trigonometry
function
inequalities unsolved
inequalities

[quote="SuperMath"]Let ABC be a triangle. Prove that:
$\large sin^{\frac{1}{3}}A +sin^{\frac{1}{3}}B + 2 sin^{\frac{1}{3}} \frac{C}{2} \leq\ 2 (4 \sqrt{2})^{\frac{1}{3}}$[/quote]

Let me typset it again
\[ \sin^{1/3}A+\sin^{1/3}+2\sin^{1/3}\tfrac12C\leq 2 (4 \sqrt{2})^{\frac{1}{3}}.  \]
I think we can use the lemma
\[{ \frac{a^3+b^3}2\geq \Bigl(\frac{a+b}2}\Bigl)^3,\;\; a,b\geq0.  \]
Then use the fact that $\sin A+\sin B\leq 2\cos\tfrac12C$. In the next step, we may need calculus.

The function $f(x)=\sin^3 x$ is concave on $[0, \pi]$. Then it follow driectly by Jensen.


trigonometry
superior algebra
superior algebra solved

Let $(a,b,c) \in R^3  or  C^3 $

$A(x)=\left (\begin{array}{ccc}b.(sinx)&-b.sinx.cosx&a.cosx\\-b.cosx.sinx&b.(cosx)&a.sinx\\c.cosx&c.sinx&0\end{array}\right )$

$B=\left (\begin{array}{ccc}0&0&a\\0&b&0\\c&0&0\end{array}\right )$

Prove that $A(x)$ and $B$ are similar for any $x \in R$

Let's try something.

$P=\left
(\begin{array}{ccc}
(\cos x)^2&\sin x\cdot\cos x&0\\
\sin x\cdot\cos x&-(\cos x)^2&0\\
0&0&\cos x
\end{array}\right)$

It can be easily checked that $A\cdot P=P\cdot B$, so if P is invertible then $A=PBP^{-1}$ and A and B are similar.
Because $detP=-(\cos x)^3$, we can conclude that P is invertible iff $\cos x\not=0$. But if $\cos x=0$ then $\sin x=1$ or $\sin x=-1$.

For case $\sin x=1$ we can take $Q=\left (\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right)$ where $Q=Q^{-1}$ and $QAQ^{-1}=B$ so A and B are similar.

For case $\sin x=-1$ take $R=\left (\begin{array}{ccc}0&-1&0\\1&0&0\\0&0&1\end{array}\right)$ then $R^{-1}=\left (\begin{array}{ccc}0&1&0\\-1&0&0\\0&0&1\end{array}\right)$ and $RAR^{-1}=B$ and that's it.

Maja


let $G$ be a $2-$connected nonbipartite graph on $2n$ vertices.Show that the vertex set of  $G$ can be split into two classes of $n$ elements each such that the edges joining the two classes form a connected ,spanning subgraph .

i think this was a Schweitzer problem as well...am not very sure,just that this problem seems familiar.
pierre,can you confirm my suspicion?


By AM-GM we have (a/b) 4 +(b/c) 4 +(c/d) 4 +(d/e) 4  \geq 4(a/e).

Sum the five similar inequalities to get the result.


A total of 119 residents live in a building with 120 apartments. An apartment is defined to be overpopulated if there are 15 or more people living there. Each day the inhabitants of an overpopulated apartment have a quarrel and each goes off to a different apartment in the building. Is it necessarily true that in finitely many days there will be no overpopulated apartments?

This is a [url=http://www.mathlinks.ro/viewtopic.php?p=1226577#1226577]chip-firing game[/url] on the complete graph with $ 120$ vertices.

It's not quite chip-firing game on the complete graph of 120 vertices, since then you would need 120 chips on a vertex to fire it off...


Hi, I don't know if you've seen this or not. This problem is a generalization from a local competition. I'm interested in seeing an elegant solution.

In a mathematical competitions, there are $2n$ problems. A student is "smart" if he solves more than half the problems. A problem is "hard" if it's solved by less than half of smart students. (NOT just half of all students).
Find the maximum number of hard problems.

Are we supposed to be given the number of students?  Also, what is the question from which this is generalized?

No, there is no information about the number of students or the number of smart students.

The original question has 8 problems instead of $2n$ and the answer is [hide]5[/hide].
I suspect, after generalization, the answer would become [hide]$2n-3$[/hide] but my proof is a little bit convoluted with the generalized case.


complex numbers
geometry unsolved
geometry

Let $ A_1A_2 \dots A_8$ be an convex octagon . $ B_1 , B_2 , B_3 , B_4$ are midpoints of $ A_1A_2 , A_3A_4 , A_5A_6$ and $ A_7A_8$ respectively . $ C_1 , C_2 , C_3 , C_4$ are midpoints of $ A_2A_3 , A_4A_5 , A_6A_7 , A_8A_1$ respectively . Prove that if the sum $ PB_1^2 \plus{} PB_2^2 \plus{} PB_3^2 \plus{} PB_4^2$ is minimal , then  $ PC_1^2 \plus{} PC_2^2 \plus{} PC_3^2 \plus{} PC_4^2$ is minimal , too . Where $ P$ is a point from the octagon's plane .

My solution don't use complex numbers. I use median theorem in triangle!
$ 4\sum PB_1^2 \equal{} \sum(2PA_1^2 \plus{} 2PA_2^2 \minus{} A_1A_2^2)$
$ \equal{} 2\sum PA_1^2 \minus{} A_1A_2^2 \minus{} A_3A_4^2 \minus{} A_5A_6^2 \minus{} A_7A_8^2$
Then $ \sum PB_1^2$ min iff and only iff $ \sum PA_1^2$ min!
Similarly then OK!

[quote="meinstein"]Could anyone give the comlex numbers solution please????[/quote]
The solution with complex numbers is very straight foward.
Let $ b_1 \equal{} \frac {a_1 \plus{} a_2}{2}$, etc. and let $ c_1 \equal{} \frac {a_2 \plus{} a_3}{2}$, etc.
$ PB_1^2 \equal{} (p \minus{} b_1)(\overline p \minus{} \overline b_1) \equal{} |p|^2 \plus{} |b_1|^2 \minus{} \overline p b_1) \minus{} p \overline b_1$, etc.
So $ PB_1^2 \plus{} PB_2^2 \plus{} PB_3^2 \plus{} PB_4^2 \equal{}$ $ 4|p|^2 \plus{} |b_1|^2 \plus{} |b_2|^2 \plus{} |b_3|^2 \plus{} |b_4|^2 \minus{} \overline p (b_1 \plus{} b_2 \plus{} b_3 \plus{} b_3) \minus{} p \overline {b_1 \plus{} b_2 \plus{} b_3 \plus{} b_4}$
In the same way $ PC_1^2 \plus{} PC_2^2 \plus{} PC_3^2 \plus{} PC_4^2 \equal{}$ $ 4|p|^2| \plus{} |c_1|^2 \plus{} |c_2|^2 \plus{} |c_3|^2 \plus{} |c_4|^2 \minus{} \overline p (c_1 \plus{} c_2 \plus{} c_3 \plus{} c_3) \minus{} p \overline {c_1 \plus{} c_2 \plus{} c_3 \plus{} c_4}$.

From here it is easy to see that $ (PB_1^2 \plus{} PB_2^2 \plus{} PB_3^2 \plus{} PB_4^2) \minus{} (PC_1^2 \plus{} PC_2^2 \plus{} PC_3^2 \plus{} PC_4^2)$ is constant, since $ b_1 \plus{} b_2 \plus{} b_3 \plus{} b_4 \equal{} c_1 \plus{} c_2 \plus{} c_3 \plus{} c_4$. And then $ PB_1^2 \plus{} PB_2^2 \plus{} PB_3^2 \plus{} PB_4^2$ is minimal whenever $ PC_1^2 \plus{} PC_2^2 \plus{} PC_3^2 \plus{} PC_4^2$ is.


calculus
integration
function
limit
derivative
calculus computations

Prove that

$ \frac {d}{dt} \int_a^b e(x,t) dx \equal{} \int_a^b \frac {\partial e(x,t)}{\partial t} dx$,

where $ a$ and $ b$ are constants (and $ e$ is continuous).

Note that $ \int_a^b e(x,t)dx$ is a function of $ t$, so let $ f(t)=\int_a^b e(x,t)dx$.

By definition \begin{eqnarray*}
\frac{d}{dt}f(t) & = & \lim_{h\to 0}\frac{f(t+h)-f(t)}{h}\\
 & = & \frac{\int_a^b e(x,t+h)dx-\int_a^b e(x,t)dx}{h} \\
 & = & \lim_{h\to 0} \frac{\int_a^b [e(x,t+h)-e(x,t)]dx}{h}\\
 & = & \lim_{h\to 0}\int_a^b\frac{e(x,t+h)-e(x,t)}{h}dx\\
 & = & \int_a^b\lim_{h\to 0}\frac{e(x,t+h)-e(x,t)}{h}dx\\
 & = & \int_a^b\frac{\partial e(x,t)}{\partial t}dx\\
\end{eqnarray*}

Just use linearity of the integral and constant multiple property for the shifting around of the inside, then the inside is a difference quotient which converges uniformly so we can switch limit and integral.

This is often called Leibniz's Rule.

This is just differentiation under the integral isn't it? You only said to assume that $ e$ is continuous, so it may not even have the partial derivative in question. Assuming that $ e(x, t)$ does have a partial derivative with respect to $ t$ and given some $ \varepsilon > 0$, choose $ h$ small enough that
\[ \left|\frac {e(x, t \plus{} h) \minus{} e(x, t)}{h} \minus{} \frac {\partial}{\partial t}e(x, t)\right| < \varepsilon/(b \minus{} a).
\]
This gives us that
\[ \left|\frac {\int^b_ae(x, t \plus{} h)\mathrm{d}x \minus{} \int^b_ae(x, t)\mathrm{d}x}{h} \minus{} \int^b_a\frac {\partial}{\partial t}e(x, t)\mathrm{d}x\right| < \ldots,
\]
and the rest follows by the choice of $ h$.

EDIT: Woops, beat me to it. I like this way better because it keeps from having to justify bringing the limit inside the integral.


inequalities
induction
USAMTS
floor function
algorithm
symmetry
geometry

Post your solutions here! Comment on anything.

edit: oops what am I thinking, we normally post solutions after the round ends, not when scores come out  :blush:  :blush:  :blush:

Can anybody tell me why i got a 2 of #5?

[hide="solution"]Consider the sequence of $A_{i}$ defined by $A_{2n+1}=A_{2n}=n+1$ and $A_{1}=1$ for all positive integers $n$. The sequence starts like 1,2,2,3,3,4,4,5,5,... We show that any $A_{i}$ is at its minimum possible value.

Lemma. $x_{n}\neq x_{n+1}$ for any positive integer $n$

Proof. This is equivalent to $(n+1)A_{n+1}-nA_{n}\neq nA_{n}-(n-1)A_{n-1}$ for any integer $n>1$. From the way we defined our sequence, either $A_{n+1}=A_{n}=A_{n-1}+1$ or $A_{n+1}=A_{n}+1=A_{n-1}+1$. In either case, rewriting all the $A_{i}$ in terms of $A_{n}$ gives the desired result.

We prove that $A_{n}$ is at its minimum possible value and $A_{n}<n+1$ using induction. Note that this holds for $n=1,2$. Now assume that $A_{i}$ is at its minimum and $A_{i}<i+1$ for all $i\leq n$ where $n\geq2$. Consider another sequence $B_{i}$ such that $B_{n+1}<A_{n+1}$.  Since $(n+1)B_{n+1}-nB_{n}>0$, we must have
\[A_{n+1}>B_{n+1}>\frac{n}{n+1}B_{n}\geq\frac{n}{n+1}A_{n}\]
Also, $n+1>A_{n}\Rightarrow nA_{n}>nA_{n}+A_{n}-(n+1)\Rightarrow\frac{n}{n+1}A_{n}>A_{n}-1$. Thus,
\[A_{n+1}>B_{n+1}>A_{n}-1\]
If $A_{n+1}=A_{n}$, we have a contradiction. Otherwise, $A_{n+1}=A_{n}+1$, and $B_{n+1}=B_{n}=A_{n}=A_{n-1}$. Then $B_{n-1}=B_{n}$ implies $(n+1)B_{n+1}-nB_{n}=nB_{n}-(n-1)B_{n-1}$ and $B_{n-1}<B_{n}$ implies $B_{n-1}<A_{n-1}$ contradicting our hypothesis, so $B_{n-1}>B_{n}$. But then,
\[nB_{n}-(n-1)B_{n-1}\leq nB_{n}-(n-1)(B_{n}+1)=B_{n}-(n+1)=A_{n}-(n+1)<0\]
a contradition. Furthermore, $A_{n+1}\leq A_{n}+1<(n+1)+1$ and our assertion is proved.

Finally, $A_{2007}=A_{2\times1003+1}=1004$[/hide]

[quote="Jiang"]Can anybody tell me why i got a 2 of #5?[/quote]
In your step,
\[A_{n+1}>B_{n+1}>\frac{n}{n+1}B_{n}\geq\frac{n}{n+1}A_{n}\]
how did you get that last inequality $\frac{n}{n+1}B_{n}\geq\frac{n}{n+1}A_{n}$? It comes out of nowhere, but it is the key to your proof.

You did not define your sequence $B_{i}$ clearly, but I can guess what you wanted. You defined the sequence $A_{i}$ as $A_{2i+1}=A_{2i}=i+1$ (and you proved it met the conditions---good work there). The sequence $B_{i}$ is supposed to be another sequence satisfying the conditions that also has $B_{n+1}<A_{n+1}$. So you are mixing a proof by induction with a proof by contradiction.

Your unproven inequality is supposed to derive from the induction hypothesis: since you have proven your induction hypothesis for $k<n$, this means that for $k=n-1$, you have proven that every sequence $B_{i}$ has $B_{n}\geq A_{_}n$; therefore, $\frac{n}{n+1}B_{n}\geq\frac{n}{n+1}A_{n}$.

But you did not mention this in your proof. In fact, you never stated an induction hypothesis. Nor did you clearly define that the sequence $B_{i}$ was a different sequence for each choice of $n$, which is necessary for the induction to work. Proofs by induction require careful setup, and you skipped that setup. Then when you mixed it with a proof by contradiction, it became impossible to tell what assumption was contradicted.

You earned two points for having the correct sequence and answer, and for validating that the sequence met the conditions. I presume that your grader could not make sense of your proof and gave you zero points for it. I would have given you one point for having a correct proof in your head, even if it did not make it onto your paper, but I am more generous about a submission's unspoken ideas than most graders are.

Erin Schram
USAMTS grader

my induction hypothesis was supposed to be "Now assume that $A_{i}$ is at its minimum and $A_{i}<i+1$ for all $i\leq n$ where $n\geq2$." The only problem i see is that there was a confusion as to what the sequence $B_{i}$ was. This sequence was supposed to be a sequence like $A_{i}$ that satisfied all the conditions stated in the problem. This sequence was made for the sake of contradition that $B_{i}$ cannot satisfy the conditions in the problem. Personally, after defining $B_{i}$, i dont see any problem with the proof and the inequality is obvious from the induction hypothesis. I think i should deserve at least a 3 and i am really desperate since i am at a 94 right now.

As long as we're asking about solutions, where is my solution for #5 unclear enough to deserve a 4? (I realize that it's not perfectly well-organized, but it seems rigorous to me...)
[hide]First, we will prove that the sequence $A_{1},A_{2},A_{3},...$ cannot have three terms in a row with the same value. Assume that $A_{k}=A_{k+1}=A_{k+2}$ for some $k$. From the first equality, we can write
\[A_{k}=A_{k+1}=\frac{x_{1}+x_{2}+...+x_{k+1}}{k+1}\]
Since we know $kA_{k}=x_{1}+x_{2}+...+x_{k}$, this implies
\[A_{k}=\frac{kA_{k}+x_{k+1}}{k+1}\]
\[(k+1)A_{k}=kA_{k}+x_{k+1}\]
\[A_{k}=x_{k+1}\]
We now have $A_{k+1}=A_{k}=x_{k+1}$. However, from $A_{k+1}=A_{k+2}$, we can write
\[A_{k+1}=A_{k+2}=\frac{x_{1}+x_{2}+...+x_{k+2}}{k+2}\]
\[A_{k+1}=\frac{(k+1)A_{k+1}+x_{k+2}}{k+2}\]
\[A_{k+1}=x_{k+2}\]
Then $x_{k+1}=A_{k+1}=x_{k+2}$, which is not allowed. So $A_{1},A_{2},...$ can have at most two equal terms in a row.

We will now prove that the sequence $A_{1},A_{2},...$ which gives the minimum possible value for $A_{n}$ is nondecreasing. If $A_{n}<A_{n-1}$, then we have
\[A_{n}-A_{n-1}<0\]
\[\frac{x_{1}+x_{2}+...+x_{n-1}+x_{n}}{n}-A_{n-1}<0\]
\[\frac{(n-1)A_{n-1}+x_{n}}{n}-A_{n-1}<0\]
\[A_{n-1}+\frac{x_{n}-A_{n-1}}{n}-A_{n-1}<0\]
\[\frac{x_{n}-A_{n-1}}{n}<0\]
Since the LHS is less than 0, we have $A_{n-1}>x_{n}$. Since the LHS is $A_{n}-A_{n-1}$, an integer, $n$ must divide $x_{n}-A_{n-1}$, implying that $A_{n-1}$ is at least $x_{n}+n$. Then
\[A_{n}=\frac{(n-1)A_{n-1}+x_{n}}{n}\geq\frac{(n-1)(x_{n}+n)+x_{n}}{n}=\frac{nx_{n}+n(n-1)}{n}=x_{n}+n-1\geq 1+n-1=n\]
So in order for $A_{n}$ to decrease from $A_{n-1}$, $A_{n}$ must be at least $n$. So no sequence where $A_{n}$ decreases from $A_{n-1}$ yields a value for $A_{n}$ less than $n$. Furthermore, if $A_{k}$ is the last term before $A_{n}$ that decreases from $A_{k-1}$, so that all terms after it DON'T decrease from the previous term, we have $A_{k}\geq k$ and the least possible values of the terms following $A_{k}$ are $A_{k+1}=k, A_{k+2}=k+1, A_{k+3}=k+1,...,A_{n}=k+\lfloor (n-k)/2\rfloor$ (recalling that no sequence for $A_{n}$ has more than two of the same terms in a row). So no sequence with a decrease anywhere will yield a lower value for $A_{n}$ than $k+\lfloor (n-k)/2\rfloor$, which is at least $k+(n-k-1)/2=(n+k-1)/2$, which is at least $(n+1)/2$, which in the case of $n=2007$ is 1004.

Now consider the nondecreasing possibilities for the sequence $\{A_{n}\}$. The sequence with the least possible value for any $A_{n}$ is clearly $1,2,2,3,3,...,\lceil(n+1)/2\rceil$ (note that $A_{2}$ can't be 1 because that would result in $x_{1}=x_{2}=1$). This sequence is achievable with the sequence $x_{n}$ as $1,3,2,6,3,9...$. When $n$ is even, we have $nA_{n}=1+3+2+6+...+n/2+3n/2=4+8+..+2n=(4+2n)(n/2)/2=n(n+2)/2$ and $A_{n}=(n+2)/2=\lceil(n+1)/2\rceil$ using the arithmetic series formula. When $n$ is odd, we have $nA_{n}=1+3+2+6+...+(n-1)/2+3(n-1)/2+(n+1)/2=(n-1)(n+1)/2+(n+1)/2=n(n+1)/2$ and $A_{n}=(n+1)/2=\lceil(n+1)/2\rceil$. This sequence gives $A_{2}007=(2007+1)/2=1004$, which is also a lower bound on what we could get with any decreasing sequence $\{A_{n}\}$. As a result, we can conclude that the least possible value of $A_{2007}$ is 1004.[/hide]

Jiang and bookaholic, let me make clear that I have no authority to change scores or suggest scores. I am a grader for the USAMTS, but the grading session is over. Besides, the opinions of graders on the score of a solution vary by a point up or down, so even at the grading table I cannot override the score given by another grader.

I am one of the most experienced USAMTS graders, having graded it since Year 6 and having missed only one grading session in that time. But that does not make me any better at assigning scores than any other grader. I am just more practiced at figuring out what a badly written solution really meant to say. Or in understanding a solution that uses such an advanced technique that it confuses an average mathematician.

[quote="Jiang"]my induction hypothesis was supposed to be $\ldots$.  I think i should deserve at least a 3 and i am really desperate since i am at a 94 right now.[/quote]

Stating your induction hypothesis now is too late to help your submission to problem 5. The grader had to go by what was on your paper.

[quote="bookaholic"]As long as we're asking about solutions, where is my solution for #5 unclear enough to deserve a 4? (I realize that it's not perfectly well-organized, but it seems rigorous to me...)[/quote]

I don't know why your solution received a 4. It is a greedy algorithm solution. The rubric says that if a solution has no other proof than use of the greedy algorithm, it earns a 3. But you actually proved that the greedy algorithm gives the correct answer, so your solution has a complete and rigorous proof, which deserves a 5.

Four points is the score for a workable proof that has holes in it. The organization of your proof is not tight, but that is not the same things as having holes.

Erin Schram
USAMTS grader

[quote="Erin J. Schram"]I am just more practiced at figuring out what a badly written solution really meant to say. Or in understanding a solution that uses such an advanced technique that it confuses an average mathematician.
[/quote]
Can we request you to be our grader :D

I guess I'll jump on the bandwagon here, and ask if anyone sees the problem with this solution (which received a 4).  I'm more wondering about how to avoid getting points off in the future than lamenting about the lost one here.  Thanks!

Label the midpoints of the sides of $ABCD$ as shown [I included the diagram, but if you can't see it, all that I added was $A'$ as the midpoint of $AB$, $B'$ as the midpoint of $BC$, $C'$ as the midpoint of $CD$, and $D'$ as the midpoint of $DA$].  We will prove that $[OA'BB'] = \frac{1}{4}[ABCD]$; the desired result will follow immediately by symmetry.

Note first of all that $BA'NB'$ is similar to $ABCD$ (if this is not clear, it follows easily from noting that $BNB' \sim BDC$ and $BNA' \sim BDA$) with scale factor $\frac{1}{2}$.  Hence,
\[[BA'NB'] = \frac{1}{4}[ABCD].\]
But note that $AC \parallel NO$, and since $A'$ and $B'$ are midpoints of sides $AB$ and $CB$ in triangle $ABC$, $AC \parallel A' B'$.  Thus,
\[NO \parallel A'B',\]
and for any point $X$ on $NO$,
\[[A'XB'] = [A' N B'].\]
In particular, let $X = O$.  Then we have that
\[\begin{array}{lll}[OA'BB'] & = & [A'B B']+[A'O B']\\ & = & [A' B B']+[A' N B']\\ & = & [B A' N B']\\ & = & \frac{1}{4}[ABCD],\\ \end{array}\]
as desired.

My only guess is that you didn't prove that $NO \parallel A'B' \Longrightarrow [A'XB'] = [A' N B']$, for any point $X$ on $NO$.

I think that that is fairly obvious and you did a good job on your solution.

My solution for this problem involved an assumption that wasn't proven anywhere in my proof, yet I still recieved a 5. I guess my grader was more lenient or less observant than yours.

[quote]Jiang and bookaholic, let me make clear that I have no authority to change scores or suggest scores. I am a grader for the USAMTS, but the grading session is over. Besides, the opinions of graders on the score of a solution vary by a point up or down, so even at the grading table I cannot override the score given by another grader.[/quote]
I'm not worried about changing my score. I just didn't want a possible flaw in my thinking/writing to stay uncorrected until I make it again. I'm willing to accept getting a point off for style, since the grading protests forum explicitly says that it is possible. Thanks so much for the help :)

[quote="gregx007"]I guess I'll jump on the bandwagon here, and ask if anyone sees the problem with this solution (which received a 4).  I'm more wondering about how to avoid getting points off in the future than lamenting about the lost one here.  Thanks![/quote]

It looks fine to me. Problem 3 was not graded at the session I attended, so I know nothing about its rubric beyond what we all can read in the Grading Issues subforum, and that rubric says little about which weaknesses can cost one point.

If I wanted to improve the proof, I would generalize from the one proven corner to all four corners at the end instead of at the beginning. The reasoning would flow better there. You used the phrase, "by symmetry," which isn't exactly correct since the sides are not symmetrical. The word "likewise" works best here: "Likewise, we can construct an identical proof with the other three pieces of ABCD, so all four pieces have area $\frac{1}{4}[ABCD]$." (The word "likewise" is a great tool for shortening a proof with nearly identical cases.)

And, as Xantos C. Guin said, you did gloss over the reason why $NO \parallel A'B'$ leads to $[A'XB'] = [A' N B']$. A sentence about the triangles having the same height above base $A'B'$ would fill in that gap.

Erin Schram
USAMTS grader

[quote="ehehheehee"]Did anybody else use vectors and cross products on #3?[/quote]
Haha, I did! What's sad is that I didn't even expand the cross products in terms of vectors A, B, C, D, etc. I somehow got the idea that cross products aren't distributive, so I actually set vec(A)=(a_1 a_2 a_3 a_4), etc., and expanded in terms of the individual components. It was horrible. I still got a 5, though.

[quote="gman"][quote="ehehheehee"]How would you know which details to gloss over (because it is self explanatory or trivial) or which details to emphasize?

Did anybody else use vectors and cross products on #3?[/quote]

I did.  Here, I attached it.  Tell me, what score do you think I deserve?[/quote]

I hadn't expected a vector solution to Problem 3. You people are so inventive.

Gman's proof is correct. It isn't perfect. The main flaw is that he said, "Medians $OA_{1}$, $OB_{1}$, $OC_{1}$, and $OD_{1}$ divide triangles $AOB$, $BOC$, $COD$, and $DOA$ into two congruent triangles each," when the paired triangles are only equal-area not congruent. He didn't use any other properties of congruence, so that could just be a bad choice of words rather than a geometric mistake. The writing would also benefit from a few more words of explanation between equations, especially at the end.

Erin Schram
Feeling way too much like an art critic

LOL love that last comment.

Just curious, how do you guys correctly solved 5?? I use induction+ bizillion contradiction.  Is that the main stream proof?

[quote="Erin J. Schram"]I hadn't expected a vector solution to Problem 3. You people are so inventive.

Gman's proof is correct. It isn't perfect. The main flaw is that he said, "Medians $OA_{1}$, $OB_{1}$, $OC_{1}$, and $OD_{1}$ divide triangles $AOB$, $BOC$, $COD$, and $DOA$ into two congruent triangles each," when the paired triangles are only equal-area not congruent. He didn't use any other properties of congruence, so that could just be a bad choice of words rather than a geometric mistake. The writing would also benefit from a few more words of explanation between equations, especially at the end.

Erin Schram
Feeling way too much like an art critic[/quote]

OK.  I admit that I messed up in my wording.  Calling triangles congruent instead of equal-area is a horrible mistake on my part.  Did my terseness and that one mis-wording really justify a docking of 3 points?

[quote="gman"]OK.  I admit that I messed up in my wording.  Calling triangles congruent instead of equal-area is a horrible mistake on my part.  Did my terseness and that one mis-wording really justify a docking of 3 points?[/quote]
I recommend that you post in the Grading Issues subforum and get an official opinion.

Erin Schram
USAMTS grader

[quote="Erin J. Schram"]I recommend that you post in the Grading Issues subforum and get an official opinion.

Erin Schram
USAMTS grader[/quote]

Thank you for your suggestion, Ms. Schram.  I just did this.


A very nice hard puzzle :)
9 people hold 5119 shares in a company. In every decision voting, any subset or all of the 9 people can participate.If a person participates in voting, he/she can either vote FOR or AGAINST the decision. The number of votes is equal to the number of shares a person holds. For every decision there shouldn't be a TIE between the two choices. In a decision, a group with smaller number of people should never win over higher number of people. What is the least number of shares a person can hold?

Let the number of shares for each person i be $a_i$ where $1\leq i \leq9$ and i integral.  WLOG assume that $a_1<a_2...<a_9$

We know that $a_i \neq a_j$ otherwise there would be a tie.  Since a smaller number of people should never outvote a higher number of people, we say that

$a_9<a_1+a_2$

But in order to minimise $a_1$ we need to maximise $a_9$ and this occurs when

$a_9=a_1+a_2-1$

By a similar argument,
$a_8=a_9-1=a_1+a_2-2$

This shows that all of $a_2,a_3...a_9$ are consecutive.  With a little experimentation, one determines that $a_9=642, a_8=641,...$, and since the sum is 5119, we conclude that the smallest number of shares is $11$


We have $f,g: \left[0,1\right]\rightarrow\left[0,1\right]$ two continuous function and $f\circ\(g=g\circ\(f$.If $f\uparrow$ then prove that there is $\xi\in\left[0,1\right]$ which $f(\xi)=g(\xi)=\xi$.

You could have said "monotonous" instead of "increasing" here.
(in what follows I replaced your $\xi$ with $a$)

1st case : $f$ decreasing.
The function $h : x \mapsto f(x)-x$ is continuous and strictly decreasing and since it's zero at least once (IVT), it's in fact zero only once for some $a \in [0, 1]$ and since $h(g(a)) = 0,$ we get that $g(a) = a.$

2nd case : $f$ increasing.
Let $S = \{x \in [0, 1] / g(x) = x \}.$ It is then easy to see that $S$ is non-empty and stable under $f.$
For $s \in S$ let's consider the recurrent sequence defined by $x_{0}= s$ and $x_{n+1}= f(x_{n}).$ This sequence is monotonous hence convergent towards some $a \in [0, 1].$
Now, the continuity of $f$ and $g$ yields $f(a) = g(a) = a.$

 :idea: If we remove the monotony hypothesis, the conclusion doesn't hold anymore. (but this is [b]very[/b] hard)


[color=darkred]Denote the [u][b]Lemoin's point[/b][/u] $S$ of the triangle $ABC\ (b\ne c)$ and $S_{a}\in AS\cap BC$.

The incircle $C(I)$ touches the sides in the points $D\in BC$, $E\in CA$, $F\in AB$.

Denote $L\in EF\cap BC$. Prove that the following sentencies are equivalently :

$\|\begin{array}{ccc}\odot & \blacktriangleright & \boxed{\ a(b+c)=b^{2}+c^{2}\ }\\\\ 1 & \blacktriangleright & OI\perp AD\\\\ 2 & \blacktriangleright & S_{a}\equiv D\\\\ 3 & \blacktriangleright & IS\parallel BC\\\\ 4 & \blacktriangleright & I\in LO\end{array}$[/color]

very,very nice problem!

first we shall prove that $a(b+c)=b^{2}+c^{2}\Longleftrightarrow D \equiv S_{a}$ :

$D \equiv S_{a}\Longleftrightarrow \frac{DB}{DC}=\frac{c^{2}}{b^{2}}\Longleftrightarrow \frac{c^{2}}{b^{2}}=\frac{p-b}{p-c}\Longleftrightarrow c^{2}a+c^{2}b+b^{3}=b^{2}a+b^{2}c+c^{3}\Longleftrightarrow a(b+c)=b^{2}+c^{2}$

now if $AD$ is the symmedian of $ABC$, then from the fact that $(LBDC)$ is harmonically, then $AL$ is tangent in $A$ to the circumcircle, so $AD$ is the polar of $L$ w.r.t to the circumcircle, therefore $LO \perp AD (\star)$

but $EF$ cuts $BC$ in $L$ as well, therefore $AD$ is the polar of $L$ w.r.t to the incircle , so $LI \perp AD (\star\star)$

from $(\star)$ and $(\star\star)$, we have $I \in LO$, and $OI \perp AD$

it remais us to prove the last equivalence: $AD$ symmedian $\Longleftrightarrow IS \| BC$

for this i have a pretty ugly vectorial solution, i will only sketch it, here it goes :

we have $ID \perp BC$, so for proving $IS \| BC$, we need to prove $ID\perp IS \Longleftrightarrow \overrightarrow{ID}\cdot \overrightarrow{IS}=0$, and now knowing $\overrightarrow{r_{S}}=\frac{a^{2}}{a^{2}+b^{2}+c^{2}}\overrightarrow{r_{A}}+...$etc,
and $\overrightarrow{r_{I}}=\frac{a}{a+b+c}\overrightarrow{r_{A}}+...$etc,
and $\overrightarrow{r_{D}}=\frac{b^{2}}{b^{2}+c^{2}}\overrightarrow{r_{B}}+\frac{c^{2}}{b^{2}+c^{2}}\overrightarrow{r_{C}}$
and following our relation $a(b+c)=b^{2}+c^{2}\Longleftrightarrow a(a+b+c)=a^{2}+b^{2}+c^{2}$ , the conclusion follows after a not very hard calculus.


Calculate $ S=2^{n}+2\cdot 2^{n-2}+3\cdot 2^{n-3}+\cdots+\left(n-1\right)\cdot 2+n$


[color=purple] Oh i have an interresting solution for a similar of your problem. That is
  $\sum\sqrt{a(b+c)}{a^{2}+bc}\geq\sqrt{2+2\sqrt{1+4\sqrt{\frac{abc\prod(a+b)}{\prod(a^{2}+bc)}}}}$
   Anh Can anh co' can` em post loi` giai? ko a? (Anh co' biet' cach' go~ tieng' Viet ko day em voi')
@ Anh Can  I think in MnF there are many hard problems about Vu Dinh Quy wiating for you you should come back MnF and kill it [/color] :D

[quote="evarist"][color=purple] Oh i have an interresting solution for a similar of your problem. That is
  $\sum\sqrt{\frac{a(b+c)}{a^{2}+bc}}\geq\sqrt{2+2\sqrt{1+4\sqrt{\frac{abc\prod(a+b)}{\prod(a^{2}+bc)}}}}$
   Anh Can anh co' can` em post loi` giai? ko a?
@ Anh Can  I think in MnF there are many hard problems about Vu Dinh Quy wiating for you you should come back MnF and kill it [/color] :D[/quote]
Oh, it is also the problem of mine (i have posted it in Mathnfriend forum, do you remember, in box "bat dang thuc doi xung" of Hoang Minh)

@: Sorry, but i don't want to return Mnf because i have some troubles with a member. Sorry. :)

[quote="evarist"][color=purple]Oh you said your solution by "Mixing Variables" didn't you? My solution for  2nd problem may be use for the 1st problem.could you give me your solution by "  "? Thanks[/color][/quote]

Sorry, but i didn't say my solution is use "Mixing variables". You should see my solution here: http://diendantoanhoc.net/forum/index.php?showtopic=22119. :lol:

[color=purple]Sorry.My computer can't read PDF file.To solve your inequalities we must prove two weaker problems.Here it is
 $\sum\sqrt{\frac{a(b+c)}{b^{2}+c^{2}}}\geq 2$
and $\sum\sqrt{\frac{a(b+c)}{a^{2}+bc}}\geq 2$
  The following solution are your aren't they?[/color]


[quote="Cartman"]A very special clock has one minute hand, and one hour hand. The hands are identical. When is it possible to tell what time it is?[/quote]

Whenever you stick around long enough to see that the minute hand is moving faster...

Or whenever you have a different clock that works. :wink:

[quote="randomdragoon"]All the time? One assignment of the minute and hour hand will usually be inconsistent with clock physics (ex. at 3:00, it can't be 12:15 because the hour hand would have moved past the 12 already)[/quote]

Except at the precise striking of the hours, in which case the hour hand would be at the numbers.

[quote="Cartman"]A very special clock has one minute hand, and one hour hand. The hands are identical. When is it possible to tell what time it is?[/quote]

Nearly allways :

Let $ t\in[0,12[$ be the time (in hours).
Let $ h\in[0,1[$ the angle made by the hour hand (in units of $ 2\pi$)
Let $ m\in[0,1[$ the angle made by the minute hand (in units of $ 2\pi$)

We have $ m \equal{}\{t\}$ (where $ \{x\}$ is the fractional part of x) and $ h \equal{}\frac{t}{12}$

So, if we see two angles $ a$ and $ b$ (without knowing which is hour and which is minute), we can't know time only if $ \exists t_{1}\neq t_{2}\in[0,12[$ such that :

$ a \equal{}\{t_{1}\}\equal{}\frac{t_{2}}{12}$ and $ b \equal{}\{t_{2}\}\equal{}\frac{t_{1}}{12}$

$ \Leftrightarrow$ $ a \equal{}\{12b\}$ and $ b \equal{}\{12a\}$, $ a\neq b$
$ \Leftrightarrow$ $ \exists m,n\in\{0,1,2,3,4,5,6,7,8,9,10,11\}$ such that $ a \equal{} 12b\minus{}m$ and $ b \equal{} 12a\minus{}n$, $ a\neq b$
$ \Leftrightarrow$ $ \exists m,n\in\{0,1,2,3,4,5,6,7,8,9,10,11\}$ such that $ a \equal{}\frac{12n\plus{}m}{143}$ and $ b \equal{}\frac{12m\plus{}n}{143}$, $ a\neq b$

So, it is possible to know exctly what time (in the half day) it is at every time except the $ 132$ instants where $ a \equal{}\frac{12n\plus{}m}{143}$ and $ b \equal{}\frac{12m\plus{}n}{143}$, $ m\neq n\in\{0,1,2,3,4,5,6,7,8,9,10,11\}$

At each of these $ 66$ positions $ (\frac{12n\plus{}m}{143},\frac{12m\plus{}n}{143})$, it is impossible to make the distinction between $ n$ hours $ 60\frac{n\plus{}12m}{143}$ minutes or $ m$ hours $ 60\frac{m\plus{}12n}{143}$ minutes

For example, with $ n \equal{} 0$ and $ m \equal{} 2$, it is impossible to distinguish between $ 0h\frac{1440}{143}min$ and $ 2h\frac{120}{143}min$


function
trigonometry
calculus
integration
ratio
logarithms
calculus computations

Honestly, I never learned this whole "series" thing and even I'm trying to learn it now, it's difficult for me. I did learn P-series while doing some AP problems though  :D . Please give explanations on these; they'll help me learn about series much better than just reading off from the book.

[hide="Answers"]
11. D
20. D [/hide]

11. The Maclaurin series for $\frac{1}{1-x}$ is $\sum^{\infty}_{n=0}x^{n}$. Which of the following is a power series expansion for $\frac{x^{2}}{1-x^{2}}$?

$\text{(A)}\ 1+x^{2}+x^{4}+x^{6}+x^{8}+\cdots$
$\text{(B)}\ x^{2}+x^{3}+x^{4}+x^{5}+\cdots$
$\text{(C)}\ x^{2}+2x^{3}+3x^{4}+4x^{5}+\cdots$
$\text{(D)}\ x^{2}+x^{4}+x^{6}+x^{8}+\cdots$
$\text{(E)}\ x^{2}-x^{4}+x^{6}-x^{8}+\cdots$

20. A function $f$ has Maclaurin series given by $\frac{x^{4}}{2!}+\frac{x^{5}}{3!}+\frac{x^{6}}{4!}+\cdots+\frac{x^{n+3}}{(n+1)!}+\cdots$. Which of the following is an expression for $f(x)$

$\text{(A)}\-3x \sin x+3x^{2}$
$\text{(B)}\-\cos(x^{2})+1$
$\text{(C)}\-x^{2}\cos x+x^{2}$
$\text{(D)}\ x^{2}e^{x}-x^{3}-x^{2}$
$\text{(E)}\ e^{x^{2}}-x^{2}-1$

22. What are all values of $p$ for which the infinite series $\sum^{\infty}_{n=1}\frac{n}{n^{p}+1}$ converges?

This problem's solution used something called Direct Comparison Test to say that:

$\frac{n}{n^{p}+1}$ is convergent since $\frac{n}{n^{p}+1}< \frac{n}{n^{p}},$ which is convergent. I understand that $\frac{n}{n^{p}}= \frac{1}{n^{p-1}}$ is convergent for $p > 2$ but how can you deduce that $\frac{n}{n^{p}+1}$ is convergent as well? I just don't really understand Direct comparison Test. Can anyone explain this?

Thanks for your help!  :)

11.  Let $f(x) = \frac{1}{1-x}$.  Then $\frac{x^{2}}{1-x^{2}}= x^{2}f(x^{2})$.

20.  Here it would be good to know that $e^{x}= \frac{x^{0}}{0!}+\frac{x^{1}}{1!}+\frac{x^{2}}{2!}+...$.  But the desired series has exponents two higher than that of $e^{x}$, and the first two terms are missing; hence the desired series is given by

$x^{2}e^{x}-x^{2}-x^{3}$

22.  $\frac{n}{n^{p}}> \frac{n}{n^{p}+1}$

The Direct Comparison Test states that for two non-negative sequences $a_{i}, b_{i}$, if $\sum a_{i}$ converges and $b_{i}\le a_{i}\forall i$ then $\sum b_{i}$ converges as well, which is obvious since

$0 \le \sum b_{i}\le \sum a_{i}$

11. Divide the top and bottom by $-x^{2}$
to get $(-1)\frac{1}{1-x^{2}}$. Can you use the geometric series to finish this off?

The direct comparison test basically states that if the terms of a series are smaller than or equal to those of a convergent series, that series converges; if the terms of a series are larger than or equal to those of a divergent series, that series converges.
(Errr... one may have to adjust that statement if one has a series with negative terms...)

11. I'm still confused on this. Sorry. I'm really new to series.  :maybe:

20. Is there any other way than using knowledge of $e^{x}= \frac{x^{0}}{0!}+\frac{x^{1}}{1!}+\frac{x^{2}}{2!}+\cdots$?

22. Ok. I think I understand it. Um then is this right?

Say I have $\sum^{\infty}_{n=1}\frac{1}{n^{2}}$, which is convergent. Then does this mean that $\sum^{\infty}_{n=1}\frac{1}{n^{2}+2}$ is convergent as well?

[quote="Zakary"](Errr... one may have to adjust that statement if one has a series with negative terms...)[/quote]
Let's get this very clear: [b]all[/b] comparison tests - and the list of comparison tests includes "direct" comparison, limit comparison, the integral test, the condensation test, the ratio test and the root test - are tests for [b]absolute[/b] converges. Or to say it another way: they are tests for convergence of positive-term series. Faced with a series of terms of mixed size, you should start by taking absolute values and asking about absolute convergence.

As for "direct" comparison: when I teach it I usually call it "one-sided" comparison.

An example: for $n\ge3,\ \ln n>n.$ Thus $\sum_{n=3}^{\infty}\frac1{n^{2}\ln n}\le\sum_{n=3}^{\infty}\frac1{n^{2}}.$

Since $\sum\frac1{n^{2}}$ converges, we conclude that $\sum{frac1{n^{2}\ln n}}$ converges.

But even though we can say that $\sum_{n=3}^{\infty}\frac1{n\ln n}\le\sum_{n=3}^{\infty}\frac1n,$ that doesn't contain any information: being smaller than a divergent series leads to no conclusion.

(We can settle this case - it diverges - by the integral test or the condensation test.)

To determine [i]exactly[/i] where $\sum{n=1}^{\infty}\frac{n}{n^{p}+1}$ converges, we need a two-sided argument. In particular, this calls for the use of the [b]limit comparison test[/b], the "all-purpose de-clutterer." As $n\to\infty,$ $\frac{n}{n^{p}+1}$ is the same size as $\frac{1}{n^{p-1}}$ ("same size" in the sense that a quotient tends to a nonzero constant), so the series converges when $p>2$ and diverges otherwise.

And as for Silverfalcon's last question: yes, $\frac1{n^{2}+2}$ - or for that matter, $\frac1{n^{2}-2}$ is the "same size" as $\frac1{n^{2}}$ so the series in question converges by the limit comparison test.

[quote="Silverfalcon"]11. I'm still confused on this. Sorry. I'm really new to series.  :maybe: [/quote]

If we are given that

$f(x) = \frac{1}{1-x}= 1+x+x^{2}+x^{3}+x^{4}+...$

Then it follows that

$f(x^{2}) = \frac{1}{1-x^{2}}= 1+x^{2}+x^{4}+x^{6}+x^{8}+...$

And therefore that

$x^{2}f(x^{2}) = \frac{x^{2}}{1-x^{2}}= x^{2}+x^{4}+x^{6}+x^{8}+x^{10}+...$

[quote="Silverfalcon"]20. Is there any other way than using knowledge of $e^{x}= \frac{x^{0}}{0!}+\frac{x^{1}}{1!}+\frac{x^{2}}{2!}+\cdots$?[/quote]

Not really.  Calculus students are expected to be familiar with several common series expansions (the other two you should know are $\sin x, \cos x$.  It would not be a bad idea to learn $\ln (1-x)$ either).

[quote]Calculus students are expected to be familiar with several common series expansions (the other two you should know are $\sin x, \cos x.$ It would not be a bad idea to learn $\ln (1-x)$ either).[/quote]
My list, issued as a handout in my second-semester calculus class, is entitles "Seven power series to memorize." All seven are series centered at $x=0.$ The list:

$\frac1{1-x}$ (The geometric series.)

$\ln(1+x)$

$\arctan(x)$

(You can think of these first three as a closely related family. In particular, we obtain the logarithm and arctangent series by integrating geometric series.)

$e^{x}$

$\cos x$

$\sin x$

(These three also form a closely related family. $\cosh x$ and $\sinh x$ could also have been included but are left off to keep the list from being too long.)

$(1+x)^{\alpha},\ \alpha$ arbritrary. (The binomial series - Newton's favorite.)

t0rajir0u, thanks. I should've gotten it.  :blush:

Professor Merryfield (I think this is more appropriate title than just Mr. Merryfield), do those seven power series refer only to college class or AP as well? I'm going to learn $\sin x, \cos x$ and obviously, $e^{x}$ but it all depends on short amount of time I have till AP.

Since we would award credit for the class I'm speaking of for the BC exam, then yes, I would hope this would apply to a high school BC course as well.

just learn them...i use the whole e^(ix)=cos x+i* sin x to remember them...

What's a good way to learn series thoroughly in... a week or two? I need to learn series so bad now because AP test is approaching and eh, I'm still not good with series.

If you know those basic series, and understand how to manipulate them into other series, and develop an intuition as to whether a series converges or not, then you'll be fine. Also, [b]know Taylor Series Formula[/b] ! I'm pretty sure the last BC Free Response is that every single year.


For x_1 \geq x_2 \geq ... \geq x_n >0  and y_1,y_2,...y_n real positive numbers such that

y_1 \geq x_1
(y_1)*(y_2) \geq(x_1)*(x_2)
..........
(y_1)*.....*(y_n) \geq (x_1)*.....*(x_n)

sum(k=1..n)y_k \geq sum(k=1..n)x_k

It looks very simple but I have spent a lot of time to find a solution.
Have you a simple solution?

[i]unknown source[/i]

Sorry, I have forgotten to write "prove that"

For x_1  \geq  x_2  \geq  ...  \geq  x_n >0 and y_1,y_2,...,y_n real positive numbers such that

 y_1  \geq  x_1
(y_1)*(y_2) \geq  (x_1)*(x_2)
..........
(y_1)*.....*(y_n)  \geq  (x_1)*.....*(x_n)

prove that

sum(k=1..n)y_k  \geq  sum(k=1..n)x_k

Don't you like this inequality?
I am waiting for a solution to compare with mine.

Ok, it's time to kill this one (though after a considerable effort):
   Let us denote a_i=y_i/x_i. Then we have a_1*...*a_k>=1 for all k and a_1+...+a_k>=k for all k. Put s_k=a_1+...+a_k. We have to prove that
a_1*x_1+...+a_n*x_n>=x_1+...+x_n, or x_1(s_1-1)+x_2(s_2-s_1-1)+...+x_n(s_n-s_n-1-1)>=0, or s_1(x_1-x_2)+...+s_n-1(x_n-1-x_n)+s_nx_n>=0 and this one is true since s_k>=k, x_k-x_k+1>=0 and
(x_1-x_2)+2(x_2-x_3)+...+(n-1)(x_n-1-x_n)+nx_n=0.


algebra
polynomial
modular arithmetic
function
number theory

... which I'm not sure is an entirely appropriate title
... that I couldn't figure out.

Prove $ 1900^{1990} \minus{} 1$ is divisible by 1991.

Prove the remainder theorem, which states that the remainder $ r$ in the identity
$ f(x) \equal{} (x \minus{} a) \cdot q(x) \plus{} r$
is equal to $ f(a)$, where $ f(x)$ is a polynomial of degree $ n$ and $ q(x)$ is a polynomial of degree $ n\minus{}1$.

Well, the first one can also be done with modular arithmetic and noting that $ 1990\equiv{\minus{}1}\pmod{1991}$...I'm guessing you can use the Remainder Theorem as well (since the second problem wants you to prove it)?

[hide="Problem 2"]
$ \frac{f(x)}{(x\minus{}a)}\equal{}q(x)\plus{}r(x)$ where $ r(x)$ is the remainder upon division.

Then, $ f(x)\equal{}(x\minus{}a)q(x)\plus{}r(x)$

If $ x\equal{}a$, then

$ f(a)\equal{}(a\minus{}a)q(a)\plus{}r(a)$

$ f(a)\equal{}0q(x)\plus{}r(a)$

$ f(a)\equal{}r(a)$

Which is true by synthetic substitution.
[/hide]

Thanks to you both. *sigh*

[hide]every time I post a question, I do so in the Basics forum, and then it gets moved to the Intermediate forum, which makes me feel smart  :) , and then I read the answer, and either I don't get it or it's so obvious now, and the makes me feel stupid  :( .[/hide]

Just for reference, anything involving a proof should probably be in intermediate forums (unless its too hard, then go to pre-olympiad). As for my solution, are you familiar with synthetic substitution? If not I can explain it.

[hide="Problem I"]Note that $ 1990 \equiv \minus{} 1\mod 1991$, so $ 1990^{1990} \equiv ( \minus{} 1)^{1990} \mod 1991 \equiv 1\mod 1991$. Now, we know that $ \minus{} 1 \equiv \minus{} 1 \mod 1991$, therefore:

$ (1990^{1990} \minus{} 1) \equiv (1 \minus{} 1) \mod 1991 \equiv 0 \mod 1991 \\
\therefore 1991 \mid 1990^{1990} \minus{} 1$.

In general, we can say that $ (n \plus{} 1) \mid n^{n} \minus{} 1$ for all even positive integers $ n$.[/hide]

[hide="Problem II"]When dividing, we know that $ \frac {x}{y} \equal{} z \plus{} \frac {r}{y}$, where $ r$ is the remainder. Applying the same concept to functions, we get that $ \frac {f(x)}{(x \minus{} a)} \equal{} q(x) \plus{} \frac {r}{(x \minus{} a)}$. Multiplying by $ (x \minus{} a)$, we get that $ f(x) \equal{} (x \minus{} a) \cdot q(x) \plus{} r$. Of course, we know that, for $ x \equal{} a$, $ f(x) \equal{} r$, because $ (a \minus{} a)$ is zero — which makes $ q(x)$ “vanish.” Thus, we say that $ r \equal{} f(a)$ when $ f(x)$ is divided by $ (x \minus{} a)$.[/hide]

MAPLE TO THE RESCUE!!!

It's 1990^1990-1


jo has an apple pie ,and $p$ or $q$(but he doesn't know),guests.he wants to devide the pie in $m$ parts(notice that the parts can be unequal), so that he
can give equal amount of pie to his guests(all of the pie must be used).
what is minimum of $m$?
(u can first assume that $p$ and $q$ are coprime.)


calculus
integration
number theory unsolved
number theory

show that $\sum_{n=1}^{\infty}\frac{1}{a^{f(n)}}$ is transcental when $Lim_{n\rightarrow \infty}\frac{f(n+1)}{f(n)} > 2$.(a is an integer)

I think it should follow from Roth's Theorem: given an algebraic number $\alpha$ and constants $A>0$, there are only finitely many rationals $\frac pq$ s.t. $\left|\alpha-\frac pq\right|<\frac A{q^{2+\varepsilon}}$. As for a proof of the theorem, I think it should be rather tough. Roth did receive the Fields Medal because of it :).

But he didn't write that $f(n)$ is an integer.

BTW, mentioned theorem is called Thue-Zigel-Roth theorem (here, in Russia, at least).
It is an ultimate improvement of Louvile's theorem.
See http://www.mathlinks.ro/Forum/viewtopic.php?highlight=algeb&t=29566

You're right, $f$ does not necessarily take integral values. Sorry about that, I rushed into it without thinking :(.

[quote="grobber"]You're right, $f$ does not necessarily take integral values.[/quote]
I think it does  :?  ;)

You mean to say that sam-n forgot to mention it, right? I also think so, because I think otherwise the fact that $a$ is an integer wouldn't make a difference. However, we'll have to wait for him :).

oh whenever I publish a problem you should expect that something is forgotten excuse me ,Grobber and Myth ,your right i forgot to mention that.

So is it really suficiently to use Tue-Zigel-Roth?


This is a bit more readable: ;)
[quote="ankur5555"]$ (b!/(b{}^{\wedge}b))\left(\sum _{k\equal{}0}^b(b{}^{\wedge}k)/k!\right)$[/quote]
I would try plugging in Stirling's formula.

See the 2 books of exercises of Polya and Szego.


$\angle BPT= \angle TPD$

Prove that $B,Z,L$ are collinear.

[quote="Altheman"]i don't really know how to solve it, but it seems like inversion in respect to O because 2 of the points are already on O, also the other circle is the perpendicular line containing P inverted in respect to O[/quote]

exactly..................

inversion will solve it...


I think Tolkien was right on this. People with a lot of power will eventually be corrupted by it. What does everyone else think?

Or is it that mankind is inherently evil, and power gives him or her more opportunity and freedom to express it?

I don't think power is necessarily inherently evil, but I do think that those who seek it tend to be the ones more easily corrupted by it.

Moderators have lots and lots of power on message boards, but they aren't evil... usually.  (My brother got his account deleted on one message board because he was making a joke about being four years old.  The moderators deleted his account because they thought he was an underage user :) )

Power is not evil unless you choose to abuse it.  Though it is rather hard to be evil without some amount of power a lot of people with power are not evil at all.

[quote="Tangratara"]I think that power itself is not overwhelmingly evil.  However, the love of power is.  In LOTR, the ring itself does not cause the corruption.  It is the ring bearer that becomes corrupted over love of the ring.

Power itself is not evil.  The evil comes from the love of that power.[/quote]

exactly what I was trying to say.

[quote="Magnara"]Of course power isn't inherently evil.  There have been a number of great honest statemen and stateswomen, but it's the evil ones who get the most attention.[/quote]
This is true, but most of them don't actually have that much power (at least in the U.S.). Before Saddam Hussein got ousted, he had the power to gas thousands of people. Hitler had the power to kill millions of people. Even if a U.S. statesperson were evil enough, which I doubt, they would not have the power to carry out these acts. Even if the President ordered the killing of thousands of innocent people for no reason, he would get impeached, and no one would obey the order anyway. In any case, statespeople serve relatively short terms, so they do not have the time to be corrupted.

[quote="Tangratara"]I think that power itself is not overwhelmingly evil. However, the love of power is. In LOTR, the ring itself does not cause the corruption. It is the ring bearer that becomes corrupted over love of the ring.

Power itself is not evil. The evil comes from the love of that power.[/quote]
The question there is does power cause love of power? In LoTR, anyone with the ring becomes obsessed with it, so Tolkien seems to be agreeing with this. I also agree that people with power love the power they have. I think the less obvious point is whether the love of power corrupts people.

If someone offered me complete control of the world and I accepted, I would love the fact that I have it. I could stop people from killing each other and instead devote the money on weapons and defense to providing food and medical care to people and researching cures to diseases. According to Tolkien, though, I would eventually start to use more and more power for my own purposes. I would start by, say, getting myself a new computer, but then I would want a better house for myself, and my family, and my friends, and so on, eventully draining the world's resources and causing poverty. The question is would I become so corrupted, or could I still continue to do good for the world?

So what you're saying Yasha is that our government was designed to limit the power of any one individual, because they understood the proclivity of tyrany by the powerful.

The whole thing about the love of power being the source of evil is really biblical. It's mistakenly said that money is the root of all evil - but actually the bible says that it is the [i]love of money[/i] that is the root of evil.

Of course, when speaking of evil one should ask, what does it mean for a person to be evil? Is it that people themselves are evil or that only actions or thoughts are evil?

Another interesting point, is brought up by the so called King's Dilema. Suppose you were an all-powerful King, or a God or Goddess, who could do whatever you wanted and there was no one to stop you. Would you behave morally? Do you have a reason to behave morally? If you could walk into a bank tomorrow and take a million dollars without being stopped, would you do it?

nah, power itself is probably not evil. as the quote goes, "guns don't kill people, people kill people"

I think actions are the only things that can be classified as evil. People are classified as evil depending on whether most of their actions are evil. Unlike evil actions, which under the same circumstances are always evil, people can't be classified as evil because they can change.

I also want to point out a difference between being an all-powerful King and an all-powerful God. If an all-powerful King wanted a huge palace, labor and money would still have to go into making it. An all-powerful God can make himself a palace without using the labor and resources others. If I could create anything I wanted for myself, I would. If getting something for myself would hurt others, I hope I wouldn't still get it, but according to Tolkien, eventually I'd become corrupted and would.

My question about power is still in the bounds of a realistic situation in which there is a finite amount of resources.

[quote="gauss202"]
Another interesting point, is brought up by the so called King's Dilema. Suppose you were an all-powerful King, or a God or Goddess, who could do whatever you wanted and there was no one to stop you. Would you behave morally? Do you have a reason to behave morally? If you could walk into a bank tomorrow and take a million dollars without being stopped, would you do it?[/quote]

This would be like whether morality is taught or in the conscience.  Or is there a conscience?  Perhaps our "conscience" is just a compilation of what was learned?

I disagree that Tolkien meant that power is inherently evil.  Tolkien was an anarcho-monarchist and believed that people should be led by great leaders (kings) and that is a form of power.

There are many forms of power used for good in the LOTR.  Bombadil is an example of a very powerful being -- so powerful that the forest obeys him and the ring has no sway over him.  The elves craft powerful possessions and use magic as do the wizards.

I think the intended lesson in regards to power in LOTR is that unified power is evil.  Even Tolkien's notion of monarchism includes a significant democratization of power as there are many leaders who act under independent motives (that sometimes coincide but sometimes do not).

The ring binds power together.  I think this represents Tolkien's fear of nation-state conglomeration wielding industry and technology for the purposes of the state, not the purposes of utility chosen by individuals.

[quote="MCrawford"]I think the intended lesson in regards to power in LOTR is that unified power is evil.[/quote]
If you have a lot of power, but many others have that same power, you actually don't have that much power after all, because if you used your power for something all the others with power didn't like, they could prevent you. Because you would have less power, you wouldn't be as corrupted by it, if at all. I think Tolkien's point is that power corrupts people, but if it is split among many people, the corruption is hardly felt, which agrees with what MCrawford said.

[quote="MCrawford"]Bombadil is an example of a very powerful being -- so powerful that the forest obeys him and the ring has no sway over him.[/quote]
I think Tom Bombadil is a very interesting character. I don't think, however, that the fact that the ring cannot control him is a measure of his power, since he cannot control the ring. I have wondered for a while what Tom Bombadil symbolizes. Anyone have any ideas? I was disappointed that they completely cut him out of the movie.

Power is certainly relative and that's part of the point.

I like the character of Bombadil because I've known people a bit like him: extremely capable, good natured, but without a clue that their might be real evil in this world.  This is why the ring has no power over Bombadil -- it's a matter of a fundamental naivety of power.

Let's see, I think that greed is something natural for the human race, and love for power is a form of greed, it is that need to do whatever you can to satisfy your desires, regardless of the consequences for someone else. I would like to make some comments about the following quote.

[quote="Yasha"] This is true, but most of them don't actually have that much power (at least in the U.S.). Before Saddam Hussein got ousted, he had the power to gas thousands of people. Hitler had the power to kill millions of people. Even if a U.S. statesperson were evil enough, which I doubt, they would not have the power to carry out these acts. [b]Even if the President ordered the killing of thousands of innocent people for no reason, he would get impeached, and no one would obey the order anyway[/b]. In any case, statespeople serve relatively short terms, so they do not have the time to be corrupted. [/quote]

Let's go step by step. You are assuming that Saddam Hussein and Adolf Hittler didn't have any reason to order the killing of thousands (even millions) or people. Well, I am totally sure that if you read history you will find that they had a lot of reasons to do it. I do not agree with those reasons, I really think that they were unnecessary masacres, and probably most of the people in this forum will agree with me, but reasons they had. On the other hand, presidents of this country (US) have ordered (and executed those orders) the killing of thousands of innocent people, for reasons that no everybody agree with. Do you believe that the bombs in Hiroshima and Nagasaki killed only bad Japaneses that were active members of the army fighting against the US!, I would say that probably 90% or 95% of the people that those bombs killed were innocent whose only sin and crime was to have been born in a country whose leaders (not the people, the leaders) decided to fight against the US.

Ahh, that is too old, well, Hittler madness is even older. But anyway, Bush senior ordered in 1991 the operation "Dessert Storm", and accordingly to AMERICAN stadistics, 130 000 iraki civilians died because of that. Do you really believe that all that people deserve to die because their dictator took the wrong decition? If you do, I don't. How many people died in Kosovo when Clinton ordred the attack in september 1999?

Now, we are talking only about executive orders (not mentioning 2003 invation!) Does any of the executors of such orders disobey the orders given to them directly? I don't think so. We aren't also mentioning all the people that were killed by third parties but with american money (or don't you remember that it was the CIA who financiated Pinochet, Bin Laden and even Hussein itself at the beginning).

I am not antiamerican, actually I believe that it is a great land where I myself have had a lot of oportunities here that probably I wouldn't have elsewhere. There is several reasons of why America is the richest and more powerful nation in the world, and one of those is the will of its people. Also, I don't believe that those decitions taken for presidents and people with a lot of power are direct consecuence of the thought of the crowd (no, it was the leaders, not the crowd), but if other countries have dirty laundry, USA also have its own! The fact that the people that such orders kill are not american doesn't mean that they are not human!

Finally, your aregument about the fact that state people serve few time, and then they don't have time to be corrupted, please, they are already corrupted when they start, because it was a way too long path to reach the elite!

Best regards,

Hmm...I agree with you.

I think what I was getting at back then is that the President alone doesn't have the power alone to kill people; he needs a relatively large amount of people to back him up. However, that doesn't really say much since a bunch of people backed up Hitler, too.

Anyways, your point seems right, and mine doesn't really. Which is why I now stick to math on this forum: It's much easier for me to make coherent arguments that way.

[quote="Yasha"]Anyways, your point seems right, and mine doesn't really. Which is why I now stick to math on this forum: It's much easier for me to make coherent arguments that way.[/quote]

Actually, there is no better way to become a good argumenter, than doing arguments. I do believe in what I wrote, but eitherway, there is no better excercise in argumentation than arguing the oppositive point of view than yours.

Best,

[quote="white_horse_king88"]You guys should read Understanding Power by Noam Chomsky. Lots of these issues are discussed.[/quote]

Thanks for the advise. I will take it into consideration.

Best,

[quote="MithsApprentice"]Bombadil is quite an interesting character indeed. I personally believe he is a Vala.[/quote]
Really?  I think he's a Maia.

Anyway, I do not believe that power is inherently evil.  What makes people think that it good or evil is what it is used for.  Power simply [i]is[/i] , and as such, cannot be either good or evil.

[quote="Danbert"]
Power simply [i]is[/i].[/quote]
and i think maybe tom bombadil is supposed to symbolize that


floor function
combinatorics unsolved
combinatorics

Let $n\geq 2$ be an integer and consider an array composed of $n$ rows and $2n$ columns. Half of the elements in the array are colored in red. Prove that for each integer $k$, $1<k\leq \dsp \left\lfloor \frac n2\right\rfloor+1$, there exist $k$ rows such that the array of size $k\times 2n$ formed with these $k$ rows has at least
\[ \frac { k! (n-2k+2) } {(n-k+1)(n-k+2)\cdots (n-1)} \] columns which contain only red cells.

Feel free to point out any mistakes.First a little lema: $\binom{n}{k}\ge n-2k+2$ (trivial by induction).Next call a $set$ ,$k$ colored squares that are in same column.We count number of $sets$, $T$ .Number columns from $1$ to  $2n$ from left to right and let in column $i$ be $d_i$ colored squares then $T=\sum_{j=1}^{2n}\binom{d_j}{k}$.Now we have $T= \sum_{j=1}^{2n}\binom{d_j}{k}\ge 2n\binom{n}{k}$ (here we used fact that $\binom{a+1}{x}+\binom{b-1}{x} \ge \binom{a}{x}+ \binom{b}{x}$ with $a-b\ge 2$).Suppose otherwise.Then in any $k$ rows there is less than $\frac {n(n-2k+2)}{ \binom{n}{k}}$ sets ( $\frac {n(n-2k+2)}{\binom{n}{k}}$ equals  $\frac { k! (n-2k+2) } {(n-k+1)(n-k+2)\cdots (n-1)} $) so $T< \binom{n}{k}\frac {n(n-2k+2)}{ \binom{n}{k}}$ .Then it follows that
 $\binom{n}{k}\frac {n(n-2k+2)}{\binom{n}{k}}> 2n\binom{n}{k}$ and this is equal to $\binom{n}{k}< n-2k+2$ which is contradiction with lemma and we are done .


number theory
prime numbers
number theory proposed

I notice that when you calculate $6n+5$ for some small $n$ you get all prime numbers except $35$ you get $5, 11, 17, 23, 29, 41, 47, 53, 59$ so this came up to my mind.

$Problem$:[b]Prove that there exist 2006 consecutive numbers $n$ s.t. numbers $6n+5$ are all composite.[/b]

I think that I have solution but I will be glad to hear your ideas... :D :D

t is well-known that for any $k$ there exists $k$ consecutive integers which are all composite.
Thus, for sufficientely large $k$ this string of composite numbers will contain at least $2006$ of them which are of the form $6n+5.$

Pierre.

I think that you didn't understand my question, I am asking proof that there exist $2006$ consecutive $n$'s s.t. numbers $6n+5$ are all composite; in other words prove that for some $n$ numbers $6n+5, 6(n+1)+5,6(n+2)+5,...,6(n+2005)+5$ are all composite

I said that you can find a string of $k = 2006 \times 6$ consecutive integers which are all composite, say $m,m+1, \cdots, m+k$.
Among the numbers $m,m+1m,\cdots, m+5$ one is of the form $6n+5$, say $m+r$. Then, just choose. The numbers $m+r + 6q$ where $q=0,1, \cdots, 2005$ are all of the form $6n+5$ and belong to the set of the $k$ initial numbers and therefore are composite.

Pierre.

Now I get it what you tried to say, well nice solution first thing that came up to my mind was Chinese Reminder Theorem but this is more elementary so it's nicer :D :D


A, B, C, D, and E represent 5 different digits. Find the digit
represented by C.

C equals 8

one of the B's are pulled down from the top, which is 7, so B equals 7
since B times A is D5, that means either B or A is 5, since B is 7, then A is 5
B times A is D5, so 7 times 5 is 35, so D equals 3
since 3B - D5 is E, 37 minus 35 is 2, so E equals 2

to find C, we multiply 1B by A and add E, so 17 times 5 plus 2 equals 85 plus 2 equals 87
C equals 8

huh whats up with the lines in the middle?

When most people do long division, they draw those lines to show the subtraction.
Like when most people learn to subtract they do this:

  9
[u]- 5[/u]
  4

(The alignment is off, but you get it.)


vector
analytic geometry
linear algebra
matrix
calculus
derivative
algebra

Let $a_{1}= (1,1,-2,1)$, $a_{2}= (3,0,4,-1)$, $a_{3}= (-1,2,5,2)$, $\alpha = (4,-5,9,-7)$, $\beta = (3,1,-4,4)$, $\gamma = (-1,1,0,1)$

a) Which of the vectors $\alpha ,\beta ,\gamma$are in the subspace of $R^{4}$spanned by the $a_{i}$?
b) Which of the vectors $\alpha ,\beta ,\gamma$are in the subspace of $C^{4}$spanned by the $a_{i}$?
c) Does this suggest a theorem?

I have determined that they all belong to the subspace and I vaguely understand the theorem but it seems rather trivial. I figure it is along these lines: Let G a subfield of a field F and let B be a basis of some vector space V over F. Let the $a_{i}$ have coordinates in G with respect to B) and let v be in the linear span of the $a_{i}$ (let v also have coordinates in G with respect to B. If there exists coordinates in F with respect to the $a_{i}$ for v then there exists coordinates in G with respect to the $a_{i}$ for v.

I THINK that's what is supposed to be going on but it doesn't make any sense to me. In this problem $\alpha ,\beta ,\gamma$have real coordinates with respect to the $a_{i}$ and therefore, obviously, these coordinates are also complex. But it seems like all this notes is that every real number is a complex number, or more generally every member of G is also a member of F?

Can someone indepently state what the theorem is and explain/prove it? Thanks..

2) Let A be an m x n matrix over the field F, and consider the system of equations AX=Y. Prove that this system of equations has a solution if and only if the row rank of A is equal to the row rank of the augmented matrix of the system.

I say the row rank of the augmented matrix cannot be less than the row rank of A clearly because the rref augmented matrix has at least as many non-zero rows as rref(A). It cannot be greater than the row rank of A because of the condition for the existence of a solution, which is that each zero kth-row in rref(A) must correspond to a zero kth-row in the rref augmented matrix. Therefore the row ranks must be the same. Rref matrices have the same row spaces as their derivative matrices and thus it is no loss of generality. Is there any gap/error in my reasoning here?

Here's the theorem:

Consider a vector space $V$ over a field $K$ with basis $B$. Let $F$ be a subfield of $K$, and let $U$ be the $F$-vector subspace of $V$ spanned by $B$.
Given any set $S\subset U$, the $F$-span of $S$ is the intersection of $U$ and the $K$-span of $S$.

One containment is trivial. For the other, look at how you find whether a vector is in the span of some vectors; what operations go into this?

Can you define what you mean by F-vector and F-span? Not familiar with that terminology.

$V$ has two vector space structures, one for each field. I'm just indicating which structure the spans are associated with.

But what is the difference between 'F-vector' and 'F-span'?

"$F$-vector" is meaningless on its own; I should probably have said "$F$-subspace".


Show that if a k-regular bipartite graph has bipartiton (X,Y) then |X|=|Y|

The number of edges is equal both to $k|X|$ and $k|Y|$.


$n=12.$

Assume that $n$ has the desired property.
Then, each of the $A_i$'s is the product of at least two (not necessarly distinct) primes. Moreover, since the $A_i$'s are pairwise coprimes,  using the pigeon-hole principle, at least one of them is the product two (not necessarly distinct) primes not less than $p_n$ (which denotes the $n^{th}$ prime in increasing order). It follows that $p_n^2 \leq (3n+1)^2$, that is $p_n \leq 3n+1$. Direct checking proves that it forces $n \leq 12.$

In another hand, $n=12$ has the desired property. Note that $(3n+1)^2 =1369$, thus i t suffices to choose the integers (each of the given factors is a prime if I did not wrong) :
$2 \times 83 = 166, 3 \times 79 = 237, 5 \times 73 = 365, 7 \times 71 = 497, 11 \times 67 = 737, 13 \times 61 = 793, 17 \times 59 = 1003, 19 \times 53 = 1007, 23 \times 47 = 1081, 29 \times 43 = 1247, 31 \times 41 = 1247, 37 \times 37 = 1369.$

Pierre.

Nice solution, but one error: n = 14  :D
.:Example: 4,9,25,49,...,43^2

[quote="chavez"]n = 14[/quote]

Computations mistakes will kill me  ;)

Pierre.


[quote="chin886"]let $ f(x) \equal{} \frac {1}{\sin\frac {\pi x}{7}}$. Prove that $ f(3) \plus{} f(2) \equal{} f(1)$.[/quote]
Multiply with $ \sin \left (  \frac{\pi}{7} \right ) \cdot \sin \left (  \frac{2\pi}{7} \right ) \cdot \sin \left (  \frac{3\pi}{7} \right )$. And use $ \sin u \cdot \sin v \equal{} \frac{1}{2} \cos (u\minus{}v) \minus{} \frac{1}{2} \cos (u\plus{}v)$, to obtain the equivalent statement:
$ \cos \frac{\pi}{7} \minus{} \cos \frac{5\pi}{7} \equal{} \cos \frac{\pi}{7} \minus{} \cos \frac{3\pi}{7} \plus{} \cos \frac{2\pi}{7} \minus{} \cos \frac{4\pi}{7}$
Using $ \cos v \equal{} \minus{} \cos \pi \minus{} v$, it is obvious, since $ \minus{} \cos \frac{5\pi}{7} \equal{} \cos \frac{2\pi}{7}$ and $ \cos \frac{3\pi}{7} \plus{} \cos \frac{4\pi}{7} \equal{} 0$ :)


Hey, it has been two years since I finnished High School and now I am currently studying for a Music Tech exam and I am having touble understanding this equasion. It's been a long while since I've done these.

The aim here is to find the value of E

30 = 20log (E1/.775)
1.5 = log (E1/.775)
10^1.5 = (E1/.775
31.6 x .775 = E

I dont understand what has been done to the log to move it from the right side to the left. How did 1.5 become 10^1.5? If you could also explain to me how this would work for any value (instead of 1.5), that woudl be most appreciated.

Thanks, MaxwellHouse

The definition of $\log$ is as follows:
$y=b^x\Longleftrightarrow x=\log_b y$

(When it simply says $\log y$, that means $b=10$.)

So in this case,
$\log_{10} y=1.5$, where $y=\frac{E}{.775}$.

Then $y=10^{1.5}$, which means $\frac{E}{.775}=10^{1.5}$.

I hope you understand my explanation...


If $ tan^{\minus{}1}\equal{}\alpha$ and $ tan^{\minus{}1}y\equal{}\beta$, show that

$ tan(\alpha\plus{}\beta)\equal{}\frac{x\plus{}y}{1\minus{}xy}$

Deduce that
$ tan^{\minus{}1}x\plus{}tan^{\minus{}1}y\equal{}tan^{\minus{}1}(\frac{x\plus{}y}{1\minus{}xy}) \plus{} k\pi$

-------------
I did part 1, but how to deduce the second part, more specifically the $ k\pi$ part

\begin{eqnarray*}
\tan(\alpha \pm \beta) & = & \frac{\sin (\alpha \pm \beta)}{\cos (\alpha \pm \beta)} \\
                                 & = & \frac{\sin \alpha \cos \beta \pm \cos \alpha \sin \beta}{\cos \alpha \cos \beta \mp \sin \alpha \sin \beta} \\
                                 & = & \frac{\frac{\sin \alpha \cos \beta \pm \cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta \mp \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} \\
                                 & = & \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta} \\
                                 & = & \frac{x \pm y}{1 \mp xy}
\end{eqnarray*}
and, $ \tan (x \pm k\pi) = \tan x \ \ (k\in\mathbb{N})$

So,

\begin{eqnarray*}
\arctan \left(\frac{x \pm y}{1 \mp xy}\right) \pm k\pi & = & \arctan (\tan(\alpha \pm \beta)) \pm k\pi \\
                                                                                & = & \arctan (\tan(\alpha \pm \beta \mp k\pi)) \pm k\pi \\
                                                                                & = & \alpha \pm \beta \mp k\pi \pm k\pi  \\
                                                                                & = & \alpha \pm \beta \\
                                                                                & = & \arctan x \pm \arctan y
\end{eqnarray*}


trigonometry calculus integration function derivative algebra domain

Why are the trig substitutions used for finding integrals valid? For example, if we do $ x \equal{} \tan{\theta}$ I can see how it works. But there are uses of $ x \equal{} \sin{\theta}$ which doesn't make sense because then $ x$ must be in the interval $ [ \minus{} 1,1]$. Do people normally omit the $ r$ in front of it because it cancels out?

When you're making that $ \sin$ substitution, there's a $ \sqrt{1\minus{}x^2}$ in there. If you want to stick to real numbers, that restricts $ x$ to $ [\minus{}1,1]$. If you don't want to stick to real numbers, $ \sin$ takes all complex values.
You do have to be careful; any formula you get is local to some region. That's about the branch points of the square root or logarithm.

Does anyone have an intuitive understanding about why trigonometric substitution works? I know it does work but the intuitive understanding was always missing for me. The variable x is linear and then you are equating it with a function that has a variable rate of increase - it's a neat trick but it just never really made complete sense to me.

What you just said applies to nearly every substitution, not just trig substitutions.

For instance, in $ \int\frac{x}{1\plus{}x^2}\,dx,$ substitute $ u\equal{}1\plus{}x^2.$ There's a variable rate of increase there.

But that's a case of what I refer to in class as a "natural" substitution, as opposed to a "forced" or "backwards" substitution. For an example of what I mean by a forced or backwards substitution:

$ \int\frac{dx}{x^{1/3}\minus{}x^{1/2}}.$

Let $ u\equal{}x^{1/6},$ but write it backwards as $ x\equal{}u^6,\ dx\equal{}6u^5\,du.$ That leaves us with

$ \int\frac{6u^5\,du}{u^2\minus{}u^3}\equal{}6\int\frac{u^3}{1\minus{}u}\,du,$ and you'll find something to do from there.

All substitutions work because of the chain rule for differentiation- they're the chain rule read backwards. Substitution is the reason why we can evaluate more integrals in closed form than sums, since no such technique exists for sums (other then the much more restrictive case of index-shifting). When the function involved in the substitution is not one-to-one, we must be extremely careful about what it's doing in the domain of our function - we might have to break the integral into pieces to insure the validity. jmerry alludes to this above.

Trig substitutions are a particular case of the idea of a forced or backwards substitution - they're not natural. And they shouldn't be your first choice. For instance, if you have $ \int\frac{x\,dx}{\sqrt{4\minus{}x^2}},$ then you should use the natural substitution, not the trig substitution.

[quote="Kent Merryfield"]What you just said applies to nearly every substitution, not just trig substitutions.

For instance, in $ \int\frac {x}{1 + x^2}\,dx,$ substitute $ u = 1 + x^2.$ There's a variable rate of increase there.[/quote]

Yeah, for some reason that's intuitively clear for me while trig substitution isn't. Even with regular integration by u-substitution it took me a while to really figure out what was going on. For a long time I couldn't understand why we can manipulate dx algebraically when it's just notation indicating what variable we're integrating with respect to. I finally figured out that making a u-substitution and solving for dx in du = u'(x)dx is just a sleight of hand way of dividing the integrand, newly expressed in terms of the variable u, by a factor of u'(x). If you were to just make a straight algebraic u-substitution and express the integrand in terms of the variable u and then go ahead and integrate with respect to u, you'd get an antiderivative F(u) but then when you differentiated this antiderivative with respect to x you'd get f(u)u'(x) instead of just f(u). That is, the chain rule introduces the factor u'(x).

For example, if you wanted to integrate $ \int\frac {x^2 - 1}{\sqrt {2x - 1}}\,dx$ and disregarded the integration sign and the notation dx and just made the straight algebraic substitution $ u = {\sqrt {2x - 1}}$ into the integrand, then you would get  $ \frac {1}{{4}}({u^{3} + 2u - \frac {3}{{u}})}$ for the new integrand and then if you went ahead and integrated this with respect to u, you would get $ \frac {1}{{4}}(\frac {u^{4}}{4} + {u^{2} - {3}\ln{u}})$ which if you converted back to a function of x (using $ u = {\sqrt {2x - 1}}$) and then differentiated, you'd get $ \frac {x^2 - 1}{{2x - 1}}$ which is greater than the original integrand by a factor of $ \frac {1}{\sqrt {2x - 1}}$ = u'(x).


The points in the plane with integer coordinates are numbered as below.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=288[/img]

What are the coordinates of the number $2003$?

[quote="Singular"]going from the origin and southwest, odd squares are displayed.  So 45^2 = 2025 is displayed.  Finally, go north 22.[/quote]
I think you mean [b]southeast[/b]. ;)

The points in the plane with integer coordinates are numbered as below.
[img]https://cdn.artofproblemsolving.com/attachments/0/2/122cb559c6fb4cb8401ffa215528a035346a3d.png[/img]
What are the coordinates of the number $2003$?


find all the positive integers a and b such that

3x (2^a) + 1 = b²

rearranging and factoring, $ (b\plus{}1)(b\minus{}1) \equal{} 3*2^a$
now consider the ways the right hand side can be factored -[hide="The Factorizations"]-it can be factored only in the form of $ (3*2^x)*(2^{a\minus{}x})$, for (nonnegative) integers x satisfying $ x<a$ or $ x\equal{}a$[/hide] and then equate factors.

P.S.-How do you latex less than or equals, greater than or equals, and the multiplication "dot"?

[code] \le ->less than or equal
 \ge ->greater than or equal
and the last one just "."[/code]

Using . to denote multiplication appears to be a common practice among some non-English speakers, but it's very confusing notation for those of us who are used to it meaning a decimal point.  What you want is \cdot.

Back to the problem, I think the only solutions are $ (a,b)\equal{}\{(3,5),(4,7)\}$?

Yes; when i solved the problem completely that is what i got also.

Testing:

$ \cdot$, $ \ge$, $ \le$- Thanks!

[hide="Another hint"]
When you know something is odd, call it $ 2k\minus{}1$.
[/hide]

[hide="Big hint"]
b+1=2+b-1, and b+1 and b-1 are divisible by 2 or one of them is 3*2^a and other is 1, or one is 2^a and other is 3
[/hide]


function
modular arithmetic
number theory

What are the last 2 digits of 9^2009? How do i find the last two digits of any random integer a^b?

The basic idea is to divide it by $ 100$ and observe the remainder. It is very often difficult to do it manually so you have to use very common trick known as [url=http://en.wikipedia.org/wiki/Euler%27s_theorem]Euler's theorem[/url]

[hide]
$ 9^{40} \equiv 1 \; (mod \; 100)$

$ 9^{2000}	\equiv 1 \; (mod \; 100)$

$ 9^{2009}	\equiv 9^9 \; (mod \; 100)$

$ 9^3	\equiv 729	\equiv 29 \;(mod \; 100) \implies 9^9	\equiv 29^3 	\equiv 24389 	\equiv 89 \; (mod \; 100)$

$ 9^{2009}	\equiv \boxed{89} \; (mod \; 100)$[/hide]

Alternatively use the Carmichael function (or CRT).

Here's a method that doesn't rely on modular arithmetic:

Apply the binomial expansion to $ (10\minus{}1)^{2009}$ to get $ \sum_{k\equal{}0}^{2009}(\minus{}1)^k\cdot\binom{2009}{k}10^{2009\minus{}k}$.

We want the last two digits, which is when the exponent of $ 10$ is either $ 0$ or $ 1$; that is, $ k\equal{}2009$ or $ k\equal{}2008$.
When $ k\equal{}2009$, we have $ \minus{}1\cdot\frac{2009!}{2009!}\cdot 10^0\equal{}\minus{}1$.

When $ k\equal{}2008$, we have $ 2009\cdot 10\equal{}20090$.
$ 20090\minus{}1\equal{}20089\equiv \boxed{89}\pmod{100}$.

Did you go to FAU Math Day too (this was one of the problems).