What are the answers to the following problems and how does one go about solving them?
1. Factor over the integers 25y^2-9-4a^2+12a

2. Factonr into two trinomials over the integers: 121x^4+41x^2*y^2+9y^4

3. Factor the follwing completely: 15+3^(x+2)-5^(x+1)-3*15^x

Factoring tricky expressions can often be aided by substituting a variable for a complicated expression or completing the square.  We discuss these techniques in depth in the Intermediate Algebra class.

Just some hints to help you out, you have to work it out yourself  :)  :

1. Did you type correctly this factorization problem?

2. Consider $ (11x^2)^2=121x^4 $, $ (3y^2)^2=9y^4 $, $ (A+B)^2=A^2+2AB+B^2 $, and $ C^2-D^2=(C-D)(C+D) $.

3. Consider $ 3^{x+2}=3^2 \cdot 3^x $, $ 5^{x+1}=5 \cdot 5^x $.

If you are still stuck, please ask your teacher for immediate help.

He did type 1 correctly.  It is:[hide]

25y^2-(2a-3)^2=(5y+2a-3)(5y-2a+3) by difference of squares[/hide]

:P  I was thinking of $y^{2-9}$. My bad. Well, I wonder if [color=red]Basketball4Life[/color] learned to factor yet...

I don't think I can get the second one without trial and error

since 41 is prime the coefficient of x^2y^2 must be an addition of two other numbers, which means there must be at least two instances where x^2 is multipled by y^2 in the brackets (or xy^2 by x, etc.)

Also there can't be a constant in each expression as they would make a constant when added

my brain is half dead at this time, but I'm sure the fact that all powers are even is an important thing as well...

I will never learn this in class, we simply arent taught this
can someone breifly explain how to do it please?

Mikey wrote:I don't think I can get the second one without trial and error

Hint: [hide]since the powers are all even, try to 'force' a perfect square...what coefficient of x^2y^2 would make the expression a perfect square?[/hide]


logarithms
calculus
calculus computations

Prove that $\sum_{k=n}^{\infty}\left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^{k}}{k+1}= (-1)^{n-1}\left( \sum_{k=1}^{n}\frac{1}{k2^{k}}-\log 2\right)$ for $n=1,2,\ldots$.

Yes. Nice quote: it is now my desktop wallpaper.

Your desktop wallpaper says, "By $\log$ do you mean $\ln$?"?
Because I just happened to say that?
...
 :rotfl:

Here, I think you'll like this [url=http://www.physicsforums.com/showthread.php?t=100114&highlight=Bad+Math+jokes]thread[/url] on http://www.physicsforums.com.

[quote="FieryHydra"][quote="benorin"]Hey, no one even tried this one!  Is it too hard?

-Ben[/quote]
Unless I am mistaken, the series on the RHS doesn't seem to converge for $n=1:$

$\sum_{k=1}^{\infty}\left(\begin{array}{c}k\\1\end{array}\right) \frac{(-1)^{k}}{k+1}= \sum_{k=1}^{\infty}(-1)^{k}\frac{k}{k+1}= \sum_{k=1}^{\infty}(-1)^{k}\frac{k+1-1}{k+1}$
$= \sum_{k=1}^{\infty}(-1)^{k}-\sum_{k=1}^{\infty}(-1)^{k}\frac{1}{k+1},$ which is the sum of a divergent series and an alternating series that converges.

Am I missing something?[/quote]

As far as I know, for n=1 the series should yield $\frac{1}{2}-\log 2$

[color=red]EDIT[/color]: My [color=red]solution[/color] can be found [url=http://www.physicsforums.com/showthread.php?t=121104]here[/url].

[quote="jmerry"]Abel's theorem has another hypothesis: we must know by some other means that the series converges. In this case, the terms don't go to zero, and the series diverges for all $n\ge1$.[/quote]

Though I see that you are correct in your asessment of the theorem and that the hypotheses are not all met, can my evaluation of the series be considered a means for summing the divergent series (after the manner of say, Abel summation or that of Ceasaro (sic), or some other)? Or is there some error in my calculations? I thought it had worked, but I must have been mistaken...  :oops_sign:

I believe your method is Abel summation. Cesaro summability is somewhat weaker; the series is Cesaro summable only for $n=1$.

[quote="jmerry"]I believe your method is Abel summation. Cesaro summability is somewhat weaker; the series is Cesaro summable only for $ n \equal{} 1$.[/quote]

Thank you for this jmerry.


Euler
linear algebra
matrix
real analysis
real analysis unsolved

I have a homework problem I'm having a hard time with

Consider the Discrete Formula Transform with n = 4

Use the Euler Formula to find z, z^2, z^3 and the 4x4 matrix

now he never taught this to us, so I'm trying to figure it out on my own

but is the 4x4 matrix:

(1    1      1       1  )
(1    z     z^2   z^3)
(1   z^2   1      z^2)
(1   z^3  z^2       1)

I am having a hard time finding the z's, could someone really help me out?

err...
I dont see logic on this problem. hauhau

Is the determinant of this matrix, dont is it?

and see euler foruma in

http://en.wikipedia.org/wiki/Euler%27s_formula

i think $z=i=\sqrt{-1}$ and the powers of $z$ are $x$ $mod$ $4$ so 4 mod 4 is 0 and $z^{0}=1$ or $i^{4}=1$ and 6 mod 4 is 2, we have $z^{6}=z^{2}$,

Should that say Fourier Transform and not Formula Transform?


Natural numbers $ a \le b \le c \le d$ satisfy $ a\plus{}b\plus{}c\plus{}d\equal{}30$. Find the maximum value of the product $ P\equal{}abcd.$

[quote="moldovan"]Natural numbers $ a \le b \le c \le d$ satisfy $ a \plus{} b \plus{} c \plus{} d \equal{} 30$. Find the maximum value of the product $ P \equal{} abcd.$[/quote]

If $ d\minus{}a>1$, then $ (d\minus{}1)(a\plus{}1)>da$
So $ d\minus{}a\le 1$ and the solution is $ (7,7,8,8)$

[hide=Solution]
We want to make the numbers $a,b,c,d$ as close as possible to maximize the product of $P.$ We see that this happens when the numbers are $7,7,8,8$ thus the maximum possible value is $\boxed{3136}$
[/hide]


superior algebra
superior algebra unsolved

Let $ L \equal{}\mathbb Q(\sqrt{3},\sqrt{5})$. Show that $ [L:\mathbb Q]\equal{}4$ and that $ {1,\sqrt{3},\sqrt{5},\sqrt{150}}$ is a basis for $ L$ over $ \mathbb Q.$Express $ (\sqrt{3}\plus{}\sqrt{5})^{\minus{}1}$ in terms of this basis. Is $ L \equal{}\mathbb Q{\sqrt{3}\plus{}\sqrt{5}}$?

It should be

Let $ L \equal{}\mathbb Q(\sqrt{3},\sqrt{5})$. Show that $ [L:\mathbb Q]\equal{}4$ and that $ \{{1,\sqrt{3},\sqrt{5},\sqrt{150}}\}$ is a basis for $ L$ over $ \mathbb Q.$Express $ (\sqrt{3}\plus{}\sqrt{5})^{\minus{}1}$ in terms of this basis. Is $ L \equal{}\mathbb Q({\sqrt{3}\plus{}\sqrt{5}})$?

[hide]
First, let $ K\equal{}\mathbb{Q}\left(\sqrt{3}\right)$ so that $ L\equal{}K\left(\sqrt{5}\right)$. Then $ \left\{ 1,\sqrt{5}\right\}$ is a basis for $ L$ over $ K$. It is certainly a spanning set, so we need to show linear independence. Suppose $ a\plus{}b\sqrt{5}\equal{}0$ for some $ a,b\in K$, not both 0. If $ b\equal{}0$, we get $ a\equal{}0$, a contradiction. Otherwise, let $ r\equal{}\minus{}a/b$, an element of $ K$ that equals $ \sqrt{5}$. So there exist rationals $ p,q$ such that
\[ p\plus{}q\sqrt{3}\equal{}\sqrt{5}\]
Squaring both sides and using the irrationality of $ \sqrt{3}$ shows that
\[ p^{2}\plus{}3q^{2}\equal{}5,2pq\equal{}0\]
But $ p\equal{}0$ implies $ q^{2}\equal{}5/3$ and $ q\equal{}0$ implies $ p^{2}\equal{}5$, not possible for rational $ p,q$. So we get linear independence.

It follows that $ \left\{ 1\cdot 1,\sqrt{3}\cdot 1,1\cdot\sqrt{5},\sqrt{3}\cdot\sqrt{5}\right\}\equal{}\left\{ 1,\sqrt{3},\sqrt{5},\sqrt{15}\right\}$ is a basis for $ L$ over $ \mathbb{Q}$, and $ [L:\mathbb{Q}\,]\equal{}4$.

Next, observe that $ \left(\sqrt{3}\plus{}\sqrt{5}\right)\left(\minus{}\sqrt{3}\plus{}\sqrt{5}\right) \equal{}2$, so that $ \left(\sqrt{3}\plus{}\sqrt{5}\right)^{\minus{}1}\equal{}\frac{\minus{}\sqrt{3}\plus{}\sqrt{5}}{2}$.

Finally, we find that $ L\equal{}\mathbb{Q}\left(\sqrt{3}\plus{}\sqrt{5}\right)$. Call the field on the right $ L^{\prime}$. Then $ L$ certainly contains $ L^{\prime}$. On the other hand, $ \sqrt{3}\plus{}\sqrt{5}$ and $ \minus{}\sqrt{3}\plus{}\sqrt{5}$ (by the above computation) are both in $ L^{\prime}$, so $ \sqrt{3}$ and $ \sqrt{5}$ are in $ L^{\prime}$, making $ L^{\prime}$ contain $ L$. It follows that $ L\equal{}L^{\prime}$.
[/hide]


Is there a formula for finding circular segments based on the length of  the chord? If therre is, please let me in on it.

don't quite understand what you mean by finding a segment, but here are some usefull tips.
consider the following picuture:
$c=2\sqrt{R^2-D^2}=2D\tan{\frac{1}{2}\alpha}=2R\sin{\frac{1}{2}\alpha}$
$A=\frac{1}{2}R^2\alpha$
if by circular segment you mean $S$, then $s=r\alpha$

[quote="amirhtlusa"]
$A=\frac{1}{2}R^2\alpha$[/quote]

Er, is A the area of the sector?

Also, is h the sagitta and is D the apothem of the circle?

Sorry, the picture is kind of unclear to me  :P

yes, $A$ is for area,
i donno what those words mean :blush:
but $D$ is the distnace from center to the chord, and $h$ is the $R-D$
hope this helps :)

Thanks.
BTW, the [b]apothem[/b] is the distance of the chord to the center and the [b]sagitta[/b] is the radius minus the apothem ;)

[quote="4everwise"]Oh, I don't think I worded this correctly. By circular segment, i meant the sector reduced by the triangle in the shaded region.[/quote]
so you mean the shaded region minus the area of the triangle?
then it would be:
$\frac{1}{2}R^2\alpha-\frac{cD}{2}=\frac{1}{2}R^2\alpha-\frac{c\sqrt{R^2-\frac{1}{4}c^2}}{2}$
is this what you want?

The formula for the area of a circular segment is $\frac{r^2}{2}(\theta-\sin\theta)$, where $r$ is the radius and $\theta$ is the central angle, in radians.

Ok, I have an extension to this problem.

What is the formula to find the area of the circular segment so that you can use the length of the chord and the radius?

ok, i didn't understand what exactly you mean by Circluar Segment.
can you draw a picture or something??

Ok. I don't know if this helps, but it's the area located of the outside of the chord in a circle.

In the diagram that was drawn earlier, it would be the area within HCS.

consider sector AOB where AB is the arc and O is the center. The area of the segment is the area of the sector minus the area of triangle AOB (connect the endpoints of arc AB and you'll see that a triangle forms)

I don't see the point in memorizing formulas.. just learn how they work. Draw an altitude and go from there.

ok, i got you know,
didn't i answered this before: the area of the portiono of the circle is $\frac{1}{2}R^2\alpha$ and there is an isoceles triangle, then you have to subtract the area of an isoceles triangle: you get:
$\frac{1}{2}R^2\alpha-\frac{cD}{2}=\frac{1}{2}R^2\alpha-\frac{c\sqrt{R^2-\frac{1}{4}c^2}}{2}$
i'm sure that this can be simplified, but there are more identities about the cord, than you can relate and derive more equations.
i'm sure there is some simple equation. but i can propabley derive it if you want.

Forgot to add it earlier.

The formula is $\frac{r^2a - \frac{1}{2}c^2 + cr}{2}$

Wait, that formula doesn't really work. I plugged in the values for some of the problems and it ended up being incorrect.

This is what I did to reach a solution.
$c$ for chord
$r$ for radius
$a$ for apothem
$n$ is half the central angle

First,
sector area - isoceles triangle
$\frac{2n}{360}\pi r^2 - \frac{1}{2}ac$
$\frac{2 sin^{-1}(\frac{\frac{1}{2}c}{r})}{360}\pi r^2 - \frac{1}{2}ac$
$\frac{sin^{-1}(\frac{c}{2r})}{180}\pi r^2 - \frac{1}{2}ac$

$a^2 = c^2 - b^2$
$a^2 = r^2 - (\frac{1}{2}c)^2$
$a^2 = r^2 - \frac{1}{4}c^2$
$a = \sqrt{r^2 - \frac{c^2}{4}}$

$\frac{sin^{-1}(\frac{c}{2r})}{180}\pi r^2 - \frac{c}{2}\sqrt{r^2 - \frac{c^2}{4}}$


[size=84]If a, b, c numbers real.
 [img]http://www.mathlinks.ro/Forum/latexrender/pictures/403db825565efebddb61f200a819e95e.gif[/img]  and [img]http://www.mathlinks.ro/Forum/latexrender/pictures/3f7c4ea658be18158b1694e7fef554e6.gif[/img]
prove [/size][img]http://www.mathlinks.ro/Forum/latexrender/pictures/a80af9a624bdd79c0e46cba5e0e3060f.gif[/img]

Manifestly false for $a=b=c=\frac 13$
 :cool:

It's false without further hypothesis (take $a=b=c=1/3$ for instance)

Are you sure that $a,b,c$ are supposed to be real numbers? In that case, the problem is trivial.

Pierre.

Easy generalization : with the condition $|a_1|=|a_2|=...=|a_n|=1$, where $a_1,...,a_n$ are complexe,  we can  write

$a_1+a_2+...+a_n= real \ \Longrightarrow \ \frac 1{a_1} + \frac 1{a_2} +...+ \frac 1{a_n} =a_1+a_2+...+a_n$
 :cool: Do you follow me ? :cool:


probability
MATHCOUNTS
geometry
geometric transformation
ratio

[hide="Umm..."]
Is it 3/10 because 1/2 chance you choose A and 3/5 chance you choose a red one? Not too sure how to do this though.  :(
[/hide]

[quote="ln(dx/dy)"]

Moderators: Feel free to bump this if it is too difficult for the Mathcounts forum.[/quote]

It's hard...but not too hard.  I'm not a moderator, but I think this type of question could be a hard state or a national question.  It's similar to the Taxi question I posted a while back.  Nice question ln(dx/dy)

Sorry about my stupid guess-like answer... I was pretty tired... but that's no excuse... anyways can someone explain how to do these reverse type probability problems.. EDIT: But it seems like a few other people have the same answer...

[quote="frost13"]

It's hard...but not too hard.  I'm not a moderator, but I think this type of question could be a hard state or a national question.  It's similar to the Taxi question I posted a while back.  Nice question ln(dx/dy)[/quote]

This one is similar to 2003-2004 mathcounts school test problem.  I remembered I did one like this.

[hide]The probability of getting a red ball from box A is 3/5 * 1/2 = 3/10, and the probability of getting a red ball from box B is 1/5 * 1/2 = 1/10, so the probability that the red ball was drawn from box A instead of box B is

3/10 over 1/10+3/10
which is 3/10 over 2/5
which is 3/10 * 5/2
which equals 3/4?

I'm not sure about my answer...

On second thought, the number of balls in each box is the same, so the number of red balls in box A over the number of total red balls should be answer. So, it's 3/4.[/hide]

I agree that it is 3/4 with the second method that PTheodora mentioned.

Billy

[hide]The probability that A happens given that B is true is

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

So the probability that box A was picked given that the ball is red is

$\frac{\frac{1}{2} * \frac{3}{5}}{\frac{2}{5}} = \frac{3}{4}$[/hide]

[quote="yakko"][quote]No, according to Ln(dx/dy), 3/4 doesn't work...[/quote]

I also got 3/4 and my solution was exactly the same as that posted by PTheodora.  Unless I'm seriously misreading the question, that answer is correct.  When it says:

[quote]A box is chosen at 50-50 and a red ball is drawn. [/quote]

I assume that means "A box is chosen randomly at 50-50, and then a ball is drawn randomly from that box, which turns out to be red."

Yakko[/quote]

You see, in my first post I posted 3/4 but then ln(dx/dy) said it was [quote]Not even close. Was that a random guess?[/quote]... so I'm just quoting him... don't mind my two edits on the first one.. I typed in something wierd for some reason and took two edits to take it out...

Everyone: STOP GUESSING $\frac{3}{4}$ :!:

Think conditional probability and B****'* *****M.

I don't think that 3/4 was a guess for everyone, that's just what everyone thinks it is, and apparently everyone's wrong, so you can calm down a bit ln(dx/dy). Can you post the way to do this problem, since most/nearly all of the AoPSers here doing Mathcounts have answered 3/4, which you have clearly stated is wrong.

Isnt it 1/2?? It is a 50-50 that either was selected

That's the other thing. However it states that a red ball was selected, so unless this is one of those trick questions, I don't think it's 1/2.

[quote="ln(dx/dy)"] A box is chosen at 50-50[/quote]
I think its right ln(dx/dy) will know

I don't know if I said this before, but I meant you might be right, but I asked why they put this in:
[quote]and a red ball is drawn[/quote]
That's why everyone is answering 3/4. See Chesspro's solution for the logic in this.

I'm going to agree with everyone and say it's 3/4 no matter how I try to solve it.

There is a 2/5 chance of getting a red ball overall, and there is a one half chance of choosing box A and a 3/5 chance to select the red ball in Box A, so there is a 3/10 chance of selecting Box A and getting the red ball in it.

3/10 divided by the 2/5 gives us 3/4.
If you do the same steps for Box B, you will get 1/4.

ln(dx/dy), maybe you should tell the correct answer and tell us how you got it... even though i didn't know how to do this before, looking at the proves, i think their way is correct...  :?

I think its 1/2......only because he said it was the wrong answer, otherwise i probably would of said 3/4 too.

There is 3/10 probability that a red ball would be selected from box A, and 1/10 probability one would be drawn from Box B. There is 4/10 a chance of getting a red ball at all. Since 3 out of the 4 tenths are from box A, the answer is 3/4, right? Considering that most posters got that for an answer, it is probably correct. In(dx/dy), I think you might have made an error posting the problem and left out something, or you have the wrong answer yourself. If you believe that we are all making a mistake, please post the answer and explanation. We are spending too much time on this problem.

I have states tomorrow, so does anyone have some last minute probability tips for me? Thanks...

Its 1/2. There is an equal chance for each!

Good luck hockey

[quote="hockeyadi23"]If 3/4 or 1/2 is wrong, then can rcv or ln(dx/dy) post the real solution please?[/quote]
Unless there is some trick wording to the question, which I do not see, the correct answer is $\frac{3}{4}$, as many people have reported.

Since some people, including the author of this question, seem to disagree, I ran a simulation to confirm the answer.  The simulation is as follows:

1.  Choose Box A or Box B with equal probability.
2.  Choose one ball from the selected box.
3.  If the ball is Red, tally a count for "Box A" or "Box B", depending on the choice made in step #1.  If the ball is blue, do not tally anything.
4.  Repeat.

After 100,000 iterations, the "Box A" tally was 29,880.  The "Box B" tally was 9899.  The total number of tallies was 39779.

This simulation confirms the probability of having chosen Box A is approximately 3/4.

RCV - hope you didn't do all those trials by hand :)

[quote="ln(dx/dy)"]Box A and Box B contain blue and red balls. Box A containes 3 red balls and 2 blue balls, and Box B contains 1 red ball and 4 blue balls. A box is chosen at 50-50 and a red ball is drawn. What is the probability that Box A was selected?

Moderators: Feel free to bump this if it is too difficult for the Mathcounts forum.[/quote]
[hide="my solution"]
1/5 chance - B
3/5 chance- A

3/4

I think that's how he got 3/4.
[/hide]

[quote="xtrememath8"][quote="biffanddoc"]Its 1/2. There is an equal chance for each![/quote]

I can see why you would think this, but it is more probable that a red ball would be drawn from Box A than Box B. RCV's probability example shows this well. The ratio of red balls to blue balls is greater in Box A. If you choose Box A, then you have a 3/5 chance of getting a red ball, and 2/5 chance of getting a blue ball and having to start over. If you choose Box B, you have a 1/5 chance of getting a red ball, and 4/5 chance of getting a blue ball and having to start over. If you stop when you get a red ball, the chances are greater you got it from Box A, because of the ratio difference from A to B. I see where your answer is coming from, and perhaps the question should have been stated more clearly, but the chances are greater from A to B.[/quote]

Exactly... are you sure (ln(dx/dy)) that your answer is correct? Cya tomorrow at states.. btw what team are you on (so I can say hi)?

[quote="TheJerk"]Define the following events:

$A$={box A is chosen}, $B$={box B is chosen}, $R$={Red ball is drawn}.

Then $P(R|A)= \frac{3}{5}$, and $P(R|B)= \frac{1}{5}$. Clearly $P(A)=P(B)= \frac{1}{2}$. Therefore $P(R \cap A)= P(R|A) \cdot P(A)= \frac{3}{5} \cdot \frac{1}{2}= \frac{3}{10}$.

Similarly, $P(R \cap B)=P(R|B) \cdot P(B)= \frac{1}{5} \cdot \frac{1}{2}= \frac{1}{10}$. Now we can find $P(R)= P(R \cap B) + P(R \cap A)= \frac{1}{10} + \frac{3}{10}= \frac{4}{10}$.

Now, given that a red ball is selected what is the probability that box A was selected? Translation: what is $P(A|R) \ ?$

$P(A|R)= \frac{P(A \cap R)}{P(R)}= \frac{P(R \cap A)}{P(R)}= \frac{3}{10} \cdot \frac{10}{4} = \frac{3}{4}$.   $\clubsuit$[/quote] I agree but I don't read the other reactions.

ln(dx/dy) has acknowledged the correct answer to this post is 3/4.  He has posted a new, harder challenge, which I have [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=184595#p184595]split to a new thread.[/url]

So the debate  (finally) is over! Always blaming the typo aren't we... no just kidding it can happen to anyone in reailty... let's give ln(dx/dy) a break...


Do there exist four real polynomials such that the sum of any three of them has a real zero while the sum of no two polynomials has a real zero?

Let \[ P_{k}(x)\equal{}\prod_{i\not \equal{}k}^{n}(x\minus{}x_{i})^{2},\] were $ x_{i}$ are distinct.
Then $ \sum_{k\equal{}1}^{n}P_{k}(x)$ had not real root, but \[ S_{k}\equal{}\sum_{i\not \equal{}k}^{n}P_{i}(x)\] had root $ x_{k}$.

How many times must you be told not to post the postal problems. This is getting too much. Mod lock this topic.

OMG!!! this is the $ \text{n^{th}}$ $ \to\infty$ someone is posting the postal problems
 :mad:  :huh:

Also, Rust's post is not a solution.
EDIT: Another topic for the same problem is http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=166926 .


i just got selected for the interview for kvpy- SA stream(XIth std.) i want to know what to do next- how do i prepare for it, when is it. and what do i do for bio-i have PCM computers.
also how is the interview- do they ask questions from anywhere or do they ask on ur preferred subject only????

congrats Rohit
They will ask u from the subjects u want but that isn't a must .u should basically be thorough with what u want them to ask u.They will ask u for choice .Well the interview at least in my case wasn't that formal .they asked easy but tricky questions.Maths was alright they asked me from where i had chosen but physics and chem. was completely arbitrary.
They asked what subjects i had chosen in +1 which didn't include bio so they never asked anything fro there  :D .
What i feel is it's mostly a informal one just to check what u want to do in future...It is the 1st round that will matter the most
u can't simply deny 500 odd people scholarship just by a single interview  :P

Lets see last year i got screwed in BioTech and genetics so learn that well!! All the Best!!!


Let $ \triangle{ABC}$,$ P\ in\triangle {ABC}$.
Call $ {L_a,L_b,L_c}$ is Lemoine point of $ \triangle {PBC},\triangle { PCA},\triangle {PAB}$.Prove that $ AL_a,BL_b,CL_c$ concurrent

[quote="livetolove212"]Please check your problem. I think $ P$ should be Lemoine point of triangle $ ABC$.[/quote]
I don't think so.$ P\in\triangle{ABC}$

[quote="livetolove212"]Then you can see the figure below:[/quote]
Sorry, $ P$ is Lemoine point :)

[quote="livetolove212"]This problem was posted by my teacher [b]Gemath[/b] on Mathlinks before. He told that it could be solved by using vector method. But now it still unsolved.[/quote]
I have sloution,but it's very long.I don't want post it :(

well, you do not have to post the complete solution
but some hints or steps are prefered
thanks

[quote="livetolove212"]This is the solution by my teacher [b]Tran Quang Hung[/b], who has nickname [b]gemath[/b] in mathlinks. It was written in Vietnamese  :)[/quote]
Thanks dear [b]livetolove212[/b]
 :)

[quote="livetolove212"]This is the solution by my teacher [b]Tran Quang Hung[/b], who has nickname [b]gemath[/b] in mathlinks. It was written in Vietnamese  :)[/quote]
Nice sloution.I use isogonal  :)

Interesting idea.Can you post your solution or some hints?


Let $ABC$ be a triangle. Find the point $D$ on its side $AC$ and the point $E$ on its side $AB$ such that the area of triangle $ADE$ equals to the area of the quadrilateral $DEBC$, and the segment $DE$ has minimum possible length.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16259][b][color=red]This[/color][/b][/url] is similar but it asks only how many segments $DE$ are there

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=336213#336213][b]Here[/b][/url] you can find a way to construct the segment, when $BC$ is the shortest side

Simply
$(ABC) = \frac{1}{2} AB \cdot AC \cdot sinA$

$(ADE) = \frac{1}{2} AD \cdot AE \cdot sinA$

$(ABC) = 2(ADE) \Rightarrow AB\cdot AC = 2 AD \cdot AE$

And the minimum length of $DE$ is when $AD=AE\Rightarrow AD^2 = \frac{AB\cdot AC}{2}$

So $AD$ is the geometric mean of $AB$ and $\frac{AC}{2}$


geometry
3D geometry
tetrahedron
trigonometry
parallelogram
geometry proposed

Let $ABCD$ is a tetrahedron. Show that
a)If $AB=CD,AC=DB,AD=BC$ then triangles $ABC,ABD,ACD,BCD$ are acute.
b)If the triangles $ABC,ABD,ACD,BCD$ have the same area , then $AB=CD,AC=DB,AD=BC$.

[quote="gemath"]They are only the fundamental properties of quasi-tetrahedron, try them in   "Problems in solid geometry" of I.F.Sharygin ![/quote]

But I haven't got the this book. Can you post your solution?

I change you notation tetrahedron $SABC$
a) With $SA=BC=a,SB=CA=b,SC=AB=c$ in triangle $ABC, bc\cos A=b^{2}+c^{2}-a^{2}$ but easy to see $SA.BC,SB.CA,SC.AB$ is three side of a triangle $\Rightarrow b^{2}+c^{2}>a^{2}\Rightarrow A<\frac{\pi}{2}$ similar to others vetex
b)$S_{SAB}=S_{SBC}=S_{SCA}=S_{ABC}$ we will show $SA=BC,SC=AB,SB=AC$
Project $ABCD$ to plane that is parallel with lines $SA,BC$ easily seen the projections of $SBC$ and $ABC$ have the same area similarly projections of $SAB,SAC$ have the same areas Thus the paralelogram with two diagonal projections of $SA,BC$ is projections of $ABCD\Rightarrow SC=AB,SB=AC$
and similarly for other vertex!
 :D

[quote="gemath"] $bc\cos A=b^{2}+c^{2}-a^{2}$ [/quote]

You are wrong! :D This is solution of a)

Let $I$ is midpoint of $BC$, we have $2IA=IA+IS>SA=BC$ therefore $\widehat{BAC}<90^{0}$,...

You are right I forgot number $2$ but in a solution you shouldn't pay some trivial things!  :rotfl:


If 120 is divided into 3 parts which are proportional to 1, $ 1/2$, $ 1/6$, what is the middle part?

If we add up $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{6}$, we get $ \frac{5}{3}$. Dividing $ 120$ by that gives $ 72$, we want $ \frac{1}{2}$ of that or $ \boxed{36}$.


[quote="elcric"][i]The following questions come from the 2005 National Sprint Round.  If you have it, please see the diagrams from there.  Sorry that I don't know how to make pictures.  Thanks in advance.[/i]

22. How many continuous paths from A to B, along segments of the figure, do not revisit any of the six labeled points?

26. When seven people go out for dinner the first time, they sit as shown in the left figure.  When the same seven people go out for diner the following time, they agree that no one will sit next to a person whom they sat next to the previous time.  Bob and Ed sit down first, as shown in the right figure.  In how many different arrangements can the other people now sit?

27. The front view and side view of Arrangement A are shown here.

Below are the front view and side view of some arrangement of unit cubes.  Let N denote the largest number of possible unit cubes in an arrangement with the two view shown below, and let n denote the least number of possible unit cubes.  What is the value of N - n?[/quote]

To get nationals, it costs money.  I doubt many people have them...you could scan teh picture.

if possible try to use LaTeX (even though i am being a hypocrite here)
click on the left interface on the AoPS pages on the link called LaTeX

Well, if anyone happens to have the 2005 National Sprint Round, would they please post the corresponding pictures?  I don't know how to.  :(  Thanks.

you could add attachments if you look at the bottom when you make a post....
just attach the pictures like that....


function
inequalities
LaTeX
integration
trigonometry
complex analysis
complex analysis unsolved

For -1 <= x <= 1, we define a function Pn(x) by

Pn(x)=(1/Pi)Sum from zero to Pi (x + i(sqrt(1-x^2))cos(theta))^n d theta
where n = 0,1,2,... . How can we show that this function satisfies the inequality  | Pn(x) | <= 1 ?

I can't tell what this question is supposed to be. Can you clarify (preferably with $\LaTeX$)?

I'm guessing $P_{n}(x)=\frac{1}{\pi}\int_{0}^{\pi}(x+i\sqrt{1-x^{2}}\cos\theta)^{n}d\theta$. If my guess is correct, then the inequality $|P_{n}(x)|\le 1$ follows from $|x+i\sqrt{1-x^{2}}\cos\theta|^{2}=x^{2}+(1-x^{2})\cos^{2}\theta\le 1$.

I believe it actually is:\[P_{n}(x)=\frac{1}{\pi}\int_{0}^{\pi}\left(x+i\sqrt{1-x^{2}}\right)\cos^{n}\theta \ \ d\theta\]
look at the parenthesis
and it makes the problem nicer a little :P
because let $x=\cos \phi$
Then \[P_{n}(\cos \phi)=\frac{1}{\pi}\int_{0}^{\pi}\left(\cos \phi+i\sin \phi\right)\cos^{n}\theta \ \ d\theta\]
which we all recognize now :wink:

Here mlok's guess is correct, and i solved the question in the same manner. It was not as tough as it seemed to me at first glance. Thanks for your interest.


A sequence $ < L_{n} >$ is defined on positive integers ,$ L_{1} \equal{} 1$ and $ L_{2} \equal{} 3$ , $ L_{n} \equal{} L_{n \minus{} 1} \plus{} L_{n \minus{} 2}$ . Show that $ L_{p} \minus{} 1$ is divisible by $ p$ if $ p$ is prime .

Try to derive the general formula for the nth term which is:

$ ((1 \plus{} \sqrt 5)/ 2)^n$ + $ ((1 \minus{} \sqrt 5)/ 2)^n$

And proceed from here
P.S. I will post my complete solution if nobody gets this problem


How many total days were there from March 1, 1998, to February 1, 2008, including the start and end days?

Let's start from February 1st. There are 3654 days. We then subtract 28 to get $ \boxed{3626}$.


Can someone explain to me why [url=http://www.artofproblemsolving.com/Wiki/index.php/2009_AMC_10A_Problems/Problem_22]this[/url] isn't $ \frac{1}{6}$, as logically predicted?

You have to factor in the fact that one dice may have 2 of one number.
For example, dice #1, you get some number n. So, for the top of dice #2, there are still 11 possibilities, since there are 11 more numbers in the bag. Of those 11 possibilities, 2 of them will add it up to 7.

[hide="Explanation"]Say that the first die has red stickers labeled 1 through 6 and the second die has blue stickers labeled 1 through 6.  Without rearranging the labels, it would be impossible to get a sum of 7 through a combination like blue 3 and blue 4 (because they're on the same die).  But by rearranging them randomly any combination is possible.

So now, rather than rolling dice, we are choosing two numbers from the set {R1, B1, R2, B2, R3, B3, R4, B4, R5, B5, R6, B6}.[/hide]


Again, I just started to learn matrices, and very confused with the following problem. Thanks in advance.

Find a 2x2 matrix A for which,

\[ A* \begin{pmatrix}2\\ 1\\ \end{pmatrix}=\begin{pmatrix}-1\\ 4 \end{pmatrix}\]

and

\[ A* \begin{pmatrix}5\\ 3\\ \end{pmatrix}=\begin{pmatrix}0\\ 2 \end{pmatrix}\]

PS THIS is suppost to be brakets not paranthasis....my coding is at beginner level

Your two conditions can be combined so it reads this:

$ A\cdot\begin{bmatrix}2&5\\1&3\end{bmatrix}=\begin{bmatrix}-1&0\\4&2\end{bmatrix}$

[Hint about $ \text{\LaTeX}$ coding: hold your mouse over that to see what the syntax looks like.]

For short, we have $ AB=C.$ As long as $ B$ is invertible (which it is, in this instance: $ \det(B)=1$), then we can solve for $ A$ as $ A=CB^{-1}.$ (Do be careful about whether you are multiplying on the left or on the right.)

Wow thanks, I would of never of though putting them together.

Here is my answer.

$ AX=B$, I will use $ X$ b/c this is how I remember the formula. Don't want to get confused.
$ AX=B$ = $ A=BX^{-1}$, just like stated previously by Kent.

Thus,
\[ \begin{bmatrix}2 & 5\\ 1 & 3\\ \end{bmatrix}=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}\]
Now solving for inverse I get the following,
\[ \begin{bmatrix}1 & 0\\ 0 & 1\\ \end{bmatrix}=\begin{bmatrix}3 &-5\\-1 & 2\end{bmatrix}\]
So, $ BX^{-1}=A$
So,
\[ \begin{bmatrix}-1 & 0\\ 4 & 2\\ \end{bmatrix}\begin{bmatrix}-3 & 5\\-1 & 2\end{bmatrix}=\begin{bmatrix}-3 &5\\ 10 & 16\end{bmatrix}\]
Thanks for the coding tip as well.

Some typos to clean up there.

$ \begin{bmatrix}-1&0\\4&2\end{bmatrix}\begin{bmatrix}3&-5\\-1&2\end{bmatrix}= \begin{bmatrix}-3&5\\10&-16\end{bmatrix}$

Checking:

$ \begin{bmatrix}-3&5\\10&-16\end{bmatrix}\begin{bmatrix}2&5\\1&3\end{bmatrix}= \begin{bmatrix}-1&0\\4&2\end{bmatrix}$


Let $ F$ be a subset of $ [0,1]$ constructed in the same manner as the Cantor ternary set except that each of the intervals removed at the nth step has lenght $ \alpha 3^{ \minus{} n}$ with $ 0 < \alpha < 1$.   Then $ [0,1] \backslash F$ (the complement) is dense in $ [0,1]$.

In other words, $ F$ does not contain any nontrivial intervals. Consider the maximal length of intervals contained in the $ n$th generation of $ F$.

Same reasoning works if you remove just the middle point of each interval.

Do you mean that $ F$ does not have interior?

[quote="LittleFermat"]I have an idea, but I do not know how to write it down.    If I have an arbitrary open interval contained in $ [0,1]$, in some moment this must be divided in the process of constructing the generalized Cantor set $ F$, so that means there are some points which cannot be in $ F$.    Well, as I said, I do not know how to write this into a proof.[/quote]

I feel sorry I am writing to many posts, but I have got some ideas about the problem.     I think I have got the proof, but there is a detail that I cannot solved, is it true that the length of the connected components of the $ nth$ generation of $ F$ approaches to zero?  if it is, can anyone give me a hint for the proof for it?

[quote="LittleFermat"]is it true that the length of the connected components of the $ nth$ generation of $ F$ approaches to zero?[/quote]
At the $ n$th stage, there are $ 2^n$ of these components, and they all have the same length. Even without calculating that length, you can see that it must get small rather quickly!


How can I show this one:

Let $V$ be a linear space over a field $K$ and $dim(V)=n$. Show that $V^{2}$ is a linear space over $K$ and $dim(V^{2})=2n$.

Can you write down a basis for $V\times V$ given a basis for $V$?

[quote="JC1"]I don't understand this cartesian product $V*V$ with linear spaces. It is a pair $(x,y)$ , where $x$ and $y$ $\in V$.
If $w=v_{1}, v_{2}, \cdots, v_{n}$ is a basis of $V$, I know I can span the space of all possible values of $x$ and $y$ with $w$, but I don't know how to represent $V^{2}$ with this basis.[/quote]

Well with that basis and definition for the product $V \times V$, we can just take $(v_{1}, 0); (v_{2}, 0); \cdots; (v_{n}, 0); (0, v_{1}); (0, v_{2}); \cdots; (0, v_{n})$, which will be a basis for $V \times V$.


Let$p$be a prime number and $a$,$n$positive integers. Prove that if$2^{p}+3^{p}=a^{n}$then $n = 1$.(Ireland 96)


Ballistics

A child throws a ball up a hill that makes an angle of 45 degree with the horizontal. The ball lands 100 feet up the hill. Its trajectory is a parabola with equation y = -x^2 + ax for some number a. Find a.

I figure I use something like y = -x^2 + ax + 100, but I don't know what to do.

Not really sure about this, but I'll give it a shot. [hide]We use the x-intercept, which is (100,0). We plug that into our equation:

0 = -(100^2) + a(100)
a(100) = 100*100
a = 100.[/hide]

Haven't done ballistics before.... Can you explain?

we must find solution to
x^2-ax=0
the first is 0
an the secon is a

the solutio are the x-intercept
O(0,0)is the point from where the child throw the ball
and A(0,a)is the point from where crached
the a=100
because as you say The ball lands 100 feet up the hill


modular arithmetic
number theory proposed
number theory

Let $ \{m_1,m_2,\dots\}$ be a (finite or infinite) set of positive integers. Consider the system of congruences

(1)   $ x\equiv 2m_i^2 \pmod{2m_i\minus{}1}$ ($ i\equal{}1,2,...$ ).

Give a necessary and sufficient condition for the system (1) to be solvable.

[quote="Albanian Eagle"]Let $ \{m_1,m_2,\dots\}$ be a (finite or infinite) set of positive integers. Consider the system of congruences

(1)   $ x\equiv 2m_i^2 \pmod{2m_i \minus{} 1}$ ($ i \equal{} 1,2,...$ ).

Give a necessary and sufficient condition for the system (1) to be solvable.[/quote][hide="Solution"]
We want $ x\equiv 2m_i^2\bmod (2m_i \minus{} 1)$, so $ x\equiv m_i\bmod (2m_i \minus{} 1)$.  This means that\[ 2x\equiv 2m_i\bmod (2m_i \minus{} 1)\equiv 1\bmod (2m_i \minus{} 1)\] We let $ p_j^{e_j}$ be the largest power of $ p_j$, dividing any of the $ 2m_i \minus{} 1$ across all possible $ i$.  We then have that $ 2x\equiv 1\bmod (2m_i \minus{} 1)$ for all $ i$ if and only if $ 2x\equiv 1\bmod p_j^{e_j}$ for all possible $ j$.  Yet, $ p_j^{e_j}$ are distinct primes powers, so they are pairwise relatively prime.  Then, by the Chinese Remainder Theorem, there exists exactly one value of $ 2x\bmod k$ that satisfies the conditions.  Notice that $ 2x\equiv 1\bmod k$ works.  This implies that $ x\equiv \frac{k\plus{}1}{2}\bmod k$.  Yet, this is only finite if $ k$ is finite, so if the sequence $ 2m_i\minus{}1$ has a finte LCM. [/hide]


It was this year on BMO, [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=19&year=2006]Problem 1[/url].


Q11 A sample of phosphorus pentachloride is placed in a sealed container, where it decomposes
into phosphorus trichloride and chlorine gas according to
PCl5 (g) PCl3 (g) + Cl2 (g)
Once the mixture is at equilibrium, a small quantity of helium gas is added, at constant
pressure and temperature, and the mixture is allowed to return to equilibrium. Which of the
following best describes the second equilibrium position with respect to the first?
A The second equilibrium position is the same as the first, because helium is much
lighter than any of the other molecules in the container and will have a negligible
effect on the reaction.
B The second equilibrium position is the same as the first, because helium does not
react with any of the other molecules in the container.
C The second equilibrium position has more PCl3 than the first.
D The second equilibrium position has more PCl5 than the first.
E It is impossible to tell what will happen without knowing the equilibrium constant
for this reaction.
[hide] the answer is c. but i don't know why [/hide]

I think it has to do with the concentration. After the helium is added, the concentration of PCl5, PCl3, and Cl2 all decrease. So the reverse reaction occurs less often because the concentration of reactants is decreased. Thus the forward reaction proceeds until enough product is formed for the reverse reaction to occur at the same rate as the forward reaction (equilibrium). Then there is more PCl3 in the second equilibrium position.

But how does helium effect the concentration?
1. It does not react with anything.
2. And the pressure of each gas will not change due assuming its room temperature and normal pressure, as the molecules are do not interact due the large amount of free space.


[quote="Pirkuliyev Rovsen"]$ f: (0;1) \rightarrow (0; \plus{} \infty)$
$ f(x)f(y) \equal{} f(\frac {xy}{2xy \minus{} x \minus{} y \plus{} 1})$
$ x,y\in(0;1)$
$ x,y\in(0;1)$
[/quote]

Let $ P(x,y)$ the assertion $ f(x)f(y) \equal{} f(\frac {xy}{2xy \minus{} x \minus{} y \plus{} 1})$

$ P(x,1\minus{}x) \iff f(x)f(1\minus{}x)\equal{}f\left(\frac12 \right)$

$ P(\frac12,\frac12) \iff f^2\left(\frac12 \right)\equal{}f\left(\frac12 \right)$

Since $ f(x)>0\ \forall x \in (0,1) \implies f\left(\frac12 \right)\equal{}1$

So $ f(x)f(1\minus{}x)\equal{}1\ \forall x \in (0,1)$

$ f(x)\equal{}1 \ \forall x \in (0,1)$

[quote="sandu2508"][quote="prester"]
So $ f(x)f(1 \minus{} x) \equal{} 1\ \forall x \in (0,1)$
[/quote]

Could you, please, explain why we get that $ f(x) \equal{} 1$ from this line. Notice that $ f(x) \equal{} a^{x \minus{} \frac12}$ also satisfy $ f(x)f(1 \minus{} x) \equal{} 1\ \forall x \in (0,1)$.[/quote]

Yes, but $ f(x) \equal{} a^{x \minus{} \frac12}$ does not match with $ f(x)f(y)\equal{}f(\frac{xy}{2xy\minus{}x\minus{}y\plus{}1})$. Or not?

However, sincerely, I am not sure that $ f(x)\equal{}1$ is the unique solution...

Thanks for your remark


Matt will arrange four identical dotless 1x2 dominoes  on a 5x4 grid. "A" is the upper  left vertex of the grid and "B" is the lower right vertex of the grid. How many distince arrangements are possible?
(in a path, consecutive dominoes must touch at their sides and not just their corners. No domino may be placed diagonally; each domino covers exactly 2 of teh unit squares shown on the grid.

notice that the fact that the dominoes are 1x2 does not hinder their "movement", in other words, every path that can be made with 1x1 dominoes can be made with 1x2 dominoes(convince yourself this is true)

that said, with 1x1 dominoes it is just combinatorics. since it is 5x4, you need to go to the right 4 times, and down 3 times, so out of 7'movements' choose 4 to be the 'to the right' option. thus the answer is 7C4=7C3=7*6*5/(3*2*1)=7*5=35.

Thanks, but, why is it 4 to the right and 3 down (why not 4 to the right and 5 down)?

This "path" formed by the dominoes is like connecting the centers of the unit squares, the equivalent of 4/3.


i dont think its factorizable, unless if u can use $i$ in ur answer.

EDIT: nevermind. Geehoon's got it.

a^3 + b^3 + c^3 - 3*a*b*c, when a = x^2, b = -2x, c = 2.
This is the shortest I could think of!

Any other shorter way to do this?

Problem appeared in Regional Maths Olympiad of Delhi, India in 1993.


Q: The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92.The common ratio of the original G.P. is

a)½                b) 2/3               c)1/3        d) -1/2

Is there a typo here?

$ \frac{a}{1\minus{}r} \equal{} 4, \frac{a^{3}}{1\minus{}r^{3}}\equal{}92$ seems to give $ 23r^{3}\minus{}r\minus{}22\equal{}0$, so $ (r\minus{}1)(23r^{2}\plus{}23r\plus{}22)\equal{}0$

Well I got that $ \frac{a}{1\minus{}r}\equal{}4$ and $ \frac{a^3}{1\minus{}r^3}\equal{}92$ which also gives $ 4(1\minus{}r)\equal{}a$ and $ 92(1\minus{}r^3)\equal{}a^3$
Substituting: $ 92(1\minus{}r^3)\equal{}64(1\minus{}r)^3$. Simplifying results in $ 23(1\plus{}r\plus{}r^2)\equal{}16(1\minus{}r)^2$ and that turns into $ 23\plus{}23r\plus{}23r^2\equal{}16\minus{}32r\plus{}16r^2$. Then we need to find the roots of the quadratic $ 7r^2\plus{}55r\plus{}7\equal{}0$....uh what did I do wrong? :|

even i got the same equation (7r2 + 55r +7)when i tried to solve it but the answer won't come by solving that equation ????


Let $ f(n)$ be the number of permutations $ a_1,...,a_n$ of the integers $ 1,...,n$ such that:

$ (i)$ $ a_1\equal{}1;$

$ (ii)$ $ |a_i\minus{}a_{i\plus{}1}| \le 2, i\equal{}1,...,n\minus{}1$.

Determine whether $ f(1996)$ is divisible by $ 3$.

1996 Canada, #3:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77672[/url]


The 9 tests used to determine the water quality index are dissolved oxygen, fecal coliform, pH, biological oxygen demand, temperature, phosphate, nitrate, turbidity, and total solids.

For my homework problems listed below, I am supposed to say which of the tests apply to each question. I'm not sure if I listed all the tests that apply to each question although I answered every question.

Which of the above can increase if fertilizer from run off gets into the river or stream?

Which of these decrease(s) due to run-off?

Which of these increase(s) due to run-off?

Which of these has to do with the murkiness of the water?

Which of these are important in relatively small amounts?

Which form of nitrogen is used by plants in the nitrogen cycle?

Which of these is important in relatively small amounts?

Which of these contributes most to the water quality index? (For this one, I know that dissolved oxygen is the has the most weight in determining the percent value of the water quality index but does it necessarily contribute the most to the water quality index?)

Which of these contributes least to the water quality index? (For this one, I know that total solids has the least weight but does it necessarily contribute the least to the water quality index?)

Thanks, for raising this question. I answered them to my best potential. I had some mental debate over 2 and 3, though. Most likely it would have to do with the weights of the compounds.

1) Fertilizer is comprised of more phosphate. Thus, phosphate would increase.

2) ---

3) ---

4) Turbidity and total solids.

5) Dissolved oxygen is my best guess.

6) Nitrate is a form of nitrogen. Just look at the prefix.
(Note: You repeated Q5 twice so, it doesn't come again after 6)

7) Fecal Coliform. I've heard about my grandfather's experiments. They say, in India, when testing the potability of water, they test mainly for Fecal Coliform.

The process: They put a container of water on a centrifuge. Let it spin for 1000 revolutions. Take the residue and count the colony count. There is a certain acceptable level of said fecal coliform

8) Temperature change. Temperature change, doesn't change the constituents of the water. So, I don't think it's that much of a factor.

-arb95

I sat in deep contemplation over the one's I could not answer. The following is an edited 1; and the answers for 2 and 3.

1) Phosphate, fecal coliform, turbidity, and total solids. Fecal coliform, bacteria from the fertilizer will increase. Of course, turbidity and total solids will increase.

2) pH will decrease

3) Dissolved oxygen, biological oxygen demand, and total solids. A lot of these come from answer 1.

-arb95


If the digits of a certain two-digit even number are reversed, the resulting number is 54 less than the original number. Find the remainder when twice the original number is divided by 7.

[hide="Solution"]OK, so the digits must be $ 6$ apart from each other.  This leaves $ {60,71,82,93}$.  However, only $ 82$ satisfies the above conditions.  So we have $ 82 \times 2$, or $ 164$.  $ 164 \equiv \boxed{3}\pmod{7}$.[/hide]

OK BEATEN

-The message is too small. Please make the message longer before submitting.

@ernie: Please hide your solutions.
@mathematigian1729: You don't need to say that you essentially got the same solution.


Let $ a,b,c$ be positive reals, no two of which are zero.Prove that:
$ \frac{2a^n\minus{}b^n\minus{}c^n}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{2b^n\minus{}a^n\minus{}c^n}{c^2\minus{}ca\plus{}c^2}\plus{}\frac{2c^n\minus{}a^n\minus{}b^n}{a^2\minus{}ab\plus{}b^2}\geq0$

[quote="Inequalities Master"]Let $ a,b,c$ be positive reals, no two of which are zero.Prove that:
$ \frac {2a^n \minus{} b^n \minus{} c^n}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {2b^n \minus{} a^n \minus{} c^n}{c^2 \minus{} ca \plus{} c^2} \plus{} \frac {2c^n \minus{} a^n \minus{} b^n}{a^2 \minus{} ab \plus{} b^2}\geq0$[/quote]
I think, it's wrong for $ n<0.$ :wink:

Sorry,my mistake,$ n$ is a positive integer.


We add all equations together to get:
$2\sum_{i=1}^5x_i=y\sum_{i=1}^5x_i$.
Which implies that either $\sum_{i=1}^5x_i=0$ or $\sum_{i=1}^5x_i=\text{anything}$ (when $y=2$).
Thus each of the $x_i$ can take on any value.

Are you sure that is an IMO problem??

is this the correct answer? it seems redundant, even to pose this question! was 1963 one of the first years of the IMO or something?

EDIT: Yeah Im wrong  :)
I meant that they won't all work: X1=2, X2=9, X3=12312341234, Y=2........ :D

[quote="me@home"]
Thus each of the $x_i$ can take on any value.
[/quote]

Big mistake.

I'm not sure what you're trying to say... I probably am wrong, but here is my logic:
the equation reduces to a simple $2a=ya$. From here we deduce that all $a$ have a solution. By definition, if $y\not=2$ then $a=0$. Is there a flaw in any of this reasoning?

sorry, I didn't see Arne's post (we posted at the same time..)
I am not very experienced with IMO problems, but what is the big mistake about that assumption? If the five x's add to zero and that is the only limiting criteria, can't we deduce that?

Compare it to this:

Let's solve the system $\left\{ \begin{aligned} xy = z \\ yz = x \\ zx = y \end{aligned} \right.$

Multiplying these equations, we get $(xyz)^2 = xyz$, so $xyz = 0$ or $xyz = 1$.

So $x$, $y$, $z$ can take any value as long as $xyz \in \left\{0,1\right\}$.

Do you see why that's wrong? (Take $x = \frac{1}{2}, y = 1, z = 2$.)

That's exactly what you did. You took some equations, derived a new equation from those and you threw away all equations you got in the beginning. So you just decided you don't need those conditions anymore... Wrong. Got it? :)

I'll continue with my system, to illustrate how this kind of problems needs to be solved.

So now, we know that $xyz = 0$ or $xyz = 1$. Suppose first that $xyz = 0$, then one of $x,y,z$ must be zero. Suppose $x$ equals zero. Then $z$ equals zero, too, and hence $y = 0$. So we get the solution $(x,y,z) = (0,0,0)$.

Now suppose that $xyz = 1$. Then $z = xy = 1/z$, so $z^2 = 1$ and $z = 1$ or $z = -1$.

Suppose that $z = 1$. Then $xy = 1$ and $x = yz = y$. Hence $x = y = 1$ or $x = y = -1$. So we get the solutions $(1,1,1)$ and $(-1,-1,1)$.  Suppose that $z = -1$. Then $xy = -1$ and $x = yz = -y$. Hence $x = 1$ and $y = -1$ or $x = -1$ and $y = 1$. Sho we get the solutions $(-1,1,-1)$ and $(1,-1,-1)$.

So now we've got all of them. Ok?

ok, thank you for explaining... I can be quite slow at times :roll:
I also noticed that the equations in the original problem aren't cyclic, does this complicate the problem?

They are cyclic! (they must be)
Xi+X[size=75]i+2 mod 5[/size]
 :)

I love this problem, since it uses extreme principle. :D
[hide]
First u monotonize the variables:$x_1\geq x_2\geq x_3\geq x_4\geq x_5$ (it doesn't have to be in this way).
Then the first equation $x_5+x_2=yx_1$ implies that $x_1+x_1=2x_1\geq yx_1$. Simplifying this, we get: $y\leq 2$.
Do the same thing to the last equation and we get:
$y\geq 2$. Hence, $y=2$.
Now sub in $y$ value to our original equations.
Looking the first equation $x_5+x_2=2x_1=x_1+x_1$, we conclude that $x_1=x_2=x_5$. Because $x_1$ was the largest among the five. In the same way, we come up with
$x_1=x_2=x_3=x_4=x_5=1$.
[/hide]
Can anyone check my solution? Thanks,

your solution should be correct, but what kind of verification is my opinion?

[quote="kimby_102"]Can anyone check my solution?[/quote]

Your solution is wrong. The set of solutions if much larger.
What went wrong? You can't divide by zero, multiplying by a negative number makes an inequality sign switch, etcetera.

[quote="kimby_102"]
Looking the first equation $x_5+x_2=2x_1=x_1+x_1$, we conclude that $x_1=x_2=x_5$.
[/quote]

That was wrong.  :)

[hide="solution"]
a trivial solution is (obviously) $x_1=x_2=x_3=x_4=x_5=0$.
adding the five equations gives:
$(x_1+x_2+x_3+x_4+x_5)(y-2)=0$.

If $y=2$ the second and third equations yield $x_1+x_4=x_2+x_3$ and by the fifth one: $x_1+x_4=2x_5$ so
$x_1+x_2+x_3+x_4+x_5=5x_5$
the fact that cyclic permutations does not alter our eqality we can conclude that:
$x_1+x_2+x_3+x_4+x_5=5x_i$ for any $i$.

It follows that all $x_i$ are equal.  it is easy to check that if $y=2$, then solution is obtained when all $x_i$ are set equal to a given constant.

when $y\not = 2$, we use elimination.  we can eliminate $x_5$ from the given equations, $x_4$ from the remaining four, and $x_3$ from the remaining 3; we find:
$(y^2+y-1)(x_2-x_1)=0$  (6)

and
$(y^2+y-1)(x_2-(y-1)x_1)=0$   (7)

if $y^2+y-1\not = 0$ then $x_1=x_2$ by (6), and hence $(y-2)x_1=0$ by (7).  Since we are assuming $y\not = 2$, this implies that $x_1=x_2=0$, the original equations yield: $x_3=x_4=x_5=0$, which is the trivial solution.

if $y^2+y-1=0$, (6) and (7) are satisfied for all values of $x_1$ and $x_2$.  so $x_1$ and $x_2$ may be given arbitary values; but by the second original equation: $x_3=yx_2-x_1$ by third: $x_4=yx_3-x_2=(y^2-1)x_2-yx_1$, and by first $x_5=yx_1-x_2$ where $y$ is either root of equation: $y^2+y-1=0$

QED ;) [/hide]


It seems like I've read this somewhere, and I looked around for it.  Oh well, I don't have much time.  When do we have to let y'all know if we're going?  I'm having a hard time getting a job.  Also, can we just e-mail somebody for the sake of expediancy.

[quote="juggle7"]It seems like I've read this somewhere, and I looked around for it.  Oh well, I don't have much time.  When do we have to let y'all know if we're going?  I'm having a hard time getting a job.  Also, can we just e-mail somebody for the sake of expeciancy.[/quote]

If you have questions, you can always email them to camp@mathcamp.org.

If you've been admitted, the deadline for registering is May 31.

[quote="aurelie"]What exactly has to be sent in by May 31? Is it all registration information except for the travel information, which is due by June 17?[/quote]

Right.

[quote="aurelie"]Thanks! And another quick question: Does the term "International Student" refer to anyone who comes from a country other than US OR Canada?[/quote]

In which context are you unsure?

[quote="aurelie"]There is a sheet entitled "Information for international students" in the registration pack, with a section on visas.  It says that students must get a visa unless they are from countries "exempt from US visa requirements"...and Canada is not one of the countries on the list... I was quite surprised, since I have been in the US as a "visitor" many times without a visa...but I just supposed that Canadians might not be considered as "international students" sice it is Canada/USA Mathcamp. :?[/quote]

Yes, you certainly do not need a visa, so whoever wrote that information sheet (not me) must have had in mind that it is not for Canadians.

[quote="math92"]what grade level is mathcamp?[/quote]High school.

Altough it is possible to attend while in middle school.


This is #21 on 2002 contest. It's pretty challenging.

Consider the sequence of numbers: $4, 7, 1, 8, 9, 7, 6, ...$. For $n>2$, the $n$th term of the sequence is the units digit of the sum of the two previous terms. Let $S_n$ denote the sum of the first $n$ terms of this sequence. What is the smallest value of $n$ for which $S_n>10,000$?

[hide]
seeing nothing better to do, I listed out some more terms:
$4,7,1,8,9,7,6,3,9,2,1,3,4,7$ :o  it repeats!  in blocks of 12 with each block adding up to 60.  Then the first 166 blocks have 1992 terms and add up to 9960.  Counting out the rest, I believe it comes out to $1999$?  hope thats one of the choices.[/hide]

I think this is called the necklace problem

Yay, I just did a practice test and did this problem today!  :D

Try finding a pattern first...

4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, ...

As one can see, there [i]is [/i]a pattern... the sequence loops over and over...

4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3,... repeats. There are 12 terms, whose sum is 60. Therefore, 10000/60 yields 166 repetitions, yielding 166*12 = 1992 terms, and the sum being 166*60 = 9960. There is still a sum of 40 to be accounted for. Therefore, we start counting out by hand until the sum first reaches over 10000. Thus, we have 4 + 7 + 1 + 8 + 9 + 7 + 6 = 42. This is 7 terms. You add this to 1992, yielding [b]1999  [/b]for n.[/code][/i]


Can someone suggest approaches to solving the following?

The five-digit number A845E is divisible by 72, and A & E are digits.
What values do A and E have?

[hide="The way I did it"]
You know that the number A845E must be divisible by both 8 and 9, as it is divisible by 72.  The divisibility rule for 8 is that the last three digits must be a multiple of 8; the divisibility rule for 9 is that the sum of all th digits in thenumber must be a multiple of 9.

The only multiple of 8 that is 45E is 456.  So you have A8456.

To make that number divisible by 9, set A as 4.  This is because 8+4+5+6=23, so the nearest multiple of 9 is 27.

The number is 48456.  This number is indeed divisible by 72: $ \frac{48456}{72}\equal{}673$
[/hide]


Um I usually use the AoPS skin, not the mathlinks one but today, I decided to use the mathlinks one and noticed that the Latex help button goes to the old page, not the new AoPS wiki latex page.

Maybe it should be changed to the new AoPS wiki one...
 :blush:


arithmetic sequence
geometric sequence
geometric series

If $ a$, $ b$, $ c$, and $ d$ are positive real numbers such that $ a$, $ b$, $ c$, $ d$ form an increasing arithmetic sequence and $ a$, $ b$, $ d$ form a geometric sequence, then $ \frac{a}{d}$ is

$ \textbf{(A)}\ \frac{1}{12} \qquad
\textbf{(B)}\ \frac{1}{6} \qquad
\textbf{(C)}\ \frac{1}{4} \qquad
\textbf{(D)}\ \frac{1}{3} \qquad
\textbf{(E)}\ \frac{1}{2}$

hello, do you meant $ a,b,d$ or $ a,b,c$ form a geometric sequence?
Sonnhard.

[hide] Let the common difference be x, so our arithmetic sequence is $ a, a\plus{}x, a\plus{}2x, a\plus{}3x$. Since a, b, and d form a geometric series, $ b^2\equal{}ad$. Substituting, we have $ a^2\plus{}2ax\plus{}x^2\equal{}a^2\plus{}3ax$ or $ ax\equal{}x^2$. Since it is an increasing aritmetic sequence, x is not zero, so we divide by x to find a=x. Hence, d=4a, and $ \frac{a}{d}\equal{}\frac{1}{4}$ so the answer is $ \textbf{(C)}\ \frac{1}{4}$.[/hide]

lets say that the common difference is x

then a, b, c, d is $ a, a \plus{} x, a \plus{} 2x, a \plus{} 3x$

by the definition of geometric sequence, $ (a \plus{} x)^2 \equal{} a(a \plus{} 3x)$

$ a^2 \plus{} 2ax \plus{} x^2 \equal{} a^2 \plus{} 3ax \implies x \equal{} a$

so the sequence becomes $ a, 2a, 3a, 4a$

thus $ \frac {a}{d} \equal{} \frac {a}{4a} \equal{} \frac {1}{4}$

answer: C

EDIT: blargh beaten

hello, we have $ b\equal{}a\plus{}d_1,c\equal{}a\plus{}2d_1$ and $ d\equal{}a\plus{}3d_1$, further $ b\equal{}aq,d\equal{}aq^2$, from here we get $ d_1\equal{}aq\minus{}a$ and $ aq^2\equal{}a\plus{}3(aq\minus{}a)$ with $ a>0$ we get $ q^2\minus{}3q\plus{}2\equal{}0$ with $ q_1\equal{}2$ or $ q_2\equal{}1$, so we get
$ d\equal{}4a$,hence $ \frac{a}{d}\equal{}\frac{1}{4}$ or $ \frac{a}{d}\equal{}1$.
Sonnhard.

Or since this is AMC you just guess 1,2,3,4 is a possible sequence, lol :P

Another way is to have $ b^2 \equal{} ad$ and $ d \minus{} b \equal{} 2x \equal{} 2(b \minus{} a) \implies 3b \equal{} 2a \plus{} d$.
Squaring both sides we have $ 9b^2 \equal{} 4a^2 \plus{} 4ad \plus{} d^2 \equal{} 9ad$.
Setting $ d \equal{} 1$ we have $ 4a^2 \minus{} 5a \plus{} 1\equal{}0$ but since $ a\neq b$, $ a \equal{} 1/4$.


Hello!
I want you to advice me(if you have) about Mathematic Olympiads. But it is better if the book is new(I mean, for example the questions of last years).


AMC
AIME
geometry
geometric transformation
rotation
algebra
polynomial

Two of the squares of a $ 7\times 7$ checkerboard are painted yellow, and the rest are painted green.  Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane of the board.  How many inequivalent color schemes are possible?

Generalizing a little, on a (2n+1)x(2n+1) board.

[hide]

There are  symmetric ./. (n+1,n+1) boards

There are  non symmetric boards.

The total number of different boards is :

n = 3 : 300 boards.[/hide]

yea, sorry in advance if you do not like people bringing old threads back..but since this was an AIME problem and AIME is coming soon, I was wondering how do you know(or prove) the generalized formula for non-symmetric and symmetric boards?

and is there another way to do this (looking at cases)

Consider an $2n+1\times 2n+1$ board. Most of the time, the four rotations of the board describe four distinct colorings. Should there be a single yellow square, we would know that this is true all the time. (The exception: the single center square, but this doesn't matter). However, less distinct colorings may be obtained when there are two squares since one square may map to the other and vice versa. Together, the main case and this side case must account for all cases, since no single square can map to itself in a distinct rotation (the special case can be ignored again).

Simply count the number of ways to choose two squares to simultaneously flip; clearly, it must be through a 180 degree rotation.

All pieces that flip upon this rotation can be found by taking an arbitrary square, flipping it 180 degrees, and marking this point as its conjugate. Then all square-conjugate distinct pairs are precisely all positions that flip.
We can pick a single, non-center square in precisely $4n^{2}+4n$ ways so there are $2n^{2}+2n$ positions that yield only 2 distinct positions throughout the four rotations.

The total number of ways to choose two squares is ${(2n+1)^{2}\choose 2}=(4n^{2}+4n+1)(2n^{2}+2n)$ thus the number of "normal" cases which yield four distinct boards upon rotations is $8(n(n+1))^{2}$

Considering the two cases, there are $n(n+1)$ and $2(n(n+1))^{2}$ coloring schemes, or $(2n^{2}+2n+1)(n^{2}+n)$ total colorings.
When $n=3$ this is $(25)(12)=\boxed{300}$ different coloring schemes.

Wait, if (n)(n+1)(2n^2+2n+1) is the generic formula for odd boards, what about even boards?

Ya, what the exact formula for even 2n by 2n boards incase there is a question on AIME this year(so I can plug in the formula :D )
thanks

Did you not read my solution? It would be stupid to memorize a "formula"; this is a really specific problem. And it certainly wouldn't be repeated... :mad:
If you read the logic and understand how the solution comes about you should be fine, this would be a 4 or 5 on today's AIMEs. Worst case, calculate it for small numbers and find a pattern. But don't memorize a formula.
If you have any questions about the solution I'd be ready to answer them though

Actually there are a couple of pairs of AIME problems that are "nearly" the same... for example (1984 AIME #9 and 1992 AIME #7) or (1993 AIME #10, 1988 AIME #10, and 2004 AIME1 #3) and there have been atleast 4 AIME problmes dealing with sports tournaments and chances that the have more wins than losses, or the chances that there is no undefeated team..and I remeber there were 2 seperate problems that talked about disjoint sets..and how many ways are there so that things are non consecutive..and so on

but nevertheless ill try and figure out this myself thanks anyways

Can someone explain the hole or flaw with my solution? (sorry for reviving and old thread, but I really need to know)

Case 1: The first yellow square is the center square. Then there are 48/4 = 12 distinct combinations.

Case 2: The first yellow square ain't the center square, and neither is the second.

There are 48 squares, but you can rotate the table 4 ways, so there are 12 ways to pick the first square. Since the second square is relative to the first square, there are 47 ways to pick the second. Thus there are 47*12 = 564 ways.

Adding it up, you get 576 ways.

EDIT: Nvm, I see it. There are those that are symmetric and have only two ways.

georgia math dude, your main problem in overcounting is that you are counting the yellow squares as distinct, when actually, they are distinct.  And yes, symettry is your second problem.

[quote="georgiamathdude"](sorry for reviving and old thread, but I really need to know) [/quote]
There is not only nothing wrong with reviving an old thread linked to from the contests section, but it is recommended.

anyone explain why this solutions wrong?

We have 3 indistinct colorings for each distinct one
Thus, $ {49\choose 2}/4 \equal{} 49*6\equal{}294$.

That's not true for every coloring.  Some colorings stay the same when they're rotated.  In terms of [url=http://en.wikipedia.org/wiki/Burnside%27s_lemma]group actions[/url], you're assuming that every orbit has size $ 4$ when in fact there are clearly orbits of size $ 2$ and $ 1$ that also need to be taken into account.  (Burnside's lemma is a great way to understand problems like this; it's quite unfortunate that it's not introduced earlier in the mathematical curriculum.)

Can somebody tell me where I went wrong in my solution?

I counted 12 placements that have an "orbit" of 2
and 49c2 - 12 placements that have an "orbit" of 4

So (12 / 2) + (49c2-12 / 4) = 297

There are, in fact, $ \frac{49\minus{}1}{2}\equal{}24$ colorings that have an orbit of size $ 2$ because each such coloring corresponds to a set of two distinct squares that are symmetric about the center square.

[hide="Polya Burnsides"]
The only symmetries we need to consider are rotations of $0^{\circ}, 90^{\circ}, 180^{\circ}, 270^{\circ}$. The 90 and 270 degree rotations are comprised of the product of $\dfrac{49-1}{4} = 12$ four-cycles, and a one-cycle (center square). The 180 degree rotation is the product of $24$ two-cycles, and a one-cycle. The identity rotation is the product of $49$ one-cycles.

Summing up, we have our characteristic polynomial $\dfrac{2f_4^{12}f_1 + f_2^{24}f_1 + f_1^{49}}{4}$. Let $x$ be yellow and $y$ be green. Substituting (with $f_i = x^i + y^i$) gives $\dfrac{2(x^4 + y^4)^{12}(x+y) + (x^2+y^2)^{24}(x+y) + (x+y)^{49}}{4}$. We desire the coefficient of $x^2y^{47}$. It is clear that $2(x^4 + y^4)^{12}(x+y)$ can't possibly give us exactly two factors of $x^2$. In $(x^2+y^2)^{24}(x+y)$ we need the $y$ from $(x+y)$ (to keep the power of $x$ even), and the coefficient of $x^2$ in $(x^2 + y^2)^{24}$ is $\dbinom{24}{1} = 24$. The coefficient of $x^2$ in $(x+y)^{49}$ is $\dbinom{49}{2} = 49 \cdot 24$. Summing up, our total coefficient is $\dfrac{24 + 49 \cdot 24}{4} = \dfrac{50 \cdot 24}{4} = \boxed{300}$. $\blacksquare$[/hide]

[hide="simpler burnside"]
So we have 4 rotations of the board we are considering. All $\binom{49}{2}$ ways we can pick 2 yellow cells are fixed under the identity, no configurations are fixed under rotation by 90 or 270 degrees because then we would need at least 4 yellow cells, and there are exactly 24 configurations that are fixed by a rotation of 180 degrees, because for every non-center cell, there is a unique cell we can pair it with to color yellow, but then we divide by 2. Thus, by burnsides lemma, we get an answer of 300.
[/hide]

How would you find the number of ways that two squares can be chosen that are symmetric about the center square?

Would you just count it like this? (letting the squares be circles, because it really doesn't matter)

Sorry for the bump, but could someone see if my solution is correct? I feel like I got lucky with the computing and that it happened to be the same answer.

[hide=Solution]
I used casework for this problem.

Case 1: The center square is chosen as one of the yellows.

This leaves 48 squares to choose from, and since you can rotate it in 4 different ways to get the same configuration, the total amount is 48/4=12.

Case 2: Both yellow squares are not the center square.

This leaves us 48 squares to choose from once again, and we can rotate it in 4 different ways again, getting us 48C2, which is 282.

Since we have "oversubtracted" the cases where the 90 degree and 270 degree rotation are the same, we know there are 6 of these cases total, so we have to add 6 to our total.

Adding up these numbers, we get 6+12+282=300.[/hide]

We take cases.

Case 1: There are two rotations. Then the two green squares are opposite the center of the square, so we just need to pick one, giving $48$ ways (can't be the center square). But this overcounts by a factor of $2$, so $\frac{48}{2} = 24$ ways. Dividing by $2$ to account for rotations gives $12$ distinct ways.

Case 2: There are four rotations. Then everything else works, so we do $\frac{\binom{49}{2} - 24}{4} = 288$ ways.

Summing gives $\boxed{300}$ ways.

Clearly the center of the square creates rotational symmetry. We claim that for every pair of yellow squares, there are $3$ other repeat outcomes created during the process, excluding when a pair is symmetrical about the origin. It is easy to see that for every square (that is not the origin), there exists exactly $1$ other square that is symmetrical about the origin. This creates $\frac{49-1}{2}=24$ overcounts (only 1 repeat is created), so the answer is $$\frac{\binom{49}{2}-24}{4} + \frac{24}{2} \implies 294+6=\boxed{300}$$

[I]Remark[/I]: This is a REALLY nice problem, recommend trying it

Silly problem, was probably challenging in 1996 though...

[hide]Burnside's Lemma. Note that for a 0 turn (do-nothing case), there are $\tbinom{49}2=1176$ ways, for a 1/4 turn there are 0 ways with only two yellows, and similarly for a 3/4 turn. For a 1/2 turn, the general formula for a $(2n + 1)$ by $(2n + 1)$ grid is $\tfrac{n^2-1}2$. Plugging in $7$, we get 24 ways for this case. Taking the average we get $$\tfrac{1176+0+24+0}4=\boxed{300}.$$[/hide]


A 20 inch picture is centered withing a 28 inch by 18 inch frame. What is the number of inches in the distance from a corner of the frame to a corresponding corner of the picture?

Im not sure what you mean by a '20 inch' picture. Do you mean that the diagnol is 20 inches long?


calculus
integration
function
trigonometry
logarithms
real analysis
real analysis unsolved

Find the value of the integral on the function 1/(x^4+1) from x=0 to x=infinity.

Bomb

[quote="Fermat -Euler"][quote="socrates"][quote="Fermat -Euler"]we have
$I_n = \int_0^{ + \infty } {\frac{{dx}}{{1 + x^n }} = \frac{\pi }{{n\sin (\frac{\pi }{n})}}}$
[/quote]

A proof??  :)[/quote]
it's very long  ;)[/quote]

Where can I find it? ;)

look here it's now proposed like an exercice
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=458283#458283]beautyful relation of integral[/url]

We can find $a,b,c,d \in \mathbb{R}$ such that : $\frac{1}{1+x^4}= \frac{ax+b}{x^2-2x+2}+\frac{cx+d}{x^2+2x+2}$. It's easier I think.


This is not the appropriate forum for that question.

However, the answer is $ b^x \equal{} y \implies \log_b y \equal{} x$.

For more information on logarithms go to [url]http://www.themathpage.com/aPreCalc/logarithms.htm[/url].

If you see something like this;

\[ log_a{b}\] $ \equal{} x$.
That equation means that $ a^x \equal{} b$.

For example:

$ \log_9 9\equal{}1$

Because $ 9^1\equal{}9$

Similarily:

$ \log_4 16\equal{}2$

etc etc.

All you need is to find what power the base has to go to to reach the number next to it.

Hope that helped!

You should get AoPS Volume 1. The very first chapter explains the basics of logarithms.

[quote="Aqpwod"]What does log mean?[/quote]

Assuming that you are familiar with exponents, those are expressions like:
\begin{align*}a\cdot a\cdot a &= a^3 \\ b\cdot b\cdot b\cdot b &= b^4\ldots \\ etc.\end{align*}

the small superscripted numbers are called exponents, and the numbers they are superscripted on, $ a$, $ b$, and so on are called bases.

$ \log$ is the abbreviation (functional notation) for [i]logarithm[/i].

Logarithms are functions whose input is a number, typically a positive number, and whose output is an exponent -- the power a particular base is raised to in order to get the input number.

For example, $ 4 =2^2 \implies \log_{2}4 =\log_{2}2^2 = 2$ and likewise $ 125 =5^3 \implies \log_{5}125 =\log_{5}5^3 = 3$.

In general then the rule for how logarithms work is this, $ a = b^{\log_{b}a}$.

Knowing how logarithms work allows you to do some really cool things like replacing multiplication with addition and replacing division with subtraction among others.

[quote="akhil0422"]This is not the appropriate forum for that question.[/quote]

@akhil0422: I am not certain that the question in itself is inappropriate in this forum. A complete answer to it might be, but as for the question, from my own experience, I was first exposed to logarithms in 7th grade science class.

The expression $ log_{2}32\equal{}?$ is asking the same thing as "To what power must $ 2$ be raised to to obtain $ 32$?"


function
inequalities
combinatorics unsolved
combinatorics

let $ I$ a finit part in $ Z$ and $ f$ and $ g$ are two functions difined on $ I$ .
$ m$ the number of couples $ (x,y)$ such that $ f(x) \equal{} g(y)$
$ n$ the number of couples $ (x,y)$ such that $ f(x) \equal{} f(y)$
and $ k$ the number of couples $ (x,y)$ such that $ g(x) \equal{} g(y)$
prove that $ n \plus{} k \geq 2m$

$ f, g$ take values in some finite set $ J$.  For each element $ j \in J$, suppose there are $ u(j)$ values of $ x$ such that $ f(x) \equal{} j$ and $ v(j)$ values of $ x$ such that $ g(x) \equal{} j$.  Then

$ n \equal{} \sum_{j \in J} u(j)^2$
$ k \equal{} \sum_{j \in J} v(j)^2$
$ m \equal{} \sum_{j \in J} u(j) v(j)$.

The desired inequality follows from AM-GM.


calculus
integration
LaTeX
trigonometry
complex numbers
calculus computations

This one is tricky. I know I'm close but my coefficients are wrong

the indefinite integral of "cos(2x)e^-x"

I'm sorry, I tried to figure out latex but I can't make it work.

What i did so far was integration by parts twice and set the original problem = to I to cancel it. It doesn't seem to work out right though.
I used a math program and then manually differentiated it, so I know this answer is right. I'm just not sure where the denominator is coming from.
-[cos(2x)e^-x+2sin(2x)e^-x]/5

Let $ u\equal{}\cos(2x)$, $ du\equal{}\minus{}2\sin(2x)\,dx$, $ dv\equal{}e^{\minus{}x}\,dx$, $ v\equal{}\minus{}e^{\minus{}x}$.  This gives $ \minus{}e^{\minus{}x}\cos(2x)\minus{}2\int\sin(2x)e^{\minus{}x}\,dx$.  Let $ u\equal{}\sin(2x)$, $ du\equal{}2\cos(2x)\,dx$, $ dv\equal{}e^{\minus{}x}\,dx$, $ v\equal{}\minus{}e^{\minus{}x}$ in the second integral.  This gives $ \minus{}2\left(\minus{}\sin(2x)e^{\minus{}x}\plus{}2\int\cos(2x)e^{\minus{}x}\right)\equal{}4\sin(2x)e^{\minus{}x}\minus{}4\int\cos(2x)e^{\minus{}x}\,dx$.  Therefore, the whole integral is $ \int\cos(2x)e^{\minus{}x}\,dx\equal{}\minus{}e^{\minus{}x}\cos(2x)\plus{}2\sin(2x)e^{\minus{}x}\minus{}4\int\cos(2x)e^{\minus{}x}\,dx$.  Rearranging gives $ \int\cos(2x)e^{\minus{}x}\,dx\equal{}\frac{\minus{}e^{\minus{}x}\cos(2x)\plus{}2\sin(2x)e^{\minus{}x}}{5}$.

That's right.

$ (e^{\minus{}x}\cos 2x)'\equal{}\minus{}e^{\minus{}x}\cos 2x\minus{}2e^{\minus{}x}\sin 2x$

$ (e^{\minus{}x}\sin 2x)'\equal{}\minus{}e^{\minus{}x}\sin 2x\plus{}2e^{\minus{}x}\cos 2x$

Elliminating $ e^{\minus{}x}\sin 2x$ gives $ \int e^{\minus{}x}\cos 2x\ dx\equal{}\frac{1}{5}e^{\minus{}x}(2\sin 2x\minus{}\cos 2x)\plus{}C.$

even i think NOra's is the best way.

Great NORA !

We all did basically the same thing, so I don't see how one way is "best."

hello, you can also write $ \int e^{\minus{}x}\cos(2x)\,dx\equal{}Ae^{\minus{}x}\sin(2x)\plus{}Be^{\minus{}x}\cos(2x)$ with real constants $ A$ and $ B$, by differentiating these equation above you can compute $ A$ and $ B$.
Sonnhard.

$ \int e^{\minus{}(1\minus{}2i)x}\ dx\equal{}\minus{}\frac{1}{1\minus{}2i}e^{\minus{}x}e^{2ix}\plus{}C$

$ \equal{}\minus{}\frac{1}{5}e^{\minus{}x}(1\plus{}2i)(\cos 2x\plus{}i\sin 2x)\plus{}C$

Taking the real part of this, we get the result.


AMC
USA(J)MO
USAMO
AIME
USAJMO
percent
AMC 12 A

I was late coming into this discussion about the addition of the new "junior" USAMO and its implications for the takers of the AMC tests. From what I have read, I don't see anything definite yet...
For 11th and 12th graders, it is claimed that essentially nothing will change, but some people brought up the point that this new system makes USAMO qualification harder for upperclassmen. (because high level underclassmen will take the AMC 12 in hopes of USAMO qualification.) Does Mr. Dunbar agree with this speculation, and will any changes be made to the system because of it?
I am not voicing my opinion on this system yet because, frankly, I am still not completely sure how this will work. I just want to know how the contests will change for upperclassmen, both theoretically and practically.

[quote]how the contests will change for upperclassmen, both theoretically and practically.[/quote]

Nothing will change either theoretically or practically for 11th and 12th grade students:
1.  Still have to take the AMC 12 A or AMC 12 B and score at least 100 points or be in the top 5% of all scores to qualify for AIME.
2.  Still have to take the AIME, on either the first date or the second date.
3.  Still have to be be among the approximately 250-260 hightest USAMO indices from all AMC 12 takers.

My estimation based on the USAMO indices of AMC 12 takers in 2008 and 2009 is that 2010 USAMO qualification will have a minimum index of about 210 points, but of course this will depend on national scores on the AMC 12 and the AIME in 2010.  Historically from 2000 through 2009, the qualification index has ranged from 197.5 to 220&233 with a mean of 213.

Steve Dunbar
MAA Driector, Mathematics Competitions

[quote="buzzer11"]Thank you, but can you clarify why the following is untrue?

[quote]
There will be a change, and it will become tougher for 11th-12th graders to qualify. In modeling the effects, you should make sure to account not for the current distribution of AMC10 and AMC12 scores, but for what will happen when most of the current top AMC10 scorers change their behavior and take the AMC12 instead.
[/quote][/quote] Well basically, no one is sure exactly what'll happen. Some people think that the new rules will cause more advanced 10- (or less) graders to take the AMC 12, thereby making it harder to qualify. However, others think that the USAJMO might be sufficient enough incentive for them and this won't happen. It really remains to be seen, and we'll have a better idea about it at the end of the contest season.

So we can only conjecture until results come out at the end of the year... That is unsettling. :/

[quote]Thank you, but can you clarify why the following is untrue?

Quote:

There will be a change, and it will become tougher for 11th-12th graders to qualify. In modeling the effects, you should make sure to account not for the current distribution of AMC10 and AMC12 scores, but for what will happen when most of the current top AMC10 scorers change their behavior and take the AMC12 instead.

[/quote]

The quoted statement is neither true nor false, it is conjectural.  It does not appear to be based on any data.  On the other hand, the information I provided was based on the scores of AMC 12 takers in 2008 and 2009, including capable 10th graders and below who qualified for the AIME and the USAMO on the basis of AMC 12 scores.  I have no reason to believe the population of capable 10th graders and below in the collection of AMC 12 takers will change substantially this year.

Based on 10 years of evidence, I expect the USAMO qualification score will be approximately 210, with a variability of about 10 points.

Steve Dunbar
MAA Director, Mathematics Competitions

[quote="AMCDirector"][quote]how the contests will change for upperclassmen, both theoretically and practically.[/quote]

Nothing will change either theoretically or practically for 11th and 12th grade students....

3.  Still have to be be among the approximately 250-260 hightest USAMO indices from all AMC 12 takers.

My estimation based on the USAMO indices of AMC 12 takers in 2008 and 2009 is that 2010 USAMO qualification will have a minimum index of about 210 points....[/quote]
Thread necro, but I'm confused. The indices for AMC 12 in 2008 and 2009 were 204 and 201 respectively. If this series of new changes really doesn't change anything for 11th and 12th graders then why does the predicted index based on 08 and 09 go up about 7 points with the new system?

Furthermore, I'm 99 percent certain that somewhere on the UNL website, something was posted about the index being roughly 211 in 2009 if USAJMO had been in place (I could be wrong, as it doesn't appear to be there anymore), as well as a file "USAMO/USAJMO selection process" (which again doesn't appear to be there anymore, maybe I was hallucinating). Does anyone else remember seeing this as well?

[quote="darkdieuguerre"]Because it changes things for the 10th graders and below who are capable of qualifying for USAMO. Previously, they could qualify with a sufficiently high USAMO index with their AMC 10 score. Now, either they must risk getting an 11+ on the AIME while taking only the AMC 10, or they must take the AMC 12 like everyone else.

Also note that fewer people qualify for USAMO, so that increases the qualifying index in any case. Basically, the pool of potential USAMO candidates has increased, but the number of USAMO candidates that will be selected has decreased. Both these factors yield to a predicted increase in the USAMO index.

Now that the AMC 12 and AIMEs have been released, it is widely speculated that the USAMO index will be in the 210-220 range, conforming with the predicted increase/[/quote]
Yes, but that seems to contradict
[quote="AMCDIrector"]
Nothing will change either theoretically or practically for 11th and 12th grade students.... [/quote]
Personally I'm not concerned about the general predictions, but rather on what makes AMCDirector feel as if the system won't change for 11th/12th graders, while he still expecting a 7 point increase from 2008 and 2009 (which he mentioned he is basing his predictions on) :)

[quote="darkdieuguerre"]A higher USAMO index won't affect the 11th and 12th graders; they only have one choice: Take the AMC 12, then take the AIME. It might affect them in that they have to study more, but with respect to the competitions they participate in, they're basically doing the same thing. They'll just be competing directly against young, talented students.[/quote]
Yes I know, but  specfically I am referring to
"3. Still have to be be among the approximately 250-260 hightest USAMO indices from all AMC 12 takers. "
since the original post asked whether USAMO was predicted to be harder to get into, to which Mr. (Dr.?) Dunbar replied that no, nothing would change, citing his list as evidence.

I am aware of the 'evidence' that USAMO cutoff will probably rise, but am not sure why the predicted cutoff by Mr. Dunbar himself is expected to rise, if I am interpreting correctly in saying that it is his belief, based on firm statistical data, that nothing will change for the index for 11th and 12th graders.

I think you're misinterpreting the statement and applying it to too generally (can you provide a link to his exact post)? Dr. Dunbar stated that nothing changes theoretically or practically for 11th and 12th graders because 11th and 12th graders can still only qualify for USAMO the way they've had to qualify since the creation of the current system; qualify for the AIME via the AMC 12, then score high enough on the AIME.

Now, for the sake of simplifying the argument, ignore that 10th graders and younger have to take the AMC 12 to qualify for USAMO (or otherwise take the AMC 10, and get an 11+ on the AIME). In the past, about 350 people have qualified for the USAMO by the AMC 12. This year, only 260 or so will qualify for the USAMO. Therefore, the USAMO index is predicted to increase, simply on that premise.

Now, add in the additional criterion. Now, you have additional students competing for USAMO qualification, further increasing the index.

Ah ok, thanks, sorry about the confusion  :oops:


[quote="borg"]What are the next three number in series: 335, 443, 554,.....[/quote]

Pretty Sure that it's

[hide]662,773,881...

I mean you're adding on the next number for the tens and hundreds digit, and for the ones, it's add 2, subtract 1...[/hide]

[quote="borg"]What are the next three number in series: 335, 443, 554,.....[/quote]
[hide="the way i think is what we are meant to think by"]we notice that we can express it ab
for the first one a=3 and b=5
the next one the last one's a is b!
so
we know that the next one
a5
then we know that the a's are going in order
therefore teh answer is
[b]665[/b][/hide]

[hide="Based on the given information it should be"]
$665$
[/hide]

[hide]665, because the first 2 digits go in increasing order, and the last one was the hundreds digit of the last number.[/hide]

[quote]it's add 2, subtract 1...[/quote]
rd5493, shouldn't it be subtract 2 add 1.

[hide]Personally I think there's limited information. I agree with rd5493, but 665, described above should make more sense.[/hide]

[quote="kohjhsd"][quote="borg"]What are the next three number in series: 335, 443, 554,.....[/quote]

There are so many possibilities unless you specify what type it is. :maybe:[/quote]

exactly. It could be a "puzzlelike" sequence as described, or it could be a quadratic, cubic, and so on.

[quote="alanchou"]You could also suddenly say the next term is $\frac{\pi}{2}$ and state that every fourth term is $\frac{\pi}{2}$[/quote]

haha, though that's not really "mathematical".
What's the technical term for that kind of "sequence"?

That would more of a discrete pattern, where every $n^{th}$ term is defined as something.
If that is that right terminology.  :maybe:

Yeah, every one knows it's 665, not need to repeat it.
No, following the samme rule, the next two (Which was what the question asked) are
[hide]776, 887[/hide]

However, the first number is an anomaly: 33[b]5[/b]  :huh:

rd5493's right, I think it's what he siad because the pattern goes in a specific order...
either that or the kid that started this didn't give enough terms in the sequence

Isn't it obvious? The next three terms are 42, 42, and 42. It's Mathnerd314's Function, which is defined

$a_{1}=335$
$a_{2}=443$
$a_{3}=554$
$a_{n}=42 \text{ for }n \ge 4$

But seriously, though, the first number being 335 throws off the 443, 554, 665, 776, 887 idea.

[quote="dragonx"]based on information $663,772,881$[/quote]
If you do it that way, it won't work, because the unit"s digit doesn't go down by 1, other wise the first 3 terms would be:335,444, and 553, which they are not.

Well, yes most people have come to an answer, so you don't need to post again. You don't know for sure if you are adding 108, then 111. It could be a whole different thing.

If this is meant to be a puzzle, then it's

[hide]one (3)two(3) three(5), four(4) five (4)six(3), seven (5)eight (5)nine(4)

# of letters in spelling of numbers

So ten eleven twelve is 366[/hide]

Oh! Clever.

Never thought of that, redcomet46. Perhaps that's the answer.

true but this probably isnt a puzzle seeing its classroom math

[quote="toadoncart"][quote="borg"]What are the next three number in series: 335, 443, 554,.....[/quote]

I think that it is 668, cuz you add 108, then 111, then it should be 114.
Another solution is probably the place values. The first was 335, then 443, then 554, right? Then it might then be 665, cuz there is a number, then a number, then a smaller number for the place values for 443 and 554. I'm not sure, but most people also but 665, too. my opinion is more on the 665 side, though, cuz majority wins...  :oops:[/quote]

If the first idea of adding 108, then 111, then 114 is correct, then we can generate a function for the nth term, but it would be second degree or higher.


calculus
integration
function
inequalities
derivative
calculus computations

From the source:

The first mean value theorem for integration states

$\bullet$ If $G : [a, b] \to \mathbb{R}$ is a continuous function and $\phi : [a, b] \to \mathbb{R}$ is an integrable positive function, then there exists a number $x$ in $(a, b)$ such that \[\int_{a}^{b}G(t)\varphi (t) \, dt=G(x) \int_{a}^{b}\varphi (t) \, dt .\] In particular for $\phi(t) = 1$, there exists $x$ in $(a, b)$ such that \[\int_{a}^{b}G(t) \, dt=\ G(x)(b-a). \,\] The second mean value theorem for integration is stated as follows (Hiroshi Okamura, 1947))

$\bullet$ If $G : [a, b] \to \mathbb{R}$ is a monotonically decreasing function and $\phi : [a, b] \to \mathbb{R}$ is an integrable function, then there exists a number $x$ in $(a, b)$ such that \[\int_{a}^{b}G(t)\varphi(t)\,dt = G(a+0) \int_{a}^{x}\varphi(t)\,dt+G(b-0) \int_{x}^{b}\varphi(t) \, dt.\] Firstly, how do you prove them?

Secondly, i was thinking about using the fundamental theorem of calculus, (for the first one) but i think that integrable refers to wider set of functions. Is this going to be a problem?

You can use the Intermediate Value Theorem.

For the first one: let $y=\frac{\int_a^bG(t)\varphi(t)\,dt}{\int_a^b\varphi(t)\,dt}$

Note that the denominator, $\int_a^b\varphi(t)\,dt,$ is positive.

Now suppose that there we never have $G(x)=y.$ By the Intermediate Value Theorem we then either have $G(x)>y$ for all $x\in[a,b]$ or $G(x)<y$ for all $x\in[a,b].$

Let's take the first case. Since $G$ is continuous on a closed bounded interval, it assumes a minimum, which must be greater than $y.$ Hence $\exists,z>y$ such that $G(t)\ge z.$

Then $\int_a^bG(t)\varphi(t)\,dt\ge\int_a^bz\varphi(t)\,dt>y\int_a^b\varphi(t)\,dt$

But since $y\int_a^b\varphi(t)\,dt=\int_a^bG(t)\varphi(t)\,dt,$ we have

$\int_a^bG(t)\varphi(t)\,dt>\int_a^bG(t)\varphi(t)\,dt,$ which cannot be true.

The case of $G(x)<y$ for all $x$ is similar.

We also can proceed like this ,
since $f(x)$ is continous its atainned both maximum and minimum value on $[a,b]$. say $M$ for maksimum and $m$ for minimum.  so
\[ m \leq f(x) \leq M \qquad x \in [a,b]  \] .
Now suppose that $\varphi(x) \not= 0$. for $\varphi=0$ the prove is trivial. we can multiply the inequality by $\varphi(x)$ and integrate from $a$ to $b$ , since it is integrable. yields
\[ m \int_{a}^b \varphi(x) \, dx \leq \int_{a}^b f(x) \varphi(x) \, dx \leq M \int_{a}^b \varphi(x) \, dx  \]
divide the inequality by $\int_{a}^b \varphi(x) \, dx$ now it  like  Kent already posted. that is
\[ m \leq \frac{\int_{a}^b f(x) \varphi(x) \, dx}{\int_{a}^b \varphi(x) \, dx} \leq M .  \] Since $m$ and $M$ are the value of the function we say that $f(s) = m$ and $f(t)=M$, so by intermediate value theorem there exist a number $\xi$ between $s$ and $t$ so that \[ f( \xi) = \frac{\int_{a}^b f(x) \varphi(x) \, dx}{\int_{a}^b \varphi(x) \, dx} \Longrightarrow \int_{a}^b f(x) \varphi(x) \, dx = f(\xi) \int_{a}^b \varphi(x) \, dx  \].
For the second part, define $F(x)= \int_{a}^x f(t) \, dt$, so by integration by part we have
\begin{eqnarray*} \int_{a}^b f(x) \varphi(x) \, dx &=& F(x) \varphi(x) |_a^b - \int_{a}^b F(x) \varphi '(x) \, dx \\ &=& F(b) \varphi(b) - F(\xi) \int_{a}^b \varphi '(x) \, dx \\ &=& F(b) \varphi(b) - F(\xi) [\varphi(b) - \varphi(a)] \\ &=& [F(b) - F(\xi)] \varphi(b) + F(\xi) \varphi(a) \\ &=& \left[\int_{a}^b f(t) \, dt- \int_{a}^{\xi} f(t) \, dt \right] \varphi(b) + \int_{a}^{\xi} f(t) \, dt \varphi(a) \\ &=& \varphi(b) \int_{\xi}^b f(t) \, dt + \varphi(a) \int_{a}^{\xi} f(t) \, dt \\ \end{eqnarray*}
where we use part I in line 2. I know that more things involved here  ;) (this is my homework about six months ago).

Maybe (I'm just going off the top of my head here) you can get around that by using Stieltjes integration.

[quote="Beginner"]Yes, and the derivative, $\varphi '(x)$ also continous on interval $[a,b]$.[/quote]
But I always thought that integrable functions can have discontinuity points...

Anyway, your solution is nice

In fact the theorem also true for any monotonic discontinue functions but it must defined for $x=a$ and $x=b$. Since we are about to calculate $\int_{a}^b f(x) \varphi(x) \, dx$.

I try to find counterexample for which $\varphi(x)$ doesn't have derivative. But I can't.
After a while (considering blahblahblah's post). I found that theorem also true for which $\varphi(x)$ doesn't have antiderivative. I've a proof :lol: .
And also keep thinking of it, I found that my previous proof has a hole (just for $\varphi(x)$ doesn't zero). But i also have fix it (its special case). Now it is analysis.
Hint:
Use this theorem
Suppose $f(x)$ and $\varphi(x)$ are complex functions of bounded variation on $[a,b]$, and $f(x)$ is also continous. Then
\[ \int_{a}^b f(x) \, d(\varphi(x)) = f(b) \varphi(b) - f(a) \varphi(a) - \int_{a}^b \varphi(x) \, d(f(x))  \]

Yes, that is the right idea.

There is a proof in Folland's 'Real Analysis: Modern Techniques and Their Applications' (lemma 8.41).  If you guys really want, I can post it.

[quote="perfect_radio"]The second mean value theorem for integration is stated as follows (Hiroshi Okamura, 1947))

    If $G : [a, b] \to \mathbb{R}$ is a monotonically decreasing function and $\phi : [a, b] \to \mathbb{R}$ is an integrable function, then there exists a number $x$ in $(a, b)$ such that

$\int_a^b G(t)\varphi(t)\,dt = G(a+0) \int_a^x \varphi(t)\,dt + G(b-0) \int_x^b \varphi(t)\,dt.$
[/quote]
Assume the integrals are all Riemann-integrals.

The following proof doesn't need the differentiability of $\varphi(x)$.

==================================

Firstly, we prove Bonnet's Formula: if $f(x)\geq 0$ and is decreasing, then there exists $\xi$ such that
$\int_a^b f(x)\varphi(x)dx = f(a)\int_a^\xi \varphi(x)dx$.

We write the integral as
\[ I = \sum_{i=0}^{n-1} \int\limits_{x_i}^{x_{i+1}} f(x)\varphi(x)dx = \sum_{i=0}^{n-1} f(x_i) \int\limits_{x_i}^{x_{i+1}} \varphi(x)dx + \sum_{i=0}^{n-1} \int\limits_{x_i}^{x_{i+1}} (f(x)-f(x_i))\varphi(x)dx=S_1+S_2  \]

Suppose $|\varphi(x)|\leq L$ and $\omega_i$ is the amplitude of $f(x)$ on $[x_i,x_{i+1}]$, then
$|S_2| \leq \sum_{i=0}^{n-1} \int\limits_{x_i}^{x_{i+1}} |f(x)-f(x_i)| |\varphi(x)| dx \leq L\sum \omega_i \Delta x_i$
Since $f$ is integrable(it is decreasing), we have $S_2\to 0$ as $\max \Delta x_i \to 0$

Now let $\Phi(x) = \int_a^x \varphi(t)\ dt$ and we can write $S_1$ as
$S_1 = \sum_{i=0}^{n-1} f(x_i) (\Phi(x_{i+1}) - \Phi(x_i)) = \sum_{i=1}^{n-1} \Phi(x_i)(f(x_{i-1})-f(x_i)) +\Phi(b)f(x_{n-1})$.

$\Phi(x)$ reaches maximum and minimal values, say $M$ and $m$, respectively. Since $f(x_{i-1})-f(x_i)\geq 0$, we have
$mf(a) \leq S_1 \leq Mf(a)$. Hence $S_1 = \mu f(a)$ for some $\mu \in [m,M]$. From the continuity of $\Phi(x)$, it equals to $G(\xi) = \int_a^\xi \varphi(x)dx$ for some $\xi$.

==========================================

Now we prove the second mean-value theorem, apply $f(x)-f(b)\geq 0$ to Bonnet's Formula that we proved above.

Is the Folland's proof different from liyi's proof, blahblahblah?
the proof  of theorem  that I've posted  before (from my  prof note) is similar with liyi's. If the proof is different i would like to see it blahblahblah? :)

[quote="liyi"]
Suppose $|\varphi(x)|\leq L$ and $\omega_i$ is the amplitude of $f(x)$ on $[x_i,x_{i+1}]$, then
$|S_2| \leq \sum_{i=0}^{n-1} \int\limits_{x_i}^{x_{i+1}} |f(x)-f(x_i)| |\varphi(x)| dx \leq L\sum \omega_i \Delta x_i$
Since $f$ is integrable(it is decreasing), we have $S_2\to 0$ as $\max \Delta x_i \to 0$
[/quote]

1) what's the amplitude?

2) "a decreasing function is always integrable". why? to me it seems that there can't be many discontinuity points, but don't know how to prove that
 :(

[quote="perfect_radio"]2) "a decreasing function is always integrable". why? to me it seems that there can't be many discontinuity points, but don't know how to prove that[/quote]
This is a classic. The proof can be summarized as follows: take an equal-width partition (partition with an arithmetic sequence). Compute upper sums and lower sums. Visualize the difference between upper and lower sums. "Stack the boxes."

More explicitly: suppose $f$ is decreasing on $[a,b].$ For positive integer $n,$ take the partition $P_n$ in which $x_k=a+\left(\frac{b-a}n\right) k$ for $0\le k\le n.$

$U(f;P_n)=\frac{b-a}n\sum_{k=1}^nf(x_{k-1})$

$L(f;P_n)=\frac{b-a}n\sum_{k=1}^nf(x_k)$

$U(f;P_n)-L(f;P_n)=\frac{b-a}n\sum_{k=1}^n\left[f(x_{k-1})-f(x_k)\right]$

This sum telescopes to

$U(f;P_n)-L(f;P_n)=\frac{b-a}n(f(a)-f(b))$

which can be made as small as we please by taking $n$ sufficiently large.

As an aside: it can be proved that $f$ has at most countably many jump discontinuities, and no discontinuities of other types. But we didn't need that here.

[quote="perfect_radio"]
1) what's the amplitude?
[/quote]
Sorry, I don't know the exact word.

I called $\sup_{x'\in E, x''\in E} |f(x')-f(x'')|$ the amptitude of $f$ on $E$.

[quote="perfect_radio"]Kent, where in your proof do you use the fact that $f$ is decreasing?

I hope the question it's not as dumb as it seems.[/quote]

The fact that the function is decreasing is used to generate the exact upper and lower sums.

[quote="Kent Merryfield"]As an aside: it can be proved that $f$ has at most countably many jump discontinuities, and no discontinuities of other types. But we didn't need that here.[/quote]

Indeed, it can quite easily be proved that $f$ has left and right limits in every $x_0 \in \mathbb R$. On the other hand, a well-known theorem tells us that any function can have at most countably many jump discontinuities.


I was trying to write up solutions in Mathcounts forum and I came across this problem.

[hide="A"]
[hide="B"]
Lala [/hide]
[hide="C"]
Lalala [/hide]
[/hide]

where $A$ is the heading of the hide (Like solution 1) and $B$ and $C$ are the subheadings of the hide (like method 1, method 2, method 3, etc).

In that same post, when I tried to have another hide, namely solution 2 (or $D$ below), I get $D$ to be under the heading $A$ like $B$ and $C$.

[hide="A"]
[hide="B"]
Lala [/hide]
[hide="C"]
Lalala [/hide]
[/hide]

[hide="D"]
LALALALA [/hide]

I think I know why it works that way (because computer understands the hide $A$ ends with the last hide) but is there any way you can use hide to have $D$ not under $A$? I'll be very appreciated if anyone can help in this.

Thanks.

it seems to be fine for me. You just need to have the {/hide} tags in the right place (replacing { with brackets of course)

so if you had:
{hide=solution 1}solution 1{hide=solution 2}solution 2{/hide}{/hide}
it would come out as:
[hide="solution 1"] solution 1[hide="solution 2"]solution 2[/hide][/hide]

However: if you had it as:
{hide=solution 1}solution 1{/hide}
{hide=solution 2}solution 2{/hide}

it would come out as:
[hide="solution 1"]solution 1[/hide]
[hide="solution 2"]solution 2[/hide]

so what's the problem with it?

The problem is that if you have a nested tag, then the whole post gets under the first tag.

[quote="Valentin Vornicu"]Peter has already submitted this "bug". Unfortunately I can't seem to find the problem yet, and I am also busy with a couple of other things, so it'll have to wait.[/quote]

Okay, I didn't know that. It's a minor problem so it doesn't matter much. Thanks.

[quote="Peter VDD"][quote="Valentin Vornicu"]Peter has already submitted this "bug". Unfortunately I can't seem to find the problem yet, and I am also busy with a couple of other things, so it'll have to wait.[/quote]Well, I don't know how "secret" you consider the inner coding of the forum, but if you want you can send me the files related to this and I'll take a look. I think I've corrected over 1000 bugs in my life, so I may have some experience in finding them. :)[/quote]Hehe, 1000 is a lot. Are you sure you didn't count typos in the code as bugs? :D We can discuss on PM about code on the forum.


Let $n\geq 2$ and $k$ be any positive integers. Prove that $(n-1)^{2}$ divides $(n^{k}-1)$ if and only if $(n-1)$ divides $k$.

I reasoned as follows, since $(n^{k}-1) = (n-1)*(1+n+n^{2}+...+n^{(}k-1))$, $(n^{2}-1)$ divides $(n^{k}-1)$ if $(n+1)$ divides $(1+n+n^{2}+...+n^{(}k-1))$. Ofcourse $(1+n+n^{2}+...+n^{(}k-1))$ can be written as $(n+1)*(1+n^{2}+n^{4}+n^{6}+...+n^{(}k-2))$ iff k is even. Hence the required condition for $(n-1)^{2}$ to divide $(n-1)^{2}$ is that $k$ be even. What's wrong with this explanation :?:

Thanks
Kris

a little generalization!

if, for a prime p, we have $p^{a}\parallel a-b$ and $p^{b}\parallel n$, we have that $p^{a+b}\parallel a^{n}-b^{n}$!