find the product on alkylation of etheylene by iso-butane  using $H_{2}SO_{4}$

yes got thsi it will be $2,4-dimethyl-butane$

but one thing
morrison&boyd says the mechanism of alkylation is

1.first the formation of carbocation by attack of acid on [b]alkene[/b]
2.attack of one carbocation ion on other alkene then capturing of hydride ion from alkane

(inbetween rearrangement allowed)

while other books say
1.first formation of carbocation by removing hydride ion from the [b]alkane[/b]
2.attack of this electrophile on the alkene and finally gain of hydride ion by the carbonium ion so formed
(inbetween rearrangement allowed)

but i don't think in general bot the mechanisms give the same end product so which is thought to be the right one?

First a remark: the name you gave, 2,4-dimethylbutane, doesn't make sense (it should be 2-methylpentane).

Anyway, the reaction you describe doesn't seem one that occur at ordinary laboratory conditions. Also, I don't see any pedagogical value on it, because no one would ever think in the laboratory in preparing an alkane by using that reaction. That type of reaction can, however, be important in oil industry. Finally, carbocation formation by hydride abstraction is certainly not a common reaction (unless perhaps at special conditions, for example high temperatures and pressures).

I cannot also see how 2-methylpentane (if that is the compound you were really referring to) could be formed by any of those mechanisms.

sorry that was a serious typo it is actually 2,3 dimethyl butane

but coming to the reaction asssuming the book says everything right the reaction conditions are
1.HF/conc. $ H_{2}SO_{4}$
2.0-10 degree celcius
so thats particularly low :wink:

The temperature is kept low in order to avoid elimination of the intermediate carbocations. Anyway this is a reaction only used in industry (according to Morrison & Boyd in the production of isooctane for gasoline), in spite of the fact that (contrary to what I said above) carbocations can be formed by hydride migration/abstraction in some reactions.

Now can you propose a mechanism for the reaction

$ ethylene+iso-butane (+H_{2}SO_{4}) \longrightarrow 2,3-dimethylbutane$?

well i can but only by using the second mechanism


geometry
power of a point
radical axis
geometry proposed

Given triangle $ABC$. Let $D$ and $E$ be two any points on side $AB$ and $AC$ respectively. Show that the radical axis of the circles with diameter $BE$ and $CD$ passes through a fixed point.

It's orthocener $H$ of  $\triangle ABC$. Let $B',C'$ be feet of altidus from $B,C$. It's easy to see that $\triangle CB'H$ and $\triangle BC'H$ are similar and $HB' \cdot HB= HC \cdot HC'$ hence $H$ lies on radical axis of circles with diametars $BE,CD$.

Slight changes with picture are made if triangle $ABC$ isn't acute, but proff holds in that case also.

Lemma:

Suppose $(C_{1}),(C_{2})$, are two circles, each point which is on the radical axis, have equal power

Toe both of the circles,

And each point which has equal power to the both circles, lies on th e radical axis.

Suppose circles with diameters $CD,BE$, intersect $AB,AC$, at$F,G$ respetively.

It is clear that: $\angle BGE=\angle CFD=90$

And it means that $F,E$ are fixed, and they are the feet of the altitudes from vertices $C,B$, respectively.

Consider the intersection of $BG,CF$    , $H$.

With the similarity of the triangles $BFH,CGH$, we have: $FH.HC=HG.HB$

As the lemma said, it means that $H$ has equal power to the two circles.

And it lies on the radical axis of them.

Then it is finished,

Because $H$ is the orthocenter, and is fixed.


Let a,b,c,d be positive reals s.t. $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$.Prove that $ \sqrt{1\minus{}a}$+$ \sqrt{1\minus{}b}$+$ \sqrt{1\minus{}c}$+$ \sqrt{1\minus{}d}$  is greater or equal than $ \sqrt{a}$+$ \sqrt{b}$+$ \sqrt{c}$+    $ \sqrt{d}$.

The problem is the following:
Let $ a,b,c,d\geq0$ s.t. $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$.Prove that:
$ \sqrt{1\minus{}a}\plus{}\sqrt{1\minus{}b}\plus{}\sqrt{1\minus{}c}\plus{}\sqrt{1\minus{}d}\geq \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\plus{}\sqrt{d}$.
This is from Mathematical Reflections. :D I have $ 2$ approaches:
$ (1)$Obviously $ a\plus{}b\plus{}c\plus{}d\leq2$. Squaring both sides we only need to prove that:
$ \sqrt{(1\minus{}a)(1\minus{}b)}\plus{}\sqrt{(1\minus{}c)(1\minus{}d)}\geq\sqrt{ab}\plus{}\sqrt{cd}$ or we need to prove that:
$ (1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)\geq abcd$,which is true but very sharp.

$ (2)$Using the fact that $ \sqrt{a}\leq\dfrac{1}{2\sqrt{1\minus{}a}}$,we need to prove that:
$ \dfrac{1\minus{}2a}{\sqrt{1\minus{}a}}\plus{}\dfrac{1\minus{}2b}{\sqrt{1\minus{}b}}\plus{}\dfrac{1\minus{}2c}{\sqrt{1\minus{}c}}\plus{}\dfrac{1\minus{}2d}{\sqrt{1\minus{}d}}\geq0$.
Anyway I can't prove the last one.Can someone HELP? :?:


inequalities
geometry
parallelogram
number theory unsolved
number theory

Let a,b,c,d are 4 positive integers such that $gcd(a,b)=gcd(c,d)=1$ and
$0<\frac{c}{d}<\frac{a}{b}<1$
Prove that : $ad-bc=1$ iff $\forall e,f\in N: \frac{c}{d}<\frac{e}{f}<\frac{a}{b}$ then $f> max(b,d)$

woohoo faery fractions!

Sorry useless post lol.

[hide]I think we just need to consider this geometrically.  Using Pick's theorem we can conclude that any fundamental parallelogram has area 1.  Thus we are essentially done just by counting lattice points and corresponding fractions to lattice points.[/hide]


trigonometry
inequalities
inequalities unsolved

but $A+B+C=\pi$ and $A\ge 0, B\ge 0, C\ge 0$ then we can't have $A=B=C$ very small!


Hi,

I am a person with absolutely NO idea of programming and computer stuff.  Of course I use internet, and I know how to use computer.  However, I pretty much have No clue about other stuff.

I would eventually like to be good at any of these computer languages.  Where should I start?  Is there any book that you would recommend??  Plus, I am planning to study for the AP computer science alone, do you have any suggestions?

THANX A LOT!!!

Well, the APCS exam is in JAVA. But this is basically proof that you should learn C++ instead, because College Board has never been known to come up with a good idea in the history of their existence, and making students learn JAVA instead of C++ is an exemplary example of this.

Basically, I think that C++ is a slightly more difficult language to start with, but the little bit of extra time you have to put in starting off will pay off immensely in the future.

Visual Basic is actually a pretty simple language to start with - now I'm talking about Visual Basic 6.0, not the latest VB.net. Unfortunately, VB 6.0 is not free and I'm not even sure if it is sold anymore now that everyone uses VB.net.

Anyway, just like JSteinhardt says, starting with C or C++ is probably the best idea. I'd actually concentrate more on C than on C++ (C++ is pretty much C with some extra features but these extra features may be a little confusing if you don't know C in the first place). Learning Java will be easy once you know C/C++ because Java has a very similar syntax. The same goes for C# - it's much like C++ and Java. And once you know C#, learning VB.net is easy because.... erm, they're both .net languages, which probably doesn't mean too much to you but trust me, they are similar :).

On the other hand, I know people who started with Java and never needed to learn C++...

But C++ is definitely the most universal language in the current computer world. Right now.

In my opinion, C++ is better language to start than C. That is because if you start learning C and then one day you find that it would be nice to learn C++, your C++ code would very probable look like C code. If you want to write web-applications, I suggest learning HTML, PHP, and SQL.

everyone, thank you so much!!!

but, ummm, where if I decide to learn C or C++ what is the best way I could start?  FYI, I don't live in a very urban area but not the most rural area either. Again, thanks!

The first step would be getting a C++ compiler and some source code editor - if you use Windows, dev-cpp is a good choice ([url]http://www.bloodshed.net/dev/devcpp.html[/url]). Later on, you may want to get some more advanced tools such as Microsoft Visual C++ or Borland C++ Builder but I wouldn't really recommend them to beginners.

As to the learning part - I learned C++ from [url=http://www.cplusplus.com/doc/tutorial/]this tutorial[/url] but I must admit that I already knew another programming language at that time so I can't tell how hard it will be for you - give it a try and if it won't work for you, consider a) finding a better web site b) getting some good C++ book c) finding somebody who knows C++ and is willing to help you learn it.

And of course, you can always post your questions here.

I hope that the rest of the people who read these forums will post some other helpful links and ideas.

Regarding an intro C++ book, I am teaching my daughter using the book [i]Accelerated C++: Practical Programming by Example[/i].  Here is a link:
[url]http://www.amazon.com/Accelerated-C%2B%2B-Practical-Programming-Example/dp/020170353X/sr=1-1/qid=1171838524/ref=pd_bbs_sr_1/103-1748889-8839828?ie=UTF8&s=books[/url].

I recommend the book.

Personally, I find dev to be pretty satisfactory on Windows. I'm not sure you will get much better, as auto-complete stuff can be more annoying than helpful, imo.

However, on Linux, obviously vim wins (uh-oh, now I'm going to start a fight over which text editor to use).

Kate is also pretty good if you prefer a more conventional editor, but after you've learned vim, you basically can type really fast. I suppose if you are just learning C++ though, stuff like dev might be good because it is really simple.

[quote="climskywell"]yes this might seem OUTof PLACE and nonsense,

but, what is C andC++ ?
Where do you "actually" use them? and how much time do you expect to spend to become fluent?[/quote]

They're programming languages. You can use them in a compiler (I use dev-cpp; link given by bus above). It takes a while to get used to the syntax, but after that you just need to code as much as possible to become fluent. Read up on algorithms and data structures after that.

Are you asking how to write computer programs in C and C++ or are you asking what is C / C++ good for?

actually both,

but it will be better if I know what they're for before I attempt to learn them.
I tried some of the tutuorial websites mentioned but they gave be no practive and I couldn't really understand the whole process

Well the main idea is that they are what you might think of as the interface between human-speak and computer-speak. Computers understand 0s and 1s, humans understand, well...mostly intuition and logic. C and C++, as well as almost any other programming language you'll find, lets you write logical instructions that can be systematically converted into 0s and 1s by a compiler such as the ones we've been talking about. So, in C/C++, the major things you can do are define varaibles, read and write input and output, and do things based on conditions. For example, ignoring syntax, a program to print the first 100 integers might look like

[code]
counter = 1
while counter <= 100:
   print counter
   counter = counter + 1
end
[/code]

Actually, the C/C++ syntax would look like:

[code]
int counter = 1;
while(counter <= 100) {
  printf("%d\n",counter);
  counter = counter+1;
}
[/code]

Here the only real major difference is we put semicolons at the end of commands, conditions are in parenthesis, and statements to be executed if a condition is true are put inside brackets. The other thing is this weird-looking printf command. Basically, the "%d\n" tells the computer

(1) I am going to give you something to be interpreted as printable characters, rather than a command (the ")
(2) First, print an integer (the %d)
(3) Now, print a newline (the \n)
(4) Go back to interpreting what I say as commands (the other ")
(5) Now I will tell you what the integer was that you should print (the ,)
(6) The integer is called counter (counter)

gosh, you people are the BEST!
and yeah, your explanation seems to make much better sense.  Do you have any specific suggestion for an EXTREME beginner?  I asked this question earlier, but what they suggested didn't really make sense.

Where should I practice???


Which of the following number is largest? \\ \\
A. $ 9.12344$ \\ \\
B. $ 9.123\overline{4}$ \\ \\
C. $ 9.12\overline{34}$ \\ \\
D. $ 9.1\overline{234}$ \\ \\
E. $ 9.\overline{1234}$ \\ \\
Please select the most appropriate letter.

its obvious to see that the 5th digit right from teh decimal point is the highest in A&B: 4, but the very next digit A will be 0 and B will be 4, so B>A, and our answer is B


[quote="Rep123max"]3. Calculate 11![/quote]

My sister, who isn't so mathematically inclined, saw this and exclaimed, "Calculate 11?  What do they want you to do?"

also, i'm guessing we have to include solutions as well as just answers, right?

...
what happpened to my answers....
did u delete it?
if u did, where should i post it?


j.b davis owns a nut shop. he sells almonds for 6 dollars per pound and walnuts for 5.20 dollars per pound. he receives a special request from a custmor who wants to purchase 30 pound of a mixture of almonds and walnuts but does not want to pay more than 165 dollars . jb used algebra to determine the amount of each to mix  so he could  mix the maximum amount of almonds without having the mixture 's value exceed 165 dollars . determine how many pounds of walnuts jb mixed ?

a) 18.75 pounds b) 11.25 pounds c) 16.65 pounds d) 10.75 pounds

e) none of these

$ \frac{165}{30}\equal{}\$5.50$ per pound. The ratio of walnuts to almonds is $ \frac35$.
$ \frac38\cdot30\equal{}11.25$. The answer is $ B$.


geometry
inequalities
inequalities proposed

Dear Gemath, I know you are excellent at Geometry Inequalities, and you may like the following

Consider every triangle with the area $1$, what kind of triangle makes the following expression minimal
\[\frac{1}{a^{4}}+\frac{1}{b^{4}}+\frac{1}{c^{4}}. \]
(a). Solve it for acute triangle.
(b). SOlve it for arbitrary triangle.

[quote="gemath"]Thanks for your problem friend  :D
For the case triangle is acute we have the following inequality base on Iran inequality
\[\frac{1}{a^{4}}+\frac{1}{b^{4}}+\frac{1}{c^{4}}\ge\frac{9}{16S^{2}}\]
Thus it gets minimal iff triangle is equilateral.

For the case triangle is non acute I think that express doesn't have minimum it has infimum only. ideed we fix point $B$ and give $C\to\infty, d(A,BC)\to 0$ we allways choose a triangle with area equal to $1$ and the side is long, thus
$\frac{1}{a^{4}}+\frac{1}{b^{4}}+\frac{1}{c^{4}}\to 0$ and there won't be equality
thus that express has infimum $0$. :)[/quote]

Yes, that is what I mean. Thank you very much.  :lol:


If x,y, and z are positive integers such that $x^{2}+y^{2}+z^{2}=174$, what is the greatest possible value of x+y+z?

[hide] Well, I just did a little guessing and checking and found that 5, 7, and 10 were the best I could do so our final answer is 22? [/hide]


Compute and express your answer as a common fraction:
$ \frac{5}{5 \plus{} \frac{5}{5\plus{}\frac15}}$

$ \frac {5}{5 \plus{} \frac {5}{5 \plus{} \frac {1}5}}$
$ \equal{} \frac {5}{5 \plus{} \frac {5}{\frac {26}{5}}}$
$ \equal{} \frac {5}{5 \plus{} \frac {25}{26}}$
$ \equal{} \frac {5}{\frac {31}{5}}$
$ \equal{} \frac {25}{31}$

So, the answer is $ \fbox{25/31}$.

You could also figure from the configuration of the $ 5$ s and derive $ \frac {5^2}{5^2 \plus{} 5^1 \plus{} 5^0} \equal{} \fbox{25/31}$.

EDIT: Sorry. See below solution.

Isn't $ 5\plus{}\frac{25}{26}\equal{}\frac{155}{26}$, not $ \frac{31}{5}$?  So instead, are we supposed to get $ \frac{26}{31}$?

[quote=jiban2013rta]answer is -8/3[/quote]

You are not serious, right?

[quote=jiban2013rta]answer is -8/3[/quote]

using the sanity check method we see that since none of the terms are negative and there is no subtraction then there is no way for it to become negative

[quote=jiban2013rta]answer is -8/3[/quote]

The answer is $\boxed{\frac{26}{31}}$.

[quote=jiban2013rta]answer is -8/3[/quote]

I don't think so....


logarithms
number theory proposed
number theory
college contests
combinatorics

Let $ 1<a_1<a_2< \ldots <a_n<x$ be positive integers such that $ \sum_{i\equal{}1}^n 1/a_i \leq 1$. Let $ y$ denote the number of positive integers smaller that $ x$ not divisible by any of the $ a_i$. Prove that \[ y > \frac{cx}{\log x}\] with a suitable positive constant $ c$ (independent of $ x$ and the numbers $ a_i$).

[i]I. Z. Ruzsa[/i]

First define $y(x)$ to be the number of positive integers smaller that $ x$ not divisible by any of the $a_i$'s (defined for all $x \in\mathbb {R}$). Now using PNT we can say for all $x$ after some stage we've at least $\frac{x}{3\log x}$ primes between $\frac{x}{2}$ and $x$.

So if \[ |\{ i \mid a_i >\frac {x}{2} \}| \leq \frac{x} {4\log x}\] then we've,

\[y=y(x) \geq \frac{x}{3 \log x}-\frac {x}{4 \log x} \geq \frac{x}{12 \log x}\]
But if \[ | \{ i \mid a_i >\frac {x}{2} \} | > \frac{x} {4\log x}\] then
\[\sum_{a_i\leq \frac {x}{2}} \frac{1}{a_i} \leq 1-\frac{1}{4 \log x}\].
And so for all $x$ after some stage we've, \[y(\frac{x}{2}) \geq \left \lceil{\frac {x}{2}}\right \rceil -\sum_{a_i\leq \frac {x}{2}} \left \lceil{\frac{x}{2a_i}}\right \rceil\ \geq \frac{x}{16 \log x}\]
And hence we can now find that $c$

Thus, you proved that for any $a_1,\dots,a_n$ satisfying the conditions, it holds: $y > \frac{cx}{\log x}$ [b]for sufficiently large[/b] $x$. But what about those $x$, which are not so large ?

Btw, one can make the version, you've proved, without the requirement: $ \sum_{i=1}^n 1/a_i \leq 1$. All of the primes less than $x$ are approximatelly $\frac{x}{\log x}$, so taking away at most some fixed number $n$ will leave at least $\frac{x}{\log x}-n$ primes, not equal to any $a_i$. Now, taking suuficiently large $x$ we can ensure this number is greater than $\frac{1}{2}\cdot\frac{x}{\log x}$.

[quote=dgrozev]Thus, you proved that for any $a_1,\dots,a_n$ satisfying the conditions, it holds: $y > \frac{cx}{\log x}$ [b]for sufficiently large[/b] $x$. But what about those $x$, which are not so large ?
[/quote]
Since there finitely many exceptions only, we can just take $c$ really small.

Well, let me explain again. We are looking for an absolute constant $c$ (not depending on $x$ or $a_i$). We have established that for any $a_i$, it holds $y(x) > \frac{1}{2}\cdot \frac{x}{\log x}$ for sufficiently large $x$ (Note that it's trivial and it can be done without using the crucial condition $ \sum_{i=1}^n 1/a_i \leq 1$). Of course, fixing $a_i$, there will be finitely many exceptions for $x$ and it also holds : $y(x) > \frac{cx}{\log x}$ for any $x > \max (x_i)$ , but now the constant $c$ would depend on $a_i$. That's the point. You should find an absolute constant. Ok?

We have:
\[ y(x)\geq (1-s)x\,,\, \forall x>1\].
where $ s= \sum_{i=1}^n 1/a_i $.
Indeed, the number of positive integers multiple of $a_j$ and smaller than $x$ is at most $\frac{x}{a_i}$, hence those of them not divisible by any of the $a_i$ are at least $x-x\cdot \sum_{i=1}^n 1/a_i=(1-s)x $.
Thus, if $s$ is not very close to $1$, we can use that estimate. The problem is what to do when $s$ is $1$ or very close to $1$.
One more preliminary fact. Using prime number theorem one can get that the number of primes in the interval $(x/2,x]$ is greater than $C_1\frac{x}{\log x}$ for some absolute constant $C_1>0$.

Let us now denote $\alpha : = \sum_{a_i\leq x/2} 1/a_i$. The idea is as follows: if $\alpha$ is "small" , we can use the above approach to estimate the integers in $[1,x/2]$ not divisible by any $a_i$. If $\alpha$ is "close" to $1$, there will be enough primes in $(x/2, x]$ that are different to any of $a_i$ 's.

[b]Case a)[/b] : $1-\alpha \geq \frac{C_1}{2}\cdot \frac{1}{\log x}$. Then:
\[y(x) \geq \# \{m : m\leq x/2\,,\, m \text{ is not divisible by any }a_i \}\geq \frac{C_1}{2}\cdot \frac{x}{\log x}\]

[b]Case b)[/b]: $1-\alpha < \frac{C_1}{2}\cdot \frac{1}{\log x}$. Then, it holds:
\[ \#\{i : x/2 < a_i \leq x \} \leq \frac{C_1}{2}  \frac{x}{\log x} \]
Indeed, assuming it's not true, we get:
\[1-\alpha \geq \sum_{a_i > x/2} 1/a_i \geq \frac{C_1}{2} \frac{x}{\log x} \cdot \frac{1}{x}=\frac{C_1}{2 \log x} \]
a contradiction.
Hence, the number of primes in $(x/2,x]$ not equal to any of $a_i$ is at least $\#\{p\in (x/2,x] : p\in \mathbb{P} \} - \frac{C_1}{2\log x}$. Thus,
\[y(x) \geq \#\{p\in (x/2,x] : p\in \mathbb{P} \} - \frac{C_1}{2\log x} \geq  C_1\frac{x}{\log x} - \frac{C_1}{2\log x}= \frac{C_1}{2\log x}\]

Combining a) and b) , the result follows.


Suppose that $p$ is an odd prime such that $2p+1$ is also prime. Show that the equation $x^{p}+2y^{p}+5z^{p}=0$ has no solutions in integers other than $(0,0,0)$.

The equation HAS a solution - $(0, 0, 0)$. We will prove that it doesn't have any others. From the Fermat Little Theorem: $p|a^{2p+1}-a = a(a^p-1)(a^p+1)$ so for every integer $a$ we have $a^p \equiv 0$ or $1$ or $-1$ mod $p$. By verifying manually all cases we can see that the only possibility is $p=3$ (since $x^p+2y^p+5z^p \equiv 0 \mod{2p+1}$). So are left with proving that the equation $x^3+2y^3+5z^3 = 0$ has no solutions in integers. To do this consider mod $9$. One can easily check that we must have $x \equiv y \equiv z \equiv 0 \mod{3}$ but in this case just divide by $27$ and repeat the process. It will terminate unless $x=y=z=0$.

[quote="TomciO"]The equation HAS a solution - $(0, 0, 0)$.[/quote]Good point. I have altered the wording.

let (x,y,z)=1,so multipling the equation and (x^p + 2y^p - 5z^2 ) and arguing on divisibility of x,y,z by 2p+1. the only case comes when p=3.(by using Fermat's little theorem). (x,y,z) = (1,1,2) or (2,1,1) module 3. now take mod 9 and you will get there's no answer.


Let $ 3$ circle $ (O_1),(O_2),(O_3)$. Common tangent of each $ 2$ circle cut other at $ A,B,C,A',B',C'$. Prove that: $ AA',BB',CC'$ concur

Let $ AA'$ and $ BB'$ intersect at point $ X$ and let $ CX$ meet circles at $ Y$ and $ Z$.From well known teorem
$ CX.XY\equal{}AX.A'X\equal{}BX.B'X\equal{}CX.XZ \Longrightarrow XY\equal{}XZ \Longrightarrow Y\equal{}Z$ $ \Longrightarrow AA',BB',CC'$ concurrent.

Can you give me a more concrete solution? What is well-known theorem? :blush:

Actually i dont know name of teorem in english.But teorem is:
İf two chords $ AB$ and $ CD$ intersect in $ X$ then $ AX.XB\equal{}CX.XD$,that's all :)

Nktp, I [i]think[/i] I know what you mean, but please rewrite your problem in an unambiguous way.

  darij

[quote="cefer"]Actually i dont know name of teorem in english.But teorem is:
İf two chords $ AB$ and $ CD$ intersect in $ X$ then $ AX.XB \equal{} CX.XD$,that's all :)[/quote]
its called the power of point $ X$ with respect to this circle...
[b][u]theorem[/u]:let chords $ AB,CD$ of circle $ w$ intersect at $ X$ then $ XA.XB\equal{}XC.XD$[/b]
the proof is pretty easy,just find two similar triangles...
also they use the following notation for [b]the power of point $ X$ with respect to circle $ w$[/w]:
$ Xw\equal{}P_{w}^{X}$

If so, see here: http://www.mathlinks.ro/Forum/viewtopic.php?p=792543#p792543

B.T.W. Have you a solution for this problem? Please post it. Thank you!

as far as I know the second problem (from Iran PPCE) is very difficult (almost open)...


find all triples $ (p,q,r)$ of prime numbers s.t.: $ (p\plus{}q)^r\equal{}(p^r\plus{}q^r)r$

by not considering the trivial cases (like $ p\equal{}q$ or $ r\equal{}2$ or $ p\equal{}r$,...),use this famous theorem
$ v_r(p^r\plus{}q^r)\equal{}v_r(p\plus{}q)\plus{}v_r(n)$
and get $ r\equal{}3,v_r(p\plus{}q)\equal{}1$.

Let
$ v_r(p \plus{} q) \equal{} \alpha \implies r^{\alpha.r} \times \left(\frac {p \plus{} q}{r^{\alpha}} \right)^r \equal{} \left(\frac {p^r \plus{} q^r}{r^{\alpha \plus{} 1}} \right) \times r^{\alpha \plus{} 2}$,
but we have
$ \gcd(\left(\frac {p \plus{} q}{r^{\alpha}} \right),r) \equal{} \gcd(\left(\frac {p^r \plus{} q^r}{r^{\alpha \plus{} 1}} \right),r) \equal{} 1$.
which leads to $ \alpha \plus{} 2 \equal{} \alpha.r \implies r \equal{} 3,\alpha \equal{} 1$.
note that i didn't considered the trivial cases ($ r \equal{} 2$,...).


Show that for each odd prime $p$, there is an integer $g$ such that $1<g<p$ and $g$ is a primitive root modulo $p^n$ for every positive integer $n$.

$g$ is a primitive root $\mod p^k$ iff it is one $\mod p^2$, thus we specify on $n=2$.

Let $S$ be the set of those residues $\mod p^2$ that are primitive roots $\mod p$ and have an representant $>1$ and $<p$.
Assume the contrary, meaning that there is no primitive root $g \in S$.

The group of invertible residues $\mod p^2$ is cyclic (by existence of primitive roots). Especially for $d|\varphi(p^2)=p(p-1)$, there are always exactly $d$ elements $x$ with $x^d \equiv 1 \mod p^2$ (so with order dividing $d$) and exactly $\varphi(d)$ many with exact order $d$.

Using $|S|=\varphi(p-1)$, we see that we already know all elements of order $(p-1)$ seen $\mod p^2$.

If $s \in S$, then we look at the order $k$ of $-s \mod p^2$:
$(-s)^k \equiv 1 \mod p^2 \ \iff \ s^k \equiv (-1)^k=\pm 1 \mod p^2$. But $s^k \equiv -1 \mod p^2$ happens for $k=\frac{p-1}{2}$ first (we don't need that it happens, but it does; and that it can't happen earlier follows from the fact that $s$ has order $p-1$).
Especially, $k=p-1$ or $k=\frac{p-1}{2}$, the first being impossible by $-s \not\in S$ (and all with order $p-1$ are in $S$).

But now we know $\varphi(p-1)$ elements of order $\frac{p-1}{2}$. This is no contradiction itself if $\varphi \left( \frac{p-1}{2} \right) = \varphi \left( p-1 \right)$ (being true iff $p \equiv 3 \mod 4$, but we will not need that). But if we just find one more, we have finally our contradiction.
We know other elements of order $\frac{p-1}{2}$, namely those $s^2$ with $s \in S$. But $s \leq p-1 \implies s^2 \leq (p-1)^2=p^2-2p+1<p^2-p$, thus $s^2 \not\equiv -t \mod p^2$ for any $t \in S$.
We thus found one (or in fact $|S|$ more) such elements and reached our contradiction.

[quote="ZetaX"]$ g$ is a primitive root $ \mod p^k$ iff it is one $ \mod p^2$, thus we specify on $ n \equal{} 2$.[/quote]Why? I think 3 is primitive mod 4 but not mod 8, or do I miss something?

For n = 1, consider a nonsquare g mod p that is not -1.  $g^\frac{p-1}{2}$ = -1 mod p. Assume $g^k$ = 1 mod p with k < p-1. We know that k | p-1 however we can also see that k $\nmid \frac{p-1}{2}$ which is only possible if k = 2. But if k = 2 then g = 1, -1 which is a contradiction. Thus g is a primitive root mod p. For n > 1. Assume primitive roots exist mod $p^{n-1}$. Then these roots to the power of $p^{n-1}-p^{n-2}$ are in the set 1, $p^{n-1}+1$, $2p^{n-1}+1$,..., $(p-1)p^{n-1}+1$ mod p^n. We can see that if g is a primitive root mod $p^{n-1}$ then $(p+g)^{p^{n-1}-p^{n-2}}$ or $g^{p^{n-1}-p^{n-2}}\neq 1$ mod $p^{n}$ by the binomial theorem. So this root raised to the power of p is congruent to 1 as can be seen by the binomial theorem. If for this same root g, there existed k < $p^{n}-p^{n-1}$ such that $g^k$ = 1 mod $p^n$ then k | $p^{n-1}-p^{n-2}$  which is a contradiction.

i'll prove that there are at least $\frac{\phi(p-1)}{2}$ of such primitive roots.
suppose that g is a primitive root modulo p. g' the inverse of g modulo p is a primitive root too.
so if for both of them we have $g^{p-1} \equiv g'^{p-1} \equiv 1 \pmod{p^2}$ we get $p^2 | (gg')^{p-1}-1$ which is impossible because we have $p | gg'-1 , p \nmid p-1 \implies p^2 | gg'-1 < p^2-1$ so we must have gg'=1 a clear contradiction. so there are at least  $\frac{\phi(p-1)}{2}$ of primitive roots which are primitive root modulo p^2 and therefor modulo p^n for all n.

This question is not related to the problem, but where did B2 go?


How many ways can you distribute $ 41$ lollipops to $ 8$ kids if an odd number of kids must receive an odd number of lollipops?

Hint: Does the restriction really matter?

Does "odd number of kids must receive an odd number of lollipops" mean that say for example kid 7, 5, 3, etc must get an odd number?


[url=http://www.icho39.chem.msu.ru/html/english/ICHO-2007/Events.htm]here[/url] are this years ICho problems

As usual, the IChO's problems are all very interesting. We could analise some of them here.

yes i feel they are better than this years IPho problems the latter seemed quite easy

[quote="BanishedTraitor"][quote="aspiringchemlete"]For question 1.1.1 it asks you to draw the structure of propanedial, but then it gives it to you in question 1.1.2 (if you look at the response sheet).[/quote]

But plus the two compounds which exist in existence with it (i.e. Keto-Enol Tautomerization).

You guys want to seriously discuss these? I have been doing these problems to practice for the Chemistry Olympiad Scholar School.[/quote]

Scholar School? You mean the study camp? or something else?

Yes, but we call them Scholar Schools.

I quite like the problems on Nanochemistry. I didn't realize that matters behaved in this manner.


function
superior algebra
superior algebra solved

Here is a quite difficult problem. Probably for those who know very well all tricks about exponential matrices it is not hard, but I did not manage to fight with it:
  Let $A,B,C$ be nxn complex matrices and $C=AB-BA$. Suppose that $C$ comutes with $A,B$. Compute for all $t$ in function of t and $C$ the value of $ exp(-t(A+B))\cdot exp(tA)\cdot exp(tB)$

$ f(t)=exp(-t(A+B))\cdot exp(tA)\cdot exp(tB)$

$f'(t)=tCf(t)$ with $f(0)=I_n$

$f(t)=\exp{\frac{Ct^2}{2}}$

Thanks, Moubi. Well, to this kind of tricks I was reffering to!


My problem is this:

I have introduced a particular theorem at some point in my document but do not wish to give the proof until later in the document. I want to restate the theorem before giving the proof but I don't know how to do this whilst retaining the original number that was assigned to it.

I am using the amsthm package.

It is possible to save the theorem counter but it's a bit of a pain as you have to keep track and put it back again. But I found this very neat solution that relies on only changing the counter inside a group - hence the {% and }% below.
[code]\begin{theorem}
	This is the first theorem
\end{theorem}
\xdef\savedtheoremnumber{\thetheorem}

\begin{theorem}
	This is the second theorem
\end{theorem}
Now repeat the first theorem

{%
  \renewcommand\thetheorem{\savedtheoremnumber}
  \begin{theorem}
	  This is the first theorem again
	\end{theorem}
}%

\addtocounter{theorem}{-1}%

Now go back to the old numbering
\begin{theorem}
	This is a new theorem
\end{theorem}[/code] which gives
[quote]Theorem 1. This is the first theorem
Theorem 2. This is the second theorem
Now repeat the first theorem
Theorem 1. This is the first theorem again
Now go back to the old numbering
Theorem 3. This is a new theorem[/quote]


conics
hyperbola
rotation
analytic geometry

If we rotate the coordinate axes clockwise through $\frac{\pi}{4}$, the new $X$- and $Y$-axes become the aysmptotes of the hyperbola. The original $x-$ and $y-$axes have equations $Y=X$ and $Y=-X$ respectively to the new axes

[url=http://imageshack.us][img]http://img297.imageshack.us/img297/8529/73491842nb8.jpg[/img][/url]

We recognise that relative to the new $X-Y$ axes, the hyperbola has equation $XY=c^{2}$ for some constant $c, c>0$. Then at $A$ and $A'$   $Y=X$ and $XY$=$c^{2}$ $=\Rightarrow$ $A,A'$ has coordinates $(c,c),(-c,-c)$

[b]Then $OA = c\sqrt{2}$[/b] and $OA =a$ ==> $c^{2}$ = $\frac{a^{2}}{2}$

Hence the hyperbola has eqn $XY=\frac{a^{2}}{2}$ relative to the new $X-Y$ axes

How did they get the distance of OA - $OA = c\sqrt{2}$

$O$ is the origin so it's $(0,0)$. The coordinates of $A$ are given as $(c,c)$. Distance formula gives that $OA = \sqrt{(c-0)^{2}+(c-0)^{2}}= c\sqrt{2}$.

:blush:

thanks man

--------

im studying rectangular hyperbola so I will post some qns later if I dont understand

thanks bye


I got the following question on FTW.

How many triangles of any type are in the figure?

[asy]size(75);
for(int i = 0; i<5; ++i)
for(int j = i+1; j<5; ++j)
draw(dir(i*72+18)--dir(j*72+18));[/asy]

How do you do this question? BTW, the answer is 35.

My first step would be to label the triangles.  Then you can slightly be more organized by starting with all the triangles that contain triangle A, then B, etc..

well, just count the number of triangles. Start with small, then go to middle, and then go to large :)

Label each segment as side length 1 (even though it might not be), and then find triangles 1x1x1, 1x2x2, etc.

My dad taught me to put triangles into type A and type B


analytic geometry
calculus
calculus computations

its  asked to graph

  $ 0\leq \theta \leq \pi \; ,r<\minus{}1$

what will be the graph?

please can anyone explain...

sorry I EDITED THE Q NOW.. PLEASE TRY

the region in quadrants 3 and 4 outside of the unit circle centered at the origin, ...maybe?

What kind of polar coordinates are you using? Mine don't like negative $ r$.

You can use negative r if your theta limits go from 0 to pi for the entire plane. I - think- that this is equivlent to:

$ \pi\leq\theta\leq\ 2\pi,r > 1$

So I agree with pluristiq

how did  u change the limits of the  angle???


logarithms
number theory unsolved
number theory

Let $ a_{1},a_{2},...,a_{10^{6}}$ be integers between $ 1$ and $ 9$, inclusive. Prove that at most $ 100$ of the numbers $ a_{1}a_{2}...a_{k}$ ($ 1\leq k\leq 10^{6}$) (this is a decimal representation, not a product) are perfect squares

[quote="freemind"]I'm quite unsure of the following, because according to my proof, there are far less perfect squares than required (at most $ 40$ or so).

Here it goes:
Set $ X_{i}=\overline{a_{1}\dots a_{i}}$.
Let $ A=X_{k}=\overline{a_{1}a_{2}\dots a_{k}}=u^{2}>10^{k-1}$ be a perfect square. Take the least $ t>0$ such that $ C=\overline{a_{1}\dots a_{k}a_{k+1}\dots a_{k+2t}}$ is also perfect square.
Let $ B=\overline{a_{k+1}\dots a_{k+2t}}$.
We have $ C=(10^{t}u)^{2}+B$, and since $ C$ is a perfect square, and $ B>0$ (the digits $ a_{i}$ are nonzero), we get $ B\geq2\cdot10^{t}u+1$, or $ B>10^{t}u$. Since $ B$ has $ 2t$ digits, $ 10^{2t}>10^{t}u$, or $ 10^{t}>u$, or $ 10^{2t}>u^{2}=A>10^{k-1}$. So $ 2t\geq k$. This means that having $ X_{k}$ a perfect square, the next $ X_{i}$ perfect square with $ i\equiv k($[i]mod[/i]$ 2)$ has index at least $ 2k$ (if $ k$ is odd, obviously this index will be at least $ 2k+1$). Further, the next perfect square with index of the same parity has index at least $ 4k$ and so on.
Let $ X_{i_{1}}, X_{i_{2}},\dots$ be the sequence of perfect squares with odd indices ($ i_{m}$ is odd, $ \forall m\geq1$) and $ X_{j_{1}},X_{j_{2}},\dots$ be the sequence of perfect squares with even indices ($ j_{m}$ is even $ \forall m\geq1$).
Then $ i_{m}\geq 2i_{m-1}$ and $ j_{m}\geq 2j_{m-1}$, $ \forall m\geq2$. This means that there are at most $ 2\log_{2}10^{6}<40$ perfect squares till the index $ 10^{6}$.[/quote]

It seems OK and very nice for me.


function
induction
algebra solved
algebra

1)Find out all the functions f:N--> N so that we have:
f(f(n)-n)=f(n)+1, and n in N.

Solution:

Lets actually prove that there isn't no such function that verifies this one!!
Lets presume that there exists such function f:N--> for which we have:
f(f(n)-n)=f(n)+1 [1]

Lets replace n with f(n)-n and we use [1] and get that:
f(f(f(n)-n)-(f(n)-n))=f(f(n)-n)+1 which is equivalent to:
f(n+1)=f(n)+2 [2]

lets denote f(0)=a. It's easy to see if we use induction that: f(n)=a+2n.
Replacing it in [1] we get that: f(a+n)=a+2n+1 which gives that a+2n+1=a+2(n+a) so a=1/2 .. so no function f:N-->N can verify that!!!

cheers!!!!!!!!!!!!!!!


Points $ D$ and $ E$ lie on sides $ AB$ and $ AC$ of triangle $ ABC$ such that $ DE \parallel BC$. Let $ P$ be an arbitrary point inside $ ABC$. The lines $ PB$ and $ PC$ intersect $ DE$ at $ F$ and $ G$, respectively. If $ O_1$ is the circumcenter of $ PDG$ and $ O_2$ is the circumcenter of $ PFE$, show that $ AP \perp O_1O_2$.

Let $ I\equiv AP\cap DE$, $ J\equiv AP\cap BC$. By Thales we have:
$ \frac {ID}{IE} \equal{} \frac {JB}{JC} \equal{} \frac {IF}{IG}\Longrightarrow ID.IG \equal{} IE.IF$, which implies that $ \mathcal {P}_{(I/(O_1)} \equal{} \mathcal {P}_{I/(O_2)}$. Let $ Q$ be second intersection of $ (O_1)$ and $ (O_2)$. Thus we conclude that $ I\in PQ$- which is the radical axes of $ (O_1)$ and $ (O_2)\Longrightarrow A\in PQ\Longrightarrow AP\bot O_1O_2$.


calculus
integration
linear algebra
matrix
calculus computations

find $ \int\int_S(x\plus{}y)^4(x\minus{}y)^{\minus{}5}dA$ where $ S$ is the square $ \minus{}1\le x\plus{}y\le 1, 1\le x\minus{}y\le 3$.

Make the substitution $ u\equal{}x\plus{}y,v\equal{}x\minus{}y.$ The whole problem is rigged to come out nicely under that substitution. Don't forget to compute the Jacobian (but the Jacobian of a linear map is constant.)

so what im confused is this:
so let $ u\equal{}x\plus{}y$ and $ v\equal{}x\minus{}y$
then the desired integral is the same as $ |det G^{\minus{}1}|\int_{\minus{}1}^1\int_1^3u^4v^{\minus{}5}dvdu$.

the inverse map is given by $ x\equal{}\frac{1}{2}(u\plus{}v),y\equal{}\frac{1}{2}(u\minus{}v)$.
what Im confused is how to find the det of this map. (this is probably a really stupid question).

$ dA\equal{}dx\,dy\equal{}\left|\frac{\partial (x,y)}{\partial (u,v)}\right|\,du\,dv.$

What you wrote can be given as $ \begin{bmatrix}x\\y\end{bmatrix} \equal{}\begin{bmatrix}\frac12&\frac12\\ \frac12&\minus{}\frac12\end{bmatrix} \begin{bmatrix}u\\v\end{bmatrix}.$

Hence, $ \frac{\partial (x,y)}{\partial (u,v)}\equal{}\det\begin{bmatrix}\frac12&\frac12\\ \frac12&\minus{}\frac12\end{bmatrix} \equal{}\minus{}\frac14\minus{}\frac14\equal{}\minus{}\frac12.$

$ dx\,dy\equal{}\frac12\,du\,dv.$


I edited a topic, but forgot to register it as a minor change.  Is there any way I can reverse that?

Don't worry about it, but try to start documenting changes.  Due to the sheer volume of changes I've been making, I haven't been perfect about it either -- particularly when the changes were minor, but I'll change that.

What's the difference whether or not we mark a change as minor?

It's a politeness that keeps other article editors/writers from having to compare recent changes as considerably.


Can anyone tell me a comprehensive congruence list of special numbers mod whatever..

something like a square is always 1 mod 4 or 0 mod 4, a prime is always 1, -1 mod 6....

any help is appreciated..

a comprehensive list is impossible to put together... because there are an infinite number of such facts.  However, it's easy to derive them.  For instance, if you want to know the possible results of squaring a number in a given modulus (called quadratic residues) then you can take all the numbers less than your modulas and square them and see which numbers you get.

For instance, in mod 6,
$0^{2}\equiv 0$
$1^{2}\equiv 1$
$2^{2}\equiv 4$
$3^{2}\equiv 3$
$4^{2}\equiv 4$
$5^{2}\equiv 1$

So now you know that 0,1,3,4 are the only possible values when you sqare something in mod 6.  The most common fact that comes up from this seems to be that the quadratic residues mod 4 are 1 and 0, but a comprehensive list of all useful facts can't be made.

But which ones are most commonly used while you regularly solve problems? I'm sure there's a number you've memorized from repeated usage.

Squares $\mod 4$ and $8$ for example.
Often $d$-th powers $\mod$ primes $\equiv 1 \mod d$, generally $\mod m$ with $d|\varphi(m)$.
And it's good to know that there are $\infty$ many primes $\equiv a \mod m$ iff $\gcd(a,m)=1$.

But there is no general rule, there are a lot of possibilities. It's a huge difference from case to case.

PS: not all primes are $\equiv \pm 1 \mod 6$, like $2$ and $3$ ;)


geometry
circumcircle
incenter
inradius
cyclic quadrilateral
geometry proposed

In $\triangle ABC$, $D$ is a point on the side $BC$. $E \in [AB], F \in [AC]$ and $AB \perp DE, AC \perp DF$. The radii of incircles of $\triangle ADE, \triangle ADF$  are  $r_1, r_2$. We know that $r_1 = r_2$ . If $\angle BAD = \alpha$, then what are the possible values of $\angle DAC ?$

In order to avoid confusion in this case distinction problem, I will denote the angle $\angle BAD = \phi$, because the interior angle $\angle A = \angle CAB$ of a triangle $\triangle ABC$ is often labeled as $\alpha$.

Since the angles $\angle AED = \angle AFD = 90^o$ are both right, the quadrilateral AEDF is cyclic, AD is a diameter of its circumcircle and the midpoint M of the diameter AD is its circumcenter. Let I, J be the incenters of the triangles $\triangle AED, \triangle AFD$ and S, T the tangency points of the incircles (I), (J) with the common side AD. Since the triangles $\triangle AED, \triangle AFD$ have the same circumradius $R = \frac{AD}{2}$ and the same inradius $r = r_1 = r_2$, their circumcenter to incenter distances $IM = JM = \sqrt{R(R - 2r)}$ are equal. Since in addition, $IS = r_1 = r_2 = JT$, the angles $\angle MSI = \angle MTJ = 90^o$ are both right and IM > IS, JM > JT, the right angle triangles $\triangle MIS \cong \triangle MJT$ are congruent by SsA. Hence, MS = MT. Now we have to distinguish 3 cases.

(1) The points S, T on the line AD are both identical with the midpoint M.
(2) The points S, T on the line AD are on the same side of the midpoint M.
(3) The points S, T on the line AD are on the opposite sides of the midpoint M.

[color=blue][b](1)[/b][/color] In the 1st case, MS = MT = 0. The incircles (I), (J) touch the common side AD of the right angle triangles $\triangle AED, \triangle AFD$ at its midpoint, which means that these 2 right angle triangles are both isosceles and congruent. Hence, the quadrilateral AEDF is a square. The angle $\angle CAB \equiv \angle FAE = 90^o$, if the point D is inside of the segment BC, $\angle CAB = 180^o - \angle FAE = 90^o$, if the point D is outside of the segment BC. Moreover, the line AD bisects the right angle $\angle FAE$, hence, it is either the internal or the external bisector of the angle $\angle CAB$. In both cases, the angle $\phi = \angle BAD = 45^o$ and the angle $\angle CAD = \angle BAD = \phi = 45^o$ or $\angle CAD = 180^o - \angle CAF = 180^o - \angle BAD = 180^o - \phi = 135^o$.

[color=blue][b](2)[/b][/color] In the 2nd case, the points $S \equiv T$ are identical. This implies that the angles $\angle DAE = 2 \angle DAI = 2 \angle DAJ = \angle DAF$ are equal, the angles $\angle ADE = 2 \angle ADI  = 2 \angle ADJ = \angle ADF$ are also equal and the right angle triangles $\triangle AED \cong \triangle AFD$ with the common hypotenuse AD are congruent by ASA. This means that the line AD bisects the angle $\angle FAE$. Since $\angle FAE \equiv \angle CAB$, if the point D is inside of the segment BC, and $\angle FAE = 180^o - \angle CAB$, if the point D is outside of the segment BC, the line AD is either the internal or the external bisector of the angle $\angle CAB$, hence, $\phi = \angle BAD = \frac{\CAB}{2}$ or $\phi = \angle BAD = \frac{\angle CAB}{2} + 90^o$. In the former case, the angles $\angle CAD = \angle BAD = \phi = \frac{\angle CAB}{2}$ are equal, in the later case, the angle $\angle CAD = 180^o - \angle CAF = 180^o - \angle BAD = 180^o - \phi$.

[color=blue][b](1)[/b][/color] In the 3rd case, DS = DM - MS = AM - MT = AT. This implies that the angles $\angle DAE = 2 \angle DAI = 2 \angle ADJ = \angle ADF$ are equal, the angles $\angle ADE = 2 \angle ADI  = 2 \angle DAJ = \angle DAF$ are also equal and the right angle triangles $\triangle AED \cong \triangle DFA$ with the common hypotenuse AD are congruent by ASA. Assume that the angle $\angle CAB \neq 90^o$ is not right. Let $P \equiv DE \cap AC,\ Q \equiv DF \cap AB$ be the intersections of these 2 pairs of lines. Depending on the position of the point D on the line BC (on opposite side of the vertex B than the vertex C or on the opposite side of the vertex C than the vertex B), either the triangle $\triangle ADP$ or the triangle $\triangle ADQ$ are isosceles, because either the angles $\angle PDA \equiv \angle EDA = \angle FAD \equiv \angle PAD$ are equal or the angles $\angle QDA \equiv \angle FDA = \angle EAD \equiv \angle QAD$ are equal. In the former case, $\angle APD = 90^o - \angle PAE \equiv 90^o - \angle CAB$,

$\angle CAD \equiv \angle PAD = \frac{180^o - \angle APD}{2} = \frac{180^o - (90^o - \angle CAB)}{2} = \frac{\angle CAB + 90^o}{2}$

$\angle BAD \equiv \angle EAD = 90^o - \angle EDA \equiv 90^o - \angle PDA = 90^o - \angle PAD \equiv 90^o - \angle CAD$ [color=white].[/color]$= \frac{90^o - \angle CAB}{2}$

In the later case, $\angle AQD = 90^o - \angle QAF \equiv 90^o - \angle CAB$,

$\angle BAD \equiv \angle QAD = \frac{180^o - \angle AQD}{2} = \frac{180^o - (90^o - \angle CAB)}{2} = \frac{\angle CAB + 90^o}{2}$

$\angle CAD \equiv \angle FAD = 90^o - \angle FDA \equiv 90^o - \angle QDA = 90^o - \angle QAD \equiv 90^o - \angle BAD$ [color=white].[/color]$= \frac{90^o - \angle CAB}{2}$

Thus in the former case, $\phi = \angle BAD = \frac{90^o - \angle CAB}{2}$, while in the later case, $\phi = \angle BAD = \frac{90^o + \angle CAB}{2}$. In both cases, the angle $\angle CAD = 90^o - \angle BAD = \phi$. If $\angle CAB < 90^o$, the point D is the intersection of the extarnal or internal bisectors of the angle $\angle CAB + 90^o$ with the line BC and if $\angle CAB > 90^o$, the point D is the intersection of the internal or external bisectors of the $\angle CAB + 90^o$ with the line BC.

If the angle $\angle CAB = 90^o$ is right, than the cyclic quadrilateral AEDF with the right angles $\angle AED = \angle AFD = 90^o$ has the angle $\angle FAE \equiv \angle CAB = 90^o$ or $\angle FAE = 180^o - \angle CAB = 90^o$ also right, henvce, it is a rectangle. Consequently, the triangles $\triangle AED \cong \triangle DFA$ are congruent and their incircles $(I) \cong (J)$ are congruent for any position of the point D on the line BC. Hence, the condition $\angle CAD = 90^o - \angle BAD = 90^o - phi$ holds for an arbitrary point D on the line BC and an arbitrary angle $\angle BAD < \text{max}(\angle ABC, 180^o - \angle ABC)$.

[color=blue][b]Conclusion:[/b][/color] Generally, 4 distinct points $D_1, D_2, D_3, D_4$ exist, such the right angle triangles $\triangle AED \cong \triangle AFD$ or $\triangle AED \cong \triangle DFA$ and their incircles incircles (I), (J) are congruent. If the triangle $\triangle ABC$ is isosceles with AB = AC, the point $D_2$, the intersection of the external bisector of the angle $\angle CAB$ with the line BC, is at infinity and we have only 3 finite points $D_1, D_2, D_4$. Likewise, if $\angle ABC - \angle BCA = 90^o$, the point $D_4$ is at infinity and if $\angle BCA - \angle ABC = 90^o$, the point $D_3$ is at infinity, hence, we have only 3 finite points $D_1, D_2, D_3$ or $D_1, D_2, D_4$. Finally, if the angle $\angle CAB = 90^o$ is right, the the right angle triangles $\triangle AED \cong \triangle DFA$ and their incircles (I), (J) are congruent for any point D on the line BC.


Find three squares in arithmetic progression. Can four squares be in arithmetic progression?

four squares can never be in AP. This can be proved by infinite descent.

On the other hand, the equation $ x^2 \plus{} y^2 \equal{} 2z^2$, of three squares in AP, has full parametric solutions the triplets $ m^2 \plus{} 2mn \minus{} n^2, n^2 \plus{} 2mn \minus{}m^2, m^2 \plus{} n^2$.


Gentlemen:

I am 78 years old, and have forgotten much of my math in solving for unkowns in equations:

Here is the formula I need help with:

This is the formula being used to determine the rate:

Payment = Principal * (Rate / (1 - (1 + Rate) ^ -Term))
Factors known:
Payment = 1140.74595
Principal = 212,500
Term = 30 years or 360 months

How do I solve for Rate???  I know that the Rate is 5%, but I need to understand the steps to solve for this value.

I get to the following point, but fail to go beyond.

[b]1140.74595 / 212500 = (Rate / ( 1 - ( 1+Rate) ^ -360))[/b]

Please help???

Is this the same formula as i=prt?
Intrest= principlex ratex time

That's a weird interest equation, hmm.

I'm assuming this is the equation:

$Payment = Principal\cdot\frac{Rate}{1-(1+Rate)^{-Time}}$

?

I am unfamiliar with tihs. Does somebody have confirmation?

Dear WindSlicer:

The formula I quoted is exactly the one you show in your reply. I simply typed it differently.

Dear WindSlicer:

[b]
Payment = Principal * (Rate / (1 - (1 + Rate) ^ -Term)) [/b]

is the same as:
[b]
Payment = Principa * \cdot\frac{Rate}{1-(1+Rate)^{-Time}}[/b]

I would like to know how to proceed in solving for [b]Rate[/b]    when Payment, Principal and Term are known????

Please outline the steps so that I may learn this topic once and for all.  I will then be able to apply in any instance where ^  power  signs are found in equations. (Hopefully that is  :roll:

Hmm, I think it's impossible to solve for Rate in your equation algebraically.

Here's what I did:

$\frac{Payment}{Principal} = \frac{Rate}{1-(1+Rate)^{-Term}}$

Let $c=\frac{Payment}{Principal}, r=Rate, t=Term$. The equation becomes:

$\frac{r}{1 - (1+r)^{-t}} = c$

First, the denominators of the fractions must be non-zero, so all valid solutions must satisfy:
$(1+r)^{t} \ne 0$, $(1+r)^{-t} \ne 1$. The latter condition is equivalent to: $(1+r)^{t} \ne 1$.

Multiply both numerator and denominator of the fraction on the left by $(1+r)^{t}$:

$\frac{(1+r)^{t} r}{(1+r)^{t} - 1} = c$

Multiplying both sides by $(1+r)^{t} - 1$:

$(1+r)^{t} r = c (1+r)^{t} - c$

And rearranging the terms:

$(r+1)^{t} (r - c) + c = 0$

Now, if you plug $t=360$, you'll get a polynomial of 361-st degree in $r$.
The roots of such polynomial (and there are 361 of them) cannot be in general expressed with arithmetical operations.

Dear Octogram:

Thanks for the response.

I must digest your treatment and get back to you with questions to refine my understanding of your solution.

I shall tackle it tommorrow morning.

Dear Octagram:

Please forgive my misspelling of your name in the prior reply.

I have examined your reply but must confess that I cannot get beyond the following steps:

Let c = Payment / Principal
C= 1140.74595 / 212500

=(1+r)^t*r * (1+r)^t -1

=(1+r)^t -1 * (1+r)^t-1

How do I multply the left  side's numerator and denominator by  (1+r)^T ???

Excel  has a formula for computing Rate as:

[b]=Rate(#periods, Higher Pmt, Orig Principal, FV, Type, Guess Rate) x 12[/b]

I can't find the derivation of this formula to be able to apply it in Visual Basic 6.0 which requires codeing of the formula.

Sorry to be such a bother for you, but I am a hound dog when it comes to understanding any unkown subject.

Thanks again for your help.

Please I need a step by step outline. The following I cannot understand at all.

[b]Now, if you plug , you'll get a polynomial of 361-st degree in .
The roots of such polynomial (and there are 361 of them) cannot be in general expressed with arithmetical operations.
[/b][quote][/quote]

Hm, where did blahblah's post go? He answered the question (I believe) unless there was a mistake.

I don't think anyone so far is going to get it - it's a pretty complicated expression to simplify - sorry I can't help you out more.

but this is no Getting Started question, maybe Economics forum?  :juggle:

Dear Octagram:

Thank you for the response, and the links.  The 2nd link will studied, and tested in Visual Basic 6.0.

Perhaps it will lead to the solution.  I understand that by iteration Excel calculates the rate.

	Rate	                                                                          5.00%
	Term in Years	                                                                30

	Monthly Payment
	 '= 212500 * (.05/12) / (1- (1+.(05/12))^-(30*12)	 1,140.75

Solve for Rate
	Monthly Payment / Principal = (Rate / (1- (1+Rate) ^ -term)
	Monthly Payment / Principal = 	                                               0.005368216

	Rate/12 / (1- (1+Rate/12)^-360)  =                               	   0.005368216

This is as far as I can get with this formula, so I will thank all that rendered assistance, and call the issue closed until I find a good solution.

I shall then post the results of my findings.

Thank again to all.

Saturday, June 11, 2005 3:28 PM ET
 :(  :(

Dear Octagram:

Your referred to links were very helpful.  Expecially the first, as you were correct that Excel functions work in VB.

I tested this formula in VB 6.0 in my application and it gave me the same result as in Excel. Although I still do not know how to solve for a 360 degree Polynomial (I will study this subject further), I am nevertheless happy that a practical solution was found with your able assistance.

Thank you, and Please  consider this topic SOLVED.
[b]APR = (Rate(TotPmts, -Payment, PVal, FVal, PayType, Guess) * 12) * 100[/b]

Tuesday, June 14, 2005 11:16 AM ET

[quote="octagram"][quote="ortizsr"]I can't find the derivation of this formula to be able to apply it in Visual Basic 6.0 which requires codeing of the formula.[/quote]
Excel doesn't know any direct formula either. They only approximate it, it's mentioned in the help.

And the function Rate seems to be also available for VB and VBA, not just for excel, check:
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/vbenlr98/html/vafctRate.asp

It seems that GNU Mono project was once working on this Rate function, too. This page has some its implementation in C#:
http://galactus.ximian.com/pipermail/mono-patches/2004-June/040700.html

If you want to code your own VB implementation, it would be a much better starting place, than my previous post.

[quote="WindSlicer"]Hm, where did blahblah's post go? He answered the question (I believe) unless there was a mistake.[/quote]
He solved for Term, not for Rate. I guess he simply deleted his post.[/quote]


geometry
trapezoid
algebra
polynomial
geometry open

At  page  17  of the above magazine , one  can  find  this  problem :
A  quadrilateral  has  verices  on a circle  with radius  R  and  its  sides  tangets  to  a  circle  with  radious  r .
Prove  that  [tex] R \geq\sqrt2\cdot r [/tex].

So  here  goes my questions :
1)If we  know  that  if  [tex]R\geq\sqrt2\cdot r [/tex]  then   can  we  contruct  a  quadrilateral  whose  vertices are  on  the  circle  with  radius  R  and its  sides  tangets  to  the  circle  with  radius  r ?
2)We  know  that   ,  if  [tex]  R\geq2r [/tex]  then we can  construct a  triagle....but  is  [tex] R\geq\sqrt2\cdot r [/tex]  enought  to  construct a  quadrilateral ?

Thank you  for  your time  :)

[quote="nickolas"]quadrilateral  has  vertices  on a circle  with radius $R$  and  its  sides  tangent  to  a  circle  with  radius  $r$ .  Prove  that  [tex] R \geq\sqrt2\cdot r [/tex].[/quote]

The theorem of Poncelet says that if such a quadrilateral exists, then there is a continuous family of such quadrilaterals; for each point on the $R$ circle, following the tangents will produce a quadrilateral tangent to the $r$ circle.  Let us say that two circles are in 4-Poncelet position if this is true. By specializing the quadrilateral (for example, to a trapezoid with the parallel sides perpendicular to the line of centers of the 2 circles) one can find a polynomial condition $P(R, r, d) = 0$ that characterizes 4-Poncelet position, where $d$ is the distance between centers.

3-Poncelet position (inscribed and circumscribed triangle), is characterized by $d^2 = R^2 - 2Rr$, a famous theorem of Euler.

I believe that for any $n-$Poncelet configuration, the maximum value of $r : R$ is attained for a regular polygon, that is, $d=0$.

[quote] 1)If we  know  that  if  [tex]R\geq\sqrt2\cdot r [/tex]  then   can  we  contruct  a  quadrilateral  whose  vertices are  on  the  circle  with  radius  R  and its  sides  tangets  to  the  circle  with  radius  r ?
2)We  know  that   ,  if  [tex]  R\geq2r [/tex]  then we can  construct a  triagle....but  is  [tex] R\geq\sqrt2\cdot r [/tex]  enought  to  construct a  quadrilateral ? [/quote]

By Poncelet's theorem, it depends on the relative position (distance) of the circles, not only the radii.


superior algebra
superior algebra solved

I'm having a problem with this group.  i would greatly and gracefully appreciate the help.

In a group prove that (a^-1)^-1 = a for all a.

thanks
the "struggler"

Remind yourself: what does $a^{-1}$ mean, anyway?

We say that $b$ = $a^{-1}$ if $b$ is an element such that $ba=ab=e$ (using $e$ to represent the identity element.)

So what would the multiplicative inverse of $a^{-1}$ be, or more to the point, what would it do?


Hi - here is a problem which appeared on a local middle school contest which is (supposedly) Much easier than Mathcounts questions, and yet - this one stumped me. I mean - I get how to get via "luck" or "guessing" but not sure if there is a method that can be used to solve both this and similar problems.  Can you solve?

Here is problem: "A book measures 8 1/2 by 5 1/2 inches by 1 1/4 inches.  How many of these books can be packed in a carton which has inside dimensions 15 inches by 11 inches by 8 inches?"

you can always divide the volume of the box with the volume of a book and floor the answer.

I think that method has flaws though, since even though the volume of that many books may fit, the dimensions of the book would not allow for the books to fit physically without damaging a few.

I would try to firstly lay down as many books on the floor of the box, and see how many layers of that floor are possible.

[quote="Fraenkel"]you can always divide the volume of the box with the volume of a book and floor the answer.

I think that method has flaws though, since even though the volume of that many books may fit, the dimensions of the book would not allow for the books to fit physically without damaging a few.
[/quote]

yeah...
i agree...
i mean, eventually, you would have to cut a few books up, but that's beyond the point...
i would go with fraenkel's idea...
try to see how many layers of the floor would fit,
a diagram would help greatly..

But couldn't you stick in some books sideways?

Not necessarily, even if you stick the books in sideways, the books will still have the same volume.

[quote="yoonzik"][quote="pingala"] 18 books=6+6+6 [/quote]

how does this work? does anyone have a final solution backed up with an explanation? And also, why wouldn't it just be 1320 divided by 58.4375 = about 22?[/quote]

The books aren't like water. They can't change shape to fit volume.

For example, a book with dimensions 1*1*8 has the same volume as a box with dimensions 2*2*2, but the book obviously won't fit.

i agree with 123456789. you can have a book that is 2x2x2 and one that is 1x1x8. they both have the same volume but they each have different shapes and sizes resulting in some of them not being able to fit.

I just do these problems by first laying down one layer, and seeing if I can maximize the base by flipping the objects...
I don't really think that there's a formula...


Three children intend to visit their grandmother for her birthday. It is an 84km journey and they have two bikes to share between them. They are not allowed to have two children on one bike at the same time or to ride one bike while wheeling the other one. Each child can walk at 5 km/h and cycle at 20 km/h, and they are able to leave bikes unattended by the roadside.

(a) What is the shortest time in which they can complete the journey and how is this achieved?

(b) Suppose they are allowed to wheel one bike while riding on the other and that they can do this and still travel at 20 km/h. Describe how thay can now reduce the time for the journey by 36 minutes.

I got a difference of more than just 36 min. I wonder which of my calculations is wrong..(maybe both are wrong)

How did you construct your working? I got 36 minutes the last time i did it.

I do have a proof, but its hard to explain.
But here is what I did for (a)
[code]
--   -- * __
-- * __ * --
__ * --   --

[/code]
the _ represents walking and the -- represents cycling the * represents a switch.
all three of them walk a third of the distance and cycle the rest.

time taken=$ \frac{\frac{84}{3}}{5}\plus{}\frac{\frac{2(84)}{3}}{20}$
              =$ 5.6\plus{}2.8$
              =$ 8.4$

for (b)
No proof, just intuition for this one.
[code]
-------*___
-------
   *---
    -------
___*-------
[/code]

this here represents the other method. Here, the second person rides forward with the first and then takes the first persons bike back to the third person.

here is the calculation.

let $ k$ be the distance that the 2nd guy cycles back and t be the total time for journey. note that we assume all reach the end at the same time. so we have,

$ \frac{\frac{84\minus{}k}{2}}{5}\plus{}\frac{k\plus{}\frac{84\minus{}k}{2}}{20}\equal{}t$
and
$ \frac{k\plus{}84}{20}\equal{}t$

solving gives
t=6.72.

the difference,
8.4-6.72>1 so my answer is wrong.

I think there's a mistake in your section b: I'll try to identify it in a while. I can't upload the BMP image i created, so just try to imagine it. The solution is in essence the same as yours.

Consider the distance-time graph. All three children begin at the origin. The line y = 84 km is where their (similar) endpoint is. Two of the children (say, J and K) cycle at 20 km/h while the other (L) walks at 5 km/h, so the gradient of J and K is 4 times that of L. After $ k$ kilometres of travelling by J and K, L has covered $ \frac {k}{4}$ kilometres.

Now J cycles back to L, wheeling K's bike. Since the relative speed between J and L is 25 km/h and the distance between them $ k = \frac {k}{4} = \frac {3k}{4}$ kilometres, they take ${ \frac {\frac {3k}{4}}{25}} = \frac {3k}{100}$ hours to meet from the time J and K part ways.

We observe that when their paths are plotted on the distance-time graph, they must form a unique parallelogram (why?). Thus when J meets L with the extra bike they must still have a extra $ k$ kilometres to complete the journey.

Then we have the total time taken as $ t = \frac {k}{20} + \frac {3k}{100} + \frac {k}{20} = \frac {84 - k}{5} + \frac {k}{20}$. Solving, $ k = 60$ and $ t = 7.8$ hours.

So the time saved is $ 8.4 - 7.8 = 0.6$ hours = $ 36$ minutes.

Now i'll peruse your method =P

I found my mistake. Its supposed to be $ t\equal{}\frac{2k\plus{}84}{20}$ and then I get the same answer as you.


A certain wedding party consists of the bride, the groom, four bridesmaids, and two groomsmen. They all know each other, except that none of the bridesmaids know any of the groomsmen. If two of the eight people are chosen at random, then what is the probability that they know each other?

I liked this problem.
[hide]The total # of ways is $ 7\cdot{8}\equal{}56.$ Using complimentary counting, there are 2x4x2=16 ways to choose 2 people who don't know each other.
$ \frac {5}{7}$.[/hide]

EDIT:Fixed

xpmath solution is right.

But I came up with very weird way of coming up with same answer..
I don't know how it worked out.. It's kind of freaky..

I did 8C2 and got 28 ways(total)
Then I did 4(# of bridesmaid)$ \cdot$2(# of groomsmen) to get the possible outcome that they do not know each other.
So $ \frac{20}{28}\equal{}\frac{5}{7}$
It's freaky...

[quote="xpmath"]I liked this problem.
[hide]The total # of ways is $ \binom{8}{2} \equal{} 56$. Using complimentary counting, there are 2x4x2=16 ways to choose 2 people who don't know each other.
$ \frac {5}{7}$.[/hide][/quote]

Umm... actually, $ \binom{8}{2} \equal{} 28$. But I suppose you did it assuming that order matters, so it would have been $ 7 \cdot 8$.


complex numbers
square
geometry
polygon
IMO
IMO 1977

In the interior of a square $ABCD$ we construct the equilateral triangles $ABK, BCL, CDM, DAN.$ Prove that the midpoints of the four segments $KL, LM, MN, NK$ and the midpoints of the eight segments $AK, BK, BL, CL, CM, DM, DN, AN$ are the 12 vertices of a regular dodecagon.

Just use complex numbers, with $ a \equal{} 1$, $ b \equal{} i$, $ c \equal{} \minus{} 1$ and $ d \equal{} \minus{} i$.
With some calculations, we have $ k \equal{} \frac {\sqrt {3} \minus{} 1}{2}( \minus{} 1 \minus{} i)$, $ l \equal{} \frac {\sqrt {3} \minus{} 1}{2}(1 \minus{} i)$, $ m \equal{} \frac {\sqrt {3} \minus{} 1}{2}(1 \plus{} i)$ and $ n \equal{} \frac {\sqrt {3} \minus{} 1}{2}( \minus{} 1 \plus{} i)$.
Now it's an easy job to calculate the twelve midpoints and to find out they are all of the form $ \frac {\sqrt {3} \minus{} 1}{2}e^{\frac {k\pi}{6}i}$, with $ k\in\mathbb{N}: 0\le k\le 11$, and the result follows.

Synthetic solution:
Clearly the four first midpoints form a square with midpoint $O$ where $O$ is the midpoint of $ABCD$.
Hence it suffices to check that the other eight points have the same distance from $O$ and that the angles at $O$ are correct.
By symmetry, it will then suffice to check that $OEF$ is equilateral where $E$ is the midpoint of $KL$ and $F$ is the midpoint of $AK$.
Being midlines in triangles $AKL$ and $AKC$, we see that $EF=\frac{1}{2} AL$ and $OF=\frac{1}{2} KC$. On the other hand, in the square $KLMN$ we clearly have $EO=\frac{1}{2} KL$. But by construction, the triangles $ABL, LBK, KBC$ are all congruent whence $AL=LK=KC$ and hence $EF=EO=OF$ as desired.

[asy]
import olympiad;
size(6cm);
defaultpen(fontsize(10pt));
defaultpen(linewidth(0.4));
dotfactor *= 1.5;

pair A = dir(225), B = dir(315), C = dir(45), D=dir(135), K=extension(A,dir(75),B,dir(105)), L=extension(B,dir(165), C,dir(195)), M=-K, N=-L, E=(K+L)/2, F=(A+K)/2, O=0;

draw(A--B--C--D--A);
draw(B--K--A, dotted);
draw(B--L--C, dotted);
draw(K--L, dashed);
draw(E--F--O--E, dashed);
draw(A--L, dashed);
draw(C--K, dashed);

dot("$A$", A, dir(225));
dot("$B$", B, dir(315));
dot("$C$", C, dir(45));
dot("$D$", D, dir(135));
dot("$K$", K, dir(90));
dot("$L$", L, dir(180));
dot("$M$", M, dir(270));
dot("$N$", N, dir(0));
dot("$O$", O, dir(0));
dot("$E$", E, dir(135));
dot("$F$", F, dir(315));
dot((M+L)/2);
dot((M+N)/2);
dot((N+K)/2);
dot((B+K)/2);
dot((B+L)/2);
dot((C+L)/2);
dot((C+M)/2);
dot((D+M)/2);
dot((D+N)/2);
dot((A+N)/2);
[/asy]


Which of the following expressions must be positive for all values of x?

A. x^3 + 2x − 5     B. x^5 +1    C. x^6 + x4 + x2     D.  5x^2 +18x + 9      E. x^4 + 3x2 + 4

Intuitively, look for the functions with only even exponents. This narrows it down to C and E. However, plugging in x=0 for C gives 0, so the answer is $ \boxed{E}$.


function
real analysis
calculus
calculus computations

Suppose that the continuous function $ f: \mathbb{R} \to \mathbb{R}$ is periodic, that is, there is a number $ p$>$ 0$ such that $ f(x \plus{} p)$ = $ f(x)$ for all x. Show that $ f: \mathbb{R} \to \mathbb{R}$ is uniformly continuous.

Hint:  what can you say about $ f$ say on $ [0,p]$?  Then use the fact that you have $ f(x) \equal{} f(x\plus{}kp)$  for any $ k$ integer.

consider $ [0,2p]$ the compact interval.

since f is continuous on such an interval, it is automatically uniformly continuous on that interval.

thus $ \forall \epsilon >0, \exists \delta >0$ such that

$ |x\minus{}y|<\delta \to |f(x)\minus{}f(y)|<\epsilon$

now consider the general $ x,y$ on the reals,

if $ |x\minus{}y|<\delta$, then there exists an n such that

$ x'\equal{}x\minus{}np, y'\equal{}y\minus{}np$ have $ x',y' \in [0,2p]$

thus $ |f(x)\minus{}f(y)|\equal{}|f(x')\minus{}f(y')|<\epsilon$

if you dont use 2p, the argument doesnt hold, do you see why?

[quote="kenn4000"]if you dont use 2p, the argument doesnt hold, do you see why?[/quote]
Strictly speaking, it didn't have to be $ 2p,$ just something a little bit bigger than $ p.$

A slightly more abstract point of view: a continuous $ p$-periodic function on $ \mathbb{R}$ can be considered to be the same thing as a continuous function on the quotient space (quotient in both the algebraic and topological senses) $ \mathbb{R}/p\mathbb{Z}.$ And $ \mathbb{R}/p\mathbb{Z},$ also known as $ \mathbb{T}$ or $ \mathbb{T}_p,$ is a compact set (a compact topological group).


Hey, does anyone know where i can find more of those little flash applets on the left hand side? (or are those the only ones)

All of the ones we have built so far are [url=http://www.artofproblemsolving.com/Gallery/AoPS_G_Gallery.php]here[/url].

There's only 10? I thought there were more like 50 (well, it helps to have click on millions of links on this site from replying to browsing topics).

I like the one on the bottom  :D It's really cool, but I dont' quite understand it  :?

Instructional "advertising." How amazing. :D

thanks.  By the way, does the creator of those intend to do anymore?  They're little gems of wisdom.


inequalities
function
inequalities proposed

Consider n real numbers whose sum is 0; let m be the smallest of these numbers, and let M be the greatest of these numbers. Prove that

[b]a)[/b] The sum of the squares of these numbers is $\leq -mMn$.
[b]b)[/b] The sum of the fourth powers of these numbers is $\leq -mMn\left(m^2+M^2+mM\right)$,

[b]PS.[/b] The proposed solution - in Russian - is avaliable at

http://kvant.mccme.ru/1993/06/resheniya_zadachnika_kvanta_ma.htm

  Darij

[b]a)[/b]  Let f be the linear (affine) function such that f(m) = m2 and f(M) = M2.  A calculation shows that f(0) = -mM.  Because x2 is convex, we have $\textstyle x^2 \le f(x)$ for all x between m and M.  Summing gives
\[\sum_i x_i^2 \le \sum_i f(x_i) = n \sum_i \frac{f(x_i)}{n} = n f(\frac{\sum_i x_i}{n}) = n f(0) = -mMn.\]

[b]b)[/b] Let g be the linear (affine) function such that g(m) = m4 and g(M) = M4.  A calculation shows that g(0) = -mM(m2 + M2 + mM).  Because x4 is convex, we have $\textstyle x^4 \le g(x)$ for all x between m and M.  Summing gives
\[\sum_i x_i^4 \le \sum_i g(x_i) = n g(\frac{\sum_i x_i}{n}) = n g(0) = -mMn (m^2 + M^2 + mM).\]


For what value of the constant $ c$ is the function $ f$ continuous on $ (\minus{}\infty,\infty)$?

$ f(x) \equal{}\begin{cases}cx\plus{}1~\text{if}~x\le3\\ cx^{2}\minus{}1~\text{if}~x > 3\\ \end{cases}$

From what I can tell, for it to be continuous, the limit of each piece at $ 3$, must be equal...and I get $ c \equal{}\frac{1}{3}$ after using the limit laws...Is this right? I'm really only seeking confirmation and maybe a better method of doing this?
:D

Also,

Is there a number that is exactly 1 more than its cube?

My work:

$ x \equal{} x^{3}\plus{}1$
$ x(x\plus{}1)(x\minus{}1) \equal{}\minus{}1$

I'm stumped after this point...my calculater found $ \minus{}1.3$ roughly, but I don't know how it arrived at this answer...

And finally,

A monk leaves a monastery at 7:00 AM and takes his usual path to the top of the mountain, arriving at 7:00 PM. The following morning, he starts at 7:00 AM at the top of the mountain and takes the same path back, arriving at the monastery at 7:00 PM. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days.

I'm really stumped on this one...

1) absolutely right.
2) "consecutive"... there are no consecutive reals, just consecutive integers. The equation has no solution in integers nor in positive reals, but in reals it has.
Using the Intermediate Value Theorem for the function y=x-x^3:
y<1 for x=0 and
y>1 for x=-2,
so there is x between -2 and 0 for which x-x^3=1. We don't have to find it, the question was "Is there?" and the answer is "Yes."
3)Let the distance between monastery and mountain be 1. Define f(x)=t if the monk was in distance x from monastery at time t in the first day. Define g(x)=t if the monk was in distance x from monastery at time t in the first day. We have f(0)=7, f(1)=19, g(0)=19, g(1)=7. So f(0)-g(0)=-12, f(1)-g(1)=12, so there is
place x between 0 and 1, in which f(x)-g(x)=0.

This Intermediate Value Theorem seems a lot more useful that I thought. Thanks a lot!


Hello,

I was an assistant coach to our mathcounts team this year and will be lead teaching next year.     :)
I welcome any input with regard to coaching strategy...
How often do you meet?
(How) do you use AoPS materials?
Do your students take AoPS classes?
How do you use your time together?
What do you assign for homework?
How do you track/assess student progress/ability?
How do you encourage students?
How do you build enthusiasm?
What are most effective tools/topics?
How do you select your team?
What qualities/skills take your students to the next level?

Any thoughts on any MathCounts topics are welcome!  Thank you!

I know I'm not a coach (im a student), so its really out of my place, but I dont think my coach (frost13) will mind.
1. We (as a team) go a bit overboard usually 2 hrs thurs, 5 sat and fri, plus opt. 2 hrs monday and wednesday
2. he just gives us the books for us to read in spare time.
3. I know I did. And it definately helps. I recommened the MATHCOUNTS Problem Series.
4. Practice tests, Tests, TESTS! (Old Sprints, Targets, Teams, and Countdowns)
5. old students packets to do at our own pace.
6. our test scores on a spreadsheet
7. call me a slacker, then give me a smile ;)
8. let me see results.
9. old tests and packets, and time
10. Whoever has the highest average from all the tests he's given us.
11. I don't think I'll say.
->Our coach has helped us win staes 4 years in a row. We're hoping to make it 5.
P.S. Having all the school, chapter, state, and national tests from 89-05 is a good idea. (ask MATHCOUNTS)
P.P.S. if you really think I shouldn't have said anything, just PM me, and I'll remove this.

I am not a coach, but the coach of my team has won state 3 times and has placed 2nd at nationals, so in my opinion, he must be a good coach. We meet every morning before school 7:10 to 7:45 and sometimes we stay until school starts at 8:15. We have an afternoon class dedicated to math competitions, not necessarily Mathcounts, but a good deal of it is. Our A team meets after school all the time for practice on team rounds. We have an Art of Problem Solving book but we normally do not refer to it. Most of our material comes from prevoius Mathcounts tests back to the 1980s. Generally, we take tests and then when we're done, grade, then figure out what we got wrong. Sometimes we will get a briefing of the test before starting. Last year, we had a points system to decide the A team. Every test we took, we'd assign a certain number of points to every problem answered correctly. We did this also with ASMA if any of yall have done that before. But sometimes there is a serious unbalance in skill in the math team, so it is possible just to pick out the four people that will make up the team. If a person slacks off, he gets kicked out.

About meeting and homework:

Our school team consists of 18 people.  only 4 were selected to go to Mathcounts. The rest of us practice and improve and hope we can go to Mathcounts next year.  We practice 6 to 8 hrs. a week, but never more than 2 hrs. a day.  The only homework we get is

a) to review the sprints, teams, or targets we did that day by yourself, with a parent, or with a sibling.
b) make sure you know [u]why[/u] you got the questions wrong
c) see if you can figure out a quicker way to do a problem you felt like you spent a lot of time on.

If we have a holiday or a week off, then we ususally don't meet, but we work on a packet that consists of around 2 sprints and 2 targets.

The actual team that is going to Mathcounts usually gets in 2 extra practices a week, but they  are only team rounds.

Make sure that you review the practices, and do chapters, states, and nationals with your team.

Dr. Doehner (my coach) only practiced us 2 hours a week, and we ended up first in NJ :? ...

We did some individual practice, but not much :P .

at the beginning of the year, we had like 16 people in Mathcounts, so we met once per week for about 1 to 1 and a half hour. Then after we decided who's going to Chapter, we started meeting 3 days for 45 minutes each during Eigth Period, then some practice after school for about an hour on like Tues and Thurs, and that's the same that goes for studying for state.

Our coach, during the summer, (yes, we had summer practices) gave our work from a geometry book she got, since they already teach algebra in our school. Then, during the practices, we mostly did team rounds to give us an idea of what it's like. When we figured out who's going to chapter, we did lots of tests from past chapter and state. and we often practiced with COuntdown.

I don't think anyone from our Mathcounts take AOPs class, but it'll probably help if you do.

Good Luck on Coaching your MathCounts team! and Don't Give TOO much Homework

 :D

WOW some of you guys practice a lot. We only had 5 90 min practices. It paid off too. Our team had 3 members in the top 10 for written and not top 3 for team.


If z/(x+y)=a

and

 y/(x+z)=b,

Is it possible to find the value of

 x/(y+z)?

$ x\neq\minus{}y , x\neq\minus{}z,$

all the ipotesi is in $ \frac{z}{x\plus{}bx\plus{}zb}\equal{}a$ so $ x\equal{}z\frac{(1\minus{}ab)}{a(b\plus{}1)}, y\equal{}b(x\plus{}z)\equal{}bz(1\plus{}\frac{(1\minus{}ab)}{a(b\plus{}1)})$

in conclusion $ \frac{x}{y\plus{}z}\equal{}\frac{\frac{(1\minus{}ab)}{a(b\plus{}1)}}{b\frac{(1\minus{}ab)}{a(b\plus{}1)}\plus{}1}\equal{}\frac{1\minus{}ab}{a\plus{}b\plus{}ab(1\minus{}b)}$

bye
paolo


algebra
polynomial
complex numbers
superior algebra
superior algebra solved

prove that there is noo root of unity ,$\alpha $, such that $Re(\alpha)=\sqrt {k+1}-\sqrt{k}$.

First, if $\alpha$ is a root of unity, $\alpha+\overline{\alpha}=2\Re(\alpha)$ is an algebraic integer (root of a monic polynomial).

Theorem: If $t=2\Re(\alpha)$ for some root of unity $\alpha$, all of the algebraic conjugates of $t$ have the same form.
Proof: Suppose $\alpha^m=1, m>1.$ Then $1+\alpha+\alpha^2+\cdots+\alpha^{m-1}=0$. If $m$ is odd, divide this by $\alpha^{(m-1)/2}$ and rewrite it as $\alpha^{-(m-1)/2}+\cdots+\alpha^{-1}+1+\alpha+\cdots+\alpha^{(m-1/2)}=0.$
Substitute $t=\alpha+\alpha^{-1}$; this becomes a monic polynomial equation with integer coefficients which $t$ satisfies.
If $m$ is even (and $m>2$), divide by $\alpha^{m/2}$ and rewrite the above as $\alpha^{-(m-2)/2}+\cdots+\alpha^{-1}+\alpha+\cdots+\alpha^{(m-2)/2}=0,$ canceling the $1$ and $-1.$ Again, after substituting $t=\alpha+\alpha^{-1}$ this becomes a monic polynomial equation with integer coefficients which $t$ satisfies.
In both cases, this polynomial must be a multiple of the minimal polynomial of $t$, so all of the conjugates of $t$ are roots of it. Since the roots of the polynomial are easily seen to be $\alpha+\alpha^{-1}$ when $\alpha^m=1$ and $\alpha\neq\pm1,$ all of the conjugates of $t$ must be of that form.

Corollary: If any of the conjugates of $t$ is not a real number between $2$ and $-2,$ $t$ is not twice the real part of any root of unity.

In this problem, $t=2(\sqrt{k+1}-\sqrt{k}).$ If $k>1,$ the minimal polynomial of $t$ is $t^4+8(2k+1)t^2+16$, which has $2(\sqrt{k+1}+\sqrt{k})$ as a root. For $k=1$, the only conjugate of $t$ is $-2(\sqrt{2}+1)$, which is outside the interval. For $k=0,$ $t=2,$ which is twice the real part of 1. We must exclude this case.


[quote="lifesapain"]how do u find the angle measures of an x-pointed star?[/quote]
I don't use a formula or anything. All I do is draw it out for an x-pointed star. Then do stuff to find the measures of the tips.

Are you talking about stars with all tips having eual measure?
If so...
[hide]the tips of a 5 point star add up to 180 degrees
360 degrees for six point star
540 degrees for seven point star
180x-720 degrees for x point star[/hide]
From here, the angle of each tip could be easily calculated (by dividing the sum of the measure of the tips by x).

[quote="kstan013"]you should support your statements with reasoing, cat, i really don't understand how you got those[/quote]
Do you me how I got the sum of the angles?
[hide]I am not so sure how well I can explain this without a picture, but here goes nothing...
to draw a x point star with all tips that are congruent, I started by drawing a regular polygon with x sides.
For example, to draw a 5 point star I would first draw a regular pentagon. Then, I would extend each side in both directions to make a star. Since I know that each interior angle of a regular pentagon is 108 degrees and that the base angle of the isoceles triangle that includes the tip of the star is supplement to 108 degrees, each tip must be 180-2(180-108) which is 36 degrees. Of course, this already answers the problem, but you can find the sum of all tips by multiplying 36 by5.[/hide]
Maybe someone else has a simpler way to explain this. Drawin a picture may also help. I'm going to try to draw one on my computer right now.


What positive number is equal to the sum of its square and its additive inverse?

$ x\equal{}x^2\minus{}x \implies x^2\minus{}2x\equal{}0 \implies x(x\minus{}2)\equal{}0 \implies x\equal{}0, 2$.  We are looking for a positive number so the answer is $ \boxed{2}$.


function
limit
calculus
calculus computations

I had a test today and I am having a dispute with my teacher over one of the answers, here is the problem:

lim cot(x)sin(x)
[size=75]x->0[/size]

I think that answer is 1 because

      cos(x)[color=white]................[/color]sin(x)
      -------[color=white].......[/color]X[color=white]......[/color]-------[color=white]...[/color]=[color=white]...[/color]cos(x)[color=white]....[/color]-->[color=white]....[/color]cos(0) = 1
      sin(x)[color=white]...................[/color]1

I might be totally wrong on how I did this but I also found several websites which agree with my answer, they were kinda  confusing in explaining the answer though.

You appear to be correct

one way to check a limit is to take a value very close to the limit
By  taking f(.001), it becomes clear that the limit is 1

AstroPhys: That logic is flawed

though it might work in that case, it will not always work

for example:
lim x-> 0

$\frac{1}{x^{2}}\cdot x$

$\frac{1}{0}\cdot 0$

$\infty \cdot 0$

but this limit is zero (from the obvious first step)

[quote="AstroPhys"]This maybe the wrong way to approach this[/quote]

Yeah thought so  :oops: .

ok thanks , so i assume that you agree that 1 is the limit then?

Yes, the limit of that is one when approaching zero.

If you're ever in doubt, just graph the function you're taking a limit of and see what the limit approaches at whatever number your approaching.

Or use a TI-89.

[quote="AstroPhys"]This maybe the wrong way to approach this

$\text{cot}\ 0 \ \cdot \ \text{sin}\ 0$
$\infty \cdot 0 \ \equiv \ 1$

Otherwise chickendude's method works better.[/quote]

I don't really think this belongs in HSB.

Note that this is really a form of L'Hopital's Indeterminate forms, so AstroPhys, you are not allowed to do that. What would you do if the limit was $10 \cdot \lim_{x\rightarrow 0}\cot(x) \sin(x)$? The answer is clearly $10$, but you get $1$ again.

top123's method is correct, but here is L'Hopitals rule:

$\lim_{x\rightarrow 0}\cot(x) \sin(x) = \lim_{x\rightarrow 0}\frac{\sin(x)}{\tan(x)}= \lim_{x\rightarrow 0}\frac{\cos(x)}{\sec^{2}(x)}= \frac{1}{1}= 1$

top123, why does your teacher disagree with you?

it's obviusly by Taylor  :lol: $cot(c)$~$\frac{1}{x}$ and $\sin(x)$~$x$

The simplest and best argument here was already stated by top123 in the very first post. There's no particular reason to elaborate any further.

[quote="1=2"]wait.. From which side does x approach 0?[/quote]
It doesn't matter in this case. This limit exists as a two-sided limit - as top123's argument shows.

oohh.. I thought that he wrote cot(x)COS(x)...

Yeah, top123 is right.

What did the teacher get?

for some reason he said infinity but when i asked him about it today he said he made a mistake and the answer was 1


Prove that $ abbcbb$ and $ abba$ aren't perfect squares.

What do you mean with $ abbcbb$ and $ abba$?

I mean the decimal representation of two numbers. The first one has six digits, and the second has four digits.

for the second one its easy to check that its divisible by $ 11$ therefore its $ (xx)^2$ which gives no solutions


geometry
3D geometry
tetrahedron
summer program
MathPath

From this year's mathpath quiz:
5. In a regular tetrahedron ABCD with edgelength 1, P is the midpoint of AB and Q is the midpoint of CD.
a)Find the lenght of PQ.
b)What angle does PQ made with plane BCD?

If it's from this year, you really aren't supposed to post the problems until after camp. It's just in case people are trying to apply late.

Yeah, but is well after the application deadline and the extension for nationals. :|

This is just what they said last year when someone posted a problem "well after the deadline".

I think it's safest to wait until after camp starts.

Lock this topic please, based on previous feedback from MathPath.

Now I really wish I was a moderator... ;)


Cum arat ca un domeniu de integritate nu poate fi izomorf cu produsul direct a 2 inele?

Nu prea stiu notiunile astea, asa ca am cautat pe Wikipedia. Sper ca am inteles bine. :blush:

Daca $A,B$ sunt inelele noastre, atunci $\left( 0_{A}, y \right) \cdot \left( x, 0_{B}\right) = \left( 0_{A}, 0_{B}\right)$ (unde $x \in A, \, y \in B$ sunt elemente nenule), deci $A \times B$ nu e domeniu de integritate. Cum proprietatea asta se pastreaza prin izomorfisme (pentru ca $f \left( 0_{A}, y \right)$ si $f \left( x, 0_{B}\right)$ sunt amandoua diferite de $0$, unde $f$ este o functie bijectiva care trimite $\left( 0_{A}, 0_{B}\right)$ in $0$), problema e gata.


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let a triangle $ ABC$. Denote $ D,E$ be two points on $ AB,AC$ respectively satisfy $ DE//BC$. $ P$ be a arbitrary point inside triangle $ ABC$. $ PB,PC$ respectively cut $ DE$ at $ F,G$. Denote $ O_1,O_2$ respectively be circumcentre of triangle $ PDG,PEF$.
Prove that: $ AP$ perpendicular with $ O_1O_2$

Easy problem:[img]http://img250.imageshack.us/img250/4720/tdlru6.png[/img]

$ AP\perp O_1O_2\Longleftrightarrow AP$ is a radical axe of $ O_1,O_2$.Denote $ R\in O_1\cap AB$ and $ N\in O_2\cap AC$.
So all we need to prove is that $ AD\cdot AR \equal{} AN\cdot AE$.
Since $ DE\|BC\Rightarrow AD\cdot AC \equal{} AE\cdot AB$.
Thus it is sufficiently to prove that
$ \frac {AN}{AR} \equal{} \frac {AB}{AC}\Longleftrightarrow$
$ BRNC$ is cyclic.
Now since $ \angle PRD \equal{} \angle PGD \equal{} \angle EGC \equal{} \angle GCB$.
Thus we must prove that $ RPNC$ is cyclic.But we know that $ BPNC$ is cyclic($ DE\|BC$),
hence it is sufficiently to prove that $ RPBC$ is cyclic.
The last one is obviously true because as we've already proved:$ \angle DRP \equal{} \angle PCB$.
Thus we are done.

Let $ J$, $ K$ be the intersections of $ AP$ with $ DE$, respectively $ BC$. By [b]Thales' theorem[/b], we have $ \frac {JF}{JG} \equal{} \frac {KB}{KC} \equal{} \frac {JD}{JE}$, it implies $ JF\cdot JE \equal{} JG\cdot JD$. Hence, we conclude that $ AP$ is the radical axis of the circumcircles of triangles $ PDG$ and $ PEF$, which completes our proof.