a native speaker of Chinese (one of my second languages), and it is very hard to come up with examples of "double negative" constructions in that language.[/quote]

Well, it was kind of an over-generalization. I guess all I meant is most (common) languages use double negatives. Anyway, it is also true for Afro-Asiatic, and those are the only two families I know much about.

Woah!
The sentences in the previous posts are so interesting!
Can you guys give me more examples in double-negative?
Cheers, Michael.

I forgot to mention:
In Mathematics the negation of a negation of a sentence is the original sentence itself.
That is to say ~(~P) = P
But in English or other languages, the negation of a negation could have some other special meanings, don't you think that it is interesting?  :oops:

This is probably rather off topic... but here's a joke I heard somewhere (tell me if it's old or not funny so I won't repeat it)

A linguistic professor was lecturing in front of his class. He says, "In English, double negatives are often considered ungrammatical, and sometimes may be interpreted as having a positive meaning. However," he continues, "In languages such as <insert languages here>, the situation is different. Double negatives in these languages are generally accepted as....."

After a while, the professor concludes, "So, double negatives may be interpreted in different ways in different languages, but in no languages does a double positive make a negative."

There is a momentary silence in the room. Then a student in the back says aloud, "Yeah, right!"

[quote="Xevarion"][quote="Buffalo"]Does not bad really mean "extremely good" in British English? In my mind, "not bad" is kind of polite way of saying "neither good nor bad".[/quote]
Really, someone who lives in Britain should answer this. I honestly don't know if it's the case now; however, there are a lot of differences between informal language in American and British English, so it is quite possible. Your meaning of "not bad" is how most people tend to use it in the USA.[/quote]
“Not bad” just means that: not bad. It can express anything from neutral to extremely good.

Buffalo is probably thinking of the difference between “not good” and “no good”. The former can mean anything from neutral to very bad, but the latter means only one thing: very bad. The latter is an example of a rhetorical figure of speech known as a [b]litotes[/b].

[quote="luimichael"]But in English or other languages, the negation of a negation could have some other special meanings, don't you think that it is interesting?  :oops:[/quote]
Yes, litotes is a good example. This is a figure of speech in which an understatement (often two negatives) is used to expresss the opposite meaning. Consider the two sentences “I’m not good!” and “I’m no good!”. In the first sentence, the speaker is just negating the statement “I am good”; he’s not implying that he’s very bad. But in the second sentence, the speaker is indeed saying that he’s the very opposite of good. The second sentence is a litotes (whereas the first sentence isn’t).


I'm really confused on which courses I should take. I'm going to both the california and texas camps.
I scored a 120 on the AMC 10 and a 3 on the AIME. I'm pretty sure that I can handle all intermediate courses, but I'm not sure how well I can handle advanced.

So I can take 4 classes, but I'm not sure which level. I want to take 1 course in each subject for sure, that way I will be well versed for the competitions. This is really the only enrichment I can get as none of the kids in my school are better problem solvers than me, and our math coach said that he probably wouldn't be much help after I start making 8+ on the AIME.

Any one who has had this camp experience that wants to help me?

Also, are there any materials I can look at to see the level of the intermediate courses to determine whether I should go forward with them or step it up and take an advanced class.

I'm thinking that I'll either take 4 intermediates, 3 intermediates and 1 advanced, or 2 of each.

Thanks in advance  :D ,
Shri

Have you seen the [url=http://www.awesomemath.org/AMSP_dailysch.pdf]Academic Curriculum[/url]?
I think you should stick with intermediate unless you normally score 9+ on practice AIMEs and have practiced some olympiad problems.

I never saw anything regarding courses, so I ended up emailing Dr. Andreescu and I told him what classes I wanted to do. Just wondering, is anyone in a similar situation as me?

shrig94, the Academic Curriculum that Max300 posted has some diagnostic problems that you can solve to find your course level placement. Furthermore, you'll have a couple of days to switch classes at the beginning of camp. Though I generally agree with Max300's recommendation (based on your credentials) that you take the intermediate courses, it's up to you to find the appropriate level.

serialk11r, students choose their courses during the first day of camp. This has been the procedure the past three years I've been at AMSP (two as student, one as assistant).

hmmmmm, based on your credentials, I'd say intermediate if you know your ready for them, and some beginner classes in areas you feel soft. I don't think you should take advanced classes until you have gotten at least 9 on AIME, because the advanced courses are for Olympiad training.

I think I'll just go all 4 intermediate then. I think all of them will be challenging enough that I will have to work hard, but not so challenging that I'll be like 'lolwut'.


use the Fibonacci sequence, so the next number is 34


Let $ABC$ be a triangle, the points $D\in (BC)$, $N\in (CA)$, $M\in (AB)$

and $P\in AD\cap MN$. Prove that if the quadrilateral $BMNC$ is cyclic,

than there is the relation $\frac{AD}{AP}\cdot MN=\frac cb\cdot DC+\frac bc\cdot DB$.

[u]Remark.[/u] Here are some particular cases (if the quadrilateral $BMNC$ is cyclic):

$1.\ P: =I\Longrightarrow \frac{a}{MN}=1+\frac{a}{b+c}$;

$2.\ P: =G\Longrightarrow \frac{a}{MN}=\frac{3bc}{b^2+c^2}$.

Nice problem!Let's take the transversal theorem \[ \frac{PD}{PA}=\frac{MB}{AM}\cdot\frac{DC}{a}+\frac{NC}{AN}\cdot\frac{BD}{a}\leftrightarrow \frac{AD}{AP}=\frac{DC}{a}\cdot\frac{c}{AM}+\frac{BD}{a}\cdot\frac{b}{AN} \].Now from similarity of $\triangle AMN$ and $\triangle ABC$ we have \[ \frac{AM}{b}=\frac{AN}{c}=\frac{MN}{a} \] so we substitute and get the desired relation. :)


Consider a triangle $ ABC$ such that $ sinAsinB \plus{} sinBsinC \plus{} sinCsinA \equal{} q$ and $ (1 \plus{} sinA)(1 \plus{} sinB)(1 \plus{} sinC) \equal{} 2(q \plus{} 1)$. Prove that the triangle $ ABC$ has a right angle

By vaf:
Substracting from the second equality the irst multiplied by 2 yields: $ (sinA\minus{}1)(sinB\minus{}1)(sinC\minus{}1)\equal{}0$. Hence $ sinA,sinB,sinC$ is $ 1$, so $ ABC$ is a right triangle.


I'm thinking to extend the 'Friend' system, we could do something like myspace where friendship is mutual and everyone can see each other's friends.

I don't think so. Mr. Vornicu said that'd he'd probably keep them private.

[quote="13375P34K43V312"] we could do something like myspace [/quote]
 :spam:

[quote="13375P34K43V312"]...something like myspace...[/quote]

I would leave and never come back.


let n>2 given natural no. and A1,A2,...,An be subsets of S=(1,2,...,n).show
there exists element x belonging to S s.t.A1-x,A2-x,...,An-x are all distinct. :?:

It's equivalent to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=kurschak&t=34747]this[/url].


ABC is an isosceles triangle with AB = AC. Suppose that
1. M is the midpoints of BC and O is the point on line AM such that OB is perpendicular to AB;
2. Q is an arbitary point on the segment BC different from B and C;
3. E lies on the line AB and F lies on the line AC such that E, Q, F are distinct and collinear.

Prove that OQ is perpendicular to EF if and only if QE = QF.

[quote="mathwizarddude"]ABC is an isosceles triangle with AB = AC. Suppose that
1. M is the midpoints of BC and O is the point on line AM such that OB is perpendicular to AB;
[/quote]

Is O an external point?

Regards
(名) 關心

Since $ OA \perp BQ$ and $ \angle ABO\equal{} \angle AQO\equal{}90$, ABOQ is a kite, so we must have $ Q\equal{}C$, so $ QF\equal{}QE$.

[quote="dgreenb801"]Since $ OA \perp BQ$ and $ \angle ABO \equal{} \angle AQO \equal{} 90$, ABOQ is a kite, so we must have $ Q \equal{} C$, so $ QF \equal{} QE$.[/quote]
Why $ \angle ABO \equal{} \angle AQO$? I see this is true iff $ Q \equal{} C$ but it is stated that $ Q$ is different from $ C$ and $ B$ in the problem.
(And yes, O is an external point.)


calculus
integration
floor function
inequalities

What is the greatest integral solution of $ |x\minus{}5| > |2x\minus{}5|$?

Except, it wants it when the inequality is true, so when the curve is above the x-axis. The largest integer satisfying this is clearly 3.

How do you solve this in 45 seconds though? I got it right just by looking at one possibility, but HOW?


probability
geometry
Pythagorean Theorem
similar triangles

In the diagram below, we have $ AB\equal{}3$ and right triangles $ ABD$ and $ ADE$. The point $ D$ bisects $ BC$. If $ AC\equal{}\sqrt{73}$, what is the value of $ AE$, in simplest radical form? Note: the diagram is not necessarily drawn to scale.

[img]http://myfiles.zapto.org/AMC10Images/Picture1.jpg[/img]

$ \textbf{(A)} \sqrt{31/2}\qquad\qquad\textbf{(B)}
\dfrac{\sqrt{39}}{73}\qquad\qquad\textbf{(C)}
\dfrac{25\sqrt{73}}{41}\qquad\qquad\textbf{(D)}
\sqrt{55}\qquad\qquad\qquad\textbf{(E)}
\sqrt{70}$

We have the letters and numbers {A,B,C,0} in the set. What is the probability that if we randomly select three of these with replacement, we will have a string in the format of LLN, where L symbolizes a letter and N a number?

$ \textbf{(A)} \dfrac{1}{64}\qquad\qquad\qquad\textbf{(B)}
\dfrac{20}{21}\qquad\qquad\textbf{(C)}
\dfrac{5}{6}\qquad\qquad\textbf{(D)}
\dfrac{3}{4}\qquad\qquad\qquad\textbf{(E)}
\dfrac{3}{64}$

There are five women and five men at a party. Each person shakes hands with their spouse and with the other people of their gender at a party. How many handshakes are there?

$ \textbf{(A)} 10\qquad\qquad\qquad\textbf{(B)}
15\qquad\qquad\qquad\textbf{(C)}
20\qquad\qquad\qquad\textbf{(D)}
25\qquad\qquad\qquad\textbf{(E)}
30$

Bob finds a weird calculator which only has two keys - [+1] and [*2]. The number ``1'' is on the screen. Bob wants to make this number 71. What is the minimal number of clicks required to reach the desired number?

$ \textbf{(A)} 9\qquad\qquad\qquad\textbf{(B)}
10\qquad\qquad\qquad\textbf{(C)}
11\qquad\qquad\qquad\textbf{(D)}
12\qquad\qquad\qquad\textbf{(E)}
13$

If the sides of a triangle are 8, 15, and 17, what is the length of the shortest altitude to the nearest tenth?

$ \textbf{(A)} 6.2\qquad\qquad\qquad\textbf{(B)}
6.9\qquad\qquad\qquad\textbf{(C)}
7.1\qquad\qquad\qquad\textbf{(D)}
7.7\qquad\qquad\qquad\textbf{(E)}
8.0$

[hide="Problem 6"][b]Brut3Forc3[/b] had an excellent solution to this problem. It is quoted below:

``Refer to the poorly made diagram below, where I have calculated some obvious side lengths with the Pythagorean Theorem.
[asy]pair A=(0,3), B=(0,0), C=(8,0), D=(4,0), E=(200/41,48/41);
draw(A--B--C--cycle); draw(A--D); draw(D--E);
label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,S); label("E",E,NE);
label("3",(0,1.5),W); label("5",(0,1.5),E); label("4",(2,0),S); label("4",(6,0),S);[/asy]
Extend $ AD$ down to point $ F$ so that $ AF\perp CF$. (Now we have this new diagram)
[asy]pair A=(0,3), B=(0,0), C=(8,0), D=(4,0), E=(200/41,48/41), F=(164/25,-48/25);
draw(A--B--C--cycle); draw(A--D); draw(D--E); draw(D--F); draw(C--F);
label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,S); label("E",E,NE); label("F",F,S);
label("3",(0,1.5),W); label("5",(0,1.5),E); label("4",(2,0),S); label("4",(6,0),S);[/asy]
Triangles $ ABD$ and $ ACD$ have the same area, so $ [ABD] \equal{} 6 \equal{} [ADE] \plus{} [CDE]$. We also have $ [ADE] \equal{} \frac {5}{2}DE$ and $ [CDE] \equal{} \frac {1}{2}DE\cdot DF$, so $ 6 \equal{} \frac {5}{2}DE \plus{} \frac {1}{2}DE\cdot DF \\
\iff 12 \equal{} 5\cdot DE \plus{} DE\cdot DF \qquad(*)$.
We also know that triangles $ ADE$ and $ AFC$ are similar, so we have $ \frac {5}{5 \plus{} DF} \equal{} \frac {DE}{FC} \Rightarrow 5\cdot FC \equal{} 5\cdot DE \plus{} DE\cdot DF$. Using $ (*)$, we have $ 5\cdot FC \equal{} 12 \iff FC \equal{} \frac {12}{5}$. By the pythagorean theorem, we know $ FC \equal{} \sqrt {16 \minus{} DF^2}$. Plugging this in and solving for $ DF$ gives $ DF \equal{} \frac {16}{5}$. Then plugging this into $ (*)$ lets you solve for $ DE$, and gives $ DE \equal{} \frac {60}{41}$. Then finding $ AE$ is just a simple use of the pythagorean theorem, and it turns out to be $ AE \equal{} \frac {25\sqrt {73}}{41}$, so the answer is $ \boxed{\textbf{C}}$.
[/hide]

[hide="Problem 7"]We note that both $ L$'s must be the same, and that an $ N$ can be chosen out of the set, representing a number. The answer is now just $ \frac{3}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}\equal{}\frac{3}{64}\implies\boxed{\textbf{E}}$.
[/hide]

[hide="Problem 8"]Each person shakes the hands of five people. This means that the answer is $ \dfrac{10\times 5}{2}\equal{}25\implies\boxed{\textbf{D}}$.[/hide]

[hide="Problem 9"]The way we can immediately see is brute force. A little dirty work results in $ \boxed{\textbf{A}}$.
[/hide]

[hide="Problem 10"]We see that the shortest altitude would be to side length 17. This is approximately $ 7.1\implies \boxed{\textbf{C}}$.[/hide]

For number 7, why did it have to be the same letter? the question states that it is a letter, not necessarily the same letter. I feel that the question was worded poorly.

[hide="Problem 6"]By Pythagoras, $ BC\equal{}8$. Drop a perpendicular from $ E$ to $ DC$ and call the foot $ P$. By similar triangles, we have $ \frac{EP}{AB}\equal{}\frac{EP}{3}\equal{}\frac{PC}{BC}\equal{}\frac{PC}{8}$ or $ EP\equal{}\frac38PC$. Also, $ \frac{EP}{BD}\equal{}\frac{EP}4\equal{}\frac{DP}{AB}\equal{}\frac{4\minus{}PC}3$. Solving, we have $ EP\equal{}\frac{48}{41}$, and again by similarity $ DE\equal{}\frac{60}{41}$. Applying Pythagoras again, we have that $ AE\equal{}\frac{25}{41}\sqrt{73}$.[/hide]

Special thanks to #H34N1 for putting so much time into this Mock AMC 10!

Sorry to revive, but in the solution to #6 why is $ [CDE]\equal{}\frac{1}{2}DE\cdot DF$?

Note that DE is the height to base DF, since it is parallel to CF


$ \lim_{x\rightarrow 0}\sqrt{|sinx|}e^{arctancos(\frac{1}{x})}$

There are so many functions.Can we just solve in this way

$ \lim_{x\rightarrow 0}\sqrt{|sinx|}\times\lim_{x\rightarrow 0}e^{arctancos(\frac{1}{x})}\equal{}0\times\lim_{x\rightarrow 0}e^{arctancos(\frac{1}{x})}\equal{}0$

[quote="azatkgz"]$ \lim_{x\rightarrow 0}\sqrt {|sinx|}e^{arctancos(\frac {1}{x})}$

There are so many functions.Can we just solve in this way

$ \lim_{x\rightarrow 0}\sqrt {|sinx|}\times\lim_{x\rightarrow 0}e^{arctancos(\frac {1}{x})} \equal{} 0\times\lim_{x\rightarrow 0}e^{arctancos(\frac {1}{x})} \equal{} 0$[/quote]
sure you can, but yuou must say $ \lim_{x\rightarrow 0}e^{arctancos(\frac {1}{x})}$ is not equal to $ \plus{}oo$ it's easy. try to  prove yourself


Prove that for any distinct rational numbers $a, b, c$, the number \[\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}+\frac{1}{(a-b)^{2}}\] is the square of some rational number.

Quite easy :grin: $\sum \frac{1}{x^2}=\frac{\sum x^2y^2}{x^2y^2z^2}=(\frac{\sum xy}{xyz})^2$(notice that $x+y+z=0$


Anyone know where I can find proofs of Newton's and Maclaurin's Inequalities? Or if they're short, would someone mind posting them here? Thanks.

If I knew what those inequalities were, I would give them a try.

try Kiran Kedlaya's inequalities packet, page 9

http://caltizzle.caltech.edu/files/ineqs.pdf

Let x_i be nonnegative reals, and
[tex](x+x_1)(x+x_2) \cdots (x+x_n) = x^n + c_1 x^{n-1} + \cdots + c_{n-1} x + c_n [/tex],
then Newton's inequality states that
[tex]\displaystyle \left(\frac{c_i}{\binom{n}{i}}\right)^2 \geq \left(\frac{c_{i-1}}{\binom{n}{i-1}}\right) \left(\frac{c_{i+1}}{\binom{n}{i+1}}\right). [/tex]
And Maclaurin states
[tex] \displaystyle \left(\frac{c_{i}}{\binom{n}{i}}\right)^{1/i} \geq \left(\frac{c_{i+1}}{\binom{n}{i+1}}\right)^{1/(i+1)[/tex].
For each i that makes sense.
These are very powerful inequalities.

zscool:

Yeah, I checked that out, but it only gives a hint, but I don't know how to use it. :(

Joel:

If s_k is the kth symmetric sum of n real numbers,

Newton's: If the n real numbers are nonnnegative, then [tex]s_{k}^{2} \geq s_{k-1}s_{k+1}[/tex] for k=1, 2, ..., n-1.

Maclaurin's: If the n real numbers are positive, then [tex]s_{1} \geq s_{2}^{1/2} \geq ... \geq s_{n}^{1/n}[/tex]

Ok, here is the proof of Maclaurin's Inequality given that Newton's Inequality is true. I still need to find a way to prove Newton's though.

Let [tex]0 < r < n[/tex], By Newton's inequality, we have

[tex]
(s_{0}s_{2}) \cdot (s_{1}s_{3})^{2} \cdot ... \cdot (s_{r-1}s_{r+1})^{r} \leq s_{1}^{2} \cdot s_{2}^{4} \cdot ... \cdot s_{r}^{2r}
[/tex]

which, after cancellations on both sides, implies [tex]s_{r+1}^{r} \leq s_{r}^{r+1}[/tex], which is Maclaurin's.

I know how, but I am just not in the right place (away from home)... when I get home, I'll post the stuff


probability
FTW
SFFT
special factorizations
AMC

Hi, I was wondering if anyone could post the answer key to the 12B.

Thanks!

If anyone posts their answers, please use hide tags, so that people who may not have seen the test yet won't be spoiled.

I'm pretty sure these are right, 12B:

[hide="Spoiler"]
E
B
D
B
D
D
E
D
A
C
D
C
D
D
E
A
D
C
A
D
A
A
A
A
C[/hide]

Is there a topic with solutions?

If not, I want to see the solutions to 11, 13 (even though I got it right), 16, 20, 23

I guess here's a solution to 11 (because i'm too lazy for the rest...yes).

[hide]Let angle A be $x$.  Thus, our sum of the angles is $x+\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=360$.  Being lazy and using our calculator, we do $360 \div (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})=173$ (about) so (D).  [/hide]

Here's 13:
[hide]
There are a total of 63 seconds (3 yellow, 30 green, and 30 red).  He can arrive up to 3 seconds before each turn.  So he can arrive 3 seconds before the light turns yellow, at any point where the light is yellow (since the period that the light is yellow is only 3 seconds), or 3 seconds before it turns green.  None of these overlap, so he can arrive for a total of 9 seconds out of the sixty-three, so the probability is $9/63=1/7$, or answer choice $D$[/hide]

I hope this was the right problem...tetrahedron right?

[hide="16"]
Case 1: All four are the same color
3 colors, 3 possibilities
Case 2: Three are the same color
Pick which color we use three times, two colors left to choose which color we use once, $3\cdot2=6$ possibilities
Case 3: Two are the same color
Subcase 3a: The other two are the same
Pick which two colors, ${3\choose2}=3$
Subcase 3b: The other two are different
Pick which color you use twice, three colors, three possibilities
So 6 in total for case 3.

$3+6+6=\boxed{15}$.[/hide]

[hide="23"]
Let the sides be $a,b$. Then

$\frac{ab}{2}=3(a+b+\sqrt{a^{2}+b^{2}})$
Then rearrange the mess of crap so the square root is on one side and the rest of the crap is on the other...
$\frac{ab}{2}-3a-3b=3\sqrt{a^{2}+b^{2}}$
Then get your hands dirty and square both sides...
$\frac{a^{2}b^{2}}{4}+9a^{2}+9b^{2}-3a^{2}b-3ab^{2}+18ab=9a^{2}+9b^{2}$
Then be like whoa stuff cancels, and everything has an $ab$ so divide both sides by that and you're left with
$\frac{ab}{4}-3a-3b+18=0$
$ab-12a-12b+72=0$
Then you be like I'm an AoPSer so I'm naturally gonna use Simon's Favorite Factoring Trick FTW
$(a-12)(b-12)=72$
And 72 has 12 factors, 'cept you can't switch around so there are 6 triangles.[/hide]

Oh... on 11 I added 1,2,3,4 and got 10D=360 then A=144
because I did it quickly in my head.... guess I didn't understand it or read it well :(
On thirteen, for some reason I thought that if she was 3 minutes late it still counts (I thought she waited 3 minutes forwarde and backwards in time... oops) but I still got a good approximation of the answer and got it right (i kept getting 15/63...)

I had to restrain myself from shouting "W00000T" when the stuff cancelled in 23.  It was a happy moment as well as a moment that improved my score by 4.5 points :D.

It's hard to believe 21-24 all happened to be A. That's odd.  :huh:

Not that that influenced my answers...

Hmm.. 4 A's in a row, and I got only 1, i misguessed the other 3.. even though 23 was easy, just simplify and solve the equation.. I was too lazy

There's a thread a little ways down by DPatrick which says "Official AMC 10B/12B Answers."  They are posted on the AMC site as well.

I think the posts asking for answer keys were before the AMC site had answers up...

and I was just asking for solutions.
Thanks everybody!

I'm so angry at the number of stupid mistakes I made!!! I completely ran out of time and was nervous and checking the clock all the time, mostly because I was in an empty room all by myself that happenned to be my calc class last year... all in all it was completely unideal conditions for me... :(
Any tips on what to do if you're the only person taking a B version test at your school?


Find all positive numbers of a pythagorean triple that form a aritmetic progression or a geometric progression and how's can i prove that if n >3 then it's part of a Pythagorean triple

To prove that there are an infinite number of pythagorean triples that form an arithmetic sequence, let $a-d,a,a+d$ be some pythagorean triple with $a>d>0$. Then,

$(a-d)^{2}+a^{2}=(a+d)^{2}\Rightarrow a^{2}=4ad\Rightarrow a=4d$

Therefore, any pythagorean triple of the form $3d,4d,5d$ forms an arithmetic sequence.

To prove that there are no pythagorean triples that form a geometric sequence, let $a,ar,ar^{2}$ be some pythagorean triple with $a,r\ge1$. Then,

$(a)^{2}+(ar)^{2}=(ar^{2})\Rightarrow 1+r^{2}=r^{4}\Rightarrow r^{4}-r^{2}-1=0$

This quartic has no rational roots (only possible rational roots are $\pm1$ and none of them are roots), therefore there are no pythagorean triples that form a geometric sequence.

Lastly, I don't understand what you are trying to ask on your last question. What is $n$? What do you want?

[quote="pkerichang"]basically, I think he's asking how to prove that if the number $n \ge 3$, then it can be a number in a pythagorean triple. (like 4 is in the 3, 4, 5 triple.)[/quote]
This isn't too hard. If $n$ is odd, then we have the triple
$n,\frac{n^{2}-1}{2},\frac{n^{2}+1}{2}$.
If $n$ is even, then we just have
$n,\frac{n^{2}-4}{4},\frac{n^{2}+4}{4}$
which is similar to the first one. (In other words, to get a triple for an even $n=2k$, just use the numbers in the triple for $k$ multiplied by $2$. For example, for $3$, we have the triple $3,4,5$. To get a triple for $6$, we just use $6,8,10$.)


I just need help understanding the problem...because I'm starting to go through old AIMEs to prepare, and I need serious help on lots of stuff like this.

The polynomial $ P(x)$ is cubic. What is the largest value of $ k$ for which the polynomials $ Q_1(x)\equal{}x^2\plus{}(k\minus{}29)x\minus{}k$ and $ Q_2(x)\equal{}2x^2\plus{}(2k\minus{}43)x\plus{}k$ are both factors of $ P(x)$?

It's asking you to find a number $ k$ such that when you stick k into the polynomials and multiply them together,  you end up with a cubic.

How does this happen if the two leading degrees of $ Q_1$ and $ Q_2$ are both 2 (give a quartic normally)?
The problem wants you to figure out how to turn them into a cubic. (hint: check shared factors).

Well $ P(x)$ has three roots $ r_1,r_2,r_3$. So two of those roots have to be roots of $ Q_1(x)$ and two of them have to be roots of $ Q_2(x)$. W.L.O.G assume that they both share $ r_2$. So from Vieta's we know that

$ r_1r_2\equal{}\minus{}k$

$ r_1\plus{}r_2\equal{}29\minus{}k$

$ r_2r_3\equal{}\frac{k}{2}$

$ r_2\plus{}r_3\equal{}\frac{43\minus{}2k}{2}$

Obviously from those equations we know that $ \minus{}2r_3\equal{}r_1$ so from there we have

$ \minus{}2r_3\plus{}r_2\equal{}29\minus{}k$

$ r_3\plus{}r_2\equal{}\frac{43\minus{}2k}{2}$

Subtracting the two equations gives that $ \minus{}3r_3\equal{}7.5$ so $ r_3\equal{}\minus{}2.5$

Now going back to $ r_2r_3\equal{}\frac{k}{2}$

$ r_2\cdot\frac{\minus{}5}{2}\equal{}\frac{k}{2}$
$ r_2\equal{}\minus{}\frac{k}{5}$

$ \minus{}2r_3\plus{}r_2\equal{}29\minus{}k$
$ 5\minus{}\frac{k}{5}\equal{}29\minus{}k$
$ \frac{4k}{5}\equal{}24$
$ k\equal{}30$

ugh...

all those solutions make me dizzy...I'm still not very good at this stuff.

I may revive this post when I try to attempt to solve it more--however you can still post hints if you see fit.


How does one get $ B_n$ the subject of this expression:

$ \frac {x}{e^x - 1} = \sum_{n = 0}^{%Error. "infin" is a bad command.
} B_n \frac {x^n}{n!}$

I tried taking $ - x$ out of the LHS and looking at the $ \frac {x^n}{n!}$ coefficient of the LHS but couldn't get the answer..

H

That depends a great deal on what your definition of $ B_n$ is.  Unrelatedly, this is a forum for intermediate high school mathematics, not including calculus.  There are other parts of the forum more appropriate for your recent posts (e.g. the calculus subfora).

Thanks for you response.

I thought this would be more of a combinatorics (enumerative) problem than strictly calculus, which is why I posted it here - I'll move it elsewhere if I can figure out how..

$ B_n$ is, for now at least, an unknown sequence. In an exam I took, we had to use the definition above to calculate the first 4 terms of $ B_n$.

Ah, okay -- then this is just a standard exercize in power series.  You can do it either by taking successive derivatives or by manually inverting the power series for $ \frac {e^x \minus{} 1}{x}$, i.e., multiply out to solve the equation $ \left(\sum_{n \geq 0} \frac {x^n}{(n \plus{} 1)!}\right)\cdot \left(\sum_{n \geq 0} B_n \frac {x^n}{n!}\right) \equal{} 1$ term-by-term for the $ B_n$.

Heh, seems so obvious now you say that. Thanks very much :)

You can also find them recursively by the relation: $ B_n \equal{} \frac{1}{n\plus{}1} \sum_{k\equal{}0}^{n\minus{}2} (\minus{}1)^{n\minus{}k} \binom{n}{k} B_k$ for $ n \geq 1$, where $ B_0 \equal{} 1$.

ahh but for that you'd have to know that they were the Bernoulli numbers !


A cat is located at C, 60 meters directly west of a mouse located at M. The mouse is trying to escape by running at 7 m/s in a direction 30º east of north. The cat, an expert in geometry, runs at 13 m/s in a suitable right line path that will intercept the mouse as quicly as possible.

a) If t is the lenght of time, in seconds, that it takes the cat to catch the mouse, determine the value of t.

b) Suppose that the mouse instead chooses a different direction to try to escape. Show that, no matter the direction it runs, all points of interception lie in a circle.

c) Suppose that the mouseis intercepted after running a distance of $ d_1$ meters in a particular direction. If the mouse would have been intercepted after it had run a distance of $ d_2$ meters in the opposite direction, show that $ d_1\plus{}d_2\geq 14\sqrt {30}$

For reference, see the attached drawing

a) From M (mouse) draw a line course 30 degs NE, take the vector MW of 7 (m/s- mouse's speed). Draw then the circle centered at W of radius 13 (m/s - cat's speed); it will intersect (CM) at A. The vector AW represent the optimum track of the cat to catch the mouse. Draw a parallel line to AW through C, it will cross the mouse track at T, the point where it is caught. To count the time, either measure MT and divide it to MW(7), or measure AM and divide 60 (distance CM) to AM.
b) See that for any colision point T there is MT/CT=7/13, therefore T is placed on the circle diameter XY on CM, the T - Apollonius circle of triangle MTC.
c) See that when the mouse runs Eastward, it is caught at Y, with MY=70m, while running Westward it is caught at X, with MX=21m. How are these figures calculated? Dividing the distance CM to the sum and difference of the speeds, respectively. So XY = 91 m, therefore the radius of the circle is 45.5 m.
If d_1 and d_2 are the segments of the chord passing through M, their minimum sum is attained when the chord is perpendicular to the diameter XY, so any chord will be greater than KL, having the value 14 sqrt(30).

Best regards,
sunken rock


integration
topology
function
real analysis
advanced fields
advanced fields unsolved

Let $ w \equal{} \frac{xdy\minus{}ydx}{x^2 \plus{}y^2}$ be the smooth 1-form deﬁned on $ R^2\minus{}\{0\}$, and $ i : S^1 \to R^2\minus{}\{0\}$
the inclusion of the unit circle.
1 Prove that $ i^* w$ is a closed 1-form on $ S^1$ .
2 Suppose $ \theta$ is a closed 1-form on $ S^1$ such that $ \int_{S^1} \theta \equal{} 0$. Prove that $ \theta$ is exact.
3 Prove that if $ \sigma$ is a closed 1-form on $ S^1$ such that $ \int_{S^1} \sigma \equal{} 2\pi c$, then $ \sigma \minus{} ci^* w$ is exact.
4 Prove that $ H^1 (S^1 , R)$ has dimension one.

kind of just not sure how approach it..

It's a straightforward chain of calculations and implications. Where do you get stuck?

i am stuck at the top. if i knew how to do one of these i could do a million, but i dont =(

You need to learn some of the language of manifolds, then.
A useful principle for the first: $ d$ commutes with pullbacks, in the sense that $ d(G\circ f)\equal{}G^*df$. You can't do it globally around the circle, but your $ w$ is "$ d\theta$" locally. From this idea, we get that the pullback of a closed form is closed, and the pullback of an exact form is exact.

okay i suppose the part of the problem im really having mental issues with is 2.

showing something is exact means its the d of some other form... i can see how the converse of the statement is true... but its not really helping me go the other way

like do i do something like..

$ \theta \equal{} \theta_1 dx \plus{} \theta_2 dy$ and
$ 0\equal{} \int_{S^1} \theta \equal{} \int_{S^1} \theta_1 dx \plus{} \theta_2 dy\equal{} ..$

and use some stoke or green theorem? or is this not even sensical..?

What is it $ d$ of? It's $ d$ of its own integral. Let $ f(x)\equal{}\int_{x_0}^x \theta$. There's more than one way to choose a path, so you need to show they're all the same.

Some hints:

[hide="1"]Any 1-form on $ S^1$ is closed since there are no nontrivial $ 2$-forms.[/hide]

[hide="2"]Define $ F(x) \equal{} \int_1^x \theta$. Show that this is a well-defined smooth function which satisfies $ dF \equal{} \theta$[/hide]

[hide="3"]This follows from 2, and the fact that $ \int_{S^1}i^*w \equal{} 2\pi$[/hide]

[hide="4"]$ H^1(S^1,R)$ is generated by $ i^*w$ [/hide]

thank you both, i think i see where my problem understanding it was.


Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3$ . Prove that :
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc\ge 5\]

Let $ a\plus{}b\plus{}c\equal{}p,abc\equal{}r$,then it becomes $ p^2\plus{}2r\geq 11$.
If $ a\plus{}b\plus{}c\geq \sqrt{11}$,it is obvious.
If $ 3\leq a\plus{}b\plus{}c\leq \sqrt{11}$,by Schur we have $ r\geq \frac{12p\minus{}p^3}{9}$,then we have to prove that
\[ 2p^3\minus{}9p^2\minus{}24p\plus{}99\leq 0\Longleftrightarrow (p\minus{}3)(2p^2\minus{}3p\minus{}33)\leq 0\]
it's true as $ 3\leq p\leq \sqrt{11}$.Done.

if we add one this inequality becomes -
$ a^2\plus{}b^2\plus{}c^2\plus{}2abc\plus{}1 \ge 2(ab\plus{}bc\plus{}ca)$
this is Darij's famous inequality. am i right?

[quote="geniusbliss"]if we add one this inequality becomes -
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \plus{} 1 \ge 2(ab \plus{} bc \plus{} ca)$
this is Darij's famous inequality. am i right?[/quote]

You are right  :)

[quote="kunny"]We can also prove the following inequality:

Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3$ . Prove that $ abc(a \plus{} b \plus{} c)\leq 3$.[/quote]

for any reals $ a,b,c,ab\plus{}bc\plus{}ca\equal{}3$, ineq holds:

$ 2(ab \plus{} bc \plus{} ca)^2 \minus{} 6abc(a \plus{} b \plus{} c) \equal{} \sum{a^2(b \minus{} c)^2}\ge0$

[quote="kunny"]We can also prove the following inequality:

Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3$ . Prove that $ abc(a \plus{} b \plus{} c)\leq 3$.[/quote]

But it is only Newton inequlity for 3 variables..

[quote="kunny"]We can also prove the following inequality:

Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3$ . Prove that $ abc(a \plus{} b \plus{} c)\leq 3$.[/quote]
Another simple way would be like this:
$ (ab\plus{}bc\plus{}ca)^2\geq 3abc(a\plus{}b\plus{}c)\implies 9 \ge 3abc(a\plus{}b\plus{}c)$(Using trivial inequality)
We are done.


I was wondering if anybody could tell me how to get access to korean physics Olympiad questions.No  matter it is in Korean language.

Do you live in Korea? Past KPhO problems are published, so you can buy it at any bookstore in Korea. The following link may help. See 자료실, please.

[url]http://kpho.kps.or.kr/info.asp[/url]


Problem: $a,b \in R^+$ and $n \in N$
If $a^{n+1}+b^{n+1}=a^{n-1}+b^{n-1}$
Prove: $a^2+b^2 \leq 2$

[hide]
This is a problem from the daily problem contest and I was given 0 points for the solution. Can anyone tell me why I was wrong?
My solution:
[quote="shobber"]
Since we already have $a^{n+1}+b^{n+1}=a^{n-1}+b^{n-1}$, so we can just set $n=1$, so:
\[a^2+b^2=a^0+b^0=1+1=2 \leq 2\].
[/quote]
[/hide]

Are you sure you did'nt copy the question wrongly, There's got to be a catch

Well, $n$ is fixed, it might even be 2005 or so :), so you can't assume that $n = 1$.

Hint: use the Chebychev inequality or prove that

[hide]\[2(a^{n + 1} + b^{n + 1}) \geq (a^{n - 1} + b^{n - 1})(a^2 + b^2)\] for all positive reals $a$ and $b$.[/hide]

but what we are trying to prove has got nothing to do with $n$.

arne has a point. the question has given n fixed to you. take it to be 3 etc.

But since we put $n=1$ we get $a^2+b^2=2$ so the problem is solved. We need not put $3,4,5,6 \cdots$ since we do not need $a^3$, $b^2$......

[quote="Arne"]I don't think you understood my post. $n$ can be any FIXED integer such as $2$, $3$, $2005$ or $10^{10^{10}}$ in which case you cannot make that assumption.[/quote]
I see. Thanks!  :)

Yeah, you can't use some n and show it for that n...but you could use induction...
But anyways, i say just multiply and subtract, so you have to show that $a^{n+1}+b^{n+1}\ge a^{n-1}b^2+a^2b^{n-1}$
which can be done


Cool, arent they?

http://www-groups.dcs.st-and.ac.uk/~history/Java/index.html

Yes, very cool =D  Thanks for the link.  My math book makes it seem as if cardioids, limacons, and lemniscates are the only ones.


1. Why are like a lot of posts stickied?
2. Is this a bug?

Obviously you haven't read the READ ME topic ... http://www.mathlinks.ro/Forum/viewtopic.php?t=28406


[quote="The Captain"]Let $f$ be a real valued differentiable function defined on $\mathbb{R}$ such that $\nexists \;x_{0}\; st f(x_{0})=f'(x_{0})=0$. Prove that $f$ has finite number of roots over $[0, 1]$.[/quote]

Let's supose that $f$ has infinitely many roots in $[0,1]$ So in virtute of Bolzano-Weierstass there is a convergent sequence of these roots $\{x_{n}\}_{n\in N^{*}}$. Let $lim_{n\rightarrow \infty}x_{n}=x_{0}$ so becouse f is continuous $f(x_{0})=0$ But $x_{0}$ is an accumulation point of these roots so
$f'(x_{0})=lim_{n\rightarrow \infty}\frac{f(x_{n})-f(x_{0})}{x_{n}-x_{0}}=0$ wich is contradiction with $\not \exists x_{0}$ so that $f(x_{0})=f'(x_{0})=0$


arent there going to be anymore mathlinks contests??
can i hope for one as soon as possible!!
_______________________________________________
myenglish sorry forit@! :(

[quote="Valentin Vornicu"]It's starting up this Monday.[/quote]
thats great!!!!!


If $a,b,c$ are nonnegative numbers such that $a+b+c=3$, then

$\frac{5-3bc}{a+1}+\frac{5-3ac}{b+1}+\frac{5-3ba}{c+1} \geq ab+bc+ca$.

Ok,I think the method I used still work :)

[quote="Vasc"]I think your method doesn't work, because $5-3bc$ isn't positive.[/quote]
Oops :blush:


calculus
function
real analysis
real analysis unsolved

Let $a,b > 0$ . $x,y,z \in R$ satisfy : $x,y,z \in [-a,b]$ and $x+y+z =0$ .
 Find the maximum and minimum value of the expression : $P=x^2+y^2+z^2$

Minimum is 0.
Lemma: if $f$ is a convex function, and $x<y$then $f(a)+f(b)<f(a+t)+f(b-t)$.  for $t<\frac{b-a}{2}$.

Using this lemma we can see that  if $(x_0,y_0,z_0)$ is a triple for which maximum is obtained, then at least two of $x_0,y_0,z_0$ should $\in {-a,b}$. Else we can move two numbers further and maximize $P$. Checking all (4) posibilities you can find out the maximum.


geometry
vector
function
inequalities
analytic geometry
AMC
AIME

I've been working through volume 2 for the sake of preparing for the amc 12. I've done a couple of tests and I feel as if I don't know enough in terms of knowledge to do some of the harder questions. This is especially true in areas such as number theory, and geometry where I barely understand the solutions because of my inexperience using some of these higher level tools. OK back to my story, I've basically read through the first nine chapters thoroughly and about half of the tenth chapter. Chapters 10,11,12 are all about vectors, matrices, and analytic methods. I don't really know how much this will help me so I would like to know which relevant chapters have the [i]most[/i] priority in terms of amc 12 preparation? The reason I'm asking this is because my time is starting to get really tight as I have to do other projects even in winter break.

Thanks in Advance!

Quoting silverfalcon:

High priority: Chapters 1, 2, 3, 4, 6, 9, 13, 15, 16, 17, 19, 23, and Volume I

Mid priority: 5, 7, 10, 11, 12, 14, 18, 24

Low: Everything else

You should definitely know high priority chapters well, and of course areas you are very weak in. But out of mid priority, I really recommend 7, 14, 12, and 24. This is because 7 is fairly useful in function manipulations, 14 is commonly seen on higher maths (and inequalities are interesting), and 24 because diophantine equations are very interesting, and relates to a lot of olympiad maths. Also, in 24, there are a few nice proofs that, just by understanding them, will improve you. 12 is basically helping you get a foundation for applying vectors and coordinates to geo, but it's not too necessary for amc 12 and aime if you are naturally good at geometry.

Analytic geo, vectors, and matrices don't come up on the AMCs too much.  Just skim those chapters.  Most geo problems on the AMC use normal Euclidean geo (especially similar triangles).  Read the combo chapters really carefully though.  There isn't much on combo in Volume 2, but combo comes up on the AMCs a lot.


$P$ is a point in $(O)$. $A,B,C,D$ are 4 points on $(O)$ that $P \in AC$, $P \in BD$ and $(\widehat{AC,BD}) \equiv const$. Find the minimum and the maximum of $S_{ABCD}$ and $P_{ABCD}$  :blush:  :blush:  :D

We all knew this case: $\widehat{AC,BD}=90^{o}$, it is easy  :wink: , but who can solve the general pro.?  :D


reciprocal sum
algebra
Extremal combinatorics
maximization

Given positive integer $n$, find the biggest real number $C$ which satisfy the condition that if the sum of the reciprocals of a set of integers (They can be the same.) that are greater than $1$ is less than $C$, then we can divide the set of numbers into no more than $n$ groups so that the sum of reciprocals of every group is less than $1$.

$C=\frac{n+1}2$. Indeed if we have $n+1$ of $\frac 12$ and divide them into $n$ groups then one will contain at least two numbers that sum to $1$.

For numbers smaller that $\frac{n+1}2$ we try to send every number one by one into some group. Suppose that at a moment we fail to place $\frac 1h$ is any group. Then the sum in any group is at least $1-\frac 1h$ so that total sum if at least $n-\frac{n-1}h \geq \frac{n+1}2$

People have mentioned that this problem is similar to Q5 of IMO 2014.[url]http://artofproblemsolving.com/community/c6h597093p3543144[/url]


I think the text I’m using to teach myself calculus (“Forgotten Calculus” by Barbara Lee Bleau) might be incorrectly setting up an optimization problem.

I will state the problem, show her way of setting it up, and then show mine. I will not consider the actual solution, as I am clear on what to do after the setup.

I’d welcome comment on who is right. If I’m wrong, guidance as to why would be very helpful.

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

The problem

A tour agency has signed up 100 people for a cruise on a ship with a maximum capacity of 150.

A ticket costs $2000. For each $5 that this price is lowered, one new passenger signs up.

Incidental costs to the agency are $500 per passenger. There are also fixed costs of $125,000 for the ship.

By how much should the price be lowered to maximize the agency’s profit?

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

Author’s setup of the total profit formula

When x (apparently; she doesn’t state it explicitly) equals the number of passengers in excess of the original 100, and 0 <= x <= 50,

Profit = Revenue – variable costs – fixed costs
= (price * quantity) – variable costs – fixed cost
= [ (2,000 – 5x)(100 + x) ] – 500x – 125,000

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

My setup of the total profit formula is the same as hers, except in the definition of variable costs.

I’d define variable costs as [500 (100 +x) ], versus her definition of 500x.

My definition multiplies the variable cost of 500 by the total number of passengers; hers seems to multiply the variable cost of 500 by only the number of passengers in excess of 100.

I find this counterintuitive. Why would the addition of any passengers over the original 100 suddenly render the original 100 “costless”?

Yeah, I also think it is counterintuitive. I set it up the same way you did when I read the problem; the author probably just made a mistake (or wanted a certain function to work with).


inequalities
invariant
linear algebra
matrix
linear algebra unsolved

Let $ A, B$ be $ n\times n$ complex matrices. For positive integer $ k\ge 2$, show $ |Tr\{(A^*B)^k\}|^2\le Tr\{(A^*A)^k\}\cdot Tr\{(B^*B)^k\}$.

This is a consequence of a unitarily invariant norm inequality,  see page 81 of X.Zhan, Matrix inequalities, Springer, 2002