A positive integer is called [i]nice[/i] if the sum of its digits in the number system with base $ 3$ is divisible by $ 3$.

Calculate the sum of the first $ 2005$ nice positive integers.

[quote="darij grinberg"]

This is not really number theory, but rather, I would say, combinatorics, and such a trivial kind of combinatorics that even I had the right idea. I lost a point because of a trivial computation mistake; indeed the problem seems to be impossible without a nasty number calculation.

  Darij[/quote]

Yeah...

Group the positive integers into sets
$A_1 = \{0,1,2,3,4,5,6,7,8\},
A_2 = \{9,10,11,...,17\},
A_3 = \{18,19,20,...,26\},
...$

or equivalently,
$A_1 = \{0_3,1_3,2_3,10_3,11_3,12_3,20_3,21_3,22_3\},
A_2 = \{100_3,101_3,102_3, ...,122_3\},
A_3 = \{200_3,201_3,202_3, ...,222_3\},
...$

Note that for any natural number $k$, all digits except the last two (in base 3) of any pair of numbers in $A_k$ are the same.

Let $S_n$ be the sum of [i]nice[/i] numbers in $A_n$ and $x_n$ be the smallest number in $A_n$. We'll prove that $S_n = 3x_n+12$ for all natural numbers $n$.

For any  natural number n, it is obvious that only one number in the set $\{x_n, x_n+1, x_n+2\}$ is nice.

1st case: $x_n$ is nice.
Since $x_n$ ends with $00$, we have that in $A_n$, only $x_n+5$, which ends with $12$, $x_n+7$, which ends with $21$, and $x_n$, are nice.
So $S_n = 3x_n+12$

2nd case: $x_n+1$ is nice.
Since $x_n+1$ ends with $01$, we have that in $A_n$, only $x_n+3$, which ends with $10$, $x_n+8$, which ends with $22$, and $x_n$, are nice.
So $S_n = 3x_n+12$

3rd case: $x_n+2$ is nice.
Since $x_n+2$ ends with $02$, we have that in $A_n$, only $x_n+4$, which ends with $11$, $x_n+6$, which ends with $20$, and $x_n$, are nice.
So $S_n = 3x_n+12$

So $S_n = 3x_n+12$ for all natural numbers $n$.

So $S_{n+1}=3x_{n+1}+12$

And since $x_{n+1}=x_n+9$ for all natural numbers $n$, we conclude that $S_{n+1}=S_n+27$ for all natural numbers $n$.

Now, all the sets $A_k$ have 3 nice integers except $A_1$, which has only 2.
So the first 2005 nice integers are all those in $A_1, ..., A_{668}$ and the first two in $A_{669}$.
Their sum is $12+(12+27)+(12+27(2))+...+(12+(27(667)))+6013+6015 = 6035050$.

Warut


Warmup:
A1. Reduce to lowest terms: $\frac{2}{4}$.

A2. Reduce to lowest terms: $\frac{2x^{2}}{4x}$.

A3. Reduce to lowest terms $\frac{x^{5}+3x^{4}+x^{3}+3x^{2}+x+3}{x^{4}+7x^{3}+16x^{2}+15x+9}$.

What's the point of posting if you're not going to answer the questions?


hameye joft tavabe $f,g: \mathbb R\rightarrow\mathbb R$ ra biabid ke:
$i$) agar $x<y$ angah $f(x)<f(y)$.
$ii$) baraye har $x,y\in\mathbb R$,$f(xy)=g(y)f(x)+f(y)$

agar $y=0$, angah: $f(0)=g(0)f(x)+f(0)$ va $g(0)f(x)=0$
pas ya $f(x)=0$ ke yek javab ast, va ya $g(0)=0$.
agar $g(0)=0$,
chon $f(x)$ increasing(so oodi) hast pas yek be yeke, pas maakoos dare, agar $y=1$:
$f(x)=g(1)f(x)+f(1)$ va dar inja agar gharar dahim $x=f^{-1}(x)$:
$x=xg(1)+f(1)$ pas $x(1-g(1))=f(1)$ pas $f(1)=0$ va $g(1)=1$.
hala agar $y=-1$, moshabehe bala bayad: $f(-1)=0$ va $g(-1)=1$.

hala yeki edame bede  :D

thanks,eiwal doroste vali baghiasham khodet bezar dige :D

in soal dar linke http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117189
ham hast, mitooni oonja ham dar moredesh bahs koni, garche ta alan kasi Reply nakarde....


Let $ a,b\in[0;1]$. Prove the inequality:
$ 1\minus{}\dfrac{a\plus{}b}{2}\plus{}\dfrac{ab}{3}\geq\dfrac{1}{1\plus{}a\plus{}b}$

[quote="Ahiles"]Let $ a,b\in[0;1]$. Prove the inequality:
$ 1 \minus{} \dfrac{a \plus{} b}{2} \plus{} \dfrac{ab}{3}\geq\dfrac{1}{1 \plus{} a \plus{} b}$[/quote]
Let $ a\plus{}b \equal{}u$ and $ ab\equal{}v$
  then  we have that : $ (a\minus{}1)(b\minus{}1) \geq 0$  $ \to$ $ u \leq  1\plus{}v$
this ineq is  equavilent to form :  $ \frac{u}{2} \minus{}\frac{u^2}{2} \plus{}\frac{v}{3} \plus{}\frac{uv}{3} \geq 0$ (*)
 case 1 :  $ u \leq 1$ thus  we have : $ u \geq u^2$   hence (*)  right .
 hold equal  if and only if $ a\equal{}0 ,b\equal{}1$ or $ a\equal{}1,b\equal{}0$
case 2 :$ 1 \leq u \leq 2$
from (*)  with note that $ LHS \geq \frac{u}{2} \minus{}\frac{u^2}{2}\plus{}\frac{u\minus{}1}{3} \plus{}\frac{u(u\minus{}1)}{3} \equal{}\frac{(u\minus{}1)(2\minus{}u)}{6} \geq 0\equal{} RHS$
hold equa if and only if $ a\equal{}b\equal{}1$  or $ a\equal{}0,b\equal{}1$  or $ a\equal{}1,b\equal{}0$

hello, the given inequality is equivalent to
$ 3(1\minus{}a)(1\minus{}b)(a\plus{}b)\plus{}ab(1\minus{}a\plus{}1\minus{}b) \ge 0$, which is true because of
$ 0\le a\le1$ and $ 0\le b\le1$.
Sonnhard.

Hong Quy, how is it always true that $ \frac{(u\minus{}1)(2\minus{}u)}{6} \geq 0$, Let's say for example a=0.1, b=0.1, then u=0.2 and that expression is negative since you have $ (\minus{}0.8\cdot1.8)/6$.

[quote="dgreenb801"]Hong Quy, how is it always true that $ \frac {(u \minus{} 1)(2 \minus{} u)}{6} \geq 0$, Let's say for example a=0.1, b=0.1, then u=0.2 and that expression is negative since you have $ ( \minus{} 0.8\cdot1.8)/6$.[/quote]
we  survey u  in  [1;2] .
 u read  solution  carelessly.

[hide="this way often works for this kinda problem :)"][quote="Ahiles"]Let $ a,b\in[0;1]$. Prove the inequality:
$ 1 \minus{} \dfrac{a \plus{} b}{2} \plus{} \dfrac{ab}{3}\geq\dfrac{1}{1 \plus{} a \plus{} b}$[/quote]
let $ f(a,b)\equal{}1\minus{}\frac{a\plus{}b}{2}\plus{}\frac{ab}{3}\minus{}\frac{1}{1\plus{}a\plus{}b}$
consider the difference between $ f(a,b)$ and $ f(1,b)$ we see that
$ f(a,b)\minus{}f(1,b)\equal{}\frac{1}{6}\frac{(b\minus{}1)(a\plus{}2a(b\plus{}1)\plus{}3b\plus{}2b(b\plus{}1))\minus{}3a}{(1\plus{}a\plus{}b)(2\plus{}b)} \geq 0$
it is left to prove that $ f(1,b) \geq 0$ which is equivalent to $ \frac{\minus{}1}{6}\frac{b(b\minus{}1)}{2\plus{}b} \geq 0 which is true  : )$
p/s :i believe f(a) is convex , has any1 tried it ? :)  [/hide]

[quote]$ \{\ a\ ,\ b\ \}\ \subset\ [\ 0\ ,\ 1\ ]\ \implies\ 1 \minus{} \ \frac {a \plus{} b}{2}\ \plus{} \ \frac {ab}{3}\ \ge\ \frac {1}{1 \plus{} a \plus{} b}$[/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \left\|\begin{array}{c} a \plus{} b \equal{} S \\
\ ab \equal{} P\end{array}\right\|$ . The proposed inequality becomes : $ \frac {1}{S \plus{} 1} \plus{} \frac S2\le\frac P3 \plus{} 1$ $ \Longleftrightarrow$

$ 3(S^2 \plus{} S \plus{} 2)\le 2(P \plus{} 3)(S \plus{} 1)$ $ \Longleftrightarrow$ $ \boxed {\ 3S^2\le (2P \plus{} 3)\cdot S \plus{} 2P\ }$ . Therefore,

$ \blacktriangleright$ $ \left\|\begin{array}{c} a^2\le a \\
\ b^2\le b\end{array}\right\|$ $ \implies$ $ a^2 \plus{} b^2\le a \plus{} b$ $ \implies$ $ S^2 \minus{} S\le 2P$ , [b]with equality iff[/b] $ \{a,b\} \equal{} \{0,1\}$ .

$ \blacktriangleright$ $ (1 \minus{} a)(1 \minus{} b)\ge 0$ $ \implies$ $ S\le P \plus{} 1$ $ \implies$ $ 2S \plus{} 1\le 2P \plus{} 3$ , [b]with equality iff[/b] $ 1\in\{a,b\}$ .

$ \blacktriangleright$ $ 3S^2 \equal{} (2S \plus{} 1)\cdot S \plus{} (S^2 \minus{} S)\le (2P \plus{} 3)\cdot S \plus{} 2P$ $ \implies$ $ \boxed {\ 3S^2\le (2P \plus{} 3)\cdot S \plus{} 2P\ }$

[b]with equality iff $ \{a,b\} \equal{} \{0,1\}$ and $ 1\in \{a,b\}$ , i.e. $ \left\|\begin{array}{c} a \equal{} 1 \\
\ b \equal{} 1\end{array}\right\|$ or $ \left\|\begin{array}{c} a \equal{} 1 \\
\ b \equal{} 0\end{array}\right\|$  or $ \left\|\begin{array}{c} a \equal{} 0 \\
\ b \equal{} 1\end{array}\right\|$ .[/b][/color]

[quote="kunny"][quote="Ahiles"]Let $ a,b\in[0;1]$. Prove the inequality:
$ 1 - \dfrac{a + b}{2} + \dfrac{ab}{3}\geq\dfrac{1}{1 + a + b}$[/quote]

Let $ a = \cos \theta ,\ b = \sin \theta (0\leq \theta \leq \frac {\pi}{2})$ and set $ \cos \theta + \sin \theta = t\ (1\leq t\leq \sqrt {2})$, the inequlaity becomes

$ 1 - \frac {t}{2} + \frac {t^2 - 1}{6} - \frac {1}{1 + t}\geq 0$

$ \Longleftrightarrow \frac {(t - 1)(t^2 - t + 1)}{6(t + 1)}\geq 0\ \because t^2 - t + 1 = (t - \frac {1}{2})^2 + \frac {3}{4} > 0$. Q.E.D.[/quote]
u're wrong,kunny.
a  and b  are't interdependent.  hence  t isn't greater 1. (if  you  let  $ a =sinx,  b= cos x$;  then u recognized  that  :$ t^2=1$
 example : $ a = 0,1$  and $ b = 0,1$  then $ t = 0,2 < 1$

[quote="Ahiles"]Let $ a,b\in[0;1]$. Prove the inequality:
$ 1 \minus{} \dfrac{a \plus{} b}{2} \plus{} \dfrac{ab}{3}\geq\dfrac{1}{1 \plus{} a \plus{} b}$[/quote]
My solution:
We have:(1-a)(1-b)>=0, which means: 1+ab>=a+b
Now clearing denominators, it's sufficient to prove: (a+b)/2 +ab/3 -(a+b)^2/2 +(a+b)ab/3>=0
We have: (a+b)/2 -(a+b)^2/2>=(-ab)(a+b)/2(since 1-a-b>=-ab and a+b>=0)
So we just need to prove: (-ab)/6.[3a+3b-2(1+a+b)]>=0, which is equal to:(-ab)/6.(a+b-2)>=0(obviously true since 0=<a,b=<1)
The equality occurs when a=1,b=0 or a=b=1(Done!)
PS:Nice inequality for 8th grade! I'm curious whether we can extend it to 3 variables with the same restriction. :)

This is equivalent to:
$$(1-a)(1-b)(1-c) \geq \sum(a-1)^2 + \sum(a^3 - 1)$$
LHS is positive while RHS is negative
Because $1-a^3 \geq (a-1)^2$ [hide=proof]$a(a+2)(a-1)\leq 0$[/hide]
Hence, done

[quote=MariusStanean][quote=sqing]$ (1\plus{}a\plus{}b)(1\minus{}\dfrac{a\plus{}b}{2}\plus{}\dfrac{ab}{3})=1+\dfrac{ab}{3}+(a+b)(\dfrac{1}{2}-\dfrac{a+b}{2}+\dfrac{ab}{3})\geq 1$
[/quote]
$(1+a+b)\left(1-\frac{a+b}{2}+\frac{ab}{3}\right)=1+\frac{ab}{3}+(a+b)\left(\frac12-\frac{a+b}{2}+\frac{ab}{2}-\frac{ab}{6}\right)\ge 1+\frac{ab}{3}-(a+b)\frac{ab}{6}=1+(2-a-b)\frac{ab}{6}\ge 1.$

If $a,b,c\in[0,1]$ prove that
$$\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\le 1.$$[/quote]
[hide=*][quote=MariusStanean][quote=MariusStanean]
If $a,b,c\in[0,1]$ prove that $$\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\le 1.$$[/quote]
or
$\frac{a}{1+b+c}\le a-\frac{ab+ac}{2}+\frac{abc}{3}$
$\frac{b}{1+c+a}\le b-\frac{bc+ab}{2}+\frac{abc}{3}$
$\frac{c}{1+a+b}\le c-\frac{ca+bc}{2}+\frac{abc}{3}$

Summing up
$LHS\le a+b+c-ab-bc-ca+abc=1-(1-a)(1-b)(1-c)\le 1.$[/quote][/hide]
$$\sum_{cyc}\dfrac{c}{1+a+b}\le\sum_{cyc}\left(c-\dfrac{ac+bc}{2}+\dfrac{abc}{3}\right)$$$$=(a+b+c)-(ab+bc+ca)+abc=1-(1-a)(1-b)(1-c)\leq 1$$
[url=https://artofproblemsolving.com/community/c6h420404p13907044]USAMO 1980 [/url]
[url=https://artofproblemsolving.com/community/c6h202747p1115333]RMO 2008 [/url]

Let $a,b,c\in[0,1].$ Prove that$$ \frac{1}{1+a+b+c}\leq 1-\frac{a+b+c}{3}+\frac{abc}{4}
$$
$$ \frac{1}{1+a+b+c}\leq 1-\frac{a+b+c}{2}+\frac{ab+bc+ca}{3}-\frac{abc}{4}
$$

[hide=*][quote=ywq233]Does this inequality hold?
$a,b,c\in[0,1]$,then $\frac1{1+a+b+c}\leq1-\frac{a+b+c}2+\frac{ab+bc+ca}3-\frac{abc}4$[/quote]
[/hide]


Perpendiculars $ MQ$ and $ MP$ are dropped from an arbitrary point $ M$ of the circle circumscribed about rectangle $ ABCD$ to the rectangle's two opposite sides;the perpendiculars $ MR$ and $ MT$ are dropped to the extensions of the other two sides. Prove that lines $ PR\perp QT$ and the intersection point of $ PR$ and $ QT$ belongs to a diagonal of $ ABCD$.

Let $ P$,$ Q$,$ R$,$ T$ be on $ DA$,$ BC$,$ AB$,$ CD$ respectively
Let $ N$ be on $ BD$ such that $ MN \perp BD$
By Simpson's line, $ P$,$ R$,$ N$ are collinear, and $ Q$,$ T$,$ N$ are collinear
Thus $ PR \cap QT \equal{} N \in BD$
Also $ MQNBR$ is cyclic, so $ \angle PNQ \equal{} \angle PMR \equal{} 90^{\circ}$


:wink:  :ninja:  :wallbash_red:  :?: Jane has a bag containing a mix of seeds for purple flowers, yellow flowers and orange flowers in the ratio of 1:1:2, respectively. There is a total of 16 seeds in the bag. Jane randomly sprinkles 3 seeds in a garden. If each of the seeds sprinkled successfully produces 1 flower, what is the probability that there are 2 purple flowers and 1 yellow flower produced? Express your answer as a common fraction. :first:

[hide]There are 4 purple, 4 yellow, and 8 orange seeds. there are 16C3 ways to choose seeds, and (4C1)(4C2)=24 ways to have 2 purple and 1 yellow flower. Thus, the probability is $ 24/16C3\equal{}3/70$[/hide]


[hide]
After rearranging and adding, we find that
$5xy=2x+6y\Rightarrow 5=\frac{6}{x}+\frac{2}{y}$

The only solution I found that fits the original equations is $(2,1)$. I'm not sure if there are any more.
[/hide]


MATHCOUNTS geometry trigonometry AMC USA(J)MO USAMO search

i know MATHCOUNTS is an answers-only test (or is it?) but where else can I put this ( please enlighten me if there is such a place in this forum...tnx)?

To what degree should you coin mathematical theorems in ur proof?  Do you have to define them or even prove them separately? and if you don't, what saves you from the chance that that very person who will check your paper is unfamiliar with the theorem and will disregard your reasoning?

(I got this problem when i thought about the Common angle Theorem, which states the relationships between the areas of two triangles sharing an angle... i also thought about the fact that very few people know it by that name...)

 :blush: please help me...

The point of the theorem is to replace extra steps, so you don't need to proove it separately unless the problem asks you to. I'm not familiar with the other part of the question, but if the person checking your paper will probably have some sort of reference to check any theorems he or she is unfamiliar with. You most likely won't be downgraded. That help? :D

Actually, that is not true. In general, citing an obscure theorem on a proof test will not be accepted without proof. If it has an obscure name, you should mention the result of the theorem, not just the name, as the result may be trivial, but go by several names. Also, some identities don't really have a widely accepted name (for example, some trig identities like $ \cos^{2}A\plus{}\cos^{2}B\plus{}\cos^{2}C\equal{}1\minus{}2\cos{A}\cos{B}\cos{C}$ or whatever it is, and the factorization $ x^{3}\plus{}y^{3}\plus{}z^{3}\minus{}3xyz\equal{}(x\plus{}y\plus{}z)(x^{2}\plus{}y^{2}\plus{}z^{2}\minus{}xy\minus{}yz\minus{}zx)$, both of which are pretty well known), so you can just cite it as a 'well-known' result.

In general, however, if you are familiar with a theorem, you should have a general idea of how to prove it, so if it doesn't take too much time (it usually doesn't), when in doubt, just prove it. If you don't know how to prove it because its proof is beyond the scope of the competition (like Combinatorial Nullensatz on IMO Problem 6), you should probably look for some other way to solve the problem.

Well there aren't many proof contests around, I mean yeah there's USAMO, and then maybe some local contests, Masters Round at Mathcounts, and others, but yeah...

the way I use a theorem is if it makes my work easier or if it is required to solve a problem.
For example, you almost always use the Pythagorean Theorem at least twice or even three times during a mathcounts exam. Like
"find the missing side of this right triangle- 3,4,_"
well, the right hints for theorem use, and there probably is no other way but  the pythagorean theorem.......... you also could guess, but its not that effective. :lol:

For Mathcounts, it is answer only, besides masters round.  for other proof required tests, I suggest u include as much as u can.  Many of proof required tests are talent searchs, that give u a certain amt of time, not a specifically timed test (like USAMO).  So, if ur sitting on the fence about proving something, I suggest u do if u no how to.  Certain super well known things don't need to be proved of course, like the pythagorean theorem.  By proving more things, u can only make ur proof clearer (if u prove them correctly)


AMC AIME calculus integration algebra polynomial conics

I think these would be numbers 4 to 10 on the AIME. What do you think?

So I'm looking for solutions. :) Just curious to see how you would do these.

1. 8712 is an integral multiple of its reverse, 2178; 8712-4*2178. Find another 4-digit number which is a (non-trivial) multiple of its reverse.

2. Find all polynomials f(x) such that the coefficients of both f(x) and [f(c)]^3 lie within the set {0, -1, 1}

3. ABC is a triangle inscribed in the ellipse with equation (x^2)/4 + (y^2)/9 =1. If its centroid is at the origin (0,0), find its area. (There's a really neat trick to this problem-- too bad I didn't figure it out .__. )

4. A man has three daughters. He knows that the product of their ages is 168 and that the sum of their ages is his street number. However he cannot determine any of their ages. What are possible ages of his oldest daughter?

I know how to do number 4 but not in an easy way. Is there any other way to count all of the factors that multiply to 168 and add to a certain sum twice? (I hope that made sense.)

5. Find all the three-digit numbers that are also the sum of the cubes of its digits.

[hide="Is this the trick for #3?"]
One vertex is at $(\pm2, 0)$.  Then, the y-components of the other two vertices cancel out, so the other two must have x-components $\mp1$.  From there, you find that the other two points must be at $(\mp1, \pm\frac{3\sqrt{3}}{2})$, so the area in question is $\boxed{\frac{9\sqrt{3}}{2}}$ [/hide]


linear algebra
matrix
linear algebra unsolved

Let $V$ be closed under linear combinations set of $5\times7$ matrices. Is it true that if $V$ contains matrices of rank $0,1,2,4,5$ then it must contain matrix of rank $3$ ?

Very bad problem. I remember my teacher gave it to me, I immediately understood that the answer is no, but I lost time to find counterexamples.


I proved the easy part  :)
We have that $ 2(a^{2}+b^{2}+c^{2})(a+b+c)=3(ab+bc+ca)(a+b+c)$ or equivalently and then Schur
$ 2(a^{3}+b^{3}+c^{3})=ab(a+b)+bc(b+c)+ca(c+a)+9abc\leq a^{3}+b^{3}+c^{3}+12abc$

and the equality doesn't  hold so QED

Now the more difficult part b)
Because the ineq and the condition are homogenous let $ a+b+c=\sqrt{7}$
Then $ ab+bc+ca=2$ and we have to prove that (using the indetity of the first part)
$ (2\sqrt{7}+1)abc\geq 2$ let a=max{a,b,c}

then we have to prove that
$ (2\sqrt{7}+1)a(2-a\sqrt{7}+a^{2})\geq 2$ or equivalently

$ (a-\frac{\sqrt{7}-1}{3})^{2}(a-\frac{4-\sqrt{7}}{1+2\sqrt{7}})\geq 0$

which holds

Yes, Silouan.
We can also prove the inequality using the theorem of Vo Quoc Ba Can from http://reflections.awesomemath.org/2007_2/3variable.pdf.
[i]If $ a,b,c$ are positive numbers such that $ a+b+c=1$ and $ ab+bc+ca=\frac{1-q^{2}}{3}$, $ q\ge 0$, then[/i]
$ 27abc \ge (1+q)^{2}(1-2q)$.


Why we must make firstly multiplication and divison at any operation?

It's just a convention, so there is no confusion.


search
number theory unsolved
number theory

A positive integer $ u$ if called [i]boring[/i] if there are only finitely many triples  of positive integer $ (n,a,b)$ such that $ n! \equal{} u^a \minus{} u^b$. Determine the all the [i]boring[/i] integers

It is China TST 2004.  Here is the [url=http://www.mathlinks.ro/viewtopic.php?search_id=1427656464&t=55883]link[/url].


Does anyone know the solutions to the last 5 problems of the AMC 12B 2006?

Take a look here:
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2006
The AMC B is at the bottom (click the numbers for the topics).

Lol, there seem to be no solutions for the 12B problems! But thanks, the 12A problems n solutions are also good.

Just click the actual number of the problem, and it will link you to the forum topic of that problem.


Let $k$ be an integer $\geq4$. Prove that there are infinitely many functions $f: \mathbf{N}^{+}\mapsto\mathbf{N}^{+}$ such that
(1) $f$ is an increasing function;
(2) $f(f(n))=kn$.

[quote="pco"][quote="Hawk Tiger"]Let the set S contains all the integer which is not disible by $k$
Suppose that $S=\{a_{1},a_{2},....\}$(We don't need $a_{1}<a_{2}...$)
then Let
For any integer $x$,
there exist $l,and ,t$ such that $x=k^{l}a_{t}$
then Let $f(x)=k^{l}a_{l+1}$ when $l$is odd,
Let $f(x)=k^{l+1}a_{l-1}$ when $l$ is even.
it is easily to check that $f(x)$ satisfy the requirement.
and becuase $S$ can be written as $\{a_{1},..\}$ in infinitely ways.so there are infinitely $f(x)$[/quote]

1) be careful on usage of l and t (seems to exist some confusion)

2) But your f(x) is not an increasing function (cf. point 1) of problem)[/quote]
Thanks very much.
I found a fuction which is bijucetive.I didn't read the problem carefully....
I didn't understand what you mean in (1) ,I think $l,t\in N$ so $l,t$ exist and only exis one($k \not|a_{t}$)

[quote="Hawk Tiger"] I didn't understand what you mean in (1) ,I think $l,t\in N$ so $l,t$ exist and only exis one($k \not|a_{t}$)[/quote]

Tou are right. There is hust a typo problem :

[quote="Hawk Tiger"] then Let $f(x)=k^{l}a_{l+1}$ when $l$is odd,
Let $f(x)=k^{l+1}a_{l-1}$ when $l$ is even.
[/quote]

I think this mean :
then Let $f(x)=k^{l}a_{t+1}$ when $t$is odd,
Let $f(x)=k^{l+1}a_{t-1}$ when $t$ is even.

[quote="Rust"]All functions, satisfyed (1-2) build by induction.
Let f defined for all $x<k^{n}$ and $Y_{n}=\{k^{n}<y<k^{n+1}| \ exist \ x: y=f(x)\}$ satisfyed $max\{y|y\in Y_{n}<k^{n+1}-k^{n}\}$. Then we define f for $k^{n}\le x<k^{n+1}$: $f(y)=kx<k^{n+1}$, were $y\in Y_{n}, y=f(x).$ For other $x\in [k^{n},k^{n+1})$ we can define any away increase function: $X_{n}=[k^{n},k^{n+1})/Y_{n}\to Y_{n+1}\ belong \ [k^{n+1},k^{n+2}-k^{n+1})$.[/quote]

I wonder how you garauntee that the functions so defined are necessarily increasing?


algebra
polynomial
quadratics
algebra unsolved

Find all values of $ n$ such that $ P(x) \equal{} x^n \minus{} 64$ is reducible over $ \mathbb{Q}$.

[hide="Is it the complete set?"]It is easy to check that $ n \equal{} 2k,3k$ are solutions. But isn't there any more solution :?: [/hide]

For integer polynomials, reducibility over $ \mathbb{Z}$ and reducibility over $ \mathbb{Q}$ are equivalent by Gauss's Lemma.

Now I think that I have a solution. Please check whether it is correct or not. :|

[hide="Solution"]
From Gauss's lemma we know that if $ P$ is reducible over $ \mathbb{Q}$, it is reducible over $ \mathbb{Z}$. So, we shall only consider the case when $ P(x)\in \mathbb{Z}[x]$

From Descates's rule of sign we have that $ P$ has at most one positive real root and calculating $ P( \minus{} x)$ we get that $ P$ has at most one negative real root if $ n$ is odd. But we can not have $ 2$ real roots when $ n$ is odd, because we must have odd number of real roots. So, in any case we have at most one real root.

Now we claim that $ n \equal{} 2k,3k$ are the only solutions. It is easy to check that this factorization works.

Now we assume that $ P(x) \equal{} f(x)g(x)$ for some non constant polynomials. So WLOG we assume that $ f$ has no real root. So the modulus of the roots are $ 64^{\frac {k}{n}} \equal{} 2^{\frac {6k}{n}} \equal{} m$ where $ \deg f \equal{} k$

Here $ m$ must be an integer as $ f(x) \in \mathbb{Z}[x]$

So, $ 2|n, 3|n, 6|n$. Hence the conclusion. :)
[/hide]

[quote="t0rajir0u"][quote="Moonmathpi496"]Here $ m$ must be an integer as $ f(x) \in \mathbb{Z}[x]$[/quote]
Nope.  This only holds if one of the factors is linear.[/quote]
Sorry, I don't understand why this should be linear  :maybe:. Here every complex root has its conjugate as root. So, we can write $ f$ as a product of $ k/2$ real quadratic polynomials.

$ f$ has only complex roots but the constant term is integer $ ( \equal{} m)$. So, all the complex roots have conjugates and from this we can see that every root has modulus $ 2^{\frac{6}{n}}$. But the constant term $ m \equal{} 2^{\frac {6k}{n}}$. So, it should be an integer?? I don't see anything wrong here.

[quote="t0rajir0u"][quote="Moonmathpi496"]Here every complex root has its conjugate as root. So, we can write $ f$ as a product of $ k/2$ real quadratic polynomials.[/quote]
Yes, and there's no reason why the constant terms of these [b]real[/b] quadratic polynomials should have [b]integer[/b] constant term.  (By the way, I believe the claim is, in fact true; it's just harder to prove than you think.)[/quote]
I am interested to continue the discussion up to the time I feel that it is constructive. I hope that you shall be interested as well. :)

First of all, I am not claiming that those [b]quadratics[/b] have integer constant terms. Rather if we multiply all the constant terms of the quadratics we shall get the constant term of $ f$ which must be [b]an integer[/b] $ \leq 64$. (In fact it must divide $ 64$).

OK probably restating my solution will make this look easier.

I assumed that $ P$ is reducible over $ \mathbb{Z}$. That is $ P(x) \equal{} f(x) g(x)$ where $ f,g$ are non constant polynomials with integer co-efficient and $ f$ has [b]integer constant term[/b]. As $ f$ has no real roots, we can write $ f$ as $ f(x) \equal{} \prod_{i \equal{} 1}^{r \equal{} \frac k2} (x \minus{} z_i)(x \minus{} \bar{z_i})$. Clearly, if we multiply all the modulus we shall get $ m \equal{} 2^{\frac {6k}{n}}$. I infer that $ m$ is [b]an integer[/b] and it is obvious because $ m$ is the constant term of $ f$ and it is an integer from definition.

So, I am still confused. :)

[quote="Moonmathpi496"]As $ f$ has no real roots,[/quote]
This doesn't follow.  For example, $ x^3 \minus{} x \minus{} 1$ is irreducible but has a real root.

[quote="Moonmathpi496"]Clearly, if we multiply all the modulus we shall get $ m \equal{} 2^{\frac {6k}{n}}$. [/quote]
This is true of $ P$, not of $ f$.

[quote="t0rajir0u"]This doesn't follow.  For example, $ x^3 \minus{} x \minus{} 1$ is irreducible but has a real root.[/quote]
Probably you haven't read my solution properly. ;) I assumed WLOG that $ f$ has no real root. As $ P$ has at most one real root by Descates's. So WLOG $ g$ has that real root.

[quote="t0rajir0u"]This is true of $ P$, not of $ f$.[/quote]
I believe that it is easy to check that all the roots of $ P$ lies on a circle on complex plane with radius $ 64^{1/n}$, i.e. every root has the same modulus. So, all $ k$ roots has modulus equal to $ 64^{k/n}$.

[u]Check:[/u] We take $ z_k\equal{}64^{\frac{1}{n}} \text{cis} \frac{2k \pi}{n}$
Where $ z_k$ are the roots of $ P$ and $ k\equal{}1,2,\cdots,n.$

Never mind, your solution looks fine.  :oops:   I'm trying to remember why I thought the generalization of this problem was hard.

No problem. :)
I have learned a lot from this problem...Thanks for this discussion.


Let $ABCDEF$ be a convex hexagon such that $AB//DE$, $BC//EF$, $CD//FA$. If $M, N, P, Q, R, S$ are the midpoints of $AB, BC, CD, DE, EF, FA$, then show that $MQ, NR, PS$ are concurrent.

Yes this is a well-known problems.
[hide="Solution 1"]
Use this lemma:Let be given a quadrilateral $ABCD(AB//CD)$.Let $E,F$ be the midpoint of $AB,CD$,$AC$ intersects $BD$ at $O$.$M$ is a point inside one of two angles $\widehat{AOB},\widehat{COD}$.Prove that : $area(MAC)=area(MBD)\Longleftrightarrow M\in EF$[/hide]
[hide="Solution 2"]
Use the Ceva's theorem in sine form[/hide]


It's a bit different from an older poll that asked percent.

I'm kind of confused to what that means.  Does that mean for each poll topic your goal is to see the amount you predicted in the poll?  Anyways, I voted 11-17.


Let $ ABC$ be a right angled triangle with right angle at $ C$. Let $ BCDE$ and $ ACFG$ be squares external to the triangle. Furthermore, let $ AE$ intersect $ BC$ at $ H$ and let $ BG$ intersect $ AC$ at $ K$. Find the measure of $ \angle DKH$.

[hide="Solution"]Let $ AC \equal{} CF \equal{} FG \equal{} GA \equal{} a$ and $ BC \equal{} CD \equal{} DE \equal{} EB \equal{} b$

$ \triangle BCK \sim \triangle BFG \implies \frac {CK}{GF} \equal{} \frac {BC}{BF} \equal{} \frac {BC}{BC \plus{} CF}$

Plugging in values of $ a$ and $ b$, we have $ \frac {CK}{a} \equal{} \frac {b}{b \plus{} a} \iff CK \equal{} \frac {ab}{a \plus{} b}$

Similarly, we have that $ CH \equal{} ED \cdot \frac {AC}{AC \plus{} CD} \equal{} \frac {ab}{a \plus{} b}$.

Thus, $ CH \equal{} CK$, and thus $ \triangle HCK$ is a 45º-45º-90º right triangle with right angle at C.

It follows that $ \boxed{\text{m} \angle DKH \equal{} 45 ^ {\circ} }$[/hide]

[b]Edit:[/b] [i]1000 posts[/i]  :)

That $ BCA$ in the second line should be $ BCK$ I think. Anyway this is sort of like my solution too. I wonder if there are solutions using angle chasing or etc.

[hide="Coordinate solution, pretty nice"]
Set $ C \equal{} (0,0)$, $ B \equal{} (b,0)$, $ A \equal{} (0,a)$, so now $ E \equal{} (b, \minus{} b)$ and $ G \equal{} ( \minus{} a,a)$. The equation for line $ AE$ is now $ y \equal{} \minus{} \frac {a \plus{} b}{b}x \plus{} a$, so the intersection with the $ x$-axis, $ H$, is $ \frac {a \plus{} b}{b}x \equal{} a$ or $ x \equal{} \frac {ab}{a \plus{} b}$. Therefore $ H \equal{} \left(\frac {ab}{a \plus{} b},0\right)$. Similar logic can be used to find that $ K \equal{} \left(0,\frac {ab}{a \plus{} b}\right)$. Thus, $ \triangle{CKH}$ is an isosceles right triangle, so $ \angle{CKH} \equal{} \angle{DKH} \equal{} \boxed{45^\circ}$.[/hide]

[quote="nayel"]Actually I found the coordinates solution as well and in my view it is not different than the first one. I prefer a purely synthetic solution by angle chasing, if it exists. But anyway thanks for sharing your solution.[/quote]
Yeah, they are essentially the same. When I see lots of right angles I immediately go for coordinates.

[b]Possibility:[/b] Could we exploit the fact that $ \triangle FDC \cong \triangle ABC$, which results in some construction symmetric about the line $ EG$?


[quote="shobber"]Hi.
In the past when I entered the topic CMO to my post, it would show as CMO, but now it becomes Cmo. If my original topic is AM-GM, it will become Am-gm. But I think CMO and AM-GM looks MUCH BETTER than Cmo and Am-gm.

Can you solve this problem, Valentin?[/quote]

This was discussed before so I'll answer for Valentin.

Valentin made it impossible to have CAPS-LOCK in the post's title except on the beginning. So, that's why CMO appears as Cmo and AM-GM appears as Am-gm. He made it in this way to prevent people from posting threads in CAPS-LOCK to gain attentions (and it's pretty annoying in sometime).

Hope this answers your question!

Wow! Thanks!

This problem has been bugging me for several months now, and I wasn't sure if it was just a problem with my browser, or something to do with the forum.


Does Anyone has the easier solution than Rust , without $ln$ can help me ?This maybe a hard with me

[quote="darij grinberg"]Two sequences of positive reals, $ \left(x_n\right)$ and $ \left(y_n\right)$, satisfy the relations $ x_{n \plus{} 2} \equal{} x_n \plus{} x_{n \plus{} 1}^2$ and $ y_{n \plus{} 2} \equal{} y_n^2 \plus{} y_{n \plus{} 1}$ for all natural numbers $ n$. Prove that, if the numbers $ x_1$, $ x_2$, $ y_1$, $ y_2$ are all greater than $ 1$, then there exists a natural number $ k$ such that $ x_k > y_k$.[/quote]

Detail solution.
We have $ x_{3} > 1 \plus{} 1 \equal{} 2$ and $ x_{n \plus{} 1} > x_{n}^{2}$. If $ z_{n} \equal{} \ln{x_{n}}$, then
$ z_{n \plus{} 1} > 2z_{n}\Longrightarrow z_{n} > 2^{n \minus{} 3}z_{3}$. It give $ x_{n} > 2^{2^{n \minus{} 3}}\ \forall n\ge 3$.
$ 2 < y_{1} \plus{} y_{2} < y_{1}^{2} \plus{} y_{2} \equal{} y_{3} < y_{4} \equal{} y_{2}^{2} \plus{} y_{1}^{2} \plus{} y_{2},$  and $ y_{4} < y_{3}^{2}$. It give $ y_{n \plus{} 1} < y_{n}^{2}\ \forall n\ge 3.$
Proove $ y_{n \plus{} 2} \minus{} y_{n \plus{} 1}^{2} \equal{} y_{n \plus{} 1} \minus{} 2y_{n \minus{} 1}^{2}y_{n} \minus{} y_{n \minus{} 1}^{4} < y_{n \plus{} 1} \minus{} y_{n}^{2} < 0.$
It give $ y_{n \plus{} 2} < 2y_{n}^{2}\ \forall n\ge 3$. If $ t_{n} \equal{} \ln{y_{n}}$ we have $ t_{n \plus{} 2} < 2t_{n} \plus{} \ln 2 \Longrightarrow t_{n} < 2^{(n \minus{} 3)/2} \plus{} \frac {n \minus{} 3}{2}\ln 2 \plus{} t_{3}.$ It give $ y_{n} < y_{3}*2^{2^{n} \plus{} 0.5n}$.
Therefore exist N: $ x_{n} > y_{n}\ \forall n > N$.

[quote="Rust"]If $ t_{n} \equal{} \ln{y_{n}}$ we have $ t_{n \plus{} 2} < 2t_{n} \plus{} \ln 2 \Longrightarrow t_{n} < 2^{(n \minus{} 3)/2} \plus{} \frac {n \minus{} 3}{2}\ln 2 \plus{} t_{3}.$ It give $ y_{n} < y_{3}*2^{2^{n} \plus{} 0.5n}$.
Therefore exist N: $ x_{n} > y_{n}\ \forall n > N$.[/quote]

How did you get it?


induction
limit
number theory
relatively prime
number theory proposed

I found this one easy but cute.
Prove that there are infinetely many naturals $n$, s.t. $n | f_n$ where $f_n$ is the fibonacci sequence.

What are your $f_0$ and $f_1$? Different conventions are adopted by different people, so please specify this.

Just for food for thought, here's a list of the first few $n$ for which $n|f_n$:

1, 5, 12, 24, 25, 36, 48, ...

Whenever I want to prove something about the Fibonacci numbers, I take a good look at [url=http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibFormulae.html#hyperbolic]this page[/url] (actually, I just browse it randomly, hoping to find something useful). I always find it inspiring :).

We can use the identity $\left(\frac{L_n+\sqrt 5F_n}2\right)^k=\frac{L_{nk}+\sqrt 5F_{nk}}2$ for $k=5$ to prove by induction that $5^n|F_{5^n}$.

Basically, $2^a3^b5^c$ works as long as either:
$a=0$ and $b=0$
$a\ge 2$ and $b\ge 1$

Theorem: If $k|F_a$ and prime $p|k$, $pk|F_{pa}$.

Proof: We have the standard identity $F_{n+a}=L_aF_n-(-1)^aF_{n-a}$ for all $n$, where $L_a$ is the $a$-th Lucas number. Define $G_n=\frac{F_{na}}{F_a}$.
Then $G_0=0,\,G_1=1,\,G_{n+1}=L_aG_n-(-1)^aG_{n-1}$. Note that the discriminant of the characteristic equation is $5F_a^2$. Solving the recursion for $G_n$ mod $p$, the discriminant is zero and $G_n\equiv n\left(\frac{L_a}2\right)^n\mod p$ for $p>2$, since the characteristic equation has a double root and $G_0=0$. Clearly, $p|G_p$ and we are done. If $p=2$ we cannot divide by 2. On the other hand, we can easily verify in this case that $L_a$ is even, and then $G_2\equiv G_0=0\mod 2$.

Applying this theorem to the known examples $5,12,60$, we get the infinite set mentioned at the beginning of my post. It is also easy to show that if $n|F_n$ and $m|F_m$ with $m,n$ relatively prime, $mn|F_{mn}$.

Nice!!!  :)
My proof is also inductive, and uses the identity : $ f_{2n}={f_n\cdot (f_{n-1} +f_{n+1})}$ .

[quote="Krishnendu"]We know $F_n>>n$ for $n$ tends to infininity i.e.$limit[n\rightarrow infinity]\frac{n}{F_n}=0$ so according to the well known  lemma $n\mid F_n$ for infinitely many $n$.[/quote]

$\mathop {\lim }\limits_{n \to  + \infty } \frac{n}{{{n^2} + 1}} = 0$,  but $n\left| {{n^2} + 1} \right.$ satisfies for $n = 1$ only.


Let P(x) be a polynomial such that when P(x) is divided by x-9, the remainder is 99, and when P(x) is divided by x-99, the remainder is 19. What is the remainder when P(x) is divided by (x-19)(x-99)?

MDK, you put x-9, instead of x-19, you know that, so he would've gotten it right if you typed it correctly.

But still, nice question and solution.


Larry bought two coats at a thrift store and then sold the coats a week
later for 135 dollars each. He made a profit of 25% on the first coat but lost
25% on the second coat. What total percent did he lose on the sale of
the two coats? Express your answer to the nearest hundredth.

Please post all of your questions in the same topic.

The first coat cost $ 108$ and the second cost $ 180$.  Total they cost $ 288$ and he sold total for $ 270$.  He lost $ \boxed{6.25}$ percent.


geometry
induction
algebra
polynomial
function
area of a triangle

1. by finding the area of a triangle in two different ways show that if $p_{1},p_{2},p_{3}$ are the altitudes of a triangle and $r$ is the radius of its inscribed circle, then $\frac{1}{p_{1}}+\frac{1}{p_{2}}+\frac{1}{p_{3}}=\frac{1}{r}$ need a hint thanks.
2. use a counting argument to prove the following, for integers $r,n,0<r<n$ $\binom{r}{r}+\binom{r+1}{r}+...+\binom{n}{r}=\binom{n+1}{r+1}$ I found the solution but how do we solve it algebraically or in other method? thanks.
3. find a general formula for the nth derivation of $f(x)=\frac{1}{1-x^{2}}$ thanks.

1 is also [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=inradius&t=107546]here[/url].

[hide="2"]
It is true that $\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}$.

Let $n=a+r$. We induct on $a$. (for each fixed $r$).
Clearly $\binom{r}{r}=\binom{r+1}{r+1}$. (This is the case $a=0$)
Now suppose that $\binom{r}{r}+\binom{r+1}{r}+\dots \binom{r+a_{1}}{r}=\binom{r+a_{1}+1}{r+1}$.
Then $\binom{r}{r}+\binom{r+1}{r}+\dots+\binom{r+(a_{1}+1)}{r}=\binom{r+a_{1}+1}{r+1}+\binom{r+(a_{1}+1)}{r}=\binom{r+a_{1}+2}{r+1}$.
By induction, it is true for all $a$ and hence all $n \geq r$.[/hide]

[quote="Rzeszut"][hide="1"]It's east to see that if $h_{a}$ is the altitude to side $a$ and $S$ is the area of triangle, then $\frac1{h_{a}}= \frac{a}{2S}$. Also it is a well-known fact that $S=\frac{(a+b+c)\cdot r}{2}$. Now combining these two formulas we get $\frac1{h_{a}}+\frac1{h_{b}}+\frac1{h_{c}}= \frac{a+b+c}{2S}= \frac{a+b+c}{2\cdot\frac{(a+b+c)\cdot r}{2}}= \frac1r$. [/hide][/quote]
thanks!
for 2-I didn't understand this row Clearly $\binom{n}{r}=\binom{n+1}{r+1}$.
paladin8, cool solution thanks!  :)

[hide="A sketch of solution to 3"]Use the formula for the $n$-th deriviative of $f\cdot g$: \[(f\cdot g)^{(n)}= \sum_{0\leqslant k\leqslant n}\binom{n}{k}\cdot f^{(k)}\cdot g^{(n-k)}\] (it is easy to show by induction) for functions $f(x)=\frac1{1-x}$ and $g(x)=\frac1{1+x}$. Their $k$-th deriviatives can be easily evaluated since $(\varphi(1+x))'= \varphi'(1+x)$ and $(\varphi(1-x))'=-\varphi'(1-x)$ for any function $\varphi$. [/hide]

tim1234133,

5choose3 is 10
6choose4 is 15

therefore (n)C(r) is not necessarily equal to (n+1)C(r+1)

[quote="tim1234133"]My Solution had a dreadful typo in it (as happens rather too often for me)  :oops:

It should read $\binom{r}{r}=\binom{r+1}{r+1}$ (because they are both 1)[/quote]
ha, easy proof  :)


[b][color=red]Can someone get scores of Mexico team?              :) [/color][/b]

Quien eres?

   Nuestros puntajes casi seguro son:

     Aldo 22
     Memo 21
     Malors  13
     Gringo 12
     Puc 11
    Campero 9

   Probablemente 2 bronces y 4 menciones.

   Saludos, Campero

Soy Pepillo, entrenador de Malors en su prepa y de Rodrigo cuando iba en secundaria, y ps la verdad es que me tienen en suspenso...

podrias colocar las puntuaciones ya seguras de cada problema de ustedes??


algorithm
combinatorics unsolved
combinatorics

Hello Mathlinkers!
I'm with some difficulty in solving the following problem from XIV Olimpíada de Mayo (2008).

In a blackboard, it's written the following expression

$ 1\minus{}2\minus{}2^2\minus{}2^3\minus{}2^4\minus{}2^5\minus{}2^6\minus{}2^7\minus{}2^8\minus{}2^9\minus{}2^{10}$

We put parenthesis by different ways and then we calculate the result. For example:

$ 1\minus{}2\minus{}\left(2^2\minus{}2^3\right)\minus{}2^4\minus{}\left(2^5\minus{}2^6\minus{}2^7\right)\minus{}2^8\minus{}\left( 2^9\minus{}2^{10}\right)\equal{} 403$ and

$ 1\minus{}\left(2\minus{}2^2 \left(\minus{}2^3\minus{}2^4 \right)\minus{}\left(2^5\minus{}2^6\minus{}2^7\right)\right)\minus{} \left(2^8\minus{} 2^9 \right)\minus{}2^{10}\equal{} \minus{}933$

How many different results can we obtain?

I tried to solve the problem with less powers of two, hoping to find a pattern. However, I hadn't much success. Maybe there is a better approach... Any ideas?

Thanks in advance

The obvious guess is that the answer is $ 2^{n \minus{} 1}$ for $ 1 \minus{} 2 \minus{} \ldots \minus{} 2^n$, since the sign of 1 and 2 are fixed but the others can vary.  So then you need to show (1) that all of the resulting expressions have different values, and (2) that you can actually achieve all possible sign combinations.

[hide="Hint for doing the first"]Suppose that two different sign sequences give you the same value.  Look at the highest-valued place at which they differ.[/hide]
[hide="Hint for doing the second"]Construct an algorithm to make the sign of each term what you want it to be, starting from the left.[/hide]

Thanks for your answer, but you forgot the cases where there are produts, such as

$ 1\minus{}2\minus{}\left(2^{2}\minus{}2^{3}\right)\left(\minus{}2^{4}\minus{}2^{5}\right)\minus{}2^{6}\minus{}2^{7}\minus{}2^{8}\minus{} 2^{9}\minus{}2^{10}$ or

$ 1\minus{}2\left(\minus{}2^{2}\minus{}2^{3}\minus{}2^{4}\minus{}2^{5}\right)\minus{}2^{6}\minus{}2^{7}\minus{}2^{8}\minus{} 2^{9}\minus{}2^{10}$

Furthermore, I think the solution without counting the products is $ 2^{n\minus{}1}$, because there are 9 ($ n\minus{}1$, where $ n\equal{}10$) signs which can vary:

$ 1\minus{}2\pm2^{2}\pm2^{3}\pm2^{4}\pm2^{5}\pm2^{6}\pm2^{7}\pm2^{8}\pm 2^{9}\pm2^{10}$

Thinkertoys


hi

i am trying to type something for mechanics in latex
i have for a derivative $\dot{q}$ but is there a way to make two dots, for a second derivative?

$\dot{\dot{q}}$ is not good  :(

I think this is what you're looking for: $\ddot{q}$ and $\dddot{q}$.


Construct a logic puzzle. Each poster adds one condition to the logic puzzle, until someone is able to figure out the logic puzzle in its entirety, or is able to prove that the puzzle cannot be solved.

If you are the one that caused the contradiction, everyone must spend 10 replies making fun of that person...  (or just ignore his post if it's blatently obvious)

Alice, Bob, Carl, Dan, and Eve are standing in a line up, though not necessarily in that order. They are from Paris, Quebec, Sacramento, Taipei, and Venice, though not necessarily in that order. Determine the order that the five are standing in the lineup, and what city they are from.

Condition #1.
Alice and Eve are females, and Bob, Carl, and Dan are males.

okay...

Alice is not 1st or last in line

Either Eve is from Venice, or Eve has a pet turtle.

There are exactly three pets among the 5 people.

The third person in the line owns a red convertible.

Carl is standing next to the person from Quebec and he is also standing next to the person from Paris. Either both of them (the two people standing next to Carl) have a pet, or both of them are female.

In the red convertable, there is a pet dog.

There is exactly one red convertible. Also, as a clarification, the "either...or" construction implies that one of the statements is false.

Either both females own a pet, or neither of them owns a pet.

I have drawn some partial conclusions so far...

The one who owns the red convertable is from paris.

The guy who recently won the game show "Who Wants to be a Millionaire" has shrapnel wounds from his service in World War II and Korea.

The college dorms that Bob lives in does not allow pets.

Carl wears a top hat on alternate Tuesdays.

please give BETTER clues. Like "Alice has a pet, but she is not third." Or something like that.

Alice is not fourth and she may or may not have a pet (she recently lent her cat to a physicist she met named Schrödinger).

You can build stuff off of certain clues... like the pets thing. Though I haven't looked at the logic lately...

There are more than 5 people in the line.

MysticTerminator: Please stop adding random things like shrapnel wounds in Korea and such -_-
The pets are a cat, a dog, and a fish.

Ok, Ok. Let's restart, because this is confusing. There are five people, named A, B, C, D, E and F. They are standing in a line. They all have pets, namely a dog, a cat, a fish, a rabbit and a bird. List the people, from first to last, and make sure that you say which pet belongs to whom.

Do NOT post things that will NOT point us in the direction that we want to go.

Condition #1. A has a pet that has no feathers, or fins and he is not third in line.

Sorry about F. Mysitc, :bomb: :bomb:  :cursing:  :cursing:  :furious: :furious:    :wallbash_red:   Stop doing that!!!! It makes no difference if he wears a top hat or not!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
*sigh* I'm back. I'm really slow to anger, too.
Now, only post HELPFUL things. None of this "He wears a top hat." , "He plays video games before school."
Things like "He walks his pet on Sundays." ,"He is not in an even position." would be GREAT.

Condition #3. B DOESN"T walk his pet on Sundays, but if he could, he would. He is also not busy on Sundays.

dude top hats are awesome what are you talking about


geometry
circumcircle
incenter
3D geometry
prism
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, 0 the circumcircle of ABC, I the incenter of ABC, A*, B*, C* the point of contact of the A, B, C-mixtilinear circles of ABC with 0,
1 the circle passing through A, I, A*,
2 the circle passing through B, I, B*,
3 the circle passing through C, I, C*.
Prove or disprove that 1, 2, 3 meet in a second point.
Sincerely
Jean-Louis

Dear Jayme:
I think your conjecture is true, and the second intersection of the three circles lies on $ OI$ line of triangle $ ABC$.

Let $ \omega$ and $ \Omega$ denote the incircle and circumcircle of triangle $ ABC$, respectively.  We invert the diagram through $ \omega$.

In "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186117]Mixtilinear incircles and somehow Poncelet's porism[/url]," yetti shows that $ A'$ and $ (A^*)'$ are diametrically opposite on $ \Omega'$.  In other words, $ A' (A^*)'$ passes through the centre of $ \Omega'$.  Similarly, $ B' (B^*)'$ and $ C' (C^*)'$ also pass through the centre of $ \Omega'$.

Therefore, the circumcircles of triangles $ AIA^*$, $ BIB^*$, and $ CIC^*$ pass through $ I$ and the image of the centre of $ \Omega'$.


geometry
parallelogram
trigonometry
3D geometry
vector
LaTeX
trig identities

In a parallelogram, there are 2 different diagonals, one longer than the other one. They are only equal when it is a rectangle. I don't know how to prove it mathematically though.

We could bash it with the law of cosines. If the sides of the parallelogram are $ a$ and $ b$, and one of the angles is $ \theta$, then we get that if the diagonals are indeed equal ($ a^2 \plus{} b^2 \minus{} 2ab \cos \theta \equal{} a^2 \plus{} b^2 \plus{} 2ab \cos \theta$, $ \cos \theta \equal{} 0$, i.e., the parallelogram must be a rectangle.

[quote="bigman"]All the diagonals of a paralleliped have equal length. Prove that this paralleliped is rectangular[/quote]

I hope you meant parallelogram, as the previous two posters assumed. If not, then your problem is quite strange, since a parallelepiped (not paralleliped) has space and side diagonals, and the space diagonals cannot be equal to all side diagonals (closest you can get is a cube, in which all side diagonals are equal, but not equal to the space diagonal).

Probably he just meant the four space diagonals of the parallelepiped are all equal.  Which also raises an interesting question: if three of the (space) diagonals of a parallelepiped are equal, must the fourth be as well?  What if we have two pairs of equal diagonals -- must they be all equal?

A vector solution may be appropriate here. Apologies for the lack of LaTeX, I'll edit it in later.
[hide="Hint"]Let one vertex be the origin. We can find vectors $ u, \; v, \; w$ such that the eight vertices are $ 0, \; u, \; v, \; w, \; u\plus{}v, \; v\plus{}w, \; u\plus{}w, \; u\plus{}v\plus{}w$

The lengths of the diagonals are then:
$ \| u\plus{}v\plus{}w \|$
$ \| u\plus{}v\minus{}w \|$
$ \| u\minus{}v\plus{}w \|$
$ \| \minus{}u\plus{}v\plus{}w \|$[/hide]

With that taken care of, we can tackle JBL's problems:
[hide="Solution to JBLs 1st Question"][b]No.[/b]

Consider $ u, \; v, \; w$ = permutations of $ (2, 1, 1)$
$ u\plus{}v\plus{}w \equal{} (4, 4, 4) \implies \| u\plus{}v\plus{}w \| \equal{} 4 \sqrt{3}$
$ u\plus{}v\minus{}w$ and etc. are permutations of $ (1, \minus{}1, 0)$, with length $ \sqrt{2}$

So three diagonals have the same length and a fourth is unequal.[/hide]
[hide="Solution to JBLs 2nd Question"][b]No.[/b]

Let's arbitrarily pick $ \| u\plus{}v\plus{}w \| \equal{} \| u\plus{}v\minus{}w \|$ and $ \| u\minus{}v\plus{}w \| \equal{} \|\minus{}u\plus{}v\plus{}w \|$
We can rewrite the second as $ \| u\minus{}v\plus{}w \| \equal{} \| u\minus{}v\minus{}w \|$

Squaring each and applying the law of cosines, we get:
$ \| u\plus{}v \| ^2 \plus{} \| w \| ^2 \plus{}2(u\plus{}v) \cdot w \equal{} \| u\plus{}v \| ^2 \plus{} |w|^2 \minus{} 2(u\plus{}v) \cdot w$
$ \| u\minus{}v \| ^2 \plus{} \| w \| ^2 \plus{}2(u\minus{}v) \cdot w \equal{} \| u\minus{}v \| ^2 \plus{} \| w \| ^2 \minus{} 2(u\minus{}v)\cdot w$

Canceling and rearranging, we get:
$ (u\plus{}v) \cdot w\equal{}0 \implies u \cdot w\plus{}v \cdot w \equal{} 0$
$ (u\minus{}v) \cdot w\equal{}0 \implies u \cdot w\minus{}v \cdot w \equal{} 0$

Solving this system, we get $ u \cdot w \equal{} 0 \; \underline{\text{and}} \; v \cdot w \equal{} 0$

We can let, for example, $ u\equal{}(3,4,0), \; v\equal{}(4,3,0), \;  w\equal{}(0,0,5)$, and we will get two pairs of equal diagonals.[/hide]

The solution to the second question should give you a good idea on how to solve the original question.


I think nobody want to solve it because:
The first, you should learn to write by LATEX.
Your ineq:
Given $ a, b, c \geq\ 0$ satisfy $ a^4\plus{}b^4\plus{}c^4\equal{}3.$
$ \frac{(4a\minus{}3b)^2}{b^2(5\minus{}2c^2)} \plus{} \frac{(4b\minus{}3c)^2}{c^2(5\minus{}2a^2)} \plus{} \frac{(4c\minus{}3a)^2}{a^2(5\minus{}2b^2)} \geq\ 1$
As you see, now, it looks nicer.
The second reason: ineed, I don't like this ineq because there are many ineqs as it.
I can creat many ineqs as it, but it's not so nice and not give the readers the ineterest.

And...
[quote="quykhtn-qa1"]Nobody can solve this problem? :D[/quote]
Are you sure?
After using Cauchy Schwarts, it's becomes:
$ \sum\ (a \minus{} b)^2(b \minus{} c)^2 \plus{} 3(abc \minus{} 1)^2 \plus{} 3 \minus{} abc \geq\ 0$
(obviously trues)


inequalities
trigonometry
inequalities proposed

Let $a,b,c,d$ be non-negative numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=4$.
(a) prove that
$3(a+b+c+d) \geq 2(ab+bc+cd+da)+4$;

(b) Find the smallest value of $p$ such that the inequality holds
$a+b+c+d-4 \geq p(ab+bc+cd+da-4)$.

Ugly solution :blush:
(1)Let $x=a+c,y=b+d$
Then we have:$4\le x^{2}+y^{2}\le 8$
We want to prove:
$3(x+y)-2xy \ge 4$
Which is equivalent to:
$3r(\cos t+\sin t)-2r^{2}\cos t \sin t \ge 4$
Let $u=\cos t+\sin t$,Then:$2\cos t \sin t=u^{2}-1$
Then the inequality is:
$3ru-r^{2}(u^{2}-1) \ge 4$ for $1 \le u \le \sqrt 2,2 \le r \le 2\sqrt 2$
And it is easy to see  the minimum of the LHS occurs when $r=2$ or $r=2\sqrt 2$
The rest is not hard :D

No, it is not ugly. It is nice, Zhaobin.

hi I don't think so because I don't like trigonometric method  :D Vasc, How to solve it? And: what do you use method?[/i]

Here is nice solution.
Let $a+b+c+d=s$ then $s^{2}\geq a^{2}+b^{2}+c^{2}+d^{2}=4 \geq \frac{s^{2}}{4}\Longrightarrow 4 \geq s \geq 2$
We must prove that
$3s \geq 2(ab+bc+ca+da)+4$
$2(ab+bc+ca+da)+4 \leq \frac{s^{2}}{2}+4 \leq 3s$
$(s-4)(s-2) \leq 0$
That's all :lol:


On the sides $AD$ and $BC$ of a rhombus $ABCD$ we consider the points $M$ and $N$ respectively. The line $MC$ intersects the segment $BD$ in the point $T$, and the line $MN$ intersects the segment $BD$ in the point $U$. We denote by $Q$ the intersection between the line $CU$ and the side $AB$ and with $P$ the intersection point between the line $QT$ and the side $CD$.

Prove that the triangles $QCP$ and $MCN$ have the same area.

[quote="nr1337"][quote="Valentin Vornicu"] We denote by $Q$ the intersection between the line $AB$ and the side $AB$...[/quote]

That'd just be AB, right?[/quote]It's $CU$ actually. I've modified the statement. Thanks for noticing.

Hint:

[hide]Similar triangles (no additional lines necessary)[/hide]

Since the triangles $\triangle QCP, \triangle MCN$ have the same altitudes from the vertices $Q, M$ to the sides $CP, CN$, respectively, equal to the distance between the opposite sides of the rhombus $ABCD$, it is necessary to show that $CP = CN$. Using a parallel projection, we can project the rhombus $ABCD$ into a square $A'B'C'D'$, while preserving all area ratios. Hence, it is sufficient to prove the proposition for a square $ABCD$. Let $a = AB$ be the side of the square and $m = \frac{AM}{AD} = \frac{AM}{a}$, $n = \frac{BN}{BC} = \frac{BN}{a}$, $p = \frac{DP}{DC} = \frac{DP}{a}$. We have to show that $n = p$ regardless of $m$.

From similarity of the triangles $\triangle BNU \sim \triangle DMU$, we have $\frac{BU}{DU} = \frac{BN}{DM} = \frac{n}{m - 1}$.

From similarity of the triangles $\triangle BQU \sim \triangle DQU$, we have $\frac{BQ}{DC} = \frac{BU}{DU} = \frac{n}{m - 1}$.

From similarity of the triangles $\triangle BCT \sim \triangle DMT$, we have $\frac{BT}{DT} = \frac{BC}{DM} = \frac{1}{m - 1}$.

From similarity of the triangles $\triangle BTQ \sim \triangle DTP$, we have $\frac{BQ}{DP} = \frac{BT}{DT} = \frac{1}{m - 1}$.

On the other hand, $\frac{BQ}{DP} = \frac{BQ}{DC} \cdot \frac{DC}{DP} = \frac{n}{m - 1} \cdot \frac 1 p$. Thus $\frac{1}{1 - m} = \frac{n}{m - 1} \cdot \frac 1 p$ or $n = p$.

you can prove it using just simillarities in the same way, without complicated stuff like parallel projections, which btw i have no idea what they are(no need to assume abcd = square)


function
algebra
polynomial
calculus
calculus computations

Let

$ \vec{f}(x,y,z)$ be a $ \mathbb{R}^3 \to \mathbb{R}$ function.

Are there any such nonzero functions so that:

$ \| \ \nabla \times \vec{f} \ \| \equal{} \nabla \cdot \vec{f}$

(i.e., so that the magnitude of the curl of a function equals the divergence of that function)?

The question doesn't make sense; you can't equate a scalar and a vector.

He did ask about the magnitude.

The answer is yes, there are such functions.

Take a globally harmonic function $ u(x,y,z)$ - that is, a function such that $ \Delta u \equal{} \nabla\cdot \nabla u \equal{} 0.$ Many examples can be found. We can even find polynomial examples. Say, $ u \equal{} 2x^2 \minus{} y^2 \minus{} z^2.$

Then let $ \vec{f} \equal{} \nabla u.$ (So in my example, that would be $ \vec{f} \equal{} 4x\vec{i} \minus{} 2y\vec{j} \minus{} 2z\vec{k}.$) We have that $ \nabla\times\vec{f} \equal{} \vec{0}$ since the curl of a gradient is always zero, and we have that $ \nabla\cdot\vec{f} \equal{} 0$ because we chose the function to satisfy this condition.

Having the divergence and the magnitude of the curl be equal but nonzero - that's going to take more work, but it should still be possible.


combinatorics solved
combinatorics
combinatorial geometry

Let E be a finite set of point(in the plane), no 3
of them colinear, no 4 of them concyclic. An unordered
pair of points {A,B} is called a good pair iff there exists
a disk which contains only A and B but no other point.
Let f(E) be the number of good pair in E.
->Prove that if E has 1003 points, then
      2003<= f(E) <=3003

Oh, Oh...

This problem sounds very familiar to me since I proposed it for the preselection of the problem proposed by France for the 2003 IMO...
It has been considered as too difficult and then has not been sent to the jury. The main reason is that it uses Euler's formula for planar graphs, which is automatically refused by some countries for an olympiad problem.

Thus, we use it for a selection test (by mail) in France for the IMO team. As far as I recall, none of the contestant has solved it (hello Alekk!!!).

This problem is connected to Delaunay's triangulation of convex polygons.
Since the solution is not so simple and little boring to give here, you may find it (in french) at :
http://boumbo.salle-s.org/xavier/maths/stmalo/

Choose the paper on graph theory. This is exercise #15.

Pierre.


LaTeX
modular arithmetic
quadratics
number theory
relatively prime
algebra
difference of squares

Find all the pairs of positive numbers such that the last
digit of their sum is 3, their difference is a primer number and
their product is a perfect square.

suppose we seek for all pair $(a,b)$ where $a>b$. let $d=\gcd(a,b)$, then $a=dm$ and $b=dn$ where $\gcd(m,n)=1$. The condition becomes
1. $d(m+n)$ ends by 3
2. $d(m-n)$ is a prime
3. $mn$ is a square
4. from (3) and condition $\gcd(m,n)=1$, we have $m$ and $n$ are both squares. Let $m=e^2$ and $n=f^2$
5. from (4), we conclude that $m-n\neq 1$
6. from (5) and (2), we have $d=1$ and $m-n$ is a prime
7. from (6), we have $(e+f)(e-f)$ is a prime. It means that $e-f=1$

The rest is looking for $e,f$ such that $e+f$ is prime, $e-f=1$ , $e^2+f^2$ ends by 3. And it's very easy to verify.

[hide]
We claim that the only such pair is $(4,9)$.

We want $x$ and $p$ such that $2x+p \equiv 3 \pmod{10}$ and $x(x+p) = y^2$. Consider $d = gcd(x,x+p)$. $d$ has to be $1$ or $p$.

If $d$ is $p$, then $\frac{x}{p}$ and $\frac{x}{p}+1$ have have product a perfect square but they are relatively prime so they are both squares. But we obviously don't have consecutive positive squares so there are no such cases.

If $d$ is $1$, then $x$ and $x+p$ are both squares. The difference of squares $p$ is a sum of consecutive odd integers,  but then $p$ is divisible by the average of these integers and cannot be prime, except in the case where $p$ is the only consecutive odd integer. So $x = z^2$ and $x+p = (z+1)^2$.

So $x+(x+p) \equiv 3 \pmod{10}$ where $x$ and $x+p$ are consecutive quadratic residues. Looking at quadratic residues $\pmod{10}$, we find $0,1,4,9,6,5,6,9,4,1$. The only consecutive ones that sum to $3 \pmod{10}$ are $2^2+3^2$ and $7^2+8^2$.

So we have the case $p = (10k+3)^2-(10k+2)^2 = 20k+5 = 5(4k+1)$. This is prime iff $k = 0$ so $p = 5$ giving us $(4,9)$.

Also we have the case $p = (10k+8)^2-(10k+7)^2 = 20k+15 = 5(4k+3)$ which is never prime. So there are no solutions.

Hence the only pair is $(4,9)$ as claimed.[/hide]

Let those numbers be $a,b$.
Suppose $\gcd(a,b)=d>1$. Write $a=da_1$ and $b=db_1$, where $\gcd(a_1,b_1)=1$.
$d(a_1-b_1)=p$
$d(a_1+b_1)\equiv 3\pmod{10}$
Since, $d>1$, we get that $a_1-b_1=1$
And thus, $$b_1(b_1+1)=x^2$$
Since $b_1$ and $b_1+1$ are co-prime, we have that $b_1=y^2$ and $b_1+1=z^2$, where $y,z$ are positive.
Hence, $(z-y)(z+y)=1$, since $y,z\geq 1$, we have $z+y\geq 2$ and since $z-y\in \mathbb Z$; a contradiction.

Thus, $\gcd(a,b)=1$, therefore we have that $a=m^2$ and $b=n^2$, where $m,n$ are just positive integers. Let $a>b$, meaning $m>n$.
Therefore, $m^2-n^2=p$.
Factorising $(m-n)(m+n)=p$, thus $m+n=p$ and $m-n=1$ or solving it, we get that $2m=p+1$ and $2n=p-1$.
Since their sum is $\equiv 3\pmod{10}$, we have that $\frac{p^2+1}{2}=\frac{2p^2+2}{4}=\frac{(p-1)^2+(p+1)^2}{4}=a+b\equiv 3\pmod{10}$.
We also conclude that $\frac{p^2+1}{2}\equiv 3\pmod{10}$.
By CRT, $\frac{p^2+1}{2}\equiv 1\pmod{2}$ and $\frac{p^2+1}{2}\equiv 3\pmod{5}$
Hence, $p^2\equiv 0\pmod{5}$, i.e. $p=5$.
We conclude that $m=3$, $n=2$, thus $a=9$, $b=4$.
Only pair is $(9,4)$.


trigonometry function logarithms geometry incenter LaTeX calculus

I've noticed a lot of questions abuot trig or trig identities have been posted lately. I think we need a topic to discuss as many of them as possible. Post the ones you know (we need to start off with the simple ones at first, of course) and post proofs as well (both trig and complex proofs are appreciated!)

To get it started, I'll prove the Pythagorean Identity:

$\sin ^2 a + \cos ^2 a = 1$

Proof 1: The identity, unsurpisingly, follows from the Pythagorean Theorem. Consider a right triangle with legs $x, y, \sqrt(x^2+y^2)$. Then we have that $\sin ^2 a = \frac{x^2}{x^2 + y^2}$ and $\cos ^2 a = \frac{y^2}{x^2 + y^2}$, and their sum is clearly 1.

Proof 2: From Euler's Theorem, $e^{ix}=\cos x + i\sin x$. Also, from the evenness/oddness of sin and cos, $e^{-ix}=\cos x - i\sin x$. So we get

$1=e^{ix}e^{-ix}=(\cos x + i\sin x)(\cos x - i\sin x)=\sin ^2x + \cos ^2x$

Furthermore, dividing each side of the identity by $\sin ^2 a$ or $\cos ^2 a$ gives:

$1 + \tan ^2 a = \sec ^2 a$
$1 + \cot ^2 a = \csc ^2 a$

Once we get $sin(a+b)=\sin a \cos b+\sin b cos a$, Note that$ \sin(90-a)=\sin(90) \cos a+\cos(90) (\sin a)=1(\cos a)+0=\cos a$.
Thus $\cos (a+b)=\sin (90-a-b)=\sin(90-a) \cos(-b)+\sin(-b) \cos(90-a)=\cos(a) \cos b-\sin b \sin a.$

SO $\cos(a+b)=\cos(a)\cos(b)-\sin(a)(\sin(b))$

Since$
\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a), $
$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(a)$

$\tan(a+b)=$
$=\frac{\sin(a+b)}{\cos(a+b)}$$=\frac{\sin(a)\cos(b)+\sin(b)\cos(a)}{\cos(a)\cos(b)-\sin(a)\sin(a)}$
$=\left(\frac{\frac{\sin(a)\cos(b)+\sin(b)\cos(a)}{\cos(a)\cos(b)}}{\frac{\cos(a)\cos(b)-\sin(a)\sin(a)}{\cos(a)\cos(b)}}\right)$
$=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$

$\cos(2a)=\cos(a+a)=\cos^2(a)-\sin^2(a)$
By Pathygorean identity, $\sin^2(a)=1-\cos^2(a),$
$\cos^2(a)-\sin^2(a)=cos^2(a)-(1-\cos^2(a))$
$=2cos^2(a)-1$

Which is very helpful to know

I'll try to bring in some less known ones later.

[size=150][b]Triangle formulas[/b][/size]

Let a, b, and c be the sides that correspond to the opposite angles A, B, and C.  s is the semiperimeter.

r is the inscribed circle's radius

\[r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\]

( I don't know why I posted that one... I have this trig sheet and it was on it)

Mollweide's Formula

\[\frac{b-c}{a}=\frac{\sin{\frac 12 (B-C)}}{\cos{\frac 12 A}}\]

(this also works if you switch the a's b's and c's around just like the law of cosines)

Newton's Formula

\[\frac{b+c}{a}=\frac{\cos{\frac 12}(B-C)}{\sin{\frac 12 A}}\]

(this also works if you switch the letters around)

Law of tangents

\[\frac{b-c}{b+c}=\frac{\tan{\frac 12 (B-C)}}{\tan \frac 12 (B+C)}\]

For these one remember that A+B+C=180(degrees)

\[\sin{A}+\sin{B}+\sin{C}=4\cos{ \frac 12 A}\cos{\frac 12 B}\cos{\frac 12 C}\]

\[\cos{A}+\cos{B}+\cos{C}=4\sin{\frac 12 A}\sin{\frac 12 B}\sin{\frac 12 C}+1\]

\[\sin^2{A}+\sin^2{B}+\sin^2{C}=2\cos{A}\cos{B}\cos{C}+2\]

[quote="joml88"][size=150][b]Triangle formulas[/b][/size]

For these one remember that A+B+C=180(degrees)

\[\sin{A}+\sin{B}+\sin{C}=4\cos{ \frac 12 A}\cos{\frac 12 B}\cos{\frac 12 C}\]

\[\cos{A}+\cos{B}+\cos{C}=4\sin{\frac 12 A}\sin{\frac 12 B}\sin{\frac 12 C}\]

\[\sin^2{A}+\sin^2{B}+\sin^2{C}=2\cos{A}\cos{B}\cos{C}+2\][/quote]

can anybody post somemore like this?

[quote="joml88"][size=150][b]Triangle formulas[/b][/size]

Let a, b, and c be the sides that correspond to the opposite angles A, B, and C.  s is the semiperimeter.

r is the inscribed circle's radius

\[r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\]

( I don't know why I posted that one... I have this trig sheet and it was on it)
[/quote]

Here's a proof:

Consider triangle ABC with incenter I as in the below diagram.  Drawing AI, BI, and CI divides ABC into three triangles AIB, AIC, and BIC.  Clearly, the three smaller triangles combined form ABC, so their areas add up to form the area of ABC.  So [AIB] + [AIC] + [BIC] = [ABC].

Since AC, BC, and AB are all tangent to the incircle, the radii are perpendicular to the sides, i.e. ID  :perp: BC, IE :perp: AC, and IF :perp: AB.  So the areas of the three small triangles may be calculated using the simple formula Area = bh/2.  So [AIB] = cr/2, [AIC] = br/2, and [BIC] = ar/2.

Thus, [ABC] = (a+b+c)/2*r = s*r, where s is the semiperimeter of ABC.  But by Heron's formula, [ABC] = $\sqrt{s(s-a)(s-b)(s-c)}$, so

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

There are the sum-to-product formulas, which convert sums of sines or cosines to products of them:

$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left (\frac{A-B}{2} \right)$
$\sin A - \sin B = 2 \sin \left( \frac{A-B}{2} \right) \cos \left (\frac{A+B}{2} \right)$
$\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left (\frac{A-B}{2} \right)$
$\cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left (\frac{A-B}{2} \right)$

We prove the first two using the sine of a sum and sine of a difference formulas:

$\sin (X+Y) = \sin X \cos Y + \cos X \sin Y$
$\sin (X-Y) = \sin X \cos Y - \cos X \sin Y$

Adding or subtracting the two equations yields
$\sin(X+Y) + \sin(X-Y) = 2 \sin X \cos Y$
$\sin(X+Y) - \sin(X-Y) = 2 \sin Y \cos X$

If we let $X = \frac{A+B}{2}$ and $Y = \frac{A-B}{2}$, then $X+Y = \frac{2A}{2} = A$ and $X-Y = \frac{2B}{2} = B$, and the last two equations become

$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left (\frac{A-B}{2} \right)$
$\sin A - \sin B = 2 \sin \left( \frac{A-B}{2} \right) \cos \left (\frac{A+B}{2} \right)$

as desired.

The last two can be proven similarly.  Challenge: can you prove them?  ([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=112795#112795]Hint: Use this[/url])

[quote="JSRosen3"]Can we have some really obscure identities posted here too? I don't know enough of those.[/quote]

In proving the trig inequality:

$ \sum\tan^2\frac{A}{2}\  + 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \geq\ 2 $.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16011

An identity is being used:

$\sum \tan\frac{B}{2}\tan\frac{C}{2}\cos{A} + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = 1$

Can somebody prove this as an exercise?

[quote="fuzzylogic"][quote="joml88"]
\[\cos{A}+\cos{B}+\cos{C}=4\sin{\frac 12 A}\sin{\frac 12 B}\sin{\frac 12 C}\][/quote]

This one is not right.[/quote]

Sorry about that.  I missed the +1 part at the end.  It is fixed now.

Law Of Sines

${a\over\sin A}={b\over\sin B}={c\over\sin C}=2R$

Law Of Tangents

${a-b\over a+b} = {\tan[{{1\over 2}}(A-B)]\over\tan[{{1\over 2}}(A+B)]}$

Law Of Cosines

$\cos C = {a^2+b^2-c^2\over 2ab}$

do you mean by posting up your TeXed pdf? or just simply all the compilation? because if you dont TeX it, but just post it here all together, i will tex it. I need some $\LaTeX$ practice!

I'm doing it in $\LaTeX$ so I will post the PDF.

[quote="tetrahedr0n"]In fact, you can use complex numbers to prove $\sin (a+b) = \sin a \cos b + \sin b \cos a$. I'll leave it to someone else to work out the details.[/quote]

$\sin(a)\cos(b)+\sin(b)\cos(a)$

$= \frac{e^{ai}-e^{-ai}}{2i} \cdot \frac{e^{bi}+e^{-bi}}{2} + \frac{e^{bi}-e^{-bi}}{2i} \cdot \frac{e^{ai}+e^{-ai}}{2}$

$=\frac{e^{(a+b)i} + e^{(a-b)i}-e^{(b-a)i}-e^{-(a+b)i}}{4i} + \frac{e^{(a+b)i} + e^{(b-a)i}-e^{(a-b)i}-e^{-(a+b)i}}{4i}$

$=\frac{e^{(a+b)i}-e^{-(a+b)i}}{2i}$

$=\sin(a+b)$

[quote="Jimmy"][quote="JSRosen3"]Can we have some really obscure identities posted here too? I don't know enough of those.[/quote]

From Euler's identity $e^{xi} = i\sin(x)+\cos(x)$, we have
$\sin(z)=\frac{e^{zi}-e^{-zi}}{2i}$
$\cos(z)=\frac{e^{zi}+e^{-zi}}{2}$

Here's an identity Kent Merryfield mentioned, but I don't know how to prove it (except by taking the derivative of both sides and assuming the normal rules hold)
$\tan^{-1}(z)=\frac{1}{2i}\ln\left(\frac{1+zi}{1-zi}\right)$
This is the complex logarithm $\ln(z)=\ln|z|+i\arg(z)$.[/quote]

I finally figured out a proof, but it involves calculus, so if anyone's interested PM me.

[quote="Jimmy"]
Here's an identity Kent Merryfield mentioned, but I don't know how to prove it (except by taking the derivative of both sides and assuming the normal rules hold)
$\tan^{-1}(z)=\frac{1}{2i}\ln\left(\frac{1+zi}{1-zi}\right)$
This is the complex logarithm $\ln(z)=\ln|z|+i\arg(z)$.[/quote]

Here's a proof of that identity that doesn't (explicitly) use calculus.
You know that $\displaystyle \sin(z)=\frac{e^{zi}-e^{-zi}}{2i}$ and $\displaystyle \cos(z)=\frac{e^{zi}+e^{-zi}}{2}$, so $\displaystyle \tan(z)=\frac{e^{zi}-e^{-zi}}{i(e^{zi}+e^{-zi})}$.

Let $\displaystyle w = \tan(z)=\frac{e^{zi}-e^{-zi}}{i(e^{zi}+e^{-zi})}$. We want to solve this for z as a function of w - that will give us $\tan^{-1}(w)$, and then we can change the w variables back to z.

Multiplying through by the denomenator we get:
$\displaystyle iwe^{zi} + iwe^{-zi} = e^{zi} - e^{-zi}$
And collecting terms:
$\displaystyle e^{zi}(1-iw) = e^{-zi}(1+iw)$
Then cross multiplying:
$\displaystyle e^{2zi} = \frac{1+iw}{1-iw}$
Until finally $\displaystyle z = \frac{1}{2i}\ln\left(\frac{1+iw}{1-iw}\right)$, where this is the complex logarithm as you mentioned above.

Therefore $\displaystyle \tan^{-1}(z)=\frac{1}{2i}\ln\left(\frac{1+zi}{1-zi}\right)$

[quote="JBL"] Although, the formulae for $\tan(x_1 + x_2 + \ldots + x_n)$ are nice looking.[/quote]

Can you post it? :?

[quote="joml88"][quote="WindSlicer"]I think someone should compile all of the ones mentioned here and either post them all in the first post, or put them into a document for downloading or something. Would be extremely useful (just thinking about the future.. taking trig in about 2 months) :P And I suppose if no one does it by then I'll do it..  but until then.. :)[/quote]

I just starting compiling something of the sort of what you are talking about.  When I get enough of it put together I will post the rough draft here for people to critique and tell me things to add to make it better.[/quote]

Awesome!

Trig class starts in a few weeks so I'll be thumbing through this thread a lot I can imagine, heh.

[quote="Psycho"]if $A, B$ and $C$ are corners of a triangle, then \[\tan A\tan B\tan C=\tan A+\tan B+\tan C\]
(i can't remember the proof, though it wasn't very difficult... i don't have much time at the moment, but will surely post it within the shortest time as possible... or if anyone else feels like it, be my guest :P )[/quote]

okay you know A+B+C = 180 degrees. Therefore, C = 180-(A+B)

Use the tangent sum identity:

$tanC = tan(180-(A+B)) = \frac{tan(180) - tan(A+B)}{1+tan(180)tan(A+B)}$.

Because tan(180) = 0, $tanC = -tan(A+B).$

Use the identity again to get:

$tanC = \frac{tan(A) + tan(B)}{tan(A)tan(B) - 1}.$

Subsitute tanC into tanA+tanB+tanC and you get:

$tan(A)+tan(B)+tan(C) = \frac{tan(A)tan(B)[tan(A) + tan(B)]}{tan(A)tan(B) - 1}$

but we use $tanC = \frac{tan(A) + tan(B)}{tan(A)tan(B) - 1},$ so

tanA+tanB+tanC = (tanA)(tanB)(tanC).

tah dah! not too hard if you think about it long enough

[quote="DPopov"]PK=pwnage  :lol:  :P[/quote]

haha, DPopov >> pk  :D

[quote="pkothari13"] tah dah! not too hard if you think about it long enough[/quote]
like i said before :P
i just had to learn for this stupid latin exam the other day  :wallbash:

[quote="shobber"][quote="DPopov"]Prove that if $A, B, C$ are angles of a triangle then $\cos{A}+ \cos{B}+ \cos{C} \leq \frac{3}{2}$

Similarly prove that if $A, B, C$ are angles of a triangle then $\sin{A}+ \sin{B}+ \sin{C} \leq \frac{3 \sqrt{3}}{2}$[/quote]
How?  :?  :spider:[/quote]

I think Jensen's + a little casework does it.


calculus
integration
calculus computations

[quote="sos440"]juset use the substitution $t = n+x$...[/quote]

oh  :D , thats nice . Thanks  :lol:


Let $ P$ be the polynomial $ P(x)\equal{}\sum_k a_kx^k$, with real coefficients, such that $ \sum_k \frac{k\plus{}2}{k\plus{}1}a_k\equal{}0$
Prove that there is $ 0 <c < 1$ such that $ P(c)\equal{}0$ . Is it hard ?
 :cool:

[quote="Diogene"]Let $ P$ be the polynomial $ P(x) \equal{} \sum_k a_kx^k$, with real coefficients, such that $ \sum_k \frac {k \plus{} 2}{k \plus{} 1}a_k \equal{} 0$
Prove that there is $ 0 < c < 1$ such that $ P(c) \equal{} 0$ . Is it hard ?
 :cool:[/quote]

Let $ Q(x) \equal{} \sum_{k \equal{} 0}^{n}a_k\frac {x^{k \plus{} 1}}{k \plus{} 1}$. Then $ xQ(x) \equal{} \sum_{k \equal{} 0}^{n}a_k\frac {x^{k \plus{} 2}}{k \plus{} 1}$ and $ Q(x) \plus{} xQ'(x) \equal{} (xQ(x))' \equal{} \sum_{k \equal{} 0}^{n}a_k\frac {k \plus{} 2}{k \plus{} 1}x^{k \plus{} 1}$

So we have $ Q(0) \equal{} 0$ and $ Q(1) \plus{} Q'(1) \equal{} \sum_{k \equal{} 0}^{n}a_k\frac {k \plus{} 2}{k \plus{} 1} \equal{} 0$. So :

either $ Q(1) < 0$ and so $ Q'(1) > 0$ and so there is a local minimum of $ Q(x)$ for some $ c\in(0,1)$ and so $ Q'(c) \equal{} 0$
either $ Q(1) > 0$ and so $ Q'(1) < 0$ and so there is a local maximum of $ Q(x)$ for some $ c\in(0,1)$ and so $ Q'(c) \equal{} 0$
either $ Q(1) \equal{} 0$ and so $ Q(0) \equal{} Q(1) \equal{} 0$ implies that $ Q'(c) \equal{} 0$ for some $ c\in(0,1)$

And since $ Q'(x) \equal{} \sum_{k \equal{} 0}^{n}a_kx^k \equal{} P(x)$, we got the required result.


modular arithmetic
number theory
least common multiple
number theory unsolved

Hello,someone help me with this
If $ a\equiv b \bmod {m_i},i\equal{}1,2,...,n$, prove that:
$ a\equiv b \bmod {m}$ where $ m\equal{}[m_1,m_2,...,m_3]$

$ a \equiv b \pmod{m_i}$ is equivalent to $ m_i \mid a\minus{}b$. Therefore $ a\minus{}b$ is a multiple for all $ m_i$.
What is $ m \equal{} [m_1,m_2,\ldots,m_n]$? The least common multiple of the $ m_i$. By its very property, it divides any other common multiple, therefore $ m \mid a\minus{}b$, i.e. $ a \equiv b \pmod{m}$.

probably it's $ lcm [m_1,m_2,...,m_n]$.
Actualy i did completely missunderstand the problem.I was thinking something else instead of $ lcm[m_1,m_2,...,m_n]$.


abstract algebra
superior algebra
superior algebra unsolved

[quote="dlhomou"]Is it true, R has no proper non zero ideals (for unitary free R-module)?  It seems true, but I can't seem to prove this.[/quote]
What is a unitary free module?

If it is some type of ring you can show that in this case it is a field. Because fields have no proper nontrivial ideals.


trigonometry algebra polynomial quadratics geometry

1. Could someone please explain how to calculate the following by hand?

$ \cos 36$
$ \cos 72$
$ \cos 54$
$ \cos 18$

I'm asking because I read that they turn out to have fairly nice values (at the first 2) and I wonder if knowing these values might be able to trivialize problems like "Find $ \cos 36 \minus{} \cos 72$"

2. In general, when you need to evaluate a trig expression with angles that almost look like special angles or seem curiously related, what strategies do you use?

3. How would you center the the four trig problems above? I'm having trouble using \begin{center} text \end{center}

[hide="1:  a geometric perspective"] Consider an isosceles triangle with angles $ 36^{\circ}, 72^{\circ}, 72^{\circ}$.  [/hide]
[hide="1:  an algebraic perspective"] Let $ \omega \equal{} e^{i \frac{\pi}{5} }$.  $ \omega$ is a root of the polynomial $ \omega^5 \plus{} 1 \equal{} 0 \Leftrightarrow (\omega \plus{} 1)(\omega^4 \minus{} \omega^3 \plus{} \omega^2 \minus{} \omega \plus{} 1)$.  The second factor factors into two quadratic polynomials; find then and use the quadratic formula. [/hide]
2:  Mess around with them algebraically.  Often the individual terms are not easy to calculate while the expression itself is because it satisfies a simple identity.

You need something to start from, but after knowing just one of the values you can find the others.

Anyway, $ cos 36^{\circ} \equal{} \frac{1\plus{}\sqrt{5}}{4}$, which I bet comes from a 5-angle formula for cosine starting with $ \cos 180^\circ \equal{} \minus{}1$, but I've never tried it myself.

Anyway, the formulas you need is:
$ \cos(a\plus{}b) \equal{} \cos a \cos b \minus{} \sin a \sin b$

For $ a\equal{}b$, you then have $ \cos(2a) \equal{} 2\cos^2 a \minus{}1$.  Inverting this, $ \cos x/2 \equal{} \pm \sqrt{\cos x \plus{} 1}{2}$, where either the plus or the minus is correct depending on the quadrant x/2 lies in.

You can use the double angle formula on $ \cos 36^\circ$ to find  $ \cos 72^\circ$, the half angle formula on 36 degrees to find $ \cos 18^degrees$, and the sum formula on the 36 and 18 degrees to find 54 degree... but for the 54 degrees you need the formula $ \cos^2 x \plus{} \sin^2 x \equal{}1$ to find the sines before you can proceed.

probably also should have mentioned $ \sin(a \plus{} b) \equal{} \sin a \cos b \plus{} \cos a \sin b$.

Anyway, where does $ cos 36^\circ$ come from?

Using the formulas I just posted, starting from $ \cos(5x) \equal{} cos(4x \plus{} x)$ I ended up with $ \cos 5x \equal{} 8\cos^5 x \minus{} 8 cos^3 x \plus{} \cos x$.  In the case of $ x \equal{} 36^\circ$, we have $ \cos 180^\circ \equal{} \minus{} 1$ and therefore are looking for the solutions to $ 8\cos^5 36 \minus{} 8 cos^3 36 \plus{} \cos 36 \plus{} 1 \equal{} 0$.  Factoring, $ 8\cos^3 36 (\cos^2 36 \minus{} 1) \plus{} \cos 36 \plus{} 1 \equal{} 0$ and then $ (cos 36 \plus{} 1)(8\cos^3 36 (\cos 36 \minus{} 1) \plus{} 1) \equal{} 0$ so we have $ \cos 36 \equal{} \minus{} 1$ (obviously false) or $ 8 \cos^4 36 \minus{} 8\cos^3 36 \plus{} 1 \equal{} 0$.

Now lets have $ y \equal{} \cos 36$ so we have $ 8y^4 \minus{} 8y^3 \plus{} 1 \equal{} 0$.  Letting $ y \equal{} z \plus{} 1/4$, we have $ 8z^4 \minus{} 3z^2 \minus{} z \plus{} \frac{29}{32} \equal{} 0$.... I think I must have done something wrong, because this has no zeros and should have one at $ z \equal{} \sqrt {5}/4$.  Does anyone see my mistake?

[quote="facis"]I ended up with $ \cos 5x \equal{} 8\cos^5 x \minus{} 8 cos^3 x \plus{} \cos x$. [/quote]

The correct formula is $ \cos 5x \equal{} 16 \cos^5 x \minus{} 20 \cos^3 x \plus{} 5 \cos x$.  Your RHS there is $ \cos x \cos 4x$.  Work from here.  (This is a modified version of the algebraic solution.)

[quote="t0rajir0u"]The correct formula is $ \cos 5x \equal{} 16 \cos^5 x \minus{} 20 \cos^3 x \plus{} 5 \cos x$. [/quote]

Thanks, t0rajir0u.  I accidently forgot a square when I simplified the $ \cos 4x$, but with the correction I agree.

So now, we have $ 16\cos^5 36 \minus{} 20 \cos^3 36 \plus{} 5 \cos 36 \plus{} 1 \equal{} 0$.  By synthetic division, we can factor this into $ (\cos 36 \plus{} 1)(16\cos^4 36 \minus{} 16\cos^3 36 \minus{} 4\cos^2 36 \plus{} 4 \cos 36 \plus{} 1) \equal{} 0$ (treating $ \cos 36$ as the variable).  Since we know $ \cos 36 \neq \minus{} 1$, we can ignore the first factor.  Now letting $ \cos 36 \equal{} z \plus{} \frac {1}{4}$ we have $ 16z^4 \minus{} 10z^2 \plus{} \frac {25}{16} \equal{} 0$, or $ 256z^4 \minus{} 160z^2 \plus{} 25 \equal{} 0$ which factors $ (16z^2 \minus{} 5)^2 \equal{} 0$ so $ z^2 \equal{} 5/16$ and $ z \equal{} \pm \frac {\sqrt {5}}{4}$, and back-substituting $ \cos 36 \equal{} \frac {1 \pm \sqrt {5}}{4}$.  But cosine is positive on the interval $ ( \minus{} 90^\circ,90^\circ)$, so the negative sign cannot be correct or else $ \cos 36$ would be negative. (Side note: the negative sign gives us $ \cos 108$ or $ \cos 252$ because 108 degrees is 1/5 of a 540 degree angle, which is coterminal with the 180 degree angle we used to set up the problem, and 252 is 1/5 of a 1260 degree angle, also coterminal with the 180 degree angle).

Now that you have a value to start from, you can use the normal trig formulas to on it to find the other values.

How do we start the geometric perspective approach? We don't have any side lengths.

Actually, do we need any side lengths? :huh:

No you don't.
The special thing with a 36 72 72 triangle is that if you bisect one of the 72 angles you get two 36 angles, so there are similar triangles. It is often convenient to call a side length 1.

Um, there is no such thing as a 36-36-72 triangle.  But with the 72-72-36 triangle, bisecting one of the 72's gives a 36-36-108 triangle and a 36-72-72... whoa, bisecting an angle produces a similar triangle!  Probably where you need to work from.

Facis, a slick way to find cos(5x) is to use Demoivre's. We have $ (cos x \plus{} isin x)^5$=(cos 5x+ isin 5x). Matching the real parts gives the answer (I know, it's still not that easy).

Anyways, here is how I find cos 36.
sin 36=sin144=2sin72cos72=$ 4sin36cos36(2cos^2(36) \minus{} 1)$, so $ 1 \equal{} 4cos36(2cos^2(36) \minus{} 1)$. Let cos 36=x, so $ 0 \equal{} 8x^3 \minus{} 4x \minus{} 1 \equal{} (2x \plus{} 1)(4x^2 \minus{} 2x \minus{} 1)$. Solving the second factor(the ans must be positive) gives x=$ (\sqrt {5} \plus{} 1)/4$

yeah, Demoivre's works but it isn't really easier.

As for how you start out, aren't you just using $ \sin 36 \equal{} \sin 144$? (sines of supplements are equal).  Anyway, I suppose that is simpler than what I did, but both give the correct answer.

Oh yeah, I didn't even think of it like that. I will change it.

[b]Follow-up:[/b]  Find necessary and sufficient conditions on an integer $ n$ such that $ n^{\circ}$ is a constructible angle.  In other words, $ \sin n^{\circ}$ is [url=http://en.wikipedia.org/wiki/Algebraic_number]algebraic.[/url]

Hint for the geometric appraoch:

look for similar triangles after splitting angles.

Is there a geometric approach for $ \cos 54^{\circ}$?  I only found a trigonometric one (relying on the result for $ \cos 36^{\circ}$).  By the way, I found the geometric approach to evaluating $ \cos 36^{\circ}$ to be much easier than the trigonometric approach.

try $ \cos 54 \equal{} \sin 36$, and $ \sin 36 \equal{} \sqrt{1\minus{}\cos^2 36}$ (pythagorean theorem).  Basically, construct a right triangle with the 36 degree angle.  You found $ \cos 36$, so just continue from there.

Yeah, that's the trigonometric approach requiring the value of $ \cos 36^{\circ}$.  I meant an approach that uses pure geometry and doesn't require previous knowledge of any trig values (just like the geometric derivation of the closed form of $ \cos 36^{\circ}$).

well, you could use the same diagram as you did to find $ \cos 36$, but drop in a perpendicular-> use pythagorean theorem


find $ fQ\minus{}>Q$ such that:
$ f(x\minus{}y)\plus{}f(x\plus{}y)\equal{}f(2x)$


A triangle is cut from the corner of a rectangle. The resulting pentagon has sides of length 8, 10, 13, 15 and 20 units, though not necessarily in that order. What is the area of the pentagon?
A. 252.5
B. 260
C. 270
D. 275.5
E. 282.5

[hide="Solution"]
[asy]size(100);
defaultpen(0.8);
pair A = (0,0), B=(20,0), C=(20,15), D=(0,15), E=(0,10), F=(8,15);
draw(A--B--C--F--E--cycle);
draw(E--D--F,dashed);[/asy]

Clearly the two sides that were cut must be shorter than the two sides that weren't cut; from there its just a process of trial and error. In fact, the side of length $ 13$ makes the $ 5\minus{}12\minus{}13 \triangle$ particularily conspicuous, and quick rearrangement gives us rectangle dimensions $ 20 \times 15$ with triangle cut out having dimensions $ 12 \times 5$. The area is $ 20 \cdot 15 \minus{} \frac{1}{2} \cdot 12 \cdot 5 \equal{} 270$.

[/hide]


We have a finite set of points in a plane, not all on the same circle.
Prove (or disprove) that there is a circle that contains exactly three points of the set.

Note: this is stronger that [url=http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem]Sylvester-Gallai theorem[/url].

Ehm, maybe we should add that no three points are collinear  :P

However I think that we can prove it in the same way...

Let's take the circle $ \Gamma_{1}$ and the point $ P_{1}$ not on the circle that minimize $ d(P,\Gamma)$ (which is the distance from a circle and a point). Let $ Q$ be the projection of $ P_{1}$ on $ \Gamma_{1}$ Let's take four point of the circle $ A,B,C,D$ such that $ A,B,Q,C,D$ are in this order on the circle. Now we draw the circumcircle of $ AP_{1}D$ (let's call it $ \Gamma_{2}$). Let's say wlog that $ AP_{1}\le P_{1}D$ We have that $ d(B,\Gamma_{2})<d(P_{1}\Gamma_{1})$, absurd...


real analysis
geometry
topology
real analysis unsolved

$f: \mathbb{R}^{2}\to \mathbb{R}^{2}$ satisfies $|f(x)-f(y)| \leq |x-y|$. Hence it is easy to see that $m^{*}(f(E)) \leq m^{*}(E)$ for arbitrary set $E$, where $m^{*}$ is the outer Lebesgue measure. Suppose that $f$ is not an isometry. Does there necessarily exist a set $E$ such that $m^{*}(f(E)) < m^{*}(E)$ ?

Obviously, yes: assume that $|f(x)-f(y)|<|x-y|=2r$. Take $E$ to be the union of two disks of radius $r$ centered at $x$ and $y$.

sorry for this idiot problem :lol:  in fact, my question was for a contraction $f: D^{2}\to D^{2}$ (I was trying to solve [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=128934]this[/url] problem with an area argument). But this proof doesn't work ..

Why doesn't it work? Just take a chain of small ajacent disks with the first disk centered at $x$ and the last centered at $y$ instead. Then you'll be able to fit the entire chain inside the unit disk if $x,y$ belong to the open unit disk. Actually, the problem you referred to has been beaten almost to death on this forum for arbitrary compact metric space (I'll try to look up the references when I have more time). If you want something really non-trivial in this venue, answer the following question.

Let $X$ be a compact metric space and $f: X\to X$. Suppose that for every two distinct points $x,y\in X$, we have $\inf_{n>0}d(f^{n}(x),f^{n}(y))>0$ ($f^{n}$ is the $n$-th iterate of $f$). Prove that $f$ is a bijection.


[quote="AndrewTom"]Show that one root of az^3 +bz^2+cz+d=0 is the reciprocal of another root if and only if a^2 - d^2 = ac - bd.[/quote]
Let $ p(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$.(Assume $ a \neq 0$..) We want to show that there exists $ q(x) \equal{} x^2 \plus{} nx \plus{} 1$ such that $ q(z) \mid p(x)$ if and only if $ a^2 \minus{} d^2 \equal{} ac \minus{} bd$. Let $ r(x) \equal{} ax \plus{} r$, such that $ r(x)q(x) \equal{} p(x)$ $ \iff$ $ (x^2 \plus{} nx \plus{} 1)(ax \plus{} r) \equal{} ax^3 \plus{} (an \plus{} r)x^2 \plus{} (a \plus{} rn)x \plus{} r \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$.
From here we easily obtain $ n \in \mathbb{R}$ and $ r \equal{} d$, so: $ (an \plus{} d)x^2 \plus{} (a \plus{} dn)x \equal{} bx^2 \plus{} cx$. So $ an \plus{} d \equal{} b$ and $ a \plus{} dn \equal{} c$.
Then we get $ n \equal{} \frac {d \minus{} b}{a}$ and $ n \equal{} \frac {c \minus{} a}{d}$ So we have the requierment: $ \frac {d \minus{} b}{a} \equal{} \frac {c \minus{} a}{d}$ $ \iff$ $ a^2 \plus{} d^2 \equal{} ac \plus{} db$.
Which is almost what the problem stated :D  (I guess there has been a small error when writing the problem)

Edit: Of course, the problem is stated correctly. It should be $ n \equal{} \frac {b \minus{} d}{a}$!

let p,q,r be the three solution st pq=1
p+q+r=-b/a
pq+qr+rp=c/a  .....(s)
pqr=-d/a
r=-d/a            ....(1)
p+q=(d-b)/a    ....(2)
use (1),(2) in (s)
u will get the answer.

Thanks. The given condition is correct: on your second-last line, you wrote n = (d-b)/a which should be    n = (b-d)/a.

Thanks kabi. I only saw your reply when I had replied to thank Mathias-DK.

No need to say thanks dude.
At least I got a question to revise those things which I had done few years ago.
So I should thank you.


Solve the functional equation : \[f(x)^{2}=1+xf(x+1) ,\] $x$ integer .

You mean that $f$ is defined on $\mathbb Z$? And what sort of values does it take? ($\mathbb {N,Z}$, something else)?

Let g(x) = f(x) - x Note that by induction g(x)>=1.

Now this implies that g(x)^2 + 2(x+1)g(x) = xg(x+1)

Then Let g(x) minimum be M and g(N) =M

then g(N-1)^2 + 2 (N) g(N-1) = N M
so g(N-1) < M contradiction
unless M=0 or N=1.

If g(N) = 0 then g(x) =0 for all x and we are done; f(x) = x+1

If N=1. Then g(1) = M is the minimum.  but then consider the same argument for all i>=2; that is consider the minimum of g(i) i>=2 the same argument implies that g(2) is the minimum of this range

So we finally get that g(1) < g(2) < g(3) < ...
So we see that f(x) >= 2(x+1)

but then it is easy to see that if f(x) >2(+1) for allx then f(x) > 3x for all x and f(x) is ingeneral bigger than Mx for any M. so a contradiction is reached again.

so the unique solution is f(x) = x+1


The program's website is here:

[url]http://www.mathzoom.com/[/url]

Has anyone ever attended this program before? If so, how was it? I was surprised to have not seen any prominent mention of it on these forums, since there are some names mentioned on the website that I recognized...

Oh, when it says, under Important Dates, Required Admission Test - April 14 - 20, does that mean that the deadline to apply is past? Because underneath it, it says Application Deadline - May 8.

We will be having a MathZoom [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Math Jam[/url]  on [b]Wednesday, April 25[/b] at a special time of [b]10 PM Eastern / 7 PM Pacific[/b].

Also, they have extended their application deadlines -- see their website for details.

April 20 is the early registration deadline; we are still accepting applications.  As long as there's space available, we will not turn away good applicants.  So if you are interested in attending, just fill the application form and submit it along with the admission test.

We had a MathZoom Winter Camp at the end of 2006.  This year's Summer Camp is our first summer program.  I spelled out a lot of details in our Math Jam:

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=203]View MathZoom Math Jam transcript[/url]

Feel free to PM or email me (kwang@mathzoom.com) for further information.


calculus
integration
function
logarithms
calculus computations

Inegrate with respect to x the following functions

Log(LogX) +/[LogX]^2

Sqrt(Sin[X-a]/Sin[X+a]) ,    where a is a constant

[quote="Michael Niland"]

Log(LogX) +/[LogX]^2

[/quote]

Do you mean $\ln(\ln x) + \frac{1}{(\ln x)^2}$ ?

Moubinool, the way you have written the first function is as I intended.

Michael

[quote="Moubinool"]$\ln(\ln x) + \frac{1}{(\ln x)^2}$ [/quote]

$\int \ln(\ln x)=x\ln(\ln x)-\int \frac{1}{\ln x}$

$\int \frac{1}{(\ln x)^2}=\frac{-x}{\ln x}+\int \frac{1}{\ln x}$

just add


This might be a stupid question.

If the cell membrane is actually composed of phosopholipid bilayers and all the hydrophobic tails are between the two outers layers. How does water flow through? If water flows in/out by osmosis wouldn't it reach a region of hydrophobicity?

Yes, it would.  "Hydrophobic" doesn't mean that the tails actively repel water.  Water is moving through the cell membrane constantly.

small water molecules can flow through channels in memb.
Hydrophobic means water soluble mol. r repelled

I believe that water flows through the phospholipid bilayer by facilitated diffusion, using a protein called an aquaporin. If this protein was not present I do not  believe water would be able to cross the membrane without ATP consumption.

Aquaporins provide channels for rapid movement of water molecules across [b]all[/b] plasma membrane. They are found in all organisms.

[quote="mathemonster"]I was correct rite?[/quote]

Partially. Water can diffuse without ATP. They can simply pass the bilayer through the holes that the shifting phospholipids create.

you guys are great rofl

and the poster above me is the future of USABO as relates to AoPS users...hehe


geometry
trapezoid
inequalities
irrational number

[b]Discussion of some 1967 AHSME problems[/b]

[hide="9"]
1967 AHSME #9

Let $ K$, in square units, be the area of a trapezoid such that the shorter base, the altitude, and the longer base, in that order, are in arithmetic progression. Then:

A. $ K$ must be an integer
B. $ K$ must be a rational fraction
C. $ K$ must be an irrational number
D. $ K$ must be an integer or a rational fraction
E. taken alone neither A nor B nor C nor D is true
[/hide]

$ \frac{(a\minus{}p)a\plus{}a(a\plus{}p)}{2}\equal{}a^2$

So the answer is E since $ a^2$ is surjective over the non-negative reals.

Good solution.

1967 AHSME #16

Let the product $ (12)(15)(16)$, each factor written in base $ b$, equal 3416 in base $ b$. Let $ s\equal{}12\plus{}15\plus{}16$, each term expressed in base $ b$. Then $ s$, in base $ b$, is:

A. 43
B. 44
C. 45
D. 46
E. 47

[hide]if shorter side,altitude,and longer side are an A.P.
let them be $ a,b,c$respectively
then $ 2b\equal{}a\plus{}c$

$ K\equal{}\frac{1}{2}*(a\plus{}c)*(b)$  using Area of Trapezoid=$ \frac{1}{2}$(sum of parallel sides)*(distance between them)

so, $ K\equal{}\frac{1}{2}*(2b)*(b)$
so $ K\equal{}b^2$

so i suppose,if b is an integer,area must be an integer.
If $ a,b,c$are irrational,area may be irrational also(not sure about this part)
so choice E is right
[/hide]

@ skand: I don't see why $ 2b\equal{}a\plus{}c$.

I love this problem!

[hide="18"]
#18 on the 1967 AHSME:

If $ x^2 \minus{} 5x \plus{} 6 < 0$ and $ P \equal{} x^2 \plus{} 5x \plus{} 6$ then

A. $ P$ can take any value
B. $ 20 < P < 30$
C. $ 0 < P < 20$
D. $ P < 0$
E. $ P > 30$
[/hide]

[hide="solution"]

Let $ F(x) \equal{} x^2 \minus{} 5x \plus{} 6$. Then $ F(x) \equal{} (x \minus{} 3)(x \minus{} 2)$. Solution for the inequality $ x^2 \minus{} 5x \plus{} 6 < 0$ is thus $ 2 < x < 3$. $ P(x) \equal{} x^2 \plus{} 5x \plus{} 6$ increases as $ x$ increases from 2 to 3, so $ P(2) \equal{} 20$, $ P(3) \equal{} 30$, so the answer is B.

[/hide]

Well, $ s\equal{}3b\plus{}13$

So therefore, since the only numbers congruent to 1 mod 3 in the list are 43,46, then we have s=10 or 11.

Since s is not equal to 10, s is equal to 11. so we have the solution to be D.

[quote="rofler"]Well, $ s \equal{} 3b \plus{} 13$

So therefore, since the only numbers congruent to 1 mod 3 in the list are 43,46, then we have s=10 or 11.

Since s is not equal to 10, s is equal to 11. so we have the solution to be D.[/quote]

Why is $ s \equal{} 3b \plus{} 13$?

[quote="ThetaPi"]@ skand: I don't see why $ 2b \equal{} a \plus{} c$.[/quote]

[hide="It is because"]$ a,b,c$ are in A.P.
$ \Rightarrow b\minus{}a\equal{}c\minus{}b$
$ \Rightarrow2b\equal{}a\plus{}c$

Alternatively,
$ b$ is the A.M. between $ a$ and $ c$
$ \Rightarrow b\equal{}\frac{a\plus{}c}{2}$

$ \Rightarrow 2b\equal{}a\plus{}c$
[/hide]

[quote="skand"][quote="ThetaPi"]@ skand: I don't see why $ 2b \equal{} a \plus{} c$.[/quote]

[hide="It is because"]$ a,b,c$ are in A.P.
$ \Rightarrow b \minus{} a \equal{} c \minus{} b$
$ \Rightarrow2b \equal{} a \plus{} c$

Alternatively,
$ b$ is the A.M. between $ a$ and $ c$
$ \Rightarrow b \equal{} \frac {a \plus{} c}{2}$

$ \Rightarrow 2b \equal{} a \plus{} c$
[/hide][/quote]

Oh ya! I didn't do the working so I must have missed out on that bit


integration
trigonometry
logarithms
LaTeX
algebra
partial fractions

$ \int \frac {\cos^4 x}{\sin^3 x} dx$

Ans- $ \minus{} \frac {\cos^3 x}{2\sin^2 x} \minus{} \frac {3\cos x}{2} \plus{} \frac {3}{4}.\log \frac {1 \plus{} \cos x}{1 \minus{} \cos x} \plus{} c$

Put $ \cos x \equal{} t$  :)  you get $ \int \frac {t^4}{(1 \minus{} t^2)^2}$ Solve by partial fractions :)

EDIT: Sorry wrong latex code.

put what =t ???? :maybe:

did you mean put cos x=t??


A line is described by the equation $ y\minus{}4\equal{}4(x\minus{}8)$. What is the sum of its $ x$-intercept and $ y$-intercept?

To make things easier, let's write this in slope-intercept form.

$ y \minus{} 4 \equal{} 4x \minus{} 32$

$ y \equal{} 4x \minus{} 28$

We easily see that the $ y$-intercept is $ \minus{}28$. The $ x$-intercept is...

$ 0 \equal{} 4x \minus{} 28$

$ 28 \equal{} 4x$

$ 7 \equal{} x$

So our answer is $ \minus{}28 \plus{} 7 \equal{} \boxed{\minus{}21}$.

I believe this is point-slope form.

$ y \minus{} 4 \equal{} 4x \minus{} 32$

$ y \equal{} 4x \minus{} 28$

The intercepts are $ \minus{}28$ and $ 7$, so the answer is $ \minus{}21$.

Oh, btw, to find intercepts, solve for $ x$ and $ y$ when the other is $ 0$.

edit: beaten to it xD


How many math/science ap classes have you taken, or do you plan to take?
Economics counts. also, physics b and physics c would count as 2. same for comp. science and calculus.

...
Okay, I voted 4, but now that I look at your list, it should be 6.

Plans:
Calculus (didn't take the course, in school, but took the AP on wednesday...next year I take other stuff like independent study/math seminar [the "special" math course for interested kids], and then the year after that, more independent study)
Economics (to fulfill my history requirement...my school only allows me to drop it after 10th grade, but I can't drop it for more than one year :( )

Honors Physics (for the Physics B AP)
AP Physics (Physics C AP)
AP Chem
AP Bio

gah, economics and APES shouldn't count . . .

k

AP Chem
AP Physics (not sure B or C or both)
AP Comp Sci A
AP Comp Sci AB
AP Calc AB (not sure)
AP Calc BC

I'm probably taking economics at a local community college

So I'm putting 6.

My school only offers 4s, AP Stats, AP Calculus, AP Biology, and AP Chemistry.  I might take the AP COmputer Science AB test if I'm motivated enough, and I'll probably take the equivilent to AP Physics C at OSU.

Calc BC, Comp Sci AB, Chemistry, Biology, Physics B, Physics C (Mech + E&M), Statistics.

I'll have them all done by the end of this year except AP Bio. I hope I can get that Siemens AP Award thing...

[quote="randomdragoon"]Anyone taking/took statistics?  :?[/quote]OH HAH I forgot about Stat.  So I voted 6 but it should actually be 7.

Hmmm...I want to take Bio next year, and I have the room. My school won't let me do it without taking Bio GBE or Anatomy first though. I can't even take the exit exam as with Physics 1 since it's the lab experience that counts. Has anyone else been in this situation and managed to convince the administrators to let it slip?

I think the question is very vague

I was misguided.

Not everyone knows what aps stands for (education systems are even different in Belgium and the Netherlands...)

What is it?

Advanced Placement. The test is given by College Board and is used to give high school students a chance to earn college credit. Usually, and AP class is taken before the test, and AP classes are considered to be the hardest courses.

College.....board?   College...credit? :huh:

It's the same uh...organization that offers SAT's. http://www.collegeboard.com/splash

sophomore right now

junior year: CS AB and Stats
senior year: Calc BC, Biology, Chemistry,  Physics C


linear algebra
matrix
linear algebra unsolved

[b]Let[/b] $ A ,B \in M_{n} \mathbb{(R)}$ ,[b] I is identity nxn matrix .[/b]
[b]Such that :[/b] $ A^{2} + I = B^{2} + B + I = 0$
[b]Prove that :[/b]
$ Lim_{x \to 0} {\frac {det^{2}(A + xI) - det^{2}(B + xI) - nx}{x^{2}} = \frac {n(1 - n)}{2}}$
[size=75]"... Do math to love math , love math must do math , do math , love math , math is my life !!!..."[/size]
:roll:

[quote="harazi"]My dear friend, let's not try to make a mathematics forum the scene of a soap opera. One can express his marvelous and surely extremely passionate feelings in different settings, but I find it pretty stupid to do it here. So, think twice before posting such comments after a mathematical problem: many users (including me) will consider your post a spam and it will surely not have the desired attention.[/quote]
Dear studen of jean, why don't you stop sending useless posts? I do not want to repeat what harazi has told you, just keep in mind that we are not here to solve your problems. This is a mathematics forum and I hope you know what a forum is. Henceforth, try to accompany your problems with your thoughts on them rather than quotations that can simply be placed in your signature, pick suitable titles for your topics, and finally, do not spam!!

[b].... Ok ! thanks you I will don't spam and do'nt write many math and show see they write do'nt today .....[/b]
[b]A nice of math :[/b]
[b]Let [/b]$ A \in M_{n} \mathbb{(R)}$ , [b]I is dentity nxn matrix [/b],
[b]Such that : [/b]
$ A^{2} + I =0$
[b]Prove that :[/b]
 If $ det(A+\sqrt {2}I)+det(A+\sqrt {3}I)=det(A+\sqrt {4}I)$
[b]Then[/b] $ n = 4$
[b]"..Math is allway nice ..."[/b]
 :P

[b]Show[/b] : $ det(A \plus{} xI) \equal{} (x^{2} \plus{} 1)^{\frac {n}{2}}$ ( n is even (You can prove eassy !!)) :lol:
[b]" ...You can read Jean Marie Monier Professor Books to find nice problem of math ..."[/b]