geometry
trigonometry
vector
geometry proposed

In a convex pentagon $ABCDE$, the points $M$, $N$, $P$, $Q$, $R$ are the midpoints of the sides $AB$, $BC$, $CD$, $DE$, $EA$, respectively. If the segments $AP$, $BQ$, $CR$ and $DM$ pass through a single point, prove that $EN$ contains that point as well.

[i]Yugoslavia[/i]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=45945 .

  darij

I think I have a pretty simple solution.
Let the four lines concur at $ X$. Let $ EX$ meet $ BC$ at $ N'$. Let $ \angle BXM \equal{} \angle DXQ \equal{} m$, $ \angle MXA \equal{} \angle PXD \equal{} n$, $ \angle AXR \equal{} \angle CXP \equal{} p$, $ \angle RXE \equal{} \angle N'XC \equal{} q$, $ \angle EXQ \equal{} \angle BXN' \equal{} r$.
Let $ [ABC]$ denote the area of $ \triangle ABC$. Then,
$ [AXM] \equal{} [MXB]$ (since they have the same base and height) $ \implies \frac {1}{2} BX \cdot MX \cdot \sin m \equal{} \frac {1}{2}MX \cdot \sin n$, or
$ BX \sin m \equal{} \ AX \sin n$. Similarly,
$ AX \sin p \equal{} EX \sin q$
$ EX \sin r \equal{} DX \sin m$
$ DX \sin n \equal{} CX \sin p$
Multiplying these,
$ BX \sin r \equal{} CX \sin q$
Thus, $ [BXN'] \equal{} [CXN']$, so $ N' \equal{} N$ is the midpoint of $ BC$.

So the problem generalizes to any convex n-gon.

Hope my solution is true :)
Let the 4 segments intersect at point $X$ and vectors $XA=a,...,XE=e$. We denote the vector product of vectors $AB$ and $CD$ as $AB\cdot CD$. It is well known that this vectors are paralel if and only if $AB\cdot CD=0$. Because $XM=\frac{a+b}{2},..., XR=\frac{e+a}{2}$, then from the conditions of the problem we deduce the following 4 relations:
\[a\times (c+d)=0\]
\[b\times (d+e)=0\]
\[c\times (e+a)=0\]
\[d\times (a+b)=0\]
From this relations we get that $e\times (b+c)=e\times b+e\times c=b\times d+ c\times a=b\times d+a\times d=-d\times (a+b)=0$ so we get that $E-X-N$.

[quote="vlad1m1r"]
From this relations we get that $e\cdot (b+c)=e\cdot b+e\cdot c=b\cdot d+ c\cdot a=b\cdot d+a\cdot d=-d\cdot (a+b)=0$ so we get that $E-X-N$.[/quote]

I think that $e\cdot b+e\cdot c = -b\cdot d - c\cdot a$, since $ b \cdot e = - b \cdot d $ and $ e \cdot c = - c\cdot a$.

Sorry...

[quote="vlad1m1r"]Hope my solution is true :)
Let the 4 segments intersect at point $X$ and vectors $XA=a,...,XE=e$. We denote the vector product of vectors $AB$ and $CD$ as $AB\cdot CD$. It is well known that this vectors are paralel if and only if $AB\cdot CD=0$. Because $XM=\frac{a+b}{2},..., XR=\frac{e+a}{2}$, then from the conditions of the problem we deduce the following 4 relations:
\[a\cdot (c+d)=0\]
\[b\cdot (d+e)=0\]
\[c\cdot (e+a)=0\]
\[d\cdot (a+b)=0\]
From this relations we get that $e\cdot (b+c)=e\cdot b+e\cdot c=b\cdot d+ c\cdot a=b\cdot d+a\cdot d=-d\cdot (a+b)=0$ so we get that $E-X-N$.[/quote]

i don't think your solution is true, since  two vectors A,B are perpendicular, instead of being paralel, if and only if $A\cdot B=0$.

We can use area method to prove it just like the Ceva theorem

Suppose the common point is $O$. Using that $XY$ bisects $ZT$ iff $S_{XYZ} = S_{XYT}$, we get $S_{EOB} = S_{DOB} = S_{DOA} = S_{AOC} = S_{COE}$, done!


If $ a, b, c$ are non-negative numbers, then

$ \frac{a^2}{b^2} \plus{} \frac{b^2}{c^2} \plus{} \frac{c^2}{a^2} \ge 2(\frac{a}{c} \plus{} \frac{c}{b} \plus{} \frac{b}{a}) \minus{}3$.

[quote="filomen"]If $ a, b, c$ are non-negative numbers, then

$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \ge 2(\frac {a}{c} \plus{} \frac {c}{b} \plus{} \frac {b}{a}) \minus{} 3$ .[/quote]
Suppose that $ c\equal{}min\{a;b;c\}$
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \ge 2(\frac {a}{c} \plus{} \frac {c}{b} \plus{} \frac {b}{a}) \minus{} 3$
<=>$ \sum_{cyc} \frac {a^2}{b^2} \minus{} 3 \geq 2(\sum_{cyc} \frac {a}{b} \minus{} 3)$
<=>$ [\frac {(a \plus{} b)^2}{a^2b^2} \minus{} \frac {2}{ab}](a \minus{} b)^2 \plus{} [\frac {(a \plus{} c)(b \plus{} c)}{a^2c^2} \minus{} \frac {2}{ac}](a \minus{} c)(b \minus{} c) \geq 0$
Easy to prove that $ \frac {(a \plus{} b)^2}{a^2b^2} \minus{} \frac {2}{ab} \geq 0$
$ \frac {(a \plus{} c)(b \plus{} c)}{a^2c^2} \minus{} \frac {2}{ac} \geq 0$
Done!

Let $ \frac{a}{b}\equal{}x$;... => $ xyz\equal{}1$
Using Schur, we have:
$ x^2\plus{}y^2\plus{}z^2 \plus{} \frac{9xyz}{x\plus{}y\plus{}z} \geq\ 2(xy\plus{}yz\plus{}zx)$
=> $ x^2\plus{}y^2\plus{}z^2\plus{}3 \geq\ 2(xy\plus{}yz\plus{}zx)$
We have q.e.d  :)


Let $ ABC$ be a triangle,  the triangle $ P_{1}P_{2}P_{3}$ is inscribed in the triangle $ ABC,$ let $ M_{1},M_{2},$ and $ M_{3}$ be the midpoints of sides $ BC,CA$ and $ AB,$ respectively. let $ N_{1},N_{2}$ and $ N_{3}$ be the midpoints of sides $ P_{2}P_{3},P_{3}P_{1}$ and $ P_{1}P_{2},$ respectively. if $ AP_{1},BP_{2}$ and $ CP_{3}$ are concurrent, then $ M_{1}N_{1},M_{2}N_{2},$ and $ M_{3}N_{3}$ are concurrent.

I generalize this problem as follows:
Let triangle $ P_{1}P_{2}P_{3}$ be inscribed in the triangle $ A_{1}A_{2}A_{3}. M_{i}$ is a point on the side $ A_{i \plus{} 1}A_{i \plus{} 2}, N_{i}$ is a point on the side $ P_{i \plus{} 1}P_{i \plus{} 2}$ such that $ \frac {A_{i \plus{} 1}M_{i}}{M_{i}A_{i \plus{} 2}} \equal{} \frac {P_{i \plus{} 1}N_{i}}{N_{i}P_{i \plus{} 2}}.$ Given that $ A_{1}P_{1},A_{2}P_{2},$ and $ A_{3}P_{3}$ concur at a point, show that $ M_{1}N_{1},M_{2}N_{2},$ and $ M_{3}N_{3}$ are also concurrent.
I wonder whether it holds.


[quote="shobber"]$x$, $y$ and $z$ are positive real numbers such that $x+y+z=1$, prove:
\[ \frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\leq \frac{\sqrt{2}}{2}  \][/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88439

This is actually a China TST problem, it has been posted allready, a classic solution above of the topic is posted by me. [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=124977[/url]

[quote="chien than"]Let $x;y;z>0$ and satisfying $x+y+z=1$ . Prove that:
$\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\le \frac{\sqrt 2}2$[/quote]
By Cauchy Schwarz inequality, we have
\[\sum\frac{xy}{\sqrt{xy+yz}}=\sum\frac{x\sqrt{y}}{\sqrt{z+x}}\le \sum\frac{x\sqrt{2y}}{\sqrt{z}+\sqrt{x}}\]
Hence, it suffices to prove
\[\sum\frac{x\sqrt{y}}{\sqrt{z}+\sqrt{x}}\le \frac{1}{2}\]
Setting $a=\sqrt{x},b=\sqrt{y},c=\sqrt{z}$, we have to prove
\[\sum\frac{a^{2}b}{c+a}\le \frac{1}{2}(a^{2}+b^{2}+c^{2}) \]
or equivalently
\[\sum a^{2}+2abc\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \ge 2(ab+bc+ca) \]
By AM-GM inequality, we have
\[2abc\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \ge \frac{9abc}{a+b+c}\]
Hence, it suffices to prove
\[\sum a^{2}+\frac{9abc}{a+b+c}\ge 2\sum ab \]
which is Schur inequality. :)
We are done.

PS:this is China 2005, not yours, Chienthan.  :wink:

[quote="Math pro"]I see that it is not very strong ineq![/quote]
No, not really actually^^
[quote="shobber"]Given three positive real numbers $ x$, $ y$, $ z$ such that $ x \plus{} y \plus{} z \equal{} 1$, prove that
\[ \frac {xy}{\sqrt {xy \plus{} yz}} \plus{} \frac {yz}{\sqrt {yz \plus{} zx}} \plus{} \frac {zx}{\sqrt {zx \plus{} xy}} \le \frac {\sqrt {2}}{2}.\][/quote]
My solution goes as follows: We have
\[ \sum \frac {xy}{\sqrt {xy \plus{} yz}} \equal{} 2 \sum \frac {x \plus{} y}{2} \sqrt {\frac {x^2y}{(x \plus{} y)^2(x \plus{} z)}} \equal{} S.\]
Since the function $ f(x) \equal{} \sqrt {x}$ is concave on $ \mathbb{R}_ \plus{}$, we have by Jensen
\[ S \le 2 \sqrt {\sum \frac {x^2y}{2(x \plus{} y)(x \plus{} z)}} \equal{} \sqrt {2}\sqrt {\frac {\sum (x^2y^2 \plus{} x^2yz)}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}}.\]
Thus, after squaring, it remains to prove $ \sum xy(x \minus{} y)^2 \ge 0,$ which is obvious.

we have that \[ \sum_{cyc} \frac{xy}{\sqrt{xy+yz}}\le \sqrt{(\sum xy(x+y))(\sum \frac{x}{(x+y)(x+z)})}\le \frac{\sqrt{2}}{2}. \] So, it's equivalent to $ \sum_{cyc} a^2(b+c)(a-b)(a-c)\ge 0$ :wink:

My solution is I guess tough for no reason :wink:

Applying jensen's to the function $f(x)=\sqrt x$
$$\frac {xy}{\sqrt {xy + yz}} + \frac {yz}{\sqrt {yz + zx}} + \frac {zx}{\sqrt {zx + xy}} \le \sqrt{\sum_{x,y,z} \frac{2x^2y^2(xy+yz+zx)}{(xy+yz)(xy+xz)}}$$
Now we are left to prove
$$\sum_{x,y,z} \frac{2x^2y^2(xy+yz+zx)}{(xy+yz)(xy+xz)} \le \frac{1}{2}$$

For this, see http://www.artofproblemsolving.com/community/c6h1160617_weird_symmetric_inequality

[quote=shobber]Given three positive real numbers $ x$, $ y$, $ z$ such that $ x \plus{} y \plus{} z \equal{} 1$, prove that

$ \frac {xy}{\sqrt {xy \plus{} yz}} \plus{} \frac {yz}{\sqrt {yz \plus{} zx}} \plus{} \frac {zx}{\sqrt {zx \plus{} xy}} \le \frac {\sqrt {2}}{2}$.[/quote]

By AM-GM,we have
\[xy\sqrt{\frac{2}{xy+yz}}=\frac{2x\sqrt{2(xy+yz)(y+z)^2}}{2(x+z)(y+z)}\]
\[\le{\frac{x[2(xy+yz)+(y+z)^2]}{2(x+z)(y+z)}}=\frac{x(2xy+4yz+y^2+z^2)}{2(x+z)(y+z)}\]
Hence,we need to prove that
\[\sum{\frac{x(2xy+4yz+y^2+z^2)}{2(x+z)(y+z)}}\le{x+y+z}\]
\[<=>x+y+z-\sum{\frac{x(2xy+4yz+y^2+z^2)}{2(x+z)(y+z)}}=\frac{\sum{zx(x-y)^2}}{2(x+y)(y+z)(z+x)}\ge{0}\]

[quote=sasukerikudo]toanhocmuonm where did you learn schur's inequality?[/quote]

Though I know the question is not directed to me,
anyways,
Secrets in Inequalities by Pham Kim Hung is a wonderful read.

[quote=Ji Chen]\[\sum_{cyc}\frac{xy}{\sqrt{xy+yz}}\leq\sum_{cyc}\frac{3x^{2}+2y^{2}+3z^{2}-4yz+20xy}{8\sqrt{2}(x+y+z)}\][/quote]
Can anyone tell me why it is true?

[quote=arqady]Use BW. You'll get something obvious.
For example, for $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$ we need to prove that:
$$24(4u^2-12uv+11v^2)x^3+4(8u^3-16u^2v-8uv^2+43v^3)x^2+$$
$$+8(u^4+48u^3v-96u^2v^2+56uv^3+30v^4)x+(2u^2-4uv+3v^2)^2v\geq0,$$ which is obvious.[/quote]

Thanks  :)

\begin{align*} \text{Observe: } \\ \frac {xy}{\sqrt {xy + yz}} + \frac {yz}{\sqrt {yz + zx}} + \frac {zx}{\sqrt {zx + xy}} \leq \frac {\sqrt {2}}{2}\\ \iff \sum_{\text{cyc}} x\sqrt{\frac{y}{x+z}} \leq \frac{\sqrt{2}}{2}\\ \iff \sum_{\text{cyc}} \frac{x+y}{2} \sqrt{\frac{x^2y}{(x+z)(x+y)^2}} \leq \frac{\sqrt{2}}{2}. \\ \text{Weighted Jensen: }\\ \sum_{\text{cyc}} \frac{x+y}{2} \sqrt{\frac{4x^2y}{(x+z)(x+y)^2}} \\ \leq \sum_{\text{cyc}} \frac{2x^2y}{(x+y)(x+y)}.\\ \text{Notice: }\\ \sum_{\text{cyc}} \frac{x^2y}{(x+y)(x+z)} \leq \frac{1}{4} \\ \iff 4\sum_{\text{cyc}} x^2y(y+z) \leq (x+y+z) \prod_{\text{cyc}} (x+y). \\ \textbf{Q.E.D.} \end{align*}


A check is written for x dollars and y cents, both x and y two-digit numbers.
In error it is cashed for y dollars and x cents, the incorrect amount exceeding
the correct amount by 17.82 dollars. Then
(A) x cannot exceed 70
(B) y can equal 2x
(C) the amount of the check cannot be a multiple of 5
(D) the incorrect amount can be twice the correct amount
(E) the sum of the digits of the correct amount is divisible by 9

[hide=" :D "]let x=10a+b; y=10c+d
original value=1000a+100b+10c+d   cents
cashed value = 1000c+100d+10a+b cents
difference
990c+99d -990a-99b=1782
10c+d - (10a+b)= 18
y - x =18

(B) y can equal 2x [/hide]


linear algebra
matrix
function
inequalities
integration
college contests

a1.
let P and Q are square matrixes, such that P^k=Q^l=O(zaro matrix) for some naturals k and l.
a)If PQ=QP, prove K=I+aP+bQ is invertible.(a,b -any real numbers).
b)If PQ=/=QP is it true that K is invertible?

a2.
Find all continuous on [0,1] functions f, which are differentiable on (0,1) and f(0)<=2,f(1)>=1 and such that for every x from (0,1):f'(x)<=2f(x)+2x-5.

a3.
let f:R->R is continuously differentiable function, such that f(0)=1, f'(0)<0 and 0<=f(x)<1 for every x in (0,1].prove that lim_(n->infinity)(n*int_(from 0 to 1)(f(x)^n)dx)=-1/f'(0).

[quote="martin1985"]
[b]a2.[/b]
Find all continuous on $[0,1]$ functions $f$, which are differentiable on $(0,1)$ and $f(0) \leq 2,f(1) \geq 1$ and such that for every $x \in (0,1) : f'(x) \leq 2f(x)+2x-5$ . [/quote]

Consider the function $g(x) = e^{-2x} \cdot (f(x)+x-2)$ .

It is $g(0) = f(0)-2 \leq 0$ and $g(1) = e^{-2} \cdot (f(1)-1) \geq 0$ .

Also $g'(x) = e^{-2x} \cdot (f'(x) -2f(x)-2x+5) \leq 0 \Rightarrow g(x)$ is a decreasing function .

For $x \in (0,1)$ we have

$\\ g(x) \leq g(0) \Rightarrow g(x) \leq 0$ and $g(x) \geq g(1) \Rightarrow g(x) \geq 0 \Rightarrow g(x) =0$ .

By continuity $g(x)=0, \, \forall x \in [0,1]$ therefore $\boxed{f(x) = 2-x}, \, \forall x \in [0,1]$

 :)

yes, this is a solution.
for a1 it resembles so much to one of the problems on vietnamese college olympiad...;)

Pick $\epsilon >0$ such that $1+f'(0)\epsilon \in [0,1)$ and $\forall x \in [0,\epsilon], |f(x)| \leq \frac{1}{2} + 1+f'(0)x$ (the last condition can be satisfied by Lagrange's inequality, since our function is continuously differentiable). Note that we then also have, from the first condition, that $\forall x \in [0,\epsilon], 0 \leq \frac{1}{2} + 1+f'(0)x$. Now we can write: $n\int_{0}^{1}{f^{n}(x)dx}= n\int_{0}^{\epsilon}{f^{n}(x)dx}+n\int_{\epsilon}^{1}{f^{n}(x)dx} \leq n \int_{0}^{\epsilon}{(\frac{1}{2} + 1+f'(0)x)^{n}dx}+n\int_{\epsilon}^{1}{K^{n}(x)dx}$, where $K$ is the least upper bound of $f$ on $[\epsilon,1]$, and, using the convexity of $x \mapsto x^{n}$ on $[0,1]$, the last expression is no more than $n(\frac{1}{2})^{n}+n \int_{0}^{\epsilon}{(1+f'(0)x)^{n}dx}+n\int_{\epsilon}^{1}{K^{n}(x)dx}$. The first and the third members clearly converge to $0$ as $n \rightarrow +\infty$, and the second member equals $\frac{n(1+f'(0)\epsilon)^{n})}{(n+1)f'(0)}-\frac{n}{(n+1)f'(0)}$, which goes to $-\frac{1}{f'(0)}$ as $n \rightarrow +\infty$.


LaTeX
geometry
3D geometry
pyramid
blogs

********************************************************************************Okay, I am beginning a new game. In this game you must submit a post after this one that has at least 20 words in it, then I will let the first twenty to post after me move on to the next round. When I submit a post with the little stars at the top, that means you must submit a post with at least 20 words, before everybody else does. Once we have twenty contestants, I will make another post with stars at the top at some random time of a random day. The first nineteen to submit a post after that post will move on. This will keep on going on and on, until there is only one person left. This person will be the winner. The object of the game is to submit a post as fast as possible starting now!!!

Okay, I guess I will start.  I don't really see the purpose of this at all.  Now I'm just typing to get to that twenty word requirement.

I'm joining.  BTW, do contractions count as one or two words? I love contractions.  Does $ \text{\LaTeX}$ count?  I think that's twenty, but I'm typing more to be sure.

Lalalalalala
I'm so happy!

I'll join i guess

blah blah blah fill up space blah blah blah fill up space blah blah blah fill up space

i think thats 20. can someone coutn for me?

Okay, so far we have three people, and we need seventeen more to post.

I'll join. By the way, does ANYTHING in $ \text{\LaTeX}$ count? I'm just asking. We need 16 more people to join.(random)
just some more...blah blah blah

this is a weird post. i guess i'll put my ctrl+ v I'm Not obsessed with humans. And I would never eat that person. I
love that person very much. ok that was weird. i think this is 20 words. i'm really not sure. yeah it is :D

20 words is a lot. I normally only use 2-5 words per post.
Thinking about what to type is annoying. Counting the number of words you have posted is also annoying. I hope this is 20.

Okay so far these are the six people that are in the competition;
1. JRav
2. AIME15
3. mz94
4. MathTwo
5. guineapiggies
6. budi713

Only 14 more.

This is just spam. People *cough*mz94*cough* have posted "blah blah blah" just to join, and they pressed enter, making the posts longer. So now people will have to scroll down even more just to get to the current post.

I'm joining!  :lol: Sounds like a pretty wierd game though.

Is there some sort of prize involved?  The glory of winning DOES NOT count as a prize.

OH 28 WORDS!  :winner_first:

ok im joining. Although this sounds weird, but it seems like this will stay at the top if everyone is rushing to post.  :lol: yay 23 words!

[quote="Wickedestjr"]Okay, thanks 1=2, that was at least twenty words, now there are seven in the competition so far, and now there are only seventeen more that need to join.[/quote]

20-7=17?!?!?!?!?!

[quote="Wickedestjr"]Okay, thanks 1=2, that was at least twenty words, now there are seven in the competition so far, and now there are only seventeen more that need to join.[/quote]

Who said that I joined???

Just popping in. Should start at 7 days max...

[hide="..."]7 days
6 days
5 days
4 days
3 days
2.5 days
2 days
1.5 days
1 day [24 hours]
20 hours
16 hours
12 hours
10 hours
8 hours
6 hours
5 hours
4 hours
3 hours
2 hours
1 hour
45 minutes
30 minutes
20 minutes
15 minutes
10 minutes
8 minutes
6 minutes
5 minutes
4 minutes
3 minutes
2 minutes
1 minute [STAY][/hide]

popping out

I'm going to shorten the amount of time as people being losing.

Can I join, I'll replace dreambold22. Please please please!!!!

#57splatyango: 1 point
dnkywin: 0 points
ernie: 0 points
Xantos C. Guin: 0 points
Bugi: 0 points
JRav: 0 points
Guinneapiggies: 0 points
mathblitz: 0 points
crazychinesemaniac: 0 points
budi713: 0 points
stevenmeow: 0 points
AIME15: 0 points
mz94: 0 points
MathTwo: 0 points
1=2: 0 points
squarerootofisoscelestria: 0 points
evilhamster: 0 points
BOGTRO: 0 points
myyellowducky82: 0 points
ra5249: 0 points
alingaling: 0 points

Sure whatever, okay dnkywin has a little over a week to post.

Should somebody tell dnkywin that he needs to post?

NO! don't do that, that's what the whole game is about. He has to notice that it is his turn on his own.

dnkywin only has until September 15th 12:16 forum time. Or he gets a point.

splatyango: 1 point
dnkywin: 1 points
ernie: 0 points
Xantos C. Guin: 0 points
Bugi: 0 points
JRav: 0 points
Guinneapiggies: 0 points
mathblitz: 0 points
crazychinesemaniac: 0 points
budi713: 0 points
stevenmeow: 0 points
AIME15: 0 points
mz94: 0 points
MathTwo: 0 points
1=2: 0 points
squarerootofisoscelestria: 0 points
evilhamster: 0 points
BOGTRO: 0 points
myyellowducky82: 0 points
ra5249: 0 points
alingaling: 0 points

ernie, you are the next chosen one, you have 14 days to post the name of another competitor and your reason why.

Er... rofl, Wickedestjr was talking to himself... THBAPSO Wickedestjr.  TY.  Wh00ps, wrong game.  :rotfl:

I choose mz94.  Haven't seen him in a while.

mz94 has until September 29th 12:47 to post the name of another user on the list and why.


probability
ceiling function
floor function
AMC

Ten women sit in $ 10$ seats in a line. All of the $ 10$ get up and then reseat themselves using all $ 10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?

$ \textbf{(A)}\ 89\qquad
\textbf{(B)}\ 90\qquad
\textbf{(C)}\ 120\qquad
\textbf{(D)}\ 2^{10}\qquad
\textbf{(E)}\ 2^2 3^8$

[hide="Solution"]
Let $ f(n)$ be the number of ways to seat $ n$ women at $ n$ seats in such a fashion.  We claim that $ f(n) \equal{} f(n\minus{}1)\plus{}f(n\minus{}2)$ for $ n>2$.
Consider the leftmost woman $ A$ in a row of $ n$ seats.  Either $ A$ goes to her original seat, or $ A$ goes to the seat on her right.  In the former case, we have $ n\minus{}1$ seats left, with $ n\minus{}1$ women left to fill them, so the number of ways in this case is $ f(n\minus{}1)$.  In the latter case, the woman $ B$ originally sitting to the right of $ A$ must switch seats with $ A$, because $ B$ is the only woman who can fill the leftmost spot.  Then we have $ n\minus{}2$ women left to arrange, which can be done in $ f(n\minus{}2)$ ways.
Since $ f(1) \equal{} 1$ and $ f(2) \equal{} 2$, we can simply calculate up to $ f(10)$ which turns out to be $ 89$.[/hide]

By the way, this is problem 9.12 in our [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]Intermediate Counting & Probability[/url] book.

Word-for-word? This is not good... unless its release date is after this contest.

Not word-for-word, but this is a "standard" Fibonacci numbers problem.  I'm sure this problem has appeared in hundreds of places.  Our book was published last year.

[hide="Solution"]
Combinatorially interpret a reseating of the $ 10$ women as a tiling of a $ 1\times10$ row by $ 1\times1$ and $ 1\times2$ tiles, which happens to be one of the most common combinatorial definitions of the Fibonacci numbers. Then notice that $ 89$ is the only Fibonacci number in the answer choices.[/hide]
:P

[hide="An alternate solution, and something cool"]
As Vworldv noted, this is the same as tiling a row of 10 squares with 1 and 2 square blocks.  Say that we decide to do this using $ k$ 2 square blocks.  Then we use $ 10\minus{}2k$ 1 square blocks.  The number of ways of placing these blocks is

$ \binom{10}{0}\plus{}\binom{9}{1}\plus{}\binom{8}{2}\plus{}\binom{7}{3}\plus{}\binom{6}{4}\plus{}\binom{5}{5}\equal{}89$.

Generalizing the problem to $ n$ women, the two methods combine to prove

$ \binom{n}{0}\plus{}\binom{n\minus{}1}{1}\plus{}\cdots \plus{} \binom{\lceil n/2\rceil}{\lfloor n/2 \rfloor}\equal{}F_{n\plus{}1}$

which is a nice result.
[/hide]

@E^(pi*i)=-1: This was the solution used by someone at my school. They did not realize that this generated Fibonacci however...I've seen this problem at least four or so times, and also the alternate solution above's result. Way too classic.

If the Fibonacci numbers have a combinatorial interpretation, then could there be combinatorial proof of binet's formula?  That would be awesome.

Euler221, I've been [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=257652]thinking about the same thing[/url].  I've discussed it with my professor, who tells me that there is indeed such a proof in the literature, but it is far from trivial.  The $ 2$ in the LHS refers to a coloring of the tiles (say, black and white) and the $ 5$ that appears in the RHS should refer to $ 5$ particular ways to fill pairs of consecutive tiles, but I haven't worked out the details.

A nice combinatorial proof of Binet's formula can be found in the book Proofs That Really Count, by Arthur T. Benjamin and Jennifer J. Quinn.


algebra
polynomial
function
algebra unsolved

find all  polynomial $P(x)$   satisfying  condition :
         $2+2P(x)=P(x-1) +P(x+1)$   $\forall x \in R$

Well...we can write our equation in a form \[ P(x+1)-P(x)=P(x)-P(x-1)+2 \] Consider a polynomial $Q(x)=P(x)-(x-1)$, we have then $Q(x)=Q(x-1)+2$ which means that $Q$ is a linear function, so $deg(Q)=1$, but as we know \[ deg(P(x)-P(x-1))=deg(P(x))-1 \], thus P(x) is a quadric function: $P(x)=ax^2+bx+c$. Bu putting this in our equation, we may easily obtain, that our polynomials are all polynomials of form $P(x)=x^2+bx+c$ .... well I think so... ;)


real analysis
function
trigonometry
probability
probability and stats

Can You give me an example of distribution which is not absolutely continuous in respect to Lebesgue measure, but $ |\phi (t)| \rightarrow 0$ when $ |t| \rightarrow \infty$ ,($ \phi (t)$ is a characteristic function of this distribution). Maybe it's not true and this distribution doesn't exist? What do you think?

You do have to avoid point masses, so you're looking for a distribution which is continuous (no point masses) but not absolutely continuous. The classic example of that would have as its cumulative distribution function $ F(x)$ the Cantor-Lebesgue "Devil's staircase" function. What can you say about the characteristic function of that?

As a hint: suppose $ X_1$ and $ X_2$ are identical independent copies of the $ X$ I just described, and let $ Y\equal{}X_1\plus{}X_2.$ Can you show that $ Y$ has an absolutely continuous distribution? If you could show that, what would that tell you about the characteristic function of $ Y?$

Well, if $ Y\equal{}X_1 \plus{} X_2$ has an absolutely continuous distribution then its characteristic function $ |\phi _Y (t)| \rightarrow 0$ when $ |t| \rightarrow \infty$ and also a characteristic function of $ X$, $ |\phi _X (t)| \rightarrow 0$  , but how to show, that $ Y$ has an absolutely continuous distribution?  Can You help me?:maybe:
I also found a characteristic function of this distribution: $ e^{\frac{i t}{2}} \prod _{i\equal{}1}^{\infty } \cos \left(3^{\minus{}i} t\right)$, but I also have a little problem to show that it aproaches $ 0$  :mad:

Any ideash how to show that $ Y\equal{}X_1 \plus{} X_2$ has absolutely continuous distribution?  :blush: :/

There is no chance with the $ 1/3$ Cantor set: if you plug $ t \equal{} 2\pi 3^n$ in, you'll get a fairly large value of the characteristic function at that point. On the other hand, for most $ \lambda$ the $ \lambda$-Cantor set will have the desired property. ;) A good question is whether we can derive the existence of such measures from some general observations without any explicit constructions. An alternative explicit construction is to start with the Riesz product $ \prod_{k\ge 1}(1 \plus{} a_k\cos 3^k x)$ with $ 0 < a_k < 1$, $ a_k\to 0$. This product represents a singular measure if (and only if) $ \sum_k a_k^2 \equal{} \plus{} \infty$ (otherwise it is locally $ L^2$). Now, to get a finite measure on the real line, you just multiply this Riesz product (which is $ 2\pi$-periodic) by some positive Schwartz function.

:arrow: [b]Kent Merryfield[/b], could you explain how to show what you said about? I'm closer to your solution. I would be appreciate  :roll:  :D

I overstepped what I know with that remark. It is true that the algebraic sum of the Cantor set with itself contains an interval, but I don't know whether that probability distribution is absolutely continuous or not.

Hi,
As said by Zurraw for Cantor Devil's Staircase we have the following Characteristic ffunction :
$ f(t)\equal{}e^{it/2}Lim_{n \to \plus{}\infty}\Pi_{j\equal{}1}^{n}cos(\frac{t}{3^j})$

For a reference Lukacs' book "Chararectiristic Functions" page 25.

But Lukacs says nothing about behaviour at inifnity of this particular characteristic function.

Though he says that characteristic functions of singular distributions may not have limit at inifinity (page 27) but that limsup at inifinity is in the interval [0,1].

As Myth would say in such case, why is nobody reading what I post? :roll: For the standard $ 1/3$ Cantor set, the characteristic function [b]does not[/b] tend to $ 0$...

Yes you are absolutly right fedja sorry about this  :roll:

I've read your post too fast.


Next year I'll be a freshman going into highschool taking AP Stats. I am horrible at probability and counting. Will this course teach me to do AMC probability and counting problems?

No.
AP Statistics is a class about, well, statistics. I haven't taken it nor do I plan on ever taking it, but I am certain that probability/counting is not required and will be not be taught in the class, unless you want to consider finding a general pattern as "probability."
If you're looking to learn C&P on your own, there is a nice text in the [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts_FAQ.php]AoPS Bookstore[/url] (look at Introduction to Counting and Probability). There are also plenty of resources on the site here. There's the [url=http://www.artofproblemsolving.com/Forum/resources.php]Contests[/url] tab at the top of your page, and going to USA, you'll be able to access AMC 8,10, or 12 questions, some of which will of course including counting. Or there's the [url=http://www.artofproblemsolving.com/Wiki/index.php/Main_Page]AoPSWiki[/url], which has its own [url=http://www.artofproblemsolving.com/Wiki/index.php/Combinatorics]combinatorics[/url] section. Also, you can find problems in High School Basics (this forum) or Intermediate, and look at other questions, or post your own. There are plenty of people willing to spend a bit of their own time to help you...

Oh, and good luck with AP Stat. I hear it's a fair bit of writing/justifying things (like why you would pick a median instead of a mean to correctly represent an average) for a math course. And be careful not to fall into the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]Calculus Trap[/url]...

AP Stats should NOT be considered a math class. The subject material is very basic statistics with little to no coverage of the math behind it. In my high school, most people just take it as an easy AP Class.


[asy]size(300);
import geometry;
import markers;

pair A=(0, 0);
pair B=(8/sqrt(cos(80*pi/180)+(sin(80*pi/180))^2), 0);
pair C=(8/(1+tan(80*pi/180))*cos(80*pi/180), -8/(1+tan(80*pi/180))*sin(80*pi/180));
pair O=(4/sqrt(cos(80*pi/180)+(sin(80*pi/180))^2), 4/sqrt(cos(80*pi/180)+(sin(80*pi/180))^2));

draw(C--A--B--cycle);
draw(O--A,StickIntervalMarker(1,2));
draw(O--B,StickIntervalMarker(1,2));

draw("$10^\circ$",arc(A,B,C,2));

label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, S);
label("$O$", O, N);

perpendicular(O,SE,O--B);
perpendicular(C,NE,C--B);[/asy]

$ \triangle OAB$ is isosceles right triangle.
$ \triangle ABC$ is right triangle.
$ AC \plus{} BC \equal{} 8$.
$ \angle ABC \equal{} 10^\circ$.

Find the area of quadrangle $ OACB$.

[b]This problem has some solver. Are you found how many?[/b]

EDIT:$ ABCD \rightarrow OACB$

[hide]I assume you meant find the area of quadrangle OABC.

The area of the quadrangle is equal to:

$ \frac{1}{2} OA^2 \plus{} \frac{1}{2} (AC)(BC)$

Now:
$ AC\plus{}BC \equal{} 8$
$ AC^2 \plus{} BC^2 \plus{} 2(AC)(BC) \equal{} 64$

By pythagoras:
$ AC^2 \plus{} BC^2 \equal{} AB^2$
$ 2OA^2 \equal{} AB^2$
$ \therefore 2OA^2 \plus{} 2(AC)(BC) \equal{} 64$
So area is

$ \frac{1}{2}OA^2 \plus{} \frac{1}{2}(AC)(BC) \equal{} 64/4 \equal{} \boxed{16}$[/hide]

[quote="Machodog"]I assume you meant find the area of quadrangle OABC.[/quote]
Thank you Machodog :)

[hide="Another algebraic solution"]Let $ AC=a$ and $ BC=b$.
So, $ AB=\sqrt{a^2+b^2}$ and $ OA=OB=\sqrt{\frac{a^2+b^2}{2}}$.
Therefore,
\begin{eqnarray*}
\frac{OA \cdot OB}{2}+\frac{CA \cdot CB}{2}
& = & \frac{a^2+b^2}{4}+\frac{ab}{2} \\
& = & \frac{(a+b)^2}{4} \\
& = & \frac{8^2}{4} \\
& = & \frac{64}{4} \\
& = & \boxed{16}
\end{eqnarray*}

But, this solution isn't solved in the range of the elementary school.[/hide]


[quote="ac1072"]If$ \{a_{n}\}$satisfy$ \lim_{n\to\infty}(a_{n}\sum_{i=1}^{n}{a_{i}}^{2})=1$,show that${ \lim_{n\to\infty}\sqrt[3]{3n}a_{n}=1}$[/quote]

we use Stolz Theorem(0/0) sovle this problem

it is easy to prove that ${ \lim_{n\to\infty}a_{n}=0}$

so from  Stolz Theorem(0/0) we have
$ \lim_{n\to\infty}1/((3na_{n}^{3})=\lim_{n\to\infty}(1/a_{n+1}^{3}-1/a_{n}^{3})/3$

so we only need to prove that$ \lim_{n\to\infty}(1/a_{n+1}^{3}-1/a_{n}^{3})=3$(1)

from $ \lim_{n\to\infty}(a_{n}\sum_{i=1}^{n}{a_{i}}^{2})=1$
we know that $ 1/a_{n}=1/a_{n+1}-a_{n+1}^{2}$,
get this into (1),and notice ${ \lim_{n\to\infty}a_{n}=0}$
we get the problem solved


In each of a $ 5\times 5$ squares, one of the numbers 1, 2, 3, 4, 5 is inserted into each of it in such a way that each row, each column, and each two diagonals contains each of the five numbers once and only once. The sum of the numbers in the four squares immediately below the diagonal from the top left to the bottom right is called the $ score$. What is the highest possible score? justify your claim and construct the corresponding $ 5\times 5$ square

I think the highest score is 17 and my solution is in the file I have attached. :D


Prove that the sum of the square of ordered positive integer is not a quadratic number

or we can say

Let $n \in \mathbb{Z}^{+}$

proof that $(n-1)^{2}+(n-2)^{2}+n^{2}+(n+1)^{2}+(n+2)^{2}$
[b]is not [/b] a quadratic number

[hide]
The perfect squares alternate between being 0 and 1 $\pmod 4$.  So the five squares are 10101 or 01010 $\pmod 4$.  But those sum up to 3 and 2, respectively, $\pmod 4$, so their sum is not a square.
[/hide]

[hide="Alternatively, even if slightly less elegant..."]
Proof by contradition. Expanding the squares, we get that that sum, say S, is $S=5(n^{2}+2)$. Suppose $S$ is a square, then $n^{2}+2$ must be a multiple of 5.
Now, $(5k\pm 1)^{2}\equiv 1\pmod 5$, while $(5k\pm 2)^{2}\equiv-1\pmod 5$; obviously $(5k)^{2}\equiv 0\pmod 5$... Respectively, we can get $n^{2}+2\equiv 3,1,2\neq 0\pmod 5$, depending on which is the form of $n$, absurd: it must be the case that S isn't a square... q.e.d.  :wink:
[/hide]


linear algebra
matrix
algebra open
algebra

Hi

I am trying to solve this set of equations

2*x1       - x2    - x3       - x4 = 1
  -x1 - 2.5*x2    - x3       - x4 = 1
  -x1       - x2 - 3*x3       - x4 = 1
  -x1        -x2    - x3 - 3.5*x4 = 1

Now I know this can be solved if the variables x1, x2 etc are negative, but is there a way to find a solution so that they are positive?  If it helps the value that the equations are equal to (i.e. 1) can be modified if this helps to find a solution.

Thanks in advance for any help you can give me.

Jon

Double posting is discouraged. The right place for this question was Matrix Algebra, Computations and Tutorials. Your other version of this topic will probably be moved there.


1.  Out of a group of 30 people, what is the chance that at least 2 share the same birthday?

2.  There is a class of 35 students.  They are assigned to each write down a time on the clock ( eg. 12:59 or 2:37 ). What is the chance that at least 2 pick the same time if they all pick randomly? No AM or PM is required.

[quote="cipos2"]1.  Out of a group of 30 people, what is the chance that at least 2 share the same birthday?[/quote]

[hide="Hint."]
=====
Find the probability that all of them have different birthdays. For example, if there were only $ 3$ people, the first person would have some birthday, the second person would have a different birthday with probability $ \frac {364}{365}$, since there are only $ 364$ other valid days to choose from, and the third person would have a different birthday with probability $ \frac {363}{365}$. Multiplying these gives the probability that they have different birthdays, so subtracting this from $ 1$ gives the probability that at least two have the same birthday.
=====
[/hide]


function
trigonometry
calculus
derivative
calculus computations

The derivative is zero.
Hence $f$ is constant on $[0;+\infty[$, on $]-\infty;-1[$ and also on $]-1; 0]$.
For $x \geq 0$,
$f(x) = f(0) = \frac{\pi}4$.

The derivative is zero? Surely that would imply the function is either zero or a constant; which it isn't.  :blush:

Consider $\tan \frac{\pi}{4}= 1$:

$f(x) = \arctan (x)+\arctan \frac{\tan \frac{\pi}{4}-\tan (\arctan x)}{ 1+tan \frac{\pi}{4}\cdot \tan (\arctan x)}$
$= \arctan x+\arctan ( \tan ( \frac{\pi}{4}-\arctan{x})$

Hence

$-1 < x \Rightarrow \frac{\pi}{4}< \arctan x \Rightarrow f(x) = \arctan x+\frac{\pi}{4}-\arctan{x}= \frac{\pi}{4}$

$x <-1 \Rightarrow \arctan x < \frac{\pi}{4}\Rightarrow f(x) = \arctan x+\arctan ( \tan (-\frac{3 \pi }{4}-\arctan x) ) =-\frac{3\pi}{4}$

$f(-1)$ does not exist. Hence $f' \equiv 0$ does not imply that $f$ is a constant

[quote="coffeym"]If

$f(x)=\tan^{-1}{x}+\tan^{-1}{\left(\frac{1-x}{1+x}\right)}$ find $f\prime(x)$.

Hence find a simple expression for $f(x)$[/quote]

Perhaps you needn't do any calculus, we can apply the addition formula for arctan:

$\tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}$

for $-1<x,y<1$ (which is formula (35) on the [url=http://mathworld.wolfram.com/InverseTangent.html]mathworld inverse tangent page[/url])

now assume that $0<x<1\Rightarrow 0<\frac{1-x}{1+x}<1$ so that the above formula is valid and gives

$f(x)=\tan^{-1}{x}+\tan^{-1}{\left(\frac{1-x}{1+x}\right)}= \tan^{-1}{\left(\frac{x+\frac{1-x}{1+x}}{1-x\frac{1-x}{1+x}}\right)}= \tan^{-1}{1}= \frac{\pi}{4}$

I need to do more work on it to say more, but maybe one of you will make use of my post...

   -Ben

[color=red]EDIT:[/color]

I did a plot of the original function at http://www.quickmath.com [they use webmathematica], here it is:

[url=http://allyoucanupload.webshots.com/v/2003426514965747786][img]http://aycu35.webshots.com/image/5274/2003426514965747786_th.jpg[/img][/url]

Looks just like Pi-J said...


I have been asked to give an example of a function that is continuous but does not attain a maximum or minimum value on (0,2). If that's the case, would a function that is continuous on [0,2] that has extrema at 0 and 2 work?

For example, would f: [0,2] ---> R where f(x) = x work?

Thanks for your help.

Oops I meant f: (0,2) --> R but you probably knew what I meant.

But it makes all the difference... over [0,2], the identity reaches a maximum value at x=2, but over (0,2), there is no such maximum value; we can get arbitrarily close to 2, but never reach a maximum value.

Let $ [a,b]$ and $ [c,d]$ be intervals in $ \mathbb{R}$. We can define a bijective function $ [a,b] \rightarrow [c,d]$. Indeed we can have have a bijective map to $ A$ iff $ f(x): \mathbb{R} \rightarrow A$ is continuous and monotonic.


It's not a complete solution but it reduces the problem to a much easier one. We have (n-2)(n-1)n(n+1)(n+2)=n(n^4-5n^2+4). Because of gcd(n,n^4-5n^2+4)|4 you have n^4-5n^2+4=k^2 or n^4-5n^2+4=2k^2 for some k. In the first case you have (n^2-3)^2<k^2<(n^2-2)^2 for n>2 and the cases for n<=2 you can check quickly. n^4-5n^2+4=2k^2 is not so easy but you have 8k^2+9 as discriminant of the quadratic equation in n^2. Now it must be shown that 8k^2+9=t^2 has no solutions.  It must be possible to show it with congruences and I'll try it tomorrow.

[quote="Misha123"]I just realized that my solution was a dead end. The last condition is not sharp enough :( , e.g. there are solutions for 8a^2+9=b^2.

Orl, what do you mean with symmetric substitution in this concrete case?[/quote]

Try to apply the subsitution $z = n-2.$ Now you have a symmetric expression. You can apply the binomial theorem afterwards.  :)

What are the restraints?

n = 0 and x = 0 is a solution, so by "numbers" do you mean positive integers?

[quote="blahblahblah"][quote="vinoth_90_2004"]prove the product of five consecutive numbers cannot be a perfect square[/quote]

This is an IMO 1983 Longlist problem, by the way.[/quote]

Hmm. I have seen the problems in so many sources. BTW which ILL1983 problem is it ? Which country proposed it ? I guess the problems has been used many times before 1983. Hence it was not included in the ISL.  :)

Hi everyone!

I can't understand Orl's hint.

There is number from n, n+1, n+2, n+3 and n+4 which are coprime with other numbers. Otherwise each of the numbers should be divisible by 2 or 3. It is clear there are at most two numbers, which are divisible by 3, and at most three numbers which are divisible by 2. It follows that thre are exactly 2 numbers which are divisible by 3, but in this case one of this numbers is divisible by 2, it implies that we have at most 4 numbers which are divisible by 2 or 3 -- contradiction.
Hence one of numbers n, n+1, ..., n+4 is a perfect square and we need to analyze five equations:
n(n+1)(n+2)(n+3)=a2
n(n+1)(n+2)(n+4)=a2
n(n+1)(n+3)(n+4)=a2
n(n+2)(n+3)(n+4)=a2
(n+1)(n+2)(n+3)(n+4)=a2

We have
1) (n2+3n)2<n(n+1)(n+2)(n+3)<(n2+3n+1)2
2) (n2+4n+1)2<n(n+1)(n+3)(n+4)<(n2+4n+2)2
3) (n2+5n+4)2<(n+1)(n+2)(n+3)(n+4)<(n2+5n+5)2
4) Suppose n(n+1)(n+2)(n+4)=a2 and n+3=b2. If n is odd then it it necessery n=4k+3 --> n+3=4(k+1)+2 - is not a perfect square. So n is even, but in this case
(n2+7n/2)2<n(n+1)(n+2)(n+4)<(n2+7n/2+1)2
5) Suppose n(n+2)(n+3)(n+4)=a2 and n+1=b2. If n is odd then it it necessery n=4k+1 --> n+1=4k+2 - is not a perfect square. So n is even, but in this case
(n2+9n/2+2)2<n(n+1)(n+2)(n+4)<(n2+9n/2+3)2

hey myth!

the numbers are only supposed to be squares, not to be fourth powers!

however, a new approach:
$(n-2)(n+2)=n^2-4$, $(n-1)(n+1)=n^2-1$. we have $gcd(n,n^2-1)=1$, $gcd(n,n^2-4)|4$, $gcd(n^2-1,n^2-4)|3$. what do we make of this? first, $n$ has to be either a $k^2$ or $2k^2$. in the first case, either $n^2-1=l^2$, $n^2-4=m^2$, contradiction, or $n^2-1=3l^2, n^2-4=3m^2$, contradiction(just look at the differences of both equations in each case). thus $n=2k^2$. $n^2-1$ is odd, thus either $n^2-1=3l^2$, $n^2-4=6m^2$ or $n^2-1=l^2$, $n^2-4=2m^2$. the second case is impossible, since $(n-l)(n+l)=1$. thus $n^2-1=3l^2$, $n^2-4=6m^2$.
the first equality gives $(2k^2-1)(2k^2+1)=3l^2$. since $gcd(2k^2-1,2k^2+1)=1$, we have $2k^2-1=3p^2$, $2k^2+1=q^2$, contradiction(look at the difference mod 8), or $2k^2-1=p^2$, $2k^2+1=3q^2$.
the second equality gives $2(k^2-1)(k^2+1)=3m^2$, thus $m=2r$, consequently $(k^2-1)(k^2+1)=6r^2$. $gcd(k^2-1,k^2+1)=2$. $k^2+1$ is not divisible by 3 and 4, consequently $k^2+1=2a^2$ or $k^2+1=a^2$. in the first case $(2a)^2=2k^2+2=2k^2-1+3=p^2+3$, contradiction. thus $k^2+1=a^2$, $k^2-1=6b^2$. then $3(2b)^2=2k^2-2=2k^2+1-3=3q^2-3$, contradiction.

i guess it's possible shorter, however.

Peter

[quote="Peter Scholze"]hey myth!
the numbers are only supposed to be squares, not to be fourth powers!
[/quote]
You are too late. I corrected my solution before you post this message.

[quote="Peter Scholze"]
i guess it's possible shorter, however.
[/quote]
i guess both solution are not too large. ;)

[quote="orl"]The problem is really folklore. Try a symmetric substitution.  :)[/quote]

Sorry, I am not at all sure about this problem, but are you sure that you have not confused it with another one? Cause I think that symmetric substitution helps on a very similar, but much easier problem: Prove that the product of [b]four[/b] consecutive natural numbers cannot be a square.

  Darij

Hi,

I was trying the problem, just before Darij posted. I was trying to solve this problem in a similar way as the problem of 'product of four consecutive integers cannot be a perfect square.' But, I am stuck at some point. Could anyone help?

Assume N = n(n+1)(n+2)(n+3)(n+4) is a perfect square.
Since (n+1)(n+2)(n+3)(n+4) cannot be a perfect square, so n isn't a perfect square.
Now,
N = n(n+1)(n+2)(n+3)(n+4) = n (n^2+5n+4) (n^2+5n+6)
Let n^2 + 5n = A
Then N = n (A+4) (A+6) = n (A^2 + 10A + 24)
So any prime p dividing n has to divide (A^2 + 10A + 24). But since p divides n, and n divides A, so p should divide 24. That means p = 2 or 3.

So, n = 2^a * 3^b
I guess 'n' cannot be only 2^a or only 3^b, as in that case, one can argue that N won't be a perfect square.

But, then what?

Kindly help if possible.

***************************
Sambit

I don't know why you're all making it that complicated... :D

Suppose $n=0 \bmod 4$: after division by $4^2$, we get $2 \bmod 4$ is not a perfect square.
Suppose $n=1 \bmod 4$: after division by $2^2$, we get $2 \bmod 4$ is not a perfect square.
Suppose $n=2 \bmod 4$: after division by $4^2$, we get $3 \bmod 4$ is not a perfect square.
Suppose $n=3 \bmod 4$: after division by $2^2$, we get $2 \bmod 4$ is not a perfect square.

(please note: $2\cdot3=2 \bmod 4$ and if $m^2\cdot n$ is a perfect square then so is $n$.)

Hi PeterVDD,

Sorry if this is trivial, but I don't get your solution.

Lets say n=0 mod 4. Then say n=4k. So after you remove 4^2 from the product, you have
k (4k+1) (4k +2) (4k+3) (k+1). Now if k=1 mod 4, then this product is 1*1*2*3*2 = 0 mod 4, and hence could be a perfect square.

Kindly clarify this.
Thanks
Sambit

[quote="darij grinberg"]
Prove that the product of [b]four[/b] consecutive natural numbers cannot be a square. [/quote]

Sorry, I got confused as Darij says. I thought the problem is really easy. BTW what about the general problem with $n$ consecutive numbers ?

Erds and Selfridge have proved in a famous paper that the product of $n \geq 2$ consecutive integers is never a power, and therefore not a square either.

Pierre.

Ok sambit, here goes my corrected attempt: :D

$A=n(n+1)(n+2)(n+3)(n+4)$
As soon as one contains a prime factor above $3$, that factor can only appear in one of the numbers so A can't be a perfect square. Either $n+1$ and $n+3$, or $n,n+2,n+4$ is odd. And since at most one of them can be a multiple of $3$, we have a contradiction.

PS: [quote="sambitnayak"]Then N = n (A+4) (A+6) = n (A^2 + 10A + 24)
So any prime p dividing n has to divide (A^2 + 10A + 24). But since p divides n, and n divides A, so p should divide 24. That means p = 2 or 3.[/quote]Here you're a little wrong, sambitnayak, just a counterexample is: 5.7 * 7 is a perfect square (35), while the prime 5 divides 5.7 but not 7. So there you got a little flaw. :)

[quote="Peter VDD"]
$A=n(n+1)(n+2)(n+3)(n+4)$
As soon as one contains a prime factor above $3$, that factor can only appear in one of the numbers so A can't be a perfect square. Either $n+1$ and $n+3$, or $n,n+2,n+4$ is odd. And since at most one of them can be a multiple of $3$, we have a contradiction.[/quote]
???

[quote="Peter VDD"]Ok sambit, here goes my corrected attempt: the solution is even simpler than what I initially said :D

$A=n(n+1)(n+2)(n+3)(n+4)$
As soon as one contains a prime factor above $3$, that factor can only appear in one of the numbers so A can't be a perfect square. Either $n+1$ and $n+3$, or $n,n+2,n+4$ is odd. And since at most one of them can be a multiple of $3$, we have a contradiction.
[/quote]

Um, what? I have no clue what you have done here.. for starters, why can't that multiple of 3 be a multiple of 9 also? The problem isn't *that* easy :P

TripleM, I think you didn't understand my solution very well (or I explained it bad :P).

Let me explain it more carefully :)

$A=(n)(n+1)(n+2)(n+3)(n+4)$

Suppose any of these has a factor above $5$:
Each digit is either of the form $2^n.3^m.(k^2)$ [i](with k integer)[/i]  -- or -- it is not. The first case has been proven impossible in previous posts, the second one cannot result in a perfect square either, since no other number in $\{n,...n,+4\}$ can contain that number as a factor. [i](like no 5-string around 35 has multiples of 5 or 7)[/i]

So that means each of $\{n,...,n+4\}$ would consist of only factors $2$ and $3$. Now here I'm not claiming that it's impossible for those to make a perfect square, I'm claiming that there are no $5$ such consequent integers. For the reason showed above, that at least $2$ are odd and that only $1$ of the odd ones can divide $3$, so we have at least $1$ number that has got other prime factors than just $3$ and $2$.

I hope this has been clear.

Peter

(Solution inspired by a  famous hindi song  which goes "kabootar kabootar ")...

Get five kabootars  and tie  one number to each of them. (n, n+1, n+2, n+3, n+4)
Get  4 kabootar-pijaras and number them   "1", "2",,"3", "6"

Now each kabootar would know that his number  can be written as   a*(x^2) where  'a'  is sqaure free. furthermore, as pointed out before ( none of the  numbers have a common prime factor greater than 3)  "a" must be one of the kabootar-pijara number.

Let every kabootar fly and get to the corrrosponding pijara.

4 kabootar-pijara's, 5 kabootars .. at least one pijara would contain two (or more) kabootars...

For n>1 , this is obviously a contradiction ( difference a*(x_1^2-x_2^2) would be greater than 4). (one can easily check n=1 case_.

QED!
P.S.  (The above solution will work even if you do not know that kabootar is pigeon in hindi and even if kabootars can't do math)

PPS - Using similar logic you can prove  the more general case of   any number (say k)  of consecutive integers.  (With somve work one can prove that all a's would be distinct..etc) ( Narumi proved that this is true   for  k's less than 200 and Erdos proved for all the numbers .. in fact in 1975 he proved, as pointed out   by pbornsztein above,  that the product is not equal to any power too.


Let be $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} \frac {1}{2}$ . Show that :

$ (6a \plus{} 1)(12b \plus{} 2)(6c \plus{} 1) \minus{} (3a \plus{} 3b \plus{} 1)(3b \plus{} 3c \plus{} 1)(3c \plus{} 3a \plus{} 1)\le 8$

When $ a \equal{} b \equal{} c \equal{} 1/6$, the expression is equal to 56.

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} \frac {1}{2}$ . Show that :

$ (12a \plus{} 2)(12b \plus{} 2)(12c \plus{} 2) \minus{} (3a \plus{} 3b \plus{} 1)(3b \plus{} 3c \plus{} 1)(3c \plus{} 3a \plus{} 1)\le 8$[/quote]


limit
calculus
inequalities
trigonometry
derivative
calculus computations

This is very well-known to equal 1.  [url=http://planetmath.org/encyclopedia/LimitOfDisplaystyleFracsinXxAsXApproaches0.html]Here's[/url] a proof.

hello, the answer is $ 1$, see also here
http://pup.princeton.edu/books/maor/chapter_10.pdf
Sonnhard.

[quote="Kono"]$ \lim_{x \to 0} \frac {\sin(x)}{x}$[/quote]

[hide="If you know a series for sin(x)"]

Let $ n$ be a positive integer.

$ \lim_{x \to 0}[sin(x)]/x \equal{}$

$ \lim_{x \to 0}[x \minus{} \frac16 x^3 \plus{} ...\plus{} \frac{(\minus{}1)^{n \plus{} 1}x^{2n \minus{} 1}}{(2n \minus{} 1)!} \plus{} ...]/x \equal{}$

$ \lim_{x \to 0}(1 \minus{} \frac16 x^2 \plus{} ...) \equal{}$

$ 1$

[/hide]

What does a beginning calculus student need to know? I'll claim that such a student needs to know this specific limit. Not a derivation of it, although you should at some point see some derivation (perhaps a geometric argument with inequalities and the squeeze theorem). But once you have that - you just have to know this one. And know it in several forms, such as "$ \sin x$ looks like $ x$ as $ x\to 0.$"

And it's not that I'm going to ask you to "just know" dozens of limits - no, it's not that at all. This one is special.

This is the formula.
\[ \boxed{\boxed{\lim_{\theta \rightarrow 0} \frac {\sin \theta}{\theta} = 1}}\]
If you know the derivation, then you study it in your text book.

As Kent Merryfield said, a geometric argument with inequalities and the squeeze theorem:

In deatil, $ \text{[Isosceles Triangle]} < \text{ [Sector] } < \text{ [Right angled Triangle]}$ gives $ \boxed{\boxed{\sin x < x < \tan x}}$ for ${ 0 < x < \frac {\pi}{2}}$.

Note that we may not prove the inequality in using differentiation. Because doing so will become circular reasoning.

The proof of $ \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1$ comes up in Japanese university entrance exam, in particulr, in medical every year.

As kunny said, one typically uses the above identity to find the derivative of sin.  In similar spirit to the identity you asked for, there's $ \boxed{\boxed{\lim_{\theta\to 0}\frac{\sin{a\theta}}{b\theta}\equal{}\frac{a}{b}}}$.


[quote="limes123"]For $ a,b,c > 0$ prove
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {27}{2}\geq 4(a \plus{} b \plus{} c) \plus{} \frac {1}{\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}}$.[/quote]
we have: $ \frac{1}{\frac{1}{a^2}}\plus{}\frac{1}{\frac{1}{b^2}}\plus{}\frac{1}{\frac{1}{c^2}} \geq \frac{9}{\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}}$
hence, $ \frac{1}{9} \left(a^2\plus{}b^2\plus{}c^2 \right) \geq \frac{1}{\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}}$

the other hand, we have:
$ \frac{8}{9}a^2\plus{}\frac{9}{2} \geq 4a$
Similar, $ \frac{8}{9}b^2\plus{}\frac{9}{2} \geq 4b$ and $ \frac{8}{9}c^2\plus{}\frac{9}{2} \geq 4c$
the problem is solved


geometry
incenter
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry unsolved

However, I didn't find it in the forum.  :roll:

In $ \triangle ABC$, the perpendicular bisectors of $ BC, AB$ meet $ AB,BC$ at $ C_1,A_1$ respectively, $ B'$ is the incenter of $ APC$ where $ P$ is the intersection point of line $ AA_1$ and $ CC_1$. Prove that $ O,I,B'$ is collinear where $ O,I$ be the circumcenter and the incenter of $ \triangle ABC$ respectively.

[quote="JG"]Hey, that's wrong...try cabri geometry and see it for yourself![/quote]

Very sorry for a typing mistake  :blush:
Now It's corrected.

The perpendicular bisector $ OC_1$ of $ BC$ cuts the circumcircle $ (O)$ at $ K, L,$ where $ K$ is on the opposite side of $ BC$ than $ A$. The perpendicular bisector $ OA_1$ of $ AB$ cuts the circumcircle $ (O)$ at $ M, N,$ where $ M$ on the opposite side of $ AB$ than $ C$. The internal bisectors $ AI, CI$ of the $ \angle CAB, \angle BCA$ pass through $ K, M,$ respectively. $ CC_1$ is a reflection of $ AB$ in $ OC_1$ and $ AA_1$ is a reflection of $ BC$ in $ OA_1.$ Therefore, the internal bisectors $ AB', CB'$ of the $ \angle CAP, \angle PCA$ pass through $ N, L.$ The intersections $ O \equiv KL \cap MN,$ $ I \equiv AK \cap CM,$ $ B' \equiv AN \cap CL$ are collinear by Pascal theorem for the cyclic hexagon $ ALMKNC.$

Wow so nice a proof!
Thanks a lot  :lol:

Posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=17766 .

  darij


[quote="djimenez"]There is one well known, I learned on middle school, but I don't think that you would be able to prove it without some knowledge of series.

Best regards,

David[/quote]

well... what is it?

[quote="jli"]come up with a generic way of converting repeating decimals into fractions

for example...

$.174747474....$

or

$.1234{567}567567.....$[/quote]
Is this question for Beginners or that you need help converting decimals to fractions?  :?

[quote="jli"][quote="djimenez"]There is one well known, I learned on middle school, but I don't think that you would be able to prove it without some knowledge of series.[/quote]

well... what is it?[/quote]

I rather not to answer because I do not consider myself a begginer!

best,


Can you solve the problem for me
  Determine all functions $ f: R \to R$ such that : $ f(f(x) \minus{} f(y)) \equal{} (x \minus{} y)^2 .f(x \plus{} y)$

$ x \equal{} y$ gives $ f(0) \equal{} 0$ and $ y \equal{} 0$ gives $ fof(x) \equal{} x^2f(x)$
let $ A \equal{} \{x: \ f(x) \equal{} 0\}$; $ \forall y\in A: \ fof(x) \equal{} (x \minus{} y)^2f(x \plus{} y)$ then $ f(x \plus{} y) \equal{} \frac {x^2f(x)}{(x \minus{} y)^2}$

if $ A \equal{} \mathbb{R}$ then $ f\equiv 0$
if $ A\neq \mathbb{R}$ we take $ x$ such that $ f(x)\neq 0$, so $ \forall a,b\in A: \ f(x \plus{} a \plus{} b) \equal{} \frac {(x \plus{} a)^2f(x \plus{} a)}{(x \plus{} a \minus{} b)^2} \equal{} \frac {(x \plus{} a)^2x^2f(x)}{(x \plus{} a \minus{} b)^2(x \minus{} a)^2}$
and like the same $ f(x \plus{} a \plus{} b) \equal{} f(x \plus{} b \plus{} a) \equal{} \frac {(x \plus{} b)^2x^2f(x)}{(x \plus{} b \minus{} a)^2(x \minus{} b)^2}$ gives $ a \equal{} b$ then $ |A| \equal{} 1\to A \equal{} {0}$

easly $ f(x) \equal{} f( \minus{} x)$ because $ f(x) \minus{} f( \minus{} x)\in A \equal{} \{0\}$
$ f(x \plus{} y)(x \minus{} y)^2 \equal{} f(f(x) \minus{} f(y)) \equal{} f(f(x) \minus{} f( \minus{} y)) \equal{} f(x \minus{} y)(x \plus{} y)^2$, so $ \forall,x,y: \ \frac {f(x \plus{} y)}{(x \plus{} y)^2} \equal{} \frac {f(x \minus{} y)}{(x \minus{} y)^2}$
that gives $ f(X) \equal{} aX^2$ for some $ a\in R$
and with remplacing in the fonctionel equation we find that a=1 or -1

$ f(x) \equal{} x^2\ or \minus{} x^2\ or\ 0$


Just asking, do you guys have any methods of squaring numbers very fast? Thanks!  :)

if it ends in a 5, or its in the form $ a5$, its $ (a\cdot (a\plus{}1))25$. for example, $ 65^2\equal{}4225$, because $ 42\equal{}6\cdot7$, and you just attach a $ 25$.

Memorize them. Even with 5's, it's faster than nikeballa's way. (no offense meant)

It took a dummy like me 3 days to mem them from 1-25. And for ten's; they're so easy you need no explanation.

AIME15's way for squaring numbers is the way I was taught at school.

In the heat of countdown though; my way can get you a few more questions correct.  :)

If you don't care about countdown; why waste time on memming stuff when you can practice other concepts?  :D

But memming squares can help you do better on Sprint (won't help on Target and Team though, go calcs!) though...

Dang It! I always think of a contradiction to my statements...

[b]CONTRADICTION --- EDITED[/b][/b]

speed is of importance when it comes to mathcounts... That's why I decided long ago to mem the squares of #1-30.

oh... You have a small grammar mistake... shouldn't it be [b]a contradiction, not an contradiction?[/b] :wink:

Another hint:

for example 16^2= 15^2 + 15 + 16

This works for all numbers :)

snowhawk's trick is quite useful. say someone asked you what $ 91^2$ was. Using his way, we could just say that it is $ 90^2$ which is $ 8100$ plus $ 90 \plus{} 91$ which is $ 8281$. Another example is $ 41^2$. its just $ 1600 \plus{} (40 \plus{} 41) \equal{} 1681$.

WHY this works is that the numbers are in form $ a$ and $ a \plus{} 1$. Lets call the $ a \plus{} 1$ $ b$ for now. Call $ b$ as 41 and $ a$ as 40. so you have $ b^2$ minus $ a^2$ which can be put in form $ (b \plus{} a)(b \minus{} a)$ or, since $ b \minus{} a$ is just $ 1$, we have $ b \plus{} a$ left. What this says is that $ (41^2) \minus{} (40^2)$ is just $ 40 \plus{} 41$. So, we are just really adding $ 40 \plus{} 41$ from $ 40^2$ to get $ 41^2$.

REMEMBER, this only works for consecutive numbers.

THANK YOU EVERYONE!!!!!!!!!! i might go with aime's and snowhawk's idea.:lol:


MATHCOUNTS
AMC
AMC 8
AMC 10
AMC 12
AIME
USA(J)MO

I've participated in Mathcounts and Science bowl (It has math questions too)

I've done Mathcounts too and am planning to do MC again and AMC8 this year.

This should be moved to the Middle School Competitions Forum...........

Just see my signature. I have previously done MOEMS and MathCounts. too old now

yay science bowl is amazing. (we got 4th at nats this year) the math questions are insanely easy though......

WOW we got 30the or something in state Gosh how do you do it?

i guess practice, talent, teamwork?

i covered the math at school and regional, but at nats, there were top mc kids, so i was toast..

Yay!
I was on ur team pink!
Midnight, what state r u in, if ur in like northern california, then a relatively low ranking isnt too bad.
This year we got kinda lucky, we were like 3rd seed in round robin and we beat like 2 1st seeds cuz one team made a buncha stupid mistakes (madison park) and the other got a whole lot better after we beat them.

Also skill talent practice and teamwork.

How do you register for the science bowl? How can I get more info?

I have been in Math Kangaroo (twice, eleventh in America), Purple Comet (alone, 6=fail), MC (5th at state), AMC8 (20, first in state), and some local competitions in which I came first.

My goals for this coming school year:
MC - Top 150 Nats
AMC8 - 25
AMC10 - 130
AMC12 - Won't take it
AIME - 3
USAJMO - Won't make it
Math Kangaroo - Help my dad set up the first place in my state to take this test, Top 5 in America
Purple Comet - 10 (alone)

For science bowl, talk to your school science department, there's pretty much at least one site for a regional science bowl in every state(there is no state science bowl).
Also, online, i think there is a full list for all the locations that hold science bowl competitions. You register through your school

I've done MathCounts and I did MOEMS in elementary school- my district doesn't offer it in middle school  :( I also did this Lehigh University High School Math Contest this year.

my goals for upcoming competitions that most AoPSers participate in:

AMC8 : 25
AMC10 : 140
AMC12 : 120+
AIME : 8+
USAMO(hopefully): 5
Mathcounts : make top 10 in state(difficult state)

of those goals I think I can achieve #1, #3, and #4.

[quote="batteredbutnotdefeated"]
Mathcounts : make top 10 in state(difficult state)

of those goals I think I can achieve #1, #3, and #4.[/quote]
Battteredbutnotdefeated, don't aim for top 10 in state.  Aim to make nats!   There's no reason not to aim too high........  Even if you don't make your goal, your better off setting the bar high because you'll probably perform better.  And do you live in Virginia?  I've seen you post in that forum, but I'm not sure....

As for me, I'm trying for the ultimate nats goal(unlikely, but still) in Indiana.  I kinda wish I lived in an easier state like Oklahoma or Minnesota...............
But with practice, patience, and a lot of luck, it can happen. :D

[quote="PowerOfPi"]You won't make USAMO with a score under 11... (USAJMO)

Does it have a website? Are you sure there is one in ND?[/quote]

The score of 11 is only applicable if you are taking the AMC 10. Otherwise, if you are taking the AMC 12, it is going to be the usual index way.

[quote="PowerOfPi"]If you set your goals so high that you don't make it, you won't feel good later. You have to set your goals just a bit above what you think will happen.[/quote]

Well, you should aim high, and then have a "more reasonable" goal too.

For me(maybach)
Stretch Goal:  Make Nats!
Reasonable Goal: Win the $4,000/year scholarship by placing in the top 10
Easy Goal: Make State countdown

Obviously, make your top goal high, and then you can still feel good if you make one of the easier goals.

What's MOP?  And what's with the colors I keep seeing about MOP (like red, black, etc.)?

also you can see [url=http://uncyclopedia.wikia.com/wiki/MOP]here[/url]

[quote="Berzerk"]I did not do any competition.[/quote]

Do some!! they'll help and they're fun

@ Cgayo:  I didn't know you could do MC in 6th grade!

And if your school doesn't host contests, then take the practice tests from the Mathcounts website, as well as say Purple Comet, Gauss 8 etc.  You can learn a lot and have fun!

Don't worry. You can do AMC 10/12 when you're older right?

[quote="polkadotman1"]In my school, there isn't even a math count team :(:(

*sob*[/quote]

convince your principle, you have to request for it!

[quote="maybach"][quote="AwesomeToad"]Of course you can do MC in 6th grade! But for some schools, middle school starts in 7th grade,[/quote]

Wow.  Epic typo!  I meant 5th grade, since cgayo15 said he was going into 5th.[/quote]
Read the first sentence, which says "Students enrolled in the 6th, 7th or 8th grade are eligible to participate in MATHCOUNTS® competitions.", in the MathCounts eligibility rules [url=https://mathcounts.org/Page.aspx?pid=287]here[/url].

You CAN'T do MC in 5th.

[quote="The greatest person on the face of Earth"]
EDIT:I meant when i go to Clark :)
namely my middle school[/quote]
[/quote]

I've done 2 district wide math competitions;
nothing big :P


when written as a decimal number,2005^2005 has D digits and leading digit L. Find D+L

[hide]I cheated. Mathematica says the first digit is 5, and that there are 6621 digits in the number, so 5+6621=6626[/hide]

[hide="hint"]There are $ 1 \plus{} (2005\cdot{\log_{10}2005}) \equal{} 6621$digits. Hey, I only used my calculator for the last part, for finding out $ \log_{10}2005$.[/hide]
Anyone have a "good" solution?


$b, m, n$ are positive reals and $m>n$. Prove that
\[\frac{b^{n}-1}{nb^{n}}>\frac{b^{m}-1}{mb^{m}}. \]

[quote="A. S. K."]$b, m, n$ are positive reals and $m>n$. Prove that
\[\frac{b^{n}-1}{nb^{n}}<\frac{b^{m}-1}{mb^{m}}. \]
[/quote]
Let$f(x)=\frac{b^{x}-1}{xb^{x}}$$\forall{x>0}$
It 's easy to see that :f'(x)>0  $\rightarrow{OK}$


Given a righ triangle $ ABC$ with $ \angle B\equal{}90$. $ M$ is on $ AB$ such that $ AM\equal{}BC$ and $ N$ on $ BC$ such that $ CN\equal{}MB$. Find angle between $ AN$ and $ CM$


Un solitario se inicia con 25 cartas en fila. Algunas están boca arriba, y otras boca abajo.

En cada movimiento se debe elegir una carta que esté boca arriba, retirarla, y dar vuelta las cartas vecinas a la que se retiró (si las hay).

El solitario se gana cuando se logra, repitiendo este movimiento, retirar las 25 cartas de la mesa.

Si inicialmente hay n cartas boca arriba, halla todos los valores de n para los cuales se puede ganar el solitario. Explica cómo se gana, independientemente de la ubicación inicial de las cartas boca arriba, y justifica por qué es imposible ganar para los otros valores de n .

Dos cartas son vecinas cuando una está inmediatamente al lado de otra, a la derecha o a la izquierda.

Por ejemplo: la carta marcada con A tiene dos cartas vecinas y la marcada con B una sola. Después de retirar una carta queda un hueco, de modo que la marcada con C tiene únicamente una carta vecina, y la marcada con D no tiene ninguna.

[img]http://www.oma.org.ar/enunciados/may15_clip_image002_0004.gif[/img]

[hide="mi respuesta"]
$ n$ tiene que ser un numero impar.
[/hide]

Yo tambien llegue a lo mismo, solo que me falto explicar bien como se gana independientemente de la posición, y en ese problema tuve 5 puntos de 10 v.v


limit
real analysis
real analysis unsolved

I don't know what to do next.
What test should I use to determine whether it converges or diverges?

(a) it is   obvious you have to use the AS test, i.e. use the fact a series given as
$ \sum_{n \equal{} 1}^{\infty} ( \minus{} 1)^n \; a_n$ where $ a_n\ge 0 \quad\forall n\ge 1$ converges  [b]iff[/b] $ \lim_{n\to\infty} a_n \equal{} 0 \qquad\text{and}$
$ \qquad a_n \quad\text{is monotonically decreasing, i.e.}\quad \dfrac{a_{n \plus{} 1}}{a_n} < 1$

once you use that, you should see that the given series converges. in fact, it is absolutely convergent.

(b) $ E_3\le \dfrac{1}{2 \plus{} \sqrt [4]{4}}$
(just put $ n \equal{} 4$ in $ a_n$)

(c) p is   $ \boxed{ \minus{} \dfrac{1}{3}}$ and q is   $ \boxed{ \minus{} \dfrac{1}{3} \plus{} \dfrac{1}{2 \plus{} \sqrt [4]{2}}}$

to misan.
Are in India such  name  as "Nasim"? What  does it mean and is it your name?

[quote="misan"]
$ \sum_{n \equal{} 1}^{\infty} ( \minus{} 1)^n \; a_n$ where $ a_n\ge 0 \quad\forall n\ge 1$ converges  [b]iff[/b] $ \lim_{n\to\infty} a_n \equal{} 0 \qquad\text{and}$
$ \qquad a_n \quad\text{is monotonically decreasing, i.e.}\quad \dfrac{a_{n \plus{} 1}}{a_n} < 1$
[/quote]
[quote]
once you use that, you should see that the given series converges. in fact, it is absolutely convergent.
[/quote]
Isn't it embarassing to display such a profound misunderstanding of this simple material while posing as a teacher of very smart high school students?

@JBL below: Of course, I am bad at writing correct things, but that isn't what it's about. Guess I'm already shitting up this thread, but seriously, on a less forgiving forum misan would have been outed as a massive liar a long time ago. It annoys me too much that he can continue to spew out tough identities that he found in papers/collections of curiosities, without providing solutions, and claiming that he is giving these as problems in a mystery unnamed school where he is a mystery teacher. misan is consistently making weird errors and claims, showing little understanding of second semester mathematics, and to me it's all too obvious that he is a sham. I suppose he's not  really a troll, but I still don't really find it acceptable; and while I suppose you will say my behaviour is not acceptable and dislike me, I'm almost certainly right, and I don't see how anyone can deny it. And by the way, this has nothing to do with thoughts that 7th graders couldn't be smart enough to understand university material.

Have you never written something wrong before?  I would have thought a correction would have sufficed in place of an off-topic slew of nastiness.

[quote="Kalle"][quote="misan"]
$ \sum_{n \equal{} 1}^{\infty} ( \minus{} 1)^n \; a_n$ where $ a_n\ge 0 \quad\forall n\ge 1$ converges  [b]iff[/b] $ \lim_{n\to\infty} a_n \equal{} 0 \qquad\text{and}$
$ \qquad a_n \quad\text{is monotonically decreasing, i.e.}\quad \dfrac{a_{n \plus{} 1}}{a_n} < 1$
[/quote]
[quote]
once you use that, you should see that the given series converges. in fact, it is absolutely convergent.
[/quote]
Isn't it embarassing to display such a profound misunderstanding of this simple material while posing as a teacher of very smart high school students?[/quote]

Your reaction is really embarassing.


Find all functions $f: \mathbb{N}_{0}\to\mathbb{N}_{0}$ such that this holds for any $m,n\in\mathbb{N}_{0}$:

$f[f^{2}(m)+f^{2}(n)]=m^{2}+n^{2}$

$f^{2}(x)$ is a square, not a composition.


I am currently thinking of conducting a research in sudoku (basically the whole thing, from generating unique sudoku boards to solving)
Does anyone know if there are any rigorous research that has been done before? If so, what are they mainly about?

I don't really see how you can do genuine research on sudoku.  But if you go to scholar.google.com or someplace similar, you can search for journals.

It's a combinatorial decision problem, which fits into theoretical computer science work on complexity. An MS student in our department just gave a presentation on this, and reported on previous work. It seems that the question of whether a Sudoku board (of arbitrary $ n^2\times n^2$ size) has more than one solution belongs to the class NP  Another Solution Problem - Complete.  This isn't quite the same as the more famous NP - Complete class of decision problems, but the arguments have a similar style.


invariant
vector
linear algebra
matrix
absolute value
linear algebra unsolved

$ W$ is a finite dim vector space, $ e_i \in Endomorphism(W)$ for $ i \equal{} 1,...,n$ , $ n$ is finite.
is it possible to define a inner product $ q$ invariant to all $ e_i$, i.e, $ q(w_1,w_2) \equal{} q(e_i \bullet w_1,e_i \bullet w_2)$ for all
$ w_1,w_2 \in W$ and all $ e_i$. If yes, does $ n$ have to be small?

What do you mean by inner product, and over which field are you working?
If you mean positive definite over $ \mathbb{R}$,
then $ e_1,\ldots,e_n$ would be [url=http://en.wikipedia.org/wiki/Orthogonal_matrix]orthogonal[/url] with respect to $ q$,
and therefore must have square-free minimalpolynomials and all eigenvalues have absolute value 1.
So clearly this doesn't work for any $ e_1,\ldots,e_n$.

If $ G$ is a finite group of endomorphisms of $ W$,
then any positive definite form $ \langle\cdot,\cdot\rangle$ gives a $ G$-invariant positive definite form $ q$ by
$ q(v,w)\equal{}\sum_{g\in G}\langle gv,gw\rangle$.


Find $ \lim_{m\to\infty} \sum_{n=1}^{m}\frac {C_{2n}^{n}}{(2n-1)4^n}$.

[i]Edited by Myth[/i]

I don't know how to solve this problem at the moment, but I can propose another one :)
Find $\lim_{m\to\infty} \sum_{n=1}^{m}\frac {C_{2n}^{n}}{(n+1)4^n}$.

 :D

:lol:  :lol:  :lol:
I think my problem can't be solved in the same way. At least, the solution I meant use a completely different idea. One can consider it as a hint.

Simply plug $\frac 14$ in the generating function for Catalan's numbers :).

Obviously, we cannot use the same method. But it suffices to look a little bit at the series developpement of $\sqrt{1+x}$.

Problem one is also the sum of all terms of the sequence for $n \ge 1$ defined by $a_{n+1}=\frac{(2n-1)}{2n+2}a_n$ where $a_1=\frac{1}{2}$ but how might you find that. :(


I've got an equation:
$ \pi^4\plus{}\pi^5\equal{}e^6$ $ (e\equal{}\lim_{n \to \infty}(1\plus{}\frac{1}{n})^n)$
Is it true?
How to prove?

No, it isn't true because
$ \pi^4 \plus{} \pi^5 \approx 403.4287\boxed{7}\ldots$
and
$ e^6 \approx 403.4287\boxed{9}\ldots$
it's just numerical approximation... but eq. looks pretty nice  :D

Very integersting, have never seen this before... :D

[quote="Extremal"][quote="tobeno_1"]I've got an equation:
$ \pi^4 \plus{} \pi^5 \equal{} e^6$
How to prove?[/quote]
How do you got this nice approach?:  :huh:
$ \pi^4 \plus{} \pi^5 \approx e^6$[/quote]
It was Castellanos who found it in 1988 :)
There are many interesting approximations e.g. :arrow:  http://mathworld.wolfram.com/PiApproximations.html

Personally I like this one $ e^\pi \minus{} \pi \approx 20$

[quote="Extremal"][quote="tobeno_1"]I've got an equation:
$ \pi^4 \plus{} \pi^5 \equal{} e^6$
How to prove?[/quote]
How do you got this nice approach?:  :huh:
$ \pi^4 \plus{} \pi^5 \approx e^6$[/quote]

You can make a new one yourself  :P

$ \pi \minus{} ln( \pi) \equal{} 2$

[quote="t0rajir0u"]Identities like this tend to fall under [url=http://en.wikipedia.org/wiki/Mathematical_coincidence]mathematical coincidence[/url].  I consider it quite likely that $ \pi, e$ are in fact algebraically independent.[/quote]
Yes, maybe it's right. But I believe that
$ e^{i\pi}\plus{}1\equal{}0$
is the most beautiful equation in maths.

It's definitely up there, but that's not what I mean by [url=http://en.wikipedia.org/wiki/Algebraic_independence]algebraic independence[/url].


ARML
geometry
inequalities
probability
summer program
PROMYS
conics

could someone post some questions from this year's contest?

I1. Let a, b, c, d, and e be distinct elements of {-10,-9,...,9,10}. Compute the least value of a+e given the following equations:

a-b=2
c-b=-3
c-d=4
e-d=-5.

I2. In the unit circle with center O, [tex]AC\cong CE\cong EA[/tex], chors AD, CF, and EB are equidistant from O, and the four inscribed circles are congruent. Compute the area of any one of the four smaller circles. (Someone should post a picture because I'm too lazy.)

I3. For 0<k<6, the graphs of [tex]\displaystyle\frac{(x-k)^2}{9}+y^2=1[/tex] and [tex]\displaystyle\frac{x^2}{9}+y^2=1[/tex] intersect at A and C and have x-intercepts at B and D respectively. Compute the value of k for which ABCD is a square.

I4. If [tex]\displaystyle \sqrt[3]{2}=a+\cfrac{1}{b+\cfrac{1}{c+\cfrac{1}{d+\cdots}}}[/tex] where a, b, c, d, ... are positive integers, compute b.

I5. Compute the area of the polygon whose vertices are the solutions in the complex plane to

[tex]x^7+x^6+x^5+x^4+x^3+x^2+x+1=0.[/tex]

Express the area in simplest form as [tex]\displaystyle \frac{a\sqrt{b}+c}{d}.[/tex]

I6. The Fibonacci sequence is defined as [tex]a_1=1,a_2=1[/tex] and [tex]a_n=a_{n-1}+a_{n-2}[/tex] for each n>2. Compute

[tex]\displaystyle\sum_{j=1}^{2004} \Bigg(i^{2004-a_j}\Bigg)[/tex]

where [tex]i=\sqrt{-1}[/tex].

I7. Compute the number of integer values of n, [tex]1\le n\le 2004,[/tex] such that for each value of n the following inequality has exactly two integer solutions:

[tex]\displaystyle\Bigg|\frac{nx}{2004}-1\Bigg|<\frac{n}{2004}.[/tex]

I8. In a game, A flips a fair coin two times. B then flips the coin until B's last two flips form a different sequence than A's. For example, if A flips HH, then B would stop flipping after each of HT, HHT, HHHT, and B's sequence would be HT. C then flips the coin until either A's sequence or B's occurs in consecutive flips. The player whose sequence occurs is the winner. Compute the probability that A wins.

thanks so much.

Is there a typo in #1?  Should it be c-d=4?

[quote="ComplexZeta"]I1. Let a, b, c, d, and e be distinct elements of {-10,-9,...,9,10}. Compute the least value of a+e given the following equations:

a-b=2
[b]c-b=-3
c-b=4[/b]
e-d=-5.
[/quote]

typo?

Aww, this year's problems aren't so cool, in my opinion. #4 and #5 are just standard stuff that people know -- no thinking/problem-solving involved.

Small world...

At ARML I ended up meeting 3 people I hadn't seen since MOP 2002, including Joel who ended up being our proctor for the super relay... then a handful of people from PROMYS and 3 people from USACO camp... and bugzpodder.

I met a few people who are on AoPS at ARML -- rruscyzk, Eugevir, Kamior, and wdragyn.

I thought the individual questions were rather easy (possibly the easiest of all the ones I have seen). The tiebreaks baffled me though. No one solved the first two tiebreaks, and then Aaron Pixton solved the third one in 25 seconds to win the individual round. Jongmin Baek solved the third one in 37 seconds to get second place.

Who were the three people who got 8s who weren't in the top three? They weren't announced at the west coast site.

aaron pixton solved the first tiebreaker. not the third.

Hm. I guess Tom or whoever it was who was announcing (I forget) made a mistake then.

T-9

Consider the sequence of numbers 24, 2534, 253534,  25353534, 2535353534, ... Let N be the first number in the sequence that is divisible by 99. Compute the number of digits in the base ten representation of N.

[quote="ComplexZeta"]

Who were the three people who got 8s who weren't in the top three? They weren't announced at the west coast site.[/quote]

u mean the top 2, right? i think they were somethng Price from TJ, ricky biggs from TJ, and anders kaseorg from i dunno where...

could someone post answers to the individuals?

is the answer to T-9 86?

There were six people in the nation who got 8s on individual round.

1. Aaron Pixton, Upstate New York
2. Jongmin Baek, SFBA A
3. Anders Kaseorg, North Carolina.

So Eric Price and Ricky Biggs from TJ are two more. But there is one more I'm looking for.

ARML individual time limit is 10 minutes per pair of questions.

I agree that number 8 was too time-consuming. However, number 4 doesn't require any prior knowledge of continued fractions (they've been on my list of things to learn for quite a while now, and I don't know them, and I got it right), just a little bit of computation (by the way, what is the continued fraction for cube root of 2?). Number five doesn't require a lot of prior knowledge (I think the roots of unity can be derived fairly quickly), as the first thing you would do with that polynomial if you hadn't seen it before in conjunction with the complex plane, which you should have if you studied for AIME with old tests) is to multiply it by x-1, to cancel out all the middle terms, and then see the connection with roots of unity. As for the octagon, I still don't have it's properties and lengths memorized, but they're not that hard to derive on the spot (I did it). The first tiebreaker didn't take that much time to solve (an estimation based on how long it would have taken me to solve it if I was sober, but I still got it right). What method did you use to solve it, ********?

By the way Complex Zeta, how many'd you get (I don't know your alterego's name, if you were one of the perfects)?

Ooh and what were those other two tiebreakers? And how long did it take the winner to solve it?

[quote]By the way Complex Zeta, how many'd you get (I don't know your alterego's name, if you were one of the perfects)? [/quote]

His "alterego" is named in his signature.

ComplexZeta's alter ego was unfortunately too old to participate in ARML However, he correctly answered seven of the eight questions in 2003 and then didn't get second place because he was not good at handling the pressure of 300 people staring at him.

(Seriously, my thought process on the first tiebreak in 2003 was something like this: The answer is half the number of positive integer factors of 800. So how many factors are there? That's a good question. Let's start writing them out: 1, 2, 4, 5, 8... Simon, you idiot! There's a formula. So what''s the prime factorization of 800? 800=245, so the number of factors is (4+1)(1+1)=10, so the answer is 5.) Obviously, the answer was 9.

[quote="ComplexZeta"]However, he correctly answered seven of the eight questions in 2003 [/quote]
thats too bad... which one did you miss?  #3 or #8?

I scored much better on this year's individual round than someone else on my team who is an indisputably better problem solver.  Why? Because he hadn't seen the factorization of x^7+x^6+...+1, and I had.  He had to derive it and ended up making an arithmetic error in the last few seconds (and he made another such error on the ellipse problem).  Someone else on my team got #4 with [i]no[/i] work, because they had done the continued fraction of 2^1/3 already.  In general, on our team people's scores were not representative of their ability, whereas something like AIME does better (I, for example, scored much lower on AIME that the previously mentioned person).  The problems could definitely have been done without the special knowledge that made them easier, but having it gave too big an advantage.  Only on #3 and #7 did I have to do any problem-solving thinking, and I do not think that the problem writers should have given me that advantage.

[quote="msmswatson"]Because he hadn't seen the factorization of x^7+x^6+...+1, and I had.[/quote]

I'll note that your teammate almost certainly had seen this factorization before - it's just a geometric series...

As for ARML as an individual competition, obviously there's tons of noise.  But it is a great team contest, and was probably the most fun math event I participated in during high school.

[quote="rrusczyk"]
[ARML] was probably the most fun math event I participated in during high school.[/quote]

Same for me.

Me also.  It seems that the team element is present more in ARML than any other competition.  Team is really the point in ARML, making the emphasis on individual questions seem somewhat inappropriate.

#5 was very doable without knowing the factorization of x^8-1.  My reasoning went something like this:
1. There is probably some trick to this, but I don't know it, so I'm going to brute-force it as best I can.
2. This problem would be insanely hard if the solutions are not on the unit circle, so I will assume they are.
3. Let's guess some likely numbers, shall we?  Hmm, -1 works, i works, -i works,  :rt2: /2+ :rt2: /2 i works (and so does its conjugate), - :rt2: /2+ :rt2: /2 i works (and its conjugate)... that's all!  Now for the easy part.
I got #5 with plenty of time to spare for #6.  Strangely enough, even after I had guessed all seven roots, I still didn't recognize the polynomial as a factor of x^8-1 until the guy reading the answers mentioned it afterwards.

By the way, did anyone else find #3 on relay 2 much too easy for third position?  There are at least three ways to solve it, none of which takes more than 30 seconds.

Is #3 in relay 2 the one about the area between the circles? I though that was extremely easy. All I did was let that segment be a diameter of both circles (since it doesnt specify) and say the outer circle has radius 3t/2 and the inner has radius t/2.

Yes, R2-3 was the one with the circles. I thought it was a good question. You guys have to remember that you aren't the "norm" - even among ARML participants. Questions that seem easy to you, aren't that easy to a lot of other people. There were probably quite a few relay teams at ARML that did not get any points on Relay 2.

I was a little disappointed in this year's ARML individuals. On #4 while Mr. Bettina read the question I said to myself, "Hey, I've seen this. This is just Euclid's Algorithm." Then after I finished 3 I said to myself, "This is ARML. It has to be something prettier than that." Then I got all screwed up and ran out of time. The questions were a little too straightforward so that any Joe off the street could get them. (Probably most Joe's couldn't, but most of the Joe's at ARML.) I only got 5, so I probably shouldn't be talking. :D

I was grading power rounds (#7, #8) at the west site. Only one team (Washington, not SFBA) got #8 right, and they found the distinct-terms solution. Nobody else was close enough to get any points. On #10, I think everyone was shut out completely- it was just too hard.


$ x,y,z$ positive number  $ x\plus{}y\plus{}z\equal{}3$                prove :
$ 2\plus{}xyz\geq xy\plus{}yz\plus{}zx$

It's not true. Try $ x\equal{}y\equal{}\frac{3}{2}\minus{}\frac{1}{9^9}$ and $ z\equal{}\frac{2}{9^9}$. :wink:


geometry
3D geometry
algebra
polynomial
factorization
sum of cubes
difference of squares

Prove that for any integer $n$ greater than 1, $n^{12} + 64$ can be written as the product of four distinct integers greater than 1.

I think we need to define 'hint' a bit better :)

[quote="mffmartinelli"]The 'hint' is so: Look to the complex roots of $z^6+1=0$.[/quote]

I think that this is your answer!!!!!

[quote="symmetricmean"]$n^{12}+64$ = $(n^2-2n+2)(n^2+2n+2)(n^4-2n^3+2n^2-4n+4)(n^4+2n^3+2n^2+4n+4)$

Ok how does one factor this? any standard techniques or just plain trial and error method?[/quote]

[hide="Factoring l33tness"]
The first two steps are relatively easy.

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ (Sum of cubes)
$n^{12} + 64 = (n^4 + 4)(n^8 - 4n^4 + 16)$

Then

$n^4 + 4 = (n^2 + 2)^2 - (4n^2) = (n^2 + 2)^2 - (2n)^2 = (n^2 + 2n + 2)(n^2 - 2n + 2)$ (this is a technique from AoPS vol. 1 - expressing as difference of squares)

Lastly, we wish to factor $n^8 - 4n^4 + 16$.  We use the same trick.  Let's investigate a possible expression we can square.

$(n^4 + 2n^2 + 4)^2 = n^8 + 4n^6 + 12n^4 + 16n^2 + 16$
$n^8 - 4n^4 + 16 = (n^4 + 2n^2 + 4)^2 - (4n^6 + 16n^4 + 16n^2) = (n^4 + 2n^2 + 4)^2 - (2n)^2 (n^4 + 4n^2 + 4) = (n^4 + 2n^2 + 4)^2 - ( 2n(n^2 + 2) )^2 = (n^4 + 2n^3 + 2n^2 + 4n + 4)(n^4 - 2n^3 + 2n^2 - 4n + 4)$

Whew.
[/hide]

[quote="symmetricmean"]Ok how does one factor this? any standard techniques or just plain trial and error method?[/quote]
A standard technique: we're looking for $a,b,c,d,e,f\in\mathbb{Z}$ such that
$n^8 - 4n^4 + 16=(n^4+an^3+bn^2+cn+4)(n^4+dn^3+en^2+fn+4)$
Polynomial identification yields to a system of equations (we only need one of its solutions). Since the coefficient of $n^7$ is $0=d+a$, we obtain from the beginning $d=-a$; then, using the coefficient of $n$, we get $f=-c$ etc. One of the solutions: $a=b=2, \ c=4,\ d=-2, \ e=2,\ f=-4$, hence the factorization above.

[quote]Polynomial identification yields to a system of equations (we only need one of its solutions). [/quote]

Aah that works for me. Thanks.  :lol:

Happy New Year :)


Hello all,

       Can anyone please help me with the following circle Q?  Urgent!  Please see the attachment.

       Thank you.

Best,
tine

Well...
[hide]$ \angle AOB \equal{} 180\minus{}\angle OBA\minus{}\angle OAB$ because the sum of the angles in AOB is 180.
Also, we know that OB bisects $ \angle ABR$ and OA bisects $ \angle BAT$:
$ \angle AOB \equal{} 180\minus{}\frac{1}{2}(\angle ABR \plus{}\angle BAT) \equal{} 180\minus{}\frac{1}{2}(180\minus{}\angle ABP\plus{}180\minus{}\angle BAP)$
Finally, since $ \angle ABP\plus{}\angle BAP\equal{}180\minus{}\angle APB$
$ \angle AOB \equal{} 180\minus{}\frac{1}{2}(180\plus{}\angle APB) \equal{} 180 \minus{} (90\plus{}\frac{1}{2}\cdot 40) \equal{} 70$[/hide]


The elements of set B are all the possible subsets of set A. Set B has 16 subsets. How many elements are in set A?

hmmm...never seen this prob, where'd u get it?

I'm not good with sets (I've never seen it either), I get confused with the subsets and elements. What's the difference?

I'm not very good with sets. but is the answer by any chance 16, since all the elements of B, which I thik are 16, are subsets of A, so is the answer 16?

If nobody gets it, I'll post the solution Monday morning.

[quote="Hamster1800"]I don't think this is possible. Since set B has 16 subsets, it has 4 elements. Say set B is {A, B, C, D}

Then according to the problem, all the possible subsets of A are {A}, {B}, {C}, and {D}. This can't be right because {} is a subset of all sets, and {A,B} and {B,C} and other combinations would also be a subset. Either it isn't worded right or I'm reading it wrong.[/quote]
i think hamster might b correct, but if he isnt, then mebbe u can get this by guess and check.

[hide="my solution"]
Set B has n elements and 16 subsets. So n=4.
Given that elements of set B are all the possible subsets of set A, so set A has 4 subsets. Then set A has[size=150] [b]2[/b][/size] elements.

I think this question has wording problem because it reads "elements=subsets" . Should be "the number of elements of B = the number of subsets of A".

[/hide]

Uhh, the number of elements = square root of number of subsets? It that true?  :?

No... the number of subsets a set has is 2 to the power of its cardinality. This is called the power set. To see why:

Each element is either in a subset or not in a subset. There are 2 possibilities for each element, and since a subset is defined by whether each member is in the subset or not, the cardinality of the power set of A is $2^{|A|}$ where the ||'s denote the cardinality.

Hope that helps.

So if set B has 16 subsets, it has 2 members. Hence there are 2 possible subsets of A (B is the power set of A) and A has only one member.

But if set B has 16 [b]MEMBERS[/b], then set A has 2 members.

Please clarify the problem.

No... you can't make sixteen distinct subsets with only two distinct members.

Oh. Duh. Sorry...

[b]4[/b] members... therefore [b]2[/b] members in A.

This is unbelievably easy if you know that the #subsets of $S$ is $2^{s}$ where $s$ is the num of elements in $S$.

Proof of the number of subsets of a set with n elements=2^n

Let's list out all the elements from 1-n

1 2 3 4 5 6 7 8 9 . . . . . n

Now, let's say there's a button above each element, and if we press it it's automatically added to the subset in question. So, for the first element we can either press the button, or not press the button, same with the second, and so on. This will cover all possibilities. Thus, because we have two choices for each element, the number of distinct subsets is 2^n

Proof 2:

The number of subsets with size $k$ of set $S$ with size $n$, is $\binom{n}{k}$. So we have

\[\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n}\]

This is commonly known to be equal to $2^n$.  :)

Proof 1 was basically what I said.

About proof 2, could you please prove the binomial coefficent theorem? I know the theorem but I don't know the proof. :(

I would take that to the Intermediate Forum, as proofs are not allowed in lower forums. ;)


Find $2^{2006}$ positive integers satisfying the following conditions.
(i) Each positive integer has $2^{2005}$ digits.
(ii) Each positive integer only has 7 or 8 in its digits.
(iii) Among any two chosen integers, at most half of their corresponding digits are the same.

My solution:
[hide]Define $S_1 = \{ 77, 78 \}$
Define the inverse of an element to be 7 ->8, 8->7
Define the $S_i'$ to be the inverse of $S_i$, e.g. $S_1' = \{88,87\}$
Define $S_i^2$ to be writing itself again, e.g. $S_1^2 = \{7777, 7878\}$
Define $S_i+S_i'$ to be writing after corresponding element, e.g. $S_1+S_1' = \{7788, 7887\}$
Define $T_i = \{x| x \in S_i or x \in S_i'\}$
Define $S_{i+1} = \{x| x \in S_i^2 or x \in S_i+S_i'\}$

Now I claim that $T_{2005}$ is the required set

First, each no. has $2^{2005}$ digits, which is obvious.
Having 7 or 8 as the only digits is trivial as well.

Now it suffices to prove that $T_{2005}$ fulfills condition 3. We will proceed by induction.

in $T_1$ obviously (iii) is fulfilled
Assume when i =k, $T_i$ is true for condition iii, then when i=k+1, $T_{k+1} = S_i^2 \cup S_i+S_i'$
Since $T_i = S_i \cup S_i'$ and it fulfills condition 3, it means that for elements in $T_{k+1}$, The first half has at most half of the digits different , the second half has at most half of the digits different.
So any two elements of $T_i$ has at most half of the digits different
Induction done.
Thus $T_{2005}$ fulfills the condition[/hide]

Construction:
To find $2^{n+1}$ positive integers satisfying the following conditions: (i) Each positive integer has $2^{n}$ digits. (ii),(iii)
Take the $2^n$ numbers satisfying the conditions
(i) Each positive integer has $2^{n-1}$ digits. (ii),(iii)

First get $2^n$ numbers by replacing each "7" by "77" and each "8" by "88". Then get another $2^n$ numbers by replacing each "7" by "78" and each "8" by "87".

For the case where the numbers have only one digit, the numbers "7" and "8" satisfy the conditions.

Proof:
Suppose we have $2^n$-digit numbers A and B with exactly half their digits in common. If we perform the same replacement on both of them (e.g. 7-->78 and 8-->87) clearly the new numbers will share exactly half their digits. If we perform different replacements on each of them (e.g. 7-78, 8-87 on A; 7-77 and 8-88 on B), the previously agreeing digits will each contribute one agreement and the previously disagreeing digits will each contribute one agreement to the new pair of numbers. Thus the new numbers will share half their digits.

Suppose we have $2^n$-digit numbers with no digits in common. Clearly if we perform the same replacement on both numbers the resulting numbers will share no digits, and if we perform different replacements, the resulting numbers will share half their digits.

For our two numers for the case n=0, no digits are in common. It follows from induction that when we carry out our construction, every pair of numbers will share exactly half their digits or none of their digits, so we can perform the construction 2005 times to get the desired set of numbers.


geometry
ratio
analytic geometry
Pythagorean Theorem
area of a triangle
Heron\u0027s formula

Let line $ AC$ be perpendicular to line $ CE$.  Connect $ A$ to $ D$, the midpoint of $ CE$, and connect $ E$ to $ B$, the midpoint of $ AC$.  If $ AD$ and $ EB$ intersect in point $ F$, and $ \overline{BC} \equal{} \overline{CD} \equal{} 15$ inches, then the area of triangle $ DFE$, in square inches, is:

$ \textbf{(A)}\ 50 \qquad \textbf{(B)}\ 50\sqrt {2} \qquad \textbf{(C)}\ 75 \qquad \textbf{(D)}\ \frac {15}{2}\sqrt {105} \qquad \textbf{(E)}\ 100$

There's probably an easier way

[geogebra]e724c6bf61b7e8e2fd141bbbf4e88cb3d32d5253[/geogebra]

[hide]
$ [AEC] = 2(15)^2, [ACD] = (15)^2 \Rightarrow [ADE] = (15)^2$

So from that we see that quadrilateral $ BFDC$ and $ \bigtriangleup AFE$ have the same area.

Also $ \bigtriangleup AFE \sim \bigtriangleup BFD$. We can see there longer sidethengths have ratio $ 2: 1$. Hence there area's have a ratio $ 2^2: 1$.

Let$ [BFD] = A \Rightarrow A + \frac {1}{2}(15)^2 = 4A \Rightarrow A = \frac {75}{2}$

$ \Rightarrow [BFDC] + [AFE] = 2(\frac {1}{2}75 + \frac {1}{2}(15)^2) = 300$

$ \Rightarrow [DEF] = \frac {2(15)^2 - 300}{2} = 75 \longrightarrow \boxed{C}$[/hide]

[hide]
$ AD$ and $ BE$ are both medians of triangle $ ACE$, so $ \frac{AF}{FD}=\frac{FE}{BF}=\frac{2}{1}$.
We can find $ AD$ using the Pythagorean Theorem:
$ AC^2+CD^2=AD^2 \Rightarrow 30^2+15^2=AD^2 \Rightarrow AD=15\sqrt{5}$.
Segment $ FD = \frac{AD}{3} = 5\sqrt{5}$.
Segment $ FE = AF = \frac{2AD}{3} = 10\sqrt{5}$.
Then by Heron's Formula:
$ [DFE]=\sqrt{(\frac {15+15\sqrt{5}}{2})(\frac {15+15\sqrt{5}}{2}-15)(\frac {15+15\sqrt{5}}{2}-5\sqrt{5})(\frac {15+15\sqrt{5}}{2}-10\sqrt{5})}$.
Simplifying, we get $ [DFE]=75 \Rightarrow \boxed{C}$.
[/hide]

Another easier way to do it without using radicals is set up the triangle in the coordinate plane, find the equations of BE and AD and their intersection point, and then use shoelace.

[quote="dgreenb801"]There's a much easier way. [AFB]=[BFC]=[CFD]=[DFE]. So [BCE]=225=3[DFE]. So [DFE]=75.[/quote]

I was thinking similarly to your way, just add the 3rd median to AE. The 6 triangles are equal (not counting the line BD) so 30*30/2*6=75.


Using Am-Gm: $ \sqrt[3]{ab}(4\minus{}ab) \leq\ 3$
So, we are easily to solve the general problem:
$ \sum\ \frac{a^2\plus{}kb^2}{ab^2(4\minus{}ab)} \geq\ k\plus{}1$  :)

Hello, first we work on a^2 + 3b^2 /ab^2 (4-ab). If we divide both numerator and denominator we get a^2 /b + 3b/ab(4-ab). Now we write a^2/b + 3b as a^2/b + b + 2b. Now we are applying am-gm inequality we get that a^2/b + b more than equal 2a.
Now we find that (a^2 + 3b^2 )/ab^2(4-ab) more than equal (2a+2b)/ab(4-ab). Now we are applying am-gm inequality again (for the numerator) we get that 2a+2b more than equal 4(ab)^1/2. So our inequality is to prove
1/(ab)^1/2 (4-ab) + ..... more than 1. For the denominator write it as (ab)^1/2 (2- (ab)^1/2) (2 + (ab)^1/2). AM/GM we get that
(ab)^1/2 (2-(ab)^1/2) less than 1.
So We want to prove that
1/(2 + (ab)^1/2) + 1/ (2+ (bc)^1/2) + 1 / (2+ (ca)^1/2) more than 1. Note that (a+b+c)^2 =9 more than 3(ab+bc+ac). And By Cauchy-Schwarz (ab)^1/2 + (bc)^1/2 + (ca)^1/2 less than 3. Now you can finish the inequality by using am-hm inequality. Equality hold iff a=b=c=1. :lol:

I solved this problem but I'm not sure
$ \frac {\frac {a^{2}}{b} + 3b}{ab.(4 - ab)}$
$ {\frac {a^{2}}{b} + b + b + b\geq 4.\sqrt {ab}}$ (AM-GM)
$ \implies$ $ \frac {\frac {a^{2}}{b} + 3b}{ab.(4 - ab)}\geq\frac {4}{\sqrt {ab}.(4 - ab)}$
$ \sqrt {ab}.(2 - \sqrt {ab})\leq 1$  $ \implies$
$ \frac {4}{\sqrt {ab}.(4 - ab)}\geq\frac {4}{2 + \sqrt {ab}}$
$ [\frac {1}{2 + \sqrt {ab}} + \frac {1}{2 + \sqrt {bc}} + \frac {1}{2 + \sqrt {ac}}]\geq\frac {9}{6 + \sqrt {ab} + \sqrt {bc} + \sqrt {ac}}$(Cauchy-Schwarz)
$ \sqrt {ab} + \sqrt {bc} + \sqrt {ac}\leq 3$  $ \implies$  $ \frac {9}{6 + \sqrt {ab} + \sqrt {bc} + \sqrt {ac}}\geq 1$ QED

$ \sum\ \frac {a^{2} \plus{} kb^{2}}{ab^{2}(4 \minus{} ab)} \equal{} \sum\ \frac {\frac {1}{b^2}}{\frac {4}{a} \minus{} b} \plus{} k \sum\ \frac {\frac {1}{a^2}}{\frac {4}{a} \minus{} b}$

From titu's lemma
$ \sum\ \frac {\frac {1}{b^2}}{\frac {4}{a} \minus{} b} \plus{} k \sum\ \frac {\frac {1}{a^2}}{\frac {4}{a} \minus{} b} \geq (k \plus{} 1)\frac {(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^2}{\frac {4}{a} \plus{} \frac {4}{b} \plus{} \frac {4}{c} \minus{} 3}$

$ (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^2 \plus{} 3\geq (\frac {4}{a} \plus{} \frac {4}{b} \plus{} \frac {4}{c})$

replace $ x\equal{}\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$
 $ (x\minus{}3)(x\minus{}1) \geq 0$

$ x \geq 3$ From $ AM\minus{}HM$

Also posted here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=440599


group theory
abstract algebra
superior algebra
superior algebra unsolved

Show that the group $ A_{5}$ has no subgroups of order $ 15$.

Using Lagrange is clearly out of the question since $ 15$ divides $ 60$ but can we use the Sylow theorems in some way to prove this?

Alternatively you can use the fact that if $ H$ is a proper subgroup of $ G$ such that $ i(H)!$ is not divisible by $ |G|$ then $ H$ contains a non-trivial normal subgroup of $ G$ - this (combined with the fact that $ A_{5}$ is simple) will show that any proper subgroup has size at most $ 12$.


function
algebra
polynomial
algebra unsolved

assume that c is complex number ,we define fc(z)=z 2 +c ,Ac is { p(z)  l  fc(p(z))=p(fc(z))} p(z) is defined in complex number.  some of the member of Ac is trivial, for example if Z0 be constant point of fc (fc(Z0)=Z0) then p(z)=Z0 is in Ac ,. also p(z)=z and p(z)=fc(fc(...(fc(z)...)  are in Ac.
at first prove that every member of Ac is even or odd functin (this part of problem is for warmup :D )then prove that under each condition which is given below the member of Ac  are  just discussed above,
1)lcl >2
2)c is in Q-Z
3) c isn't algebric(  like  \pi  or e )
4)c isn't in this interval [-2,1/4]
( I hope that u understood the  problem)

Can anybody help me? I love this problem, I tried to solve it for some time, but didn't manage. This is such a nice problem, it doesn't deserve to stay in the place where noone can see it!

Prove that with this condiitions if $p(z)$ is such polynomial and $deg(p)>1$ then $\sqrt {p(z)-c}$ is polynomial and have the problem conditions.

[quote="sam-n"]Assume that $c$ is complex number, we define $f_c(z)=z^2+c$, $A_c$ is $\{ p(z) \mid f_c(p(z))=p(f_c(z))\}$ $p(z)$ is defined in complex numbers. Some of the members of $A_c$ is trivial, for example if $Z_0$ be constant point of $f_c$ ($f_c(Z_0)=Z_0$) then $p(z)=Z_0$ is in $A_c$. Also $p(z)=z$ and $p(z)=f_c(f_c(\ldots(f_c(z)\ldots)$  are in $A_c$.
At first prove that every member of $A_c$ is even or odd functin (this part of problem is for warmup :D )then prove that under each condition which is given below the member of $A_c$  are  just discussed above,
1)$|c| >2$
2)$c \in \mathbb{Q-Z}$
3)$c$ isn't algebraic (like $\pi$  or $e$)
4)$c \notin [-2, \frac 14]$
[/quote]

 :)