I'm lost on how to figure this out.

Basically I'm given $1,000 over 7 years with 8% yearly increases, what are the year totals?

I'm totally lost on figuring this out.  Any help?

So you want to find out how much $ \$1000$ will become in $ 7$ years, if there's an $ 8\%$ increase each year? If so,

[hide="Total after 7 years"]Each year, you multiply the current amount by $ 100\% \plus{} 8\% \equal{} 108\% \equal{} 1.08$. Thus, after $ n$ years, the amount will be $ \$1000\cdot(1.08)^n$.

In this particular case, $ n \equal{} 7$, so $ A \equal{} \$1000\cdot(1.08)^7\approx\$1713.82$.[/hide]

If you meant to ask for the increase each year relative to the previous year, then

[hide="Increase each year"]For any year $ n$, the increase from the previous year will be $ \$1000\cdot(1.08)^{n \minus{} 1}\cdot(0.08)$. Notice that here, we first find the total after the previous year, and then multiply by $ 8\%$ to find how much this year will bring.

The first year will increase $ \$1000$ by $ \$1000\cdot(1.08)^0\cdot(0.08) \equal{} \$80$. After $ 1$ year, the total will be $ \$1080$.

The second year will increase $ \$1080$ by $ \$1000\cdot(1.08)^1\cdot(0.08) \equal{} \$86.40$. The new total will be $ \$1166.40$.

And so on...[/hide]

Thank you for your reply, but I'm afraid I worded my question poorly.

What I'm looking to answer is:

You will be given $1000 total over a 7 year period with 8% annual increases.
Provide the yearly amounts paid.
Year 1
$
Year 2 = Year 1 $ * 1.08

and so on.

Hopefully that is more clear.

Wouldn't you just use the formula I provided, $ f(n) \equal{} \$1000\cdot(1.08)^n$, where $ n$ is the year? It gives the total amount after $ n$ years.

Ravi B you have it exactly right.

year 1 + year 2 ... with a sum total of $1000

[quote="Ravi B"]1000 dollars is not the amount given in the first year.  1000 dollars is the total amount given over all 7 years.  At least that's how I interpret the problem.[/quote]
If that is indeed how the problem should be interpreted, then...

[hide="Solution"]Instead of multiplying by $ 1.08$, we now divide by $ 1.08$: The original amount was $ \$1000\cdot(1.08)^{ \minus{} 7}\approx\$583.49$.

If $ 7$ years ago is our reference point (i.e. year $ 0$), then after $ n$ years, the amount will be $ \left(\$1000\cdot(1.08)^{ \minus{} 7}\right)\cdot(1.08)^n \equal{} \$1000\cdot(1.08)^{n \minus{} 7}$. Notice that the exact value $ \$1000\cdot(1.08)^{ \minus{} 7}$ is used instead of the approximation $ \$583.49$.[/hide]

I think the confusion I'm causing here is because I have not clarified that this is not a compound interest problem.

The closest thing I can relate it to is a contract.

"We are going to pay Johnny $1000 total over a period of 7 years to build a dock"
"We are not going to just pay Johnny $142 a year, but instead give him 8% annual raises"

How much would that be per year?

Now [i]I'm[/i] confused. :D  :|  :(

So what you're saying is:

"You get some amount $ x$ one year, then $ x\cdot(1.08)$ the next year, then $ x\cdot(1.08)^2$ then next, etc., and this goes on for $ 7$ years. If  the sum $ x\plus{}x\cdot(1.08)\plus{}\ldots$ is $ \$1000$, then what are $ x$, $ x\cdot(1.08)$, etc.?"

Is that correct?

[quote="i_like_pie"]Now [i]I'm[/i] confused. :D  :|  :(

So what you're saying is:

"You get some amount $ x$ one year, then $ x\cdot(1.08)$ the next year, then $ x\cdot(1.08)^2$ then next, etc., and this goes on for $ 7$ years. If  the sum $ x \plus{} x\cdot(1.08) \plus{} \ldots$ is $ \$1000$, then what are $ x$, $ x\cdot(1.08)$, etc.?"

Is that correct?[/quote]

Exactly!

I've just got no idea how to calculate it.

Ah, ok then. At least we're on the same page now. :)

[hide="Solution"]Let the amount of money paid in the first year be $ x$.

We know $ x + x\cdot(1.08) + x\cdot(1.08)^2 + \ldots + x\cdot(1.08)^6 + x\cdot(1.08)^7 = \$1000$.

Notice that the LHS is a geometric series, and recall that the sum of the first $ n$ terms is $ a_1\cdot\frac {r^n - 1}{r - 1}$, where $ a_1$ is the first term and $ r$ is the common ratio.

Thus, we can express the LHS as the sum of a geometric sequence, and then solve for $ x$:
\begin{align*}x\cdot\frac {(1.08)^7 - 1}{1.08 - 1} & = \$1000 \\
 \\
x & = \$1000\cdot\frac {1.08 - 1}{(1.08)^7 - 1} \\
 \\
x & = \$\frac {244140625000}{2178418789} \\
 \\
x & \approx\$112.07\end{align*}
To get accurate numbers for the other six years, use either $ x = \$1000\cdot\frac {1.08 - 1}{(1.08)^7 - 1}$ or $ x = \$\frac {244140625000}{2178418789}$, which are both accurate.[/hide]

Thank you very much!

It certainly makes sense now.  Couldn't have gotten there without your help.


algebra
polynomial
vector
function
abstract algebra
geometry
geometric transformation

Let $ V$ be the vector space of all functions of one variable. Suppose $ P: V\to\mathbb{R}$ is a linear map such that for all $ f,g\in V$ with $ P(fg) \equal{} 0$ we have that $ P(f) \equal{} 0$ or $ P(g) \equal{} 0$. Prove that there exist real numbers c and $ x_0$ such that $ P(f) \equal{} cf(x_0)$ for all $ f\in V$.

Let $ K$ be the kernel of $ P$.  Note that given any two functions $ f, g$ we have $ P(g) f \minus{} P(f) g \in K$.  We will be using this fact repeatedly; it is in fact equivalent to the condition that $ P$ is linear.

[b]Case:[/b]  $ 1 \in K$.  Then the constant functions are in $ K$, and if $ f(x) \in K$ then so is $ f(x) \plus{} C$ for any $ C$.  Suppose there existed $ g$ such that $ P(g) \neq 0$.  Then $ P(g) g^2 \minus{} P(g^2) g \in K$.  But since $ g \not \in K$ it follows that $ P(g) g \minus{} P(g^2) \in K \implies P(g) g \in K \implies g \in K$; contradiction.  Hence there exists no such $ g$ and $ P(f) \equal{} 0$ everywhere.

[b]Case:[/b]  $ 1 \not \in K$.  Normalize $ P$ so that $ P(1) \equal{} 1$.  Then given any $ f(x) \not\in K$ we have $ f \minus{} P(f) \in K$ (and this is the unique constant $ P(f)$ for which this is true).  Let $ P(x) \equal{} x_0$.  Then $ x \minus{} x_0 \in K$.

For any $ f$ and any $ g$ with $ P(g) \neq 0$ we therefore have $ P(g) fg \minus{} g P(fg) \in K \implies P(g) f \minus{} P(fg) \in K$.  But the unique translation of $ P(g) f$ in $ K$ is given by $ P(g) f \minus{} P(g) P(f)$.  It follows that $ P(fg) \equal{} P(f) P(g)$ when $ P(g) \neq 0$.  When $ P(g) \equal{} 0$ clearly $ P(g \plus{} 1) \equal{} 1 \neq 0$ and $ P(fg \plus{} f) \equal{} P(f) P(g \plus{} 1) \implies P(fg) \equal{} P(f) P(g)$.  Hence $ P$ is multiplicative.  In particular, any function of the form $ (x \minus{} x_0) q(x)$ is in $ K$.

Now, for any $ f$ write $ f(x) \equal{} f(x_0) \plus{} (x \minus{} x_0) q(x)$ where $ q(x)$ is uniquely determined except for its value at $ x_0$, which is irrelevant.  By the above discussion $ P(f) \equal{} f(x_0)$ as desired.

[b]Remark:[/b]  What we have essentially done is shown that $ K$ is a prime ideal of $ V$.  In fact, all the prime ideals of $ V$ take this form.  I believe the above argument remains valid if $ \mathbb{R}$ is replaced by an integral domain; what happens if we lose commutativity?

The "one-variable functions" run from where to where with what properties?

The solution makes no assumptions other than that the functions take $ \mathbb{R}$ to $ \mathbb{R}$.  No other assumptions are necessary.  In particular, the "division algorithm" $ f(x) \equal{} f(x_0) \plus{} (x \minus{} x_0) g(x)$ where $ g(x)$ is uniquely determined except for its value at $ x_0$ works with no assumptions.

The more interesting case is probably when $ V$ is restricted to the set of polynomial functions, in which case what we are doing is the easiest kind of [url=http://en.wikipedia.org/wiki/Algebraic_geometry]algebraic geometry[/url].

Thanks, it wasn't clear to me that the functions were real to real.


Source: The Art of Problem Solving - A Resource for the Mathematics Teacher - Page 240 - Exercise 1-3

1) A certain locksmith makes keys using anyone of 15 brands of keys. Each
brand uses a different length and thickness. No matter what brand is
chosen, there are 8 different positions from which metal can be removed
or left alone. For the first 5 positions, there are 3 different depths at which
the metal can be removed. For the remaining positions, there are 4
different depths. How many different keys are possible if metal must be
removed in at least one position?

Here is my solution... Is this correct:
[hide]
((4^5*5^3)-1)*15 = 1,919,985 different keys are possible.

(3 depths + 1 no cut for 5 positions.. 4 depths + 1 no cut for 3 positions) and removing 1 for all positions are not cut. Multiplied by number of brands 15.
[/hide]

Please let me know. Thanks.

2) How many subsets does a set of 15 elements have?

My solution:

[hide]2^15 = 32,768[/hide]

3) Many different kinds of desserts have been put out for people to choose
from at a party. Judy is watching her weight and decides to have at least
one piece of fruit and no other type of dessert. If the fruit bowl contains
exactly 6 apples, 5 plums, 3 oranges, 4 bananas, and 5 pears, how many
different combinations of desserts can Judy have?

My solution:

[hide]
(7x6x4x5x6) - 1 = 5,039 different combinations she can have
[/hide]

Please advise if the solution is correct in all 3 exercises. Thank you.

i think there is problem in 3 answer. its a question of permutation and i love solving them


How many four-digit positive integers are there with thousands digit 2?

I think that the wording was awkward, once I viewed the answer I got it, but the wording confused me.

It's not much difficult. Just 4-digit numbers with 2 in the thousands digit.

[hide="Alternate solution"]

All the numbers between $2000$ and $2999$ have $2$ as a thousands digit.

$2999-2000=999$, but we still have to account for the number $2999$, so we add $1$ to get ${\boxed{1000}}$.[/hide]

I think that they should change how they wrote the problem because I did not understand it at first.

[hide=Solution]We use the formula b-a+1.

Plugging in, we find 2999-2000=999

999+1 is ${\boxed{1000}}$[/hide]

[hide=Solution] Since the only 4-digit numbers with thousands digit 2 are from 2,000 to 2,999, we could add. SInce we want 2,00 included, we start at 1,999. 1,999 to 2,999 is 1,000, so our solution is is $\boxed{1,000}$.

[quote=TethysTide][quote=fidgetboss_4000]Bruh that’s easy the answer is 1000[/quote]

Please hide your solutions.[/quote]

[color=#9a00ff]Ummm... Ur smart....SOme people r not so smart, (me). :) :([/color]

[quote=BluRose11][quote=wowacow][quote="snee"]Overly complicated lol[/quote]

its not that hard im in 5th grade and that problem took me 2 seconds to find out its super easy[/quote]

[color=#9a00ff]Well, you must be Super  smart. :) I'm not even as smart as u. :( I'm in 6th.  [/color][/quote]

Please don't revive a nearly 2-year old post to reply to a nearly 9-year old post about a topic that has nothing to do with the problem. From your posts I'm inferring that you're relatively new and so I just wanted to tell you that it's not a good idea to bump a topic if you have nothing constructive to add to it. :)

[hide=Solution]All of the numbers with thousands digit 2 are the numbers from $2000$ through $2099$. Thus, including these numbers, there are $\boxed{1000}$ numbers with thousands digit $2$.

[hide=Solution]   We can use permutations. There is 1 choice for the first number and 10 for the last three numbers. There will be $1\cdot10\cdot10\cdot10 = 1000$ numbers that have thousands digit 2.[/hide]

[quote=math154][hide]All numbers satisfying this must be of the form $ \overline{2abc}$ where $ 0\le a,b,c\le9\implies10^3\equal{}\boxed{1000}$.[/hide][/quote]

me to expect me not in 5th grade

[quote=jcl-12]this problem is easy. How is it level 6 prealgebra?
[hide=Answer.2]The answer is 1000.[/hide][/quote]

Ummmm... it is level 10 prealgebra. Also, here is [hide=My solution]We make a list: 2000, 2001, 2002 ... 2997, 2998, 2999. We subtract 1999 from each number so the first number is one. we get: 1, 2, 3, ... 998, 999, 1000. So there are $\boxed{1000}$ integers.


LaTeX
algebra
binomial theorem
Pascal\u0027s Triangle

[quote="sean2win2004"](x+y)^10
I believe thats how you write it, otherwise its (x+y) to the tenth power.  I have to write out the product.  It says there is a way to do it w/o multiplying it out the long way, but I'm stuck.  Thank you for the help.[/quote]

The binomial theorem

http://mathworld.wolfram.com/BinomialTheorem.html

http://www.answers.com/binomial+theorem?gwp=12&method=2

Thanks for the reply.

I already know from FOIL that (x+y)^2 is x^2 + 2xy + y^2

Do you happen to have the complete answer? The first link was a little complicated for me.

By FOILing you get $(x+y)^2=x^2+2xy+y^2$.  However, if you look at the coefficients of $x^2$, $xy$ and $y^2$ in the expansion, you see that they are 1, 2, 1.  But 1, 2, 1 are the numbers of the second row of Pascal's triangle (the top row is the "zeroth" row, by definition, since any number to the power (except sometimes zero itself) is one).  If you look down the triangle a little bit, and glance at the tenth row, you'll find the coefficients for the expansion of $(x+y)^{10}$.  As easyas3.14159... wrote:

[quote="easyas3.14159..."]You don't need to multiply.The coefficients are of the numbers of the tenth row of pascal's triangle, so this is what the product looks like:
$(x+y)^{10}=x^{10}+10x^9y+45x^8+y^2+110x^7y^3+190x^6y^4+232x^5y^5+190x^4y^6+110x^3y^7+45x^2y^8+10xy^9+y^{10}.$[/quote]

I would advise finding a friendly site (or person) that will describe the binomial theorem for you.

Pascal's triangle. Or a ton of really annoying multiplication. I recommend the former.

Perhaps one of you could explain why Pascal's triangle works for this problem ;)

$(x+y)^{10}$, when fully expanded, will have a whole bunch of terms.  In fact, it'll have $2^{10}$ (not necessarily) distinct terms (because each $(x+y)$ contributes either an $x$ or a $y$ to each term), all with coefficient $1$.

Now, a bunch of these have the same power (for example, there will be a lot of $x^5 y^5$), so they get added up.  But let's keep them separate for now because it's easier to count them that way.  Each of them takes either $x$ or $y$ from one of the $10$ factors, so each of them has a total power of $10$ - that is, they're all of the form $x^{a} y^{10-a}$.

How many of $x^{10} y^{0}$ are there?  This is equivalent to asking how many ways we can pick ten $x$s out of ten factors, which is ${10 \choose 10} = 1$.

How many of $x^{9} y^{1}$ are there?  This is equivalent to asking how many ways we can pick nine $x$s out of ten factors, which is ${10 \choose 9} = 10$.

...

And so forth.  It turns out that this exactly gives the tenth row of Pascal's triangle because the $m^{th}$ entry in the $n^{th}$ row of Pascal's triangle is ${n \choose m}$.

Short demonstration why:  Pascal's triangle is formed by repeated adding of two entries to give the entries immediately below it.  Similarly, given the complete expansion of $(x+y)^{n}$, if we multiply it by $(x+y)$ each $x^{k} y^{n+1 - k}$ term will get some value from the entry above and to the left of it (that is, $x^{k-1} y^{n+1-k} \cdot x$) and some value from the entry above and to the right of it (that is, $x^{k} y^{n-k} \cdot y$).

(Apologies if this is really hard to see without a visual demonstration, but once you understand the connection you'll like combinatorics a lot more.  :) )

Is there anything like that for $(x-y)^n$?

Just plug in $-y$ for $y$ - every term with an odd power of $y$ is just negative what it was before.  :)

FOIL is a method for multiplying binomials.  You multiply the [b]F[/b]irst terms of each binomial, then the [b]O[/b]uter terms of each binomial, then the [b]I[/b]nner terms of each, and finally the [b]L[/b]ast terms of each.

ok so basically its like this

This is a triangle

$1$
$1,1$              coeficients for $(x+y)^1$
$1,2,1$           coeficients for $(x+y)^2$
$1,3,3,1$        coeficients for $(x+y)^3$
$1,4,6,4,1$     coeficients for $(x+y)^4$

How can we generate this triangle further one line ?

Well..
if we denote by $A_{L,C}$ as the number on line L in column C.
then we find that $A_{L,C}=A_{L-1,C-1}+A_{L-1,C}$
otherwise said if you want to find the numebers on the next line
a number is the sum of the number above him and the number to the
left of the number above him.

so that $(x+y)^n=A_{n+1,1}x^{n}+A_{n+1,2}x^{n-1}y+A_{n+1,3}x^{n-2}y^2+...$
so that the kth term in the sum is
$A_{n+1,k}x^{n-k+1}y^{k-1}$

So now you know how to write $(x+y)^n$ in unfolded form as the guys reffer to..

to find a number in Pascal's triangle, just add the two number directly above the number you are trying to find.  See the diagram below.

Hi everybody... this is my first reply so I hope it helps to generalize the idea.
As t0rajir0u said. The coefficients can be found using combinatorics. After simplifing we could come up with the following theorem, called the binomial theorem:

(a+b)^n=a^n+na^(n-1)b+{[n(n-1)]/2!}a^(n-2)b^2+
{[n(n-1)(n-2)]/3!}a^(n-3)b^3... + nab^(n-1) + b^n.

(If you don't understand please tell me how can I make this look better, I guess using the code or something but I don't know, so please tell me)

[quote="Jhonny Ruiz"]
(If you don't understand please tell me how can I make this look better, I guess using the code or something but I don't know, so please tell me)[/quote]\

Often, putting dollar signs around a simple formula can make it look pretty.  However, your formula is more complex, and you will need some knowledge of Latex code.  For a nice tutorial on Latex, visit http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php

i'm trying the latex code out please delete this

\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1}}}}

\sqrt{x+\frac{1}{2}

\sum_{i=1}^{\infty}\frac{1}{i}

\[ n^2 + 5 = 30\text{ so we have }n=\pm5 \]

Here we go. This is the theorem...

$(a+b)^n=a^n+na^{n-1}b^1+\frac{n(n-1)}{2!}a^{n-2}b^2+ \frac{n(n-1)(n-2)}{3!}a^{n-3}b^3+... + na^1b^{n-1} + b^n$

I'm sure you can understand it now. Hope it helps
Thanks


logarithms
integration
trigonometry
real analysis
real analysis unsolved

Dear everyone.

find the value of      $ \frac {1}{1\cdot 2\cdot 3\cdot 4} \plus{} \frac {1}{4\cdot 5\cdot 6\cdot 7} \plus{} \frac {1}{7\cdot 8\cdot 9\cdot 10} \plus{} \cdots$
Is this calculation possible?  :(  please help me~

$ \sum \frac{1}{n(n\plus{}1)(n\plus{}2)(n\plus{}3)} \equal{} \sum (\frac{1}{2n(n\plus{}3)} \minus{} \frac{1}{2(n\plus{}1)(n\plus{}2)})$
$ \equal{} \sum (\frac{1}{6n} \minus{} \frac{1}{6(n\plus{}3)} \minus{} \frac{1}{2(n\plus{}1)} \plus{} \frac{1}{2(n\plus{}2)})$

notice which terms cancel with previous terms in sum leaving

$ \frac {1}{6} \plus{} \frac{1}{12} \plus{} \frac{1}{18} \minus{} \frac{1}{4} \equal{} \frac{1}{18}$

philg you are wrong his sumation doesnt include 1/2.3.4.5 but yours does

$ \sum_{n\equal{}1}^{\infty }\frac{1}{(3n\minus{}2)(3n\minus{}1)3n(3n\plus{}1)}\equal{}\frac{\pi }{12\sqrt{3}}\allowbreak \plus{}\frac{1}{6}\minus{}\frac{\ln 3}{4}$

Oops! I should have read more carefully  :blush:

[quote="IndoChina"]$ \sum_{n \equal{} 1}^{\infty }\frac {1}{(3n \minus{} 2)(3n \minus{} 1)3n(3n \plus{} 1)} \equal{} \frac {\pi }{12\sqrt {3}}\allowbreak \plus{} \frac {1}{6} \minus{} \frac {\ln 3}{4}$[/quote]

Please show your work on this :lol:

We have:

$ S \equal{} \sum ^\infty _ {n \equal{} 0} \frac {1}{(3n \plus{} 1)(3n \plus{} 2)(3n \plus{} 3)(3n \plus{} 4)} \equal{}$

$ \equal{} \frac {1}{81} \sum ^\infty _ {n \equal{} 0} \frac {1}{(n \plus{} \frac {1}{3})(n \plus{} \frac {2}{3})(n \plus{} 1)(n \plus{} \frac {4}{3})}$

$ \equal{} \frac {1}{162} \sum ^\infty _ {n \equal{} 0} \left(\left(\frac{9}{n\plus{}\frac{1}{3}}\minus{}\frac{9}{n\plus{}\frac{1}{3}\plus{}1}\right)\plus{}\left(\frac{27}{n\plus{}1}\minus{}\frac{27}{n\plus{}\frac{2}{3}}\right)\right)$

Now, the first two elements in the sum cancel each other out:

$ S \equal{} \frac{1}{6}\plus{} \frac{1}{6}  \sum ^\infty _ {n \equal{} 0}\left(\frac{1}{n\plus{}1}\minus{}\frac{1}{n\plus{}\frac{2}{3}}\right)$

Now, we can define:

$ f(x)\equal{}\sum ^\infty _ {n \equal{} 1}\left(\frac{x^n}{n}\minus{}\frac{x^{n\minus{}\frac{1}{3}}}{n\minus{}\frac{1}{3}}\right)$

This makes $ S \equal{}\frac{1}{6}\left(1 \plus{} f(1)\right)$. We have:

$ f'(x)\equal{}\sum ^\infty _ {n \equal{} 1}\left(x^{n\minus{}1}\minus{}x^{n\minus{}\frac{4}{3}}\right) \equal{}\frac{x}{1\minus{}x}\left( \frac{1}{x} \minus{} \frac{1}{x^{\frac{4}{3}}}\right) \equal{} \frac{1}{1\minus{}x}\left(1\minus{}\frac{1}{\sqrt[3]{x}}\right)$

This gives us:

$ f(x)\equal{}\int \frac{1}{1\minus{}x}\left(1\minus{}\frac{1}{\sqrt[3]{x}}\right) \,dx$

Using the substitution $ x \equal{} u^3$ we get:

$ f(u) \equal{} \int \frac{1}{1\minus{}u^3}\left(1\minus{}\frac{1}{u}\right) 3 u ^2\,du \equal{} \minus{}3 \int \frac{u}{1\plus{}u\plus{}u^2} \,du$

We can write:

$ \frac{u}{1\plus{}u\plus{}u^2} \equal{} \frac{1}{2}\frac{2u\plus{}1}{1\plus{}u\plus{}u^2} \minus{}\frac{1}{2}\frac{1}{u^2\plus{}u\plus{}1}$

And integrating the first term gives us:

$ f(u) \equal{}\frac{3}{2} \int \frac{du}{u^2\plus{}u\plus{}1} \minus{} \frac{3}{2} \ln \left(1\plus{}u\plus{}u^2\right)$

Finally, we have:

$ \int \frac{du}{u^2\plus{}u\plus{}1} \equal{} \int \frac{du}{\left(u\plus{}\frac{1}{2}\right)^2\plus{}\frac{3}{4}} \equal{} \frac{2}{\sqrt{3}}\tan^{\minus{}1}\left(\frac{1\plus{}2u}{\sqrt{3}}\right) \plus{} C$

So, all this rounds up to:

$ f(x)\equal{}\sqrt{3} \tan^{\minus{}1}\left(\frac{1\plus{}2\sqrt[3]{x}}{\sqrt{3}}\right)\minus{}\frac{3}{2} \ln \left(1\plus{}\sqrt[3]{x}\plus{}\sqrt[3]{x^2}\right) \plus{} C$

Since from the sum-involving definition of $ f$ we get $ f(0) \equal{}0$ we have $ C \equal{} \minus{}\sqrt{3} \tan^{\minus{}1}\left(\frac{1}{\sqrt{3}}\right) \equal{}\minus{} \frac{\pi}{2 \sqrt{3}}$. So, for $ f(1)$ we have:

$ f(1)\equal{}\sqrt{3} \tan^{\minus{}1} \sqrt{3}  \minus{} \frac{3}{2}\ln 3 \minus{} \frac{\pi}{2 \sqrt{3}} \equal{} \frac{\pi}{2 \sqrt{3}} \minus{} \frac{3}{2}\ln 3$

This gives us:

$ S \equal{} \frac{1}{6}\left(1\plus{}f(1)\right) \equal{} \frac{1}{6}\left(1\plus{}\frac{\pi}{2 \sqrt{3}}\minus{} \frac{3}{2}\ln 3 \right) \equal{} \frac{1}{6} \plus{} \frac{\pi}{12\sqrt{3}}\minus{}\frac{\ln3}{4}$


calculus
integration
logarithms
limit
calculus computations

1)  find the limit
lim          (e^(1/x))/x
x->0-

2)
integral sign (0 to 2)    |x-1||x+1| dx

3)
suppose f(0) = 0 and integral sign (0 to 2) f'(2t)e^(f(2t))  dt = 5 . Find the value of f(4)

the answer the teacher told me b4 was ln9, but i don't know how to get this, also can you explain the steps near the end.

Solution for number 2:

   $ \int_{0}^{2}|x \minus{} 1\parallel{}x \plus{} 1|dx \equal{} \int_{0}^{2}|x^2 \minus{} 1|dx \equal{} \int_{0}^{1}1 \minus{} x^2 dx \plus{} \int_{1}^{2}x^2 \minus{} 1 dx \equal{} .....$.

Btw,for problem 1 use the substitution $ u \equal{} \frac {1}{x}$.

I think new_member forgot a factor $ 2$ in his solution of number 3: this comes from $ (e^{f(2t)})' \equal{} e^{f(2t)} f'(2t) \cdot 2$, that factor $ 2$ by the chain rule.

I then get $ \int_{0}^{2}(e^{f(2t)})'dt \equal{} 10$ hence $ e^{f(4)} \minus{} e^{f(0)} \equal{} 10$.

As we have $ f(0) \equal{} 0$, we get $ e^{f(0)} \equal{} 1$ so $ e^{f(4)} \equal{} 11$ which gives $ f(4) \equal{} \ln 11$.

for problem 1, i use u = 1/x, but i dont get what is the next step, i cant really use L'H rule, i have a ue^u

for problem 2 i get 2/3 when the answer is 2

Note that in problem 1, that is $ \lim\limits_{x \to 0^ \minus{} }$, i.e. from below!

By substituting $ u \equal{} \frac1x$, the limit then becomes $ \lim\limits_{u\to \minus{} \infty} u e^u \equal{} 0$ (or are you having problems with this last equality here?).

For problem 2, we get using new_member's partial solution that

$ I \equal{} \int_{0}^{2}|x \minus{} 1| \,\, |x \plus{} 1|dx \equal{} \int_{0}^{2}|x^{2} \minus{} 1|dx \equal{} \int_{0}^{1}(1 \minus{} x^{2}) \, dx \plus{} \int_{1}^{2}(x^{2} \minus{} 1)\, dx \equal{} 1 \minus{} \frac13 \plus{} \frac13(2^3 \minus{} 1^3) \minus{} (2 \minus{} 1) \equal{} \frac23 \plus{} \frac73 \minus{} 1 \equal{} 2.$

yeah i am having problems with that last equality
i dont see why it equals 0

i get something like -infinity*0 = 0?, is this even right or is this indeterminate?

also just because i am curious , is'nt    1^infinity = 1, because no matter how many times you multiply 1 with 1, you will always have a 1?


superior algebra
superior algebra unsolved

Is there a simplex with integer edges and volume 1?

Basically I've got nothing on this problem, but since I find it funny, I'd like to revive it..

So, for $n = 1,$ needless to say it's obvious; for $n = 2,$ it's not too hard either, thanks to Hero's formula.

But what about $n \geq 3?$  (where $n$ is the dimension of our Euclidean space)


calculus
search
geometry
3D geometry
algebra
factorization
sum of cubes

what is the sum of $ 1^k\plus{}2^k\plus{}3^k\plus{} ... \plus{}n^k$ for any natural number k and n?

It has appeared many times on the forum before (notably in the high school forums, although within the past week in the Calculus forums) -- try a search, e.g. for "sum of cubes" or "power sum" or something similar and a few minutes work will lead to several threads on the subject.


Evaluate the sum: $ \displaystyle\sum_{n\equal{}1}^{1994}(\minus{}1)^n \frac{n^2\plus{}n\plus{}1}{n!}.$

hello, you can also use this formula
$ \sum_{n\equal{}1}^k(\minus{}1)^n\frac{n^2\plus{}n\plus{}1}{n!}\equal{}\frac{(\minus{}1)^k\plus{}2(\minus{}1)^kk\plus{}(\minus{}1)^kk^2\minus{}(k\plus{}1)!}{(k\plus{}1)!}$.
Sonnhard.


Hi
is there any one who has this book:
[b]Functional Equations and How to Solve Them [/b]by [b]Christopher G. Small [/b]by [b]springer verlag [/b]
If you have please attach here.
thanks
Bye.

Sorry, I know that this is dead, but I am trying to help out as much as possible.

@glog, This probably isn't the best place to post these questions. You should probably post them in Questions, Suggestions, and Announcments.

Here:   http://www.artofproblemsolving.com/Forum/viewforum.php?f=144
P.S. how do you do the thing that It says click here, and here is the website???

So it would be [url=http://www.artofproblemsolving.com]AOPS[/url]

i tried it like 1,000,000 times im confused :?

[code]It is [url=WEBSITE HERE]WHAT YOU WANT IT TO SAY HERE[/url][/code]
[url=http://www.artofproblemsolving.com]awesomeness[/url]

[url=http://www.wimp.com]Funny internet videos (not youtube)[/url]

Wait, are you linking to that Old Spice parody?

ya its that website and its the video of the day on [url=http://www.artofproblemsolving.com/Forum/blog.php?u=88632]My blog[/url]
its there and there is a different video every day


limit
integration
function
trigonometry
calculus
Support
real analysis

Show that for every $\varphi \in C_{0}^{1}(\mathbf R)$,
\[\varphi(0) = \lim_{n\to\infty}\frac 1\pi \int_{-\infty}^\infty \frac{\sin(nx)}x \varphi(x) dx.\]

Define a function $H(x)=\int_{0}^{x}\frac{\sin t}t\,dt.$

$H$ is clearly an odd function, and we'll assume that we already know this:

$\lim_{x\to\infty}H(x)=\frac{\pi}2.$

Now note that $\int \frac{\sin nx}x\,dx=H(nx)+C.$ Split the integral into two pieces and integrate each by parts:

$\int_{0}^{\infty}\frac{\sin nx}x\,\varphi(x)\,dx= \left.\left(H(nx)-\frac{\pi}2\right)\varphi(x)\right|_{0}^{\infty}-\int_{0}^{\infty}\left(H(nx)-\frac{\pi}2\right)\varphi'(x)\,dx$

$\int_{-\infty}^{0}\frac{\sin nx}x\,\varphi(x)\,dx= \left.\left(H(nx)+\frac{\pi}2\right)\varphi(x)\right|_{-\infty}^{0}-\int_{-\infty}^{0}\left(H(nx)+\frac{\pi}2\right)\varphi'(x)\,dx$

Evaluate the boundary terms and we get this:

$\frac1{\pi}\int_{-\infty}^{\infty}\frac{\sin nx}x\,\varphi(x)\,dx= \varphi(0)-\frac1{\pi}\int_{-\infty}^{\infty}\left(H(nx)-\frac{\pi}2\,\text{sgn}(x)\right)\,\varphi'(x)\,dx.$

Now consider what happens to the last integral as $n\to\infty.$ Because of the compact support of $\varphi,$ this may be taken to be the integral over a bounded interval. $H(nx)-\frac{\pi}2\,\text{sgn}(x)$ tends pointwise to zero, and is uniformly bounded. Hence (by a weak form of the Dominated Convergence Theorem), the integral tends to zero, proving our result.

Note that this proof functions as a proof of the pointwise convergence of the Fourier integral representation of a sufficiently regular function.


Solve the equation
\[ \sqrt {abx(x - a - b)} + \sqrt {bcx(x - b - c)} + \sqrt {cax(x - c - a)} = \sqrt {abc(a + b + c)} \]
[hide="Solution"]\[x = - \frac {abc}{ab+bc+ca \pm 2\sqrt{abc(a+b+c)}}\][/hide]

What a nice example of how geometry can facilitate things which would be arkward with algebra or calculus! I've found it. :D
[hide="hint: Herons formula"] Putting $ y \equal{} \minus{} x$, we have to prove that
\[ \sum\sqrt {aby(a \plus{} b \plus{} y)} \equal{} \sqrt {abc(a \plus{} b \plus{} c)}\]
is solved by $ y \equal{} \frac {abc}{\sum ab\pm 2\omega}$. By Heron's formula, the equation just sums up areas of triangles, where $ y$ is the radius of the small red inscribed circle. On the other hand, [url=http://en.wikipedia.org/wiki/Descartes'_theorem]Descartes' theorem[/url] tells us the solution(s) for $ y$. [hide]For the solution with the minus sign, corresponding to the circumscribed circle, some of the roots have to be taken negative of course.[/hide] [/hide] :)

Thank you spanferkel for your nice approach. Now, I present a detailed solution

$\alpha= \sqrt{bcx(x-b-c)} \ , \ \beta= \sqrt{cax(x-c-a)}$

$\gamma=\sqrt{abx(x-a-b)} \ , \ \delta=\sqrt{abc(a+b+c)}$

By Heron's formula, we figure out that the numbers $\alpha,\beta,\gamma,\delta$ represent twice the area of the triangles with side lengths $(b+c,x-b,x-c),$ $(c+a,x-c,x-a),$ $(a+b,x-a,x-b),$ $(b+c,c+a,a+b),$ respectively. Now, we consider the triangle $\triangle ABC$ with side lengths $(b+c,c+a,a+b).$ From $\alpha+\beta+\gamma=\delta,$ we deduce that there exists a point $P,$ such that $PA=x-a,$ $PB=x-b,$ $PC=x-c.$ Using the Briggs formulas, we get

$\sin \frac{_1}{^2}\widehat{CAP}=\sqrt{\frac{a(x-a-c)}{(x-a)(c+a)}} \ , \ \cos \frac{_1}{^2}\widehat{CAP}=\sqrt{\frac{cx}{(x-a)(c+a)}}$

$\sin \frac{_1}{^2}\widehat{BAP}=\sqrt{\frac{a(x-a-b)}{(x-a)(a+b)}} \ , \ \cos \frac{_1}{^2}\widehat{BAP}=\sqrt{\frac{bx}{(x-a)(a+b)}}$

$\sin \frac{_1}{^2}\widehat{CAB}=\sqrt{ \frac{bc}{(a+b)(a+c)}}$

Substituting these values into the trigonometric identity

$\sin \frac{_1}{^2}\widehat{CAB}=\sin \frac{_1}{^2}\widehat{CAP} \cdot \cos \frac{_1}{^2}\widehat{BAP}+\cos \frac{_1}{^2}\widehat{CAP} \cdot \sin \frac{_1}{^2}\widehat{BAP}$

yields $bc(x-a)=b\beta+c\gamma.$ By similar reasoning, we get the cyclic expressions $ca(x-b)=c\gamma+a \alpha$ and $ab(x-c)=a \alpha+b\beta.$ From these three we obtain

$x(bc+ca+ab)-abc=2a \alpha=2a \sqrt{bcx(x-b-c)}$

$[(bc+ca+ab)^2-4bca^2]x^2+2abc(bc+ca+ab)x+a^2b^2c^2=0$

$[x(bc+ca+ab)+abc]^2=4\delta^2x^2 \ \Longrightarrow$

$ x =-\frac{abc}{ab+bc+ca\pm 2\delta}$


inequalities
rearrangement inequality
inequalities proposed

If $ a,b,c,d$ are non-negative real numbers such that $ a + b + c + d = 4$

$ a\sqrt {bc} + b\sqrt {cd} + c\sqrt {da} + d\sqrt {ab} \leq 2 + 2\sqrt {abcd}$

P.S.

I tried to solve it this way, but I failed

Let $ a=x^2$,$ b=y^2$,$ c=z^2$,$ d=t^2$ ; now our problem is: if $ x^2+y^2+z^2+t^2=4$ then

$ x^2yz+y^2zt+z^2tx+t^2xy \leq 2+2xyzt$

Let $ (X,Y,Z,T)$ a rearrangement of $ (x,y,z,t)$ such that $ X\geq Y \geq Z \geq T$

Then by rearrangement inequality we have

$ x^2yz+y^2zt+z^2tx+t^2xy = x(xyz) + y(yzt) + z(ztx) + t(txy) \leq X(XYZ) + Y(XYT) + Z(XZT) + T(TYZ)
= X^2YZ + Y^2XT + Z^2TX + T^2YZ$ as $ XYZ \geq XYT \geq XZT \geq TYZ$

so it remains to prove that

$ X^2YZ + Y^2ZT + Z^2TX + T^2XY \leq 2 + 2XYZT$ with $ X\geq Y \geq Z \geq T$ and $ X^2+Y^2+Z^2+T^2 = 4$

I tried to prove this using the fact that

$ T^2YZ \leq XYZT=xyzt$ and $ Z^2TX \leq XYZT=xyzt$ but unfortunately it doesn't work. :(

I suspect Hungkhtn uses a very nice trick.

[quote="manlio"]If $ a,b,c,d$ are non-negative real numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$

$ a\sqrt {bc} \plus{} b\sqrt {cd} \plus{} c\sqrt {da} \plus{} d\sqrt {ab} \leq 2 \plus{} 2\sqrt {abcd}$

P.S.

I tried to solve it this way, but I failed

Let $ a \equal{} x^2$,$ b \equal{} y^2$,$ c \equal{} z^2$,$ d \equal{} t^2$ ; now our problem is: if $ x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \equal{} 4$ then

$ x^2yz \plus{} y^2zt \plus{} z^2tx \plus{} t^2xy \leq 2 \plus{} 2xyzt$

Let $ (X,Y,Z,T)$ a rearrangement of $ (x,y,z,t)$ such that $ X\geq Y \geq Z \geq T$

Then by rearrangement inequality we have

$ x^2yz \plus{} y^2zt \plus{} z^2tx \plus{} t^2xy \equal{} x(xyz) \plus{} y(yzt) \plus{} z(ztx) \plus{} t(txy) \leq X(XYZ) \plus{} Y(XYT) \plus{} Z(XZT) \plus{} T(TYZ) \equal{} X^2YZ \plus{} Y^2XT \plus{} Z^2TX \plus{} T^2YZ$ as $ XYZ \geq XYT \geq XZT \geq TYZ$

so it remains to prove that

$ X^2YZ \plus{} Y^2ZT \plus{} Z^2TX \plus{} T^2XY \leq 2 \plus{} 2XYZT$ with $ X\geq Y \geq Z \geq T$ and $ X^2 \plus{} Y^2 \plus{} Z^2 \plus{} T^2 \equal{} 4$

I tried to prove this using the fact that

$ T^2YZ \leq XYZT \equal{} xyzt$ and $ Z^2TX \leq XYZT \equal{} xyzt$ but unfortunately it doesn't work. :(

I suspect Hungkhtn uses a very nice trick.[/quote]
We have
\[ X^2YZ \plus{} Y^2XT \plus{} Z^2TX \plus{} T^2YZ \equal{}YZ(X^2\plus{}T^2)\plus{}XT(Y^2\plus{}Z^2)\minus{}2XYZT\plus{}2XYZT
\equal{} YZ(X\minus{}T)^2\plus{} XT(Y^2\plus{}Z^2)\plus{}2XYZT \le \frac{(Y^2\plus{}Z^2)}{2}(X\minus{}T)^2\plus{}XT(Y^2\plus{}Z^2) \plus{}2XYZT\equal{}\frac{(X^2\plus{}T^2)(Y^2\plus{}Z^2)}{2}\plus{}2XYZT \le \frac{(X^2\plus{}Y^2\plus{}Z^2\plus{}T^2)^2}{8}\plus{}2XYZT\equal{}2\plus{}2XYZT\]
:)

[i]    You are really intelligent Cẩn

   Very nice solution

  Congratulation[/i] :lol:

You are really great, can_hang2007!!!  :)


search
geometry
projective geometry
geometry theorems

Can somebody (maybe Darij  :D ) explain for me about Polar and Polar reciprocal, polar line of a point respect to a circle, and all it stuff. I'm not sure about "polar and polar reciprocal" that I use, I'll explaine more detail about it.
Give a circle (O) and a point M in plan. If the circle with diameter MN is [u]|[/u] with (O) < angle of circle (MN) and circle (O) is 90 > then N is on polar line (d) of M respected circle (O). And M called polar of the line (d) respect circle (O).
Sorry coz my English is so bad to explain my idea, but if you can understand...

There are some good sources on poles and polars on the internet.

First, you should check out http://www.cut-the-knot.org/Curriculum/Geometry/PolePolar.shtml for main properties and some applications of poles and polars.

Then, there are the topics http://www.mathlinks.ro/Forum/viewtopic.php?t=31812 and http://www.mathlinks.ro/Forum/viewtopic.php?t=5501 on ML.

Then, you can search for "polar" in the geometry section of ML in order to find lots of applications of poles and polars in solving geometry problems.

  Darij


For the non-negative numbers $ a,b,c$ prove the inequality
$ 4(a\plus{}b\plus{}c) \geq 3(a\plus{}\sqrt{ab}\plus{}\sqrt[3]{abc})$
When does equality hold?

Becouse of $ AM\geq GM$ inequality we have:

$ \frac {3a}{4} \plus{} 3b\geq 3\sqrt {ab}$  and  also: $ \frac {a}{4} \plus{} b \plus{} 4c\geq3\sqrt [3]{abc}$

so after summing these two inequalities with $ 3a \equal{} 3a$ we get desired inequality.

Equality is achieved when $ \frac {a}{4} \equal{} b \equal{} 4c$ so for any triple $ (t,4t,16t)$

(which is in fact  equivalnet to $ a$, $ b$ and $ c$ being (in this order) in geometric progreession in which $ q \equal{} 4$)


Let $a, b, c>0$ such that $ab+bc+ca=3abc.$ Prove that the following inequality holds:

   \[ \frac{1}{1+\sqrt{ab}}+\frac{1}{1+\sqrt{bc}}+\frac{1}{1+\sqrt{ca}}\leq\frac{3}{2}.  \]

Very nice Cezar!! :)

Is the trick to write
$\sum\frac{\sqrt{ab}}{1+\sqrt{ab}}\geq\frac{3}{2}$??

[quote="cezar lupu"]Yes, it is. :)   Keep going....  ;)[/quote]

Then we write

$\sum\frac{1}{1+\frac{1}{\sqrt{ab}}}$

si applicam ineqalitatea din Cauchy....

Do you catch me?? :)

Hint:
 [hide]
 Yes, Cauchy and AM-GM. Try to post a full solution. ;)
 [/hide]

[hide]$\sum\frac{1}{1+\frac{1}{\sqrt{ab}}}\geq\frac{9}{3+\sum\frac{1}{\sqrt{ab}}}\geq\frac{3}{2}$

since trivial ineq yields

$\sum\frac{1}{a}\geq\sum\frac{1}{\sqrt{ab}}$ (and $\sum\frac{1}{a}$=3 which follows from the given relation )[/hide]

Why do you care about full solutions?? :D

Actually, I don't care. I have found a solution too. In fact, my solution is almost the same with yours.:P


we know,if a prime p=4k+1,there exit integers a,b so that p=sqr(a)+sqr(b).but can we define there is only one solution of (a,b)???

Yes, up to sign and permutation.  Otherwise $ a \plus{} bi$ would be composite in the Gaussian integers.

thanks,but can anyone explain it clearly?

Here's an elementary proof.  Suppose $ a, b, c, d$ are positive integers with $ \{ a, b \} \neq \{ c, d \}$ such that $ p \equal{} a^2 \plus{} b^2 \equal{} c^2 \plus{} d^2$.  Then

$ a^2 \minus{} c^2 \equal{} d^2 \minus{} b^2 \Leftrightarrow (a \minus{} c)(a \plus{} c) \equal{} (d \minus{} b)(d \plus{} b)$.

This implies (and this is a nice exercise) that there exist positive integers $ x, y, z, w$ such that

$ a \minus{} c \equal{} xy, a \plus{} c \equal{} zw, d \minus{} b \equal{} xw, d \plus{} b \equal{} yz$

hence

$ a \equal{} \frac {zw \plus{} xy}{2}, c \equal{} \frac {zw \minus{} xy}{2}, d \equal{} \frac {yz \plus{} xw}{2}, b \equal{} \frac {yz \minus{} xw}{2}$.

We then compute that

$ a^2 \plus{} b^2 \equal{} c^2 \plus{} d^2 \equal{} \frac {(zw)^2 \plus{} (xy)^2 \plus{} (yz)^2 \plus{} (xw)^2}{4} \equal{} \frac {(x^2 \plus{} z^2)(y^2 \plus{} w^2)}{4}$

which contradicts $ p$ prime.


Hi every1,

Can we devise an algorithm using dynamic programming which can find nth fibonacci number in O(log n)..

This is given as a problem in this pdf file---

 activities.tjhsst.edu/sct/lectures/dp1.pdf

[quote="mahbub"]yes it is possible. the idea is matrix power. u know if a matrix is: A then A^n can be computed y some short procedure, may be nlgn. u have to use that idea, and surely u have to make a useful matrix A.[/quote]

Well, finding A^n requires at least O(A) time because it is somewhat necesarry to create the matrix of size A, which is itself O(A). [O(A) refers to O(size of A)].

A cheap shot would be this: solve the linear recurrence equation then use numeric methods. That is certainly log(n).

EDIT: Nvm, maybe I should've read the source first. O(A) would be O(4) lol. So idea is just DP, calculate A^1, A^2, A^4, A^8, etc., that's log(n), then you pick the necesarry numbers to get to A^n. (Like for n = 13 you'd pick 1, 4, 8).

your search is over
the method you seek for is here
http://www.cs.utexas.edu/users/EWD/ewd06xx/EWD654.PDF


limit
trigonometry
algebra
polynomial
calculus
calculus computations

In a math test, because of the lack of time and I'm not lucid at that time , I use this approximation when $ x\rightarrow 0$ in this limit finding problem :
$ \lim_{x\rightarrow 0}\frac {\sqrt [3]{cosx} \minus{} \sqrt {cosx}}{{x}^{2}} \equal{} \lim_{x\rightarrow 0}\frac {\sqrt [3]{1 \minus{} \frac {{x}^{2}}{2}} \minus{} \sqrt [2]{1 \minus{} \frac {{x}^{2}}{2}}}{{x}^{2}} \equal{} \lim_{x\rightarrow 0}\frac {(1 \minus{} \frac {{x}^{2}}{2}*\frac {1}{3}) \minus{} (1 \minus{} \frac {{x}^{2}}{2}*\frac {1}{2})}{{x}^{2}} \equal{} \frac {1}{12}$
And this way gives the right result but the TA [b]refuse to give me any score[/b] for this problem despite all of my best efforts  :(  . She said "we couldn't approximate $ cosx \equal{} 1 \minus{} \frac {{x}^{2}}{2}$ when we find the limit of a sum , we just can use this in a product limit . I had knew that there're some cases that we couldn't approximate like that , but I also think that there're many cases that we can approximate like that way when compute the limit ,with $ x\rightarrow 0$ .
I really need of a good explanation for this solution :why it should give the exact result, a little bit why and when I could approximate like that  .. in order to get some marks on this problem .
A generalisation for this problem is : if $ x\rightarrow 0$ => $ f(x)\approx u(x)$ and $ g(x)\approx v(x)$ , when we could approximate like this :
$ \lim_{x\rightarrow 0}\frac {f(x) \plus{} g(x)}{h(x)} \equal{} \lim_{x\rightarrow 0}\frac {u(x) \plus{} v(x)}{h(x)} ???$
I have found some other solutions after the test but it's too late to submit it . Hope you can help me an explanation for this solution . Thanks a lot .

This approach is correct in the sense that it gives the correct coefficients of the Taylor expansion of the numerator, but ideally you should've used l'Hopital's rule to rigorously justify that you 1) had the right coefficients, and 2) had enough of them.

Yah, when I go home, I think I should have used L'Hopital's rule or using make the "1" appear like : $ ( \sqrt [3]{cosx} \minus{} 1 ) \minus{} ( \sqrt {cosx} \minus{} 1 )$but it's too late  :(  .
Hey your suggestion about Taylor expansion is great . I will use that to explain . Thanks .

The equals sign in your work would bother me because it's not literally true that $ \cos x\equal{}1\minus{}\frac{x^2}2.$ You have to deal with the extent to which that is in error.

I might try justifying it by writing my Taylor polynomials in big-O style:

The binomial series: $ (1\plus{}u)^{\alpha}\equal{}1\plus{}\alpha u\plus{}O(u^2)$

The cosine series: $ \cos x\equal{}1\minus{}\frac{x^2}2\plus{}O(x^4).$

Hence, $ \sqrt[3]{\cos x}\equal{}1\minus{}\frac{x^2}{6}\plus{}O(x^4)$ and $ \sqrt{\cos x}\equal{}1\minus{}\frac{x^2}4\plus{}O(x^4).$

Hence we have

$ \lim_{x\to 0}\frac{\left(1\minus{}\frac{x^2}{6}\plus{}O(x^4)\right)\minus{}\left(1\minus{}\frac{x^2}4\plus{}O(x^4)\right)}{x^2}\equal{}\lim_{x\to0}\frac{\frac{x^2}{12}\plus{}O(x^4)}{x^2}$

$ \equal{} \lim_{x\to0}\frac1{12}\plus{}O(x^2)\equal{}\frac1{12}.$

One measure of the limitations of your method: what if you had been asked this?
\[ \lim_{x\to0}\frac{12\sqrt[3]{\cos x}\minus{}12\sqrt{\cos x}\minus{}x^2}{x^4}\]

Of course, my work above does not solve this, but it tells me what I need to do - to use additional terms of the Taylor polynomials to replace $ O(x^4)$ with more exact details.

(I would have given you substantial partial credit - I would have made some deduction for the sloppy notation, but acknowledged the idea.)

Yah, so all the reasons behind these approximations is the Taylor polynomials  :lol:  . Thanks .


Let $ B(N)$ be the number of 1s in the binary expression of the integer $ N$. Find $ \displaystyle\sum_{n\equal{}1}^{\infty}\frac{B(n)}{n(n\plus{}1)}$.

I might have made some error, but at least this will get you started if you solved it yourself...

I'm using B to represent B(n)...

B/(n(n+1)) = B/(n+1) - B/n

now simply plug values for upto n=13 or so and you'll see a trend in the series. Add every term in B/(n+1) to the same denominator term in B/n (which is negative) and the result will have numerator 1, except for every 4th term....i.e it will look something like  0 + 1/3 +1/4+1/5+0+1/7+1/8+1/9+0+1/11.......

and thus you can express the sum as

SUM(n=1 to inf) [(1/(4n-2)) - (1/n)]

Now, I'm not quite sure if you can simplify that to a better expression (there must be a way, but I can't remember it). But at least that takes care of the main problem.

Please feel free to comment/correct/provide a better solution.

(Sorry for the sloppy notation, but I'm not sure how to represent math expressions properly in this forum.)


parameterization algebra function domain calculus derivative logarithms

What does $y$ equal in this problem?

is $y=f(x)$?

$y$ is an arbitrary parameter- presumably, this is intended to be true for all values of $x$ and $y$.

The domain of $f$ should also be specified- I would guess that it's the positive real numbers.

[quote="JBL"]I'm working on the assumption that $f$ is to be a differentiable function over ${\bf R}_{> 0}$, and the functional equation is meant to hold for all $x, y \in{\bf R}_{>0}$.  [hide]Then make the substitution $x \mapsto xy$ and $y \mapsto \frac{x}{y}$ to get $2f(xy) = f(x^{2})+f(y^{2})$ for all positive reals $x, y$.  Since this is a functional identity in $x$, we can take a derivative on both sides and get $2yf'(xy) = 2xf'(x^{2})$ for all real $x, y$.  Then setting $x \mapsto1$ gives $2yf'(y) = 2f'(1) = 2$, so $f'(y) = \frac{1}{y}$ for all $y$.  It follows immediately that $f(x) = \log x$ is the only solution.[/hide][/quote]

MY dear  JBL, I'm sorry ur soln is wrong .you are not given that f is diff over R or R>0

If he's wrong, it's only because he couldn't tell what the exact question you wanted to ask is. Please clarify what the domain of $f$ is, and [b]all[/b] assumptions about it.

[quote="jmerry"]If he's wrong, it's only because he couldn't tell what the exact question you wanted to ask is. Please clarify what the domain of $f$ is, and [b]all[/b] assumptions about it[/quote]
 I think , you dont need the domain of $f$ and you are not provided with the differentiablity of $f$ on whole of its domain.


induction
combinatorics unsolved
combinatorics

n signals are equally spaced along a rail track. No train is allowed to leave a signal whilst there is a moving train between that signal and the next. Any number of trains can wait at a signal. At time 0, k trains are waiting at the first signal. Except when waiting at a signal each train travels at a constant speed, but each train has a different speed. Show that the last train reaches signal n at the same time irrespective of the order in which the trains are arranged at the first signal.

Can you say me if I understood the problem?

According to my interpretation, there are $ k$ trains in a rail track, and all they are in the first sign. The size of the trains is $ 0$. The trains are in a certain order. Each train moves if and only if it is authorized, i.e., if it is the first train waiting at the sign (the order of the trains is always the initial order) and there is no other train between the sign in which the train is and the next one. We need to prove the time to the last train reaches the last sign does not depend on the initial order.

I am sorry if this sounds stupid, but I don't think I am allowed to use intuition in a mathematical problem... Specially because the trains have size $ 0$!

I think I have a solution (according to my interpretation).

Let the trains be $T_1,T_2,\ldots$ and $T_k$. Let $t_i$ be the time the $i$th train takes to go from one signal to the next, with $t_1\leq t_2\leq\ldots\leq t_k$. Consider the case the foremost train is $T_1$, the second one is $T_2$, etc. It is easy to see the trains move in an essentially independent way, and that the total time is $t_1+t_2+\ldots+t_{k-1}+(n-1)t_k$ (observe that there are $n-1$ "rail pieces" between the signals).

Thus, the slowest train is of particular interest. Now consider any order of the trains, and define $S=\{s\text{ such that }T_s\text{ is ahead of }T_k\}$, so that the trains indexed by $S$ leave before $T_k$. It is easy to see the total time spent before the appearence of $T_k$ equals $\sum_{s\in S}t_s$. After that, the trains $T_s, s\in S,$ don't affect $T_k$ anymore.$(*)$

The time until the arrival of $T_k$ at the $n$th signal is therefore $\sum_{s\in S}t_s+(n-1)t_k$. If, at this moment, the remaining trains were in the $(n-1)$th signal, it is clear the total time needed would be the same. This is not always the case, but we can show there is no wasted time, i.e., after the end of $T_k$'s movement, each time a train arrives at the $n$th sign, there is another train waiting for it in the $(n-1)$th sign.

This is guaranteed with the induction hypothesis (we use induction on $n+k$). In fact, for any train $T_i$ after $T_k$, we can assume the total time for $T_i$ to go to signal $n-1$ is a sum of $t_j$'s, where $T_j$ is before $T_i$, $j\neq k$, and a particular term $(n-2)t_k$. This time is smaller than the time for the train before $T_i$ to go to signal $n$, which is found by the same formula (the time difference is $t_k-t_i$).

$(*)$ In reviewing, I realized this is probably not so obvious. But we can use the induction hypothesis to see the ordering of the trains indexed by $S$ doesn't matter. And if the trains are ordered by speed, they all move independently.


[hide="2"]
$ (x\plus{}y\plus{}z)^2\equal{}x^2\plus{}y^2\plus{}z^2\plus{}2(xy\plus{}yz\plus{}zx)\Longrightarrow xy\plus{}yz\plus{}zx\equal{}0$
and by the identity:
$ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}zx)$
$ 7\minus{}3xyz\equal{}7\times 49$
hence we have $ xyz\equal{}\minus{}112$.
[/hide]

[quote="therobz"]1. Find$ x^2 \plus{} y^2$if x, y are natural numbers and $ xy\plus{}x\plus{}y\equal{}71, x^2y\plus{}xy^2\equal{}880$.

2. If $ x^2\plus{}y^2\plus{}z^2\equal{}49$ and$ x\plus{}y\plus{}z\equal{}x^3\plus{}y^3\plus{}z^3\equal{}7$, find$ xyz$.

3. Determine the triples of integers $ x,y,z$ satisfying:
$ x^3\plus{}y^3\plus{}z^3\equal{}(x\plus{}y\plus{}z)^3$[/quote]
I think it very easy :oops:

[quote="quykhtn-qa1"][quote="therobz"]1. Find$ x^2 \plus{} y^2$if x, y are natural numbers and $ xy \plus{} x \plus{} y \equal{} 71, x^2y \plus{} xy^2 \equal{} 880$.

2. If $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 49$ and$ x \plus{} y \plus{} z \equal{} x^3 \plus{} y^3 \plus{} z^3 \equal{} 7$, find$ xyz$.

3. Determine the triples of integers $ x,y,z$ satisfying:
$ x^3 \plus{} y^3 \plus{} z^3 \equal{} (x \plus{} y \plus{} z)^3$[/quote]
I think it very easy :oops:[/quote]

Er, could you post your solution? :P

1.

The first equation gives xy = 880/(x+y). Substituting this into the second equation gives 880/(x+y) + (x+y) = 71 which implies x+y = 16 or 55. Then xy = 55 or 16, respectively abd so x^2 + y^2 = (x+y)^2 - 2xy = 146 or 2993.

[quote="therobz"]1. Find x^2 + y^2 if x, y are natural numbers and xy+x+y=71, x^2y+xy^2=880.

2. If x^2+y^2+z^2=49 and x+y+z=x^3+y^3+z^3=7, find xyz.

3. Determine the triples of integers x,y,z satisfying:
x^3+y^3+z^3=(x+y+z)^3[/quote]
3 is very easy$ x^3\plus{}y^3\plus{}z^3\equal{}(x\plus{}y\plus{}z)^3$
$ \Leftrightarrow (x \plus{} y)(y \plus{} z)(x \plus{} z) \equal{} 0$ :)

I had (x+y+z)(xy+yz+zx) = xyz, which is equivalent to TRAN THAI HUNG's condition. AM-GM shows that all three of x, y and z cannot be positive. I also had (0, 0, 0), (n, 0, 0), (n, -n, 0), (m, n, -n) and permutations.

This gives 49 >= 49^2 so x, y and z cannot all be positive.


What's the units digit of a^5 - a, where a is a positive integer?

[hide]$a^5$ always has a units digit of the units digit of a, so $a^5-a$ has a units digit of 0.[/hide]

[hide]
The answer is $0$, which means $a^5-a$ is always a multiple of 10.

Prove by induction.

(I) For $a=1$, $1^5-1=0$, so the statement is true.

(II) Assume true for $a=n$. Then $n^5-n=10k$ for some positive integer $k$.

(III)For $a=n+1$,

$(n+1)^5-(n+1)$
$=n^5+5n^4+10n^3+10n^2+5n+1-n-1$
$=(n^5-n)+5(n^4+n)+10(n^3+n^2)$
$=10k+5n(n^3+1)+10(n^3+n^2)$
$=10k+5n(n+1)(n^2-n+1)+10(n^3+n^2)$

Note that $n(n+1)$ is always even because either $n$ or $n+1$ is even. So, we can let $5n(n+1)(n^2-n+1)=5(2m)(n^2-n+1)$ for some positive integer $m$.

Then, the expression becomes
$10k+10m(n^2-n+1)+10(n^3+n^2)$

Therefore, the statement is true for $a=n+1$.

From (I), (II), (III), $a^5-a$ is a multiple of 10 for any positive integer $a$.
[/hide]

Another way to prove this is by casework:

[hide]Consider two cases:

Case 1.) If $a$ is even:

Notice $a^5 - a$ factors into $a(a^4-1)$, so it is sufficient to show that for all $a$, $5 | a^4 - 1$.

If $a \equiv 0 \pmod{5}$, we are done. Since $a$ is even it must be divisible by 10 also.
If $a \equiv 1 \pmod{5}$, $a^4 \equiv 1 \pmod{5}$.
If $a \equiv 2 \pmod{5}$, $a^4 \equiv 16 \pmod{5} \equiv 1 \pmod{5}$.
...
And so on for $a$ equivalent 3 and 4 modulo 5. We find that $a^4 \equiv 1 \pmod{5}$ for all $a$, and $a^4 - 1 \equiv 0 \pmod{5}$.

Case 2.) If $a$ is odd:

$a(a^4 - 1) = a(a^2-1)(a^2+1)$

We want to show that either $a^2-1$ or $a^2+1$ is equivalent to $0\pmod{10}$ for all $a$. Notice $a$ can only be equivalent to 1, 3, 5, 7, or 9 modulo 10.

If $a \equiv 1\pmod{10}$, $a^2-1 \equiv 0\pmod{10}$, so we're done.
If $a \equiv 3\pmod{10}$, $a^2+1 \equiv 0\pmod{10}$.
If $a \equiv 5\pmod{10}$, we are done because both $a^2-1$ and $a^2+1$ are even.
If $a \equiv 7\pmod{10}$, $a^2 + 1 \equiv 0\pmod{10}$.
If $ a\equiv 9\pmod{10}$, $a^2 - 1 \equiv 0\pmod{10}$.

So we are done with casework.[/hide]

I really appreciate your solutions :)
Although my pick goes for frt's one  :P


But how can we prove it is impossible to tile without purely 3 by 5s?

See $n=1, 2, 4, 7$ are not achievable. Now, are achievable those $n's$ that can be written as $n=3a+5b$ with $a, b$ non negative integers. Do your cases and see that $n=3, 5, 6, 8, 9, 10 ...$ are achievable; now I'll prove if $n$ is achievable, $n+1$ is achievable. If $b \ge 1$ and $n=3a+5b$ we can achieve $n+1=3(a+2) + 5(b-1)$, if $b=0$, since we proved all cases until $10$, we have $a > 3$ so we can achieve $n+1=3(a-3)+5(b+2)$ and we are done, $n=1, 2, 4, 7$ are not achievable, every other $n$ is achievable.

[quote=Sepp]Clearly we can cover every $ 5 \times 3$ or $ 3 \times 5$. So we can also cover a $ 15 \times n$ with:
$ n \equiv 0 \pmod 3$ (all $ 3 \times 5$ pieces),
$ n \equiv 2 \pmod 3$ and $ n > 2$ (a column of $ 5 \times 3$ and all the others $ 3 \times 5$),
$ n \equiv 1 \pmod 3$ and $ n > 7$ (two columns of $ 5 \times 3$ and all the others $ 3 \times 5$).
Then it's easy to see we cannot cover the board if $ n = 1, 2, 4, 7$.[/quote]

[quote=Sepp]Then it's easy to see we cannot cover the board if $ n = 1, 2, 4, 7$.[/quote]

HOW do you prove that $n=4,7$ are not possible?


limit
real analysis
real analysis unsolved

Let an infinite sequence of measurable sets be given on the interval $ (0,1)$ the measures of which are $ \geq \alpha>0$. Show that there exists a point of $ (0,1)$ which belongs to infinitely many terms of the sequence.

$ \mu\left(\bigcap\limits_{i \equal{} 1}^{\infty}\bigcup\limits_{j \equal{} i}^{\infty}A_j\right) \equal{} \lim \mu(\bigcup\limits_{j \equal{} i}^{\infty}A_j)\geq \alpha$, the equality holds because measure of the union is bounded by 1.

[hide=To clarify Yustas' solution]
We use
[hide=Rudin RCA, Theorem 1.19 (e)]
Given measurable sets $B_1 \supset B_2 \supset B_3 \supset \cdots$ such that $m(B_1)$ is finite, we have $m(B_i) \to m(B_1 \cap B_2 \cap B_3 \cap \cdots)$ as $i \to \infty$
[/hide]
Let $B_i = A_i \cup A_{i+1} \cup A_{i+2} \cup \cdots$. Since $m(A_i) < 1$, we know that $m(B_1) < 1 < \infty$, and we also know that $B_1 \supset B_2 \supset B_3 \supset \cdots$ so the theorem applies and we know that there exists $0<x<1$ such that $x \in B_i$ for all $i$. Now it is easy to construct a sequence $\{k_i\}^{\infty}_1$ such that $x\in A_{k_i}$ for all $i$, which finishes the proof.
[/hide]


calculus
topology
integration
derivative
complex analysis

I'm just curious: what kind of background is needed to begin learning complex analysis? I'm currently going through standard Calc2 right now. Is multivariable needed? (I'm assuming yes)

Also, this is all outside the classroom so I would have no textbook or any access to one. Is there anyway I'd be able to begin complex analysis without one?

I'd say multivariable is needed.  Here's a link to an online complex analysis book I found by googling for "complex analysis" (so you may have found this one already).

http://www.math.gatech.edu/~cain/winter99/complex.html

You can do it without multivariable calculus using the right book (even if it is used, it's normally only 2 variables and very few stuff about differentiation). But some experience with analytical terms (continuity, metric topology, integrals, etc) is necessary.
A good book I've read was "Complex Analysis" by Freitag and Busam. It uses not much about multivariable calculus. The first half are the classical things (e.g. residue theorem), the second half shows more advanced things (like modular forms and analytical number theory). I've read it with knowing only what here is called "Analysis 1" (Differentiation+Integration in one variable, related concepts); the book needs a bit more, but if one is either willing to accept the missing bits or just looks them up, then it's no problem to read it.


[quote]If a scuba diver is to remain submerged for 1 hour, what pressure must be applied to force sufficient air into the tank to be used? Assume .5 Liters of air per breath a standard atmospheric pressure, a respiration rate of 38 breaths per minute, and a tank capacity of 30 Liters.[/quote]

How would you go by solving this problem? I tried dimensional analysis but I got stuck.

Try
P1V1=P2V2

i think its similar 2 an NCERT  Question(india)

Well, I'd start by thinking about how much air the diver is going to consume in that hour.  Then it's just just your standard $ PV \equal{} nRT$ problem.

Well what would you plug in for each variables?

My Chemistry teacher is expecting the students to learn the gas laws by ourselves...but I'm having trouble understanding it.

That's a relatively difficult problem to pose so early in the game, I think.  In any case, with regards to gas laws, mostly all you need is the ideal gas law, $ PV \equal{} nRT$.  This is a relationship that is true (at least in theory, in practice it's just an approximation) for all gases.  This means that whenever we multiply pressure and volume, we get the same result as multiplying the number of moles and temperature, up to a constant ($ R\approx .0821 \text{ L atm/K mol}$).

Ok so how does that help?  First of all, using dimensional analysis, can you find the total volume of breath (at normal, atmospheric pressure) needed?  Can you find the final volume that you want to compress all this breath into?

Now we can think about plugging things in: we know the beginning volume/pressure, we know the final volume and want to find the final pressure.  In both cases the $ nRT$ is the same, so we have that $ P_\text{beginning} V_\text{beginning} \equal{} nRT \equal{} P_\text{end}V_\text{end}$.  It's just plug and play from here.

Let me know if any questions.

Well I know that the diver will need 1140 liters of air at standard atmospheric pressure and that the tank has a capacity of 30 Liters. So would the beginning Volume = 30 L and the ending Volume = 1140 L and the beginning Pressure = 101.3 KPa?

that's rite--but where does 101 Kpa come?????

101.3 KPa is the standard atmospheric pressure.


function
search
algebra proposed
algebra

Find all functions $f: \mathbb R^{>0} \to  \mathbb R^{>0} $ such that
\[f(x)f(yf(x))=f(x+y)\]
holds for all reals $x$ and $y$.

I know a solution, but what is your solution?  :lol:

This is from IMC, 2000 : http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1113833585&t=58521

[quote="dreammath"]Find all functions:R* to R* (R* means the set of positive reals.)
$ f(x)f(yf(x))=f(x+y)$.
It 's very interesting for anybody who want to solve it. :lol:[/quote]

[u]Claim 1[/u] : $ f(x)\leq 1$ $ \forall$ $ x>0$
Let $ x_{0}>0$ such that $ f(x_{0})>1$. Then let $ y_{0}=\frac{x_{0}}{f(x_{0})-1}>0$.
$ f(x_{0})f(y_{0}f(x_{0}))=f(x_{0}+y_{0})$ $ \implies$ $ f(x_{0}) f(\frac{x_{0}f(x_{0})}{f(x_{0})-1}) = f(\frac{x_{0}f(x_{0})}{f(x_{0})-1})$ and, since $ f(x)>0$ $ \forall x$ : $ f(x_{0})=1$, which is a contradiction. So Claim 1 is true.

[u]Claim 2[/u] : $ f(x)$ is a non increasing function
Obvious since $ y>x$ $ \implies$ $ f(x) f((y-x)f(x))=f(y)$ and so $ f(y)\leq f(x)$ since $ f((y-x)f(x))\leq 1$ (with Claim 1)

[u]Claim 3[/u] : if it exists $ x_{0}>0$ such that $ f(x{0})=1$, then $ f(x)=1$ $ \forall$ $ x>0$
Just put $ x=x_{0}$ in the original equation and you get $ f(y+x_{0})=f(y)$ $ \forall$ $ y>0$. and so $ f(x)$ is a constant (since it is a non increasing function (claim 2) and so $ f(x)=f(x_{0})=1$ $ \forall$ $ x>$ and Claim 3 is true.

[u]Claim 4[/u] : if $ f(x)$ is non constant, then $ f(x)$ is an injective function (and so is strictly decreasing)
Let $ f(u)=f(v)$ with $ u>v$. Then $ f(v) f((u-v)f(v))=f(u)$ and so $ f((u-v)f(v))=1$ and so $ f(x)=1$ (with claim 3). Q.E.D.

Solutions :
$ f(x)=1$ is a solution.
Consider now $ f(x)$ non constant, that's to say $ f(x)$ injective (claim 4) :
Let $ y=\frac{z}{f(x)}$. Then the initial equation becomes $ f(x)f(z)=f(x+\frac{z}{f(x)})$
But we also have $ f(z)f(x)=f(z+\frac{x}{f(z)})$
So  $ f(x+\frac{z}{f(x)})=f(z+\frac{x}{f(z)})$ and, since $ f(x)$ is injective : $ x+\frac{z}{f(x)}=z+\frac{x}{f(z)}$ and so (dividing by $ xz$) $ \frac{1}{z}+\frac{1}{xf(x)}=\frac{1}{x}+\frac{1}{zf(z)}$.

So $ \frac{1}{z}-\frac{1}{zf(z)}=\frac{1}{x}-\frac{1}{xf(x)}=a$ and $ f(x)=\frac{1}{1-ax}$

And it is easy to verify that this expression is a solution of initial equation as soon as $ a\leq 0$ (since we want $ f(x)>0$)

So the general solution is $ f(x)=\frac{1}{1+ax}$ for any $ a\geq 0$

[quote="aviateurpilot"]it's easy,
$ f(x)=\frac{|2-x|+2-x}{(2-x)^{2}}\ if\ x\neq 2\ and\ f(2)=0$

or $ f(x)=\frac{Max(2-x,0)}{(2-x)+[10^{-|x-2|}]}$  :wink:[/quote]
sorry [b]N.T.TUAN[/b], here it's solution for another problem (easy) where $ f(2)=0$ and $ f(x+y)=f(x)f(yf(x)),\forall (x,y)\in R^{+}^{2}$
and $ f(x)=0\ in [0,2[$  :rotfl: .

joke :P  is it a problem from IMO? pco are right!


Two buses start at the same time from A and B to B and A respectively, each with constant velocity. They reach their destinations respectively after 1 hour and 4 hours of passing each other. What is the ratio of the velocities of the two buses?

[hide="Bus"]Let the faster bus have velocity a, and the slower bus have velocity b.  In addition, let the distance between A and B be d, so the time it takes for the two buses to pass each other is given by d/(a+b).  The distance that the faster bus has traveled in this time is given by [d/(a+b)]a.
Since d-[d/(a+b)]a is the distance left to travel, we have:
[d-da/(a+b)]/a=1. Solving for d gives d=a(a+b)/b.  Going on to b, we find that the distance b goes before it passes the faster bus is given by
[d/(a+b)]b, so the distance it has left is d-[db/(a+b)].  Bus b can cover this time in [d-[db/(a+b)]]/b=4 hours.  Solving for d gives d=4b(a+b)/a.  Since d=d, we have:
4b(a+b)/a= a(a+b)/b==>this gives 4b^2=a^2.  Taking square roots give 2b=a, or a/b=[u]2[/u][/hide]

After filling a page with complicated looking expressions, I came up with this
[hide]Let s be the velocity of the slower bus, R the ratio of the velocity of the faster bus to the velocity of the slower bus, and d the distance traveled by the slow bus when the buses meet.
When the buses meet, the slow bus has traveled a distance d, while the fast bus has traveled a distance Rd.
After that, it takes 4 hours for the slower bus to cover the distance Rd, while it only takes 1 hour for the faster bus to cover distance d, at velocity Rs
4s = Rd
1Rs = d    or    Rs = d
Substituting Rs for d in the first equation:
4s = RRs
R=2

Would a generalization be that the ratio of the remaining travel times is the square of the ratio of the velocities?[/hide]

Actually the ratio of their velocities is -2 because velocity is a vector quantity. :wink:

Nice solutions guys. I had to do some messy work too to get a solution.

[hide="My solution"]

[asy]size(200);
pair A,B,M;
A=origin;
B=(180,0);
M=(120,0);
dot(A);
dot(B);
dot(M);
label("A",A,W);
label("B",B,E);
label("M",M,S);
draw(A--B);[/asy]

In the diagram, let $ M$ be the point where the two buses meet, and let $ AB \equal{} s, AM \equal{} x$. Assume that the bus from $ A$ takes $ t$ hours to reach $ M$. Let $ v_A$ and $ v_B$ be the velocities of the two buses. Then it's clear that $ v_A \equal{} \frac xt \equal{} \frac {s \minus{} x}{1}\ (1)$ and $ v_B \equal{} \frac x4 \equal{} \frac {s \minus{} x}{t}\ (2)$.

$ (1)\implies t \equal{} \frac {x}{s \minus{} x}$

$ (2)\implies t \equal{} \frac {4(s \minus{} x)}{x}$

Thus $ \frac {x}{s \minus{} x} \equal{} \frac {4(s \minus{} x)}{x}\implies \left(\frac {x}{s \minus{} x}\right)^2 \equal{} 4$. Hence $ \frac {x}{s \minus{} x} \equal{} 2$. But $ \frac {x}{s \minus{} x} \equal{} \frac {\frac xt}{\frac {s \minus{} x}{t}} \equal{} \frac {v_A}{v_B}$. Therfore $ \boxed{\frac {v_A}{v_B} \equal{} 2}$ [/hide]

[quote="KMST"]Would a generalization be that the ratio of the remaining travel times is the square of the ratio of the velocities?[/quote]

Yes, I think.


The sum of the two 5-digit nmbers $AMC10$ and $AMC12$ is 123422. What is $A+M+C$?

$\left(A\right) 10$
$\left(B\right) 11$
$\left(C\right) 12$
$\left(D\right) 13$
$\left(E\right) 14$

[hide]
Looking at the last two digits we can see that $10+12=22$ so that nothign carries over. So, we can ignore the last two digits. We know:
$\begin{eqnarray*}AMC+AMC &=& 2 \cdot AMC = 1234 \\ AMC &=& \frac{1234}{2}= 617 \\ A+M+C &=& 6+1+7 = 14 \implies \boxed{\text{E}}$
[/hide]


Prove that there are infinitely many composite numbers $ n$ such that $ n\,|\,3^{n\minus{}1}\minus{}2^{n\minus{}1}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=150409
Look there

Daniel


Though a whole bunch of thread regarding favorite piece(s) already. I still wonder which Beethoven's Sonata leaves you deepest impression.

I like to know which Sonata(out of the 32) you love to listen to the most recently(not says you can play the best). Try to select only one and give reasons why you prefer it if possible. Thanks :)

For me it's No.29, Op.106(Hammerklavier)... Well, simply because that's the most BEETHOVEN-type Sonata.

Op.57 "Appasionata" is the most moving one for me, although they are all good, as Beethoven=genius.

[quote="CatalystOfNostalgia"]I completely agree. Hammerklavier is, in my opinion, the most techincally and musically demanding in the standard repertoire by far. Not to mention the stamina needed...

I think all of the late ones are the best though (106, 109, 110, 111), I think their darkness best defines Beethoven. And all of them have fugues in the last movement![/quote]

Doesn't op.111 just have themes replaced with fugues in the first movement?

I like the pathetique sonata the most. :lol:

Yeah, but I don't know much about Beethoven's violin works :blush: Any recommendations?

I like the sonata No. 4 and No. 5. 5 is overplayed, but for a reason! :)

All of Beethoven Sonatas are amazing!  Beethoven was a genius!

My favorite would have to be the Appasionata though.  I find the Pathetique the most fun to play.


geometry
Circumcenter
circles
IMO Shortlist

A quick angle chase reveals that $CA_1QA_2$ is cyclic, meaning that we can forget about the $B_i$'s: we're looking for the locus of the circumcenter of $A_1A_2Q$.

Since $\angle PA_iQ$ are constant angles, it means that the triangles $A_1A_2Q$ are all similar, and this means that the triangles $A_1QO$ are all similar, where $O$ is the circumcenter of $A_1A_2Q$. This means that the locus of $O$ is the image of $S_1$ through the spiral similarity of center $Q$ which turns $A_1$ into $O$. In other words, this locus is a circle, Q.E.D.

Comment: it can be shown fairly easily now, by choosing particular positions of $A_1$, the the circle is, in fact, the circumcircle of $QO_1O_2$, where $O_i$ is the center of $S_i$.

If we regard $ A_1A_2PB_1B_2C$ as a complete quadrilateral the problem becomes trivial once we know that the circumcenters of the $ 4$ "small" triangles of the complete quadrilateral and the miquel point of this complete quadrilateral (which in this case is point $ Q$) lie on a circle. Is this a well-known fact? If so, where can I find a proof to this fact?

Hmm. This problem did not involve the introduction of a new point other than the center of the indicated circle.

[b]Solution[/b]

Let the centers of $S_1$ and $S_2$ be $O_1$ and $O_2$. First note the following.

\[\angle{CA_1 Q} + \angle{CA_2 Q} = \angle{QPB_2} + \angle{CA_2 A_1} +\angle{PA_2 Q} = \angle{PQB_2}+\angle{PB_2 Q} +\angle{QPB_2}=180\]
Therefore $QA_1 C A_2$ is cyclic. Let $\gamma$ and $\omega$ denote the perpendicular bisectors of $A_1 Q$ and $B_2 Q$ and let $X=\omega \cap \gamma$. Note that $X$ is the circumcenter of $QA_1 C A_2$.

Since $O_1 \in \gamma$ and $O_2 \in \omega$, we have that $\angle{O_1 X O_2}=180-\angle{A_1 Q A_2}$. However, since $Q$ is the center of the spiral similarity mapping $S_1$ to $S_2$, it follows that $\angle{O_1 Q O_2}=\angle{A_1 Q A_2}$.

This yields that $O_1 X O_2 Q$ is cyclic and since $O_1$, $O_2$ and $Q$ are fixed, that $X$ always lies on one circle.

Actually, a similar problem to this has appeared in two different UK texts, however they were both published after 2002, so it's OK! It also reminds of this years British MO round 1 Q5, which is of the same essence, posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=381274]here[/url]. On these grounds I'm guessing this was proposed by the UK  :D

[hide][img]http://oi53.tinypic.com/wri7hc.jpg[/img]

[color=#FF0000][mod edit: http://oi53.tinypic.com/wri7hc.jpg][/color][/hide]
$O_1,O_2,O_3$ are the centres of $S_1,S_2$ and $\odot(A_1QA_2)$.

As previously shown, we only need to study the locus of $O_3$. $O_1O_3$ and $O_2O_3$ are perpendicular to $A_1Q$ and $A_2Q$ respectively as by definition all of $\triangle A_2QO_2,\triangle A_2QO_3,\triangle A_1QO_1,\triangle A_2QO_3$ are isosceles. Let the midpoints of $QA_1,QA_2$ be $M_1,M_2$ respectively. To show that the quadrilateral $QO_1O_2O_3$ is cyclic it suffices to prove $\angle O_2QM_2=\angle O_1QM_1$ as then it would follow $\angle QO_1M_1=\angle QO_2M_2$ if we recall the right angles $\angle M_1,\angle M_2$.

There is a fairly quick angle chase:
[hide]$\angle O_2QM_2=\alpha\implies\angle QO_2A_2=180^{\circ}-2\alpha$ $\implies QPA_2=90^{\circ}-\alpha\implies\angle A_1PQ=90^{\circ}+\alpha$. Then $A_1Q$ subtends an angle of $90^{\circ}-\alpha$ (at least on this diagram) so $\angle A_1O_1Q=180^{\circ}-2\alpha$ $\implies\angle M_1O_1Q=\frac{1}{2}\angle A_1O_1Q=90^{\circ}-\alpha=\angle QO_2M=180^{\circ}-\angle O_3O_2Q$ so $QO_i$ is cyclic.[/hide]

But a more appealing argument is that no matter where we place $A_1$ on $S_1$, $\triangle A_1A_2Q$ has fixed angles. So if we consider the diagram, but with another $A_n$ diametrically opposite $Q$ on $S_1$ then easily $\angle A_nQA_{n+1}=\angle A_1QA_2\implies\angle O_1QO_2=\angle A_1QA_2$, since $O_1,O_2$ lie on $QA_n,QA_{n+1}$ respectively. So this rotation must leave an equal against the common $\angle A_1QA_2$, i.e. $\angle O_1QM_1=\angle M_2QO_2$. Then the locus of $O_3$ is the circumcentre of $O_1QO_2$.

Do I feel guilty reviving a 7 year old thread? No.

[hide="solution"]
We use directed angles.
$\angle B_1CB_2=\angle A_1CA_2 = \angle CA_1P+\angle PA_2C = \angle B_1QP+\angle PQB_2=\angle B_1QB_2$$\implies Q\in (B_1CB_2)$. Let $O_1, O_2, O$ be the circumcenters of $S_1, S_2, (B_1QB_2C)$. By spiral similarity, $\angle B_1QB_2$ is constant. Since $OO_1\perp B_1Q$ and $OO_2\perp B_2Q$, then $\angle O_1OO_2 = \angle B_1QB_2$ is constant; thus, $O\in (O_1QO_2)$ as desired. $\Box$
[/hide]

My solution:
Let's show that $O_1,Q,O_2,O$ are concyclic,where $O_1,O_2,O$ be the centers of $S_1,S_2,(A_1A_2C),$ respectively.
Let $S=OO_1\cap A_1Q$ and $T=OO_2\cap A_2Q.$
We know $\angle CA_1Q=180-\angle B_1PQ=\angle QPB_2=\angle QA_2B_2=180-\angle QA_2C\rightarrow A_1,Q,A_2,C$ are cyclic.
Then $OS\perp AQ,OT\perp QA_2\to S,Q,T,O$ are cyclic.
$Claim:$ $\angle A_1QA_2=\angle O_1QO_2.$
$Proof:$ We have $\angle QPA_2=180-\angle QPA_1=180-\frac{\angle A_1O_1Q}{2}=90+\angle O_1QA_1.$
Also we know $\angle QPA_2=180-\frac{\angle QO_2A_2}{2}=90+\angle O_2QA_2,$ then we know $\angle O_2QA_2=\angle O_1QA_1\to \angle O_1QO_2=\angle O_1QA_1+\angle A_1QO_2=\angle O_2QA_2+\angle A_1QO_2=\angle A_1QA_1.$ As desired.
Also we know
$$\angle O_1OO_2=\angle SOT=180-\angle SQT=180-\angle A_1QA_2=^{\text{Claim}}180-\angle O_1QO_2\implies O_1,QO_2,O$$ are concyclic.

[b]Lemma: [/b] This problem is trivial.
[b] Proof [/b]: For any $B$, by a trivial angle chase, we have $\angle A_1CA_2 = \angle A_1QA_2$, so we put the $B$ aside the picture, and focus on the locus of the circumcentre of $\Delta A_1QA_2$, as $A_1$ varies. We claim the locus is on $\omega_{O_1QO_2}$. Indeed, inverting around $Q$, the problem becomes trivial, because taking a homothety of factor $.5$ w.r.t $Q$ maps $O_1^{*} O_2^{*}$ to the $Q$-simson line of $\omega_{A_1^{*}P^{*}A_2^{*}}$, hence the desired result.

As a corollary, we have:
[b] Lemma [/b]: 10 year or more old IMOSL problems are trivial.
(Sketch of proof is trivial, just also note that JMO P1 was 1998 G8, then trivial by induction)

Let the centers of $S_1, S_2$ be $O_1, O_2$ resp.\\
Some angle chasing shows that $\angle A_1QA_2=\angle A_1QP+\angle PQA_2=\angle AB_1B_2+\angle CB_2B_1=180-\angle A_1CA_2$\\
$\Longrightarrow A_1CA_2Q$ is cyclic.\\
Thus the problem is equivalent to find the locus point of the circumcenter of $\triangle QA_1A_2$ which doesn't depend on $B_1$ so we can ignore $B_1$ in the problem.\\
We will prove that $O\in (O_1QO_2)$ which is a fixed circle where $O$ is the circumcenter of $\triangle QA_1A_2$.\\ We have \\$(A_1CA_2Q)\cap S_2=A_2Q$, $(A_1CA_2Q)\cap S_1=A_1Q$\\ $\Longrightarrow OO_2\perp A_2Q,\ OO_1\perp A_1Q\Longrightarrow \angle O_1OO_2=180 -\angle A_1QA_2$.\\
But $\angle A_1QO_1=\frac{180-2\angle A_2PQ}{2}=90-\angle A_2PQ=\angle A _2QO_2\Longrightarrow \angle A_1QA_2=\angle O_1QO_2$.\\
$\Longrightarrow \angle O_1OO_2=180 -\angle O_1QO_2\Longrightarrow O\in (O_1QO_2)$ which is a fixed circle as desired.

[hide=Solution]
[quote=ISL 2002 G4]Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$.  Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.[/quote]
[b][color=#000]Solution:[/color][/b] Let $O_1,O_2,O$ be centers of $S_1,S_2, \odot (A_1A_2C)$. Now, $QO_1O_2O$ is cyclic, because $\overline{A_1O_1}$ $\mapsto $ $\overline{A_2O_2}$ (because: $\Delta QO_1A_1$ $\sim$ $\Delta QO_2A_2$) and, $\angle BA_1Q$ $=$ $\angle B_1PQ$ $=$ $\angle CA_2Q$. Since, $Q,O_1,O_2$ are fixed $\implies$ $O$ moves on a fixed circle $\qquad \blacksquare$
[/hide]

Adding an Inversive Solution. :)
[quote=ISL 2002 G4]Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$.  Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.[/quote]

Note that $$\angle AQA_2=\angle AQP+\angle PQA_2=\angle AC_1B_2+\angle B_1B_2C=180^\circ-\angle A_1QA_2\implies A,Q,A_2,C\text{ are concyclic.}$$So the Circumcenter of $\triangle A_1A_2C$ is same as the Circumcenter of $\triangle A_1QA_2$. So now we can safely delete the points $C,B_1,B_2$ from the Diagram. So, the problem can now be restated as
[Quote=Reduced Problem]$\omega_1$ and $\omega_2$ are two circles and $\omega_1\cap\omega_2=\{P,Q\}$. A line through $P$ intersects $\omega_1$ and $\omega_2$ at $A_1,A_2$ respectively. Then Prove that the locus of the Circumcenters of such triangles $A_1QA_2$ is a circle.[/quote]

For this Invert around $Q$ and let this map be denoted as $\Psi$. So, $\Psi:P\leftrightarrow \omega_1'\cap\omega_2'=P'$ where $\omega_1'$ and $\omega_2'$ are the Inverted Image of $\omega_1$ and $\omega_2$ respectively during this transformation $\Psi$.

We have to prove that the circles passing through $P',Q$ intersecting $\omega_1'$ and $\omega_2'$ at points $\{A_1',A_2'\}$ respectively, then the reflections of $Q$ on $A_1'A_2'$ are collinear.

Drop Perpendiculars from $Q$ onto $\omega_1'$ and $\omega_2'$, So by Simson Line we get that the Projections of $Q$ on such lines $A_1'A_2'$ are collinear, so by a Homothety at $Q$ with a scale factor of $2$ we get that the reflections of $Q$ of such lines $A_1A_2$ are also collinear, and the reflections of $Q$ on such lines $A_1A_2$ are the points where the circumcenters of such triangles $QA_1A_2$ map!

Hence Inverting back we get that the Locus of the Circumcenters of such triangles $QA_1A_2$ which are the circumcenters of $\triangle A_1A_2C$ will lie on a circle. $\blacksquare$

What troll!
[quote=orl]Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$.  Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.[/quote]
Let $O_1,O_2$ be center of $S_1,S_2$ respectively.We have $Q$ is the miquel point of  complete quadrilateral $PB_2CA_1A_2B_1$.Hence center of $\odot{A_1A_2C}$ lies on the miquel circle $\odot{QO_1O_2}$ which is fixed.$\square$

[hide=Solution]
Just note that \(Q\) is the Miquel point of \(CA_1PB_2\). It is well known that the centers of \((CA_1A_2),(PB_2A_2),(PB_1A_1)\) form a cyclic quadrilateral with \(Q\); the desired center always lies on this circle.
[/hide]

Easy G4.

Notice that $\angle A_1QA_2=\angle A_1B_1B_2+\angle CB_2B_1=180^{\circ}_\angle C,$ so $A_1CA_2Q$ is cyclic and the problem reduces to showing that the circumcenter of $(A_1A_2Q)$ lie on a fixed circle (the $B$'s were just a distraction!).

We finish with

Claim: Denote the center of $S_1$ by $O_1$ and that of $S_2$ by $O_2$. Then the desired circle is $(O_1QO_2)$.

Proof: Denote the midpoints of $A_1Q$ and $A_2Q$ by $M_1, M_2,$ respectively. Then the circumcenter of $\triangle A_1A_2Q$ is the intersection of the perpendicular at $M_1$ and $M_2$ to $A_1Q$ and $A_2Q$. Notice that $\angle AO_1Q=2\angle A_1PQ=180^{\circ}-\angle B_2PQ=\angle QO_2A_2,$ and this easily implies the result.

We are done.

Obviously $Q$ is the Miquel point of $A_1CB_2P,$ and so circumcenter of $\triangle A_1A_2C$ lie on circle passing through $Q$ and centers of $S_1,S_2,$ which is fixed.

Let $O,O_1,O_2$ be the centers of $S_1,S_2,(A_1A_2C),$ respectively. Notice that $Q$ is the Miquel point of $A_1B_1CA_2B_2P$ so $OO_1QO_2$ is cyclic. $\square$

No one:
EGMO Ch. 11:

[img width=50]https://media.discordapp.net/attachments/925784397469331477/952410078668009472/Screen_Shot_2022-03-12_at_8.37.24_PM.png[/img]

Let $O_1,O_2$ be centers of $S_1,S_2.$ Let $O$ be center of $(A_1CA_2).$ We claim that $O_1OO_2Q$ is cyclic. First, we claim that $CA_1QA_2$ is cyclic. Note that $\angle B_1A_1Q=\angle B_1PQ=\angle CA_2Q.$ This implies the claim.

Thus, $O$ is the center of $(QA_1A_2).$ Note that $OO_1$ is the perpendicular bisector of $A_1Q$ and $OO_2$ the perpendicular bisector of $A_2Q.$ Thus, $\angle OO_2Q=\frac{1}{2}\overset{\huge\frown}{QPA_2}=180^\circ-\angle QPA_2$ and $\angle OO_1Q=\frac{1}{2}\overset{\huge\frown}{QPA_1}=180^\circ-\angle A_1PQ.$ These sum to $180^\circ$ so we are done.

Wut
Observe that $Q$ is miquel point of quad $A_1PB_2C$ which means the circumcenter of $\triangle A_1A_2C$ belongs to $\odot(Q0_1O_2)$ where $O_1$ and $O_2$ are the centres of $S_1$ and $S_2$ which are obviously fixed $\blacksquare$

[b]Claim: We can simply consider the circumcenter of $A_1QA_2$[/b]

Proof:

Observe that $Q$ is the miquel point of quadrilateral $A_1PB_2C$

Therefore, $Q$ lies on $(A_1A_2C)$

As such, the circumcenter of $(A_1A_2C)$= the circumcenter of $(A_1QA_2)$

[b]Rephrased problem:[/b] the circumcenters of $(A_1A_2Q)$ lie on a circle as $A_1A_2$ varies

[b]Claim: $\angle A_1QA_2$, $\angle A_2A_1Q$, $\angle A_1A_2Q$ remain invariant over all $A_1$,$A_2$[/b]

Proof:
Fix one $A_1$,$A_2$,

Now, suppose we take another pair, say $B_1,B_2$

Note that $Q$ is the center of the spiral sending $A_1 \rightarrow A_2$, $B_1 \rightarrow B_2$

Which implies that $\triangle A_1QA_2 \sim \triangle B_1QB_2$

[b]Claim: If $O$ is the circumcenter of $(A_1QA_2)$, $\triangle A_1QO$ remains similar as $A_1$ varies [/b]

Proof:

Note $Q$ is the spiral center sending all $A_1 \rightarrow O$

As such, the possible positions of $O$ are simply a fixed scaling and translation from every possible position of $A_1$ with respect to the spiral center $Q$

This means that all possible positions of $O$ must cover a circle

Viewing $CA_1PB_1$ as a complete quadrilateral, $CA_1QA_2$ is cyclic. Therefore, we can delete $C,B_1,B_2$ from the diagram and focus on the center of $(A_1A_2Q)$ as $A_1$ varies around $S_1$. As $A_1$ varies around $S_1$, $\triangle QA_1A_2$ is directly similar to a fixed triangle for spiral similarity reasons, so $\triangle QA_1O$ is directly similar to a fixed triangle as well, where $O$ is the (variable) circumcenter of $\triangle QA_1A_2$. Therefore, it suffices to prove the following "independent" lemma.

[b][color=#00f]Lemma:[/color][/b] Let $P$ be a point and suppose that $\triangle PAB$ is directly similar to a fixed triangle as $A$ varies around a circle. Then $B$ varies around a circle too.
[color=#00f]Proof:[/color] We use complex numbers. WLOG let $p=0$. Then there exists some fixed complex number $z \neq 0$ such that $b=za$ always holds, and the rest is clear. $\blacksquare$

wow how did I not see the Miquel point

We first do some angle chasing:
\begin{align*}
\angle A_1QA_2 &= \angle A_1QP + \angle PQB_2 + \angle B_2QA_2 \\
&= \angle A_1B_1P + \angle PA_2B_2 + \angle B_2PA_2 \\
&= \angle A_1B_1P + \angle A_1PB_1 + \angle A_1A_2C \\
&= \angle CA_1A_2 + \angle A_1A_2C \\
\angle A_1QA_2 &= 180^\circ - \angle A_1CA_2
\end{align*}
From here, we see that $CA_1QA_2$ is a cyclic quadrilateral. Obviously, the circumcenter of $CA_1A_2$ is the same as the circumcenter of $A_1QA_2$, a point we will label as $O$.
The centers of $S_1$ and $S_2$ will be called $O_1$ and $O_2$, respectively. Now, we claim that $OO_1QO_2$ is a cyclic quadrilateral.
In order to prove this, construct the perpendicular bisectors to $A_1Q$ and $A_2Q$. Denote the midpoints of $A_1Q$ and $A_2Q$ as $M$ and $N$, respectively.
Visibly, $OMQN$ is a cyclic quadrilateral, which means that $\angle MQN + \angle MON$.
It is easy to see that the perpendicular bisector of $A_1Q$ passes through both $O_1$ and $O$, while the perpendicular bisector of $A_2Q$ passes through both $O_2$ and $O$. As a result, $\angle MON = \angle O_1OO_2$.
We finish with a short angle chase:
\begin{align*}
180^\circ &= \angle MQN + \angle MON \\
&= \angle A_1QA_2 + \angle O_1OO_2 \\
180^\circ &= \angle O_1QO_2 + O_1OO_2
\end{align*}
Thus, $OO_1QO_2$ is a cyclic quadrilateral, and $O$ always lies on $(O_1QO_2)$, a fixed circle. $\blacksquare$


[quote="nayel"]Determine the ordered systems $(x,y,z)$ of positive rational numbers for which $x+\frac{1}{y},y+\frac{1}{z}$ and $z+\frac{1}{x}$ are integers.[/quote]

If we write $x=\frac{p}{q}$, $(p,q)$ coprime, $y=\frac{r}{s}$, $(r,s)$ coprime, $z=\frac{t}{u}$, $(t,u)$ coprime, we have :

$\frac{p}{q}+\frac{s}{r}=\frac{pr+qs}{qr}$ integer $\implies$ $q|pr$ and so $q|r$ and also $r|qs$ and so $r|q$ and so $q=r$ and $q|(p+s)$
So we have $q=r$, $s=t$ and $u=p$ and :
$q|(p+s)$, and so $q|(p+q+s)$
$s|(q+p)$, and so $s|(p+q+s)$
$p|(s+q)$, and so $p|(p+q+s)$

And so $pqs|(p+q+s)$ and very few solutions (since normally $pqs>p+q+s$) : $(1,1,1)$, $(1,1,2)$ and $(1,2,3)$

And so the solutions for $(x,y,z)$ :
$(1,1,1)$, $(1,\frac{1}{2},2)$, $(\frac{1}{2},2,1)$, $(2,1,\frac{1}{2})$, $(\frac{1}{2},\frac{2}{3},3)$, $(3,\frac{1}{2},\frac{2}{3})$, $(\frac{2}{3},3,\frac{1}{2})$, $(\frac{1}{3},\frac{3}{2},2)$, $(2,\frac{1}{3},\frac{3}{2})$, $(\frac{3}{2},2,\frac{1}{3})$

nice solution. also the same as mine. can this be solved in any other way?


Find the number of square units in the area of the region. The dots are equally spaced vertically and horizontally.
[asy]size(50);
for(int i = 0; i < 5; ++i){
 for(int j = 0; j < 5; ++j){
   dot((i,j));
  }
}
draw((0,1)--(0,2)--(3,4)--(1,1)--(0,1),linewidth(0.7));[/asy]

Can you explain...? The solution makes no sense

$ S \equal{} 2 \plus{} \frac {4}{2} \minus{} 1 \equal{} \boxed{3}$.

[b]Remark[/b]
[url=http://mathworld.wolfram.com/PicksTheorem.html]Pick's Theorem[/url] - MathWorld

In easier words, Find the number of dots inside the figure (2), then add that to half of the number of dots that enclose the figure (4x1/2) then subtract that by 1.

Alternatively, realize that there are two triangles of length 3 base, and 1 height.

$3\times1\div2\times2=3$


geometry
circumcircle
inradius
inequalities solved
inequalities

Let S,R,r be respectively the area,the circumradius and the inradius of
triangle ABC.Let BC=a,CA=b,AB=c .Prove that
2S/R<=(b+c-a)sin(A/2)+(a-b+c)sin(B/2)+(a+b-c)sin(C/2)<=S/r
When does one equality occur?

I think this one is easy. :D  :D
 The right one you use Tchebychev's inequality.
 The left one by using b+c-a =8Rcos(A/2)sin(B/2)sin(C/2) , you can solve it easily. :)

yeah i had got the left one but not the right one.
 thanx treegoner


These questions came on our test, and i want to verify if the solution given is right :mad:

1) $ \ce {CH_3COCH(CH_3)_2}+\ce {HCHO} + \ce {R_2NH} \rightarrow ?$

2) $ \ce {(CHO)CH_2CH_2(CHO)} + \ce {CH_3NH_2} + \ce {CH_3COCH_3} \rightarrow ?$

Both reactions in presence of HCl

Number 2) is the classical Robinson's 1917 synthesis of tropinone. For the first, what is [b]your[/b] answer?

Isn't the crux of problem 1 the acidity of the alpha-hydrogens?

I think the CH3CO- hydrogen is more acidic compared to the one on the other side.

The crux of the problem is the formation of the Eschenmoser salt/iminium ion as a result of reacting the formaldehyde and dialkylamine.


Three circular arcs of radius 5 units bound the region shown. Arcs
$ AB$ and $ AD$ are quarter-circles, and arc $ BCD$ is a semicircle.
What is the area, in square units, of the region?
[asy]/* AMC8 2000 #19 Problem */
draw((0,0)..(1,1)..(2,0));
draw((0,0)..(.7,-.3)..(1,-1));
draw((1,-1)..(1.3, -0.3)..(2,0));
label("$A$", (1,-1), SW);
label("$B$", (0,0), W);
label("$C$", (1,1),N);
label("$D$", (2,0),E);[/asy]

[hide]if cut the region bounded by BD and the two circular arcs BA and AD, and you move the peices up to the top of the semicircle, you will get a rectangle of dimensions 5 by 10, for an area of [b]50[/b]

for a more rigorous solution, we have pi*5^2/2+2(5^2-pi*5^2/4)=pi*25/2+50-25*pi/2=50.[/hide]


inequalities
LaTeX
special factorizations

if a^2- 4a+b^2- b/2+65/16=0 then what is a^2- 4 root b

Completing the square, this reduces to $ a^2\minus{}4a\plus{}b^2\minus{}\frac b2\plus{}\frac{65}{16}\equal{}(a\minus{}2)^2\plus{}\left(b\minus{}\frac14\right)^2\equal{}0$. Assuming $ a,b\in\mathbb R$, $ (a\minus{}2)^2,\left(b\minus{}\frac14\right)^2\ge0$ by the Trivial Inequality, so the only solution is $ (a,b)\equal{}\left(2,\frac14\right)\implies a^2\minus{}4\sqrt b\equal{}4\minus{}4\sqrt{\frac14}\equal{}4\minus{}2\equal{}\boxed2$.

if you don't mind, next time, please LaTeX your question so it's easier to read
find out how to on the LaTeX thingy on the sidebar


I'm at university and beginning to type up my final year maths project using Latex, which is new to me, and I'm not sure which document style to use. Report, article, amsart and amsbook seem to be the most likely candidates, but I'm not sure. The obvious approach would be to try them all and see the differences, but I just thought I'd ask for advice from someone with a bit more experience of Latex. If it makes any difference, the project will be about 40-50 pages long with around 5 chapters/sections and various subsections for each of these.

Additionally, are there any specific packages that I could / should use to make things look nicer?

I think that you may be approaching things the wrong way round. Many universities insist that write-ups are presented in a certain way; some are very specific and others are more relaxed. Some provide their own packages to use. So I would find out what the university requires or thinks is acceptable. You could also look at other previous projects to see how they are presented. Once you know what you need then you can decide which document class will do the job.
[quote="allego"]Additionally, are there any specific packages that I could / should use to make things look nicer?[/quote]LaTeX output always looks 'nice'; you may want to use packages to do a certain job (eg geometry for margins, fancyhdr for headers and footers) but which depends on the requirements.

The big advantage of LaTeX is that you can get on with typing your project immediately and can change the class at any time.


So does anyone know him? Just curious...

for anyone who [i]didn't[/i] know this, he's a ninth grader who made USAMO...u'll probably see him at ARML Karthik

Okay, I knew his name sounded familiar, so I did a google search and found that he, as well as Vova, tied me in the Georgia Math League in 8th grade; however, Vova and I are both in 10th grade and were both apparently in 8th grade at the same time as Sagar, so how did he end up in 9th grade this year? :huh:
Idea #1: He was allowed to take the 8th grade GAML as a 7th grader (which seems kind of weird).
Idea #2: He somehow managed to misbubble his grade on his AMC and AIME and is really in 10th grade.
Idea #3: He got held back (a USAMO qualifier getting held back???).
Idea #4: Some sort of time travelling maybe...
Link: http://www.mathleague.com/reports/2004_05/grade678/GA_8.HTM

The one I want to know about is that 4th grader up in Canada!

i heard about that too, but don't you have to be a resident of the US to take the AMC and AIME?

[quote="gtf"]My guess (and it's only a guess. I'm too tired to actually look it up.) is that Canada uses the AMC and AIME the same way we do. They may not use USAMO and I'm almost positive that they can't get invited to MOSP. They probably have something similar.[/quote]
U.S. students and legal residents of the U.S. or Canada may qualify for the USAMO (there are forty-odd Canadian qualifiers this year).  However, Canadians may not go to MOP :  MOP is for members and potential future members of the USA team.  Canadian team selection is apparently based on the Canadian Math Olympiad, the USAMO, and the APMO.  They apparently have some seminars for people likely to be on the team, and they qualify for the Canadian Math Olympiad by their own qualification exam.


integration
trigonometry
calculus
logarithms
real analysis
real analysis unsolved

Someone posted an interesting problem. But unfortunately I cannot find his message. The problem is as follows.

Prove that $\int\limits_0^\frac\pi4\log(4\cos^2\theta-1)d\theta=\dfrac23\int\limits_0^\frac\pi4\log(4\cos^2\theta)d\theta$.

When I saw that problem (2-4 days ago), it hasn't been solved. Maybe it is solved already. It is very interesting to see a solution.

You've lost a factor of $\frac23$ on the RHS.  This integral has something to do with Catalan's constant, but I don't know anything more than that.

I have the following "semi-solution" which I devided into 2 parts.

1) $\int\limits_0^{\frac\pi4}\log(4\cos^2\theta-1)d\theta =$

$\dfrac12\int\limits_0^{\frac\pi2}\log(2\cos\theta+1)d\theta = \dfrac12\int\limits_0^{\frac\pi2}\log\left(2\left(\cos\theta+\cos\dfrac\pi3\right)\right)d\theta = \int\limits_0^{\frac\pi4}\log\left(4\cos\left(\theta-\dfrac\pi6\right)\cos\left(\theta+\dfrac\pi6\right)\right)d\theta = \int\limits_{-\frac\pi6}^{\frac\pi{12}}\log(2\cos\theta)d\theta+\int\limits_{\frac\pi6}^{\frac{5\pi}{12}}\log(2\cos\theta)d\theta = \int\limits_{-\frac\pi{12}}^\frac{5\pi}{12}\log(2\cos\theta)d\theta = \dfrac12\int\limits_{-\frac\pi{12}}^\frac{5\pi}{12}\log(4\cos^2\theta)d\theta$.

Thus we have to prove that $I: =\int\limits_{-\frac\pi{12}}^\frac{5\pi}{12}\log(4\cos^2\theta)d\theta=\dfrac43\int\limits_0^\frac\pi4\log(4\cos^2\theta)d\theta$.

2) $I=2\int\limits_0^\frac\pi{12}\log(4\cos^2\theta)d\theta+\int\limits_\frac\pi{12}^\frac{5\pi}{12}\log(4\cos^2\theta)d\theta$.

Now $\int\limits_\frac\pi{12}^\frac{5\pi}{12}\log(4\cos^2\theta)d\theta = \int\limits_{-\frac{5\pi}{12}}^{-\frac\pi{12}}\log(4\sin^2\theta)d\theta = \int\limits_\frac\pi{12}^\frac{5\pi}{12}\log(4\sin^2\theta)d\theta$.

Hence $\int\limits_\frac\pi{12}^\frac{5\pi}{12}\log(4\cos^2\theta)d\theta = \dfrac12\left(\int\limits_\frac\pi{12}^\frac{5\pi}{12}\log(4\cos^2\theta)d\theta+\int\limits_\frac\pi{12}^\frac{5\pi}{12}\log(4\sin^2\theta)d\theta\right) = \dfrac12\int\limits_\frac\pi{12}^\frac{5\pi}{12}\log(4\sin^22\theta)d\theta=\dfrac14\int\limits_\frac\pi6^\frac{5\pi}6\log(4\sin^2\theta)d\theta = \dfrac14\int\limits_{-\frac\pi3}^\frac\pi3\log(4\cos^2\theta)d\theta = \dfrac12\int\limits_0^{\frac\pi3}\log(4\cos^2\theta)d\theta$.

So the problem is reduced to proving that $2\int\limits_0^\frac\pi{12}\log(4\cos^2\theta)d\theta+\dfrac12\int\limits_0^\frac\pi3\log(4\cos^2\theta)d\theta=\dfrac43\int\limits_0^\frac\pi4\log(4\cos^2\theta)d\theta$.

[b]P.S.[/b] I devided this "semi-solution" into the 2 parts because I consider the part 1 as the key idea and the part 2 as one of the possible contunuations of 1.


real analysis
calculus
integration
function
real analysis unsolved

Let $ g$ be a monotone increasing absolutely continuous function on $ [a,b]$ with $ g(a) \equal{} c, g(b) \equal{} d$. Let $ f$ be an integrable function on $ [c,d]$. Let $ F(y) \equal{} \int_{c}^{y} f$ and set $ H(x) \equal{} F(g(x))$.

Show that $ \int_{c}^{d} F(y) \; dy \equal{} \int_{a}^{b} F(g(x))g'(x) \; dx$.

[In previous parts to the question, I have shown that $ H'(x) \equal{} F'(g(x))g'(x)$ a.e. except on a set $ E$ where $ g'(x) \equal{} 0$ and also $ H'(x) \equal{} f_{0}(g(x))g'(x)$ a.e. where $ f_{0}$ is defined by $ f_{0}(y) \equal{} f(y)$ if $ y \notin g[E]$ and 0 otherwise. Can these be used here?]

Is the expression still true if $ g$ is not assumed monotone? What if $ g$ is bounded and $ f$ is integrable on an interval containing the range of $ E$?

The hypothesis you can't drop is absolute continuity of $ g$. Without that, $ g'$ is too badly behaved.

I came across the following example:
Let $ f(x)\equal{}1/\sqrt{x}$ for $ x>0$ and $ f(0)\equal{}0$. Let $ g(t)\equal{}t^{2} \sin(1/t)$ for $ t \neq 0$ and $ g(0)\equal{}0$. Then $ f$ is integrable over $ [g(0),g(\pi/2)]$ and $ g$ is absolutely continuous on $ [0,\pi/2]$ but the expression is not valid in this case.

Any comments on this?

Okay, we need the hypothesis that $ g([a,b])\subset [c,d]$, or that $ f$ is nice on the range of $ g$.

For the case where $ g$ is bounded on $ [a,b]$ and $ g([a,b]) \subset [c,d]$, how would the proof be modified from the initial case where $ g$ was assumed to be monotone?

I would like to revisit the original post in this thread. How do I actually show the equality of the two integrals?


Prove that for each $n\geq 3$ we can cut equilateral triangle into $n$ pieces in such way, that each piece is equilateral or isosceles triangle.

Do you mean cutting with straitedge and compass or just to show that such cuts do exist ?

...

[hide]Thee type of cuts: Let ABC be the traingle, O is center, DEF midpoints, and  G,H on BC such that BG=GH=HC = a/3
X..  make a small triangle joing the mid-points (DEF)  of the sides - 4 parts, each a (smaller) equilateral triangel.
Y. Join each angle to O ,  center of triangle   - 3 equal parts
Z .  Join OA, OB, OC, OG, OH  - 5 triangles, one (OGH) equilateral.

You can always  divide one equilateral trinagle into 4 by method A.
So   X gives N = 1,4,7,....  all  of the form  (3k+1)
 X with  one equilateral triangle using Y will give  6,.. we still have one equilateral triangle so can get 9,12,...
X with  two  equilateral tringle   using Y will give  8 .. we will stiall have one equilateral traingel so can get 8,11,..

So you have covered everything except  n=2...[/hide]

Your solutions seems to be right. Just one thing, when you're describing Z, OGH is not equilateral (it also is not a triangle that you cut (maybe typo)), but it doesn't change anything  :lol:

[quote]OGH is not equilateral[/quote]
What am I missing OG = GH = HO =  1/3 of the length  of triangle. right.
You can  get 8 by   "Y" cutting two triangles in X, then X cutting another triangle in X will give 11,14..... I should have said that.

Oh, yes, sorry, I've read your definition of O,G,H wrong  :blush:


Below is the problem of SMO (Open section, Final round) on July 2nd, 2005

Time allowed : 4.5 hours

1. An integer is square-free if it is not divisible by $a^2$ for any integer $a>1$. Let $S$ be the set of positive square-free integers. Determine, with justification, the value of\[\sum_{x\epsilon S}\left[\sqrt{\frac{10^{10}}{k}}\right]\]where $[x]$ denote the greatest integer less than or equal to $x$.

2. Let $G$ be the centroid of triangle $ABC$. Through $G$ draw a line parallel to $BC$ and intersecting the sides $AB$ and $AC$ at $P$ and $Q$ respectively. Let $BQ$ intersect $GC$ at $E$ and $CP$ intersect $GB$ at $F$. If $D$ is the midpoint of $BC$, prove that triangles $ABC$ and $DEF$ are similar.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=43304]Problem 3[/url]
Let $a,b,c$ be real numbers satisfying $a<b<c,a+b+c=6,ab+bc+ac=9$. Prove that $0<a<1<b<3<c<4$

4. Place $2005$ points on the circumference of the circle. Two points $P,Q$ are said to form a pair of neighbours if he chord $PQ$ subtends an angle of at most 10 degrees at the centre. Find the smallest number of pairs of neighbours.


Hello !!

prove that

$ \lim_{x\rightarrow \plus{}\infty}(e^{\sqrt{x\plus{}1}} \minus{} e^{\sqrt{x}})^ {\frac{1}{\sqrt{x}}}\equal{}e$

This should be moved in appropriate forum.

Solution removed because [b]kunny[/b] wants this topic locked.

I will not re-post until he tells why he wants this topic locked -- is it from a journal/contest?

This should be locked.

Japanese Communities Moderator

kunny


Let  ABC be a triangle. For any line l not parallel to any side of the triangle, let Gl be the centroid of the degenerate triangle whose vertices are the intersection of l with BC,CA,AB (i.e. the vector average of these three points). Determine the union of the Gl over all l.

This has been posted by both Moubinool and Pierre. Try looking for "Concours General". I think I can remember solving it the second time it was posted (by Pierre), because it had some additional details which made the whole thing easier.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Concours+General&t=14871]Here[/url]'s the one posted by Pierre.

There are lots of things missing from that um.. "solution" of mine :). I seem to have skipped over most of it, but the main idea is there, I think.


[quote="darij grinberg"]Just wondering - the problem you called [b]France 4[/b] in your file [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=4831]was posted on MathLinks by grobber[/url], where he claimed that it was proposed by [b]Romania[/b]. And actually I find it more plausible to be a Romanian problem, since it doesn't really fit with the "global remark" below France 5. And, by the way, your file doesn't contain any problem marked Romanian, so maybe there are some Romanian problems listed with incorrect country names.

Ah yes, and please correct $cos^2 \left(\frac{\alpha+\beta}{2}\right)$ into $\cos^2 \left(\frac{\alpha+\gamma}{2}\right)$ in problem [b]Socialists Republic of Czechoslovakia 5[/b]. Also please check Poland 6 (it seems to contain at least two typos).

  Darij[/quote]

I will have a look into it. So the task for [b]grobber[/b] is to take care of the issue. He is from Romania. And he should have the best knowledge of the early IMO years. So we are
waiting for your contributions, [b]grobber[/b] !!!  ;)


integration
calculus
calculus computations

So is the solution completely correct? Anyone can help me to check? Thanks