I am pwned, I am still in School. :(
Never had any school breaks.
It is true that I am homeschooled.

Yeah, it's kinda hard to accept that summer vacation is finally over. No more 5 hours of video games every day anymore  :| .

summer has actualy been quite boring 4 me. basicaly, 5 hours of video games is 5 hours of math. btw does anyone here play runescape? ive been doing it over the summer for the heck of it because the arithmatic game and all has become less adicting what with me playing it 100 times a day

I used to play Runescape. I now realize that I was completely wasting my time, and now I put my time to better use.

I used to play runescape...it got boring though...I still like hanging out with friends and laughing at noobs  :D

getting off topic here...

Runescape? Okay, personally I think its a waste of time that could've been spent on math, sports, and better games. What I've been doing is beating Super Paper Mario and Twilight Princess in a total of 3 weeks (and I got a lot of math done of course).

well...then...my life is a waste of time because i have nothing better to do than play runescape. we have a wii with 2 games though. and nerd of the ages, who plays paper mario brothers.

I think you thought I meant Super Mario Brothers. But no, I meant Super Paper Mario, the 3rd Paper Mario game.

http://en.wikipedia.org/wiki/Super_paper_mario

no, i just dont know what it is. never played it before. the name just sounds dumb

[quote="D3m0n Shad0w"][quote="abacadaea"]i used contradictiong to proove im not going back to school. i know this topic shouldnt be on this forum but i dont know which one to put it in.

here it is

assume i am going back to school. the definition of school is a place to learn. i am learning at home. learning to have fun. therefore i am already in school. if i am already in school i cant go to school. i cant go to a place i am already at. if i were going back to school, the statement would be i can go to a place i am already at. those two statements contradict each othder. therefore i am not going back to school.[/quote]

Good one. I've got to use it sometime,  :D[/quote]

same. i'll print this out/copy it and give it to my teachers :lol:

PM me his/her reply. especially your math teacher's. :rotfl:

[quote="abacadaea"]i used contradictiong to proove im not going back to school. i know this topic shouldnt be on this forum but i dont know which one to put it in.

here it is

assume i am going back to school. the definition of school is a place to learn. i am learning at home. learning to have fun. therefore i am already in school. if i am already in school i cant go to school. i cant go to a place i am already at. if i were going back to school, the statement would be i can go to a place i am already at. those two statements contradict each othder. therefore i am not going back to school.[/quote]

Thanks, you just got me out of school!

Yeah, try giving that to your principal! ;)
1=2, your avatar [b][i][u]Really[/u][/i][/b] supports what you said! :rotfl:

im tired of being boosed around by the teacher :rotfl:

And then AoPS mods then declared: Thou shall not revive old threads, unless you have a reason.

well he was the one to revieve it. he posted it 4 days after the other guy did

whatever....appearently mods have been inactive....

What mods?

MithsApprentice hasn't posted anything since May 2007, and MathFiend hasn't posted anything since December 2005, so the only active mod is Klebian.

And not that Kelbian doesn't do anything, but in my opinion, not enough.


[quote="Anonymous52111"][quote="ckck"][quote="Anonymous52111"]Solve this equation and include step-by-step. Thanks.

1/3a - 1/3 = -2[/quote]
[hide]Ok so we have $\frac{1}{3}a-\frac{1}{3}=-2$.
Add $\frac{1}{3}$ on both sides.
$\frac{1}{3}a=-\frac{5}{3}$.
Then multiply by three
$a=-5$.[/hide][/quote]

Okay, I understand how you did solve the problem. I have another question that has been rephrased: 2/3a - 2/3 = -2[/quote]
[hide]Just do the same steps as I did for the first problem. So add 2/3 to both sides to obtain 2/3a=-4/3. Multiplying by 3/2 you get a=-2.[/hide]

[quote="ckck"][/quote][quote="Anonymous52111"][quote="ckck"][quote="Anonymous52111"]Solve this equation and include step-by-step. Thanks.

1/3a - 1/3 = -2[/quote]
[hide]Ok so we have $\frac{1}{3}a-\frac{1}{3}=-2$.
Add $\frac{1}{3}$ on both sides.
$\frac{1}{3}a=-\frac{5}{3}$.
Then multiply by three
$a=-5$.[/hide][/quote]

Okay, I understand how you did solve the problem. I have another question that has been rephrased: 2/3a - 2/3 = -2[/quote][quote="ckck"]
[hide]Just do the same steps as I did for the first problem. So add 2/3 to both sides to obtain 2/3a=-4/3. Multiplying by 3/2 you get a=-2.[/hide][/quote]

Thanks.

[quote="Anonymous52111"]Solve this equation and include step-by-step. Thanks.

1/3a - 1/3 = -2[/quote]

edited.
[hide]
$\frac{1}{3}a=-2+\frac{1}{3}$

multiply by 3 to eliminate the fractions:

$a=-6+1$

$a=\boxed{5}$
[/hide]

[quote="xxxxx"][quote="Anonymous52111"]Solve this equation and include step-by-step. Thanks.

1/3a - 1/3 = -2[/quote]

[hide]
$\frac{1}{3}a=-2+\frac{1}{3}$

multiply by 3 to eliminate the fractions:

$a=6+1$

$a=\boxed{7}$
[/hide][/quote]
$a=6+1$ should be $a=-6+1$.


function
inequalities
inequalities unsolved

Let $ f: R \to R$ be a function which satisfy for every $ t \in (0,1)$ and $ x_1,x_2 \in R$
        $ f(tx_1\plus{}(1\minus{}t)x_2)$ $ \le$ $ tf(x_1)$+$ (1\minus{}t)f(x_2)$
    $ a_1$ , $ a_2$, . . . , $ a_{2004}$ be real numbers such that
       $ a_1$ $ \ge$ $ a_2$ $ \ge$ . . . $ \ge$ $ a_{2003}$  and  $ a_{2004}\equal{}a_1$

Prove that

$ \displaystyle\sum_{k\equal{}1}^{2003} \ f(a_k)a_{k\plus{}1}$  $ \le$ $ \displaystyle\sum_{k\equal{}1}^{2003} \ f(a_{k\plus{}1})a_k$

I don't think this inequality holds...

If we consider it just for three variables, we have to prove, for $ a_1\geqslant a_2\geqslant a_3$,
\[ a_1f(a_2)\plus{}a_2f(a_3)\plus{}a_3f(a_1)\geqslant a_2f(a_1)\plus{}a_3f(a_2)\plus{}a_1f(a_3)\].

For the function $ f(x)\equal{}x^2\minus{}4x\plus{}6$, which is clearly convex, and the values $ (2,1,0)$ the inequality holds, but if we take $ f(x)\equal{}x^2$ and the triple $ (3,2,1)$, it does obviously not hold.

Furthermore, I'm sure we can construct convex functions and decreasing $ n$-tuples of real numbers such that the inequaltiy does not hold for any number of variables...


Find all functions $ P: \mathbb{R} \to \mathbb{R}$ satisfys
\[ P(a\plus{}b\plus{}c) \equal{} P(a\plus{}b\minus{}c\plus{}a^2) \plus{} P(a\minus{}b\plus{}c\plus{}c^2) \plus{} P(\minus{}a\plus{}b\plus{}c\plus{}b^2) \plus{} 2ab \plus{} 2bc \plus{} 2ac\]
for all $ a,b,c \in \mathbb{R}$.

[quote="mathson"]Find all functions $ P: \mathbb{R} \to \mathbb{R}$ satisfys
\[ P(a \plus{} b \plus{} c) \equal{} P(a \plus{} b \minus{} c \plus{} a^2) \plus{} P(a \minus{} b \plus{} c \plus{} c^2) \plus{} P( \minus{} a \plus{} b \plus{} c \plus{} b^2) \plus{} 2ab \plus{} 2bc \plus{} 2ac
\]
for all $ a,b,c \in \mathbb{R}$.[/quote]

None :

$ a\equal{}2, b\equal{}\frac{7}{2}, c\equal{}2$ $ \implies$ $ P(\frac{15}{2})\equal{}P(\frac{15}{2})\plus{}P(\frac{9}{2})\plus{}P(\frac{63}{4})\plus{}36$ $ \implies$ $ P(\frac{9}{2})\plus{}P(\frac{63}{4})\equal{}\minus{}36$

$ a\equal{}\minus{}\frac{9}{2}, b\equal{}0, c\equal{}0$ $ \implies$ $ P(\minus{}\frac{9}{2})\equal{}P(\frac{63}{4})\plus{}P(\minus{}\frac{9}{2})\plus{}P(\frac{9}{2})\plus{}0$ $ \implies$ $ P(\frac{63}{4})\plus{}P(\frac{9}{2})\equal{}0$

Hence a contradiction.


calculus
integration
function
calculus computations

Prove that
[url=http://img39.imageshack.us/my.php?image=29935650.gif][img]http://img39.imageshack.us/img39/3245/29935650.gif[/img][/url] [url=http://img200.imageshack.us/my.php?image=51626456.gif][img]http://img200.imageshack.us/img200/6664/51626456.gif[/img][/url]

Integrating this seems easy enough but when i put the limits I keep getting  -3e^(-8) + 1/4 (incorrect)

hello, by integration by parts we get
$ \int xe^{\minus{}2x}\,dx\equal{}x\left(\frac{\minus{}1}{2}e^{\minus{}2x}\right)\plus{}\frac{1}{2}\int e^{\minus{}2x}\,dx$
Sonnhard.

No this is the not problem. I can integrate it just fine but you've put the limits on it and then compare it to the answer.

[quote="mathema*"]Hi :) !!!

i didn't understand what you want exactly but i give that:

$ \int xe^{ax}dx \equal{} \frac {1}{a}xe^{ax} \minus{} \frac {1}{a^2}e^{ax} \plus{} cte \ \ \ \ \forall a \in \mathbb{R}^*$[/quote]

What i want you to do is to find the definite integral. First you need to integrate the function and then you have to put the upper limit and the lower limit.

Now Ive been trying to match it with the real answer for hours but nothing so far.

You can also use $ \int^4_0x(e^{ \minus{} 2x})'dx \equal{} [xe^{ \minus{} 2x}]^4_0 \minus{} \int^4_0e^{ \minus{} 2x}dx$.
But $ \int^4_0x(e^{ \minus{} 2x})'dx \equal{} \minus{} 2\int^4_0xe^{ \minus{} 2x}dx$
I've already checked the result... the expotent of e must be -8!

hello, as i posted above the result is $ \int_{0}^4xe^{\minus{}2x}\,dx\equal{}\frac{1}{4}\minus{}\frac{9}{8}e^{\minus{}8}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, as i posted above the result is $ \int_{0}^4xe^{ \minus{} 2x}\,dx \equal{} \frac {1}{4} \minus{} \frac {9}{8}e^{ \minus{} 8}$.
Sonnhard.[/quote]

Then apparently the answer book has it all wrong :mad:


Let $ a,\ b,\ c$ be positive numbers.
Prove that $ a^4\plus{}b^4\plus{}c^4\geq abc(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}).$

$ \left(4,0,0\right) \succ \left(\frac{3}{2},\frac{3}{2},1\right)$
By Muirhead

by chev' $ a^4 \plus{} b^4 \plus{} c^4 \ge \frac {a^3 \plus{} b^3 \plus{} c^3}{3}\cdot(a \plus{} b \plus{} c)$
by AM-GM,$ \frac {a^3 \plus{} b^3 \plus{} c^3}{3} \ge abc$ and $ a \plus{} b \plus{} c \ge \sum ab^{\frac{1}{2}}$
isn't too easy for this forum?

[quote="kunny"]Let $ a,\ b,\ c$ be positive numbers.
Prove that $ a^4 \plus{} b^4 \plus{} c^4\geq abc(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}).$[/quote]
This problem is similar to Canada's competition topic
$ a^4 \plus{} b^4 \plus{} c^4\geq abc(ab \plus{}bc \plus{}ca).$

Oh, really? I didn't know it.

What are Muirehead and chev'?

That's all Greek to me! :wink:

well maybe you know it like this Tchebychef's Inequality if not see the aops wiki

[quote="kunny"][quote="nguyenvanhuyen"][quote="kunny"]Let $ a,\ b,\ c$ be positive numbers.
Prove that $ a^4 \plus{} b^4 \plus{} c^4\geq abc(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}).$[/quote]
This problem is similar to Canada's competition topic
$ a^4 \plus{} b^4 \plus{} c^4\geq abc(ab \plus{} bc \plus{} ca).$[/quote]

This is a famous problem in Japanese university entrance exam, say, Nagasaki University entrance exam in 1970's.

Could anyone solve the original problem in using more basic tools, if possible, without Muirehead or Tchevichev or

Rearrangement Inequalities for Japanese highschool students who don't know those inequalities?[/quote]
By AM-GM: $ \frac{3a^4\plus{}3b^4\plus{}2c^4}{8} \ge abc\sqrt{ab}$

[quote="Mathias_DK"][/quote][quote="kunny"][quote="nguyenvanhuyen"][quote="kunny"]Let $ a,\ b,\ c$ be positive numbers.
Prove that $ a^4 + b^4 + c^4\geq abc(\sqrt {ab} + \sqrt {bc} + \sqrt {ca}).$[/quote]
This problem is similar to Canada's competition topic
$ a^4 + b^4 + c^4\geq abc(ab + bc + ca).$[/quote]

This is a famous problem in Japanese university entrance exam, say, Nagasaki University entrance exam in 1970's.

Could anyone solve the original problem in using more basic tools, if possible, without Muirehead or Tchevichev or

Rearrangement Inequalities for Japanese highschool students who don't know those inequalities?[/quote][quote="Mathias_DK"]
By AM-GM: $ \frac {3a^4 + 3b^4 + 2c^4}{8} \ge abc\sqrt {ab}$[/quote]

Exactly same as my solution. :lol:

Thank you!

how about this -
By AM-GM,
$ \frac{a^3}{bc} \plus{}\frac{a^3}{bc} \plus{}\frac{b^3}{ca} \plus{}\frac{c^3}{ab} \ge 4a$
add the cyclic inequalities to get -
$ \frac{a^3}{bc} \plus{} \frac{b^3}{ca} \plus{}\frac{c^3}{ab} \ge a\plus{}b\plus{}c \ge \sqrt{ab} \plus{}\sqrt{bc} \plus{}\sqrt{ca}$
now multiply by $ abc$ to get the question! :P

We are done with this inequality once we prove $ \sum a^4 \ge abc(\sum a) \ge abc(\sum \sqrt{ab})$.
By the Power Mean, $ \sum a^4 \ge (\frac{\sum a}{3}) ^3 (\sum a) \ge abc(\sum a )$ by AM GM.

isnt that wat i did?? although i used chev'

[quote="kunny"][quote="geniusbliss"]how about this -
By AM-GM,
$ \frac {a^3}{bc} \plus{} \frac {a^3}{bc} \plus{} \frac {b^3}{ca} \plus{} \frac {c^3}{ab} \ge 4a$
add the cyclic inequalities to get -
$ \frac {a^3}{bc} \plus{} \frac {b^3}{ca} \plus{} \frac {c^3}{ab} \ge a \plus{} b \plus{} c \ge \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}$
now multiply by $ abc$ to get the question! :P[/quote]

Nice solution! :lol:[/quote]
thnx friend

Here is another my solution:

$ f(a): \equal{} a^4 \plus{} b^4 \plus{} c^4 \minus{} abc(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca})$, W.L.O.G. $ a\geq b\geq c > 0.$

$ f'(a) \equal{} 4a^3 \minus{} bc(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}) \minus{} abc \cdot \frac {\sqrt {b} \plus{} \sqrt {c}}{2\sqrt {a}}$

$ \equal{} 4a^3 \minus{} \frac 32\sqrt {a}bc(\sqrt {b} \plus{} \sqrt {c}) \minus{} bc\sqrt {bc}$

$ \geq 4a^3 \minus{} \frac 32\sqrt {a}\cdot a\cdot a(\sqrt {a} \plus{} \sqrt {a}) \minus{} a^2\sqrt {a\cdot a}$, yielding $ f'(a)\geq 0$.

$ \therefore f(a)\geq f(b) \equal{} 2b^4 \plus{} c^4 \minus{} 2b^2c\sqrt {bc} \minus{} b^3c: \equal{} g(b)$,

$ g'(b) \equal{} 8b^3 \minus{} 3b^2c \minus{} 5b^{\frac 32}c^{\frac 32} \equal{} 3b^2(b \minus{} c) \plus{} 5b^{\frac 32}(b^{\frac 32} \minus{} c^{\frac 32})\geq 0$

$ \therefore g(b)\geq g(c) \equal{} 3c^4 \minus{} 3c^4 \equal{} 0$

$ \therefore f(a)\geq f(b) \equal{} g(b)\geq g(c) \equal{} 0$ Q.E.D.


Given a six digit number N = abcdef which is divisible by 7. Show that  another six digit number M = edfabc is also divisible by 7.

Did you mean M = defabc?  Because I don't think edfabc works, but defabc does...

:blush: Sorry! i made a mistake!!!!!
it is M = defabc

[hide]Since 999999= 142857*7, N=abcdef :equiv: 10(abcdef)-999999a=bcdefa=M (mod 7). So N :equiv: M (mod 7). Repeat with bcdefa to get to cdefab and then to defabc.[/hide]

Sorry I can not understand your solution. Could you please write it more deliiberately?

[hide]
$abcdef = 1000(abc)+def$
$defabc = 1000(def)+abc$

We know $7|(1000(abc)+def)$

Add up $1000(def)+abc$ and $1000(abc)+def$ to get $1001(abc+def)$.

Since $1001$ is divisible by $7$, the sum of the two numbers is divisible by $7$. And since $abcdef$ is divisible by $7$, $defabc$ must also be.[/hide]

[quote="limkimsrun"]thank you paladin8. You are far smarter than me.[/quote]

Haha. Thanks, I guess, but believe me, you'll find quite a few people around here that are several leagues beyond me.

yea...I have to say...I appreciate it when I see manipulations like that...


Each horse on a ranch in North Dakota needs at least 30 acres of land for grazing. How many horses could graze on a 5000-acre ranch in North Dakota?

We take the floor of 5000/30.

5000/30 = 166.666666......

So the number of horses which can graze is $ \boxed{166}$

EDIT: Thanks jxl28. My mistake.  :)

I got a different answer...
\begin{align*} \\
\lfloor \frac {5000}{30}\rfloor & = \lfloor \frac {500}{3}\rfloor \\
& = \frac {498}{3} \\
& = \boxed{166}\end{align*}
I think you did $ \lfloor \frac{5000}{33}\rfloor$


Find all solutions in positive integers, $a,b,c$ to the equation

$a!b! = a!+b!+c!$

[quote="chesspro"][quote="vishalarul"]I don't know if this is helpful, but...
[hide="Hint"]$c!+1 = (a!-1)(b!-1)$[/hide][/quote]

This is helpful because you can see that $c$ is the greatest of the three integers. Also, examine the units digit in the RHS to establish an upper bound for $a$ and $b$.[/quote]

[hide] $a,b\ge2$. Since $c$ is greater than $a$ and $b$, $c\ge3$. So, $c!+1\equiv7,5,1\bmod{10}$. $x!-1\equiv1,5,3,9\bmod{10}$.

Case 1: $c!+1\equiv7\bmod{10}$, so $(a!-1)(b!-1)\equiv3\cdot9\bmod{10}$. Either $a$ or $b$ must be 4 and the other must be greater than 4. But $c=3$. Contradiction.

Case 2: $c!+1\equiv5\bmod{10}$, so $(a!-1)(b!-1)\equiv5\cdot(1,5,3,9)\bmod{10}$. $a=3$, $c=4$. $4!+1=(3!-1)(b!-1)\Longrightarrow25=5(b!-1)\Longrightarrow b=3$. $\boxed{(3,3,4)}$.

Case 3: $c!+1\equiv1\bmod{10}$, so $(a!-1)(b!-1)\equiv1\cdot1\bmod{10}$. $a=2$, $b=2$, $c\ge5$, which is impossible. [/hide]

[b]lotrgreengrapes7926[/b], your last case is still satisfied if $a!-1,b!-1\equiv 9\pmod{10}$.


hello!
i want to learn mechanics,right from the start.
i know maths for it pretty well,so i need a good
physics book that explains the physical phenomena
in mechanics,i saw that for college they start
analysing non-linear movement of some stuff like
pendulum etc,i want to learn that part of mechanics
also.my question is if on the net there is something
i could read,or a common book you guys would
suggest for learning.
thank you!

Basically any college-level text will do.  Personally, I recommend the book by Young & Freedman.


geometry
circumcircle
geometric transformation
inradius
quadratics
trigonometry
trapezoid

From the excenter $I_{a}$ drawn the tangent $I_{a}T$ to the circumcircle $(O)$ of $ABC$. If $I_{a}P$ cut $(O)$ at $S,Q$ where $P$ is the Bevan point of $ABC$ and $K$ is the touch point of the excircle $(I_{a})$ with $BC$, prove that $<KTS=<PTQ$

Maybe I can't make the translation right but what do you mean by' the Beaven point of the triangle ABC'? :huh:

Let $R,\ r,\ r_{a},\ s,\ \triangle$ be the circumradius, inradius, A-exradius, and semiperimeter, and area of the triangle $\triangle ABC.$ The Bevan point P of this triangle is the circumcenter of its excentral triangle $\triangle I_{a}I_{b}I_{c},$ (O) is the 9-point circle of the excentral triangle, hence $I_{a}P = 2R.$ Since $I_{a}K \perp BC,\ I_{a}A \perp I_{b}I_{c}$ are isogonals of $I_{a}I_{b},\ I_{a}I_{c}$ in the excentral triangle and $I_{a}A$ its altitude, $P \in I_{a}K.$ In order to show that TP, TK are isogonals of TQ, TS in the triangle $\triangle TQS,$ we have to show that

$(?)\ \ \ \frac{KQ}{KS}\cdot \frac{PQ}{PS}= \frac{TQ^{2}}{TS^{2}}$

Since $I_{a}T$ is a tangent from $I_{a}$ to (O), a circle centered at $I_{a}$ and with radius $I_{a}T$ is perpendicular to (O), hence it is the T-Apollonius circle of the triangle $\triangle TQS$ and consequently, $\frac{TQ^{2}}{TS^{2}}= \frac{I_{a}Q}{I_{a}S}.$ Thus we have to show that

$(?)\ \ \ \frac{KQ}{KS}\cdot \frac{PQ}{PS}= \frac{I_{a}Q}{I_{a}S}$

Assume the points $I_{a}, S, K, Q$ follow on the line $I_{a}K$ in this order and $KQ > 0,\ KS < 0.$ Then $PQ = I_{a}Q-I_{a}P = I_{a}Q-2R > 0,$ $PS = I_{a}S-I_{a}P = I_{a}S-2R < 0,$ which allows to elliminate PQ, PS:

$(?)\ \ \ \frac{KQ}{KS}\cdot \frac{I_{a}Q-2R}{I_{a}S-2R}= \frac{I_{a}Q}{I_{a}S}$

Power of $I_{a}$ to (O) is $I_{a}S \cdot I_{a}Q = I_{a}O^{2}-R^{2}= 2r_{a}R,$ which allows to elliminate $I_{a}S:$

$(?)\ \ \ \frac{KQ}{KS}\cdot \frac{I_{a}Q^{2}-2R \cdot I_{a}Q}{2r_{a}R-2R \cdot I_{a}Q}= \frac{I_{a}Q^{2}}{2r_{a}R}$

Reducing by $\frac{I_{a}Q}{2R},$

$(?)\ \ \ \frac{KQ}{KS}\cdot \frac{I_{a}Q-2R}{r_{a}-I_{a}Q}= \frac{I_{a}Q}{r_{a}}$

Power of K to (O) is $KQ \cdot KS = KB \cdot KC =-(s-b)(s-c),$ which allows to elliminate $KS:$

$(?)\ \ \ \frac{KQ^{2}}{(s-b)(s-c)}\cdot \frac{I_{a}Q-2R}{I_{a}Q-r_{a}}= \frac{I_{a}Q}{r_{a}}$

Finally, $I_{a}Q = I_{a}K+KQ = r_{a}+KQ,$ which allows to elliminate $I_{a}Q:$

$(?)\ \ \ \frac{KQ \cdot(KQ+r_{a}-2R)}{(s-b)(s-c)}= \frac{KQ+r_{a}}{r_{a}}$

Thus we have to show that $y_{1}= KQ > 0,\ y_{2}= KS < 0$ are the positive and negative roots of the quadratic equation

$y^{2}-[2R-(r_{a}-r)]\ y-(s-b)(s-c) = 0$

where we substituted

$\frac{(s-b)(s-c)}{r_{a}}= \frac{(s-a)(s-b)(s-c)}{\triangle}= \frac{\triangle}{s}= r$

Simplifiying,

$2R-(r_{a}-r) = 2R-a \tan \frac{A}{2}= 2R\left(1-\sin A\ \frac{1-\cos A}{\sin A}\right) = 2R \cos A$

$y^{2}-2R \cos A \cdot y-(s-b)(s-c) = 0$

But it is clear that $KQ \cdot KS =-(s-b)(s-c)$ from the power of K to (O) and $KQ+KS = 2R \cos A$ (where $KQ > 0,\ KS < 0$) just from reflecting the line BC in the triangle circumcenter O.

[color=darkred][size=117][b]Thank you, Yetti, very nice, congratulations ![/b][/size][/color]

Thanks, but I could have made it simpler. Multiplying out the 3rd equation in question,

$(?)\ \ \ \frac{KQ}{KS}\cdot \frac{I_{a}Q-2R}{I_{a}S-2R}= \frac{I_{a}Q}{I_{a}S}$

$(?)\ \ \ KQ \cdot (I_{a}Q-2R) \cdot I_{a}S = KS \cdot (I_{a}S-2R) \cdot I_{a}Q$

Power of $I_{a}$ to (O) is $I_{a}S \cdot I_{a}Q = I_{a}O^{2}-R^{2}= 2Rr_{a}.$ Substituting this,

$(?)\ \ \ KQ \cdot (2Rr_{a}-2R \cdot I_{a}S) = KS \cdot (2Rr_{a}-2R \cdot I_{a}Q)$

Reducing by 2R,

$(?)\ \ \ KQ \cdot (r_{a}-I_{a}S) = KS \cdot (r_{a}-I_{a}Q)$

Finally, substituting $I_{a}Q = I_{a}K+KQ = r_{a}+KQ,$ $I_{a}S = I_{a}K+KS = r_{a}+KS,$ where $KQ > 0,\ KS < 0,$

$(?)\ \ \ KQ \cdot KS = KS \cdot KQ$

which is an obvious identity. I did not even need the power of K to (O).

$\triangle ABC$ and its circumcircle $(O,R)$ become orthic triangle and 9-point circle of its excentral triangle $\triangle I_aI_bI_c$ $\Longrightarrow$ circumcenter of $\triangle I_aI_bI_c$ (Bevan point P of ABC) is on perpendicular $I_aK$ from $I_a$ to $BC$ and $PI_a=2R.$ Power of $I_a$ WRT $(O,R)$ is ${I_aT}^2=2R \cdot r_a=I_aK \cdot I_aP$ $\Longrightarrow$ $\odot(TPK)$ is internally tangent to $(O)$ at $T$ $\Longrightarrow$ $T$ is exsimilicenter of $(O) \sim \odot(TPK).$ Thus if $TP,TK$ cut $(O)$ again at $P',K',$ we have $PK \parallel P'K',$ i.e. $SQK'P'$ is isosceles trapezoid with circumcircle $(O)$ $\Longrightarrow$ arcs $SP'$ and $QK'$ are equal $\Longrightarrow$ $\angle STP'=\angle QTK',$ i.e. $TP,TK$ are isogonals WRT $\angle STQ$ $\Longrightarrow$ $\angle KTS=\angle PTQ.$

[quote="Luis González"]$\triangle ABC$ and its circumcircle $(O,R)$ become orthic triangle and 9-point circle of its excentral triangle $\triangle I_aI_bI_c$ $\Longrightarrow$ circumcenter of $\triangle I_aI_bI_c$ (Bevan point P of ABC) is on perpendicular $I_aK$ from $I_a$ to $BC$ and $PI_a=2R.$ Power of $I_a$ WRT $(O,R)$ is ${I_aT}^2=2R \cdot r_a=I_aK \cdot I_aP$ $\Longrightarrow$ $\odot(TPK)$ is internally tangent to $(O)$ at $T$ .[/quote]

So $\angle PKT = \angle {I_a}TP$,$\angle SQT = \angle {I_a}TS$, hence $\angle QTK = \angle {I_a}TP - \angle {I_a}TS = \angle PTS$, and we have $\angle KTS = \angle PTQ$.


function
calculus
derivative
logarithms
number theory
relatively prime
number theory unsolved

Let a,b be positve integer of numbers and a>b.Solve the equation:
                (a-b)^ab=a^bb^a

[hide="Solution"]Rewrite this as $a-b=a^{1/a}b^{1/b}$.

For $n\geq 5$ the function $f(n)=n^{1/n}$ is strictly decreasing (its derivative is $f'(n)=n^{1/n}{{1-\ln n}\over {n^2}}$, which is negative for $n\geq 3$). Hence $a>b\geq 5\implies a^{1/a}b^{1/b}<5^{1/5}\cdot 5^{1/5}=5^{2/5}<2$, since $5^2<2^5$. Therefore, $a-b<2$ and can only be equal to $1$. But for $a=b+1$ we get $1=(b+1)^{1/(b+1)}b^{1/b}\implies 1=(b+1)^bb^{b+1}$ which is clearly impossible for $b\geq 5$.

For $b=4, a\geq 5$ we have $a^{1/a}b^{1/b}\leq4^{1/4}5^{1/5}<2$ since $5^2<2^5\implies 5^4<2^{10}\implies 5^44^5<2^{20}$

For $b=3, a\geq 5$ we have $a^{1/a}b^{1/b}\leq3^{1/3}5^{1/5}<2$ since $3^5<2^8\land 5^3<2^7\implies 3^55^3<2^{15}$

For $b=2, a\geq 5$ we have the same as for $b=4$ since $2^{1/2}=4^{1/4}$

For $b=1$ and any $a$ the equation becomes $(a-1)^a=a$, which has no solution since $a$ and $a-1$ are relatively prime and can't have the same prime factors.

Therefore, $a,b\in\lbrace 2,3,4\rbrace$. Check the three possibilites - the only solution is $a=4, b=2$.[/hide]


function
trigonometry
calculus
calculus computations

Is every bounded function has infimum and supremum?
 help me with an example?

I recommend you answer the following:

1. What does it mean for a function to be bounded from above and from below?

2. What is the definition of an infimum, a supremum?

If f:R---->R ,then f(x) is said to be bounded above means ,f(x)<=M for all x belongs to R
  If f(x) is bounded below means f(x)>=M  for all x belongs to R

 supremum and Infimum
  If M is said to be supremum ,then all elements of f are less than or equal to M and if any other upper bound for f ,then it should be greater than M

 If m is said to be infimum ,then all elements of f are greater than or equal to M and if any other lower bound for f ,then it should be less than m

Also clarify this problem
 If f(x)=x , for all x belongs to (0,1) is continous and bounded but attains neither the infimum nor the supremum

Explain this problem..

There is a theorem that says a continuous function on a CLOSED interval attains both maximum and infimum.

For additional counter example: Consider $ f(u) \equal{} \sin u$ on an open interval.

akech is referring to the [url=http://en.wikipedia.org/wiki/Extreme_value_theorem]extreme value theorem[/url], which is one of the "3 hard theorems" of calculus.

what  are the   3 difficult theorems of calculus?

I don't think "difficult" is the appropriate word.  They're called "hard theorems" because their proofs require new techniques and their results are fundamental to the rest of calculus.  The three "hard theorems" are:
1) [url=http://en.wikipedia.org/wiki/Intermediate_value_theorem]Intermediate value theorem[/url]
2) [url=http://en.wikipedia.org/wiki/Extreme_value_theorem]Extreme value theorem[/url]
3) [url=http://en.wikipedia.org/wiki/Extreme_value_theorem]Boundedness theorem[/url] (same page, just scroll down)

They're pretty intuitive statements, but their proofs require us to introduce the idea of supremums and infimums.

bharathisid u r really a f.....
for the example u gave 0 and 1 are resp infimum and supremum

Yes, and the function attains neither of those values, as barathisid wrote.


Does there exist a non constant polynomial p(x) such that

p(n) = k!

(for k a positive integer) if n is any positive integer?

[i]Hungary 2002[/i]

Maybe you wanted to say:
  For all n there exists k_n such that p(n)=k_n!. In this case the problem is allmost trivial. Suppose p is not constant. Then it is clear that there exists M such that for all n>M we have p(n+1)>p(n).Since lim p(n+1)/p(n)=1, there exits N such that for all n>N we have p(n+1)<p(n). Thus for all n>max{M,N} we have k_n!<k_n+1!<2k_n!, which is clearly a contradiction.


If f (x) = x(x(x(x +1) +1) +1) +1 then when simplified, f(x) equals

A. 4x^4 +1
B. x^4 + 2x^3 + 3x^2 + 2x +1
C. 4x^4 + 4
D. x^4 + x^3 + x^2 + x +1
E. none of these


You can post the problems you are unsure about on aops, I'm sure that the first 20 or so your students can agree on answers and for the rest you can see which answer is most common and dispute.

Though this indeed isn't as convenient as have the answer keys, I agree with that.

[quote="life1"]I have state mathcounts tests for 84, 92, 93, 96, 97, 98, and 99 but NO ANSWERS!  I can't have kids I'm coaching do tests without knowing the answers. Can  you please help with any or all of these? Thanks very much.[/quote]

chess64 might have answers for some of them

but normally what I would do in you position is if no one can agree on an answer then post it here on AoPS for a definite one

I think some of those problems are in the 150 greats math counts problem books so the solutions would be there too

yah, but that might take MORE time then posting them on aops

just bug someone who got top 30 at nats for them :P

no, really, do bug them. please.

jorian

1. Post each problem in a separate thread for discussion.

2. Post in the thread called "MATHCOUNTS Resources," it was created for a reason.

Thanks.


inequalities
logarithms
function
geometry
similar triangles

If the arithmetic mean and the geometric mean of two numbers are equal, what is the relationship between the numbers? Justify.

[hide="Huge hint"]OK so now what if $ \frac{a\plus{}b}2\equal{}\sqrt{ab}$? Simplify this expression.[/hide]

[quote="10000th User"][hide="Huge hint"]OK so now what if $ \frac {a \plus{} b}2 \equal{} \sqrt {ab}$? Simplify this expression.[/hide][/quote]
simplify so that a and b are on one side?

Did you even try to do the work? Simplify as much as you can... another hint is: the final equation is a VERY simple looking equation. Another giveaway: solve for $ a$ and you will see a relationship between the numbers.

[hide="Can not resist..."]
Actually, since you're not proving the whole theorem, you could technically use the equality condition of AMGM.  :D

[hide="Real solution"]
$ \frac {a \plus{} b}2 \equal{} \sqrt {ab}$

$ \Rightarrow a \plus{} b \equal{} 2\sqrt {ab}$

$ \Rightarrow a^2 \plus{} 2ab \plus{} b^2 \equal{} 4ab$

$ \Rightarrow a^2 \minus{} 2ab \plus{} b^2 \equal{} 0$

$ \Rightarrow (a \minus{} b)^2 \equal{} 0$

$ \Rightarrow \boxed{a \equal{} b}$

[/hide]
[/hide]

That solution is too easy! C'mon, that theorem is so simple that the solution has to be as hard as possible :D

[hide]
1. Take rearreangement inequality which yields $ \sqrt a\sqrt a \plus{} \sqrt b \plus{} \sqrt b\ge \sqrt a\sqrt b \plus{} \sqrt a\sqrt b$ with equality for $ \sqrt a \equal{} \sqrt b$.
2. $ \log x$ is strictly concave so $ \log ab \equal{} \frac {\log a \plus{} \log b}2 \equal{} \log\left(\frac {a \plus{} b}2\right)$ iff $ a \equal{} b$.
3. Divide by $ b$ and set $ x \equal{} \frac {a}{b}$, then the statement is equivalent to that the minimum of $ 1 \minus{} 2\sqrt x \plus{} x$ is 0, so let's differentiate it: $ 1 \minus{} \frac 1{\sqrt x} \equal{} 0$, so $ x \equal{} 1$ and we get the minimum $ (1,0)$.

Anyone else has useless proves? ;)
[/hide]

[quote="bertram"]Anyone else has useless proves? ;)
[/quote]

I have this one:
[hide]By Cauchy, $ (a \plus{} b)(b \plus{} a)\ge (\sqrt {ab} \plus{} \sqrt {ab})^2\Rightarrow a \plus{} b\ge 2\sqrt {ab}$ with equality iff $ \frac ab \equal{} \frac ba\Rightarrow a \equal{} b$.[/hide]

Consider a right triangle inscribed in a semicircle. Draw the altitude to the hypotenuse, let it split the hypotenuse into $ a$ and $ b$. We have (by similar triangles) that $ \sqrt{ab}$ is the length of the altitude, and that the radius is half the hypotenuse, in other words $ \frac{a\plus{}b}{2}$. However, in order for our altitude to equal the leg length, we can see that the angles in the triangle must be 45, implying that the triangle is isosceles, implying that $ a\equal{}b$.

Somewhat lengthier, but it can be a proof without words :)

Or we can use karamata's inequality with the sequences $ x$, $ y$ and $ \frac{x\plus{}y}{2}$, $ \frac{x\plus{}y}{2}$ where $ x\geq y$ and the function $ \ln x$ for a one-liner :D

[quote="diophantient"]
Somewhat lengthier, but it can be a proof without words :)
[/quote]
[img]http://www.cut-the-knot.org/pythagoras/means.gif[/img]

There are more on some old thread on AoPS...


function
search
real analysis
real analysis unsolved

How do you prove that exp ra=(exp a)^r if r is rational?

It depends how you define the function $\exp(x) = e^{x}$.  You should try a search of the forum -- there was a long discussion on this subject not too long ago.


In a class of 23 students, what is the probability that at least two people have the same birthday, not considering February 29.

[hide]23/365*22/364... not in the mood of simplifying that.... I think this is wrong.. :( [/hide]

Here's how you can do it:

[hide]Instead, find the probability no two people have the same birthday and subtract that from 1.
Thus, we have 365 choices for the first birthday, 364 for the next and so on all up until 343.

So number of ways for everyone to have different birthdays is:

365*364*363...*344*343.

Total number of birthdays is 365^23 since every person has a total of 365 possible days to have a birthday.

Thus our total probability is (365*364*363...*344*343)/365^23

And the answer to the question is 1 minus this, or

[b]1 - (365*364*363...*344*343)/365^23[/b]

[/hide]

Of course, that assumes that all the dates are equiprobable (that is, if you take a random person from the population, with a probability of 1/365 his birthday would be, let's say, July 4th) but that is, in general, not true. It depends on many aspects, but for example, I have heard that usually from late July to early October (specially on September) ther usually is a peak on the births in Latin America. I think that should be true, because most of the birthday parties when I was in HS used to be during september, specially early september  :D

23 is the first number of people that the probability that 2 people have the same birthday is greater than 1/2 right?

If you assume that the distribution of birthdays is uniform, yes, it is. The distribution is uniform if you assume that for any date of the calendar, 1/365 of the population was born that day.

Best,

Please express your answer as a percent to the nearest tenth.


analytic geometry
graphing lines
slope
conics
parabola
calculus
calculus computations

Find the equations of all tangent lines to y=x^2 that pass through the point (0,10)

A line is tangent to both f(x)=1-x^2 and g(x)=4-(4-x)^2. Find the slope of this line

Please someone help me out wanna see if I solved this the right way

First Problem:
The point (0, 10) lies 'inside' the parabola, no tangent can pass through this point.

Second:
Let the line $ y=mx+c$ touches both $ y = 1-x^{2}$ and $ y=4-(4-x)^{2}$.  Then both the equations $ mx+c = 1-x^{2}$ and $ mx+c=4-(4-x)^{2}=-x^{2}+8x-12$ must have equal roots.  So, the discriminants of the equations must be zero.
Making the discriminants equal to zero, we get $ m^{2}-4c+4=0$ and $ (m-8)^{2}-4c-48=0$
Eliminating c, we get the value of m as $ 3/4$.

[quote="vidyamanohar"]

Second:
Let the line $ y=mx+c$ touches both $ y = 1-x^{2}$ and $ y=4-(4-x)^{2}$.  Then both the equations $ mx+c = 1-x^{2}$ and $ mx+c=4-(4-x)^{2}=-x^{2}+8x-12$ must have equal roots.  So, the discriminants of the equations must be zero.
Making the discriminants equal to zero, we get $ m^{2}-4c+4=0$ and $ (m-8)^{2}-4c-48=0$
Eliminating c, we get the value of m as $ 3/4$.[/quote]

Your answer is correct, vidyamanohar.

The vertices of the parabola $ y=f(x),\ y=g(x)$ are $ A(0, 1), B(4,\ 4)$ respectively. Since these curves are congruent. Thus the line $ AB$ are pararell to  the line $ y=mx+c$, yielding $ m=\frac{3}{4}$.


$ P(x)\equal{}x^{3}\minus{}12x^{2}\plus{}47x\minus{}60$
the roots are $ x\equal{}3,4,5$
find the boundaries for the dintinct coefficients.

Sorry, could you please explain what you mean by "boundaries for the distinct coefficients"?

sorry (The Zuton Force) i had to explain more,

$ p(x)\equal{}ax^{n}\plus{}bx^{n\minus{}1}\plus{}\cdots\plus{}z$

Let $ a,b\cdots$ be distinct coeficient ($ a\neq b\neq\cdots$) of $ P(x)$

consider $ x^{3}\minus{}12x^{2}\plus{}47x\minus{}60\equal{}(x\minus{}3)(x\minus{}4)(x\minus{}5)$

$ a(3)^{3}\plus{}b(3)^{2}\plus{}3c\plus{}d$
$ a(4)^{3}\plus{}b(4)^{2}\plus{}4c\plus{}d$
$ a(5)^{3}\plus{}b(5)^{2}\plus{}5c\plus{}d$

how many and what values can $ a$,$ b$,$ c$,$ d$ take, to make the roots of $ ax^{3}\plus{}bx^{2}\plus{}cx\plus{}d$ equal to $ 3,4,5$
sorry i dont know how to put it in words

The question appears to be,

"How many choices are there for the coefficients of a cubic polynomial $ p(x) \equal{} ax^{3}\plus{}bx^{2}\plus{}cx\plus{}d$ if it has roots $ 3, 4, 5$?"

yes that is true, thanks JBL

How many choices are there for the coefficients of a cubic polynomial $ p(x) \equal{} ax^{3}\plus{}bx^{2}\plus{}cx\plus{}d$ if it has roots $ 3, 4, 5$?

$ b \equal{}\minus{}12a, c \equal{} 47a, d \equal{}\minus{}60a$.  This follows trivially from the Factor Theorem.  In other words, up to multiplication by a constant, $ x^{3}\minus{}12x^{2}\plus{}47x\minus{}60$ is the unique cubic polynomial with roots $ 3, 4, 5$.


A centipede climbs a 40-foot tree. Each day he climbs 5 feet, and each night he slides down 2 feet. In how many days will the centipede reach the top of the tree?

Because the centipede climbs 5 feet and drops 2 feet, it then accomplishes 3 feet every day and night. So 40/3=13 1/3. That means on the 13th day, it reaches the top.


The 1999 and 2000 exams are on the main ARML site, but where can I find other past exams (outside of 1989-1994, which are in a book, I believe)?

[color=cyan]There is a bood of 1983-1988 ARML-NYSML contests.  I don't know how available it is.  There is a website whose name I'll post on monday that might ahve it.[/color]

Unfortunately, I believe the book Joel mentioned is out of print, and so it may be quite difficult to get it. I seem to remember that it is spiral bound, but I could be wrong about that. A few years ago, Ted Alper, coach of the SFBA ARML team, said he was interested in making a book of more recent ARML problems, but I don't know what happened to that project. He may still be working on it, in which case it is possible that it will come out soon.

[color=cyan]Yes, it is spiral bound.  Actually, I'm not too surprised it's out of print.  Too bad.

It's the "Green Book."  The newer one is the "Blue Book."  :)[/color]

[quote]There is a bood of 1983-1988 ARML-NYSML contests. I don't know how available it is. There is a website whose name I'll post on monday that might ahve it.[/quote]

whats that website?

[color=cyan]Make that Tuesday.  Sorry.  I'll try not to forget.  Sometime this week, anyhow.  Long, if you read this, remind me.[/color]

Has anyone discovered any more websites of ARML problems, like the Nebraska Team Page Fierytycoon found some problems on?

[color=cyan]I haven't checked this myself, but the website http://www.mathpropress.com has a pretty good selection of problem books of all sorts.  You might check it out.  Also, you don't need authentic ARML problems to practice for AMRL -- AMC problems (wihtout looking at the multiple choices, of course) and easier AIME questions are approximately the same level.  And many of the littler contests, such as Mandelbrot.  Also, http://www.mandelbrot.org has a "Problem Corner" which has some good problems most of which are of a reasonably difficulty.[/color]

These aren't actual ARML problems, but are from a contest that Duke puts on for anyone interested:

http://www.math.duke.edu/dumu/contests/problems/index.html

Also, the contest is Nov 15 for anyone who can round up a school team for it.

http://www.math.duke.edu/dumu/FallContest03/

[quote="darkquantum"]OK, sure. I have ARML POwer 1987, 1991, and 1994 if anyone is interested...[/quote]

I'm interested.  Could you please post those (one problem at a time for legal reasons?) or something to that effect?

You can't really post an ARML power one question at a time -- it doesn't really work.  And that shouldn't change the status of the copywrite violation anyhow, I don't think.

Oh.  Well, then, what does Darkquantum mean by "interested"?

PM him.  My guess would be that he would e-mail it to you.

i dont know if you want to, but if you really care you could buy the contest books, one which has all questions from 89-94 of ARML and NYSML, and another book is coming out soon with 95-99 or something like that.  But they're kind of expensive... $20 per book i think.


${\left(x_{1}\right)^{\frac{n}{k}}+\left(x_{2}\right)^{\frac{n}{k}}+\left(x_{3}\right)^{\frac{n}{k}}\text{ }\geq \text{ }\left(x_{1}+x_{2}+x_{3}\right)^{\frac{n}{k+1}}}$
${x_{1}>0, x_{2}>0, x_{3}>0, n >0 , }$
${4\geq k\geq 1 }$

Dump. How can I proove or disproove this?
$n,k$ are not necessarily integers.

if you mean this holds for all reals $n$, then taking $x_{i}=1$ implies $3\ge 3^{\frac{n}{k+1}}$ which is not true for all n. :P

[quote="yagaron"]$n,k$ are not necessarily integers.

but

$n,k$ are not necessarily Reals.

I just dont know what kind of numbers $n,k$ need to be for the inequality becomes true.
It is what I thougth "not necessarily integers" :D[/quote]

Dear Yagaron,

Your inequality is obviously wrong. It is not homogeneous. So think about it again, nayel shows you a counter-example.

hm...

If $n < k+1$ it become true?

there is no way to it become true!?

:( sorry for being dunce


trigonometry
inequalities
geometry
3D geometry
prism
algorithm
number theory

I don't know if this is in the right forum (so move it if it's not).

What is different between a law and a theorem?  Like as in the [i]Law[/i] of Cosines, but the Pythagorean [i]Theorem[/i].  When, if it can, a theorem become a law?

I think it's based on what it was named originally, or the naming convention that effected afterwards.

In other words, it has nothing to do with importance.

Some others of high importance are Cramer's Rule, Hensel's Lemma, Bertrand's Postulate (which is a theorem), the AM-GM Inequality, the Sum-to-Product Identities, Steiner's Porism, etc.

Or the Euclidean Algorithm (also a theorem).  And, of course, there was Fermat's Last Theorem, which was given that name when it was still a conjecture.

For math, these naming conventions don't matter, although sometimes the chosen name has some meaning.  For instance, the Euclidean Algorithm is a theorem, but it's a theorem which provides an algorithmic result, so calling it an algorithm tells you something useful about it.

In most cases, a law is something that is well known and applies in at least many cases, if not all known cases, such as the law of conservation of energy and mass.  One can also refer to the law of conservation of mass, knowing that it strictly doesn't apply to atomic reactions.

A theorem expresses the results of a theory or an idea that one thinks is true.

While it is true that many things once named don't have the name upgraded from a theorem to a law, if the convention were followed a law would start out as a theorum.


Kiran Kedlaya has released his work thus far on Geometry Unbound, a book on olympiad-level geometry (and beyond). The book is free, under the GNU Free Documentation license. It's very challenging, but those of you who are comfortable getting into olympiad geometry should check it out. More info, and the book itself, here: [url]http://math.mit.edu/~kedlaya/geometryunbound[/url]

If anyone takes the time to add the diagrams to it, please let us know!

[quote="rrusczyk"]
If anyone takes the time to add the diagrams to it, please let us know![/quote]

I did! :)
I've just finished drawing the diagrams with this program:
http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/doc_en/index.html

You can download them here:
http://andfog.altervista.org/math/geometry-unbound-figures.zip

This is just the first version, some diagrams should be improved or added in the future... and it would be better to put them directly into the book.
I've exported each diagram in three formats:
- ZuL's native format (.zir), so you'll be able to edit the constructions
- .png, compressed raster images
- .svg, vectorial images

That's all. Bye bye (I hope you understand my poor English  :blush: )

I can't download the diagrams. Could you relink them? Thanks.

Also, .eps is usually the easiest format to include into a LaTeX file.

Thank you for the feedback!
I tried to add .eps files (but they don't look so good  :oops: )

I've put the diagrams ad an attachment.

Very nice, thanks. I'll try to add them in when/if I go through GU.

Hey y'all...anyone know where one could find the solutions to the problems? At the back of the book, it said that the solutions are on some website, but it didn't list where. Thanks!

A TeX file of the solutions for the original MOP notes is [url=http://web.mit.edu/rwbarton/Public/mop/kiran.tex]here[/url], I think.

Has anyone gone through and added the figures?

BUMP!
Yeah... so has anyone added the figures?

So, I tried opening the solutions, but it first took me to some link that asked me to select a program to open this file, and when I chose Notepad, it just took to a bunch of source code. Does anyone have a solution to this? I would really appreciate it.

Thanks for trying to do this.

It's a [b].TeX[/b] file, you need [url=http://www.texniccenter.org/]TexnicCenter[/url] (or some other program) to open the files.

When I'm trying to add in the figures and compile the .tex version of Kedlaya's Geometry Unbound, TeXnic Center gives me the error of "File: 'custombooks.cls' not found."

It seems like there needs to be a custombooks.cls file in ~\MiKTeX 2.9\tex\latex\base - am I supposed to create that myself as an empty file or is there somewhere to download that? I haven't been able to find such a file anywhere, so far.

If I change the document class to something like book, it gives me another error which involves something about the .tex file's .ind file. Does anyone know how to fix any of these issues?

Edit: Well I managed to resolve the above after renaming the .ind file to geometry-unbound.ind not gu-060118.ind, changing the document class to "book" rather than "custombooks," and compiling twice to fix the errors in the table of contents. There's still a problem though - I can't seem to include the figures. When I put in "\includegraphics{fig1.6.1.svg}", it loads [b]that[/b] into the text and not the figure.

Second Edit: That was resolved by putting in the preamble "\usepackage{graphicx}". New problem: LaTeX does not recognize .svg file type (and .png looks just ugly... so that's not an option :P).

Just wondering if the solutions are anywhere. I can't seem to find them, and the previous link doesn't work :(


i was wondering if anyone knew anygood sites for ti-89 titanium games/apps
thanks :)


algebra
polynomial
vector
invariant
complex numbers
superior algebra
superior algebra solved

could we consider the number of ideals I such that the quotient of the ring mod I is isomorphic to C[x] ? It seems like there are 2 in the first ring and 3 in the second ring, so they can't be isomorphic.

There are $\infty$ such ideals, e.g. all those generated by linear terms $x-a$ etc..

One could solve this by using the notion of transcendence degree, which is like the dimension of normal vector spaces an invariant:
Let $K \subset L$ be fields. Take a set $M \subset$ such that all the elements of $M$ are algebraical independent and $M$ is maximal for that (thus adding any more element gives an algebraic dependence).
Then $|M|$ is called the transcendence degree of $L$ over $K$, written $\mathrm{trg}(L|K)$ (or some similar notations).

By definition is $L$ algebraic over $K(M)$, and as for vector spaces one shows:
- such $M$ exists (using AC).
- $|M|$ is invariant, thus only defined by $K,L$.

I don't think I understand how you suggest the concept of the transcendence degree be applied to these rings (or vector spaces)... Can you elaborate a little on that?

If the rings would be ismorphic, then their field of quotients would be so. Thus we show that the field of quotients are not ismorphic.

To prove invariance of $\mathrm{trg}$, one has the following easy lemma (note that for vector spaces we have a very similar result again):

Let $x_{1}, ... , x_{n}$ be algebraical independent over $K$. Let $t$ be algebraical over $K(x_{1},...,x_{n})$ and transcendent over $K$. Then there is an $i$ such that we can replace $x_{i}$ by $t$, and still $x_{1}, ... x_{i-1}, t , x_{i+1}, x_{n}$ are algebraical indpendent (and $x_{i}$ is then algebraical over these others).

[quote="hedgehuogis"]Prove the ring of polynomials in two variables over complex numbers is not isomorphic to the ring of polynomials in three variables over complex numbers. I'd be happy if the proof were as elementary as possible (powerful algebraic weapons that I haven't encountered in my life always make me a bit sick ;))[/quote]

Here is another possible approach:

If there were an isomorphism $f$, then it would send maximal ideals to maximal, thus $f(<x,y>) = <x-a,y-b,z-c>$ for some $a,b,c \in \mathbb{C}$.  Composing $f$ with an automorhism of $\mathbb{C}[x,y,z]$, which sends $a,b,c$ to $0$, we can assume that $a=b=c=0$.
So $f : <x,y>$   $\to$   $<x,y,z>$. Then $f : <x,y>^{2}$   $\to$   $<x,y,z>^{2}$, and $f : <x,y> / <x,y>^{2}$   $\to$   $<x,y,z> / <x,y,z>^{2}$.

But since $(x,y)^{2}= <x^{2},xy,y^{2}>$, $<x,y> / <x,y>^{2}$ considered as a vector space over $\mathbb{C}$ has dimension $2$.
In the same way $<x,y,z> / <x,y,z>^{2}$ considered as a vector space over $\mathbb{C}$ has dimension $3$..

Contradiction with that our $f$ is an isomorphism.

Thank you guys! None of the approaches made me sick, though I prefer the last one;)

Here's yet another approach, due to my instructor; I've managed to sort it out for myself only now:

Suppose f: C[X,Y,Z] -> C[X,Y] is an isomorphism. Then
deg(f(polinomial of degree N)) is less or equal to N*max{deg(f(x)), deg(f(y)), deg(f(z))} = N*const. Now take the basis for polynomials of degree not bigger than N in C[X,Y,Z], then it has cardinality of order N^3, and thus is the cardinality of its image, which has to be linearly independent, however, maximal LI set for polynomials in two variables of degree const*N is of order N^2, which leads to a contradiction for N big enough


The time we land in is full of three generations of robots, well the newest ones or the third generation robots (G3) are the most productive, the second generation (G2) have medium productivity and the first generation robots (G1) are the least productive of the three. The original productivities of the three robots expressed in whole numbers on a scale of 100 are 90, 60 and 30 respectively. Based on this productivity ratio, they were assigned jobs of their own, of 90, 60 and 30 units respectively, but due to some confusion the distribution was mistaken and a few robots ended up doing someone else's job, though not completely. No robot did meet the actual specified productivity and didn't complete the expected unit of work.

G2s did half of the job that G3s would have left, if only G3s had done their set of work and no one else's. G1s had done half the amount of work that G2s would have left of G2s originally assigned chore, if only if G3s would have done some of G2s work and completed one-third of G3s original total workload. G3s did the work that G1s would have left if only if G2s had done G1s work and completed one-third of G2s workload too.

Tell me how many units of work G2s did(just tell the numeric value)?

When I prepare an exam for my students I NEVER DID an incredible and confusing kind of problem, but maybe someone try to solve, but I am sure the answer is between 0 and $\infty$  :P  :P  But in my opinion you need to write in a better and beautiful way to people TRY to solve it as Myth said.

Best regards.
RX the Crazy Math


Find all pairs $(a,\, b)$ of positive integers such that $2a-1$ and $2b+1$ are coprime and $a+b$ divides $4ab+1.$

$a+b | 4ab+1$ $\longrightarrow$ $a+b | 4a^{2}-1$ $\longrightarrow$ $a+b|(2a-1)(2a+1)$ , $a+b|(2b-1)(2b+1)$ , if there exist $p$ such that $p|2a-1$ and
$p|a+b$ then $p|(2b-1)(2b+1)$ , but $2a-1$ and $2b+1$ are coprime $\longrightarrow$ $p|2b-1$ $\longrightarrow$ $p|2a-2b$,$p|2a+2b$ $\longrightarrow$
$p|4a$ but $p|2a-1$ . contradiction .
now we have $(a+b,2a-1)=1$ and we have $(a+b,2b+1)=1$ $\longrightarrow$ $a+b|2a+1$ , $a+b|2b-1$ $\longrightarrow$ $2a+1=m(a+b)$ and $2b-1=n(a+b)$ and $m,n$ are natural numbers because $a$ and $b$ are natural . $\longrightarrow$ $(2a+1)+(2b-1)=(m+n)(a+b)$ $\longrightarrow$ $m+n=2$
$\longrightarrow$ $m=n=1$ $\longrightarrow$ $b=a+1$
$(a;b)=(n;n+1)$

The answer $(a,b)=(n,n+1)$ for every integer $n$.

It is clear from the fact $a+b|4ab+1$ that $a+b|(2a+1)(2b+1)$ and $a+b|(2a-1)(2b-1)$

Now let $gcd(2a-1,a+b)=g$.Hence $g|2(a+b)$ and $g|2a-1$ implies that $g|2b+1$ which contradicts our assumption that

 $gcd(2a-1,2b+1)=1$.So $gcd(a+b,2a-1)$.So $a+b|2b-1$.But $2b+1<2(a+b)$.Hence $2B+1=2a+2b$.And thus $b=a+1.$.Now

it is clear that then $a+b|4ab+1$.Again $gcd(2a-1,2b+1)=gcd(2b+1,4)=1$.Thus every pair of this form satisfies the condition.

So answer is $(a,b)=(n,n+1)$ $\forall$ $n$.

[hide=Solution]
It would be nice to relate the GCD condition with anything appearing in the divisibility question, so using the euclidean algorithm
\[(2a-1,2b+1)=(2a+2b,2b+1)=(a+b,2b+1)=1\]
(The last equality follows because $2$ doesn't divide $2b+1$). Aha! we just came up with an $a+b$ so maybe we can rewrite $4ab+1$ in such a way that $2b+1$ appears and then use the GCD to guide us through the divisibility condition. Notice $4ab+1=(2b+1)(2a+1)-2(a+b)$ so for $a+b$ to divide $4ab+1$ we need $a+b|(2b+1)(2a+1)$ but because $a+b$ and $2b+1$ are relatively prime we have $a+b|2a+1$ and this happens iff there is $k \in \mathbb{N}$ such that
\[(a+b)k=2a+1 \rightarrow bk-1=(2-k)a \]
since $bk-1$ is nonnegative $k$ can only be $1$ or $2$, the latter will give us $b=1/2$ so we rule it out, therefore $k=1$ and the solutions are $(a,a+1)$ .


Find the number of odd numbers among the following

$ \binom{n}{0}$,$ \binom{n}{1}$,$ \binom{n}{2}$,........$ \binom{n}{n}$

where $ n\in N$


What integer x satisfies the equation: $ 2^{x \plus{} 2} \equal{} 32$ ?


i have completely messed up in my solution and i can't seem to delete it, so i'll just replace it with this
plz just ignore this post
(i can't delete it)

1) The total volume of those two blocks of clay is $ 2(10^3)$ or $ 2000$. The volume of a single sphere with radius $ 5$ is $ \frac {4}{3}\pi(5)^3$ or roughly $ 524$. Someone correct me if I'm wrong, but the answer is $ 2000 \div 524 \approx \boxed{3}$.

@vallon: The formula is $ \frac{4}{3}\cdot \pi \cdot r^3$

2) Let's call the amount of money she gets each month $ x$. So we have:

$ 12x \minus{} (\frac {1}{2}(12x) \plus{} \frac {1}{6}(12x) \plus{} \frac {1}{12}(12x) \equal{} 7777$

$ 12x \minus{} 9x \equal{} 7777$

$ 3x \equal{} 7777$

$ x \equal{} \frac {7777}{3}$

In one year, she gets $ \$31108$. (That's a lot of money!)

izzy, u are correct for the first one, but the second question askes how much she gets every month
therefore, the answer is
$ \frac {31108} {12} \equal{} 2592 \frac {1} {3}$ dollars every month

PS. i gotta remember the formula for the volume of a sphere....

vallon: read her 2nd to last line. She said x is the money the person makes each month, and then she proceeded to solve for x. Her solution was correct; the last line was just saying that that's how much money she makes in a year.

hmmm

ok. she has 7777 dollars left at the end of teh year. this is 3/4 of her total before she bought toys. so 4/3*7777. this represents her 2/3 money before she bought games. so now we have 4/3*3/2*7777=2*7777. thsi represents 1/2 of her money before she bought clothes. so you have 4*7777 dollars a year. divide that by 12 to gegt 7777/3 dollars a month. my thought is that you may ahve made a typo somewhere in writing the problem

If the answer is 2222, maybe she started with 6666 dollars?

ending up with 6666 dollars makes more sense and fits the answer

[quote="fishythefish"]Um...What exactly is your basis for concluding that the answer is 2222? Try reading either

a) mz94's post

or

b) izzy's post.[/quote]

his answer booklet most likely  :wink:


function
integration
calculus
derivative
calculus computations

Find all continuous functions s. t.
$ y(t) \equal{} \int_0^1\sinh{(x \plus{} t)}y(x)dx$, $ t \in \mathbb{R}$.

Wicker Man, we'd appreciate having an argument to go with that. You can start with showing that $ y$ satisfies the differential equation $ y''\equal{}y.$

$ y' \equal{} \int_0^1\cosh{(x \plus{} t)}y(x)dx$ by Leibniz Integral Rule
$ y'' \equal{} \int_0^1\sinh{(x \plus{} t)}y(x)dx \equal{} y$ by Leibniz Integral Rule
$ y'' \equal{} y \Longleftrightarrow y \equal{} ae^t \plus{} be^{ \minus{} t}$
$ ae^t \plus{} be^{ \minus{} t} \equal{} \int_0^1\sinh{(x \plus{} t)}(ae^x \plus{} be^{ \minus{} x})dx \Longleftrightarrow a \equal{} b \equal{} 0$.

What's the leibniz integration rule? It appeared to me you considered the derivative of an antiderivative as the function itself (FCT1), then re-wrote the statement as an equivalent integral.

No, what he's talking about is this: for sufficiently well-behaved functions,

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

Can you explain that in words? I've never seen that symbol before.

That's just the symbol for a partial derivative - the derivative of $ f(t,x)$ with respect to $ t$ if you treat $ x$ as constant.

The result is just the statement the in favorable circumstances, we can bring the differentiation (with respect to some variable that is not the variable of integration) inside the integral.

for more details: http://planetmath.org/encyclopedia/DifferentiationUnderIntegralSign.html


(i) Solve   $ 6 \sin 2x \equal{} 5 \cos x$; fotr$ 0 < x < 90^{\circ}$.

(ii) $ 8 \cos x \csc^{2} x \equal{} 3$; find the exact value of $ \cos x$.

[hide="Problem 1"]$ 6\sin 2x \equal{} 12 \sin x \cos x \equal{} 5 \cos x$ implies that $ \sin x \equal{} \frac{5}{12}$. Thus, the only solution is $ \boxed{x \equal{} \sin^{\minus{}1} \frac{5}{12}}$.[/hide]

[hide="Problem 2"]$ 8 \cos x \equal{} 3 \sin^2 x$.

$ 8 \cos x \equal{} 3(1\minus{}\cos^2 x)$

$ 3\cos^2 x \plus{} 8\cos x \minus{} 3 \equal{} 0$

$ (3\cos x \minus{} 1)(\cos x \plus{} 3) \equal{} 0$

$ \boxed{\cos x \equal{} \frac{1}{3}}$

[/hide]


What is the formula for finding the sum of an infinite number of numbers in an arithmetic series? And what's the difference between a sequence and a series...?  Is it like one's finite and one's infinite?  What's convergent and divergent, relating to limits?

a series is finite and a sequence goes on forever

Okay. Haha, I just realized that the first one was a really stupid question because in an arithmetic sequence goes on forever...then it's either infinty or negative infinity.  But I would still like to know what are convergent and divergent.

well all i know is that in science, convergent is when two tectonic plates push into one another and divergent is when two tectonic plates move away from eachother.

I'm not very good with limits and stuff yet, but:

Convergent:  the value converges to a specific value, for example $.3+.03+.003....$ is convergent because the value is $\frac{1}{3}$.

Divergent: the value does not converge, it is $\infty$ or $-\infty$.  for example $1+2+3+4+5+.....$ is divergent because it does not approach a specific value.


modular arithmetic
number theory
prime numbers
divisibility tests

[b]Osculator Method/Seed of a number[/b]

Osculator/seed of a number can be negative and positive

Let the number be n,then if nk+1=10m then positive osculator of n is m
And if nk-1=10m then the negative osculator of n is m
Thus if the number is 7
We have 7(3)=21 and 21-1=10(2) so the negative osculator of 7 is 2
We have 7(7)=49 and 49+1=10(5) so the positive osculator of 7 is 5

This osculator is used to check for the divisibility by 7 , 13 and other prime numbers for which there are no straight forward rules.

Lets see whether 7896834 is divisible by 7 or not?
The negative osculator of 7 is 2
We will do the following operations on the number
789683 -4(2)=789675
78967-5(2)=78957
7895-7(2)=7881
788-1(2)=786
78-6(2)=66
6-6(2)=-6 which is not divisible by 7
Hence the number is NOT divisible by 7

Infact the thing which we are doing is this
7896834mod7=>(7.10^6 + 8.10^5 + 9.10^4 + 6.10^3 + 8.10^2 + 3.10^1 + 4.10^0 mod7)
Now gcd(7,10)=1
So we can find integers r and s such that 7r+10s=1
We know 7(3)-1=10(2)=>  the negative osculator of 7 is 2
7(3)+10(-2)=1 hence the value of s=-2

[b]Can any one give me the proof behind this Osculator/Seed of a number.[/b]
Let abcde be a 5digit natural number.
Is abcde mod 7  is the same as
[a(-2)^4 + b(-2)^3 + c(-2)^2 + d(-2)^1 + e(-2)^0] mod 7..I dont think so..

Are you studying Vedic Arithmetic techniques also?  :)
I'm not sure about a proof of why the osculator method works...

First, explain to me why in $ 789683\minus{}4(2) \equal{} 789675$ the number $ 4$ was picked.

MellowMelon,

"We will do the following operations on the number
789683[b]4[/b]
789683 -[b]4[/b](2)=78967[color=red]5[/color]
78967-[color=red]5[/color](2)=7895[color=green]7[/color]
7895-[color=green]7[/color](2)=7881
..."

In fact, what you said is not true. Here's an explanation of what [b]is[/b] true.

We want to find whether $ 10a\plus{}d$ is divisible by $ 7$ for $ 0\leq d\leq 9$
We do this by considering whether or not $ a\minus{}2d$ is divisible by $ 7$.

Now notice that $ (10a\plus{}d)\plus{}4(a\minus{}2d)\equiv 14a\minus{}7d\equiv 0\pmod{7}$
Thus, $ 10a\plus{}d\equiv 3(a\minus{}2d)\pmod{7}$

Notice that we do [b]not[/b] have that $ 10a\plus{}d\equiv a\minus{}2d\pmod{7}$. However, $ 10a\plus{}d\equiv 0\pmod{7}$ if and only if $ a\minus{}2d\equiv 0\pmod{7}$. That is why the test works.

For example, in the case of $ 1002$ we get
$ 100\minus{}2(2) \equal{} 96$
$ 9\minus{}2(6) \equal{}\minus{}3$

But $ 1002\equiv 1\not\equiv\minus{}3\pmod{7}$.

THnx a lot.That's what I had been thinking about..Seed cant give us a Remainder.It can only check for the divisibility by a number  :) Thnx for the xplanation


It's really annoying to look at a topic and see that a lot of the text in the responses has disappeared. Can someone tell me why?

These are spoilers - people hiding their solutions so others can try the problem.  You can reveal them by highlighting them just like you highlight text in a text editor.  Or you can set your options to always reveal them by clicking the Options button above.

I would answer this  too but  put the answer in spoiler        (To see my response follow the above mentioned advice)

[hide]So you have found the secret and now can read this too! Have a nice day and  have fun   < sorry could not resist it >  [/hide]

As Mr. Rusczyk said, the spoilers are for letting people who haven't solved the problem and still want to try to solve it get a fair chance without seeing your solution. It'd be nice to use it in most situations, except with latex.

[quote="RC-7th"]A related question would be: Can moderators edit your post but not let people know?[/quote]

They can; in other words, that is technically possible. I ordinarily only impose moderator edits on offensive content or on confusing misspellings in thread TITLES.

My biggest use of moderation power on the two forums on which I have it used to be to "promote" threads from Getting Started Forum to a higher-level forum. Recently people are getting used to the new five-level division of forums, and the problems in Getting Started are mostly at the correct level for beginners.

Why do you ask?

[quote="RC-7th"]I noticed the little italicized text ot the bottom of edited posts, but never bothered to read it. I wondered if it was automatically generated. Looking at a few more threads, the answer seems obvious.[/quote]

Heh, you're right. It's not automatic. Us mods just do that sometimes to let people know we're actually doing our jobs. It gives us a nice warm feeling inside.  :lol: Anyways, we could just edit your message without anyone knowing...

Of course, then we'd lose the position. Don't worry, we're not that mean. (or are we?)

Yes, I've noticed this.  I came back the next day to a post i made and all MY message said was something along the line of "please do not use offensive language in the forums."  I don't remember using any offensive language and it didn't say anything about who edited it.

Well, you should trust the moderators to do their jobs. I'm sure the moderator who editted your message had a good reason to do it. Maybe you didn't notice putting in offensive language.


Given a triangle ABC with circumcircle (O) and three sides BC=a;AB=c;CA=b.Denete by $ A_1;B_1 and C_1$ the midpoints of BC;CA and AB  respectively and $ A_2;B_2;C_2$ are the midpoint of the ares BC(wich does not contain A);CA;AB.   Draw the cirdes ($ O_1);(O_2);O_3)$ whose diemeters are $ A_1A_2;B_1B_2;C_1C_2$  respectively.  Prove that:
 $ P_{A/(O_1)}\plus{}P_{B/(O_1)}\plus{}P_{C/(O_1)} \geq  \frac{(a\plus{}b\plus{}c)^2}{3}$

$ P_{A/(O_1)}\equal{}AM^2  with OM\equal{}R_1 and <AOM\equal{}90^0$


inequalities
algebra
function
domain
search
inequalities proposed

let a,b and c be positive real numbers such that $2a+b+c=1$
we suppose that $a\geq b$
$A=a^{3}+ba^{2}-b^{3}-b^{2}a-b^{2}c+ca^{2}$

determine the least positif real number K such that

$A\leq K(a+1)^{3}$

:) really it's a nice solution but you can prove it only by using AM-GM but you must be attentive.

[quote="BoesFX"]Not attentive enough. I haven't post why when $K<\frac{1}{27}$ the inequality is wrong. :P[/quote]

Can you outline how you would go about doing that? Thanks.

My approach is that I'm going to search counterexample for $K<\frac{1}{27}$
$a,b,c$ would be set into functions of $K$ such that it will contradict the inequality, but it is a bit too hard work.

If $a,b,c$ nonnegative it should be much easier. :P

From $c=1-2a-b,\ a\geq b>0$ and $1-a=a+b+c>0,$ using A.M.-G.M. we have $A=(a-b)(a+b)(1-a)\leq \left\{\frac{(a-b)+(a+b)+(1-a)}{3}\right\}^{3}=\frac{1}{27}(a+1)^{3},$ yielding $K<\frac{1}{27}.$

Because if the equality holds, we have $a-b=a+b=1-a,$ yielding $b=0,$ which contradicts.

Oops... Is that easy? :oops:

I haven't bother factoring it again so yeah :roll:

[quote]Oops... Is that easy?

I haven't bother factoring it again so yeah [/quote]
so am i


modular arithmetic
function
algorithm
number theory
prime factorization

These are just two random questions I thought of when reviewing; I have no idea how to approach this, so if anyone here has a solution to or an idea on how to start on any of these, please do reply! I give you my thanks :)

1. Given arbitrary integer p < k, under what conditions will $ p^n \equiv \minus{}1 \pmod k$ be satisfied?

(When p is congruent to -1 (mod k) and n is odd, this is satisfied, but what other solutions exist, if any?)

2. A more general question:

Given arbitrary integer p < k and q < k such that $ p^n \equiv q \pmod k$, find q in terms of p, k, and n.

1.  It's hard to even give a good answer for $ p^n \equiv 1 \bmod k$.  All you can say is that $ n$ must be divisible by a quantity known as the order of $ p \bmod k$, and it is not known how to compute this number in closed form.  What is known is that it must divide the value of the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url] $ \lambda(k)$.

If $ k$ is prime, $ p^n \equiv \minus{} 1 \bmod k$ if and only if $ p^{2n} \equiv 1 \bmod k$ and $ 2n$ is not divisible by the order of $ p \bmod k$, and the previous applies.  The problem is that if $ k$ isn't prime, this is no longer a sufficient condition, and I can't think of a particularly succinct way to describe the other possibilities.  This is how the [url=http://en.wikipedia.org/wiki/Miller-Rabin_primality_test]Miller-Rabin[/url] test works.

Edit:  If you just want solutions when $ n$ is even, take $ k$ a prime congruent to $ 1 \bmod 4$; then $ p^{ \frac{n}{2} }$ needs to be congruent to a square root of $ \minus{}1$, which exists here.  For example, you can take $ p$ to be a primitive root and $ n \equal{} \frac{p\minus{}1}{2}$.

2.  What does "given $ q$, find $ q$" mean?  If you meant "find $ q$," the standard algorithm is [url=http://en.wikipedia.org/wiki/Exponentiation_by_squaring]exponentiation by squaring[/url].  If you meant "find $ p$," it exists uniquely if and only if $ \gcd(n, \varphi(k)) \equal{} 1$.  If this is true, you need to find an integer $ m$ such that $ nm \equiv 1 \bmod \varphi(k)$; then

$ p \equiv q^m \bmod k.$

In general, you can't do this without knowing the prime factorization of $ k$, and this is how [url=http://en.wikipedia.org/wiki/RSA]RSA[/url] works.

Thank you for the answer! I had suspected that there was no particularly simple answer to that question, so I will have to look into it later.

(As for 2, I think it would have been more clear if it was phrased as:

Given a positive integer $ p$ such that $ p < k$, $ k \in \mathbb{N}$, find a closed-form formula describing $ q$, the residue of $ p^n$ modulo $ k$, where $ n$ is a positive integer, in terms of the variables $ k$, $ n$, and $ p$.

This is stronger than 1., so it probably has no solution either.)

What would a closed-form formula even look like besides "$ p^n \bmod k$"?  Modular arithmetic just doesn't work that way.  If there really were a closed-form formula, Miller-Rabin would run in constant time.  As it is, exponentiation by squaring is quite practical for cryptographic applications.


The operation $ *$ is defined for non-zero integers as follows: $ a * b \equal{} \frac{1}{a} \plus{} \frac{1}{b}$. If $ a\plus{}b \equal{} 9$ and $ a \times b  \equal{} 20$, what is the value of $ a*b$? Express your answer as a common fraction.

$ a$ and $ b$ must be $ 4$ and $ 5$.  Thus, the answer is $ \boxed{\frac{9}{20}}$.

Another way to do this is that
1/a+1/b=(a+b)/ab.  Thus, the answer is simply 9/20.

[quote="ernie"]That's what I did, I just didn't write out the equation. :P

Oh, and Jongao typoed, so what he really means is:

$ \frac{1}{a} \plus{} \frac{1}{b} \equal{} \frac{a \plus{} b}{ab}$[/quote]

hehe you typoed too :) "Jongao"

The reason $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$ is because you can get the LCM of $a$ and $b$, which is obviously $ab$ because we don't know what they are. That gives $ab$ as our denominator. We multiply $b$ to $\frac{1}{a}$ and vice versa. That gives us $b\left(\frac{1}{ab}\right)+a\left(\frac{1}{ab}\right)=\frac{a+b}{ab}$.

hello, Jongy's username used to be Jongao.


geometry
parallelogram
rhombus
geometry unsolved

Let $ABCD$ be a parallelogram. The circle with diameter $AC$ intersects the line $BD$ at points $P$ and $Q$. The perpendicular to the line $AC$ passing through the point $C$ intersects the lines $AB$ and $AD$ at points $X$ and $Y$, respectively. Prove that the points $P,Q,X$ and $Y$ lie on the same circle.

Prolongate $ PQ$ until it meets $ XY$ at $ O$ then it is clear that $ OP\cdot OQ = OC^2$. On the other hand since $ \triangle{DYC}$ similar to $ \triangle{BCX}$ we get that: $ \frac {DY}{BC} = \frac{DC}{BX}$, also from $ \triangle{ODC}$ similar to $ \triangle{OBX}$ and $ \triangle{ODY}$ similar to $ \triangle{OBC}$ $ \rightarrow$ $ \frac {OY}{OC} = \frac {DY}{BC}$ and $ \frac {OC}{OX} = \frac {BC}{BX}$ then we get $ OC^2 = OY\cdot OX$ $ \Rightarrow$ $ PQXY$ is cyclic.

+ case when ABCD is rhombus (trivial). Nice and easy solution :) any other ideas?

Here's another (less nice) solution:
[hide="Solution"]
[b]Lemma[/b]: Consider right $ \triangle ABC$ with $ \angle BAC = 90$.  Let $ P$ and $ Q$ lie on $ AB$ and $ AC$ respectively such that $ PQ\parallel BC$.  Let the line through $ B$ perpendicular to $ BC$ be $ b$ and define the line through $ C$ perpendicular to $ BC$ as $ c$.  Let $ CP$ meet $ b$ at $ X$ and $ BQ$ meet $ c$ at $ Y$.  Then, $ \angle XAB = \angle YAC$.

[i]Proof[/i]: Let $ XC$ and $ BY$ meet at $ Z$, $ XB$ and $ AC$ meet at $ L$, $ YC$ and $ AU$ meet at $ M$, the feet of the perpendiculars from $ X$ to $ AB$ and $ AC$ be $ R$ and $ T$ respectively, and the feet of the perpendiculars from $ Y$ to $ AC$ and $ AB$ be $ S$ and $ T$ respectively.  Now, it suffices to show that $ \angle XAT = \angle YAU$, so we want for $ \triangle XAT\sim \triangle YAU$.  Thus, it suffices to have $ \frac {YU}{XT} = \frac {AU}{AT}$.  Now, from $ UY\perp AM$, we have that $ UY\parallel AC$, so $ \frac {AU}{YC} = \frac {AM}{MC}$.  Since $ \angle BCM = 90 = \angle BAC$, we have that $ \frac {AM}{MC} = \frac {AC}{BC}$, so $ \frac {AU}{YC} = \frac {AC}{BC}$.  Similarly, $ \frac {TA}{XB} = \frac {AB}{BC}$.  Dividing, we get that $ \frac {AU}{TA} = \frac {YC}{XB}\cdot \frac {AC}{AB}$.

Furthermore, $ \frac {PA}{XT} = \frac {PC}{XC}$ and $ \frac {AQ}{YU} = \frac {BQ}{BY}$.  Dividing, we get that $ \frac {UY}{XT} = \frac {AQ}{AP}\cdot \frac {BY}{BQ}\cdot \frac {PC}{XC}$.  Notice that $ \frac {PA}{QA} = \frac {AB}{AC}$.  Now, recall that we want for $ \frac {AU}{TA} = \frac {YU}{XT}$, so $ \frac {BY}{BQ}\cdot \frac {PC}{XC} = \frac {YC}{XB}$.  Notice that $ \frac {PC}{BQ} = \frac {ZC}{ZB}$ from $ PQ\parallel BC$.  Furthermore, we have that $ \frac {YC}{XB} = \frac {ZC}{ZX}$ and $ \frac {BY}{XC} = \frac {BZ}{ZX}$ both from $ YC\parallel BX$.  This gives us that $ \frac {BY}{XC}\cdot \frac {PC}{BQ} = \frac {BZ}{ZX}\cdot \frac {ZC}{ZB} = \frac {ZC}{XZ} = \frac {YC}{XB}$, as desired.

Now for the problem.  Allow for $ AP$ to meet $ XY$ at $ T$, $ AQ$ to meet $ XY$ at $ U$, let the line parallel to $ XY$ through $ A$ meet $ CP$ at $ Z$, let the midpoint of $ AC$ be $ M$, allow for $ CZ$ to meet $ AP$ at $ L$, and $ AX$ to meet $ PC$ at $ K$.  Since $ M$ is the midpoint of $ AC$, we have that\[ 1 = \frac {PL}{AL}\cdot \frac {AM}{MC}\cdot \frac {KC}{PK}\implies \frac {PL}{LA} = \frac {PK}{KC}\implies LK\parallel AC\]by Ceva's Theorem.  Apply the lemma to right $ \triangle APC$ to get that $ \angle APZ = \angle XPC$.  Reflecting everything into $ M$, we see that the $ XY$ maps to $ AZ$, $ P$ maps to $ Q$, $ A$ maps to $ C$, and $ B$ maps to $ D$, so $ Y$ maps to $ Z$, so $ \angle XPC = \angle APZ = \angle CQY$.  Now, since $ \angle QUC = 90 - \angle QAC = \angle PAC = \angle APQ$, we have htat $ QPTU$ is cyclic.  Then, $ \angle QPX = \angle QPT - \angle TPX = 180 - \angle TUQ - \angle UQY = 180 - \angle QYC$, establishing that $ PQYX$ is cyclic.  [/hide]

[quote]Prolongate $ PQ$ until it meets $ XY$ at $ O$ then it is clear that $ OP\cdot OQ \equal{} OC^2$.[/quote]
After that it can be easily noticed that $ ODCY$ figure is homothetic to $ OBXC$ figure with homothetic center $ O$ because $ BC\|\|DY$ and $ BX\|\|DC$. Thus $ \frac{XO}{CO}\equal{}\frac{CO}{YO}$ and $ XO\cdot YO\equal{}CO^2$, from which $ OP\cdot OQ\equal{}XO\cdot OY$ follows. But that means $ P$, $ Q$, $ X$, $ Y$ are concyclic, q.e.d.

I think that could be a little bit shorter, however, expressing the same idea.


Since I am in Canada, we have the Canadian Open Math Challenge and the Euclid contest. They are both 2.5 hours long.

we generally have 10 questions, and the last two questions are fairly hard.

If you only have 30 minutes left, do you go back and check your work in case you made stupid mistakes, or do you try to finish the last two problems.

if you are absolutely certain that you can do the last two, then try them. if you can't, then just check your work first and then attempt the last 2.


It depends what the base is. There could be 3 different bases for a triangle. This rule works for all 30-60-90 triangles. The shortest leg or the leg opposite of the 30 degrees angle is x. The hypotenuse would be 2x. Then the other leg would be x times the square root of 3. :)

[quote="7h3.D3m0n.117"]say there was a right triangle (just picture one)

the degrees are  $60^{\circ}$, $30^{\circ}$, and $90^{\circ}$.

Say that the base of the right triangle was x. What would be the other lengths of the triangle if you only knew x? I thought it could be $\sqrt 3$ and $2x$ or something?[/quote]

it could also be x, x(3^1/3)/3, and 2x(3^1/3)/3

[quote="7h3.D3m0n.117"]say there was a right triangle (just picture one)

the degrees are  $60^{\circ}$, $30^{\circ}$, and $90^{\circ}$.

Say that the base of the right triangle was x. What would be the other lengths of the triangle if you only knew x? I thought it could be $\sqrt 3$ and $2x$ or something?[/quote]
The side opposite the $30^{\circ}$ angle is the shortest, the side opposite the $60^{\circ}$ angle is longer than the first one, and the last side, opposite the $90^{\circ}$ angle, is the longest.

The ratio of short : medium : long is $1: \sqrt{3}: 2,$ where $2$ is the hypotenuse.

The side lengths of a 30-60-90 triangle.

Oh ok, I havn't done these type of questions be4....


geometry
geometric transformation
rotation
group theory
abstract algebra
superior algebra
superior algebra solved

Need help, thx.

Find all conjugacy classes and their sizes:

1. D8
2. Q8
3. A4

1/in D8, every element is either a rotation, either a symetry ... not very hard from here
2/ you know that all subgroup of Q8 are normal. So there are just a few computations ..
3/ for A4, use the cycle decomposition ...

[quote="alekk"]3/ for A4, use the cycle decomposition ...[/quote]

This is not a (near complete) proof. $A_4$ is not $S_4$. $(123)$ and $(132)$ are not in the same conjugation class in $A_4$

[quote="amfulger"]You can check by hand that in $A_4$ the conjugacy classes are (given by a representant): $e$, $(12)(34)$, $(123)$ and $(132)$.[/quote]

Why do you write $e$ and not $1$, or $(1)$, or $id$? Where does the $e$ come from?

Usage of $e$ is hardly nonstandard - my algebra professor uses that notation as well, along with at least one of my algebra textbooks (it's an older version though, so the newer ones may have changed over to $1$)

I usually write $e$ as wel. As there would basicly be many ways to write it, so there would be no unique notation any more. (like you could write $(1)$ but also $(3)$ or whatever)

[quote="mchis68"]
Why do you write $e$ and not $1$, or $(1)$, or $id$? Where does the $e$ come from?[/quote]

The questions are rhetorical. Each of $1$, $(1)$(this is by the way kind of standard for permutations) or $id$ is better justified then $e$(except if you are German - $e$ comes from "Einheitselement"= unit element, so one should better use even $u$ then $e$).

No more $e$'s in Algebra!

I don't see what's wrong with $e$. I like it :)


calculus
integration
trigonometry
real analysis
real analysis unsolved

a. int (0 to infinity) (sinx)/x dx
b. int (0 to 1) (x-1)/lnx dx

I'm guessing he meant "Tonelli," not "Torricelli."

Fubini's Theorem: $ \int_A\int_Bf(x,y)\,dy\,dx \equal{} \int_B\int_Af(x,y)\,dx\,dy$ provided $ \int_A\int_B|f(x,y)|\,dy\,dx < \infty.$

Tonelli's Theorem:  $ \int_A\int_Bf(x,y)\,dy\,dx \equal{} \int_B\int_Af(x,y)\,dx\,dy$ provided $ f(x,y)\ge 0.$

(Both of these theorems have additional hypotheses about measurability and $ \sigma$-finite measure spaces that I've assumed as background.)

The two theorems are obviously closely related. If you were to give a single name to the combination of both, you'd probably use "Fubini."

I've written about (a) on numerous occasions on this forum. If you were good enough at searching, you could probably find some post I've made about it.

Here's a first stab:

[b]Tentative proposed argument:[/b]

$ \int_0^{\infty}\frac {\sin x}{x}\,dx \equal{} \int_0^{\infty}\sin x\left[\int_0^{\infty}e^{ \minus{} tx}\,dt\right]\,dx$

$ \equal{} \int_0^{\infty}\int_0^{\infty}e^{ \minus{} tx}\sin x\,dx\,dt \equal{} \int_0^{\infty}\frac {1}{t^2 \plus{} 1}\,dt \equal{} \frac {\pi}2.$

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

Now, there's a problem with that argument. There is a step in there that is [b]not[/b] justified by either Fubini's Theorem or Tonelli's Theorem. Can you see why not? Can you find another path that creates a change of order of integration that you can justify?


Find a positive integer n :ge: 2 such that the quadratic mean of the first n natural numbers is an integer.

(From Prof. J. Barbeau's book, "Pell's Equation", Exercise 2.2.2)

I used the formula [hide]n(n+1)(2n+1)/6[/hide] for the sum of the squares, which made the mean [hide](n+1)(2n+1)/6[/hide]  and the quadratic mean the square root of that.  I then went to Excel for some brute force, finding [hide]337[/hide], but am still looking for an easier way and a general solution.  I can prove n must be odd...

this was on one of the correspondence issues a year or so back

check out 154: http://cms.math.ca/Competitions/MOCP/2002/sol_jun.html

[hide]

As tomclymer said, we want (n+1)(2n+1) = 6a^2 for some integer a. Now if we expand that and use the quadratic formula.. 2n^2 + 3n + (1-6a^2) = 0 and

n = (-3[pm]:sqrt:(48a:^2: + 1))/4.

Now we just need to find solutions for 48a^2 + 1 = b^2 (and in fact, check that this solution has b congruent to 3 mod 4 so that n gets to be a positive integer).

This is just a Pells equation (hinted by the title), b^2 - 48a^2 = 1. Now we can see one solution is b=7, a=1 (but this gives n=1 so we want to find more). This means we can get the recursion b[n+1] = 7b[n] + 48a[n] and a[n+1] = b[n] + 7a[n]. So we get (7,1), (97,14), (1351, 195)... I think these alternate between b being 3 mod 4 and 1 mod 4.. anyway, using this last pair and subbing in, n=(-3+1351)/4 = 337.

[/hide]

[quote="darkquantum"]you realize
this has absolutely nothing to do with Pell Equations
ur question title is a misnomer, and so is Pell Equations
hmm
interesting[/quote]

:?: What do you mean :?:

TripleM's solution involves a Pell Equation.


How exactly is everything scored? Like how many points per team problem, how are the relays scored, etc.

Team round:

10 questions, 20 minutes, short answer, one answer sheet for the entire team. 4 points per question. No partial credit. Maximum possible credit: 40 points.

Power round:

A linked series of questions (often 10, but with various parts to each question), 60 minutes. Proofs and examples. One set of answers (which could run to 20 pages) for the entire team. Possible partial credit. Maximum possible credit: 40 points.

Individual round:

8 questions, given in sets of 2, with 10 minutes per pair of questions. Each student is on his own. 1 point per question. No partial credit. Maximum possible credit: 120 points.

Relay round:

Each 15-member team forms five 3-member relay teams. Two relays are contested. A correct answer handed in at 3 minutes with no answer at 6 minutes earns 4 points. A correct answer handed in at 6 minutes (no matter what was or wasn't handed in at 3 minutes) earns 2 points. Maximum possible credit: 40 points.

Maximum total credit: 40 + 40 + 120 + 40 = 240.

Super relay:

A single relay involving the whole team. Contested for a bucket of candy or a sweatshirt or something like that. Does not count in the standings.

Slight additions: The super relay is simply a race.  Each team gets one shot to submit an answer, and the first team to get it right wins.  It's done separately at each site, so each site has its own winner, and the times are not compared.

The power round is often--but certainly not always--written as 10 questions worth 4 points each.  Some of the questions are then split into 2, 3, or 4 parts.

Super-slight bonus addition: in 2000, the power event concluded with a "bonus question", worth up to 4 additional points. So that year, the power event was worth a potential of 44 points. That was my senior year, and I totally don't recall us getting that problem right.

::gets back to working on the russian imperial::

Interesting.  There was also a bonus point on the 2006 power round, but I was told that any score of "41" would be converted to 40.  It wasn't an issue, though, as the top score was 39.

hmm how come I don't remember doing the superrelay last year? :huh:

how much time is given for the super relay?
i might take one by myself for fun

The super relay at Vegas got canceled last year because of the heat.  They did that game with picking the numbers instead.

Basically, the super relay is a relay involving the whole team, where you pass answers inward from the ends to the middle, and the middle person turns in the final answer.  There really isn't a time limit, and the first team to get it right wins.  You only get one chance to turn in an answer.


Julian and Joshua each maintain a constant speed as they run laps around a 400-meter track. In the time it takes Julian to complete two laps, Joshua completes three laps. Julian runs each mile in 12 minutes. How many minutes does it take Joshua to run exactly one mile?

Julian runs 800 meters in the same time that it takes Joshua to run 1200 meters. Thus, Joshua's speed is 3/2 times Julian's.

If Julian runs a mile in 12 minutes, Joshua runs a mile in $ \frac{2}{3} \cdot 12 \equal{} 8$ minutes.


Since it is the start of February, I would like to remind people of some things:
1. Good luck with your  chapter and state contests
2. THERE SHOULD BE ABSOLUTELY NO DISCUSSION ABOUT ANY ASPECTS OF THE CHAPTER TEST UNTIL (see mathcounts website for exact dates) LIKE SECOND WEEK OF MARCH! ALSO, NO DISCUSSION OF THE STATE TESTS UNTIL SECOND WEEK OF APRIL
I just wanted to remind everyone about this....

ooo,

are you taking the test also?

My school doesn't accept MATHCOUNTS.. :(


MATHCOUNTS
quadratics
geometry
algebra
quadratic formula
similar triangles

I got for homework this diagram, and I was to find x, y, and z.
My answer seems to not work
Can someone help?
I'm putting it in this forum because it's somewhat similar to some mathcounts problems I've seen...

My work:
$x^2 + z^2 = 144$
$z^2 + 256 = y^2$
$y^2 + 144 = x^2 + 32x + 256$
$z^2 + 400 = x^2 + 32x + 256$
And then we can get
$144 - x^2 + 400 = x^2 + 32x + 256$
$2x^2 + 32x - 288 = 0$
$x^2 + 16x - 144 = 0$
Using the quadratic formula,
$x = 4\sqrt{13} - 8$
Then z would = $\sqrt{64\sqrt{13} - 64}$
But I'm pretty sure thats wrong...

Can someone tell me where I messed up?

umm...

[hide][hide="solution"]$x^2+z^2=144$
$y^2=256+z^2$

from similar triangles $\frac{z}{x}=\frac{16}{z} \implies z^2=16x$

substituting
$x^2+16x=144$
by quadratic formula
$x=4\sqrt{13}-8$

therefore $z^2=16(4\sqrt{13}-8)=64\sqrt{13}-128 \implies z=\sqrt{64\sqrt{13}-128}$

therefore $y^2=256+64\sqrt{13}-128 \implies y=\sqrt{128+64\sqrt{13}}$
[/hide][/hide]
Those answers look ugly  :| .I think your answer for z is wrong.

I think that the two right triangles in the diagram are simillar, something I learned in a past geometry lesson ;) It might not be the case, however, if it is,
 $\frac{z}{12}=\frac{x+16}{y}$  This means that $yz=12x+192$  Try to take it from there.  Hope this helped :?

[quote="Virgil Nicula"]Know you the theorem of the cathetus or the theorem of the altitude in a rightangle triangle ? If yes, then it is simply.[/quote]

No.... what is it?

From the proportion $\frac{12}{x+16}=\frac{x}{12}$, x is also a very unfriendly number, and the values of the other variables are similar to that of pianoforte's.

Are you sure that it isn't the entire hypotenuse that is 16?

[hide="Some remarkable theorems in a right-angled triangle ABC."]
Denote: $AB\perp AC$; $D\in BC$, $AD\perp BC$, $AD=h$;

$BC=a$, $CA=b$, $AB=c$. From $\triangle ADB\sim \triangle CDA\sim\triangle CAB$ result:

$1.\blacktriangleright \frac{h}{DC}=\frac{DB}{h}=\frac cb\Longrightarrow \boxed {h^2=DB\cdot DB}$ ([u]the theorem of the altitude[/u]) .

$2.\blacktriangleright \frac{DC}{b}=\frac hc=\frac ba$ and $\frac hb=\frac{DB}{c}=\frac ca\Longrightarrow \boxed {h=\frac{bc}{a}\ \ \wedge\ \ \frac{DB}{DC}=\left(\frac cb\right)^2}$

and $\boxed {b^2=a\cdot DC\ \ \wedge\ \ c^2=a\cdot DB}$ ([u]the theorems of the cathetus[/u]).

$3.\blacktriangleright$ From the sum of the last two relations we obtain the [u]Pitagora's theorem[/u]: \[ \boxed {\ b^2+c^2=a^2\ }\ .  \][b]Some exercises.[/b]

$1).\ \ a=2R\ \ ,\ \ r=p-a\ \ ,\ \ r_a=p\ \ ,\ \ r_b=p-c\ \ ,\ \ r_c=p-b\ ;$
$2).\ \ S=\frac{bc}{2}=\frac{ah}{2}=p(p-a)=(p-b)(p-c)\ .$
$3).\ \ r+r_a+r_b+r_c=2p\ \ ;\ \ 2R+r=p\ .$

\[ \mathrm {You\ are\ the\ immediate\ future\ of\ this\ planet\ .} \] \[ \mathrm {I\ dare\ to\ hope\ still\ .} \][/hide]

1. The triangles are simmilar. To prove this, label the angle between 16 and y as p and solve for the other angles.

2. Your math is correct, the numbers come out very ugly.

You guys know that you can always use the geometric means theorems.  Therefore, 12/x=x+16/12  After cross multiplying you get a simple equation.


Suppose that $ f(x)\in\mathbb Z[x]$ be an irreducible polynomial. It is known that $ f$ has a root of norm larger than $ \frac32$. Prove that if $ \alpha$ is a root of $ f$ then $ f(\alpha^3\plus{}1)\neq0$.

suppose that$ \alpha$ and $ \alpha^3\plus{}1$ are two root of $ f$
and also suppose $ \beta$is also a root we can show for each $ \beta$,$ \beta^3\plus{}1$ is also a root of $ f$

we knoew $ (x^3\minus{}\alpha^3)|f(x^3\plus{}1)$ so $ \alpha$ is a root of $ f(x)$ but we know that $ f(x)$ is minimal polynomial that contains $ \alpha$ as a root so we have $ f(x)|f(x^3\plus{}1)$ so if we put $ \beta$ we will see that $ \beta^3\plus{}1$ is a root.

if we put $ \beta$ with norm greater than $ \frac{3}{2}$
we have $ |\beta^3\plus{}1|>|\beta^3|\minus{}1>|\beta|$ because root of $ g(x)\equal{}x^3\minus{}x\minus{}1$ with maximum norm is not greater than $ \frac{3}{2}$
so we reach unlimited root for $ f$ which contradicts.

I think I have the same solution as sumita, but I'll write a bit more clearly.

I will use the following

Lemma: If $f \in Z[x]$ is irreducible and $\alpha \in C$ s.t $f(\alpha)=0$, then $f$ is the minimal polynomial of $alpha$ (up to a constant factor).

Suppose $\alpha$ and $\alpha^3+1$ are both roots of $f$ for some $\alpha \in C$. Let $g(x)=f(x^3+1)$. Since $g(\alpha)=0$ and $g \in Z[x]$, we have $f|g$. Therefore if $f(r)=0$ then $g(r)=f(r^3+1)=0$.

We also need to use

Lemma: If $r \in C$ and $|r|>1$ then $|r^3+1|>|r|$

Let $r_1 \in C$ denote the root of $f$ with norm larger than $3/2$ (i.e $|r_1|>3/2$). By the above lemma, $r_1^3+1$ is a root of $f$ distinct from $r_1$. $(r_1^3+1)^3+1$ is yet another distinct root of $f$, and so on.

Thus $f$ has infinitely many distinct roots, so it is the zero polynomial, contradiction.

[quote=sumita]suppose that$ \alpha$ and $ \alpha^3\plus{}1$ are two root of $ f$
and also suppose $ \beta$is also a root we can show for each $ \beta$,$ \beta^3\plus{}1$ is also a root of $ f$
[/quote]

I don't understand this part ,can anyone help me?thanks! :blush:

[quote=viperstrike][quote=Stranger8][quote=sumita]suppose that$ \alpha$ and $ \alpha^3\plus{}1$ are two root of $ f$
and also suppose $ \beta$is also a root we can show for each $ \beta$,$ \beta^3\plus{}1$ is also a root of $ f$]

I don't understand this part ,can anyone help me?thanks! :blush:[/quote][/quote]

He's trying to prove the statement by contradiction, so he supposes $\alpha$ and $\alpha^3+1$ are both roots. He then says he will prove that if $b$ is any root of $f$, then $b^3+1$ is also  a root.[/quote]

But how to prove if $b$ is any root of $f$, then $b^3+1$ is also a root.   I don't know this sorry  :blush:  :blush:  :blush: can you help me please?

He says it here:

we knoew $ (x^3\minus{}\alpha^3)|f(x^3\plus{}1)$ so $ \alpha$ is a root of $ f(x)$ but we know that $ f(x)$ is minimal polynomial that contains $ \alpha$ as a root so we have $ f(x)|f(x^3\plus{}1)$ so if we put $ \beta$ we will see that $ \beta^3\plus{}1$ is a root.

I also say it here:

[quote=viperstrike]

Lemma: If $f \in Z[x]$ is irreducible and $\alpha \in C$ s.t $f(\alpha)=0$, then $f$ is the minimal polynomial of $\alpha$ (up to a constant factor).

Suppose $\alpha$ and $\alpha^3+1$ are both roots of $f$ for some $\alpha \in C$. Let $g(x)=f(x^3+1)$. Since $g(\alpha)=0$ and $g \in Z[x]$, we have $f|g$. Therefore if $f(r)=0$ then $g(r)=f(r^3+1)=0$.
.[/quote]

Oh I get it now, I didn't  know how to use "irreducible "condition  , so I didn't understand your solution,now I realize the importance of minimal polynomial which helps me understand it well,thank you very much! :blush:  :-D

[quote=viperstrike]I think I have the same solution as sumita, but I'll write a bit more clearly.

I will use the following

Lemma: If $f \in Z[x]$ is irreducible and $\alpha \in C$ s.t $f(\alpha)=0$, then $f$ is the minimal polynomial of $\alpha$ (up to a constant factor).

Suppose $\alpha$ and $\alpha^3+1$ are both roots of $f$ for some $\alpha \in C$. Let $g(x)=f(x^3+1)$. Since $g(\alpha)=0$ and $g \in Z[x]$, we have $f|g$. Therefore if $f(r)=0$ then $g(r)=f(r^3+1)=0$.

We also need to use

Lemma: If $r \in C$ and $|r|>1$ then $|r^3+1|>|r|$

Let $r_1 \in C$ denote the root of $f$ with norm larger than $3/2$ (i.e $|r_1|>3/2$). By the above lemma, $r_1^3+1$ is a root of $f$ distinct from $r_1$. $(r_1^3+1)^3+1$ is yet another distinct root of $f$, and so on.

Thus $f$ has infinitely many distinct roots, so it is the zero polynomial, contradiction.[/quote]

[hide=storage] $f$ is irreducible so $f$ is the minimal polynomial for $\alpha$ and  $\alpha$ is a root of$ f(x^3+1)$ so $f(x) \mid f(x^3+1)$
let $z$ be a root of $f$ such that $\mid z\mid  \geq \frac{3}{2} \implies \mid z^3+1\mid  \geq \mid z^3 \mid - 1 \geq_{\bigstar}  \mid z \mid$
So we have an infinite growth in roots of $f$ so contradicion.
Notice the $\bigstar$:
$\mid z^3 \mid -1 \geq \mid z \mid $ comes from $ \mid z^3 \mid - \mid z \mid -1 \geq 0$
Notice roots of $x^3 -x -1$ by puting $1.5$ and checking roots of the $3x^2 -1$ we get $x^3-x-1 \geq 0 $ for $ x \geq 1.5 $ $\blacksquare$


geometry
circumcircle
trigonometry
geometric transformation
reflection
cyclic quadrilateral
projective geometry

Let $ABCD$ be a convex quadrilateral, let $O$, $R$ be the circumcenter and the circumradius of $\triangle ABC$ respectively. If $N=AB \cap CD$, $M=AD \cap BC$, $\alpha= \angle MON$ show that $ABCD$ is cyclic if and only if
\[
MO \cdot NO \cdot \cos \alpha  = R^2
\]

Assume the quadrilateral $ABCD$ is cyclic, having the circumcircle $(O)$ with radius $R$. Construct the associated tangential quadrilateral $PQRS$ with $PQ, QR, RS, SP$ tangent to the circle $(O)$ at the points $A, B, C, D$. The diagonals $AC, BD$ of the cyclic quadrilateral $ABCD$ and the diagonals $PR, QS$ of the tangential quadrilateral $PQRS$ intersect at the same point $X$ and the points $M, N$, lying on the diagonals $QS, PR$, are poles of the diagonals $PR, QS$ with respect to the circle $(O)$ (by repeated use of Pascal's theorem for the cyclic quadrilateral $ABCD$). The angle $\angle QXR = \angle MON = \alpha$, because the angle formed by the lines $PR, QS$ shows as the angle from their poles $M, N$ to the center $O$ of the polar circle $(O)$. The lines $MO, NO$ connecting the poles $M, N$ with the center $O$ are perpendicular to their polars $PR, QS$. Let $M', N'$ be the intersections of $MO \cap PR$, $NO \cap QS$. On one hand, the points $M', N'$ are feet of the altitudes $MM', NN'$ of the triangle $\triangle MON$ and the side $M'N'$of its orthic triangle is equal to

$M'N' = MN\ cos(\widehat{MON}) = MN\ cos(\alpha)$

On the other hand, the points $M', N'$ are reflections of the points $M, N$ in the circle $(O)$, i.e.

$M'N' = MN \frac {R^2} {OM.ON}$

Combining the two equations yields the result

$OM.ON\ cos(\alpha) = R^2$

Look mom, no Lagrange multipliers!

Dear Yetti, your proof can be simplified. Moreover in the following solution I will prove also the viceversa, that is if

$(1) \ \ \  MO \cdot NO \cdot \cos \alpha  = R^2$

then the point $D$ belongs to the circumcircle of $\triangle ABC$.

[b]Solution.[/b]
First of all, using Carnot theorem,  it is easy to show that  $(1)$ is equivalent to:

$(2) \ \ \  MN^2=MO^2 +NO^2 - 2 \cdot R^2$

$\left(  \Rightarrow  \right)$ Let $L = AC \cap BD$, let $M',p$ be the inverse of $M$ wrt $(O)$ and the line through $M'$ perpendicular to $OM$ respectively (i.e. $p$=polar of $M$ wrt $(0)$).  The point $N$ belong to $p$ since $\triangle LMN$ is autopolar ($\triangle LMN$  is the diagonal triangle of complete quadrilateral $ABCD$). Therefore $NM' \bot OM$ and Carnot theorem yields

$MN^2  = OM^2  + ON^2  - 2 \cdot OM \cdot OM' = OM^2  + ON^2  - 2R^2 $

$\left(  \Leftarrow  \right)$ Let $M'$ be the orthogonal projection of $N$ onto the line $OM$.  From $MN^2  = OM^2  + ON^2  - 2R^2$ follows $OM \cdot OM' = R^2 $, i.e.  $M'$ is the inverse of $M$ wrt $(O)$. Then $N$ belong to the polar of $M \ \ \Rightarrow \ \  DM' \bot OM$ because $DM'=NM'$. If $D' = CM' \cap \left( O \right)$ then $D'M' \bot OM$ for the part $\left(  \Rightarrow  \right)$ because $ABCD'$ is cyclic. From $DM' \bot OM$ and $D'M' \bot OM$ follows $D=D'$ for the uniqueness of the perpendicular from a point to a line. Thus $ABCD$ is cyclic $\blacksquare$

[b]P.S.[/b]
It is possible to give a solution without using the theory of polarity (see for example the solution of problem 3 of Chinese Mathematical Olympiad 1997, http://mathdb.org/notes_download/elementary/geometry/ge_G3/e_control.htm )

The problem can be resolved also by means of the complex numbers.  [color=red]I invite the others mathlinkers to supply one such solution.[/color]

[quote="Leon"]Dear Yetti, your proof can be simplified. Moreover in the following solution I will prove also the viceversa, that is if

$(1) \ \ \  MO \cdot NO \cdot \cos \alpha  = R^2$

then the point $D$ belongs to the circumcircle of $\triangle ABC$.
[/quote]

The proof I offered can be easily reversed as well. Assume that the quadrilateral $ABCD$ is not cyclic. Let $D'$ be the intersection of the line $DO$ with the circumcircle $(O)$ on the opposite side of the line $AC$ than the vertex $B$. Let $M', N'$ be the reflections of the intersections $M \equiv AB \cap CD$, $N \equiv BC \cap DA$ in the circumcircle $(O)$, i.e.,

$OM' \cdot OM = ON' \cdot ON = R^2$,

$\frac{OM'}{ON} = \frac{ON'}{OM}$,

$M'N' = MN \frac{R^2}{OM \cdot ON}$,

and let $M, N$ be the feet of altitudes of the triangle $\triangle OMN$ from the vertices $M, N$, i.e.,

$\frac{OM}{ON} = \frac{ON}{OM} = \cos \alpha$.

$MN = MN\ \cos \alpha$.

Consequently,

$\frac{OM}{OM'} = \frac{ON}{ON'}$,

the triangles $\triangle OM'N' \sim \triangle OMN$ are similar and the sides $MN \parallel M'N'$ are parallel. Unless both the points $M \equiv M'$ and the points $N \equiv N'$ are identical,

$M'N' \neq MN$

$MN \frac{R^2}{OM \cdot ON} \neq MN\ \cos \alpha$

$R^2 \neq OM \cdot ON\ \cos \alpha$

As the point $D$ moves on the line $OD$, staying on the opposite side of the line $AC$ than the vertex $B$ (for the quadrilateral $ABCD$ to remain convex), the segment $MN$ moves parallel to the segment $M'N'$, both segments in the same direction. They can coincide for one position of the point $D$ only and they coincide when the point $D$ falls on the circumcircle $(O)$ and coincides with the point $D'$, i.e., when the quadrilateral $ABCD$ becomes cyclic, as we proved before.


we can make a proportion

$ \frac {AB} {BE} \equal{} \frac {FC} {CD} \Rightarrow \frac {28} {x} \equal{} \frac {x} {58} \Rightarrow 1624 \equal{} x^2$
where $ x \equal{} \text {side of square}$

since we are trying to find the area of a square, the formula would be $ A\equal{}s^2$

we have $ 1624 \equal{} X^2$ where $ x$ is the side length, so $ 1624$ is our answer

We're looking for the area, not the side length.

well, you could find the root, to get the side length, but you would need to square it again to find the area, so it's kinda pointless


[quote="AIME15"]The ratio of the areas of two concentric circles is $ 1: 3$.  If the radius of the smaller is $ r$, then the difference between the radii is best approximated by:

$ \textbf{(A)}\ 0.41r \qquad \textbf{(B)}\ 0.73 \qquad \textbf{(C)}\ 0.75 \qquad \textbf{(D)}\ 0.73r \qquad \textbf{(E)}\ 0.75r$[/quote]
[hide]If the ratio of the areas of two circles is $ \frac{1}{3}$, then the ratio of the radii of the circles is $ \sqrt{\frac{1}{3}}\equal{}\frac{\sqrt{3}}{3}$. Setting this fraction equal to $ \frac{r}{R}$, we see that $ R\sqrt{3}\equal{}3r$, so $ R\equal{}\sqrt{3}r$. Since we want to approximate $ R\minus{}r$, a little substitution brings us $ \sqrt{3}r\minus{}r$, or $ r(\sqrt{3}\minus{}1)$. $ \sqrt{3}$ is approximated around 1.73 so the final answer would be $ .73r$ or B.[/hide]

[quote="ckck"][hide]so the final answer would be $ .73r$ or B.[/hide][/quote][hide="Final answer is"]$ \bf (D)$[/hide]