okay, i don't know which forum this should be in.  It has to do with math, but not really.  Ever created strange names for even stranger conjectures? Well, we did at the mathy camp i'm in right now.  It's full of nerdy people.  Well, jk.  There's a couple.  Credit to Kitty and Jeremy as well.  oh, and me.  Well, if you find this boring, too bad.  Post some of your own if you're bored!  Note: we didn't make the conjectures.  We made the names.

Conjecture 1: The Distributive Property of Unification

A x (cross, not times) (B :cup: C) = (A x B)  :cup: (A x C)

Conjecture 2: The Distributive Property of Intersection

A x (B  \cap C) = (A x B)  \cap (A x C)

and last but not least:

Conjecture 3: The Property of Intercross Exchange

(A x B)  \cap (C x D) = (A   \cap C) x (B  \cap D)
or,
(A x B)   :cup:  (C x D) = (A    :cup:  C) x (B  :cup: D)

Sure the last \cap doesn't need to be a :cup: ?


Given $\triangle ABC$, if $\frac{\tan A}{\tan B}=\frac{BC}{AC}$, what kind of triangle is $\triangle ABC$?

[hide="Answer"]Lemma: $<A=<B$.

Proof: $\frac{\tan{A}}{\tan{B}}$
$=\frac{\frac{\sin{A}}{\cos{A}}}{\frac{\sin{B}}{\cos{B}}}$
$=\frac{\frac{BC}{2R\cos{A}}}{\frac{AC}{2R\cos{B}}}$
$=\frac{\frac{BC}{\cos{A}}}{\frac{AC}{\cos{B}}}$

By equating $\frac{\tan{A}}{\tan{B}}= \frac{BC}{AC}$, we have:
$\frac{\frac{BC}{\cos{A}}}{\frac{AC}{\cos{B}}}=\frac{BC}{AC}$
$\frac{\cos{B}}{\cos{A}}=1$
$\cos{A}=\cos{B}$

Now $<A,<B$ are less than $180^{\circ}$, so $<A=<B$. [b]QED.[/b]

Thus the triangle is iscoscles. (Sorry, I felt like using a lemma) [/hide]

\[ \frac{\tan{A}}{\tan{B}}=\frac{\sin{A}}{\sin{B}}\cdot \frac{\cos{A}}{\cos{B}} \]
Since $\frac{\sin{A}}{\sin{B}}=\frac{BC}{AC}$, so we have $\cos{A}=\cos{B}$, which gives $A=B$.


Let $s_{n}$ be a sequence of numbers which are sums of two squares of positive integers, arranged in increasing order.

It's not very difficult to show that exists positive constant $C>0$ such that $s_{n+1}-s_{n}\leqslant Cs_{n}^{\frac14}$ for all $n\in\mathbb{N}$.

Question: is there a subsequence $s_{n_{k}}$ such that $s_{n_{k}+1}-s_{n_{k}}\geqslant cs_{n_{k}}^{\frac14}$ for some positive constant $c>0$ and for all $k\in\mathbb{N}$?

Oh, I'm very sorry! I did stupid mistake in the title of topic: not "without", but "between"  :blush:

However, who thinks something about it?

[mod.: changed title]


Let a,b,c>0 and  $ a \plus{} b \plus{} c \le 3$
Prove that  2(ab+bc+ca)-abc $ \leq$5

Hint: Show that you can homogenize (careful its not trivial)

Let $ c$ be the smallest of $ a,b,c$. Let $ d\equal{}3\minus{}a\minus{}b\minus{}c\ge 0$ so $ a\plus{}b\plus{}c\plus{}d\equal{}3$. Now we homogenize,

\[ F[a, b, c, d] \equal{}
 \frac{5}{27} (a \plus{} b \plus{} c \plus{} d)^3 \plus{} a b c \minus{}
  \frac{2}{3} (a \plus{} b \plus{} c \plus{} d) (a b \plus{} b c \plus{} c a)\]

[code]In[2]:= Factor[F[a, b, c, d] - F[(a + b)/2, (a + b)/2, c, d]]

Out[2]= 1/12 (a - b)^2 (2 a + 2 b - c + 2 d)[/code]

Since $ c$ is the smallest, that difference is positive. So we can consider \[ F[x,x,y,d]\equal{}\frac{1}{27}\cdot ( d (5 d^2 \plus{} 30 d x \plus{} 42 x^2 \plus{} 15 d y \plus{} 24 x y \plus{} 15 y^2)\plus{}(x \minus{} y)^2 (4 x \plus{} 5 y))\]

So $ F[a,b,c,d]\ge 0$ with equality iff $ d\equal{}0,a\equal{}b\equal{}c\equal{}1$.


geometry
circumcircle
incenter
geometry unsolved

Let ABC be a triangle and its circumcenter O. The lines AB and AC cut the circumcircle again OBC in B1 different B, C1 different C, respectively, the lines BA and BC cut the circumcircle of OAC in the A2 different A and C2 different C, respectively, and the lines CA and CB cut the circumcircle of OAB in the A3 different A and B3 different B, respectively.
Prove that the lines A2A3, B1B3 and C1C2 go through the same point.

Let the tangents at A, B, C to the circle (ABC) determine the triangle $ A_1B_2C_3$, with $ A_1$ opposite to A, etc. Obviously, $ A_1$ belongs to (BOC), being diametrically opposite to O, etc. Since $ \angle B_2A_1C_3 \equal{} \pi \minus{} 2A$ and O is the incenter of $ \triangle A_1B_2C_3$, if we denote  $ {A_4} \equal{} A_3O \cap BC$, it's easy to find that $ A_4$ and $ C_2$ coincide, etc, hence all the subject lines pass through O.

Best regards,
sunken rock


find all $ f: \mathbb{R}\to\mathbb{R}$ such that:

$ f(x\plus{}f(x)f(y))\equal{}f(x)\plus{}xf(y)$

[quote="reason"]find all $ f: \mathbb{R}\to\mathbb{R}$ such that:

$ f(x \plus{} f(x)f(y)) \equal{} f(x) \plus{} xf(y)$[/quote]

$ x\equal{}y\equal{}\minus{}1$ $ \implies$ $ f(\minus{}1\plus{}f(\minus{}1)^2)\equal{}0$. Let then $ u\equal{}f(\minus{}1)^2\minus{}1$. We got $ f(u)\equal{}0$
$ x\equal{}u$ $ \implies$ $ 0\equal{}uf(y)$ and so either $ f(x)\equal{}0$ $ \forall x$ (first solution), either $ u\equal{}0$ and so $ f(\minus{}1)^2\equal{}1$

If $ f(x)$ is a solution, then $ \minus{}f(x)$ is too a solution. So WLOG say $ f(\minus{}1)\equal{}\minus{}1$

So $ f(0)\equal{}0$ and $ f(\minus{}1)\equal{}\minus{}1$. Let then $ a\equal{}f(1)$

$ x\equal{}1$ and $ y\equal{}\minus{}1$ $ \implies$ $ f(1\minus{}a)\equal{}a\minus{}1$
$ x\equal{}\minus{}1$ and $ y\equal{}1\minus{}a$ $ \implies$ $ f(\minus{}a)\equal{}\minus{}a$
$ x\equal{}1$ and $ y\equal{}\minus{}a$ $ \implies$ $ f(1\minus{}a^2)\equal{}0$ and so $ a^2\equal{}1$ but $ a\equal{}f(1)\equal{}\minus{}1$ is impossible since then $ f(\minus{}a)\equal{}\minus{}a$ would imply $ f(1)\equal{}1$
So $ a\equal{}f(1)\equal{}1$

$ x\equal{}\minus{}1$ and $ y\equal{}v$ $ \implies$ $ f(\minus{}f(v)\minus{}1)\equal{}\minus{}f(v)\minus{}1$
$ x\equal{}1$ and $ y\equal{}\minus{}f(v)\minus{}1$ $ \implies$ $ f(\minus{}f(v))\equal{}\minus{}f(v)$
$ x\equal{}\minus{}1$ and $ y\equal{}\minus{}f(v)$ $ \implies$ $ f(f(v)\minus{}1)\equal{}f(v)\minus{}1$

Let then $ v\equal{}x\minus{}f(x)$ and $ w\equal{}f(v)\minus{}1$ : we get $ f(w)\equal{}f(x\minus{}f(x))\minus{}1$ and so $ \minus{}1\minus{}f(w)\equal{}\minus{}f(x\minus{}f(x))$

But : $ y\equal{}\minus{}1$ $ \implies$ $ f(x\minus{}f(x))\equal{}f(x)\minus{}x$ and so $ \minus{}1\minus{}f(w)\equal{}x\minus{}f(x)$
But : $ x\equal{}\minus{}1$ and $ y\equal{}w$ $ \implies$ $ f(\minus{}1\minus{}f(w))\equal{}\minus{}1\minus{}f(w)$ and so $ f(x\minus{}f(x))\equal{}x\minus{}f(x)$

So $ x\minus{}f(x)\equal{}f(x)\minus{}x$ and so $ f(x)\equal{}x$, which indeed is a solution.

Hence the three solutions :
$ f(x)\equal{}0$
$ f(x)\equal{}x$
$ f(x)\equal{}\minus{}x$


[quote="mathgeniuse^ln(x)"][quote="bpms"][quote="janieluvsmusic"]How many positive three-digit even multiples of 3 are comprised of three distinct digits?[/quote]
[hide="maybe, but probably not"]$9\cdot 9 \cdot 8=648$ one sixth of this is $108$[/hide][/quote]

I doubt that is right?

I think the best way of doing this is to notice that the digit sum is 3, 6, 9, ...21, and then use some delicate casework, to figure it out.

This seems tedious.[/quote]

I did casework and got 113.

This is how I did it:
[hide="casework"]
First you know that the last digit must be 0, 2, 4, 6, or 8; this gives you the 5 cases to consider. Also, for it to be divisible by three, the sum of the digits must be 3. The format of the list given below is: (sum of the two left most digits given the right most digit) to (number of possible arrangements of the two left most digits given that all digits are distinct). i.e. for the 0th case, it says "3 to 2", it means that the two left most digits must sum to 3 and there are 2 ways to do this, 12 and 21 for it to be distinct (the 20 case does not work because the right most digit is already 0).

last digit is 0:
3 to 2, 6 to 4, 9 to 8, 12 to 6, 15 to 4.
Total cases: 24

last digit is 2:
1 to 1, 4 to 2, 7 to 4, 10 to 6, 13 to 6, 16 to 2.
Total cases: 21

last digit is 4:
2 to 1, 5 to 3, 8 to 7, 11 to 6, 14 t o 4, 17 to 2.
Total cases: 23

last digit is 6:
3 to 3, 6 to 4, 9 to 7, 12 to 6, 15 to 2.
Total cases: 22

last digit is 8:
1 to 1, 4 to 3, 7 to 7, 10 to 6, 13 to 4, 16 to 2.
Total cases: 23

Answer: 24+21+23+22+23=113
[/hide]

[quote="janieluvsmusic"]The answer is close to 113, but not that. The problem's supposedly doable quickly[/quote]

I don't think so; it seems to me at best that we fix two of the digits, and then it's casework $\bmod 3$ for the third.

[quote="janieluvsmusic"]The answer is close to 113, but not that. The problem's supposedly doable quickly, so there's probably another way besides casework.[/quote]

my answer
[hide]124 ---- I did direct calculation.[/hide]

Well, it was a problem at a math competition, and the answer is 115.

Pretty much Case Work..
[hide]
First of all we notice that all these numbers are divisible by 6 and all the three digit numbers that are divisible by 6 are divisible by three (obviously)..so there are [b]150[/b] such numbers

So lets find all the numbers that do not have distinct digits..

[b]Case 1: (Ends with 0)[/b]
300, 330, 600, 660, 900, 990....$\implies 6$

[b]Case 2: (Ends with 2)[/b]
222, 252, 282, 522, 552, 822, 882,....$\implies 7$

[b]Case 3: (Ends with 4)[/b]
114, 144, 414, 444, 474, 744, 774,....$S\implies 7$

[b]Case 4: (Ends with 6)[/b]
336, 366, 606, 636, 666, 696, 966, 996,....$\implies 8$

[b]Case 5: (Ends with 8)[/b]
228, 288, 558, 588, 828, 858, 888,....$\implies 7$

So summing up the mini-totals we get $35$ and so all the distinct values $\implies 150-35=\boxed{115}$
[/hide]

I got 115 but i had to do some careful case work like i set the last digit to either (0,2,4,6,8) and then I set another digit equal to the last digit or to another digit being placed twice (e.g. 552) that would result it in being divisible by three...so thats how i found all the values ... Is there a better way to do it other then case work..

[hide]
There are $\lfloor 1000/6\rfloor-\lfloor 100/6\rfloor =150$ 3-digit multiples of 6 in total, now just subtract the ones with repeated digits. If $b$ appears at least twice and $a$ is the other digit, then the possible 3-digit numbers are $100a+11b$, $10a+101b$, and $a+110b$.

[b]Case 1:[/b] $a$ is ones digit. Then $a\equiv 4b\pmod{6}$ so $a$ is even. Let $a=2m$, so $m+b$ is divisible by 3. But $m+b\leq 4+9=13$, so the possible sums are 3, 6, 9, 12. The number of solutions in each case is 3, 5, 5, 2, respectively. (For the first one, (0,3), (1,2) and (2,1) are options for $(m,b)$; for the second and third, each value of $m$ produces a solution; for the fourth, only $(3,9)$ and $(4,8)$ work.) This gives a total of 15.

[b]Case 2:[/b] $a$ is tens digit. Now $4a\equiv b\pmod{6}$; let $b=2m$. The only difference between the ordered pairs $(m,b)$ above and the ordered pairs $(m,a)$ here is that above, $m$ could be 0, now $a$ could be 0. Anyway, we get 3+4+4+2=13 more.

[b]Case 3:[/b] $a$ is hundreds digit. Again $4a\equiv b\pmod{6}$. This is just like Case 1, with 15 more solutions.

Of course, we twice overcounted the 4 numbers 222, 444, 666, 888. Our final answer is $150-15-13-15+2\cdot 4=\boxed{115}$. whew.
[/hide]

Yeah, I must have messed up some part of my casework, I got 113, but 115 is the answer then.

[quote="JavaMan"]That's wrong, not sure where you made a mistake though.

EDIT: wow, new posts, this one was directed at xxxyyy :P

[hide]
$\lfloor\frac{1000}{6}\rfloor-\lfloor\frac{100}{6}\rfloor= 150$ 3 digit numbers are divisible by 6.

Let's count non-distinct 3 digit numbers divisible by 6 then subtract them from 150.

Forms:
1. aaa - 4
2. aab - 11
3. aba - 9
4. baa - 11
Total: 35
Here's the casework
1. 222, 444, 666, 888
2.
b = 0 - a = 3, 6, 9
b = 2 - a = 5, 8
b = 4 - a = 1, 7
b = 6 - a = 3, 9
b = 8 - a = 2, 5

3.
a = 2 - b = 5, 8
a = 4 - b = 1, 7
a = 6 - b = 0, 3, 9
a = 8 - b = 2, 5

4.
a = 0 - b = 3, 6, 9
for cases a = 2, 4, 6, 8 it's the same as (3) except with a = 6, b can't be 0.

$\boxed{115}$
[/hide][/quote]

I made the mistake... found the mistake in my calculation. Thanks for pointing it out.


Do you read manga/watch anime?

If yes, which ones do you follow?

I started Prince of Tennis but gave up on it b/c it was a bit repetitive.
I started Katekyo Hitman Reborn but gave up because I was waiting for the next one to come out and I forgot half the storyline (even though I can just read it on one-manga)
I started Full Moon and borrowing it from a friend (even though I can just read it on one-manga)

how about you?

See:
http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=204284

For mine.

YES FINALLY MANGA!!! awesome mangas:
Shaman King
InuYasha
Rurouni Kenshin

GO ASAKURA HAO!!!!!

Well i'm not azn

but i saw death note once. it was pretty cool...

hmm....wait...mangas like a comic book right?

and anime's like....a cartoon?

Err the only manga I follow is the webcomic [url=http://www.megatokyo.com]MegaTokyo[/url], but since the author rare updates to schedule I only check every year or so to get a coherent story.

YAY i've finished all the manga that they have out for Shaman King now! (approx. 290 chapters) next, i dunno what i'll read... any recommendations?

I don't even know why i'm here, but... for me, the number... I don't remember. Too many to count. XD


[quote="white_horse_king88"]Each of 3 circles are internally tangent to an equilateral triangle and the other two circles. If each circle has radius of three, then what is the radius of the triangle's incircle?

A circle of radius r is inscribed in a right isosceles triangle, and a circle of radius R is circumsribed about the triangle. What is R/r?[/quote]

A picture would help...

[hide="solutions"]
1.

$\displaystyle{2(\frac{1}{3^2} + \frac{1}{3^2} + \frac{1}{3^2} + \frac{1}{x^2} = (\frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{x})^2}$

$\displaystyle{\frac{2}{3} + \frac{2}{x^2} = (1 + \frac{1}{x})^2 = 1 + \frac{2}{x} + \frac{1}{x^2}}$

$x^2 + 6x - 3 = 0$

$x = \frac{-6 \pm \sqrt{36 + 12}}{2} = \pm 2\sqrt{3} - 3$

The negative solution gives the radius of the outer circle that is tangent to the three inner circles, which is the incircle of the equilateral triangle.  So:

$\displaystyle{x = -(2\sqrt{3} + 3) \rightarrow R = 2\sqrt{3} + 3}$

2.

Denote the legs of the right isosceles triangle as length $a$.

$\displaystyle{\frac{a^2}{2} = rs = r(\frac{2a + \sqrt{2}a}{2}) \rightarrow r = \frac{a}{2 + \sqrt{2}}}$

$R = \frac{\sqrt{2}a}{2}$

$\frac{R}{r} = \sqrt{2} + 1$
[/hide]

On #2, there is a much easier way.
 Draw the line from the midpoint of the hypotenuse to the opposite vertice and get the equation: $r+r\sqrt2=R$, divide by $r$ to get $\frac{R}r=\sqrt2+1$

Number 2:

Call the length of each of the legs of the $45,45,90$ triangle $2$, therefore, the radius of the bigger circle would be $\sqrt{2}$ because the hypotenuse of an inscribed right triangle is the diameter of the circle. Then, you need to find the inradius of the circle inscribed in the triangle, so use $rs=A \longrightarrow r(2+\sqrt{2})=2 \longrightarrow r=\frac{2}{2+\sqrt{2}}$, so $\frac{R}{r}=\boxed{\sqrt{2}+1}$.

Number 1:

Construct an equilateral triangle within the larger equilateral triangle with the radii of the three circles, thus you can say that each side of the equilateral triangle has length of $6+2x$, now, you can find the area of this triangle by two ways, first by using two  formulas, so by setting the two ways equal, we can find $x$:

$\frac{(6+2x)^2\sqrt{3}}{4}=\sqrt{(9+3x)(3+x)(3+x)(3+x)}$, but that's too complicated
 :?

[quote="236factorial"]Iverson, what's with the editing above my post? I thought it was number one first. It was edited 4 times..   :?[/quote]

No, it was number $2$, I accidentally put number $1$, so I edited it, and I made some mistakes with the latex so I edited it.

[quote="236factorial"][hide="number 2 also"]Let the sides be 1, 1, and $\sqrt2$. The inscribed circle has radius $2-\sqrt2$. Also the hypotenuse is the diameter of the circle inscribing it, so its radius is $\frac{1}{2}\sqrt2$

R/r=$\frac{\frac{1}{2}\sqrt2}{2-\sqrt2} = \frac{\sqrt2}{4-2\sqrt2}$.

Multiplying the numertor and denominator by $4+2\sqrt2$ to get $\frac{4\sqrt2+4}{4} = \sqrt2+1$[/hide][/quote]
you really don't need to find the value of each. Just find a place where the radii coincide, and make a simple equation, like what I did before

Very fast solution of nat mc. I give one for problem 1

Connect 3 centers of small incircles and get an equilateral triangle with side 6. Radius of large incircle $R$ is sum of 2/3 height of equilateral triangle (same center) and a radius of small incircle: $R=\frac23(3\sqrt3)+3=2\sqrt3+3$


What is the minimum possible product of three different numbers of
the set $ \{\minus{}8, \minus{}6, \minus{}4, 0, 3, 5, 7\}$?

$ \minus{} 8 \times 7 \times 5 \equal{} \boxed{ \minus{} 280}$.

these problems you need to remember that  -*+*+ is negative

so you want to use as many big numbers as you can so we get   -8*7*5=$\boxed{-280}$


It's certainly been a while.

[u][b]Formal Instructions:[/u][/b]
The aim of the game is to be the last player remaining.  Every round, there is a time limit, usually about 2 or 3 days.  Before the time limit ends, all players must submit via PM a number between 0 and 100 to me.  There is a specific submission format written below and you must abide by this format or else your submission [i]will not be counted[/i].  At the end of the time limit, I tally up the submissions and scores in this fashion:
- I post all the submissions (a fail to submit counts as 100).
- I post what half the average submission was.
- I post all players who submitted under half the average.  These players are eliminated.
- Everybody gains points equal to their distance from half the average.  [i][u]Note:[/u] you do not want points![/i]
- If anybody has over 100 points total, they are eliminated.
When there is one player remaining, they win.

[u][b]Submission format:[/u][/b]
Submission Title: "[u]Number[/u], [u]Username[/u] (Round [u]#[/u])"
Submission Body: Any comments.  If you would like me to read the body, put a * after the Submission Title.

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I will take 15-20 sign-ups, but please only sign up if you think you will be able to make most of the 2-3 day time limits.

I'm moving this to Games.  Moderator, please lock/delete this thread.


Let $ f: \mathbb{R}\to\mathbb{R}$ such that there exist $ f'(x)\forall x\in\mathbb{R}$ and $ |f(x)|\leq p,|f'(x)|\leq q\forall x\in\mathbb{R}$ for some $ p>0,0<q<1$. Prove that $ (x_{n})$ is conveger, here $ x_{0}=0,x_{n+1}=f(x_{n})(n=1,2,...).$

The assumption $ |f(x)|\le p$ is not needed. Just use the Mean Value theorem to get $ |f(x)-f(y)|\le q|x-y|$, and the rest follows.

[quote="N.T.TUAN"]Not need that hypothesis if $ f: [a,b]\to [a,b]$.[/quote]
I don't understand why we need that hypothesis if $ f: R\to R$ or we have diffrent solution?


Let be $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}1$ . Show that :

$ \frac{a^5\plus{}bc}{a}\plus{}\frac{b^5\plus{}ca}{b}\plus{}\frac{c^5\plus{}ab}{c}\ge abc$

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 1$ . Show that :

$ \frac {a^5 \plus{} bc}{a} \plus{} \frac {b^5 \plus{} ca}{b} \plus{} \frac {c^5 \plus{} ab}{c}\ge abc$[/quote]
$ \sum (a^2)^2 \plus{} \sum\frac {bc}{a} \geq abc$

But $ \sum (a^2)^2\geq \sum a^2b^2 \geq abc(a \plus{} b \plus{} c) \equal{} abc$

$ Q.E.D$

\[ \begin{gathered}
  \frac{{a^5  \plus{} bc}}
{a} \plus{} \frac{{b^5  \plus{} ca}}
{b} \plus{} \frac{{c^5  \plus{} ab}}
{c} \geqslant abc \hfill \\
   \Leftrightarrow a^4  \plus{} b^4  \plus{} c^4  \plus{} \frac{{bc}}
{a} \plus{} \frac{{ca}}
{b} \plus{} \frac{{ab}}
{c} \geqslant abc \hfill \\
  Now, \hfill \\
  \left( {\frac{{a^4  \plus{} b^4  \plus{} c^4 }}
{3}} \right)^{\frac{1}
{4}}  \geqslant \left( {\frac{{a \plus{} b \plus{} c}}
{3}} \right) \equal{} \frac{1}
{3} \hfill \\
   \Leftrightarrow a^4  \plus{} b^4  \plus{} c^4  \geqslant \frac{1}
{{3^3 }} \equal{} \left( {\frac{{a \plus{} b \plus{} c}}
{3}} \right)^3  \geqslant abc \hfill \\
   \Leftrightarrow a^4  \plus{} b^4  \plus{} c^4  \plus{} \frac{{bc}}
{a} \plus{} \frac{{ca}}
{b} \plus{} \frac{{ab}}
{c} > abc\left\{ {\because a,b,c > 0} \right\} \hfill \\
\end{gathered}
\]


The base of a triangle is twice as long as a side of a square and the areas of the triangle and the square are the same. What is the ratio of the altitude of the triangle to a side of the square?

(A) 1:4 (B) 1:2 (C) 1:1 (D) 2:1 (E) 4:1

If the square's length is $ x$, then its area is $ x^2$.

Then the base of the triangle has length $ 2x$. Its area is also $ x^2$.

Altitude of triangle is $ \frac{2x^2}{2x}\equal{}x$.

Thus the ratio of the altitude of the triangle to a side of the square is $ \boxed{1: 1}$, or answer choice $ \boxed{\text{C}}$.


For instance I am about to explain how to find the flux density around a current carrying wire. I need to include some artwork to graphically explain the vectors and what not. Is there a simple way to do this?

I was thinking about using adobe photo shop to generate an image and then insert that image into my posts using HTML code, but is there a simpler way?

29 views and not a single reply oh well

I'll try to hand draw it then scan that and run it through adobe PS then get my JPG file Oh I wish there was a computere tech geek on this site.

Use Printkey 2000.

Edit. Also, you can use asymptote or Geometer Sketchpad or Corel Draw if you have any of those.

Asymptote is free.

And then you can generate an image using asymptote and then using the

[code][img]...[/img][/code]

tags.

You can't use HTML tags here. You'd have to go [code][img]http://imgurl[/img][/code]


:maybe:  :maybe:  :maybe:
Anybody knows?

[quote="cdymdcool"]:maybe:  :maybe:  :maybe:
Anybody knows?[/quote]

probably trig, almost all year long its what you focus on in precalc

TRIG most definitely unless you count intro to calculus. and it always helps to be able to know what's going on in (i.e. find equations of) functions and graphs


Hello Everyone,

I would like to invite you all to "CodeCraft 08", a programming contest conducted by IIIT, Hyderabad as part of their technical fest "Threads". The contest is scheduled on 27th January. The contest is open to everyone. For more details and to register for the event, please visit [url]http://felicity.iiit.ac.in/codecraft/[/url].

The problemset is more challenging and interesting as compared to CodeCraft 07. I hope you all will participate and enjoy the contest.

Wow, IIIT sure organizes a lot of contests. Too bad I won't be able to do this one either :(. It's mostly holidays here right now (at least for people who finished their exams already).


$ \Delta U (lattice) \equal{} \minus{} 107000 \frac {v|z_ \plus{} \parallel{}z_ \minus{} |}{r_ \plus{} \plus{} r_ \minus{} }$

Does anyone how this is derived? Any applications? Margin of errors compared to experimental data?

Huh.

What are the units in that equation?  What do those symbols mean?

Thanks.

i dont know abt much kapustinski eq, but i know abt born-lande eq, and its derivation. the margin of error for the born-lande eq  is 3%, and i think that shud be the margin of error for kapustinski, unless allowances are made for the van der waals forces and zero point energy.

this may not be useful but i think those eqns are in jd lee if i'm not wrong  :maybe:

yep, born-lande is there, but there is nothing abt kapustinski


find all n,p such that:

                                        n^8_p^3=p^5+n^2

i don't understand it can u re-write it with latex..i mean n^8_p^3 what does it means? n^8 - p^3? or n^8(p^3)..please let me know.

:-) really nice problem and i believe that it was posted before :)

anyway here is what i do remember :

$n^8-p^3=p^5+n^2$  is equivalent to $n^2(n^6-1)=p^2(p^3+1)$

Obviously for $p=2$ we have no solutions, thus p is odd.

1. Suppose that $n>p$, then we get a trivial contradiction which is very easy to see,
$n^2(n^6-1)=n^2(n^3-1)(n^3+1)>p^2(p^3+1)$ so no solutions when in this case.

2. Suppose that $p>n$, we have $p^2|n^2(n-1)(n+1)(n^2n+1)(n^2-n+1)$, since $p>n$
we see that $p^2|(n^2+n+1)$ this implies that $p^2<n^2+n$(*).

Also we have $n^2(n^2-1)|(p^3+1)$ this implies that $p^3+1>n^4-n^2$ ... $p^3>n^4-n^2$(**)

Combining (*) and (**) we get $(n^4-n^2)^2<(n^2+n)^3$ ... $n^4(n-1)^2(n+1)^2<n^3(n+1)3$
so it approachs to $n(n-1)^2<n+1$ ... $n^3-2n^2+n<n+1$ ... $n^3<2n^2+1$
resulting, we have only one possible value for n which is 2 because for $n\geq3$ we get the several contradictions
anyway lets prove this, too :

$n^3<2n^2+1$ holds if and only if $n=2$.
[u]Proof:[/u] Lets take a number n which is bigger than 2.
Thus, $2n^2+1<n^3+1$ but we have $2n^2+1>n^3$.

Finally we get the answer : $(n,p)=(2,3)$

davron

[quote="Davron"]
$n^8-p^3=p^5+n^2$  is equivalent to $n^2(n^6-1)=p^2(p^3+1)$
[/quote]

That last equality is wrong.  It should be $p^3(p^2+1)$.

i said that i had seen this problem :-)
in that problem p was a prime and i am not saying that there everything is correct i just wanted to share with knowledge thats all :-)

davron


Can some one post their solution to problem 9 in admission test C. Thnx :wink:  :D

[quote="t0rajir0u"]It would certainly help if we knew what the problem was.[/quote]
Here's a link:
http://awesomemath.org/admissiontestc.pdf

[quote="t0rajir0u"][quote="Awesomemath Admission Test C"] 9.  Let $x_{n}= \sqrt{n+\sqrt{n^{2}-1}}$ for all $n \ge 1$.  Express $\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{49}}$ as $a+b \sqrt{2}$ for some integers $a, b$.  [/quote]

[hide="Hint"] Can you find a nice way to rewrite $\frac{1}{x_{n}}$? [/hide][/quote]
[hide]
well, $\frac{1}{x_{n}}=\frac{1}{\sqrt{n+\sqrt{n^{2}-1}}}$.
now rationalize it and then multiply $\frac{n-\sqrt{n^{2}-1}}{n-\sqrt{n^{2}-1}}$..[/hide]

[quote="mysmartmouth"]Is the test over?  It says May on it, so it seems like I can help, but I just want to make sure...[/quote]
Yes, the admission test ended almost a month ago. The program starts in about a week, in fact.  :)


Let $O,O'$ be two circles intersect each other in $X,Y$ .
Let the common tangent of two circles intersect $O$ in $A$ and $O'$ in $B$ . Draw the tangent of $O$ in $X$ intersect $O'$ in $C$ . Let $AX$ intersect $BC$ in $Z$ . Prove that $BZ,BX$ are tangents of $B$ for circumcircle of $\triangle XYZ$ .

We have $\angle ZXC=\angle BAX$ and $\angle ZCX=\angle ABX$, so $XAB, ZXC$ are similar. From here we also easily get $\angle BZX=\angle BXZ=\angle AYB\ (*)$, so $AYZB$ is cyclic. This means that $\angle AYX=\angle BAX=\angle BYZ$, so $\angle AYB=\angle XYZ$. From this and $(*)$ we get the result.


real analysis
probability
advanced fields
advanced fields unsolved

Does anyone know a good reference to see how Solovay shows that if we "deny" the Axiom of Choice, then there does not exist Lebesgue non-measurable sets? I'm guessing that this is his result. I can't seem to find it anywhere, only things that allude to it.

No, that's in general wrong. Your axiomes may not include the AC but it could hold nevertheless in a model.

Solovay showed the following: If there is a inaccessible cardinal, there is a model of ZF where every subset of the reals is lebesgue-measurable.

Is there a book or article with this in it? Maybe JSTOR or something. I'm very interested in seeing the full thing.

Also, does this mean that even to provide a non-constructive proof of the existence of non-measurable sets one needs the AC? Like, would it be possible to show that the probability of randomly choosing a set that is non-measurable is strictly greater than 0?

Well, I don't remeber the books name at the moment.

But the answer to your question is: No, you can't show this only with ZF, you need some form of the AC, maybe not the full AC, but something like DC or so is needed.


[quote="Little Gauss"]See this.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=137885[/url][/quote]
I read, but that isn't this problem.


linear algebra
matrix
linear algebra unsolved

Let T be a linear operator on a finite dimensional inner product space V.
Suppose that B is an ordered basis for V such that [T]_B is an upper triangular matrix. Let Y be the orthonormal basis for V obtained by applying the Gram-Schmidt orthogonalization process to B and then normalizing the resulting vectors. Prove that [T]_Y is an upper triangular matrix.

Well, I know that when you multiply two upper triangular matrices... you get an upper triangular matrix. So I have to show that the Gram-Schmidt process is like a Q (Change of basis matrix) such that it is upper triangular so that when it multiplies with [T]_B, it will get [T]_Y which is also upper triangular.

I am not sure how to show this. Any help is appreciated. Thanks!

If you study the proof of the Gram-Schmidt process, you will notice (if the basis before is $ \{v_1,\ldots,v_k\}$ and the resulting basis is $ \{e_1,\ldots,e_k\}$) that $ \text{span}(v_1,\ldots,v_i)\equal{}\text{span}(e_1,\ldots,e_i)$ for all $ i$.

Now, an operator $ T$ being upper-triangular with respect to a basis $ v_1,\ldots,v_k$ is equivalent to $ T(v_i)\in \text{span}(v_1,\ldots,v_i)$ (this is important, and should be in any linear algebra book).

Do you see how to proceed form here?

I'll expand. I'm assuming the facts I mentioned in my previous post.

Let $ v_1,\ldots,v_k$ be the original basis and let $ e_1,\ldots,e_k$ be the basis obtained from Gram-Schmidt.

Now, since $ T$ is upper triangular with respect to $ v_1,\ldots,v_k$, we have that $ T(v_i)\in\text{span}(v_1,\ldots,v_i)$. Since $ e_i\in \text{span}(e_1,\ldots,e_i)\equal{}\text{span}(v_1,\ldots,v_i)$‚ we have that $ T(e_i)\in\text{span}(v_1,\ldots,v_i)\equal{}\text{span}(e_1,\ldots,e_i)$. Hence the criteria for $ T$ being upper triangular for $ e_1,\ldots,e_k$ is fulfilled.


Hi Everyone..

Here are two more challenges in chess. Nice one .  :)

[url]http://oraclearticles.googlepages.com/chessprob.htm[/url]

Position: White to move and mate in two moves...

Thanks!

In case you haven't seen it, there is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193263]Chess Problems Marathon[/url]. Of course, you are still welcome to post problems as separate topics. :)

Also, can you please post images of the board (along with the source link)? It's very annoying for me to go to that site and as I'm concentrating on the problem the ad pops up  :mad: .


$ f: \mathbb{N}^\plus{} \longrightarrow \mathbb{N}^\plus{}$ is a function such that:
$ f(n)\equal{}n\minus{}f(f(n\minus{}1))$ for all $ n > 1$
$ f(1)$ is odd

Find the value for f(4198)

[quote="Gabuntu94"]Hi Arrange your tan!

Thank you very much for the answer (I don't know how to do this problem, so I'm looking for good ideas from who knows more than me about maths)...

Just one thing... I really can't understand this:  :blush:
4198 - f(f(4197)) ==> 4198 - [4197 - f(f(4196))] =[/quote]

Note:  It was stated that this doesn't make sense.

Recall in your problem that f(n) = n - f(f(n))

We started out with n = 4198.
So n - 1 is 4197.

It's f(4198) = 4198 - f(f(4197)

Now we're working with f(f(4197)).

Then n = 4197 and n - 1 = 4196.

So f(4197) = 4197 - f(f(4196)

Bring down the "4198 -" in front of it.

And so on.

Ok... but so why isn't it

4198-f(4197-f(f(4196)))

instead of

4198 - 4197 + f(f(4196))

??

I hope this is not a dumb question  :blush:

[quote="Gabuntu94"]$ f: \mathbb{N}^ \plus{} \longrightarrow \mathbb{N}^ \plus{}$ is a function such that:
$ f(n) \equal{} n \minus{} f(f(n \minus{} 1))$ for all $ n > 1$
$ f(1)$ is odd

Find the value for f(4198)[/quote]

Note :  It was stated that this doesn't make sense.

Notice in the formula that

$ f(n)$ does not equal $ f(n \minus{} f(f(n \minus{} 1))$;

it equals the integer $ n$ minus $ f(f($the next lower integer$ ))$,

so it equals $ 4198 \minus{} f(f(4197))$ when $ n \equal{} 4198$.

That's all I will post on that.

[quote="Gabuntu94"]I hope this is not a dumb question  :blush:[/quote]  No, you're absolutely right and A.y.t.'s solution doesn't make any sense.  (From $ f(n) \equal{} n \minus{} f(f(n \minus{} 1))$ it follows that $ f(f(4187)) \equal{} f(4187) \minus{} f(f(4187) \minus{} 1)$, which is totally useless.)

Here's a better beginning: that condition "$ f(1)$ is odd" is really weird and annoying, so let's try to get rid of it.  Let $ f(1) \equal{} a$.  Then $ f(2) \equal{} 2 \minus{} f(a) > 0$, so $ f(a) \equal{} 1$ and thus $ f(2) \equal{} 1$.  Now $ f(3) \equal{} 3 \minus{} f(f(2)) \equal{} 3 \minus{} f(1) \equal{} 3 \minus{} a > 0$, so $ f(1) \equal{} 1$ (it's given to be odd) and thus $ f(3) \equal{} 2$.  Now you can actually compute some values of the function in order to form a conjecture.  (There's still plenty of work to be done.)

If you just want the answer, it's [url=http://www.research.att.com/~njas/sequences/A005206]A005206[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

[quote="JBL"][quote="Gabuntu94"]I hope this is not a dumb question  :blush:[/quote]  No, you're absolutely right and A.y.t.'s solution doesn't make any sense.  (From $ f(n) \equal{} n \minus{} f(f(n \minus{} 1))$ it follows that $ f(f(4187)) \equal{} f(4187) \minus{} f(f(4187) \minus{} 1)$, which is totally useless.)

[/quote]

Actually, I think it's worse, as in another "f" applied at the end:

If $ n \equal{} f(4187)$ (as opposed to the given 4196), then

from $ f(n) \equal{} n \minus{} f(f(n \minus{} 1))$ it follows that $ f(f(4187)) \equal{} f(4187) \minus{} f(f(f(4187) \minus{} 1))$.

I must have been thinking of a different formula that was not given in the
first post, such as:

$ f(n) \equal{} n \minus{} f(n \minus{} 1)$ with, say, $ f(0) \equal{} 0$.

For instance,

$ f(3) \equal{}$

$ 3 \minus{} f(2) \equal{}$

$ 3 \minus{} [2 \minus{} f(1)] \equal{}$

$ 3 \minus{} 2 \plus{} f(1) \equal{}$

$ 3 \minus{} 2 \plus{} 1 \minus{} f(0) \equal{}$

$ 3 \minus{} 2 \plus{} 1 \minus{} 0 \equal{}$

$ 2$

$ f(3) \equal{} 2$

And this sequence (not the sequence the OP gave in the formula) starts
out with these values for $ f(n)$, where $ n$ begins with $ 0$:
$ 0, 1, 1, 2, 2, 3, 3, ...$

The fourth entry, $ f(3) \equal{} 2$, as it does in my calculation.


Prove that for any positive real numbers $ a_1,...,a_n$ we have the following inequality:
   $(n-2)\cdot\sum_{i=1,n}{\frac{1}{a_i}}+\frac{n^2}{a_1+...+a_n}\geq\sum_{1\leq i<j\leq n}{\frac{2}{a_i+a_j}}$

I think that you mean:
[tex] (n-2) \cdot \sum{\frac{1}{a_i}}+\frac{n^2}{\sum{a_i}} \geq \sum \frac{4}{a_i+a_j} [/tex]
If so, this is my solution for it:
[tex] \sum\frac{1}{a_i}-\frac{n^2}{\sum{a_i}}=\sum\frac{(a_i-a_j)^2}{{a_{i}} \cdot {a_{j}} \cdot\sum{a_i}} [/tex]
[tex] (n-1)\cdot\sum\frac{1}{a_i}-\sum\frac{4}{a_i+a_j}=\sum \frac{1}{a_i}+\frac{1}{a_j}-\frac{4}{a_i+a_j}=\frac{(a_i-a_j)^2}{{a_{i}} \cdot {a_{j}}\cdot(a_i+a_j)} [/tex]
Hence:
[tex] (n-2) \cdot \sum{\frac{1}{a_i}}+\frac{n^2}{\sum{a_i}}-\sum \frac{4}{a_i+a_j} [/tex]
[tex] =\sum(\frac{(a_i-a_j)^2}{{a_i} \cdot {a_j} \cdot(a_i+a_j)}
-\frac{(a_i-a_j)^2}{{a_{i}} \cdot {a_{j}} \cdot \sum{a_i}}) \geq 0 [/tex]. :)


You can find free latex books at http://lib.homelinux.org/_djvu_index_tables.html
username and password is [i]deleted by moderator[/i].
I hope you will find the books helpful

There is such disclamer (in Russian) there:

[quote]I ask you not to publish login and password anywhere, otherwise I will be forced to change them more often.[/quote]
 ;)

[quote="birkof"]yes they do change the password but in order to have the password you don't need to do anything. When you visit the site there is a hint for the password and usually it is very easy to guess it.[/quote]Please -- if there's a notice on the site not to publicly post the password, then you don't publicly post the password!  It's as simple as that.

How can I have password ? If someone can help us ...

Thanks !

[quote="birkof"]yes they do change the password but in order to have the password you don't need to do anything. When you visit the site there is a hint for the password and usually it is very easy to guess it.[/quote]

Site? Hints? I'm afraid I don't follow. There is no site, unless you mean http://lib.homelinux.org/ or http://homelinux.org/ ; the former is also protected; the latter has no relevant information.

also probably the thing in russian is the hint how are people who don't know russian supposed to understand that or are they not supposed  :o  .

I am locking this thread started 2 years ago. The site referred to is clearly not meant to be open to the public so users should not be encouraged to get into it.


geometry
geometric transformation
rotation

Similar to the one started by isabella2296, this POTD series will begin on Monday, and hopefully run over the summer and into next school year.  PowerOfPi and I will be running it, and we will be tag-teaming with the problems.  The majority of these problems will be posted by PowerOfPI, and I will post some others.  You should PM whoever has posted the problem to avoid confusion, and we will work together to keep the scoreboard current and the solutions together with the previous day's problem.  Hopefully this thread will be as active as the pervious contest's was.  Anyway, see all of you Monday with the first problem, which will be posted by me.  Good luck!

Just wondering: Generally, what time is the Problem of the Day II posted?

Late afternoon, usually...

[size=167][b]Day 1[/b][/size]

A ball is dropped from a window $ 40$ feet from the ground.  After the ball reaches the ground is bounces back into the air at $ \frac{4}{5}$ the height.  Each proceeding bounce and drop is $ \frac{4}{5}$ the height of the previous bounce.  How far will the ball have traveled when it finally comes to a rest?

PM your answers to me.

it won't really come to a stop really, right?

No, but the bounce's height will become so minimal as to assume that it has come to a stop.  Really, that fact does not change the answer to the problem, sorry for the confusion.

[size=167][b]Day 2[/b][/size]

My friend and I were raking leaves in the backyard over the weekend.  He raked leaves into $ \sqrt{75}*\sqrt{3}$ piles and I raked them into $ 6^2\minus{}4!$ piles.  If we put it all together, how many piles will we have?

PM your answers to me.

So PoTD II and PoTD are totally separate now?

Pretty much. I will be posting the problems (PoTD 2) Friday to Sunday because rd5493 won't be able to.

[hide="Submissions"]7[/hide]
[hide="Correct"]0[/hide]
[hide="Solution"]Notice that if you put all the piles together, you have $ 1$ big pile.  It seems as if you guys focused so much on the computational side that you overlooked the logical common sense.[/hide]

[hide="Scoreboard"]$ \begin{tabular}{|rl|r|}\hline & Name & Points \\
\hline 1. & XK \minus{} 39 & 1 \\
\hline 1. & alex960531 & 1 \\
\hline 1. & ytrewq & 1 \\
\hline 1. & arb95 & 1 \\
\hline 1. & abcak & 1 \\
\hline\end{tabular}$[/hide]

NOTE: I will no longer let you know if you are right or wrong as I have been from now on.

EDIT:  I kinda screwed up and now the Day 1 and Day 2 solution posts look the same.

That's a really ambiguous problem - one interpretation would yield the answer is $ \sqrt{75} \times \sqrt{3} \plus{} 6^2\minus{}4!\equal{}15\plus{}12\equal{}\boxed{27}$.

Doesn't matter.  When you put them all together as I said, there would only be $ 1$.

[b][size=167]Day 3[/size][/b]

How many zeroes would $ 40!*20!*10!*5!\plus{}10!\plus{}5!$ end with?

So the old problem of the day thread is stopping? Or will it also continue?

I will run the old one, and rd5493 will run the new one. I will post the problem this Friday to Sunday, because rd5493 won't be able to.

[hide="Submissions"]1[/hide]
[hide="Correct"]0[/hide]
[hide="Solution"]There are only possible distributions of the records are
$ 3\minus{}0,2\minus{}1,1\minus{}2,0\minus{}3$ which has $ 4!\equal{}24$ possible ways of distributing the teams.
$ 3\minus{}0,1\minus{}2,1\minus{}2,1\minus{}2$ which has $ \frac{4!}{3!}\equal{}4$ possible ways of distributing the teams.
$ 2\minus{}1,2\minus{}1,2\minus{}1,0\minus{}3$ which has $ \frac{4!}{3!}\equal{}4$ possible ways of distributing the teams.
$ 2\minus{}1,2\minus{}1,1\minus{}2,1\minus{}2$ which has $ \frac{4!}{2!*2!}\equal{}6$ possible ways of distributing the teams.

This gives us a total of $ 24\plus{}4\plus{}4\plus{}6\equal{}\boxed{38}$ ways.[/hide]

The updated scoreboard will be posted on Sunday after I get back.

[quote="rd5493"]Doesn't matter.  When you put them all together as I said, there would only be $ 1$.[/quote]

You never said how the piles were "put together" :wink:

[b]If we put it all together[/b], how many piles will we have?

This was in my original question.  You didn't even submit, y r u trying to disprove me so much?

You never defined "it".

Anyway, I'm sure that many people answered 27, and it's not fair that they don't get credit simply because of an ambiguous problem.

OK, because of AIME15's dauntless protesting, I will award anyone who submitted an answer of $ 27$ to Day $ 2$'s problem the point.  If you answered this, please PM me so I can put the points in the scoreboard.  I know that there were only $ 4$ or $ 5$ that answered it right, and I deleted those PM's I got so I wouldn't get them confused with other days.  Please be honest with your submission.  Thank you, and I'll try not to post such "ambiguous" problems in the future.

[hide="Submissions"]1[/hide]
[hide="Correct"]0[/hide]
[hide="Solution"]There are only possible distributions of the records are
$ 3 \minus{} 0,2 \minus{} 1,1 \minus{} 2,0 \minus{} 3$ which has $ 4! \equal{} 24$ possible ways of distributing the teams.
$ 3 \minus{} 0,1 \minus{} 2,1 \minus{} 2,1 \minus{} 2$ which has $ \frac {4!}{3!} \equal{} 4$ possible ways of distributing the teams.
$ 2 \minus{} 1,2 \minus{} 1,2 \minus{} 1,0 \minus{} 3$ which has $ \frac {4!}{3!} \equal{} 4$ possible ways of distributing the teams.
$ 2 \minus{} 1,2 \minus{} 1,1 \minus{} 2,1 \minus{} 2$ which has $ \frac {4!}{2!*2!} \equal{} 6$ possible ways of distributing the teams.

This gives us a total of $ 24 \plus{} 4 \plus{} 4 \plus{} 6 \equal{} \boxed{38}$ ways.[/hide]

The updated scoreboard will be posted on Sunday after I get back.

[size=150][b]Day 4[/b][/size]

Anna wrote a 2-digit number. Ben created a 4-digit number by copying Anna's number twice. Then Anna divided Ben's number by her number. What was the result she got?

[b]PM your answers to me.[/b]

Wait- do you have to submit solutions too?

No, not when you submit it to me. I don't think so for the rest of the days too.

[hide="Submissions"]4[/hide]
[hide="Correct"]4 (XK-39, abcak, yaofan,  alex960531, arb95[/hide]
[hide="Solution"]Let the 2-digit number be $ a$. Then we have the two numbers $ a$ and $ 100a\plus{}a$. When you simplify the second number, you get $ 101a$. So dividing the second number by the first, we get $ \frac{101a}{a}\equal{}\boxed{101}$.[/hide]
[hide="Scoreboard"]$ \begin{tabular}{|rl|r|}\hline & Name & Points \\
\hline 1. & abcak & 2 \\
\hline 1. & XK \minus{} 39 & 2 \\
\hline 1. & alex960531 & 2 \\
\hline 1. & arb95 & 2 \\
\hline 2. & ytrewq & 1 \\
\hline 2. & yaofan & 1 \\
\hline \end{tabular}$[/hide]

[size=150][b]Day 5[/b][/size]

In how many ways can 4 keys (including Key A and B) be arranged around a circle if Key A and Key B can't be together? Rotations and flips count as the same arrangement.

[b]PM your answers to me.[/b]

[hide="Submissions"]1[/hide]
[hide="Correct"]1 (alex960531)[/hide]
[hide="Solution"]1, because we fix Key A, and then we know Key B has to be opposite it. Then we have the two other ones to put. We can put them there in two ways, but one is the flip of the other, so we only have one.[/hide]
[hide="Scoreboard"]$ \begin{tabular}{|rl|r|}\hline & Name & Points \\
\hline 1. & alex960531 & 3 \\
\hline 2. & abcak & 2 \\
\hline 2. & XK \minus{} 39 & 2 \\
\hline 2. & arb95 & 2 \\
\hline 3. & ytrewq & 1 \\
\hline 3. & yaofan & 1 \\
\hline \end{tabular}$[/hide]

[size=150][b]Day 6[/b][/size]

How much greater, in square inches, is the area of a circle of radius 20 inches than a circle of diameter 20 inches? Express your answer in terms of $ \pi$.

[b]PM your answers to me[/b]

[hide="Submissions"]4[/hide]
[hide="Correct"]4 (Urc, yaofan, Canton, cgyao15)[/hide]
[hide="Solution"][quote="Urc"]
Let the circles mentioned be circle 1 and 2, respectively.

Since the area of a circle is $ \pi$$ r^2$, we can calculate circle 1's area by plugging in $ r \equal{} 20$, to get $ 400\pi$ as the area of circle 1.

however with circle two, we have to note that the radius is half of the diameter. thus $ r \equal{} d/2 \equal{} 10$, and plugging in $ r \equal{} 10$, we get the area of circle two to be $ 100\pi$.

thus the difference between their areas is $ 400\pi$ - $ 100\pi$ = $ 300\pi$, and circle one has an area $ 300\pi$ greater than circle two.
[/quote][/hide]
[hide="Scoreboard"]$ \begin{tabular}{|rl|r|}\hline & Name & Points \\
\hline 1. & alex960531 & 3 \\
\hline 2. & abcak & 2 \\
\hline 2. & XK \minus{} 39 & 2 \\
\hline 2. & arb95 & 2 \\
\hline 2. & yaofan & 2 \\
\hline 3. & ytrewq & 1 \\
\hline 3. & Urc & 1 \\
\hline 3. & Canton & 1 \\
\hline 3. & cgyao & 1 \\
\hline \end{tabular}$[/hide]

I think this Problem of the Day is going to stop. If anyone wants to continue it, then you may.

hmm...
maybe my score was not included?  :)
and i will continue.

So does that mean the other problem of the day is going to stop also?

rd5493 and I are going to continue the other one. I think cgyao15 will continue this one.

This is from mathcounts.
Find all perfect squares divisible by 8 and less than 500.
plz pm ur answers, i'm sorry but we're starting over.

[hide="submissions"]
1
[/hide]

[hide="correct answers"] [b] 0![/b]
[/hide]
Answer, 0,16, 64, 144, 256, 400.
Day 8
How many perfect 3-digit squares are divisible by 3 and 4.
PM ur answers.

Since i have noticed that [hide="this"]
This is dying :)[/hide]

Yea it is... the other one is doing well though


I'm not sure if this is appropriate for this section, but I couldn't think of where to post it.

Since it is nearing the end of the school year, I wanted to have a fun class activity with cards. Please post some interesting card tricks that you know and also explain the mathematics behind it.

I have quite a few good ones, I used to do lots of magic. But I can't divulge any _really_ good ones because that would be against the rules :roll: If I don't have time tomorrow morning, though, I'll do it after I get back from VBS.

Ok, so I don't have much time, so I'll try and type quickly. This is an ace trick, but it is not at all cliché.
[hide]
You start out with the four aces on the top of the deck, and if you know any false shuffles, do them. Then, get a student volunteer and emphasize the fact that you will no longer touch the cards at ALL, knowing that the four aces are still on the top. Now, get him or her (I will now use the politically incorrect "him" from now on for sake of simplicity) to make four piles by dropping off a fourth of the deck in one area, moving a tad to the right, then drop a third of the remaining deck from the bottom, moving a smidge to the right, drop one half of the remaining deck, and then drop the rest of the deck right next to that. You (well, he) just made four about equal piles, 1, 2, 3, and four, with the four aces in pile number 4. It should look like this:

Pile One :: :: Pile Two :: :: Pile Three :: :: Pile Four <-- with four aces on the top

Okay. Next, tell him to take three cards from pile number 1 and place them on the bottom of that pile. Next, he takes three cards from the top of that same pile and places one on each of the other three piles. All of this is done upside down.

Okay. Next, tell him to take three cards from pile number 2 and place them on the bottom of that pile. Next, he takes three cards from the top of that same pile and places one on each of the other three piles. All of this is done upside down.

Okay. Next, tell him to take three cards from pile number 3 and place them on the bottom of that pile. Next, he takes three cards from the top of that same pile and places one on each of the other three piles. All of this is done upside down.

Okay. Next, tell him to take three cards from pile number 4 and place them on the bottom of that pile. Next, he takes three cards from the top of that same pile and places one on each of the other three piles. All of this is done upside down.

Lol that was all C&P'ed so it may be a bit repetitive. But basically, what he did was place three cards on top of the pile with the aces, then take those three random cards and put them on the bottom, making the aces be on the top again. Then, he took the top three cards and placed them on the other piles, making the four aces be the top card on each pile. Then you get the volunteer to flip over the top card of each card, and they're totally amazed. It sounds easy, but you all have minds set on figuring things out, so it's actually a very good effect.

Tell me what you think of it (I didn't make it up lol). But if you fool someone then I'd like to hear about it lol.[/hide]

Thanks breakdancer. I'll let you know about it.

I just found a site with tons of cool tricks so I'm alright by myself now. Thanks to all who contributed.

Ooh do you have a link? I wouldn't mind seeing it myself  :D.

http://web.superb.net/cardtric/

Check out the link for Mathematical card tricks.

i knew that site as well... especially [url=http://web.superb.net/cardtric/tricks/intermed/sun.htm]the sun[/url] is a nice and easy trick which can impress people a lot :P


hello,i'm new to this blog
i'm in 11th
i hav heard people telling dat kvpy needs concepts.can u give 1 or 2 sample questions 4 better knwldge

thanks ernie 4 ur suggestion

simple one (twelthies plz don't answer) that came last yr
a cube has 12 edges( :rotfl: ) each of resistance R.Calculate net resistance betw opp vertices.
I don't think u need options
i'll post more later

and yes 1 more (in bio section)
What is a product called that has all rights reserved? or sth
i don't remember the options :maybe:  :(  :P

I am having difficulties imagining the cube :(
how can i solve then???

I mean was the shape of cube given as a picture????

Its too tough for me
I remember doing a similar question on cubes in 10(that required junction laws,new topic for us at that time)

about bio question,I can't get head-tail of question :(

so it proves i am not going to be selected :(

if u hav read junction law in 10th u shudn't worry abt kvpy clearance :)

the bio ques was a general 1-the easiest
see the options
a  library
b  comittee
c   patent
d   don't remember :P
 :rotfl:

for whole current electricity,i learnt only to simplify circuits only(i don't remember anything about heating and magnetic effects)

I read it and wheatstone bridge just for NTSE(there was no question from it  :( )
it has nothing to do with my kvpy

And your kvpy bio 'easy questions' are way tough for me.I am not understanding anything in them  :(

P.S:Is the answer for cube question $ \frac{5R}{6}$ (this is for diagonally opposite points)  :huh:

I misinterpreted the cube having 12 'faces' instead of edges  :oops: and thus was having difficulties in imagining cube.

[quote="skand"]
P.S:Is the answer for cube question $ \frac {5R}{6}$ [/quote]do u think it can b wrong?? :oops: anyways which method did u use?? :D

abt bio ques,well its not a bio ques,its a typical english one :D
and u kno after exam we were discussing y only bio suffers with non relevent quesn :rotfl:

honestly speaking memory  :rotfl:
(i remembered three ques on resistances for NTSE)
1 was this(i mean cube)
another was wheatstone bridge and next was complex one(that renders wheatstone bridge when you pull it)

but the soln goes like that:
considering current $ I$ passing through one point.It divides in three parts(among which two are equal).also using junction law,we can prove no current exists in 2-3 edges.and then the question resumes to normal calculation of resistances  :)

P.S:Nirlipta seem to have lost her way

[quote="skand"]
considering current $ I$ passing through one point.It divides in three parts(among which two are equal).also using junction law,we can prove no current exists in 2-3 edges.and then the question resumes to normal calculation of resistances  :)
[/quote]i don't think u can b wrong :oops:
but as long as i remember the solution,current is thru all edges in this case
and the best way to solve it is by finding equipotential pts :D

don't remember,actually my teacher had given me the answer  :oops: (i am not that smart)

Will try it again

and please post some of other questions

the questions are really interesting............but just tell the cube problem.................how to solve......

and ya........
pl post some more ques........


let A(n,n) be a triangular matrix by bolcs prove that the determinant of A is the product of the determinant of its diagonal matrix.

It's sort of hard to understand that question.  Could you by any chance put it into syntax?

Here's the matrix syntax:

\begin{pmatrix}
a_{11}&a_{22}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}

Just copy and paste it, substituting your entries for a_{11}, a_{12},...

First: if a matrix is block diagonal with all but one of its blocks the identity, then its determinant is the determinant of that block. That is

$ \det\begin{pmatrix}I&0&0\\0&A&0\\0&0&I\end{pmatrix}\equal{}\det A.$

This can be shown by computing the determinant by minors.

Second: if a matrix is block diagonal, then its determinant is the product of the determinants of the blocks. We can see this by factoring the matrix:

$ \begin{pmatrix}A_1&0&\cdots&0\\0&A_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&A_k\end{pmatrix}\equal{} \begin{pmatrix}A_1&0&\cdots&0\\0&I&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&I\end{pmatrix}\cdots \begin{pmatrix}I&0&\cdots&0\\ 0&I&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&A_k\end{pmatrix}$

Third: An arbitrary square matrix $ A$ can be factored as $ A\equal{}RT,$ where $ R$ is invertible and $ T$ is upper triangular. This is simply the Gaussian elimination algorithm.

We put this together by applying Gaussian elimination to each block separately in a block upper triangular matrix:

$ \begin{pmatrix}A_1&*&\cdots&*\\0&A_2&\cdots&*\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&A_k\end{pmatrix}\equal{} \begin{pmatrix}R_1&0&\cdots&0\\0&R_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&R_k\end{pmatrix} \begin{pmatrix}T_1&*&\cdots&*\\0&T_2&\cdots&*\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&T_k\end{pmatrix}.$

The first matrix in this product is block diagonal, hence its determinant is $ \det(R_1)\det(R_2)\cdots\det(R_k).$ The second matrix in the product is upper triangular in the ordinary sense, which means that its determinant is the product of the elements on the diagonal, which is $ \det(T_1)\det(T_2)\cdots\det(T_k).$ Multiply those together to get that the determinant of the original matrix is $ [\det(R_1)\det(T_1)]\cdots[\det(R_k)\det(T_k)]\equal{} \det(A_1)\cdots\det(A_k).$

There's an extra assumption here: the blocks are such that each diagonal block is square.

Reminds me of a question from my final:

Prove that if for a square matrix $ M$ with matrices $ A$, $ B$, $ 0$ (zero) and $ I$ (identity), that
$ \det(M)\equal{}\begin{vmatrix}
A&B\\
0&I\end{vmatrix}
\equal{}A$
which of course requires expansion along the identity matrix.

many thanks Kent Merryfield but as JRav said how to prove this easily $ \det(M) \equal{} \begin{vmatrix} A & B \\
0 & I\end{vmatrix} \equal{} A$?

Gaussian elimination using the rows containing the rows of the identity matrix $ I$ reduces each entry of $ B$ to zero and then we just have (edit:) Kent Merryfield's first lemma.  You mean $ \det(A)$.

I didn't post any lemmas- that would be Kent's.

There are formulas to expand along many columns at once, called Laplace formulas I think. They can be proven by appealing to some kind of uniqueness of certain types of operators on matrices (exterior algebras).

[quote]You mean $ \det(A)$.[/quote]
Yes I do.
The way I tackled this problem, is to expand along the last row.  You end up with
$ (\minus{}1)^{2n}\times\det(\tilde{M})$
You continue the process for $ \tilde{M}$ and eventually, once all of the ones have been exhausted, you wind up with $ \det(A)$


LaTeXRender has limits on the number of rows you can create with tabular (currently, it's 8). If you want that to show up, you have to separate it into groups of 8.


It appears that a lot of people on AoPS GFF are within a month of the same age.  To see what percentage this is, I'm taking a poll.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148778]Reason.[/url]

Where are i_like_pie, footballrocks41237, funcia, and mathnerd314 when you need them?

Hey, about ppl born in the same month, there are 20000 users or more born in years 1984 to 1993. Then, by PHP in one of these months in this range we got at least ceiling of ${20000/10/12}$ which is [b]167[/b] people! :wink:

well then...

at least one of us came!!!

we are so same we come on at the same hour to vote(well 3/4 of us)

[quote="ra5249"]nov. 94  :P[/quote]
that's just scary...that's asmallgoat's


Three people take the AMC, and get a given problem right with probability p (i.e. they have equal chance of getting any two problems correct). If at the end of the exam, they all compare answers and all 25 of their answers match up, what is the probability that they actually got everything correct?

Well, the condition that their answers match up means we have to find the probability that their answers match up, and perform conditional probability, so the probability they get them all right is
$ \frac{\text{probability all right}}{\text{probability answers match up}}$

However, I don't think we know the probability that the answers match up because their are five answer choices (A,B,C,D,E), but we only know the probability of getting it right.

[quote="1=2"]We know that they match up, we need the probability that all of them got perfect, and if that guy*points to one of them* got perfect, then the rest got perfect, and the probability that he gets all of them right is $ p^{25}$.[/quote]

I think that the fact that they all got the same answer is tripping me up. Since they all got the same answers, wouldn't there be a higher probability that they all got the right answers?

My logic is very intuitive, and I'm not sure if it is correct.

There are a couple of assumptions we need to make here (or just one):

- If a person gets a question wrong, he chooses the other answer choices with equal probability i.e. $ (1 \minus{} p)/4$

The probability is not simply $ p^{25}$, your logic implies one person did the test and the other two people copied off his paper. The three people are doing the work independently, so if they all get the same answer for a question, chances are it's the right answer (unless $ p$ is too small).

[hide="Solution"]
For any one question, the probability that all three people got it right given that all three of their answers match up is $ \dfrac{p^3}{p^3 \plus{} 4\cdot(\frac {1 \minus{} p}{4})^3} \equal{} \dfrac{p^3}{p^3 \plus{} \frac {(1 \minus{} p)^3}{16}}$
(If you don't believe me, check out [url=http://en.wikipedia.org/wiki/Conditional_probability]conditional probability[/url])

The probability that this happens for all 25 questions is

$ \displaystyle\left(\dfrac{p^3}{p^3 \plus{} \frac {(1 \minus{} p)^3}{16}}\right)^{25}$

Could probably be simplified...
Note plugging in p=0.5 results in a probability of approximately 0.220
[/hide]


AMC
AIME
AMC 12
geometry
USAMTS
ARML
AIME I

I am helping out with the math team at a local high school.
Basically the teacher who is in charge has given me free reign
to cover whatever I want.  Right now there are 3 kids (juniors).  Last year they all qualified for the AIME (I think), but none of them qualified for the USAMO.  That's how I did when I was in high school as well.  I think one way to summarize the situation is that they (and I) are good math students, but haven't been taught much about how to prepare for contest math.

I have two goals: I want to help them improve their performance on the AMC12 as much as possible, and I want to make math team fun.  The thing is, we really don't have that much time.  Right now, math team only meets once a week at lunch (35 minutes).

A typical meeting consists of working problems from old AMC12s.  Unfortunately, it's hard to get through very many problems (we usually only look at problems starting around number 10 or 15), and I don't think it emphasizes the general concepts as much as is desirable.

Finally, I'm not really sure what topics to focus on.  Two possibilities I have thought about are combinatorics and geometry.
It seems that most high school students I know are not very good at geometry and most high schools I know don't teach hardly any combinatorics.  Just a few thoughts.

I would appreciate any ideas.

--Petra

Showing them subjects they haven't seen before/won't see in high school is a good idea, particularly because there are many fun areas you can show them - counting, number theory, challenging geometry.  Follow up, if you can, any class by giving them some problems to do on their own.

There's no substitute for doing problems.  There's also no substitute for doing work on your own.  Find a way to give them sufficient incentive to study on their own.  Unfortunately, this may be really hard to do.  Are there local/state events you can take them to?  Getting your school to be part of a regional math culture goes a long way towards getting students to allocate some of their spare time to preparing.

I'd also recommend getting them involved in Mandelbrot & USAMTS, and working towards an ARML team if your area has one.  If you have a long-term view (as opposed to just this year), think of ways to build a local culture - maybe have the school host a math event (which also can serve as an excellent fund-raiser).


limit
logarithms
real analysis
real analysis unsolved

[hide]$ \lim \frac{n(n^{1/n}\minus{}1)}{\ln n} \equal{} \lim \frac{n(e^{\ln n/n}\minus{}1)}{\ln n} \equal{} \lim \equal{} \frac{n(1 \plus{} \frac{\ln n}{n} \plus{} O[(\frac{\ln n}{n})^2] \minus{} 1)}{\ln n}$

$ \equal{} \lim \left(1 \plus{} O \left(\frac{\ln n}{n}\right)\right) \equal{} 1$.[/hide]

... or you can calculate this limit by using the well known: $ \lim_{x_n \rightarrow 0} \frac{a^{x_n}\minus{}1}{x_n} \equal{} lna$.
We write your limit in the form $ \lim_{n \rightarrow \infty} \frac{e^{\frac{\ln n}{n}}\minus{}1}{\frac{\ln n }{n}}$ and so the answer is $ 1$.


H is the radical center of the three circles

[quote="ridgers"]In the acute triangle ABC and in its side AB and AC are taken the points M and N respectively (M is on AB and N is on AC). The circles with diameters BN and CM intersect in points P and Q. H is the intersection point of the altitudes. Prove that P, Q and H are collinear points.[/quote]
There is also a remark that if $ P$ is the intersection of $ MN$ and $ BC$ then the circle with diameter $ AP$ and the circles with diameters $ BN$ and $ CM$ are coaxal circles with $ PQ$ is the radical axes of them.

[quote="mathVNpro"]Let $ B_1$, $ C_1$, respectively be the projections of $ B,C$ onto $ AC,AB$. It is easy to note that $ B_1\in$ circle with diameter $ BN$ and $ C_1\in$ circle with diameter $ CM$. But $ \overline {HB}.\overline {HB_1} \equal{} \overline {HC}.\overline {HC_1}\Longleftrightarrow \mathcal {P}_{H/(BNB_1)} \equal{} \mathcal {P}_{H/(CNC_1)}\Longrightarrow H\in PQ$, which is the radical axes wrt $ (BNB_1),(CNC_1)$, the result is lead as follow.
Our proof is completed then.[/quote]

I dont understant very well this step $ \overline{HB}.\overline{HB_{1}}\equal{}\overline{HC}.\overline{HC_{1}}\Longleftrightarrow\mathcal{P}_{H/(BNB_{1})}\equal{}\mathcal{P}_{H/(CNC_{1})}\Longrightarrow H\in PQ$
Actually what's this $ \mathcal{P}_{H/(BNB_{1})}\equal{}\mathcal{P}_{H/(CNC_{1})}$

[quote="ridgers"][quote="mathVNpro"]Let $ B_1$, $ C_1$, respectively be the projections of $ B,C$ onto $ AC,AB$. It is easy to note that $ B_1\in$ circle with diameter $ BN$ and $ C_1\in$ circle with diameter $ CM$. But $ \overline {HB}.\overline {HB_1} \equal{} \overline {HC}.\overline {HC_1}\Longleftrightarrow \mathcal {P}_{H/(BNB_1)} \equal{} \mathcal {P}_{H/(CMC_1)}\Longrightarrow H\in PQ$, which is the radical axes wrt $ (BNB_1),(CMC_1)$, the result is lead as follow.
Our proof is completed then.[/quote]

Actually what's this $ \mathcal{P}_{H/(BNB_{1})} \equal{} \mathcal{P}_{H/(CMC_{1})}$[/quote]
I made a typo, sorry, it is correct now.

I think that this problem is from the classic "geometry revisited".


geometry
circumcircle
inradius
geometry solved

Given $\triangle ABC$ with any $I$ inside triangle, denote $r'$ is circumradius of pedal triangle of $I$ then $r' \ge r$ ($r$ is inradius of triangle $ABC$) seem it is obvious but I can't give a proof! Can any body help me? Thanks  :)  :)

This is not obvious, but not hard either. More generally, the radius of every circle that intersects all three sidelines of the triangle is $\geq$ to the inradius of the triangle. I gave one possible proof (I know now that it is not the shortest) at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=17210#17210]http://www.mathlinks.ro/Forum/viewtopic.php?t=5292 post #5[/url].

  darij

You can also regard the 3 stretches with factors 0 < k_1, k_2, k_3 < 1, and whose centers are the vertices of the triangle, which transform the pedal circle into the incircle.

Benjamin


Pascal\u0027s Triangle
combinatorics proposed
combinatorics

prove this equation..

$ C^{2n}_{n}\equal{}\sum_{k\equal{}0}^n(C^n_k)^2$

if you're still confused with my country's writing style,,..
$ C^{n}_{k}\equal{}(^n_k)\equal{}\frac{n!}{(n\minus{}k)!k!}$

This is [url=http://en.wikipedia.org/wiki/Vandermonde%27s_identity]Vandermonde's identity[/url] with $ n \equal{} m \equal{} r$.  The combinatorial proof is very nice; there is another proof by counting the number of paths from the top of Pascal's triangle to the center of the $ 2n^{th}$ row by counting the number of paths to each entry in the $ n^{th}$ row.  Finally, there is the straightforward proof

$ {2n \choose n} \equal{} [x^n] (x \plus{} 1)^{2n} \equal{} [x^n] (x \plus{} 1)^n \cdot (x \plus{} 1)^n \equal{} \sum_{a \plus{} b \equal{} n} {n \choose a} {n \choose b}$.

Edit:  I should note that it is important to recall that $ {n \choose k} \equal{} {n \choose n\minus{}k}$.


The arithmetic mean of a set of positive real numbers is greater than their geometric mean. What do you mean, "how do I use it"?

maybe he wants some tricky examples of its usage  :D   :maybe:

the spelling of 'definately' is definitely not to be rude but see this  :)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193667

[quote="madness"]
the spelling of 'definately' is definetly not to be rude but see this  :)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193667[/quote]
 :what?:  :rotfl: D-E-F-I-N-I-T-E-L-Y

sorry he had me confused there too  :rotfl:   :oops:

AM-GM is defined as arithmetic-mean-geometric-mean inequality. For details, click [url=http://www.artofproblemsolving.com/Wiki/index.php/Arithmetic_Mean-Geometric_Mean_Inequality]here[/url] for wiki


A function $ f: X \rightarrow Y$ is said to have a left inverse g whenever $ g: Y \rightarrow X$ is a function satisfying $ g \circ f \equal{} I_X$. Assuming that $ X \neq \emptyset$, prove that f has an inverse if and only if it is an injection. Give an example for f and g for which $ g \circ f \equal{} I_X$ but $ f \circ g \neq I_X$.

That's incorrect. $ f : X \to Y$ must be a bijection. I'll give you the proof for that.

If: Take any $ y$ in $ Y$. Since $ f$ is surjective, there exists some $ x$ in $ X$ such that $ f(x) \equal{} y$; since $ f$ is injective, that $ x$ is unique. Thus, every $ y$ in $ Y$ defines a unique $ x$ such that $ f(x) \equal{} y$. Define a function $ g : Y \to X$ such that $ g(y)$ is equal to that unique $ x$. To show that $ g \circ f \equal{} I_X$, take any $ x$. Since by definition, $ g(f(x))$ is defined to be the unique $ y$ such that $ f(y) \equal{} f(x)$, and $ f$ is injective, it follows that $ y \equal{} x$, i.e., $ g(f(x)) \equal{} y \equal{} x \equal{} I_X$. Thus, if $ f$ is bijective, it has an inverse.

Only if: Let $ g$ be an inverse of $ f$. To show that $ f$ is injective, take any $ x$ and $ y$ such that $ f(x) \equal{} f(y)$. Thus, $ g(f(x)) \equal{} g(f(y))$, so $ x \equal{} y$. To show that $ f$ is surjective, take any $ y$ in $ Y$. Let $ x \equal{} g(y)$. Since $ f(x) \equal{} f(g(y)) \equal{} y$, by definition $ f$ is surjective. Thus, since $ f$ is both injective and surjective, it is bijective.

If $ f$ is bijective, there is no counterexample in your case. If $ f$ is only injective, we may assume $ g$ and $ f$ have different domains; in this case, let $ f$ be $ e^x$ and let $ g \equal{} \ln(x)$. $ \ln(e^x) \equal{} I_X$ for all $ x \in X$ (here, $ X$ is $ \mathbb{R}$); however, $ e^{\ln(x)} \neq I_X$ for all $ x \in X$ (since the expression is undefined when $ x \leq 0$.)

I think the question was for whether $ f$ has a left inverse.

[quote="Zhero"]Since $ f(x) \equal{} f(g(y)) \equal{} y$, by definition $ f$ is surjective. [/quote]

But it isn't given that $ f \circ g \equal{} I_X$, and $ g$ might not be injective; for example, $ f(x) \equal{} \sqrt{x}, g(x) \equal{} x^2$ for $ X$ being $ \mathbb{R}^{\plus{}}_0$, $ Y$ being $ \mathbb{R}$, satisfies the conditions.


geometry
parallelogram
ratio
geometry solved

Let $G$ be the centroid of the triangle $ABC$ in which the angle at $C$ is obtuse and $AD$ and $CF$ be the medians from $A$ and $C$ respectively onto the sides $BC$ and $AB$. If the points $\ B,\ D, \ G$ and $\ F$ are concyclic, show that $\dfrac{AC}{BC} \geq \sqrt{2}$. If further $P$ is a point on the line $BG$ extended such that $AGCP$ is a parallelogram, show that triangle $ABC$ and $GAP$ are similar.

Let's slightly extend the problem (some of the conditions are unnecessary):

[color=blue][b]Problem.[/b] Let G be the centroid of a triangle ABC, and let D and F be the midpoints of its sides BC and AB.

[b](a)[/b] Let P be a point such that the quadrilateral AGCP is a parallelogram. Prove that this point P lies on the line BG.
[b](b)[/b] Assume that the points B, D, G and F are concyclic. Show that the triangles ABC and GAP are similar.
[b](c)[/b] If the points B, D, G and F are concyclic and the angle C of triangle ABC is obtuse, then prove that $\frac{b}{a}>\sqrt2$.[/color]

[i]Solution.[/i] [b](a)[/b] We have denoted by D and F the midpoints of the sides BC and AB of triangle ABC; now denote by E the midpoint of the side CA. The segments AD, BE and CF are the medians of triangle ABC and thus concur at its centroid G.

Since the quadrilateral AGCP is a parallelogram, its diagonals GP and CA bisect each other. In other words, the midpoint of the segment GP is the midpoint of the segment CA, i. e. the point E. Hence, the point P lies on the line GE, i. e. on the line BG. Thus, problem [b](a)[/b] is solved.

[b](b)[/b] Pure angle chasing: Since the points B, D, G and F are concyclic, < BGD = < BFD. Since the points F and D are the midpoints of the sides AB and BC of triangle ABC, we have FD || CA, and thus < BFD = < BAC. Thus, < AGP = < BGD = < BFD = < BAC. Since the quadrilateral AGCP is a parallelogram, AP || CG, so that < APG = < CGP. On the other hand, since the points B, D, G and F are concyclic, < BGF = < BDF, and since FD || CA, we have < BDF = < BCA. Thus, < APG = < CGP = < BGF = < BDF = < BCA. Now, from < AGP = < BAC and < APG = < BCA, it follows that the triangles ABC and GAP are similar. This solves problem [b](b)[/b].

[b](c)[/b] Since the points B, D, G and F are concyclic, the intersecting chords theorem yields $CG\cdot CF=CD\cdot CB$. Now, since D is the midpoint of the side BC of triangle ABC, we have $CD=\frac12\cdot CB$. Since G is the centroid of triangle ABC, and the centroid of a triangle divides the medians of the triangle in the ratio 2 : 1, we have $CG=\frac23\cdot CF$. Thus, $CG\cdot CF=CD\cdot CB$ becomes $\left(\frac23\cdot CF\right)\cdot CF=\left(\frac12\cdot CB\right)\cdot CB$. In other words, $\frac23\cdot CF^2=\frac12\cdot CB^2$; equivalently, $4\cdot CF^2=3\cdot CB^2$. But the segment CF is the C-median of triangle ABC, and thus, by the well-known formula for the length of a median in a triangle, $4\cdot CF^2=2a^2+2b^2-c^2$. Hence, this becomes $2a^2+2b^2-c^2=3\cdot CB^2$; in other words, $2a^2+2b^2-c^2=3a^2$. This simplifies to $2b^2=c^2+a^2$. Now, since the angle C of triangle ABC is obtuse, $c^2>a^2+b^2$. Hence, $2b^2=c^2+a^2>\left(a^2+b^2\right)+a^2$. Subtraction of $b^2$ yields $b^2>2a^2$, so that $\frac{b^2}{a^2}>2$. Taking the square root, we get $\frac{b}{a}>\sqrt2$. Hence, problem [b](c)[/b] is solved.

  darij

[quote=ilovepizza2020][quote=Ajjubhai94]Where is equality[/quote]

The inequality is strict?[/quote]

No, it's non-strict. (the original problem)


Does the osmotic pressure when going from artery to capillary increase, decrease, or remain constant?

Thanks!

Thanks, for posting this question.

Now as you can see from the below image, an example of RBCs leading from artery to capillary, is from the arm to the fingers.

Evidently, the blood is leading from a larger vessel to a smaller. When a liquid passes through, something larger to smaller, there is greater resistance, and thus more osmotic pressure.

The answer to your question: When RBCs pass from artery to capillary, there is an increase in osmotic pressure.

-arb95
[img]http://library.thinkquest.org/05aug/01883/introtocircman.gif[/img]

Sorry to have to correct you here, but osmotic pressure isn't really the same thing as blood pressure. Osmotic pressure is the tendency of a solution to gain or lose water based on its solute concentration versus its environment's solute concentration--the two have to be separated by a semipermeable membrane. Cells will gain or lose water until their solute concentration is the same as their environment's.

[url=http://en.wikipedia.org/wiki/Osmotic_pressure]Here's[/url] its Wikipedia article.

So, to answer your question: unless there's a significant change in environment from the artery to its capillaries (such as in a kidney), the osmotic pressure on a red blood cell should remain the same.

Also, in terms of blood pressure from an artery to capillaries, you have to remember that one artery can split into dozens of arterioles that can split into hundreds of capillaries. What this does is increases the surface area through which the blood travels, lowering the blood's speed and pressure. Blood pressure is actually lower in a capillary than in an artery. Arteries must have thick walls and must be deep under the skin to counteract their pressure; in contrast, capillaries can be at the surface of the skin and can have very thin membranes because of their low pressure.

Your logic was good, though, arb95--pressure does increase if a given amount of liquid or gas is put into a smaller volume. It's just that arteries branch off into so many smaller vessels that the blood is spread out, not just confined to one vessel that gets smaller and smaller.

Again, sorry to correct you. I hope this helps, g1sk8terz.


vector
linear algebra
matrix
linear algebra unsolved

Prove that a set of all orthogonal matrices $SO(n,\mathbb{R})$ with determinant $1$ over $\mathbb{R}$ of size $n \times n$ is irreducible.
(A subset of linear applications $W$ of $L(E)$ is iireducible if the only subspaces of $E$ which are stable by all the elements of $W$  are ${0}$ and $E$ itself.)

you must impose some conditions on this set I think. (or just take $\{Id\}$ and it does not work :) )
Or maybe you mean that $SO_n$ is irreducible, but it is transitive ?

Yes, Alekk, i meant $SO_n$ and this transitivity property: for every pair of vectors $x,y$ of equal norm $||x||=||y||$ there exists $A \in SO(n,\mathbb{R})$ such that $Ax=y$.
At the moment of posting this i didn't know the solution, but know i have a solution: we can use Hausholder matrix $U_{\omega}=I-tww^*, t=2(w^*w)^{-1}$ with $w=x-y$ which would also be hermitian.


[quote="Myth"]And one more similar problem:
Is there a function $f$ from natural numbers to natural numbers such that $f(f(n)) = n^2$?[/quote]

Let this function $f$ be defined on $\mathbb{R}^+ \mapsto \mathbb{R}^+$ as $f(f(x)) = x^2$ or $f \circ f = (id_{\mathbb{R}^+})^{2}$, respectively, prove that

a.) $f$ is bijective and give an example for it,
b.) for each $x \in \mathbb{R}^+$ we have $\sqrt{f(x)} = f(x).$
c.) How many continuous solutions does $f \circ f = (id_{\mathbb{R}^+})^{2}$ $(1)$ have. What do the functions look like satisfying this relation ?
d.) Does $f \circ f = (id_{\mathbb{R}})^{2}$ $(2)$ have continuous solutions. What do the functions look like satisfying this relation ?
e.) Can you construct a solution for $(2)$ from each solution for $(1).$ If so, in this way is it possible to construct all solutions for $(2)$.

a) It is clear that $f$ is bijective, since $Im f\circ f=R^+$ and if $f(x)=f(y)$ then $x^=f(f(x))=f(f(y))=y^2$, so $x=y$. A trivial example is $f(x)=x^{\sqrt{2}}$.

b) What do you mean here? I think you wanted to write $\sqrt{f(x^2)}=f(x)$. :)

c) We can construct $f(t)$ on $[2,2^\sqrt 2)$ in such a way that $f(t)\geq t^\sqrt 2$, $f(t)$ is increasing, $f(2)=2^\sqrt 2$. And then iterate it, using b) :?

[quote="orl"]
b.) for each $x \in \mathbb{R}^+$ we have $\sqrt{f(x)} = f(x).$
[/quote]

Sorry for the wrong problem statement. It should be like that for b.)

b.)  for each $x \in \mathbb{R}^+$ we have $\sqrt{f(x)} = f(\sqrt{x}).$


You are right shalex, acutally your result is stronger than mine but I got this problem from an open problem proposed by Ji Chen in Crux (I don't remember the number)
Given $x,y,z,t>0,x+y+z+t=2$ prove that
$\frac{x^{2}}{(x^{2}+1)^{2}}+\frac{y^{2}}{(y^{2}+1)^{2}}+\frac{z^{2}}{(z^{2}+1)^{2}}+\frac{t^{2}}{(t^{2}+1)^{2}}\leq \frac{16}{25}$
And you can see $\frac{16}{25}<\frac{6\sqrt{3}}{16}$ it is really hard with me, hope I can see nice solution from everybody!

Oh I see why you get $39/100$. And I guess the hard one will be terrible. I have seen a problem, that forall $a,b,c\in\mathbb{R}$ with the property that $a+b+c=1$ there is
$\frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+\frac{c}{c^{2}+1}\leq\frac{9}{10}$
It's very hard because $a,b,c\in\mathbb{R}$. But even if $a,b,c>0$ it's not easy. And now
$\frac{x^{2}}{(x^{2}+1)^{2}}+\frac{y^{2}}{(y^{2}+1)^{2}}+\frac{z^{2}}{(z^{2}+1)^{2}}\leq\frac{27}{100}$
is obviously much stronger! I have no ideas about it now.
Ji Chen, who is from NingBo University, said he is the best one in China to solve inequalities. Perhaps he's right ('cause he does so well), and he's really famous.

I've solved the problem that forall $x,y,z\in\mathbb{R_{+}}$ that $x+y+z=1$ we have
$\frac{x^{2}}{(x^{2}+1)^{2}}+\frac{y^{2}}{(y^{2}+1)^{2}}+\frac{z^{2}}{(z^{2}+1)^{2}}\leq\frac{8}{25}$
as well as Ji Chen's problem.
Consider
$f(x)=\frac{x^{2}}{(x^{2}+1)^{2}}$
then
$f''(x)=2\cdot\frac{3x^{4}-8x^{2}+1}{(x^{2}+1)^{6}}$
We know the roots of
$3x^{4}-8x^{2}+1=0$
are
$a=\sqrt{\frac{4-\sqrt{13}}{3}}=0.36\cdots,\ b=\sqrt{\frac{4+\sqrt{13}}{3}}=1.59\cdots$
Now we need to prove
$f(x)+f(y)+f(z)\leq\frac{8}{25}$
We assume $x\geq y\geq z$
1) if $y\leq a$,
  A) when $y+z\geq a$, with Karamata Inequality we have
$f(x)+f(y)+f(z)\leq f(x)+f(a)+f(y+z-a)=f(x)+f(1-a-x)+f(a)$
but with the method before, we have
$f(x)+f(1-a-x)\leq\frac{3\sqrt{3}}{16}x+\frac{3\sqrt{3}}{16}(1-a-x)=\frac{3\sqrt{3}}{16}(1-a)<\frac{8}{25}-f(a)$
so
$f(x)+f(y)+f(z)\leq\frac{8}{25}$
  B) when $y+z<a$, with Karamata Inequality we have
$f(x)+f(y)+f(z)\leq f(x)+f(y+z)+f(0)$
and
$f(x)+f(1-x)+f(0)\leq\frac{8}{25}\Longleftrightarrow(1-4t)(7+14t-5t^{2}-2t^{3})\geq0\ \left(t=x(1-x)\leq\frac{1}{4}\right)$
2) if $y>a$, with Jensen we get
$f(x)+f(y)+f(z)\leq2f\left(\frac{1-z}{2}\right)+f(z)$
and with $z\leq\frac{1}{3}$ there is
$2f\left(\frac{1-z}{2}\right)+f(z)\leq\frac{8}{25}\\ \Longleftarrow\frac{2\left(\frac{1-z}{2}\right)^{2}}{\left[\left(\frac{1-z}{2}\right)^{2}+1\right]^{2}}+z^{2}\leq\frac{8}{25}\\ \Longleftarrow 25z^{5}(z-4)+18z^{3}(19z-26)+z(713z-240)\leq0$
Therefore forall $x,y,z\geq 0$ that $x+y+z=1$ we get
$f(x)+f(y)+f(z)\leq\frac{8}{25}$
The equality holds when $x=y=\frac{1}{2},\ z=0$.
Actually for Ji Chen's problem, that forall $x,y,z,w\in\mathbb{R_{+}}$ that $x+y+z+w=2$ there is
$\sum\frac{x^{2}}{(x^{2}+1)^{2}}\leq\frac{16}{25}$
I have similar ideas.
Take
$f(x)=\frac{x^{2}}{(x^{2}+1)^{2}}$
and $a,b$ defined like before. We assume that $x\geq y\geq z\geq w$.
1) if $z\geq a$ then $x\leq b$ (or we'll have
$x+y+z+w\geq x+2z>b+2a>2$)
so that $x,y,z\in[a,b]$. With Jensen we get
$f(x)+f(y)+f(z)+f(w)\leq 3f\left(\frac{2-w}{3}\right)+f(w)$
And now we just need
$3f\left(\frac{2-w}{3}\right)+f(w)\leq\frac{16}{25}$
which is not so hard (but annoying).
2) if $0\leq w\leq z<a$,
  A) when $z+w\geq a$, with Karamata we have
$f(z)+f(w)\leq f(a)+f(z+w-a)$
while with the method before there is
$f(z+w-a)+f(x)+f(y)\leq\frac{3\sqrt{3}}{16}[(z+w-a)+x+y]=\frac{3\sqrt{3}}{16}(2-a)<\frac{16}{25}-f(a)$
So
$f(x)+f(y)+f(z)+f(w)\leq\frac{16}{25}$
  B) when $z+w<a$, with Karamata we have
$f(z)+f(w)\leq f(0)+f(z+w)=f(z+w)$
therefore the problem comes down to this one
$f(x)+f(y)+f(z+w)\leq\frac{16}{25}$
    i) if $y\leq a$, with Karamata we can similarly lead $y$ or $z+w$ to $0$ or $a$, while keep $x+y+z+w$ a constant $2$. Then the problem is almost done.
    ii) if $y>a,\ x\leq b$, with Jensen we can let $x,y$ go to $\frac{x+y}{2},\frac{x+y}{2}$, the work left isn't difficult.
    iii) if $y>a,\ x>b$, anyway we notice that $f$ is increasing in $[0,a]$ and decreasing in $[b,2]$, so we're able to lead $z+w$ to $a$, or $x$ to $b$, while keep $x+y+z+w$ a constant $2$; also, the work nearly comes to an end.
Concluding from all above, we shows
$f(x)+f(y)+f(z)+f(w)\leq\frac{16}{25}$
whose equality holds when $x=y=z=w=\frac{1}{2}$. :wink:


Having problems how to solve this?

If cos theta = .3 , find value of : cos theta + sin (90 degrees - theta)

hello, do you mean $ \cos(\theta)\equal{}0.3$, find the value of $ \cos(\theta)\plus{}\sin(90^{\circ}\minus{}\theta)$?
Sonnhard.

yes that is what i mean.  dont know where to find the theta and other symbols.

hello, it is well-known that $ \sin(90^{\circ}\minus{}\theta)\equal{}\cos(\theta)$ so we have
$ \cos(\theta)\plus{}\cos(\theta)\equal{}2\cdot 0.3\equal{}0.6$, this is the result.
Sonnhard.


Hi I am trying to find some math circles in the Dallas/Fort Worth area I know UT Arlington's math circle and  2 math circles in dallas does any one know of any more?

howdy guys,

sorry for a late bump, but a math circle is pretty much a group of people who get together and do math, talk about math, think about math, etc etc etc.

I know there is a rather large one at UTD run by Dr. Titu Andreescu.
As a former member of the Circle, I highly recommend it.

DFW is very fortunate to have two first class math circles.  Dr. Titu Andreescu's Metroplex Math Circle will begin its fourth year September 19th at UT Dallas.  Information about the circle can be found at the following web site and you are welcome to join the Yahoo list and Facebook group.

[url]http://www.metroplexmathcircle.org[/url]

The second math circle was started last year at UT Arlington by Dr. Dimitar Grantcharov.  Dates for the new season of the Mid Cities Math Circle have not been announced.

[url]http://www.midcitiesmathcircle.org[/url]

These math circles are free and open to students interested in both competitive problem solving and interesting mathematical concepts.  Check out the sites for more information or feel free to post questions here.


Hey, i think it's a great that Silverfalcon teaches like a school lesson

I think this is better than the school math

Hey Silverfalcon,

do you make like a lesson plan for this forum?

That's right, KBabe.

To shinwoo, FYI, I probably would introduce some important lessons of the getting started math. I think I did this on Getting Started, on Even and Odd and Introduction to the Sum.

I did lead the math chat after joml88 was busy, and I think I did that for about 7 sessions. And three of them were middle-schoolers. :)

Of course, my math chat did not go to high so it wasn't that hard for any of them to follow up.

looks like a [i]Cyberchase[/i] problem
an episode was talking about probability, I think it's great

[quote="Silverfalcon"]Unfortunately, the normal textbook that is used to teach the middle schoolers for the probability is very weak. But in the reality, the probability takes a lot of part on the matheamatics, especially if you are wishing to attend a math competition.

So, I'll briefly introduce what the probability is. It's very simple but the questions that come out from this simple idea makes hard.

[b]Probability[/b]

Probability is the fraction of the chances that can happen over the total chances of the thing. (I know there is another way to express this but I like my way). so if you have 5 marbles and each has different color, the probability that you're picking a puple one is:

1/5

Since there is one purple and total of 5 marbles.

Make sense, doesn't it? :)

Let's try two questions and then move on.

There are five marbles in a bag. The colors are green, red, and yellow. There are 3 yellow marbles, what is the probability of picking green?

There are 12 marbles in a bag. Suppose one you picked is the green. If there are eleven marbles left with only two colors and 6 of them are red, what is the probability that you'll pick the one that's not red?

[Think on the second one! This is not hard!]

I'll see how this goes and maybe add the multiplication of the probability.[/quote]

Maybe this is the answer

Green=$ G$
Red=$ R$
Yellow=$ Y$

$ R: G: Y \equal{} (5 \minus{} (G \plus{} Y)): (5 \minus{} (G \plus{} R)): 3$
$ R: G: Y \equal{} (5 \minus{} (G \plus{} 3)): (5 \minus{} (3 \plus{} R)): 3$

I think $ R$ and $ G$ same $ (1)$, because it is only the solution

$ R: G: Y \equal{} (5 \minus{} (3 \plus{} 1)): (5 \minus{} (3 \plus{} 1)): 3$
$ R: G: Y \equal{} (5 \minus{} 4): (5 \minus{} 4): 3$
$ R: G: Y \equal{} 1: 1: 3$

So $ G$ is $ \frac {1}{5}$


algebra
polynomial
integration
real analysis
real analysis unsolved

One can compute sums of the form

$S = \sum_{n=1}^{\infty} P(n) \cdot x^n$

for a polynomial $P$ and $|x| < 1$ by telescoping it with itself in the expression $(1-x)\cdot S = P(1)x + \sum_{n=2}^{\infty} (P(n) - P(n-1))x^n$ obtaining a similar expression in $Q(n) = P(n) - P(n-1)$ plus a residue such that $\mbox{deg}(Q) < \mbox{deg}(P)$; this will degenerate to geometric series. Out of curiosity, I wondered about instances where a polynomial appears in the denominator, as in

$S = \sum_{n=1}^{\infty} \frac{1}{n 2^n}$

I have found no corresponding elementary solution, and can only solve this with calculus. [hide]In particular,

$f(r,k) := \sum_{n=1}^{\infty} \frac{r^n}{n^k}$
$\frac{\partial f(r,k)}{\partial r} = \frac{\partial}{\partial r} \sum_{n=1}^{\infty} \frac{r^n}{n^k} = \sum_{n=1}^{\infty} \frac{\partial}{\partial r}(\frac{r^n}{n^k} ) = \sum_{n=1}^{\infty} \frac{r^{n-1}}{n^{k-1}} = \frac{1}{r}\cdot f(r,k-1)$
$\frac{d f(r,1)}{dr} = \frac{1}{r} \sum_{n=1}^{\infty} \frac{r^n}{n^0} = \frac{1}{r} \cdot \frac{r}{1-r} = \frac{1}{1-r}$
$f(r,1) = \int\frac{1}{1-r}dr = -\mbox{ln}(1-r) = \mbox{ln}(\frac{1}{1-r}) + f(0,1)$

Since by inspection $f(0,1) = 0$, I get $f(\frac{1}{2},1) = \mbox{ln}(2)$. [/hide]

My question: does anyone see a generalization of this or some other approach to deal with less particular cases of

$\sum_{n=1}^{\infty} \frac{x^n}{P(n)}?$

You can factor $P(x)=(x+a_1)...(x+a_k)$ and then use it as follows:
if $f(x)=\sum_{n\geq 1} \frac{x^n}{P(n)}$ then $(x^{a_1}f(x))'=\frac{1}{x^{a_1-1}}\sum_{n\geq 1}\frac{x^{n}}{Q(n)}$, where $Q(x)=(x+a_2)...(x+a_k)$, so $\deg Q<deg P$.
Finally, you will obtain a certain differential equation.

I am not especially good at solving integrals. I can't even solve the case $P(n) = n + r$ where $r \notin \mathbb{Z}$, which gives

$\frac{d}{dx}\left(x^r\cdot f(x)\right) = \frac{d}{dx}\left(x^r \cdot \sum_{n\ge 1} \frac{x^n}{n+r}\right)$

$= \sum_{n\ge 1} \frac{d}{dx}\left(\frac{x^{n+r}}{n+r}\right) = \sum_{n\ge 1} x^{n+r-1} = \frac{x^r}{1-x}$

where $f(0) = 0$.

Am I missing something, or could it be that there does not exist a closed form expression for

$\int \frac{x^r}{1-x} dx$


Generalise the given sequence....

$120 , 720 , 2520......$

Couldnt you just use finite differences?

Another method I found(maybe its right) is that each term in the sequence is of the form $n (n+1) (n+2) (n+3) (n+4)$


Consider a fair, 6-sided die. Person A rolls 3 dice, Person B rolls 2 dice. Order their rools as $a_1\ge a_2 \ge a_3$ and $b_1 \ge b_2$.
If $a_1>b_1$, then A gets a point. If $a_1\le b_1$, then B gets a point.
similarly, if $a_2>b_2$, then A gets a point, and if $a_2\le b_2$, then B gets a point
What is the probability that A gets 2 points? 1 point? 0 points?

Before I post a solution, can you tell me if this is right for the case A gets 2 points?
[hide]2/7[/hide]

hmm, well i found a website that just gave the probabilities, and they are not pretty, they brute forced all $6^5$ situations with a computer

There aren't $6^5$ unique situations. If we consider A's rolls, for any set of $3$ numbers rolled, there is only one way to sort them in ascending order. All we care about to solve this problem are the number of ways to choose $3$ numbers from $1$ to $6$, including sets with repeated numbers. That is, $6C3+30+6=56$ rolls. Similarly, for player B, we need only consider the $6C2+6=21$ unique rolls. If this is right, I can post the rest of my solution.

i suppose that is correct, but the difficulty comes in where you have to weight the probability differently, ex, 6,5,4 occurs more often than 6,6,5, etc, also you can go ahead and post your solution (why would you be reluctant to share your thoughts?) but i am pretty sure the answer is 2890/7776

my comp while bruteforcing tells me that chance on 2 points is 3205/7776, chance on 1 is 2716, and chance on 0 is 1855.

[quote="arjunlandes"][quote="pkerichang"]can you explain why 6,5,4 should appear more than 6,6,5??? (I believe both of them has 1/36 probability...)[/quote]
I agree that in the context of this problem, both $(6,5,4)$ and $(6,6,5)$ have the same probability, but that probability is $P=\frac{1}{56}$ for each outcome.[/quote]

sorry, i was not clear, i was thinking of the [i]sorted[/i] triple, ex

665
656
566

654
645
546
564
456
465

so the the second occurs 2 times more, etc


hola  :lol: ... tengo una pregunta (question?).. es de mi clase de espanol pero no se (with accent on e).

does this sentence makes sense:

Lo siento pero a cocineros se les olvidaron las patillas.

(I'm sorry but the cooks forgot the watermelons)

I know that the translation is right but I'm not sure if it makes sense. It's from the conversation in the restaurant, and well, I'm trying to say that the restaurant doesn't have watermelons for customers because they forgot them.

[color=blue]Más apropiado sería decir:
[b][i]Lo siento, pero los cocineros olvidaron las sandías.[/i][/b]
Tiene sentido lo que dices (make sense = tener sentido), pero surgen algunos errores gramaticales, y el verdadero significado de [b]watermelon[/b] es [b]sandía[/b].
Espero que te sirva de ayuda.[/color]

Acabo de ver esto hoy, y tengo una pregunta (soy estudiante de español también):

¿No se puede decir "lo siento, pero a los cocineros se les olvidaron las sandías"? Me parece que la traducción R.G.A.M. es correcta, pero ¿la de arriba es correcta tambien?

[quote="236factorial"]Acabo de ver esto hoy, y tengo una pregunta (soy estudiante de español también):

¿No se puede decir "lo siento, pero a los cocineros se les olvidaron las sandías"? [/quote]

Si, es correcta :lol:

Tu nivel de español es bueno  :lol:


Let a,b be odd integers such that (a^2 - b^2 +1) divides (b^2-1). Prove that (a^2 - b^2 + 1) is a perfect square

[quote="aniketmishra"]Let $ a,b$ be odd integers such that $ (a^{2}-b^{2}+1)$ divides $ (b^{2}-1)$. Prove that $ (a^{2}-b^{2}+1)$ is a perfect square[/quote]
Use dollar signs around your math notation, for example [code]Let $a,b$ be...that $(a^2 - b^2 +1)$ divides...[/code] and [u]welcome to the forum! [/u] :)

[b]Here is just an idea:[/b]
Now, $ (a^{2}-b^{2}+1)$ divides $ (b^{2}-1)$ implies that $ \frac{b^{2}-1}{a^{2}-b^{2}+1}\in\mathbb{Z}$.  I.e.
\[ -\bigg(1-\frac{a^{2}}{a^{2}-b^{2}+1}\bigg) = \frac{b^{2}-1}{a^{2}-b^{2}+1}\in \mathbb{Z}\implies \frac{a^{2}}{a^{2}-b^{2}+1}\in \mathbb{Z}.\] Altogether, $ (a^{2}-b^{2}+1) \mid \gcd(a^{2},b^{2}-1)$ which forces $ a^{2}-b^{2}+1 =\gcd(a^{2},b^{2}-1)$.  If $ \gcd(a^{2},b^{2}-1)=1$, then $ a^{2}-b^{2}+1=1$ and we are done.  Otherwise, I am not sure how to show that this $ \gcd$ is always a square.  I think it turns out to always be 1, but maybe someone else can help prove that (I am stuck).

As b^{2}-1 is divisible by a^{2}-(b^{2}-1), [b^{2}-1]/[a^{2}-(b^{2}-1]=k, where k is integer number. k+1=a^{2}/[a^{2}-(b^{2}-1] is also integer. Let's n=a^{2} and m=b^{2}-1. Then, m,n are also integers. As m/(m-n)=p, where p is integer, m=np-mp, or m(p+1)=np. As (p+1,p)=1, k+1=np,
where p is integer. We got: mp=np-1, or (n-m)p=1. So, n-m=-1 or n-m=1, which is the same as:
a^{2}-b^{2}+1=1 or a^{2}-b^{2}+1=-1. If a^{2}-b^{2}+1=1, we are done, because 1 is perfect square. If a^{2}-b^{2}+1=-1, a^{2}-b^{2}=2,impossible for integers (a,b). So, a^{2}-b^{2}+1 is perfect spuare, Q.E.D.

I see what I was missing.  Thanks.
Also use dollar signs around your math notation so it is easier to read  :) .

For example, you could use
[code]We got: $mp=np-1$, or $(n-m)p=1$. So, $n-m=-1$ or $n-m=1$, which is...$a^{2}-b^{2}+1$[/code]
to write [quote="Lasha"]We got: $ mp=np-1$, or $ (n-m)p=1$. So, $ n-m=-1$ or $ n-m=1$, which is...$ a^{2}-b^{2}+1$[/quote]

[quote="venatrix"]just because this has been said to be a "too easy exercise", does anybody can explain me why lasha's reasoning does not work for $ a=105, b=99$?[/quote]
You are correct.  In my attempt to decipher what lasha wrote, I missed an error.  Note in some spots he wrote $ p$ and meant $ k$ and other minor typos occurred.  Let me attempt to explain what he wrote:

As $ b^{2}-1$ is divisible by $ a^{2}-(b^{2}-1)$, $ [b^{2}-1]/[a^{2}-(b^{2}-1]=k$, where $ k$ is integer number. $ k+1=a^{2}/[a^{2}-(b^{2}-1]$ is also integer. Let's $ n=a^{2}$ and $ m=b^{2}-1$. Then, $ m,n$ are also integers. AS $ m/(n-m)=k$, WE GET $ m(k+1)=nk$. As $ (k+1,k)=1$.  HE MEANT we get $ k+1=n'$ and $ k=m'$ where $ m'=m/d$, $ n'=n/d$, and $ d=\gcd(m,n)$.  So, $ 1=n'-m'$ implying $ d=n-m$ which is exactly what I wrote in my post.

SORRY, BUT WE STILL DON'T HAVE A PROOF.  :(

I think we can restate the problem as follows:
let $ a^{2}-b^{2}+1=k$
let square free part of $ k = y$
Hence we may write $ k=x^{2}y$ where $ x$ is some integer
Then $ a^{2}$ must be of the form $ x^{2}y^{2}z^{2}$ where $ z$ is another integer
then we get $ b^{2}-1=x^{2}y(yz^{2}-1)$
so the problem can be restated as :
If $ x^{2}y^{2}z^{2}-x^{2}y+1$ is a perfect square then prove that $ y=1$

Maybe http://www.mathlinks.ro/Forum/viewtopic.php?t=5935 can help (I don't have the time to check number theory proofs right now, and even if I had, I am trying to solve this problem for a long time and don't want to read spoilers)...

  darij


Can someone write proof of theorem:
 There exists at least 2 prime numbers in (n/2 , n] for any natural n>6.

Thank you.

This is called Bertrand's postulate.
Here you go:
http://www.nd.edu/~dgalvin1/pdf/bertrand.pdf
http://en.wikipedia.org/wiki/Proof_of_Bertrand's_postulate (although I know it's not a fully trustworthy source, copy and paste the link - don't just click it)

Hope it helps!


Prove that for any prime number $ p$, any $ i\in \{1,2,...,p\minus{}1\}$,

$ p$ divides $ \sum a_1a_2...a_i$, where the sum is taken over all combinations $ a_1a_2...a_i$ in the set $ \{1,2,...,p\minus{}1\}$

Here you go
http://www.mathlinks.ro/viewtopic.php?t=141274
If you don;t understand something, ask it in that thread,  I see know that i didn't explain
my solution well enough.
bye


Let n be a positive integer and let $ k_1, k_2, k_3,\ldots,k_n$ denote an arbitrary order of the integers 1 to n. What is the largest possible value of the expression

$ (1\minus{}k_1)^2\plus{}(2\minus{}k_2)^2\plus{}(3\minus{}k_3)^2\plus{}...\plus{}(n\minus{}k_n)^2$

of n terms?

This doesn't belong in this forum.

[hide="solution outline"]$ (1 \minus{} k_{1})^{2} \plus{} (2 \minus{} k_{2})^{2} \plus{} (3 \minus{} k_{3})^{2} \plus{} ... \plus{} (n \minus{} k_{n})^{2}$

$ \equal{} k_1^2 \plus{} k_2^2 \plus{} ... \plus{} k_n^2 \minus{} 2(k_1 \plus{} 2k_2 \plus{} 3k_3 \plus{} ... \plus{} nk_n) \plus{} 1^2 \plus{} 2^2 \plus{} ... \plus{} n^2$

$ \equal{} 2(1^2 \plus{} 2^2 \plus{} ... \plus{} n^2) \minus{} 2(k_1 \plus{} 2k_2 \plus{} 3k_3 \plus{} ... \plus{} nk_n)$.

Thus, the problem reduces to minimizing $ k_1 \plus{} 2k_2 \plus{} 3k_3 \plus{} ... \plus{} nk_n$ which is at least $ 1(n) \plus{} 2(n \minus{} 1) \plus{} 3(n \minus{} 2) \plus{} ... \plus{} n(1)$ by rearrangement. Now you can hopefully simplify that.[/hide]

I'm not good at all with inequalities, so I probably forgot something in rearrangement.


Is it true that for any integer $d$ such that $d$ is not divisible by 2 and 5 there exist infinitely many integers $n$ such that the number $\sum_{i=0}^{n-1}{10^i}$ is divisible by $d$?

yup it is

It is known as the theorem of Plateau in some literature :
Joseph Plateau:
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Plateau.html

At least that is the name I found in a book written by Dickson (http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Dickson.html)
(Actually I didn't read that book, I was just pageturning in a librarie)
Dickson did not mention a proof. Any suggestions?

Well, very straightforward:
The desired sum is $=\frac{10^n-1}{9}$, so it's enough to have $10^n \equiv 1 \mod 9d$ whis is clearly true for $n=\phi (9d)$.


Inspired by a problem in this [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=154985&sid=ef027de5ec4e9261a32e0c63b2736566#154985]topic[/url], I'm not sure whether this problem was posted in the forums or not, here we go:

The three sides of a triangle have positive integer lengths $a,b,c$ where $a\leq b\leq c$. Find the number of (noncongruent) triangles that can be formed when $c=n$, for some given positive integer $n\in\mathbb{Z}^+$.

If noncongruent means not having same sides then:
(n-2)+(n-4)+(n-6)+... (to 0 or 1)
So for even n it's n(n-2)/8
And for odd n it's (n-1)(n-3)/8

Are these values all right?