inequalities
function
inequalities solved

If $ a$, $ b$, $ c$ are positive reals, then prove that
$ \frac{a^{3}}{b^{2}\minus{}bc\plus{}c^{2}}\plus{}\frac{b^{3}}{c^{2}\minus{}ca\plus{}a^{2}}\plus{}\frac{c^{3}}{a^{2}\minus{}ab\plus{}b^{2}}\geq\frac{3\left(ab\plus{}bc\plus{}ca\right)}{a\plus{}b\plus{}c}$.

Rewrite our ineq as

Sum (b+c)a^3/(b^3+c^3) >= 3(ab+bc+ca)/(a+b+c)

Our problem reduced to more general result

"If a, b, c, x, y, z > 0 then   Sum a(y+z)/(b+c) >= 3Sum xy/Sum x" (*)

which once had been proven in forum. But the sln was very complicated, so I will give another proof (for (*)).

L.H.S(*) + 2 Sum x = (Sum a)*(Sum (y+z)/(a+b)) >= (1/2)(Sum sqrt(x+y))^2 (Cauchy)
We will show that
      (1/2)(Sum sqrt(x+y))^2 >= 3Sum xy/Sum x + 2 Sum x
It is equivalent to
     (Sum x + Sum sqrt(x^2+xy+xz+yz))*Sum x >= 3Sum xy + 2(Sum x)^2
By Minkowsky
     Sum sqrt(x^2+xy+xz+yz) >= sqrt((x+y+z)^2+9(xy+yz+zx))
it is sufficient to show that
     (Sum x)*sqrt((Sum x)^2 + 9 Sum xy) >= 3Sum xy + (Sum x)^2
which is trivial by squaring two sides.

Namdung

Splendid, magnificient, extraordinary! I can't believe, that super-hard problem from Crux has such an elegant solution. Namdung, you're great!
    I also have a solution for this problem, but I cannot say anything now, except:
   We will prove the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6551]stronger inequality[/url] sum a^3/(b^2-bc+c^2)>=sum a.
    By Cauchy we have sum a^3/(b^2-bc+c^2)>=(sum a^2)^2/[ sum a(b^2-bc+c^2)]. So, it's enough to prove that this quantity is at least sum a. This can be written sum a^4>= sum a(b+c)(b^2+c^2)-3abc(a+b+c) and also sum a^4+abc(a+b+c)>=sum a^3(b+c), or sum a^2(a-b)(a-c)>=0, which holds by Schur.

Really a WONDERFUL and GENIAL solution Namdung, for this very hard inequality; it is impossible not to congratulate you for this MATHEMATICAL GEM you discovered by your extraordinary mathematical skill and intuition.
Fantastic!!!!

[hide]
We multiply by $a/a, b/b, c/c$ and apply the Cauchy Schwarz lemma:

$\frac{a^4}{a(b^{2}-bc+c^{2})}+\frac{b^4}{b(c^{2}-ca+a^{2})}+\frac{c^{4}}{c(a^{2}-ab+b^{2})} \ge \frac{(a^2+b^2+c^2)^2}{\sum a^2b-3abc} \ge \frac{(a^2+b^2+c^2)(ab+bc+ca)}{\sum a^2b-3abc}$.

Now we show that $\sum a^2b-3abc \le \frac{1}{3}(a^2+b^2+c^2)(a+b+c)$.
Expand and rearrange terms to get
$\frac{2}{3}\sum a^2b \le \frac{1}{3}(a^3+b^3+c^3)+3abc$.
But $2(a^3+b^3+c^3) \ge \sum a^2b$ by rearrangement, so we need to show
$\frac{4}{3}(a^3+b^3+c^3) \le \frac{1}{3}(a^3+b^3+c^3)+3abc$, or
$a^3+b^3+c^3 \ge 3abc$ which is trivial by the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$

Thus we have
$\frac{(a^2+b^2+c^2)(ab+bc+ca)}{\sum a^2b-3abc} \ge \frac{(a^2+b^2+c^2)(ab+bc+ca)}{\frac{1}{3}(a^2+b^2+c^2)(a+b+c)} = \frac{3(ab+bc+ca)}{a+b+c}$.[/hide]

[quote="paladin8"]

$\frac{4}{3}(a^3+b^3+c^3) \le \frac{1}{3}(a^3+b^3+c^3)+3abc$, or
$a^3+b^3+c^3 \ge 3abc$ which is trivial by the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$
[/quote]

This is flipped?

[quote="paladin8"]Now we show that $\sum a^2b-3abc \le \frac{1}{3}(a^2+b^2+c^2)(a+b+c)$...

Expand and rearrange terms to get
$\frac{2}{3}\sum a^2b \le \frac{1}{3}(a^3+b^3+c^3)+3abc$.
But $2(a^3+b^3+c^3) \ge \sum a^2b$ by rearrangement, so we need to show
$\frac{4}{3}(a^3+b^3+c^3) \le \frac{1}{3}(a^3+b^3+c^3)+3abc$, or
$a^3+b^3+c^3 \ge 3abc$ which is trivial by the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$...
[/quote]

Actually, this lemma is reversed for $a$, $b$, $c$ sides of a triangle. For a quick counter-example, consider $a = b = 1, c = \epsilon$, giving $\frac{4}{3} \le \frac{2}{3}.$

Well, let's give a correct solution. We have to prove the inequality

$\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{c^2-ca+a^2}+\frac{c^3}{a^2-ab+b^2}\geq\frac{3\left(bc+ca+ab\right)}{a+b+c}$

for any three positive reals a, b, c. First we go sure that the denominators in this inequality are all positive: In fact, $b^2-bc+c^2=\left(b-c\right)^2+bc>0$, and similarly $c^2-ca+a^2>0$ and $a^2-ab+b^2>0$. Now, we straighten a bit the right hand side of the inequality: Since $\left(a+b+c\right)^2\geq 3\left(bc+ca+ab\right)$ (this is because

$\left(a+b+c\right)^2-3\left(bc+ca+ab\right)=\frac12\left(\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2\right)$,

what is clearly $\geq 0$), we have $a+b+c\geq\frac{3\left(bc+ca+ab\right)}{a+b+c}$. Thus, in order to prove the inequality

$\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{c^2-ca+a^2}+\frac{c^3}{a^2-ab+b^2}\geq\frac{3\left(bc+ca+ab\right)}{a+b+c}$,

it is enough to establish

$\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{c^2-ca+a^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c$.

But this inequality seems to be a rather sharp one, and the only proof I have is quite complicated:

We rewrite the inequality in question using the $\sum$ notation as follows:

$\sum\frac{a^3}{b^2-bc+c^2}\geq a+b+c$.

But $a+b+c=\sum\frac{b+c}{2}$; thus, our inequality becomes

$\sum\frac{a^3}{b^2-bc+c^2}\geq\sum\frac{b+c}{2}$.

This is equivalent to

$\sum\left(\frac{a^3}{b^2-bc+c^2}-\frac{b+c}{2}\right)\geq0$.

But

$\sum\left(\frac{a^3}{b^2-bc+c^2}-\frac{b+c}{2}\right)=\frac12\cdot\sum\left(\frac{a^3-b^3}{b^2-bc+c^2}-\frac{c^3-a^3}{b^2-bc+c^2}\right)$
$=\frac12\cdot\left(\sum\frac{a^3-b^3}{b^2-bc+c^2}-\sum\frac{c^3-a^3}{b^2-bc+c^2}\right)=\frac12\cdot\left(\sum\frac{a^3-b^3}{b^2-bc+c^2}-\sum\frac{a^3-b^3}{c^2-ca+a^2}\right)$
$=\frac12\cdot\sum\left(\frac{a^3-b^3}{b^2-bc+c^2}-\frac{a^3-b^3}{c^2-ca+a^2}\right)$
$=\frac12\cdot\sum\left(\frac{a^2+ab+b^2}{\left(b^2-bc+c^2\right)\left(c^2-ca+a^2\right)}\left(a-b\right)^2\left(a+b-c\right)\right)$.

Thus, we have to show that

$\sum\left(\frac{a^2+ab+b^2}{\left(b^2-bc+c^2\right)\left(c^2-ca+a^2\right)}\left(a-b\right)^2\left(a+b-c\right)\right)\geq 0$.

Upon multiplication with $\left(b^2-bc+c^2\right)\left(c^2-ca+a^2\right)\left(a^2-ab+b^2\right)$ (this is > 0, since we know that $b^2-bc+c^2>0$, $c^2-ca+a^2>0$ and $a^2-ab+b^2>0$), this inequality becomes

$\sum\left(\left(a^2+ab+b^2\right)\left(a^2-ab+b^2\right)\left(a-b\right)^2\left(a+b-c\right)\right)\geq 0$.

In other words, we have to prove the inequality

$\left(b^2+bc+c^2\right)\left(b^2-bc+c^2\right)\left(b-c\right)^2\left(b+c-a\right)+\left(c^2+ca+a^2\right)\left(c^2-ca+a^2\right)\left(c-a\right)^2\left(c+a-b\right)$
$+\left(a^2+ab+b^2\right)\left(a^2-ab+b^2\right)\left(a-b\right)^2\left(a+b-c\right)\geq 0$.

But this follows from Lemma 2 in http://www.mathlinks.ro/Forum/viewtopic.php?t=6138 post #4, if we WLOG assume that $a\geq b\geq c$, define the function $g:\mathbb{R}^+_0\to\mathbb{R}^+_0$ by the equation $g\left(t\right)=t^2$ (it is clear that this function is monotonically increasing on $\mathbb{R}^+_0$) and define the nonnegative reals $x=\left(b^2+bc+c^2\right)\left(b^2-bc+c^2\right)$, $y=\left(c^2+ca+a^2\right)\left(c^2-ca+a^2\right)$ and $z=\left(a^2+ab+b^2\right)\left(a^2-ab+b^2\right)$ (we have $x\leq y\leq z$ since $a\geq b$ yields $b-a\leq 0$, so that

$x-y=\left(b^2+bc+c^2\right)\left(b^2-bc+c^2\right)-\left(c^2+ca+a^2\right)\left(c^2-ca+a^2\right)=\left(a+b\right)\left(a^2+b^2+c^2\right)\left(b-a\right)$
$\leq 0$

and thus $x\leq y$, and similarly we can show $y\leq z$).

This completes the proof of our inequality.

I could bet there is a simpler way, but I don't really feel like searching for it now. Anyway, I am glad to have found another application of Siuhochung's nice inequality which he called an "ugly generalization" in http://www.mathlinks.ro/Forum/viewtopic.php?t=6138 .

  Darij

nice!!
It's really nice to see that my generalization has some application...
yesterday both yptsoi and me cannot think of any use of it..

Just a remark,
\[
\begin{gathered}
  \sum {\frac{{a^3 }}
{{b^2  + c^2  - bc}}}  \geqslant \frac{{\left( {\sum {a^2 } } \right)^2 }}
{{\sum\limits_{sym} {a^2 b}  - 3abc}} \hfill \\
  \frac{{\left( {\sum {a^2 } } \right)^2 }}
{{\sum\limits_{sym} {a^2 b}  - 3abc}} \geqslant a + b + c \hfill \\
   \Leftrightarrow \frac{1}
{2}\sum\limits_{sym} {a^4 }  + \frac{1}
{2}\sum\limits_{sym} {a^2 bc}  \geqslant \sum\limits_{sym} {a^3 b}  \hfill \\
   \Leftrightarrow \sum {a^2 (a - b)(a - c)}  \geqslant 0 \hfill \\
\end{gathered}
\]
Which is schur.

Oh yes, this is indeed much shorter! I actually had the same idea of applying Cauchy-Schwarz, but saw that paladin8 made a mistake with the same ansatz, so I gave it up...

By the way, I now see that a stronger inequality has been discussed before: http://www.mathlinks.ro/Forum/viewtopic.php?t=6551

  Darij

Strange, I thought to solve it using Cauchy-Schwarz(a very similar but easier inequality has a such solution) but couldn't finish it. Thank you all, guys.


[quote="buzzer11"]Is there a shortcut to this other than adding it up?  :help:[/quote]
i think there is. but i don't know the shortcut.


Princeton
college
summer program
PROMYS
HCSSiM

Hi,

I'm a college student at Princeton University and am interested in being a counselor for high school students at a summer math program, such as ones like PROMYS, math camp, and HCSSIM in the states.

However, I'm having trouble finding programs such as these. If the names of some programs in the UK could be provided, that would be greatly appreciated.

Thanks

you mean in the uk there aren't any summer mathematics camps for high school pupils? :O

Not really. Summer programmes aren't really a big thing over here.
There is however Headstart. They do engineering courses, etc. They're launching their first maths course this summer (I'm going) so yeah.
But there isn't anything really for maths. For other subjects though, yes.

Thanks :)

However, I saw that Headstart was for people currently in Year 12. I will be entering Year 12 [b]this[/b] fall...


You have 5 bales of hay. Instead of being weighed individually, they were weighed in all possible combinations of two: bales 1 and 2, 1 and 3, 1 and 4, 1 and 5, bales 2 and 3, and bales 2 and  4 etc.
The weights of each of these combinations were written down and  arranged in numeric order, without keeping track of which weight matched which pair of bales. The weights in kilos are 80,82,83,84,85,86,87,88,90,91.

How much does each bale weigh? Is there more than one possible set of weights?  Explain your answer.

HOW TO DO but NOT solution
use substitution. Find C, and then use substitution to figure out any other equations, such as smallest, second to smallest, heaviest, and second to heaviest.

[hide=""if u click here then um yeah u'll see the answer only use to chec ur answer.""]39,41,43,44,47[/hide]

Hope this helps.

[hide="how you do it"]you set them all up as equations and use deductive reasoning set up all the equations and find the [hide="answer"]39, 41, 43, 44, 47[/hide][/hide]

You start with the equations $B4 + B5 = 91$ and $B3 + B5 = 90$ and $B1 + B2 = 80$ and $B1 + B2 = 82$ and find out the difference between different numbers like B4 and B3 and B2.  Then you can make more equations on what you know about B4, B3, and, B2 to figure out the relationship between B1, B2, B3, B4, and B5 and you should be able to find out each number.

that made no sense to me. Can you re-explain in a simpler manner
Thanks

Trying to cheat your way to the top eh? I be seeing you at school eh. and im also going to be a cool icon eh  :starwars:  :ddr:


What is the value of $x$ when $4^3+6^2=10^x.$

[quote="math92"]use the hide thing...
[hide]
$4^3+6^2=10^x
4^3+6^2=100
100=10^x
\boxed{x=10}$[/hide][/quote]
You wrote it wrong, its supposed to be $\boxed{x=2}$ :P


$ a,b,c\geq 0$ prove:

$ \sum\frac{1}{a^2\plus{}b^2}\geq\frac{10}{(a\plus{}b\plus{}c)^2}$

[quote="zaya_yc"]$ a,b,c\geq 0$ prove:

$ \sum\frac {1}{a^2 \plus{} b^2}\geq\frac {10}{(a \plus{} b \plus{} c)^2}$[/quote]
It's obviously true after expanding.

[i]I think we can solve in that way ( excuse me , I don't have time to give in more details )

   Assume : $ 0 \leq c \leq b \leq a$

   denote : $ x \equal{} a \plus{} \frac{c}{2} ; y \equal{} b \plus{} \frac{c}{2} \rightarrow x ;y > 0$

  So $ a^2 \plus{} b^2 \leq x^2 \plus{} y^2 ; a^2 \plus{} c^2 \leq x^2 ; b^2 \plus{} c^2 \leq y^2$

   $ \rightarrow   (a\plus{}b\plus{}c)^2 \sum \frac{1}{a^2 \plus{}b^2 } \geq (x\plus{}y)^2 \left( \frac{1}{x^2 \plus{}y^2 } \plus{} \frac{1}{x^2} \plus{} \frac{1}{y^2 } \right)$

  So we need to prove : $ (x\plus{}y)^2 \left( \frac{1}{x^2 \plus{}y^2 } \plus{} \frac{1}{x^2} \plus{} \frac{1}{y^2 } \right) \geq 10$

   Can anybody help me to finish that solution  :lol: [/i]

[quote="nguyenvuthanhha"][i]
   Can anybody help me to finish that solution  :lol: [/i][/quote]

I'm not sure if it was sarcastic - but I'll finish the solution because I really like the substitution.  :)

Let $ M(t) \equal{} t \plus{} 1/t.$

Expanding one gets

$ \frac {2xy}{x^2 \plus{} y^2} \plus{} 2M(x/y) \plus{} M(x^2/y^2) \plus{} 3 \equal{} 2M(M(x/y)) \plus{} M(x^2/y^2)$

combining the first two terms.

M(t) has minimum (for t>0) at t=1 and increases for t>1, so $ M(x^2/y^2)\geq 2$ and $ 2M(M(x/y))\geq 2M(2) \equal{} 5$.  Adding 3,5,2 gives 10.

-evo

[i]Now , I have a few time to finish it  :lol: [/i]

 [i]  We need to prove $ (x\plus{}y)^2 \left(  \frac{1}{x^2 \plus{} y^2 } \plus{} \frac{2}{xy} \right) \geq 10$

  $ \iff \frac{2xy}{x^2 \plus{} y^2 } \plus{} \frac{2(x^2 \plus{} y^2 )}{xy} \geq 5$

  $ \iff \frac{2(x^2 \plus{} y^2 ) \minus{} 4xy}{xy} \minus{} \frac{x^2 \plus{} y^2 \minus{}2xy}{x^2 \plus{} y^2 } \geq 0$

    $ \iff  \frac{(x\minus{}y)^2 ( 2x^2 \minus{} xy \plus{} 2y^2 )}{xy(x^2 \plus{} y^2 )} \geq 0$ [/i]   [i]Done  :lol: [/i]

[quote="nguyenvuthanhha"][i]Now , I have a few time to finish it  :lol: [/i]

 [i]  We need to prove $ (x \plus{} y)^2 \left( \frac {1}{x^2 \plus{} y^2 } \plus{} \frac {2}{xy} \right) \geq 10$

  $ \iff \frac {2xy}{x^2 \plus{} y^2 } \plus{} \frac {2(x^2 \plus{} y^2 )}{xy} \geq 5$

  $ \iff \frac {2(x^2 \plus{} y^2 ) \minus{} 4xy}{xy} \minus{} \frac {x^2 \plus{} y^2 \minus{} 2xy}{x^2 \plus{} y^2 } \geq 0$

    $ \iff \frac {(x \minus{} y)^2 ( 2x^2 \minus{} xy \plus{} 2y^2 )}{xy(x^2 \plus{} y^2 )} \geq 0$ [/i]   [i]Done  :lol: [/i][/quote]

or
$ \iff \frac {2xy}{x^2 \plus{} y^2 } \plus{} \frac {2(x^2 \plus{} y^2 )}{xy} \geq 5$

$ \iff \frac {2xy}{x^2 \plus{} y^2 } \plus{} 4 \frac {(x^2 \plus{} y^2 )}{2xy} \geq 5$
the last one is obvious by AM-GM  for 5 numbers and then AM-GM  for 2 numbers :)


I want the best step-by-step guide to algebra on the net. I have a very basic knowlegde of algebra , however, I want to improve it to a advanced level by using one all encompassing all guide. I would be grateful for any help.

Personally, I think books are the best way to learn a topic. There aren't very many algebra(not abstract algebra, I am guessing) textbooks free on the internet. You'll have to shell out the money for a textbook. The art of problem solving volume 2 covers a bit of algebra, so you could try that. Or you can just borrow a book from some middleschool math teacher.

[quote="Whizz_kid"]Kalle : I'm looking for a link to a great algebra guide... did you even read my post?[/quote]

Obviously he did, otherwise he wouldn't be able to give such an accurate response :D Really, if you seriously want to learn it I think you're going to want a book.

Ok...I will make this nice and easy for you.

[b]DO YOU HAVE ANY LINKS TO GREAT ALEGBRA GUIDES?[/b]

http://www.google.com

It's the best link.. :)

Oh, and, DON'T USE CAPS!

you're in one right now, so i don't know what youre complaining about

there is no tell all site, you have to do the work in books and doing problems (like on this site) to get better, if you have really basic knowledge, you ought to go to getting started, or middle school forums and do problems there, like the daily contests


Let $ABCD$ be a convex quadrilater ${O}=AC\cap BD$ and ${E}=AD\cap BC$. We know that there exist a line $MN$, $M\in[AB],N\in[CE]$ and $M,O,N$ are colinear so that the triangles DON and $CEM$ have the same centroid.
a) Prove that $OA>OC$
b) Prove that if $OD>OB$, then $OA>2OC$
c) prove that if $AM>MB$ then $OA>3OC$
hope it's correct!

Hint:
[hide] if they have the same centroid, it means that $\overrightarrow{OD}+\overrightarrow{ON}=\overrightarrow{OC}+\overrightarrow{OE}+\overrightarrow{OM}$[/hide]
Tell me if you want to write the solution

Solution to the problem. I expect some comments from you.


geometry
parallelogram
trigonometry
similar triangles
geometry unsolved

Let $ K$ be a point inside the triangle $ ABC$. Let $ M$ and $ N$ be points such that $ M$ and $ K$ are on opposite sides of the line $ AB$, and $ N$ and $ K$ are on opposite sides of the line $ BC$. Assume that $ \angle MAB \equal{} \angle MBA \equal{} \angle NBC \equal{} \angle NCB \equal{} \angle KAC \equal{} \angle KCA$. Show that $ MBNK$ is a parallelogram.

The problem is very easy to solve with a little trig.

First, notation. Denote$ AB,BC,CA$ by $ c,a,b$ respectively. Denote $ AK\equal{}r$. As the triangles erected on each side of $ ABC$ are all similar, we easy get $ AM\equal{}\frac{cr}{b}$ and $ BN\equal{}\frac{ar}{b}$. We will prove that $ MK\equal{}BN$ and $ MB\equal{}NK$, using the cosine law.

$ MK^2\equal{}(\frac{cr}{b})^2\plus{}r^2\minus{}2\frac{cr^2}{b}\cos\hat{A}$
$ MK^2\equal{}r^2(\frac{c^2}{b^2}\plus{}1\minus{}\frac{2c}{b}\cos\hat{A})$

Now to prove that $ MK\equal{}BN$ we only need to verify that

$ \frac{c^2}{b^2}\plus{}1\minus{}\frac{2c}{b}\cos\hat{A}\equal{}\frac{a^2}{b^2}$

Clearing the denominators yield the cosine law for $ a$. Similarly we can prove that $ MB\equal{}NK$ so we are done. Note that we don't need $ K$ to be in the interior of $ ABC$.

Note the problem is very old and even stronger: On the sides of a triangle ABC one erects similar triangles ABM, BCN and ACK, with $ \angle MAB \equal{} \angle NBC \equal{} \angle KAC,  \angle MBA \equal{} \angle NCB \equal{} \angle KCA$ and B and K on the same side of AC, N and A on opposite sides of BC, C and M on opposite sides of AB. Then MBNK is a parallelogram.
The proof is extremely easy, just see some spiral similarities...

Best regards,
sunken rock


geometry
3D geometry
number theory
prime factorization

Find an n Natural, divisible by 17, that complies with the following conditions:

n/2 is a power of 2

n/3 is a power of 3

n/4 is a power of 5

Could someone tell me what do I need to learn to be able to solve this problem, and maybe solve it?
Thanks.

The problem statement is incorrect; $ \frac{n}{2}$ should be a square (that is, a second power) rather than a power of $ 2$, and so forth.

What you need to do is to become very familiar with [i]prime factorization.[/i]  You can look up this subject almost anywhere.

[hide]Are you sure the problem is correct? If $ \frac n2$ is a power of two, then we can write $ \frac n2 \equal{} 2^k$, with $ k$ a positive integer. That means $ n \equal{} 2^{k \plus{} 1}$. But now $ \frac n3$ can never be a power of $ 3$, $ \frac n4$ can never be a power of $ 5$ and most of all, $ n$ cannot be divisible by $ 17$.[/hide]

EDIT: Sorry, didn't see t0rajir0u's post there.

I am sorry I am translating from spanish, I meant a square, and a cube and I dont how to say the same for 5.

[hide="Hint"]If $ n$ is your number, let $ n \equal{} p_1^{e_1} p_2^{e_2} p_3^{e_3} \cdots$, where $ p_i$ is the $ i$th prime. $ n$ is a perfect $ k$th power if and only if $ k$ divides all $ e_i$.
[hide="Another hint"]This fact should allow you to conclude that $ n$ has $ 17^{30}$ in its prime factorization, among other things.[/hide][/hide]


What is the wavelength of the characteristics X-ray produced when an electron in the third orbit of a tungsten atom(Z=74) drops from the third shell to the first?

I guess, assume other electrons don't have any effect on this electron, thus, calculate the potential energy difference from the 3rd orbit to the 1st one. And use: E=hf=hc/wavelength


prove that n*(sin 180/n)=pi when n tends toward infinity

The limit is equivalent to
$\lim\limits_{n\to\infty} \frac{\sin(\pi/n)}{\pi/n} = 1$

Do you know that $\sin x / x \to 1$ as $x\to 0$?

let's turn the : "lim n-> infinity" in "lim N->0", with: N=1/n.

so, n sin (180/n) becomes: sin(180N)/N, which is the same as:

[sin (180N) - sin 0 ]/(N-0).

We can recognize, for lim N->0 a formula of the cosinus:

if f(x)=sin x, then f'(x)=cos x

Here,  lim N->0  [sin (180N) - sin 0 ]/(N-0) = cos 0 =pi

I apologize for my English and i can't give further explanations in English, i don't know the mathematic vocabulary.

[quote="icare"]We can recognize, for lim N->0 a formula of the cosinus:

if f(x)=sin x, then f'(x)=cos x

[/quote]

dear icare, this is based on $(\sin x)/x \to 1$ as $x\to 0$.

Yes, but your answer was a bit short for me, i just wanted to complete it. Don't you think it's too easy to solve a problem just by using one formula ?

:blush:  :blush:
My proof is invalid, indeed.

the limit shows the definition of $\pi$... but I assumed that $\pi = 180^\circ$.

[quote="beta"]Take a regular n-gon with radius r and side s. Since [tex]\frac{s}{2}=r\sin{\frac{180}{n}}[/tex]. [tex]s=2r\sin{\frac{180}{n}}[/tex]. Perimeter[tex]=ns=2rn\sin{\frac{180}{n}}[/tex]
Perimeter/Diameter [tex]=n\sin{\frac{180}{n}}[/tex]
As n approaches infinity, n-gon becomes a circle, Circumference/diameter[tex]=\pi[/tex]. QED[/quote]I got a very similar proof.  sin180/n (when n is a big number)is almost a straight line.  Therfore producing n of them will get a circle.  Using the circle formula, (2 :pi: r, we can conclude that the half circle generated is pi.


5. Show that if p is an odd prime and p|(x^(2^r)   +    1), then p = 1 (mod 2^(r+1))


This problem is in my ODE book and I don't see how it's supposed to be solved. It's in the section on separable DE's.

A substance contains two Radon isotopes, 210Rn (with half-life of 2.42 hours) and 211Rn (with half-life 15 hours). At first 20% of the decays come from 211Rn. How long must one wait until 80% of the do so?

Any help would be much appreciated.

Let $ x_{1}(t)$ and $ x_{2}(t)$ be the amounts of isotopes at time $ t$. You want to find $ t$ such that $ \frac{x_{2}(t)}{x_{1}(t)}\equal{}\frac{80}{20}\frac{x_{2}(0)}{x_{1}(0)}$. Solve the (separable) ODE for $ x_{1}$ and $ x_{2}$ and plug the solutions into the above equations, then solve for $ t$.


find all functions that F(F(X+Y))=F(X+Y)+f(X)F(Y)-XY

This is not High School Basics. Try Pre-Olympiad forum.

$ f(f(x\plus{}y))\equal{}f(x\plus{}y)\plus{}f(x)f(y)\minus{}xy$.
this problem is discussed before.
but I can`t find the proper link.

[url=http://www.mathlinks.ro/viewtopic.php?t=312067]Here[/url] it is. :)


Four seconds pass between the first and third flash of a strobe light.
The rate at which the strobe flashes is constant. How many seconds
will pass between the first and the twelfth flash of the same light?

first - second - third => 2 margins

so 1 margin = 2 seconds

1~12 = 11 margins

2*11=22

answer : 22


Mafia
geometry
geometric transformation
reflection
\/closed

1) It would be helpful to split up the existing "Main Games" Forum into the following:

> Marathon Games, games that have no defined end (See Genetic Mutation, Scrabble, The Word Game, Revival - Mental Dictionary Game)
> Game Discussions, discussion about existing games (See Super Smash Bros. Brawl Discussion Thread, eRepublik - A New World), game development, and theoretic aspects about existing forum games.
> Theme (? A better word is appreciated) Games, games with signups and have ends (See Half The Average, Beat the Mod, Big Brother, The Mole)
> The existing Mafia extension
> The existing FF extension, which could possibly be subdivided into marathon and discussion groups yet again (theme games are almost never frivolous)

The Replacements Sticky would be relocated to the Theme Thread.

2) A more active moderation.

Topics like "ULTRA HARD Problem Marathon" need to be relocated to the Middle School Tournaments page, also removing some frivolity in the title.
Topics like "The Pi Game" need to be relocated to the Fun Factory, as this is a pretty frivolous game.
Conversely, I believe "The Fight against The Dark Lord" could be moved into the Games forum; it is a fairly organized game.
The Mafia forum is at a standstill, as well. There should be more moderation.

I think the current (de facto) 3 division system is fine.

Mafia: for mafia games
Fun Factory: for frivolous games
Games: for nonfrivolous games

Game discussion isn't frequent enough to really deserve any special recognition, but currently it goes in "games" (in fact "game discussion" is the official description of the "games" forum, but of course that doesn't really reflect reality).

I agree that new local mods are needed.

I totally agree with James. Now, the whole Games forum is turning into FF, and who wants that?
And, I think that all marathon games should be dispatched to FF.

About Mafia: we could get a new moderator there (see mafia mods list what's the problem), I recommend iLord or Brut3Forc3 if they want to...

No, I believe you're really only using relative comparison here. Only meaningless or near-meaningless marathons belong in the fun factory. Also, without regards to the forum it is in, the Fun Factory definitely has much more that would be defined as "spam" in a regular forum than the Games Forum.

[quote="Bugi"]Now, the whole Games forum is turning into FF, and who wants that?
And, I think that all marathon games should be dispatched to FF.[/quote]
[quote="monkeygirl12"]I have to agree with the people above games is turning into fun factory again there is too many marthon threads in games all marthon games belong in fun factory, these are not thread from before the division of fun factory but new and recent threads. I do have to say fun factory has little spam these days and it is well monitered.[/quote]
And these problems are solved by having active moderators. Not more subdivisions.


geometry incenter trigonometry circumcircle vector rectangle ratio

Given a triangle ABC with [tex]\angle B=30^{\circ}[/tex] and AC is the shortest side. Points D and E are on BA and BC respectively such that DA=EC=AC. Prove that OI is perpendicular and equal to DE, where O and I are the circumcentre and incentre respectively.

I have solved it using the formula [tex]d^2=R(R-2r)[/tex] and so use sine law to deduce the two lines are perpendicular. However, I think it is too troublesome and I believe that there is a simpler proof. Would someone work it out?

I think this was a Crux problem some years ago. I will prove a more general result:

[i]Let ABC be an arbitrary triangle, and D and E two points on the halflines AB and CB such that AD = CE = CA. Prove that $OI\perp DE$ and $DE=2\sin B\cdot OI$, where O is the circumcenter and I is the incenter of triangle ABC.[/i]

Here is a quite smart proof of this (not from me; Peter surely will remember my loooong proof using vectors, but this one is much simpler):

We will assume that b = CA is the shortest side of the triangle ABC, but in all other cases the proof will work as well, with some slight modifications.

The circumcenter O of the triangle ABC lies on the perpendicular bisectors of the segments BC, CA, AB. Let D' and E' be the orthogonal projections of the point I on the perpendicular bisectors of the segments BC and AB. Also, let A' and C' be the midpoints of the segments BC and AB; and let X and Z be the orthogonal projections of the point I on the lines BC and AB, i. e. the points where the incircle of triangle ABC touches the sides BC and AB.

It is well-known that BX = s - b, where $s=\frac{1}{2}\left( a+b+c\right)$ is the semiperimeter of the triangle ABC. Thus,

$XA^{\prime }=BX-BA^{\prime }=\left( s-b\right) -\frac{1}{2}a=\left( \frac{1}{2}\left( a+b+c\right) -b\right) -\frac{1}{2}a=\frac{1}{2}\left( c-b\right) $.

Since the quadrilateral IXA'D' is a rectangle (it has three right angles: at X, at A' and at D'), we have ID' = XA', so $ID^{\prime }=\frac{1}{2}\left( c-b\right) $. Now we have BD = AB - AD = AB - CA = c - b; therefore, $ID^{\prime }=\frac{1}{2}\left( c-b\right) =\frac{1}{2}\cdot BD$. Similarly, $IE^{\prime }=\frac{1}{2}\cdot BE$. Finally, we have $D^{\prime }I\perp A^{\prime }D^{\prime }$ and $A^{\prime }D^{\prime }\perp BC$, i. e. $A^{\prime }D^{\prime }\perp EB$, so that D'I || EB. Similarly, E'I || DB. Hence, the angle < D'IE' is equal to < DBE, since the sides of the two angles are respectively parallel. Together with $ID^{\prime }=\frac{1}{2}\cdot BD$ and $IE^{\prime }=\frac{1}{2}\cdot BE$, this shows that the triangles DBE and D'IE' are similar, with the ratio of similitude $\frac{1}{2}$. Hence, $D^{\prime }E^{\prime }=\frac{1}{2}\cdot DE$.

Now, since < OD'I = 90° and < OE'I = 90°, the points D' and E' lie on the circle with diameter OI. In other words, the circumcircle of triangle D'IE' has the segment OI as diameter. Therefore, after the extended law of sines, $D^{\prime }E^{\prime }=OI\cdot \sin \measuredangle D^{\prime }IE^{\prime }$. Hence we have

  $\frac{1}{2}\cdot DE=D^{\prime }E^{\prime }=OI\cdot \sin \measuredangle D^{\prime }IE^{\prime }=OI\cdot \sin \measuredangle DBE=OI\cdot \sin B$,

and thus $DE=2\cdot OI\cdot \sin B=2\sin B\cdot OI$. This already proves one part of our assertion. Remains to show $OI\perp DE$.

Now, we will work with directed angles modulo 180°. Actually, the triangles DBE and D'IE' are not just similar, they are inversely similar (since the angles < DBE and < D'IE' are oppositely equal, if we consider them as directed angles). Hence, < ID'E' = - < BDE. But since the points D' and E' lie on the circle with diameter OI, we have < IOE' = < ID'E', so < IOE' = - < BDE. Hence,

  < (OI; OE') = < IOE' = - < BDE = < EDB = < (DE; AB),

and thus

  < (OI; DE) = < (OI; OE') + < (OE'; AB) + < (AB; DE)
             = < (OI; OE') + < (OE'; AB) - < (DE; AB)
             = < (OE'; AB) = 90°

(since $OE^{\prime }\perp AB$). Therefore, $OI\perp DE$, qed..

  Darij

Note that in the case where <B=30 (not in the general case!), we can also prove that I is the orthocentre of ODE. This case is much easier than the general case, mainly because here we have OAC is equilatrel.
First, the triangles EIC, AIC and AID are congruent. Since <AIC=105, then <AID=<CIE=105 and <DIE=45. For proving IO is perpendicular to DE, it is sufficient to prove OE is perpendicular to ID, i.e. <OEI=45. Since O is the circumcentre of ABC, <AOC=2*<ABC=60 and AOC is equilatrel and AO=AC=OC=EC=DA so OEC is iscoceles. <OCE=<OCB=90-A, so
<CEO=(90+A)/2. But <IEC=<IAC=A/2, so <IEO=(90+A)/2 - A/2=45. So it follows O is the orthocentre of IDE. Proving IO=DE is equivalant to the following result:
Let XYZ be a triangle, H its orthocentre, with <YXZ=45. Then XH=YZ.
This is easy since HA=a/tanA.

I just got a short proof of the indentity darij grinberg posted.

Let $AI$ intersects the circumcircle of $ABC$ at $A'$ and AA' cuts CD at F. It's obvious that $OA'$ is perpendicular to $BC$ and AA' is perpendicular to CD. So

$<OA'B=<OA'C=<BA'C/2=(180-<BAC)/2=<ADC=<ACD$.

Hence

$<OA'I=<OA'B-<IA'B=<ACD-<ACB=ECD$ --- (1)

Also since

$\frac{EC}{OA'}=\frac{b}{R}=2\sin B=2\sin <FA'C=2\frac{CF}{CA'}=\frac{CD}{A'I}$ --- (2)

, from (1) and (2) we get triangle $ECD$ is similar to triangle $OA'I$.

Since $OA'$ is perpendicular $EC$, so $DE$ is perpendicular to $OI$ and $DE=2\sin B\cdot OI$.

[quote="Leonhard Euler"][quote="darij grinberg"]I think this was a Crux problem some years ago. I will prove a more general result:

$ XA^{\prime }=BX-BA^{\prime }=\left( s-b\right)-\frac{1}{2}a=\left( \frac{1}{2}\left( a+b+c\right)-b\right)-\frac{1}{2}a=\frac{1}{2}\left( c-b\right)$.

[/quote]

Does $ XA^{\prime }=BX-BA^{\prime }$ always true?[/quote]

Actually, it's rather $ XA^{\prime }=\left|BX-BA^{\prime }\right|$. I wrote the above referring to my sketch, where $ BX\geq BA^{\prime}$. If $ BX\leq BA^{\prime}$, then the proof needs some slight modifications (but is analogous). One day I will learn Asymptote and start attaching graphics to my posts...

  Darij

[color=darkred][b][u]Enunciation.[/u][/b] Let $ ABC$ be a triangle with the circumcircle  $ w=C(O,R)$ and the incircle $ C(I,r)$.

Define the points $ \{\begin{array}{c}D\in (AB\ ,\ E\in (CB\\\ AD = CE = CA\end{array}$. Prove that $ OI\perp DE$.[/color]

[color=darkblue][b]Proof.[/b] Denote $ p_{w}(X)$- the [b]power of the point $ X$ w.r.t. the circle $ w$[/b]. Thus, $ \{\begin{array}{c}p_{w}(D)\equiv OD^{2}-R^{2}=-b(c-b)\\\ p_{w}(E)\equiv OE^{2}-R^{2}=-b(a-b)\end{array}\|$ $ \implies$ $ \boxed{OD^{2}-OE^{2}=b(a-c)}$.

The relations $ IA^{2}=\frac{bc(p-a)}{p}$ a.s.o. are well-known. Apply the [b]Stewart's theorem[/b] to the rays $ (ID$, $ (IE$ in $ \triangle IAB$, $ \triangle IAC$ respectively :

$ \{\begin{array}{c}b\cdot IB^{2}+(c-b)\cdot IA^{2}=c\cdot ID^{2}+bc(c-b)\\\ b\cdot IB^{2}+(a-b)\cdot IC^{2}=a\cdot IE^{2}+ab(a-b)\end{array}\|$ $ \implies$ $ \boxed{ID^{2}-IE^{2}=b(a-c)}$. In conclusion, $ OD^{2}-OE^{2}=ID^{2}-IE^{2}$, i.e. $ OI\perp DE$.[/color]


Prove That:
$ \frac {x^2}{(x\minus{}1)^2}\plus{}\frac {y^2}{(y\minus{}1)^2}\plus{}\frac {z^2}{(z\minus{}1)^2}\geq 1$
for all real numbers $ x, y, z$, each different from $ 1$, and satisfying $ xyz \equal{} 1$.

This is 2008 IMO Problem 2.

Should be locked.

See here. [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=215222[/url]

Japanese Communities Moderator

kunny


algebra
polynomial
quadratics
algebra solved

Let a, b, c be roots of polynomial  $ x^3 - 9x + 9 $. Show that $a^2 + a - 6$ is equal to b or c.

=====
Namdung

Very nice and very neat proof. Congratulations!

Actually the most direct way is to define

$f\left( x\right) =x^{3}-9x+9$

and then see (by factoring) that

$f\left( x^{2}+x-6\right) =\left( x^{3}+3x^{2}-6x-17\right) \left( x^{3}-9x+9\right) $.

Thus, if a is a zero of f(x), then $a^{2}-9a+9=0$, and hence $f\left( a^{2}+a-6\right) =\left( a^{3}+3a^{2}-6a-17\right) \left( a^{3}-9a+9\right) =0$, too, so that $a^{2}+a-6$ is also a zero of f(x). Hence, $a^{2}+a-6$ is one of the numbers a, b, c. But it can't be a, otherwise we would have $a^{2}+a-6=a$, so $a^2 = 6$, what is not possible (since $\sqrt{6}$ and $-\sqrt{6}$ are not zeros of f(x)). Hence, $a^{2}+a-6$ must be either b or c.

I am interested in a more general problem: For which cubic polynomials $p\left( x\right) =x^{3}+ax^{2}+bx+c$ does there exist a quadratic polynomial $q\left( x\right) =x^{2}+Ax+B$ such that p(q(x)) is divisible by p(x) ?

  Darij


geometry
circumcircle
angle bisector
geometry proposed

Let the points $D$ and $E$ lie on the sides $BC$ and $AC$, respectively, of the triangle $ABC$, satisfying $BD = AE$. The line joining the circumcenters of the triangles $ADC$ and $BEC$ meets the lines $AC$ and $BC$ at $K$ an $L$, respectively. Prove $KC = LC$.

Copy and past:

Assume that the circumcircles of the triangles $ADC$ and $BEC$ meet at $C,P$. The problem is to show that the line $KL$ makes equal angles with the lines $AC$ and $BC$. Since the line joining the circumcircles of thr triangles $ADC$ and $BEC$ is perpendicular to the line $CP$, it suffices to show that $CP$ is the angle-bisector of $\angle ACB$. Since the points $A,P,D,C$ are concyclic, we obtain $\angle EAP=\angle BDP$. Analogously,we have $\angle AEP=\angle DBP$. These two equalities together with $AE=BD$ imply that triangles $APE$ and $DPB$ are congruent. This means that the distance from $P$ to $AC$ is equal to the distance from $P$ to $BC$, and thus $CP$ is the angle-bisector of $\angle ACB$ ,as desired.


Let $ C$ be a circle with center $ O$ and $ P$ be a point outside that circle. Let $ N\in C$ be a mobile point. Let $ M\in C$ be the point such that line $ OP$ is a symmedian, but not a median of $ \triangle PMN$ [provided that such a point exists]. Prove that the circumcircle of $ \triangle PMN$ passes through a fixed point.

we know that A-symedian pass through the tangents line at B and C to the circumcircle of triangle ABC.
back to the problem cause OP is symmedian and $ OM\equal{}ON$ hence OM and ON are tangent to circumcircle of $ MNP$ let OP intersect with circumcircle of MNP at Q we have:$ OM^{2}\equal{}OP.OQ$ hence Q is constant point.


algebra
partial fractions
calculus
calculus computations

Find the following inverse Laplace transforms:

$ F(s) \equal{} \frac {1}{(s \plus{} 2)^3}$

$ F(s) \equal{} \frac {s}{s^2 \plus{} 4s \plus{} 5}$

Is there an easy way to find this inverse?
formual etc?

There's a table and some basic formulas.

The inverse Laplace transform of $ \frac1{s\minus{}a}$ is $ e^{at}.$

The inverse Laplace transform of $ \frac1{(s\minus{}a)^k}$ is $ \frac{t^{k\minus{}1}e^{at}}{(k\minus{}1)!}.$

For your second example, use partial fractions to split it into to terms you can handle.


I"m searching some book named  [b]'Equations and Inequalities'[/b] for [b] PDF[/b] file
where can I find it????????????????????????

try in gigapedia.com...i remember it's a djvu file and not pdf...and BTW, I'm not authorized to post here the full link of the ebook 'coz it's highly discouraged here in mathlinks..just continue searching or "googling"...before, I only have the ebook but now I've my own book..


So I am studying graph and modelling now and I found this equation:

yx-3x+1=0

How do I make y as the subject? I need it to sketch the hyperbola graph.

Thanks.

Hint
[hide]This is $ y \equal{} 3 \minus{} \frac{1}{x}$. Once you know how to draw $ y\equal{}\frac{1}{x}$, you should be able to draw this.[/hide]


geometry
3D geometry
trigonometry
AMC
AIME
trig identities
Law of Cosines

A right circular cone has a base with radius 600 and height $200\sqrt{7}$. A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is 125, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is $375\sqrt{2}$. Find the least distance that the fly could have crawled.

[hide]its helpful to picture the cone as part of a circle folded up.
the radius of this circle would be $200* \sqrt{3^2+\sqrt{7}^2} = 800$.
Since the cone has a base of radius $600$, the circumference is $1200\pi$.
However, the original circle had a circumference of $1600 \pi$
Thus we have $\frac 34$ of the original circle.  (refer to diagram)
Two points will be on opposite sides of the cone if one of them lies along one of the pieces that fold together, and the other one is exactly halfway between those pieces.
The fly starts at point $A$, and ends up at point $B$.  We wish to find the distance $AB$.
Since we know it is exactly $\frac 34$ of a circle, angle $AOB$ is $135^\circ$, and we are given $AO = 125$ and $BO = 375\sqrt{2}$.
Applying law of cosines, $AB^2 = AO^2 + BO^2 - 2(AO)(BO)\cos{AOB}$.
Plugging in values: $AB^2 = 125^2 + (375\sqrt{2})^2 - 2(125)(375\sqrt{2})\cos{135^\circ}$.
pull out as many 25's as possible: $AB^2 = 25^3(1 + 2*3*3 + 2*3\sqrt{2} * \frac{\sqrt{2}}{2}) = 25^4$.
Thus $AB = 25^2 = \boxed{625}$.[/hide]

is it just me, or was this the easiest 10-15 AIME problem this decade?

Problem of minimum distance travelled is one of the most interesting calculations. The problems given here is very difficult to solve. In neutral geometry, the distance between two points x1 and x2 is given by the equation d^2=x2^2-x1^2. The distance between two points are given by analytic geometry. We can solve distance problems using different equations. There are different methods to solve the distance problems. Algebraic distance describes different distance solving equations and methods.


linear algebra
matrix
Euler
function
group theory
abstract algebra
number theory

Let $ m$ be a positive integer, and define $ G$ as the set of all matrices of the form:
$ \begin{bmatrix}1 & b\\ 0 & a\end{bmatrix}$
where $ a$ and $ b$ are elements of the ring of integers ( mod $ m$), and $ a$ is relatively prime to $ m$. Show that:

$ (a)\;\;G\;$is a group under matrix multiplication, and $ G$ has order  $ m\phi(m)$, where $ \phi(m)$ is the Euler $ \phi$-function;

$ (b)$ if $ m$ is an odd prime, then each automorphism of $ G$ is an inner auto morphism;

$ (c)$ if $ m \equal{} 8$, then the mapping of $ G$ into itself defined by

$ \begin{bmatrix}1 & b\\ 0 & a\end{bmatrix}\equal{}\begin{bmatrix}1 & b\plus{}\frac{1}{2}\left(a^{2}\minus{}1\right)\\ 0 & a\end{bmatrix}$

is an outer automorphism of $ G$ which maps each conjugacy class of $ G$ onto itself.

For a),b).
Let $ A_{a,b}\equal{}\begin{pmatrix}1&b\\0&a\end{pmatrix},\phi_{a,b}: z\in\mathbb{Z}_{m}\rightarrow{az\plus{}b}\in\mathbb{Z}_{m}$ where $ a\in\mathbb{Z}_{m}^{*}$ and $ b\in\mathbb{Z}_{m}$.
$ A_{a,b}A_{c,d}\equal{}\phi_{c,d}\phi_{a,b}$ then $ G$ is isomorphic to the group of invertible affine transformations over $ \mathbb{Z}_{m}.$ $ card(\mathbb{Z}_{m}^{*})\equal{}\phi(m)$ (it isn't the same $ \phi$) then $ G$ has $ m\phi(m)$ elements.
If $ m$ is prime then $ G$ is the Frobenius group $ F_{m(m\minus{}1)}(m)$, cf:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137762.
We consider $ G$ as a subgroup of $ S_{m}$.
$ G\equal{}KH$ where: 1)the complement  $ H$ is the subgroup fixing $ m$ and has $ m\minus{}1$ elements.$ H\equal{}[\tau]$ with $ \tau\equal{}\{1,\alpha,\cdots,\alpha^{m\minus{}2}\}$ and  $ \alpha$ a generator of $ \mathbb{F}_{m}^{*}$.
2) The kernel $ K\equal{}[\sigma]$ ($ \sigma\equal{}(1,\cdots,m)$) contains $ id$ and the elements which aren't conjugate to $ H$-elements. $ K$ has $ m$ elements.
Remark 1: $ Z(G)\equal{}\{id\}$ then $ Inn(G)$ is isomorphic to $ G$.
Remark 2: $ K\minus{}\{id\}\equal{}\bigcup_{x\in{H}}{x\sigma{x^{\minus{}1}}}$.
          Let $ u\in{Aut(G)}$; $ u$ is defined by $ u(\sigma),u((\tau)$:
$ u(\tau)$ is in a $ H$-conjugate; then $ u(\tau)\equal{}e\tau{e^{\minus{}1}}$. Let $ E\equal{}\{f|e\tau{e^{\minus{}1}}\equal{}f\tau{f^{\minus{}1}}\}$. If $ f\in{E}$ then $ e^{\minus{}1}f$ commutes with $ \tau$ then fixes $ m$; thus $ e^{\minus{}1}f\in{H}$ and $ f\equal{}e\tau^{r}$.
$ u(\sigma)$ is of order $ m$ then $ u(\sigma)\equal{}\sigma^{s}\not\equal{}{id}.$
It remains to prove that there exists $ r\in\{0,\cdots,m\minus{}2\}$ s.t. $ e\tau^{r}\sigma(e\tau^{r})^{\minus{}1}\equal{}\sigma^{s}$ or s.t. $ \tau^{r}\sigma\tau^{\minus{}r}\in{K\minus{}\{id\}}$.
We conclude with Remark 2.

It remains to prove that $ u(\tau)\equal{}e\tau{e^{\minus{}1}}$. I have only $ u(\tau)\equal{}e\tau^{k}{e^{\minus{}1}}$ where $ gcd(k,m\minus{}1)\equal{}1$.
In fact I don't see why we couldn't choose for example: $ u(\sigma)\equal{}\sigma,u(\tau)\equal{}\tau^{\minus{}1}$.

What is a Frobenius group? Is it defined as the group of x^q -2 over the rationals?

Yes it's isomorphic to the Galois group of $ x^{m}\minus{}2$ or to the group of the affine transformations (cf the last post of the reference). The definition of this group is given also in the reference; it's also a semi direct product of $ F_{m}^{*}$ and $ F_{m}$.
cf. wikipedia for the details.

It's easier if we consider  that $ G\equal{}\{\phi_{a,b}|a\in\mathbb{F}_{m}^{*},b\in\mathbb{F}_{m}\}$. Let $ \alpha$ be a $ \mathbb{F}_{m}^{*}$-generator and $ u$ be a $ G$-automorphism. Let $ u(\alpha{z})\equal{}\gamma{z}\plus{}\delta$; $ u(\alpha^{k}z)\equal{}u(\alpha{z})^{k}\equal{}\gamma^{k}z\plus{}(1\minus{}\gamma^{k})(1\minus{}\gamma)^{\minus{}1}\delta\not\equal{}z$ if $ k\in\{1,\cdots,m\minus{}2\}$ and $ u(\alpha^{m\minus{}1}{z})\equal{}u(z)\equal{}z$. Thus $ \gamma^{m\minus{}1}\equal{}1,\gamma^{k}\not\equal{}1$ and $ \gamma$ is a $ \mathbb{F}_{m}^{*}$-generator.
Let $ u(z\plus{}1)\equal{}\epsilon{z}\plus{}\mu$; then $ u(z\plus{}m)\equal{}u(z)\equal{}z\equal{}u(z\plus{}1)^{m}\equal{}\epsilon^{m}{z}\plus{}?\equal{}\epsilon{z}\plus{}?$; then $ \epsilon\equal{}1,\mu\not\equal{}0$. Thus $ u(z\plus{}c)\equal{}z\plus{}c\mu$.
$ u(\phi_{\alpha,0}\phi_{1,1})\equal{}u(\phi_{\alpha,0})u(\phi_{1,1})$; therefore $ \gamma{z}\plus{}\delta\plus{}\alpha\mu\equal{}\gamma{z}\plus{}\gamma\mu\plus{}\delta$ and $ \gamma\equal{}\alpha$
Conclusion: $ u(az\plus{}b)\equal{}u(\alpha^{k}z\plus{}b)\equal{}u(\alpha{z})^{k}\plus{}b\mu\equal{}\alpha^{k}z\plus{}(1\minus{}\alpha^{k})(1\minus{}\alpha)^{\minus{}1}\delta\plus{}b\mu\equal{}az\plus{}(1\minus{}a)\delta(1\minus{}\alpha)^{\minus{}1}\plus{}b\mu.$
$ u$ is the inner automorphism: $ \phi_{a,b}\rightarrow{Y^{\minus{}1}\phi_{a,b}Y}$ with $ Y(z)\equal{}\mu^{\minus{}1}z\minus{}\mu^{\minus{}1}\delta(1\minus{}\alpha)^{\minus{}1}$.

Remark: the gap in my first post ($ u(\tau)$ and $ \tau$ are conjugate) is equivalent to prove that $ u(\alpha{z}\plus{}b)$ is in the form $ \alpha{z}\plus{}?.$

Hi sinajackson.
for c):   $ 1/2(a^{2}\minus{}1)$ isn't well defined; for example if $ a\equal{}5$ then $ 1/2(a^{2}\minus{}1)\equal{}0$ or $ 4$.

Hi loup blanc, well now that you say this, not sure about part c), but nice solution for part b) :) , another solution for b):

In this case, $ G$ has order $ p(p\minus{}1)$, and the matrices whose diagonal entries are equal to $ 1$ form a characteristic subgroup of order $ p$. Put

$ u\equal{}\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix},\;\;\;\;,v\equal{}\begin{bmatrix}1 & 0\\ 0 & a\end{bmatrix},\;\;\;\;\;\text{and}\;\;\;\;w\equal{}\begin{bmatrix}1 & 0\\ 0 & 2\end{bmatrix}$

and suppose that $ \alpha\in\text{Aut }G$. Then we can write:

$ u^{\alpha}\equal{}\begin{bmatrix}1 & k\\ 0 & 1\end{bmatrix}\;\;\;\;\text{and}\;\;\;\;v^{\alpha}\equal{}\begin{bmatrix}1 & t(a)\\ 0 & a^{\prime}\end{bmatrix}$

for some integer $ k$, $ 1\leq k\leq p\minus{}1$, and some integers $ t(a)$ and $ a^{\prime}$ with $ p\not|a^{\prime}$. Applying $ \alpha$ to the identities $ uv\equal{}vu^{\alpha}$ and $ vw\equal{}wv$, we find that $ a^{\prime}\equal{}a$ and $ t(a)\equal{}(a\minus{}1)t(2)$. Thus, by direct calculation, $ x^{\minus{}1}ux\equal{}u^{\alpha}$ and $ x^{\minus{}1}vx\equal{}v^{\alpha}$ where

$ x\equal{}\begin{bmatrix}1 &\minus{}t(2)\\ 0 & k\end{bmatrix}.$


[quote="Rust"]It followed from ${n+1\choose k+1}={n\choose k+1}+{n\choose k}$ by induction.[/quote]
Please, post a complete solution ! Thanks/Alex


This forum is for AMC and I feel kind of stupid because there isn't much information about AMC8.....

well,

Could anyone give me some tips to do good in AMC8?

and could anyone make mockAMC8 ?

( I'm actually posting it because I'm probably the only person in AoPS who got 13 out of 25 in AMC8)

This website has quite a few AMC 8 questions: http://www.unl.edu/amc/e-exams/e4-amc08/archive8.html.

You'll also probably want to work through all the problems in the AoPs books.

Hm....

I solved all the practice problems before....

so, I guess I will have to work on AoPS textbook...

It's alittle difficult for me...

lol, if someone can do all of the problems without difficulty, I'm guessing they should be working on their AIME score, not their AMC8 score.  :)


I don't know if this has been done before.

Where's the combinations of two of those? I have 4 hard disks with linux installed, 1 with os x only, I admin a windows machine, and I am installing Solaris and BSD as soon as I have time and curing that os x only disk with a dose of open-source goodness.

[quote="hunter34"]Where is mozilla firefox?[/quote]
Mozilla Firefox is a web browser, not an operating system.

If you have a combination, please list your favorite.

[quote="#H34N1"]PC == Windows?[/quote]Not literally, but Windows are commonly referenced to as PC's, yes.

Who ever said you can't have Linux on a PC? I don't have it yet, but eventually will.

Exactly as i_like_pie said.

I have 3 HDD's (gonna get a fourth one), and 1 has windows, 1 suse linux, and 1 redhat.


IMO
equation
number theory
Diophantine equation
Divisibility
IMO 1982

Prove that if $n$ is a positive integer such that the equation \[ x^3-3xy^2+y^3=n \] has a solution in integers $x,y$, then it has at least three such solutions. Show that the equation has no solutions in integers for $n=2891$.

If $x, y$ is a solution then so is $y-x$, $-x$. Hence also $-y$, $x-y$.
If the first two are the same, then $y = -x$, and $x = y-x = -2x$, so $x = y = 0$,
which is impossible, since $n >0$. Similarly, if any other pair are the same.

$2891 = 2 (mod.9)$ and there is no solution to $x^3 - 3xy^2 + y^3 = 2 (mod 9)$.
The two cubes are each $-1$, $0$ or $1$, and the other term is $0$, $3$ or $6$,
so the only solution is to have the cubes congruent to $1$ and the other term congruent to $0$.
But the other term cannot be congruent to $0$, unless one of $x$, $y$ is a multiple of $3$, in which case its cube is congruent to $0$, not $1$ or $-1$.

$ (x \plus{} \omega y)^3 \equal{} x^3 \minus{} 3xy^2 \plus{} y^3 \plus{} \omega(3x^2y \minus{} 3xy^2)$

But $ a^3 \equal{} (a\omega)^3 \equal{} (a\omega^2)^3$, so, if there is a solution, we obtain other two.

For $ n \equal{} 2891$, we see the equation $ mod 11$. We must have $ x^3 \minus{} 3xy^2 \plus{} y^3 \equiv 2891 \equiv 0$. If $ 11 \nmid x, y$ there is a solution for the equation $ t^3 \minus{} 3t \plus{} 1 \equiv 0$ (take $ t \equal{} x.y^{\minus{}1}$). Testing cases, we see this is impossible. So $ 11 \mid x, y$. This implies $ 11^3|x^3 \minus{} 3xy^2 \plus{} y^3 \equal{} 2891$, contradiction.

Feliz's solution is false, since 11 doesn't divide 2891...

[quote=claserken]What's the motivation behind the transformation $(x,y)\to (y-x,-x)$? I understand why one would want to go from $(x,y)\to (ax+by,cx+dy)$, but is there a better way to find $a,b,c,d$ instead of plugging that in, equating coefficients, and solving a system of equations?[/quote]

Just write $y=x+k$ and see what happens.

I would like to show another way to prove the second part of the problem and to do this I will prove more general result In fact there is no solution to $n=4k+2$ suppose to the contrary that there is a solution $(x,y)$ then discuss the following cases:
1) $x$ is even then clearly $y$ is even and Hence :
$(x,y)=(2a,2b)$ so $8(a^3-3ab^2+b^3)=4k+2$ which is wrong.
2) $x$ is odd and Hence:
a) $y$ is even then $n$ must be odd which is wrong.
b) $y$ is odd wthen $n$ must be odd which is wrong.
so when $n=4k+2$ no solution but $1982=4*495+2$ Hence no sloution and we are done :-)

[b]Detailed Solution as #2:[/b]
WLOG, Assume that $x=y+m$,
Then we have,
$$(y+m)^3-3y^2(y+m)+y^3=n$$
$$m^3-3y^2m+(-y)^3=n$$
So, we see that if $(x,y)$ is solution then $(-y,x-y)$ is also a solution.
Similarly, letting $y=x+k$ gives.
$$k^3-3x^2k+(-x)^3=n$$
So, $(-x,y-x)$ is also a solution.
[b]Claim 1: All the solution pairs are distinct[/b]
assume to the contrary that $(x,y)=(-y,x-y)\implies y=-2y \implies x=y=0$.
 [i]Contradiction since $n$ is a positive integer.[/i]
[b]This proves our claim.So, we must have three different solution.[/b]

[b][i][color=red][Side note: In the case of cubes check $\pmod{7} $ or $\pmod{9}$ for relevant results][/b][/i][/color]
Suppose for the sake of contradiction there exists a solution pair $(x,y)$ in integers for equation:
$$x^3-3xy^2+y^3=2891$$
We have, $x^3 \equiv 0,1,8 \pmod{9}$,
[b]Now we work on the basis of: $2891 \equiv 2 \pmod{9}$.[/b]
[b]Case I: $x,y$ are multiple of $3$.[/b]
$$x^3-3xy^2+y^3 \equiv 0 \pmod{9}$$
[i]Contradiction.[/i]
[b]Case II: $x$ or $y$ is multiple of three.[/b]
$$\implies x^3-3xy^2+y^3 \equiv 0 \pmod{9}$$
[i]Contradiction.[/i]
[b]Case III: $x,y$ are integers not multiple of 3. [/b]
$$\implies x^3-3xy^2+y^3 \equiv 3,6 \pmod {9}$$
[i]Contradiction.[/i]
So, there are no solutions when $n=2891$. And, we are done.$\blacksquare$

We can use quadratic residues for solving this.

For the first part of the problem, notice that if $(x, y)$ is a solution, then $(y-x, -x)$ and $(-y, x-y)$ are also solutions, which means that all solutions come in distinct triplets of three, which is enough.

For the second part, notice that $2891 = 7^2 \cdot 59$. Now, setting $a=yx^{-1} \pmod 7$, we have $$a^3 - 3a^2 + 1 \equiv 0 \pmod 7.$$ If $a^3 \equiv 1 \pmod 7$, then $a^2 \equiv 3 \pmod 7$ is not a quadratic residue. If $a^3 \equiv -1  \pmod 7$, then $3a^2 \equiv 0 \pmod 7$, contradiction as $7 \nmid a$. This shows that there are no solutions.


geometry
geometric transformation
reflection
trigonometry
angle bisector
geometry solved

prove or disprove:
We have triangle ABC and CL is the bisector. Points M and N are on CL such that angles CAM and BAN are equal. Prove that angles ABN and CBM are equal, too.

[quote="Beat"]We have triangle ABC and CL is the bisector. Points M and N are on CL such that angles CAM and BAN are equal. Prove that angles ABN and CBM are equal, too.[/quote]

This is a consequence of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=359003#p359003]http://www.mathlinks.ro/Forum/viewtopic.php?t=58872 post #1[/url] Theorem 2. In fact, that theorem yields that the reflections of the lines AM, BM, CM in the angle bisectors of the angles CAB, ABC, BCA concur at one point. Denote this point by Q. Now, since the line AQ is the reflection of the line AM in the angle bisector of the angle CAB, while the line AB is the reflection of the line AC in this angle bisector, the angle between the lines AB and AQ must equal to the angle between the lines AC and AM (since reflections preserve angles); in other words, < BAQ = < CAM. Comparing this with < CAM = < BAN, we get < BAQ = < BAN, and thus the point Q lies on the line AN. On the other hand, since the line CM is the line CL, i. e. the angle bisector of the angle BCA, the line CQ, being the reflection of the line CM in the angle bisector of the angle BCA, must be the angle bisector of the angle BCA, as well; thus, the point Q lies on the angle bisector of the angle BCA, i. e. on the line CL, i. e. on the line CN. So, altogether, we see that the point Q lies on the lines AN and CN. Thus, Q = N. Now, similarly to how we showed that < BAQ = < CAM, we can find that < ABQ = < CBM; since Q = N, this becomes < ABN = < CBM, and the proof is complete.

Sorry for the messy explanation, I hope it was, at least, understandable.

  Darij

Thanks a lot. But is there a simple proof only for this fact? Just asking. If there is please post it.. Thank you

[b]A interesting Lemma.[/b] $\triangle ABC,\ M\in BC\Longrightarrow\frac{MB}{MC}=\frac{AB}{AC}\cdot \frac{\sin \widehat {MAB}}{\sin \widehat {MAC}}\ \ (1)\ .$
[b]Proof.[/b] Denote $m(\widehat {MAB})=x$, $m(\widehat {MAC})=y$. From the product of the relations
$\frac{MB}{MA}=\frac{\sin x}{\sin B},\ \frac{MA}{MC}=\frac{\sin C}{\sin y},\ \frac{\sin C}{\sin B}=\frac{AB}{AC}$ we obtain the relation $(1)\ .$

[b]The Steiner's theorem.[/b] Let $ABC$ be a triangle and let $M,N\in BC$ be two points
so that the rays $[AM$ and $[AN$ are isogonals in the angle $\widehat {BAC}$,i.e. the angles
$\widehat {BAC}$ and $\widehat {MAN}$ have same bisector. Then $\frac{MB}{MC}\cdot \frac{NB}{NC}=\left(\frac{AB}{AC}\right)^2\ \ (2)\ .$
[b]Proof.[/b] Denote $m(\widehat {MAB})=m(\widehat {NAC})=x$, $m(\widehat {MAC})=m(\widehat {NAB})=y$.
Using the above Lemma, from the product of the relations
$\frac{MB}{MC}=\frac{AB}{AC}\cdot \frac{\sin x}{\sin y},\ \frac{NB}{NC}=\frac{AB}{AC}\cdot \frac{\sin y}{\sin x}$
results the [u]Steiner's relation[/u] $(2)\ .$

[b]A solution of the proposed problem.[/b] Denote $m(\widehat {ABN})=x$, $m(\widehat {MBN})=z$,

$m(\widehat {CBM})=y.$ From the product of the relations

$\left(\frac{AL}{AC}\right)^2=\frac{NL}{NC}\cdot \frac{ML}{MC}\ \ \ ;\ \ \ \frac{NL}{NC}=\frac{BL}{BC}\cdot \frac{\sin x}{\sin (y+z)}$

$\frac{ML}{MC}=\frac{BL}{BC}\cdot \frac{\sin (x+z)}{\sin y}\ \ \ ;\ \ \ \left(\frac{BL}{BC}\right)^2=\left(\frac{AL}{AC}\right)^2$

we obtain the relation $\sin x\sin (x+z)=\sin y\sin (y+z)\Longleftrightarrow$

$\cos z-\cos (2x+z)=\cos z-\cos (2y+z)\Longleftrightarrow2x+z=2y+z\Longleftrightarrow x=y\ .$

Virgil Nicula Thanks !  :omighty:
I`m sure that Darij`s solution is nice, too but i don`t have time for it. this problem is my homework for tomorrow  :)
Thanks again both of you


Is [0, 1, 1] a linear combination of $ \vec{v_1}$ = [1, 2, 3], $ \vec{v_2}$ = [1, -1, 0] and $ \vec{v_3}$ = [2, 1, 3]?

Normally the answer to these kinds of questions (in my experience, at least) is yes, but I am getting no. Can anyone confirm this? By the way they are supposed to be column vectors, not row vectors.

To notate column vectors, use either matrix environments as in $ \vec{v}_1\equal{}\begin{bmatrix}1\\2\\3\end{bmatrix}$ or transposes as in $ \vec{v}_1\equal{}\begin{bmatrix}1&2&3\end{bmatrix}^T$

In this case, you have the wrong answer. It is a linear combination. Since $ v_1,v_2,v_3$ are linearly dependent, it is a linear combination in many different ways.

A systematic approach to finding such combinations:
You're trying to solve a system of linear equations $ av_1\plus{}bv_2\plus{}cv_3\equal{}w$. Write this in matrix form: $ \begin{bmatrix}v_1&v_2&v_3\end{bmatrix}\cdot\begin{bmatrix}a\\b\\c\end{bmatrix}\equal{}w$. Here, that's $ \begin{bmatrix}1&1&2\\2&\minus{}1&1\\3&0&3\end{bmatrix}\cdot\begin{bmatrix}a\\b\\c\end{bmatrix}\equal{}\begin{bmatrix}0\\1\\1\end{bmatrix}$
We put this in an augmented matrix and run Gaussian elimination:
$ \left(\begin{array}{ccc|c}1&1&2&0\\2&\minus{}1&1&1\\3&0&3&1\end{array}\right),$ then $ \left(\begin{array}{ccc|c}1&1&2&0\\0&\minus{}3&\minus{}3&1\\0&\minus{}3&\minus{}3&1\end{array}\right),$ then $ \left(\begin{array}{ccc|c}1&1&2&0\\0&1&1&\minus{}\frac13\\0&0&0&0\end{array}\right)$ and finally $ \left(\begin{array}{ccc|c}1&0&1&\frac13\\0&1&1&\minus{}\frac13\\0&0&0&0\end{array}\right).$

From this, it is clear that one solution is $ a\equal{}\frac13,b\equal{}\minus{}\frac13.$ More generally, we can have $ c\equal{}t,a\equal{}\frac13\minus{}t,b\equal{}\minus{}\frac13\minus{}t.$

Ahh okay, that's what I got, but I just got my general solution a little mixed up. Thanks a lot for clearing that up!


geometry
analytic geometry
function
algebra
domain
algorithm
number theory

How do I create a program for the TI-89 that allows me to compute the area of a polygon given its coordinates? I know the formula:

If the coordinates are:
\[(x_{1}, y_{1})\]
\[(x_{2}, y_{2})\]
\[\vdots\]
\[(x_{n}, y_{n})\]

then the formula is:

\[A=\frac{1}{2}((\sum_{i=1}^{n-1}x_{i}\cdot y_{i+1}-x_{i+1}\cdot y_{i})+x_{n}\cdot y_{1}-x_{1}\cdot y_{n})\]

I don't know how to put this on there.

Is this the right forum for this? I don't know where to put it...

does anyone know where i can find one for the TI-83?

when I follow the above site to a TI-83 Plus,
whatever I put in for X, it gives Invalid Dimension error.

[quote="i_like_pie"]Here's a site that has the TI-81 and TI-82 programs written:

http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf[/quote]

Thanks, i_like_pie, but I have a TI-89 (as you already know). Is it possible to modify that program so that it works for the TI-89?

Like 8parks11 said, it doesn't work on the TI-83, so I doubt that it works on the TI-89

I think maybe something like this

You will need to input the x- and y-coordinates as a list (i.e. in the form {x1, x2, x3, x4, ...} and {y1, y2, y3, y4, ...})

I haven't tested this program yet but it should work

:shoelace()
:Prgm
:0→area
:Input "Input x-coords as a list:",xc
:Input "Input y-coords as a list:",yc
:If dim(xc)≠dim(yc) Then
:Disp "Invalid input!"
:Else
:For a, 1, dim(xc)-1
:area+(xc[a]*yc[a+1]-xc[a+1]*yc[a])/2→area
:EndFor
:area+(xc[dim(xc)]*yc[1]-xc[1]*yc[dim(xc)])/2→area
:Disp "Area is:"
:Disp area
:EndIf
:EndPgrm

=================================
BONUS: For you TI-83 users
=================================

:0→B
:Disp "INPUT X-COORDS"
:Disp "AS A LIST:"
:Input XC
:Disp "INPUT Y-COORDS"
:Disp "AS A LIST:"
:Input YC
:If dim(XC)≠dim(YC)
:Then
:Disp "INVALID!"
:Else
:For (A, 1, dim(XC)-1)
:B+(XC(A)*YC(A+1)-XC(A+1)*YC(A))/2→B
:End
:B+(XC(dim(xc))*YC(1)-XC(1)*YC(dim(XC)))/2→B
:Disp "Area is:"
:Disp B
:End
:Return

EDIT: Hopefully the 83+ code is fixed now

no ti-89 tit. programs?

i feel left out :(

where did you guys learn to program the ti-89... What guide do you reccomend for learning how to use a ti-89?

[quote="b-flat"]yay, a chance to recommend my program to everyone.

It includes a function to find the area, on the 83+ series.

[url=http://www.ticalc.org/archives/files/fileinfo/392/39245.html]linkie[/url]

There are a lot of other useful functions as well.

I'll post the actual code, if needed.[/quote]

Your functions are great, and I really need them.

But...
I have a TI-89 (NOT Titanium)

Can you make it work for the TI-89?

When I try i_like_pie's program, it says
"Error: Missing )"  :maybe:

HELP!

[quote="8parks11"]can u specify
For A, 1, dim(xc)-1

because the For command always comes as
For(  )  in the TI-83Plus

I did For(A,1,dim(XC)-1

but anyways i dont know if thats the problem but it gives SYNTAX error[/quote]

For(A,1,dim(XC)-1) I think. Sorry, I haven't used my 83 in a while.

[quote="vishalarul"]randomdragoon,

How am I supposed to input the numbers?[/quote]
{1, 3, 6, 9, 3}
{2, 6, 10, 2, 0}

Where the first list you input represents the set of x-coordinates, and the second list is the set of their respective y-coordinates.

This applies to both the 89 and 83 programs.

randomdragoon, it tells me "Domain Error" once I do the x-coords.

[quote="b-flat"]@vishalarul:  I may or may not be able to convert my program to the 89, depending on if my mom lets me use her 89 or not.

However, maybe I could post the source code & you could use the general algorithms?[/quote]

OK, you can post the source code.

Alright, the code is quite long, but here it is.

Just say the features you want & I'll break up the code so you can get just those features.

Also, I might be able to convert a few of the functions to the 89 Basic from memory...

[hide="code"][code]:SetUpEditor∟MATH
:4→dim(∟MATH
:If not(∟MATH(3
:2→∟MATH(3
:If not(∟MATH(4
:Unarchive∟PRIME
:ClrHome
:DelVar θ1+∟MATH(1→∟MATH(1
:∟MATH(2→Z
:If not(Z
:1→Z
:While θ≠45
:¦ Z+(θ=26 and Z<3)-(θ=24 and Z>1→Z
:¦ ClrHome
:¦ Disp "Math Program 2.4
:¦ Output(8,1,"[CLEAR] to Exit
:¦ If Z=3
:¦ Then
:¦ ¦ Output(2,2,"Base Conversion Mod Simplifier  Encode
:¦ ¦ Output(5,2,"Decode
:¦ ¦ If θ=21
:¦ ¦ Then
:¦ ¦ ¦ If X=4 or X=5
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ "ABCDEFGHIJKLMNOPQRSTUVWXYZ .→Str1
:¦ ¦ ¦ ¦ " →Str2
:¦ ¦ ¦ ¦ Repeat 28=length(Str2
:¦ ¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ ¦ Input "Code?",Str2
:¦ ¦ ¦ ¦ ¦ If 28≠length(Str2
:¦ ¦ ¦ ¦ ¦ Disp "Incorrect Length","Remember to add","  and .
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ If X=5
:¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ Str2→Str3
:¦ ¦ ¦ ¦ ¦ Str1→Str2
:¦ ¦ ¦ ¦ ¦ Str3→Str1
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ Input "Text?",Str4
:¦ ¦ ¦ ¦ " →Str3
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ For(I,1,length(Str4
:¦ ¦ ¦ ¦ ¦ sub(Str4,I,1
:¦ ¦ ¦ ¦ ¦ If inString(Str2,Ans
:¦ ¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ ¦ Str3+sub(Str1,inString(Str2,Ans),1→Str3
:¦ ¦ ¦ ¦ ¦ ¦ Else
:¦ ¦ ¦ ¦ ¦ ¦ Str3+"?→Str3
:¦ ¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ¦ Output(1,1,iPart(100I/length(Str4
:¦ ¦ ¦ ¦ ¦ Output(1,3,"?
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ sub(Str3,2,length(Str3)-1
:¦ ¦ ¦ ¦ Pause Ans
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=4
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ Str1→Str3
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=3
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ Repeat J>1 and fPart(J)=0 and fPart(K)=0 and K and L>1 and L,J)=1
:¦ ¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ ¦ Disp "JX=K (mod L)
:¦ ¦ ¦ ¦ ¦ Prompt J,K,L
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ Repeat not(fPart(K/J
:¦ ¦ ¦ ¦ ¦ K+L→K
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ K/J→K
:¦ ¦ ¦ ¦ If K≥L
:¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ Repeat K<L
:¦ ¦ ¦ ¦ ¦ ¦ K-L→K
:¦ ¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ Disp "X=
:¦ ¦ ¦ ¦ Output(1,3,K
:¦ ¦ ¦ ¦ Output(1,5+iPart(log(K)),"(mod
:¦ ¦ ¦ ¦ Output(1,10+iPart(log(K)),L
:¦ ¦ ¦ ¦ Output(1,11+iPart(log(K))+iPart(log(L)),")
:¦ ¦ ¦ ¦ Pause
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=2
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ DelVar A
:¦ ¦ ¦ ¦ Repeat A>1 and A<63 and not(fPart(A
:¦ ¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ ¦ Disp "Starting Base
:¦ ¦ ¦ ¦ ¦ Input "{2,62}?",A
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ "123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz→Str4
:¦ ¦ ¦ ¦ sub(Str4,1,A-1→Str2
:¦ ¦ ¦ ¦ 1→C
:¦ ¦ ¦ ¦ Repeat not(E
:¦ ¦ ¦ ¦ ¦ 2→E
:¦ ¦ ¦ ¦ ¦ Input "Number?",Str1
:¦ ¦ ¦ ¦ ¦ If E=2
:¦ ¦ ¦ ¦ ¦ DelVar E
:¦ ¦ ¦ ¦ ¦ For(I,1,length(Str1
:¦ ¦ ¦ ¦ ¦ ¦ If sub(Str1,I,1)≠"0" and A≤inString(Str4,sub(Str1,I,1
:¦ ¦ ¦ ¦ ¦ ¦ 1→E
:¦ ¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ Repeat C>1 and C<63 and not(fPart(C
:¦ ¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ ¦ Disp "New Base
:¦ ¦ ¦ ¦ ¦ Input "{2,62}?",C
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ DelVar B
:¦ ¦ ¦ ¦ For(I,1,length(Str1
:¦ ¦ ¦ ¦ ¦ B+A^(length(Str1)-I)(inString(Str2,sub(Str1,I,1→B
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ " →Str3
:¦ ¦ ¦ ¦ For(I,iPart(log(B)/log(C)),0,‾1
:¦ ¦ ¦ ¦ ¦ If C^I≤B
:¦ ¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ ¦ Str3+sub(Str4,iPart(B/(C^I)),1→Str3
:¦ ¦ ¦ ¦ ¦ ¦ Else
:¦ ¦ ¦ ¦ ¦ ¦ Str3+"0→Str3
:¦ ¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ¦ B-C^IiPart(B/(C^I→B
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ sub(Str3,2,length(Str3)-1→Str3
:¦ ¦ ¦ ¦ Pause Str3
:¦ ¦ ¦ End
:¦ ¦ End
:¦ End
:¦ If Z=1
:¦ Then
:¦ ¦ Output(2,2,"*Prime Factor   *Test Prime
:¦ ¦ Output(4,2,"*Radical Simp
:¦ ¦ Output(5,2,"Euclid's
:¦ ¦ Output(6,2,"LCM
:¦ ¦ Output(7,2,"*Geometric Mean
:¦ ¦ If θ=21
:¦ ¦ Then
:¦ ¦ ¦ If X=5 or X=6
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ Input "1st?",A
:¦ ¦ ¦ ¦ Input "2nd?",B
:¦ ¦ ¦ ¦ AB→D
:¦ ¦ ¦ ¦ Repeat A=B or not(fPart(A/B
:¦ ¦ ¦ ¦ ¦ If A<B
:¦ ¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ ¦ A→C
:¦ ¦ ¦ ¦ ¦ ¦ B→A
:¦ ¦ ¦ ¦ ¦ ¦ C→B
:¦ ¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ¦ A-BiPart(A/B→A
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ If not(fPart(A/B
:¦ ¦ ¦ ¦ B→A
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ If X=5
:¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ Disp "GCD:
:¦ ¦ ¦ ¦ ¦ Else
:¦ ¦ ¦ ¦ ¦ D/A→A
:¦ ¦ ¦ ¦ ¦ Disp "LCM:
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ Disp A►Frac
:¦ ¦ ¦ ¦ Pause
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=4 or X=7
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ If X=7
:¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ .5→A:.5→B
:¦ ¦ ¦ ¦ ¦ Repeat not(fPart(AB
:¦ ¦ ¦ ¦ ¦ ¦ Input "1st?",A
:¦ ¦ ¦ ¦ ¦ ¦ Input "2nd?",B
:¦ ¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ¦ AB→A
:¦ ¦ ¦ ¦ ¦ Else
:¦ ¦ ¦ ¦ ¦ Input "Radicand?",A
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ 1→C
:¦ ¦ ¦ ¦ 1→Y
:¦ ¦ ¦ ¦ Repeat A=1 or A<∟PRIME(Y)2
:¦ ¦ ¦ ¦ ¦ If not(fPart(A/∟PRIME(Y)2
:¦ ¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ ¦ A/∟PRIME(Y)2→A
:¦ ¦ ¦ ¦ ¦ ¦ C∟PRIME(Y→C
:¦ ¦ ¦ ¦ ¦ ¦ Else
:¦ ¦ ¦ ¦ ¦ ¦ Y+1→Y
:¦ ¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ Output(1,1,C
:¦ ¦ ¦ ¦ If A>1
:¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ Output(1,8,"√(
:¦ ¦ ¦ ¦ ¦ Output(1,9,A
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ Pause
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=3
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ DelVar A
:¦ ¦ ¦ ¦ Repeat not(fPart(A
:¦ ¦ ¦ ¦ ¦ Input "Number?",A
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ If A≤0
:¦ ¦ ¦ ¦ 1→A
:¦ ¦ ¦ ¦ 1→Y
:¦ ¦ ¦ ¦ Repeat A<∟PRIME(Y)2 or not(fPart(A/∟PRIME(Y
:¦ ¦ ¦ ¦ ¦ 1+Y→Y
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ If A=1 or not(fPart(A/∟PRIME(Y
:¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ Output(1,1,"Not Prime
:¦ ¦ ¦ ¦ ¦ Else
:¦ ¦ ¦ ¦ ¦ Output(1,1,"Prime
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ Pause
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=2
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ .5→A
:¦ ¦ ¦ ¦ Repeat not(fPart(A
:¦ ¦ ¦ ¦ ¦ Input "Number? ",A
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ A→B
:¦ ¦ ¦ ¦ 1→Y
:¦ ¦ ¦ ¦ 1→W
:¦ ¦ ¦ ¦ ClrListL1
:¦ ¦ ¦ ¦ If A<0
:¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ ‾1→L1(1
:¦ ¦ ¦ ¦ ¦ 2→dim(L1
:¦ ¦ ¦ ¦ ¦ 2→W
:¦ ¦ ¦ ¦ ¦ abs(A→A
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ Repeat A=1 or A<∟PRIME(Y)2
:¦ ¦ ¦ ¦ ¦ If not(fPart(A/∟PRIME(Y
:¦ ¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ ¦ ∟PRIME(Y→L1(W
:¦ ¦ ¦ ¦ ¦ ¦ A/∟PRIME(Y→A
:¦ ¦ ¦ ¦ ¦ ¦ 1+W→W
:¦ ¦ ¦ ¦ ¦ ¦ Y-1→Y
:¦ ¦ ¦ ¦ ¦ ¦ 1+dim(L1→dim(L1
:¦ ¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ¦ Y+1→Y
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ If A<∟PRIME(Y)2
:¦ ¦ ¦ ¦ A→L1(W
:¦ ¦ ¦ ¦ Pause L1
:¦ ¦ ¦ ¦ ClrListL1
:¦ ¦ ¦ End
:¦ ¦ End
:¦ End
:¦ If Z=2
:¦ Then
:¦ ¦ Output(2,2,"Average
:¦ ¦ Output(3,2,"Settings
:¦ ¦ Output(4,2,"Roman->Decimal  Clock
:¦ ¦ Output(6,2,"Area
:¦ ¦ Output(7,2,"Fast Graph
:¦ ¦ If θ=21
:¦ ¦ Then
:¦ ¦ ¦ If X=7
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ AxesOn
:¦ ¦ ¦ ¦ Disp "Clear Graph
:¦ ¦ ¦ ¦ Input "{Yes/No}?",Str1
:¦ ¦ ¦ ¦ If inString(Str1,"Y
:¦ ¦ ¦ ¦ ClrDraw
:¦ ¦ ¦ ¦ Input "Y=",Str1
:¦ ¦ ¦ ¦ ClrDraw
:¦ ¦ ¦ ¦ Input "Quality?",N
:¦ ¦ ¦ ¦ For(X,Xmin,Xmax,(Xmax-Xmin)/96/N
:¦ ¦ ¦ ¦ ¦ Pt-On(X,expr(Str1
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ Pause
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=6
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ Input " Coordinates?",A
:¦ ¦ ¦ ¦ A+1→dim(∟X
:¦ ¦ ¦ ¦ Ans→dim(∟Y
:¦ ¦ ¦ ¦ For(I,1,A
:¦ ¦ ¦ ¦ ¦ Prompt X,Y
:¦ ¦ ¦ ¦ ¦ X→∟X(I
:¦ ¦ ¦ ¦ ¦ Y→∟Y(I
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ ∟X(1→∟X(A+1
:¦ ¦ ¦ ¦ ∟Y(1→∟Y(A+1
:¦ ¦ ¦ ¦ DelVar B
:¦ ¦ ¦ ¦ For(I,1,A
:¦ ¦ ¦ ¦ ¦ B+∟X(I)∟Y(I+1)-∟X(I+1)∟Y(I→B
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ Pause abs(B/2
:¦ ¦ ¦ ¦ DelVar XDelVar Y
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=5
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ Input "HOUR?",H
:¦ ¦ ¦ ¦ Input "MINUTE?",M
:¦ ¦ ¦ ¦ 90-30H-M/2
:¦ ¦ ¦ ¦ ClrDraw
:¦ ¦ ¦ ¦ AxesOff
:¦ ¦ ¦ ¦ ZStandard
:¦ ¦ ¦ ¦ ZSquare
:¦ ¦ ¦ ¦ Degree
:¦ ¦ ¦ ¦ Circle(0,0,6
:¦ ¦ ¦ ¦ Line(0,0,3.5cos(Ans),3.5sin(Ans
:¦ ¦ ¦ ¦ 90-6M
:¦ ¦ ¦ ¦ Line(0,0,5cos(Ans),5sin(Ans
:¦ ¦ ¦ ¦ Pt-On(0,0,2
:¦ ¦ ¦ ¦ Text(55,5,abs(30H-5.5M),"o
:¦ ¦ ¦ ¦ Pause
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=3
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ Disp "Math Program 2.4","Used:","Auto-Archive:","Penguin Studios","","Main Programmer:"," Bryce Taylor
:¦ ¦ ¦ ¦ Output(8,4,"(C) 6-17-2006
:¦ ¦ ¦ ¦ Output(2,6,∟MATH(1
:¦ ¦ ¦ ¦ Output(2,8+iPart(log(∟MATH(1))),"Times
:¦ ¦ ¦ ¦ 0
:¦ ¦ ¦ ¦ Repeat Ans=105
:¦ ¦ ¦ ¦ ¦ If Ans=21
:¦ ¦ ¦ ¦ ¦ Archive∟PRIME
:¦ ¦ ¦ ¦ ¦ If Ans=22
:¦ ¦ ¦ ¦ ¦ Unarchive∟PRIME
:¦ ¦ ¦ ¦ ¦ ∟MATH(4)+(Ans=26)-(Ans=24→∟MATH(4
:¦ ¦ ¦ ¦ ¦ Output(3,14,sub("On Off",6abs(fPart(∟MATH(4)/2))+1,3
:¦ ¦ ¦ ¦ ¦ getKey
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=4
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ Input "ROMAN? ",Str1
:¦ ¦ ¦ ¦ Str1+" →Str1
:¦ ¦ ¦ ¦ DelVar V
:¦ ¦ ¦ ¦ For(I,1,length(Str1)-1
:¦ ¦ ¦ ¦ ¦ sub(Str1,I,1→Str2
:¦ ¦ ¦ ¦ ¦ If Str2="I
:¦ ¦ ¦ ¦ ¦ 1→Y
:¦ ¦ ¦ ¦ ¦ If Str2="V
:¦ ¦ ¦ ¦ ¦ 5→Y
:¦ ¦ ¦ ¦ ¦ If Str2="X
:¦ ¦ ¦ ¦ ¦ 10→Y
:¦ ¦ ¦ ¦ ¦ If Str2="L
:¦ ¦ ¦ ¦ ¦ 50→Y
:¦ ¦ ¦ ¦ ¦ If Str2="C
:¦ ¦ ¦ ¦ ¦ 100→Y
:¦ ¦ ¦ ¦ ¦ If Str2="D
:¦ ¦ ¦ ¦ ¦ 500→Y
:¦ ¦ ¦ ¦ ¦ If Str2="M
:¦ ¦ ¦ ¦ ¦ 1000→Y
:¦ ¦ ¦ ¦ ¦ sub(Str1,I+1,1→Str2
:¦ ¦ ¦ ¦ ¦ If Str2="I
:¦ ¦ ¦ ¦ ¦ 1→W
:¦ ¦ ¦ ¦ ¦ If Str2="V
:¦ ¦ ¦ ¦ ¦ 5→W
:¦ ¦ ¦ ¦ ¦ If Str2="X
:¦ ¦ ¦ ¦ ¦ 10→W
:¦ ¦ ¦ ¦ ¦ If Str2="L
:¦ ¦ ¦ ¦ ¦ 50→W
:¦ ¦ ¦ ¦ ¦ If Str2="C
:¦ ¦ ¦ ¦ ¦ 100→W
:¦ ¦ ¦ ¦ ¦ If Str2="D
:¦ ¦ ¦ ¦ ¦ 500→W
:¦ ¦ ¦ ¦ ¦ If Str2="M
:¦ ¦ ¦ ¦ ¦ 1000→W
:¦ ¦ ¦ ¦ ¦ If Y≥W
:¦ ¦ ¦ ¦ ¦ Y+V→V
:¦ ¦ ¦ ¦ ¦ If Y<W
:¦ ¦ ¦ ¦ ¦ V-Y→V
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ Disp V
:¦ ¦ ¦ ¦ Pause
:¦ ¦ ¦ ¦ DelVar X
:¦ ¦ ¦ End
:¦ ¦ ¦ If X=2
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ Disp "√(2)-1 TO STOP
:¦ ¦ ¦ ¦ DelVar BDelVar CDelVar D
:¦ ¦ ¦ ¦ While B≠√(2)-1
:¦ ¦ ¦ ¦ ¦ Input "NUMBER?",B
:¦ ¦ ¦ ¦ ¦ B+C→C
:¦ ¦ ¦ ¦ ¦ 1+D→D
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ D-1→D
:¦ ¦ ¦ ¦ C-√(2)+1→C
:¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ If not(fPart(C/D
:¦ ¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ ¦ Output(1,1,C/D
:¦ ¦ ¦ ¦ ¦ Else
:¦ ¦ ¦ ¦ ¦ Input "PLACES?",A
:¦ ¦ ¦ ¦ ¦ round(C/D,A→B
:¦ ¦ ¦ ¦ ¦ ClrHome
:¦ ¦ ¦ ¦ ¦ Output(1,1,B
:¦ ¦ ¦ ¦ End
:¦ ¦ ¦ ¦ Pause
:¦ ¦ ¦ End
:¦ ¦ End
:¦ End
:¦ If θ≠21
:¦ Then
:¦ ¦ ∟MATH(3→X
:¦ ¦ Repeat max(θ={21,45,26,24,105
:¦ ¦ ¦ getKey→θ
:¦ ¦ ¦ X-(θ=25 and X>2)+(θ=34 and X<7→X
:¦ ¦ ¦ Output(X,1,"
:¦ ¦ ¦ If X>2:Output(X-1,1,"
:¦ ¦ ¦ If X<7:Output(X+1,1,"
:¦ ¦ End
:¦ ¦ If θ=105
:¦ ¦ 21→θ
:¦ ¦ X→∟MATH(3
:¦ ¦ Else
:¦ ¦ DelVar θ
:¦ End
:End
:ClrHome
:ZSquare
:Disp "Thanks for","using!
:Z→∟MATH(2
:AxesOn
:FnOn
:ClrDraw
:If not(∟MATH(4
:Archive∟PRIME
:DelVar ADelVar BDelVar CDelVar DDelVar EDelVar FDelVar GDelVar HDelVar IDelVar JDelVar KDelVar LDelVar MDelVar NDelVar ODelVar PDelVar QDelVar RDelVar SDelVar TDelVar UDelVar VDelVar WDelVar XDelVar YDelVar ZDelVar θDelVar Str1DelVar Str2DelVar L1DelVar L2DelVar ∟XDelVar ∟YDelVar Str3DelVar Str4[/code]
All coding and tables ©2004-2007 Kerm Martian
[/hide]

I generated the code using Kerm Martian's Source Coder 2

http://www.cemetech.net/projects/basicelite/sourcecoder2.php

Here is what I need:

7. Average
9. Roman->Decimal
10. Clock
11. Area
13. Base Conversion
14. Mod Simplification

If you give me the code, how do I take it from computer to calculator?

[quote="jli"]is there a shoelace theorem for 3d planes?[/quote]

learn about cross product

[quote="randomdragoon"]
[quote="vishalarul"]randomdragoon,

How am I supposed to input the numbers?[/quote]
{1, 3, 6, 9, 3}
{2, 6, 10, 2, 0}
[/quote]

Input "XCOORDS? ",$L_{1}$
Input "YCOORDS? ",$L_{2}$

should work.


Αφιερωμένη στα παιδιά που θα διαγωνιστούν σε λίγες μέρες ...

Αν  x,y,z θετικοί ακέραιοι ναβρεθούν οι λύσεις της εξίσωσης

$ x^2 \plus{} y^2  \equal{} xyz\minus{}1$

Για όσους θέλουν να την προσπαθήσουν
[hide]Η γνωστή εργασία  :wink:  (Fermat Infinite Descent)

$ z\equal{}3$ και $ (x,y)\equal{}(F_{2n\plus{}1},F_{2n\minus{}1})$ όπου $ F$ ή ακολουθία Fibonacci .[/hide]


1.Bissector angle A
2.Perpendicular to bissector through A
3.paralell to AB with distance Rc that cut line item 2 in O
4.Projection of O on AB determines T
5.Circle center A radius AT determines S on AC
6.Circle center S radius (b-c) determines on AC point U
7.Perpendicular to AC through U that cut line of item 2 in V
8.External tangent to Circles centers O radius OT and V radius UV will be the line BC side of triangle ABC


Euler
function
geometry
geometric transformation
rotation
group theory
abstract algebra

I'm trying to understand [url=http://www.springerlink.com/index/X558J42T8K1615G6.pdf]this paper[/url], the only thing  I can't understand is the derivation of the generating function with fraction.

[hide="if you cant read the whole paper,i explain the sufficient terms"]
In a general term, it's actually the enumeration of euler circuit, with a certain repeat (there is a vertices crossed more than one).

Suppose that we are given a digraph $ G_R$, we can construct a main pattern graph of $ G_R$, denoted by $ G_P$ by drawing a circle such that :
i) Take any vertex , say $ X$. Find its degree, say $ d$. Then place $ d$ points in the circle and label them $ X$. connect them with $ d$ lines so that they form a $ d\minus{}gon$ inscribed in the circle.
ii) The clockwise order preserves the adjecency of $ G_R$. That is when there is an arrow from $ X$ to $ Y$ in $ G$, then $ Y$ places next to $ X$ in the circle.
for example:
[url=http://img691.imageshack.us/i/graphkh.jpg/][img]http://img691.imageshack.us/img691/4721/graphkh.th.jpg[/img][/url]

Notice that when we add a chord in $ G_P$ such that it does not intersect any other chord, then it means that we add a vertex in $ G_R$ with loop, hence we add a repeat in the euler circuit.

[b]Lemma [/b]
[i]Given a main pattern graph $ G_p$ with no repeat as below,
[url=http://img337.imageshack.us/i/1con.jpg/][img]http://img337.imageshack.us/img337/3513/1con.th.jpg[/img][/url]

the generating function of the number of the reduced graph with $ n$ repeats obtained from $ G_P$ (by adding $ n\minus{}2$ chords) is given by \[ \frac{x^2C(x)^4\plus{}x^2C^2(x^2)\plus{}2x^2C(x^4)}{4}\]
where $ C(x)\equal{}\frac{1\minus{}\sqrt{1\minus{}4x}}{2x}$ the generating function of Catalan Numbers $ C_n \equal{}\frac{1}{n\plus{}1} \binom{2n}{n}$.[/i]

[hide="the proof in the paper is still inside my understanding"]
Notice that we can add chord in the 4 surrounding area. And rotating the $ G_P$, $ 90k$ degrees where $ k\equal{}0,1,2,3$ give the same reduced graph, so this rotation group is a symetric subgroup of $ S_4$. The cycle index of this group is $ \frac{x_1^4\plus{}x_2^2\plus{}2x_4}{4}$. It is quite well-known that the number of the arrengement of $ n$ chords in the circle without intersecting each other is given by $ C(x)$. Since we already have $ 2$ chords, we need to add $ n\minus{}2$ more to get $ n$ repeats.
So by Generalized Polya's Enumeration Theorem the generating function is \[ x^2 \left(\frac{C(x)^4\plus{}C^2(x^2)\plus{}C(x^4)}{4}\right)\]
[/hide]

[/hide]

Using the above method, i was able to derive almost all the generating functions in the paper, except those with the fraction, one is below:

[hide="cant understand this one"]The main pattern
[url=http://img148.imageshack.us/i/2con.jpg/][img]http://img148.imageshack.us/img148/4707/2con.th.jpg[/img][/url]
has generating function
\[ \frac{x^4C^8(x)}{2\minus{}2xC^2(x)} \plus{} \frac{x^4C^4(x^2)}{2\minus{}2xC^2(x^2)}\]
[/hide]

I searched all the literature about generating function i have (Concrete mathematics, generatingfunctiology, JSTOR papers).. none of them has something about emerging a fraction in the generating function. I also think about using $ \frac{1}{1\minus{}xC^2(x)}\equal{}1\plus{}xC^2(x)\plus{}x^2C^4(x)\plus{}\cdots$ , but it give me less control to the coefficients.

any ideas?

thanks before.
[dont worry, this is not an assignment]

nevermind, i've got it now. :lol:
i should not too depended on  cycle index. :blush:


Fie numerele reale pozitive x,y,z astfel incat x+y+z>$\sqrt3$.Demonstrati ca avem atunci urmatoarele inegalitati:
a) $\ x^2+y^2+z^2$>1.
b) xyz<$\sqrt3$/9.
c) x+y+z>9xyz.

Poate ceva nu este exact ???

$\forall x,y,z \in \left( {0, + \infty } \right)$ cu $x + y + z > \sqrt 3 \mathop \Rightarrow \limits_? xyz < \frac{{\sqrt 3 }}{9}$

De ce oare nu inteleg ...
:(
Sunt date trei numere pozitive "mari de tot" chiar de suma mai mare decat $\sqrt 3$
si totusi produsul lor este mai micut decat $\frac{\sqrt 3}{9}$  ??

Chiar ca e adevarat ca e mai usor sa compui probleme decat sa le rezolvi.Doar o indicatie: poti lua x=$\tan A/2$.

Acum pentru mine este mai clar.
este undeva o gresala ...........
 :mad:

z=?
asa mare ??

E clar ca e greseala!! $x + y + z > \sqrt 3 \mathop \Rightarrow \limits_? xyz < \frac{{\sqrt 3 }}{9}$ e falsa.

Da, mi-am dat seama.Mai trebuiau adaugate cateva conditii, ca sa mearga usor cu tangenta.Puteti sa le intuiti si singuri?Daca va dati seama de niste proprietati ale functiei tangenta, atunci problema merge.

Fiindca problema nu ofera alte conditii de existenta in afara de faptul ca numerele sunt reale si cu suma $\me\sqrt3$ rezulta ca egalitatea are loc pentru toate numerele ce satisfac aceste conditii,si evident pentru $x=y=z=2$.Ori pentru aceste valori egalitatzile de la pctele b) si c) nu au loc. Rezulta contradictie.Problema are o greseala.

De fapt, mah andyciup, avem $\ x+y+z$>$\sqrt3$.

Citeste te rog codul scris de mine in dreptul lui $\sqrt3$ din mesajul meu.Am vrut sa scriu [code]\ mai mare sau egal[/code] si am uitat ca simbolul nu era [code]\me[/code]

[quote="spix"]Ma eu pe tine te bat!  :mad:[/quote]

Cred ca ma bag si eu. Eu zic sa-l tragem doar de urechi asa putin... mai mult, ca-i inca mic. :D

Nu cred ca merit sa fiu batut.Dar poate mi-o iau la anu', ca am sanse destul de mari sa merg la nationala si...
 :stretcher:

si.... care e problema? :)

Gata..... enough spamm here. ;)

ce dur e baiatu asta :) ....vai ce-mi place de pureblack :)

[quote="pureblack2003"]de ce nu te-ai calificat ma de anu asta ? sau te-ai calificat ? de unde stii ca ai sanse asa de mari ? nu propui decat niste probleme care au rezolvarea de 1 rand, la judeteana si mai ales la nationala se dau chestii ceva ceva mai mult de un rand . si cred ca pana la anu daca o tii tot asa, probabil innebunesti ... so ... get a life . la varsta ta fumam si beam ca porcu :) nu zic ca e bine asta, dar mai lasa-le si tu , fa altceva ... sheesh . scuzati-ma pt offtopic[/quote]
    Eu sunt de-acord cu pohaotza.Esti dur, mai omule.Nu cred ca fumai si beai cand erai de varsta mea, dar... :roll:
Cat despre problemele propuse de mine, te sfatuiesc sa te uiti mai atent.Unele nu au rezolvarea chiar de un rand.
Crezi tu ca postez toate problemele propuse de mine pe mathlinks?Iti spun eu ca nu.Dar nu mai zic...tu te-ai calificat
la ONM, dar ce-ai facut acolo? ;) Macar spune-mi si mie numele tau adevarat sa vad si eu cat ai luat...sau poate intr-o buna zi te voi vedea la ONM.La noi in judet merg anul viitor primii trei, asa ca zic si eu, dupa umila mea opinie, ca am ceva sanse, oricat ar fi ele de mici.


I have thought of a name. Mods, please delete this topic.


The USAMTS solutions form says to put AoPS username, while I read one should include the USAMTS User ID. Those are different for me. Which one should I put?

[quote="joml88"]There should be a spot for both.[/quote]

As far as I can tell there isn't.

Something like this?

[code]
%% Enter your AoPS username here
%% Example: \username{DPatrick}
%% IMPORTANT
%% If your username contains one of the following characters:
%%      # ~ _ ^ % { } &
%% then this character must be preceded by a backslash \
%% for example: if your user name is math_genius, then the line below should be
%% \username{math\_genius}

\username{}

%% Enter your USAMTS ID# here
%% Example: \usamtsid{9999}
\usamtsid{}
[/code]

The ID's a number that's in your USAMTS profile.   :D

Oh I see, I was thinking of USAMTS username.

It should be your USAMTS username.  I assume that for most people, your AoPS username and your USAMTS username are the same, but if not, please use your USAMTS username for the USAMTS.

Sorry for the confusion!


geometry
circumcircle
Asymptote
power of a point
radical axis
perpendicular bisector
geometry proposed

Let $ABC$ be an acute triangle. The points $M$ and $N$ are midpoints of $AB$ and $BC$ respectively, and $BH$ is an altitude of $ABC$. The circumcircles of $AHN$ and $CHM$ meet in $P$ where $P\ne H$. Prove that $PH$ passes through the midpoint of $MN$.
[i]V. Filimonov[/i]

[quote="silouan"][quote="Sailor"]the radical axis (which happens to be $PH$)[/quote]
Colud you explain this one please ?Thank you[/quote]

The radical axis of two intersected circles in the plane always pass trought the midpoint of the commom tangent!

Sailor: Yes, seems easy to ARO..

[quote="Sailor"]Are your sure that's from ARO? ... it just seems too easy.
$MN$ is a common tangent to the circumcircles of $\triangle{AMH}$ and $\triangle{HNC}$.
Thus the midpoint of $[MN]$ lies on the radical axis (which happens to be $PH$) of the above mentioned circles.[/quote]

[b] I post a figure to make reading the solution easier.[/b]

[quote="silouan"][quote="Sailor"]the radical axis (which happens to be $PH$)[/quote]
Colud you explain this one please ?Thank you[/quote]

Well, basically I meant that the midpoint of $MN$ has equal power w.r.t. to the circles given in the statement (i.e if you extend $MN$ to meet the circumcircle of $\triangle{AHN}$ at $N_{1}$ then $M$ is the midpoint of $NN_{1}$, etc. ).

 :!:  Also, it is interesting to notice that if the circumcircles of $\triangle{AMH}$ and $\triangle{CHN}$ meet each other (for the second time) at $L$ then $BL$ is a symmedian.

[quote="Sailor"] Well, basically I meant that the midpoint of $MN$ has equal power w.r.t. to the circles given in the statement (i.e if you extend $MN$ to meet the circumcircle of $\triangle{AHN}$ at $N_{1}$ then $M$ is the midpoint of $NN_{1}$, etc. ).[/quote]

Great !  :)  Very nice solution solution .Thank you for this and the explanation.
(Also the noting for the symmedian is also nice)

[quote="Sailor"]

 ... if the circumcircles of $\triangle{AMH}$ and $\triangle{CHN}$ meet each other (for the second time) at $L$ then $BL$ is a symmedian.[/quote]

[color=red] Do you have any simple explanation for this nice result ?[/color]
 Thanks !

First of all notice that $BMLN$ is cyclic.
Then, if $K$ is the midpoint of $[MN]$ and $B_{1}$ is the midpoint of $[AC]$, an easy angle chasing shows that
$KH$ and $KB_{1}$ are symmetric w.r.t to the perpendicular bisector of $[AC]$ and the conclusion follows.

Let $(O1)$ and $(O2)$ be the circumcircles of triangle $AHN$ and $CHM$
$MN$ meets $(O1)$ $(O2)$ at $E,F$
We see $O2N \bot AC$ and $MN//AC$ so $O2N \bot MN$. Then $MN=NF$. Similarly $MN=ME$.
$HP$ meets $MN$ at $K$. We have $KM*KF=KH*KP=KN*KE$
so $KM/KN=KE/KF=ME/NF=1$ $q.e.d$

[quote="anonymous1173"]am i reading the question wrong it says circumcircles of AHN and CHM are intersecting on P not AHM and CHN[/quote]

           [color=red]Which solution are you trying to read ?[/color]

Assume that $ MN$ cuts $ (AHN)$ and $ (CHM)$ respective at $ H$ and $ K$.
We have: $ MN \equal{} MH$ and $ MN \equal{} NK (1)$
Call the midpoint of MN is I. from $ (1)$ we get: $ IN.IH \equal{} IM.IK$
Hence, $ I$ is in the axis radical of $ (AHN)$ and $ (CHM)$. So $ PH$ pass through $ I$
Denote that $ (XYZ)$ is the circumcircle of  $ \delta XYZ$

Also very easy by barycentric coordinates!
[hide=Solution]Let $\omega_1$ and $\omega_2$ denote the circumcircles of $AHN$ and $CHM$, respectively. Then
$
\omega_1:\ -a^2 yz-b^2 zx-c^2 xy+\left(\frac{S_B-S_C}{2} y+S_C z\right)(x+y+z)=0,\\
\omega_2:\ -a^2 yz-b^2 zx-c^2 xy+\left(S_A x+\frac{S_B-S_A}{2} y\right)(x+y+z)=0.
$
Hence, the radical axis of $\omega_1$ and $\omega_2$ is $PH:\ S_A x+\frac{S_C-S_A}{2} y-S_C z=0.$
Clearly, the point $(\tfrac{1}{4},\tfrac{1}{2},\tfrac{1}{4})$ lies on $PH$. $\blacksquare$
[/hide]

Let $D$ be the midpoint of $\overline{MN}$. We are asked to show $D$ has equal power wrt $(AHN)$ and $(CHM)$, which by linpop is equivalent to $\mathrm{Pow}_{(AHN)}(M)=\mathrm{Pow}_{(CHM)}(N)$. Extend $\overline{MN}$ to hit $(AHN)$ again at $X$ and $(CHM)$ again at $Y$. Since the center of $AHN$ lies on the perpendicular bisector of $\overline{AH}$, which passes through $M$, $MN=MX$, and likewise $NM=NY$, hence these powers are both equal to $-MN^2$. $\blacksquare$


logarithms
LaTeX
calculus
function
geometry
3D geometry
inequalities

I would be  :lol:  if a starter can solve this problem at a glance... :D

they should be able to solve it, but not from a glance :P

I would say knowing the rule for resolving this problem is (barely) within the scope of what a person getting started in problem-solving should know. Try it out, everyone, and put your solution in hidden text.

Heres mine:
[hide]
$3^{4^{5^{6^{7^{8}}}}} = x\\
8(7(6(5(4 \log 3)))) = \log x\\
3206.254 = \log x\\
10^{3206.254} = x$

$4^{5^{6^{7^{8^{3}}}}} = x\\
3(8(7(6(5\log 4)))) = \log x
3034.382 = \log x\\
10^{3034.382} = x$

So ${3^{4^{5^{6^{7^8}}}}}$ is greater.
I dunno, not very fast, there is probably an easier way.  This required a calculator too.[/hide]

Silverfalcon's Note: I don't usually do this but this person's formula was not neat. Use Latex please.

I don't usually do this but I edited SnowStorm's post to make it readable.

Snowstorm, use Latex. If you don't know, learn it. ;)

[quote="SnowStorm"]Silverfalcon's Note: I don't usually do this but [u]this person's formula was messed up [b]REALLY BAD[/b]. Use Latex.[/u][/quote]
Hmm ... please give some encouragement to other users. No one is expected to know everything.
Your last post has MUCH BETTER words of encouragement than the one mentioned in my quote. :)  I like tokenadult's words:
[quote="tokenadult"]Try it out, everyone, and put your solution in hidden text.[/quote]

:blush: Sorry.

Here, you can learn Latex in here:

http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf

Also:

http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php

It takes sometime to load it but when it's done, it's a very good site with the very good information.

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

Back to the topic of the problem. :)

[quote="LynnelleYe"][hide="Um."] $7^8$ is much larger than $8^3$. So $6^{7^8}$ should be much larger than $7^{8^3}$. Then you go all the way down the exponent tower until you get that the first expression is much larger. I think. Not a very rigorous solution though, is it? SnowStorm's is much better.
[/hide][/quote]
Lynnelle, you might want to check your answer. The reasoning may have worked in your particular case but the deduction is not free from contradiction. I can come up with a counterexample right-off-the-bat.
[hide=""THE TRUTH""]The truth is that SnowStorm's solution applied logarithms inappropriately!!![/hide]
That's why I said that if a starter can solve this problem at a glance I would be happy because writing a full solution to this problem is a lot harder than just glancing it and let our intuitions guide to the answer.

SnowStorm -

You may like to check your  method, (a^b)^c i[i]s not the same as [/i](a^(b^c).

For example if you  take  log of google which is 10^(10^2), it  is   100, and [b]not   2(10) log (10) which is just 20 [/b]

Hope this helps...

Oh. Sorry about that. I didn't look too carefully at SnowStorm's solution, unfortunately. Yeah, I know my solution is bad, but I really don't know how to do the problem properly. Hint?  :?
And I had thought I was somewhat Intermediate by now  :(

Yes you are somewhat intermediate, it's just that the problem is easy at the intuitive level, like 1+1=2. But to see an axiomatic proof of 1+1=2, I recommend consulting the book A.F.Whitehead & B.Russell's book Principia Mathematica. Or something like '$\sqrt{2}$ is not rational' seems easy to say but requires steps to give a rigorous proof.

Keep up the hard work. I have faith in you Lynnelle  :)

This is very interesting area of mathematics, if I had made the base integer a 2 it would have been much more difficult to justify. Keep trying!

[quote]I had made the base integer a 2 it would have been much more difficult to justify.[/quote]

Yes,  Aren't we lucky that none of the  numbers are less than  e. (=2.71828..)

Some what related item,   a  hint  but a bit off the subject  ...

[hide]Sorry,   it is   related to the orignal problem, but a bit advnace  to be put  in getting started section,  but a classic  problem  is to ask

which is greater  e^(pi) or (pi)^e ?

(Ask this question to  some one who has taken calculus and see if (s)he can answer it without a calculator)

Hint: when x is positive, maximum value of x^(1/x)  is reached when  x=e  .. and the function becomes samller as x increases. In other words,   if   a > b  > e (=2.71)  then  b^a  > a^b  (which is same as saying that  (b^(1/b) > (a^(1/a)) Hope this  makes things easy ..[/hide]

Thanks Gyan, I didn't know that before.  But why is that true?  Or would it be too advanced for getting started?

[quote="RC-7th"]$3^{4^{5^{6^{7^{8}}}}}=x$

$4^{5^{6^{7^{8}}}}\log 3=\log x$

$5^{6^{7^{8}}}\log (4\log 3)=\log(\log x)$
.....

[/quote]

Hi -  a small point .. this  should look like ..
<Edited Later ... the correction is already made in the original>

Google is actually spelled googol, for any of you who are spelling freaks like me  :lol:

[quote="DPopov"]Which is greater:

\[3^{4^{5^{6^{7^8}}}}\]

or

\[4^{5^{6^{7^{8^3}}}}\][/quote]

I do know that the first expression is greater (actualy MUCH GRAETER}  than :
\[3^{3^{3^{3^{3^8}}}}\]

And yet [b] the above [/b] is much much .. greater than  this:

\[(999999999999.....)^{9^{9^{9^{9^3}}}}\]

[b](Which, in turn   is much greater  than the second expression)[/b]

Even if  you put as many 9  as you "can" in the base numebr.. If you want,  you can put  a googol number of 9's or even more .. You can use world's fastest computer and convert all the starts in the sky for the ink material.. and shrink the size of "9" to less than an atom.. and fill the whole universe with 9's . ... and still the number will be  LESS ... MUCH MUCH  MUCH..   GOOGOL..googol  googol ...  TIMES ..LESS than  the first expression (in the problem) ... and greater than the second expression (in the problem)...

Sweet dreams...!!!!

you are correct, where did you learn this?

May be a slightly simple proof, if one is interested .. (of course it is like  trying to prove   size of the Universe is bigger than an atom )..

[hide]Noticing the brilliant  fact:   3^2  = 9  If we   assume

b  >  0  … and
a    >    2b + 1
we can show easily that:

 3^a         >      3^(2b+1)    =  3*9^b   =   2*9^b +9^b > 2*9^b + 1

So, obviously 3^a > 9^b     but we also have a stronger relation of   the form  x > 2y+1 where x=(3^a) and y=(9^b) all over again ..

Hence:
3^(3^a )   >   2*(9^(9^b)) + 1   >  (9^(9^b))
keep doing that a few times and we get:

3^(3^(3^(3^(3^(a))))) >  9^(9^(9^(9^(9^(b)))))
if we put a = 8, and b =3  we get (obviously 8 > 2*3+1)
3^(3^(3^(3^(3^(8))))) >  9^(9^(9^(9^(9^(3)))))
The expression given in the original problem   has to be true too, since, LHS is greater than the LHS of the above relation, and RHS  of the original expression is less than the LHS of the above…

Hope this is helpful..

Some more comments:

[hide]Of course, the number on the left is much..much..bigger than the right .. when we  use a rather larger number (9^9^..  and consider it just greater than 1..  even a slightly  stronger  condition will allow you  to make the lower bases much much larger than 9.. ..
eg A^B^C^D^9^3) where  A,B,C   can be gigantic  numbers.. yet the relationship will hold....A,B, can be larger than googol still the number will be much smaller than the one given on   LHS, as commented previously..)   ....
[/hide]

[/hide]

another related  fun problem...

As you see   n^n^n.... starts gettting bigger very fast but not always....

what is \[\sqrt{2}^{\sqrt{2}^\sqrt{2}^\sqrt{2}^\cdots}}}}\]

(P.S if some one from the kindness of their heart, wan to format it to  look pretty.. please go ahead .. I  will learn latex some time...

[quote="Gyan"]
(P.S if some one from the kindness of their heart, wan to format it to  look pretty.. please go ahead .. I  will learn latex some time...[/quote]

Done.

Joml88 - Thanks.

The answer : It converges  rather slowly to:

[hide]2[/hide]

Snow S, Lynelle Y etc - Hope the solution provided above be a little helpful
[/hide]

Neat solution, I think I understand now...but why is it important that no number within the expressions be less than e?
I can come pretty close to demonstrating the convergence, now that I see the answer...
[hide]Let $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}}$ be equal to x. Then the exponent of the first $\sqrt{2}$ is also equal to x. So we get the equation
$x=\sqrt{2}^x$
Take the log base 2 of both sides.
$\log_2 x=x \log_2 2^{1/2}$
$\log_2 x=\frac 12 x$
$\log_2 x^2=x$
$2^x=x^2$
Now I'm in trouble. First of all, because I can't figure out how to further algebraically manipulate the last line, and secondly because both 2 and 4 work and I'm not even sure how to show that no other numbers work. Help again!
[/hide]

[quote]but why is it important that no number within the expressions be less than e? [/quote]

Sorry,   if  it was not clear  The comment was just to point out some fun fact:

  when a>b>e  you can  prove  b^a > a^b

Also when both  a and b are smaller than e (But still  positive) , similar  result .
(That is   if    a<b<e     and (a,b >0) then a^b   < b^ a)
If  one number is greater than  e and other less than  e,  then you  can't say which is greater without checking the actual value .  (They can even be equal  eg 2^4 = 4^2).

In all cases   (e^ someting)   is always greater than  or equal to   (something)^e   (As long as "something"  is positive  -- and the "equal to" occurs only when something = e)

This was  just for fun .. NO need to remember it  of give it more importance  than  a mere fun  fact ...

Cool...now help me with the root 2 to root 2 to root 2 thing! :P

Looks like  this will be harder, (for me, at least)   to explain.. getting much more advanced than this level.
(If  you take a   few terms and see the behaviour - as mentioned before - the sequence  converges to 2 ..   remaining less than 2 but slowly approaching it. )

Anyway  the thing does converge to 2, (and not 4)  but I am at loss to explain in simple terms why. ..

 May be some one else can  give a better  explaination without using  advanced math..

In fact a  *VERY HARD* problem can be made by replacing    sqrt(2) with  cube_root(3).. where   the obvious answer (3) would be wrong.. .. May be  I  will ask the problem in the Pre-Ollympiad section.. :)

Look at the sequence

\[\sqrt{2}, \sqrt{2}^{\sqrt{2}}, \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}, \ldots\]

[hide="Point number 1"]
Its increasing.
[/hide]

[hide="Point number 2"]
If $x<2$, then $\sqrt{2}^x < 2$. Thus all terms of the sequence are less than 2.
[/hide]

Therefore it converges to 2.


Our first competition of the year, the 2005 Fall Startup Event, will take place on Thursday, September 29th.  This 100-problem test is a terrific way to start the year; it is an early contest to get your team focused sooner than usual and is a powerful assessment tool to help you learn the strengths and weaknesses of your team members.  Because the contest lasts just 30 minutes, the problems are fast and easy for those familiar with them, requiring students to quickly recall common equations and relations, and to skip problems that they do not immediately know how to do.  Richard Rusczyk describes this contest as being analogous to a musician's need to practice scales.

     The Fall Startup Event has a registration fee of just fifty dollars, which allows unlimited participation by students at your school.  Schools that register for all five of our contests pay a discounted registration fee of just two-hundred dollars.  Participants receive testing materials approximately one week before the test date by both postal and electronic mail, but schools whose registrations are received within two weeks of the test date will receive materials by e-mail only.

     More information on our contests is available on our website, or you can contact me at clymer@natassessment.com or by phone at (206) 433-7320.

How did everyone do?  I got 60 out of the 67 I put answers down to.  Fall start up's always a fun contest :D

I got to number 74, but I skipped around 5 or so.  I know I missed at least two, but I haven't had it graded yet.  ;)

52

Made a couple stupid mistakes and guessed 1 to all the ones I couldn't do. Actually got me 2 points or so this way. lol, very glad with that.

someone told me I got a 54, haha I did lots of stupid things.    :P But that's what happens on a speed thing like this.

76. Was really careless. Another at my school made 81.

46. probably missed around 10. i'm kinda slow with these fast problems.

just wondering...
when will the results come out?
really want to know what was the highest score. :)

They used to put the "Results out" date on their site last year. Guess not.

For your question about the highest score, last year i believe was like 81-82ish and two years ago someone was near 90. Some amazing scores there.

on their web site it says:

The 2005 Fall Startup Event occurred on Thursday, September 29th, 2005. Results will be available on this site by Monday, October 24th, 2005.

i did a whole bunch, missed a whole bunch, and got a 64. id been out of practice for 2 months, so im happy with that result.


modular arithmetic
number theory open
number theory

Conjecture, For p>2, the $2^{(p-1)}$ th square triangular number is divisible by $2^{p} - 1$ (labeled $M_{p}$) if and only if $M_{p}$ is prime.  I checked this for 2<p<27.  For instance the first four square triangular numbers are 0,1,36 and 1225 and the fourth is divisible by $M_{3}$.
PS In fact it appears that if $M_p$ is prime then for any starting integers $S_{1}, S_{2}$ and the recursive relation $S_{n} = 6S_{n-1}-S_{n-2}$,  $S_{2^{p-1}} = S_{1} \mod M_{p}$

[quote="ramsey2879"]Conjecture, For p>2, the $ 2^{(p \minus{} 1)}$ th square triangular number is divisible by $ 2^{p} \minus{} 1$ (labeled $ M_{p}$) if and only if $ M_{p}$ is prime. I checked this for 2<p<27. For instance the first four square triangular numbers are 0,1,36 and 1225 and the fourth is divisible by M_{3}.[/quote]
You probably meant $ (2^{(p \minus{} 1)} \minus{} 1)$-th square triangular number (they are indexed from 0, so 1225 is the third not fourth square triangular number).
It is easy to prove the "if" part of your conjecture. The "only if" part may be incorrect - it is similar to many other pseudoprimality tests but counterexamples are hard to find since it deals with rather sparse set of Mersenne numbers.

The $ (2^{(p \minus{} 1)} \minus{} 1)$-th square triangular number is
\[ N \equal{} \left( \frac {(1 \plus{} \sqrt {2})^{2^p \minus{} 2} \minus{} (1 \minus{} \sqrt {2})^{2^p \minus{} 2}}{4\sqrt {2}} \right)^2.\]
If $ M_p > 3$ is prime then $ 2$ is a square modulo $ M_p$, implying that both $ 1 \pm \sqrt {2}$ modulo $ M_p$ can be viewed as elements of $ \mathbb{Z}_{M_p}$. Hence,
\[ (1 \pm \sqrt {2})^{M_p \minus{} 1} \equiv 1\pmod{M_p}\]
implying that
\[ N \equiv 0\pmod{M_p}.\]
The same arguments apply to your PS as well.


I need to express the vector in terms of I and j, and round to the nearest tenth if necessary.  QUESTION: A plane with an airspeed of 450mph if flying in the direction N35degreesW

i doubt this is right but this is what i came up with: -450sin35i + 450cos35j

can someone tell me if this is right, and if not can someone show me the right answer, thanks so much :)

Yes, your answer is correct. Of course, you still need to round it to the nearest tenth.