Solve equation:
$ 20^x\plus{}11^x\plus{}2009x\equal{}2\plus{}1019\log_2(2\plus{}29x\minus{}27^x)$

hello, this equation has only the solutions $ x\equal{}0$ and $ x\equal{}1$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, this equation has only the solutions $ x \equal{} 0$ and $ x \equal{} 1$.
Sonnhard.[/quote]

Why? :o  :o  :o


let $x_1$ and $x_2$ be two polatts. polatts are polygons whose vertices have interger co-ordinates on a Cartesian plane. $x_1$ and $x_2$ share a common edge $PQ$ but no other points, i.e. their intersection is the edge $PQ$.

let $i_1$, $i_2$, $b_1$, $b_2$ be the number of interior and boundary lattice (interger co-ordinate) points of $x_1$ and $x_2$ respectively.

let $k$ be the number of lattice points on $PQ$.

Let $x$ be the union of $x_1$ and $x_2$ (the set of ponts belonging to $x_1$ or $x_2$ (or both)).

(i) if $i$ is the number of interior lattice points of $x$ show that:  $i$ $=$ $i_1$ $+$ $i_2$ $+$ $k$ $-$ $2$.

(ii) if $b$ is the number of boundary lattice points of $x$, express $b$ in terms of $b_1$, $b_2$ and $k$.

in this next one, $f(x, y)$ $=$ $x$ $+$ $y/2$ $-$ $1$

(iii) with $i$, $b$, $i_1$, $b_1$, $i_2$ and $b_2$ as defined as above, show that:

$f(i,b)$ $=$ $f( {i_1}, {b_1}) $ $+$ $f({i_2}, {b_2})$

Just a remark : Then prove the $f$ is the aera of the polygon. You will have a proof of the so-called Pick's formula. ;)

Pierre.


Let $ x,y,z\ge0$ with $ xyz\equal{}1$. Find the minimum value of
\[ \frac{x}{y\plus{}z}\plus{}\frac{y}{x\plus{}z}\plus{}\frac{z}{x\plus{}y}\]

[quote="worthawholebean"]Let $ x,y,z\ge0$ with $ xyz \equal{} 1$. Find the minimum value of
\[ \frac {x}{y \plus{} z} \plus{} \frac {y}{x \plus{} z} \plus{} \frac {z}{x \plus{} y}
\]
[/quote]
This is Nesbitt's inequality, and the condition that $ xyz \equal{} 1$ is superfluous.
[hide="Solution"]\[ \frac {x}{y \plus{} z} \plus{} \frac {y}{x \plus{} z} \plus{} \frac {z}{x \plus{} y} \equal{} \frac {x^2}{yx \plus{} xz} \plus{} \frac {y^2}{yx \plus{} yz} \plus{} \frac {z^2}{xz \plus{} yz}
\]By Titu's Lemma,\[ \frac {x^2}{xy \plus{} yz} \plus{} \frac {y^2}{yz \plus{} yx} \plus{} \frac {z^2}{zx \plus{} zy}\ge \frac {(x \plus{} y \plus{} z)^2}{2xy \plus{} 2yz \plus{} 2xz} \equal{} 1 \plus{} \frac {x^2 \plus{} y^2 \plus{} z^2}{2xy \plus{} 2yz \plus{} 2xz}\]Yet, $ (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (x \minus{} z)^2\ge 0$, so $ x^2 \plus{} y^2 \plus{} z^2\ge xy \plus{} yz \plus{} xz$, so $ \frac {x^2 \plus{} y^2 \plus{} z^2}{2xy \plus{} 2yz \plus{} 2xz}\ge \frac {1}{2}$, so we conclude that
\[ \frac {x}{y \plus{} z} \plus{} \frac {y}{x \plus{} z} \plus{} \frac {z}{x \plus{} y}\ge 1 \plus{} \frac {1}{2} \equal{} \frac {3}{2}\][/hide]

$ (x(y\plus{}z)\plus{}y(z\plus{}x)\plus{}z(x\plus{}y))(\frac{x}{y\plus{}z}\plus{}\frac{y}{x\plus{}z}\plus{}\frac{z}{x\plus{}y})\ge (x\plus{}y\plus{}z)^2$

$ \frac{x}{y\plus{}z}\plus{}\frac{y}{x\plus{}z}\plus{}\frac{z}{x\plus{}y} \ge \frac{(x\plus{}y\plus{}z)^2}{2xy\plus{}2yz\plus{}2xz}$
$ \equal{}1\plus{}\frac{\frac{x^2\plus{}y^2\plus{}z^2}{xy\plus{}xz\plus{}yz}}{2} \ge 1\plus{}1/2\equal{}3/2$

This is y method :)
$ S \equal{} \frac{x}{{y \plus{} z}} \plus{} \frac{y}{{x \plus{} z}} \plus{} \frac{z}{{x \plus{} y}}$
$ S \plus{} 3 \equal{} \frac{x}{{y \plus{} z}} \plus{} \frac{y}{{x \plus{} z}} \plus{} \frac{z}{{x \plus{} y}} \plus{} 3$
$ S \plus{} 3 \equal{} \frac{{x \plus{} y \plus{} z}}{{x \plus{} y}} \plus{} \frac{{x \plus{} y \plus{} z}}{{x \plus{} z}} \plus{} \frac{{x \plus{} y \plus{} z}}{{y \plus{} z}}$
$ S \plus{} 3 \ge (x \plus{} y \plus{} z) \times \frac{9}{{2(x \plus{} y \plus{} z)}} \equal{} \frac{9}{2}$
$ S \ge \frac{3}{2}$


Let $f: \mathbb R \to \mathbb R$ be a function with the property that
\[f(f(x)\minus{}f(y)) \equal{} f(f(x)) \minus{} y\]
holds true for all reals $x$ and $y$. Prove that $ f$ is an odd function.

Easy to check that $ f(x)$ is inject.
Suppose exist $ a,b$ satisfy $ f(a)\equal{}f(b)$
Let $ y\equal{}a,y\equal{}b$ then
$ f(f(x)\minus{}f(a))\equal{}f(x)\minus{}a$
$ f(f(x)\minus{}f(b))\equal{}f(x)\minus{}b$ s
but $ f(a)\equal{}f(b)$ then $ a\equal{}b$
Let $ y\equal{}0$ then
$ f(f(x)\minus{}f(0))\equal{}f(f(x))$ so $ f(0)\equal{}0$
Let $ x\equal{}y$ then
$ f(f(x))\equal{}x$
Let $ x\equal{}0$ then
$ f(\minus{}f(y))\equal{}\minus{}y$
Let $ y\equal{}f(y)$ then
$ f(\minus{}f(f(y))\equal{}\minus{}f(y)$
But from $ f(f(y))\equal{}y$ we have
$ f(\minus{}y)\equal{}\minus{}f(y)$
It mean that $ f(x)$ is a odd function.

[quote="stergiu"]If function $ f$ has the propeprty

  $ f(f(x) \minus{} f(y)) \equal{} f(f(x)) \minus{} y$

for every reals $ x , y$ , prove that $ f$ is an odd function.[/quote]

Let $ P(x,y)$ be the property $ f(f(x) \minus{} f(y)) \equal{} f(f(x)) \minus{} y$

$ P(x,x)$ implies $ f(f(x))\equal{}x\plus{}f(0)$ and then $ f(x)$ is bijective.
$ P(0,0)$ implies $ f(f(0))\equal{}f(0)$ and, since $ f(x)$ is bijective, $ f(0)\equal{}0$

So we have $ f(f(x))\equal{}x$ and $ P(0,f(x))$ gives $ f(\minus{}x)\equal{}\minus{}f(x)$

Q.E.D.

Let $ x\to f(x),y\to f(y)$ then
$ f(x\minus{}y)\equal{}f(x)\minus{}f(y)$ so $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and more :
$ f(f(x))\equal{}x$
But with conditon we can find out it.

[quote="silouan"]Could we find all the functions
 $ f$ with propeprty

  $ f(f(x) \minus{} f(y)) \equal{} f(f(x)) \minus{} y$    :)  :?:[/quote]

We have $ f(f(x))\equal{}x$ and $ f(\minus{}x)\equal{}\minus{}f(x)$ so $ f(f(x)\plus{}f(y))\equal{}x\plus{}y$ and so $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$

So the solutions are :

Let $ \mathbb{A}$ and $ \mathbb{B}$ two supplementary $ \mathbb{Q}$-vectorspaces of $ \mathbb{R}$ ($ \mathbb{A}\cap\mathbb{B}\equal{}\{0\}$ and $ \mathbb{A}\plus{}\mathbb{B}\equal{}\mathbb{R}$

Let $ a(x)$ and $ b(x)$ the projections of $ x$ on $ \mathbb{A}$ and $ \mathbb{B}$  (We have in a unique manner $ x\equal{}a(x)\plus{}b(x)$)

Then $ f(x)\equal{}a(x)\minus{}b(x)$

Note:
Without Axiom of Choice, the two only possibilities for $ \mathbb{A}$ and $ \mathbb{B}$ are ( $ \mathbb{A}\equal{}\mathbb{R}$ and$ \mathbb{A}\equal{}\{0\}$) or ($ \mathbb{A}\equal{}\{0\}$ and $ \mathbb{B}\equal{}\mathbb{R}$)

which give the two solutions $ f(x)\equal{}x$ and $ f(x)\equal{}\minus{}x$

With Axiom of Choice, we have infinitly many others.


linear algebra
matrix
linear algebra unsolved

Let $M \in GL_{2n}(K)$, represented in block form as \[ M = \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right] , M^{-1} = \left[ \begin{array}{cc} E & F \\ G & H \end{array} \right] \]
Show that $\det M.\det H=\det A$.

[hide="My solution"]$\det M.\det H$=$\det M$ $\det \left[\begin{array}{cc} I & F \\ 0 & H \end{array}\right]$=$\det\[ \left[ \begin{array}{cc} A & o \\ D & I \end{array} \right] =\det A$ [/hide]


I have this code that a professor posed to use in class. He found it on the internet and has been unable to solve it. The only clue he gave us was A = 1.

Here it is. Both parts are decoded seperately:

Part 1: [66-9-13-18-1-19-95-12-3-14-5-12-15-7]
Part 2: [1-12-14-56-18-21-1-23-15-5-20-18-13-18-22-9-25-15-21-15-15-14-4-20]

(the prof. is a big TechTV geek so it could have something to do with that!)

Try googling? :D

There are already other discussions about this going on in other places on the Internet. I typed in the first code in quotations (to search for the exact phrase) and got a few matches.

Or if what you find doesn't satisfy you, you could resort to letter frequency or (eek) a program that exhausts the possibilities and displays everything it finds. The display would be really long, of course, but...


Denote by $d(n)$ the largest odd divisor of positive integers $n$ and define $D(n),\ T(n)$ as follows.
\[ D(n)=d(1)+d(2)+\cdots +d(n),\ \ T(n)=1+2+\cdots +n.  \]
Prove that there existed infinitely positive integers $n$ such that $3D(n)=2T(n).$

[quote="kunny"]Denote by $ d(n)$ the largest odd divisor of positive integers $ n$ and define $ D(n),\ T(n)$ as follows.
\[ D(n) \equal{} d(1) \plus{} d(2) \plus{} \cdots \plus{} d(n),\ \ T(n) \equal{} 1 \plus{} 2 \plus{} \cdots \plus{} n.\]
Prove that there existed infinitely positive integers $ n$ such that $ 3D(n) \equal{} 2T(n).$[/quote]
I solved it this way:
After writing $ T(n),D(n)$ up for the first few $ n$ i find that it holds for $ n\equal{}2,6,14$. So it seems that $ n\equal{}2^r\minus{}2$ is a solution for $ r \ge 2$.
I know that $ T(n) \equal{} \frac{n(n\plus{}1)}{2} \Rightarrow 2T(2^n\minus{}2) \equal{} (2^n\minus{}2)(2^n\minus{}1)$.
From here I got the idea to prove it by induction.
$ D(2^n\minus{}2) \equal{} d(1)\plus{}d(2)\plus{}d(3)\plus{}...\plus{}d(2^n\minus{}2) \equal{}$
$ (d(1)\plus{}d(3)\plus{}d(5)\plus{}...\plus{}d(2^n\minus{}3))\plus{}(d(2)\plus{}d(4)\plus{}d(6)\plus{}...\plus{}d(2^n\minus{}4))\plus{}d(2^n\minus{}2)$
Now see that $ d(2n)\equal{}d(n)$ so $ d(2)\plus{}d(4)\plus{}...\plus{}d(2^n\minus{}4) \equal{} d(1)\plus{}d(2)\plus{}..\plus{}d(2^{n\minus{}1}\minus{}2) \equal{} D(2^{n\minus{}1}\minus{}2)$.
$ d(1)\plus{}d(3)\plus{}...\plus{}d(2^n\minus{}3) \equal{} 1\plus{}3\plus{}5\plus{}..\plus{}(2^n\minus{}3) \equal{} \left ( \frac{(2^n\minus{}3)\plus{}1}{2} \right)^2 \equal{} (2^{n\minus{}1}\minus{}1)^2$
So $ 3D(2^n\minus{}2) \equal{} 3D(2^{n\minus{}1}\minus{}2) \plus{} 3(2^{2n\minus{}2}\minus{}2^{n\minus{}1})$
$ 2T(2^n\minus{}2) \equal{} (2^n\minus{}1)(2^n\minus{}2) \equal{} 2^{2n}\minus{}3 \cdot 2^{2n}\plus{}2 \equal{} (2^{n\minus{}1}\minus{}1)(2^{n\minus{}1}\minus{}2) \plus{} 3(2^{2n\minus{}2}\minus{}2^{n\minus{}1}) \equal{} 2T(2^{n\minus{}1}\minus{}2) \plus{}3(2^{2n\minus{}2}\minus{}2^{n\minus{}1})$
So: $ 2T(2^n\minus{}2) \equal{} 3D(2^n\minus{}2) \iff$
$ 2T(2^{n\plus{}1}\minus{}2)\minus{}3(2^{2n}\minus{}2^n) \equal{} 3D(2^{n\plus{}1}\minus{}2)\minus{}3(2^{2n}\minus{}2^n) \iff$
$ 2T(2^{n\plus{}1}\minus{}2) \equal{} 3D(2^{n\plus{}1}\minus{}2)$. And the rest follows by induction :)


A sequence of cubes is given ,each containing a color pattern (see the attach).

Below each cube there are two codes.
1) Pattern code that indicates the arrangment of the pattern on each cube.

2)Transition code that indicates the changes in the patterns between adjacent elements.

Determine logic behind the transition codes and find the unique pattern code that completes the sequence and matches last transition code.

Have fun

[quote="Hexaditidom"]This looks like a mensa  :o[/quote]

Maybe just looks like   :alien:  ;)

Ok,with regard to the pattern code,it's a quite trivial.
3 shown sides of each cube can be viewed as 3 Cartesian planes with 3X3 facelets elements.
"Rows" are denoted by letters{A,B,C,D,E,F,G,H,I} and "italics" by numbers {1,2,3}.
In that way every facelet element has a unique coordinate.
Of course,overall pattern codes are given in increasing alphanumerical order.

Figuring out the transition code logic is another fish in a pond.
Leaving that part for brainteasers aficionados' entertainment.

Yeah, I had figured that out, but can't get the other code without some thought. I'm not exactly sure how to approach it, so I'll leave that up to someone else.

Any mathematicians that are puzzlers at the same time on this subforum?

  :)

Yeah, when I first saw it, but I can't think of a relationship.

Let the top be side $A$, the left be side $B$, and the right be side $C$. Let the ordered triple $(A_n,B_n,C_n)$ be the number of red squareson sides $A$, $B$, and $C$ respectively in the $n^{th}$ stage. Then the transition code under the $k^{th}$ stage is $(\left|A_k-A_{k-1}\right|,\left|B_k-B_{k-1}\right|,\left|C_k-C_{k-1}\right|)$. As for the last one. It's going to have one block on the left side (side $B$), but I can't figure out where. It won't be in the center, and it won't be on a corner, but that's all I can figure out.

Hey all

Mmmm. That's a very interesting problem. It does have a lot more to do with puzzles and even encryption than maths itself. Besides that, we all like to play a bit and for us math problems are our fun ;) Speaking of which, I recommend [url=http://www.fetchbook.info/math_puzzles.html]this place[/url] for finding great and fun math and non-math problems, that can enrich your world and just be hell of a lot of fun.

Best,  :)
Jason


A quadrilateral  $ABCD$ with perpendicular diagonals $AC , BD$ is inscribed in a circle with center $O$. If  $OH \perp CD$ , prove that $AB = 2 OH$.

[u]Babis[/u]

[hide]I suppose that you mean $H \in CD$ (so $H$ is the midpoint of $CD$)

$OH^2 + HD^2 = R^2 \Rightarrow$

$(2OH)^2 + CD^2 = 4R^2$

But from http://www.mathlinks.ro/Forum/viewtopic.php?t=58491 we know that $AB^2+CD^2 = 4R^2 \Rightarrow$

$AB = 2OH$[/hide]

Does this link really help? I think that  , that problem says something else. Any way , the solution is good.I will use it as well. The official solution is based on producing DO to meet the circle at E .  Then  $arc CE = arc AB$. But $CE = 2OH$.

 [u]Babis[/u]

 Ah..., good luck in EME 2005(Thalles)

[quote="stergiu"]Does this link really help? [/quote]

Sorry... I mixed up the topics  :roll:

I mean [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56241]this[/url] (post #4)

[quote="stergiu"]Ah..., good luck in EME 2005(Thalles) [/quote]
By the way, I'll not take part because I'm not a student anymore ...
Good luck to you!  :)

Now it is OK. I was sure that you wanted to write an other link. As you saw , the techniq in your link works also for the new problem. Just an observation !!!

  I wish you .. good Studies in the UNIVERSITY.

 [u]Babis[/u]


How many positive integers less than 2008 have an even number of divisors?

We can see that the only integers that have an odd number of divisors are the perfect squares. There are $ 44$ perfect squares less than $ 2008$, so we have $ 2008\minus{}44\equal{}\boxed{1964}$.

[quote="Doink"]I also got 1964, but alcumus said I was wrong and it was 1963!!?? :mad:  :o[/quote]

Well, see it says LESS than 2008. So you cannot include 2008. The only numbers that have an odd number of divisors are the square roots, and there are 44 square roots less than 2008. So instead of 2008, you have $2007-44=\fbox{1963}$  integers.

I don't understand the solution.  Can anyone help please?  Thank you.

Every integer less than $2008$ has an even number of divisors or an odd number of divisors.  Those with an odd number of divisors are all perfect squares.  Since there are $44$ perfect squares less than or equal to $2007$, we just subtract $44$ from $2007$, and we get $\boxed{1963}$ integers.

Take $27$

27 has 1,27,3,9

The only reason that $x^2$ has even is because when you make pairs, a square number has a repeat!

If you look at $x^4 and x^8$ they are already included in squares.

[hide][img]http://data.artofproblemsolving.com/aops20/latex/images/eb1bb087edb454f5fe091e4e7a135abe67142f58.png[/img][/hide]

in the above hidden text, what do they mean by "d and n/d pair up except for......"?

Is this the official solution?

The proof that $n$ is a perfect square $\Leftrightarrow$ $n$ has an odd number of positive divisors could be worded better, but what it's saying is for each divisor $d$ of $n$, $n/d$ is also a divisor. If $n$ is not a perfect square, one can divide the set of divisors into two sets such that the divisors in one are less than $\sqrt{n}$ and the divisors in the other are more than $\sqrt{n}$. For example, the pairs for $18$ are $\{1,18\},\{2,9\},\{3,6\}$.

On the other hand, if $n$ is a perfect square, there is no counterpart divisor to $\sqrt{n}$. For example, there are pairs $\{1,36\},\{2,18\},\{3,12\},\{4,9\}$ but we are left with a lone divisor, $6$.

[quote="AwesomeToad"][b]Is this the official solution?[/b]

The proof that $n$ is a perfect square $\Leftrightarrow$ $n$ has an odd number of positive divisors could be worded better, but what it's saying is for each divisor $d$ of $n$, $n/d$ is also a divisor. If $n$ is not a perfect square, one can divide the set of divisors into two sets such that the divisors in one are less than $\sqrt{n}$ and the divisors in the other are more than $\sqrt{n}$. For example, the pairs for $18$ are $\{1,18\},\{2,9\},\{3,6\}$.

On the other hand, if $n$ is a perfect square, there is no counterpart divisor to $\sqrt{n}$. For example, there are pairs $\{1,36\},\{2,18\},\{3,12\},\{4,9\}$ but we are left with a lone divisor, $6$.[/quote]

Yes, that was the official solution.

I think the answer should be 1964, because 1 is counted as a perfect square but it doesn't have an even number of divisors.

You do not have to bump every single thread.

What does "the divisors $d$ and $n/d$ pair up, except for when $d=n/d$, or $n=d^2$?" Can someone explain?

For MasterProblemSolver123
[hide=don't click me][hide=don't click me][hide=don't click me]n is the number, d is a divisor and n/d is another divisor, d*n/d=n, when n=d^2(it is a square) n has an odd number of divisors, the middle one is d, so technically d=n/d so d pairs up with itself[/hide][/hide][/hide]

[hide=Solution]All positive integers have an even number of divisors. For example, $9$ has six divisors: $1,3,9,-1,-3,-9$. Therefore, the answer is $\boxed{2007}$.

[quote=montresor]Why would this not be 1964...[/quote]

The answer is not $1964$ because there are only $2007$ integers less than $2008$.

@above I agree with you that Alcumus is inconsistent with its definition of “divisor”.

[hide=Proof]Notice a number with $n$ divisors can be formed with $x^{n-1}$, or any other factors. Notice the factors of an odd $n$ is odd, so then we have even powers. Therefore, every number with an odd number of divisors is a square.[/hide]

[quote name="annwoo" url="/community/p13520525"]
Doesn't this problem mean factors and not divisors?
[/quote]

They're essentially the same.

@everyone, I checked their profile, they're clearly active so don't accuse me of replying to an over 1y old post.

I got james bond on this problem :| my clock said it was 9:08pm pst instead of 9:07 though

[hide=shruggie]so we guess a number whose square is around 2008

we guess 45

45·45=2025, oops slightly too big

so there are 44 squares under 2008, so we subtract 2007-44 to get 1963 :}[/hide]


geometry
circumcircle
cyclic quadrilateral
geometry solved

The diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ meet at a point $X$. The circumcircles of triangles $ABX$ and $CDX$ meet at a point $Y$ (apart from $X$). Let $O$ be the center of the circumcircle of the quadrilateral $ABCD$. Assume that the points $O$, $X$, $Y$ are all distinct.  Show that $OY$ is perpendicular to $XY$.

Let $P=AB\cap CD,Q=AD\cap BC$, and consider the inversion of pole $P$ which invariates the circle $(ABCD)$. Let $O'$ be the image of $O$. $O'$ must lie on the polar of $P$ wrt the circle (draw a circle, an exterior point, its polar, and you'll see why :)). This polar is $XO'=QX\perp PO=OO'\ (*)$, so $PO\perp O'X$. Now try to show that the image of $X$ through this inversion is $Y$ (just consider the power of $P$ wrt the circles $(ABX),(CDX)$, which must be the same as its power wrt $(ABCD)$), so $XYOO'$ is cyclic, and from $(*)$ it now follows that $XY\perp OY$.

Grobber,
that's brilliant!  How do you come up with these solutions?

[Moderator edit: See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=151499#151499]http://www.mathlinks.ro/Forum/viewtopic.php?t=4919 post #7[/url] for a solution of this problem.]

There was a time when basicly all I did was solve problems such as the ones you've posted here lately :).

Well, it certainly seems to have paid off.

Firstly, AYOD is cyclic and also BYOC is cyclic. Then <XYO=<XYD-<OYD=180-<ACD-<OAD=180-<ACD-(180-2<ACD)*1/2=90.


function
induction
combinatorics theorems
combinatorics

I always get puzzled when proving combinatorics equalities. It's said that generating function may be useful...but, how to find them? Were any formulas to proving combinatorics equalities posted here?
For example,
$ \sum_{k\equal{}m}^nC_n^kC_k^m\equal{}2^{n\minus{}m}C_n^m$

Hi
I know what to do. Let's look at your express (identity).
Then, if you're an amateaur then apply induction method on the identity you have posted; so easily:
i) ...
ii)......
Finished. And if you've been busy with combinatorics before, let's go with Koh-Keeh Meng's method... minimazing the upper to the lower. So we have.


calculus
integration
algebra
function
domain
abstract algebra
superior algebra

If $ R$ is a domain and $ I$ is an ideal of $ R$, define the integral closure of $ I$ in $ R$ to be the set of elements $ s\in R$ satisfying an equation of the form $ s^n\plus{}r_1s^{n\minus{}1}\plus{}...r_n\equal{}0$ where $ r_j\in I^j$, the $ j$th power of $ I$ for each $ j$. Show that $ s$ is integral over $ I$ if and only if there is a finitely generated $ R$-module $ N$, not annihilated by any element of $ R$ such that $ sN\subset IN$. Use this to show the integral closure of an ideal is an ideal.

Ok so my idea for "$ \Leftarrow$" is to use calely hamilton theorem treating multiplication by $ S$ as an endomorphism. It works perfectly fine; the problem is in the other direction. I've been playing around with trying to show things like $ I[s]$ have the desired properties but i cant seem to find one that satisfies the property that $ sN\subset IN$. Any ideas? After finding such modules, I was thinking of just showing that there was a module with that property corresponding to $ ss'$. If anyone has suggestions for that step too, input is very welcome =) Thanks in advance

WLOG assume that $ I$ is finitely generated (otherwise, replace $ I$ by the subideal of $ I$ generated by those finitely many elements of $ I$ that are used to have $ r_j\in I^j$ for every $ j$).
The condition $ s^n \plus{} r_1s^{n \minus{} 1} \plus{} ... \plus{} r_n \equal{} 0$ where $ r_j\in I^j$ rewrites as $ s^n\in s^{n \minus{} 1}I^1 \plus{} s^{n \minus{} 2}I^2 \plus{} ... \plus{} s^0I^n$.
Now let $ N \equal{} s^0I^{n \minus{} 1} \plus{} s^1I^{n \minus{} 2} \plus{} ... \plus{} s^{n \minus{} 1}I^0$ (of course, $ I^0$ means $ R$). Then, $ sN\subseteq IN$ since $ ss^kI^{n \minus{} 1 \minus{} k}\subseteq Is^{k \plus{} 1}I^{n \minus{} 1 \minus{} \left(k \plus{} 1\right)}$ for all $ k\in\left\{0,1,...,n \minus{} 2\right\}$ and $ ss^{n \minus{} 1}I^0\subseteq IN$ because of $ ss^{n \minus{} 1} \equal{} s^{n \minus{} 1}I^1 \plus{} s^{n \minus{} 2}I^2 \plus{} ... \plus{} s^0I^n \equal{} I\left(s^{n \minus{} 1}I^0 \plus{} s^{n \minus{} 2}I^1 \plus{} ... \plus{} s^0I^{n \minus{} 1}\right) \equal{} IN$.
The $ R$-module $ N$ is not annihilated by any element of $ R$ since $ N$ is an ideal of $ R$, and $ R$ is a domain (this is the only place where we use that $ R$ is a domain).
Since $ I$ is finitely generated, so is $ N$.

  darij


geometry
inequalities
topology
calculus
calculus computations

Let $ W$ e an open set contained in : $ \{(x,y,z): x^2\plus{}y^2\plus{}z^2\le R^2\}$
Show that $ vol(W)\le \frac{R}{3}area(\partial W)$

notation: $ \partial W$ is the boundary of the set.

By the isoperimetric inequality, $ V^{2/3}\le (4\pi)^{\minus{}1/3}3^{\minus{}2/3}A$.
Also, $ V^{1/3}\le (4\pi)^{1/3}3^{\minus{}1/3}R$ because $ W$ is contained in a ball of radius $ R$.
Multiply these inequalities.


Solve the equation $3^{x}+1=x^{2}y$ in  natural numbers.

[quote="yeppyyep"][quote="e.lopes"]suppose x>1.
let p the smallest prime divisor of x. let x = pA.

we have that $p|3^{2pA}-1$ and $p|3^{p-1}-1$, so, $p|3^{k}-1$, where k = gcd(2pA,p-1).

but k = 2, because gcd(p-1, pA) = 1. so, p = 2.

now the right side is a multiple of 4, and the left side is $\equiv 2 (mod.4)$.  :wink:[/quote]
solution $(1,4)$ works no?  :idea:[/quote]

in the begin i say: 'suppose x>1'

I shall remark in ZetaX's name that this thread hasn't good a title (also look at http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 ) but this thread won't be deleted because you don't have enough posts. (he is at my house that moment)


Let $A\in M_n(R)$.Then $A$ is nilpotent if and only if there exists a sequence of matrices similar to $A$ with nul limit.

Using well known result : $\forall k,Tr(A^k)=0 \Rightarrow A$ is nilpotent

sketch of proof : by Cayley-Hamilton and linearity of $Tr$, $det(A) = 0$, i.e $0$ is an eigenvalue, take a stable hyperplane and proceed by induction.

If A is nilpotent then just take $A_n = A$ as sequence

If $A = lim A_n = {P_n}^{-1} \times A \times {P_n}$ with $lim A_n= 0$
Then $A^k = lim \ A_n^k$
Since $tr(A^k) = tr({A_n}^k) \rightarrow 0$, $tr(A^k) = 0$ and $A$ is nilpotent.


inequalities
geometry
trigonometry
inequalities proposed

Let $a,b,c$ be three positive real numbers with $a+b+c=3$. Prove that \[ (3-2a)(3-2b)(3-2c) \leq a^2b^2c^2 . \]
[i]Robert Szasz[/i]

This is equivalent to

$(a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) \leq 27a^2b^2c^2$

Let $x = -a + b + c, y = a-b+c, z = a+b-c$ and note that at most one of $x,y,z$ can be negative (since the sum of any two is positive).  If $x < 0$ then $(a+b+c)^3xyz \leq 0 < 27a^2b^2c^2$.  If $x,y,z > 0$, note $x+y+z = a+b+c, x+y = 2c$ etc so our inequality becomes

$64xyz(x+y+z)^3 \leq 27(x+y)^2(y+z)^2(z+x)^2$

Note that $9(x+y)(y+z)(z+x) \geq 8(x+y+z)(xy+yz+zx)$ and $(xy+yz+zx)^2 \geq 3xyz(x+y+z)$.  Combining these completes our proof!

I think this can also be proved using triangle.

As noted in the last post, we may assume $a,b,c$ are the sides of a triangle. Multiplying LHS by $a+b+c$ and RHS by $3$, the inequality becomes $16\Delta^2 \le 3a^2b^2c^2$, where $\Delta$ is the area of the triangle. That is equivalent to $R^2\ge \frac{1}{3}$ since $\Delta = \frac{abc}{4R}$, where $R$ is the circumradius. But this is true since $R=\frac{a+b+c}{2(\sin{A}+\sin{B}+\sin{C})}$ and $\sin{A}+\sin{B}+\sin{C} \le \frac{3\sqrt{3}}{2}$ by Jensen.

Let x=3-2a y=3-2b ve z=3-2c than inequality is

xyz is less than or equal to (x+z)^2(y+z)^2(x+y)^2 / 8

But we now from AM-GM   xyz is less than or equal to (x+z)(y+z)(x+y) / 8 and (x+z)(y+z)(x+y) is less than or equal to (x+y+z)^3 / 3^3 .

And we now that x+y+z=a+b+c=3, so that (x+z)(y+z)(x+y) is less than or equal to 1.

Conclusion follows....

Sorry for my poor english........

[quote="Valentin Vornicu"]Let $ a,b,c$ be three positive real numbers with $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ (3 \minus{} 2a)(3 \minus{} 2b)(3 \minus{} 2c) \leq a^2b^2c^2 .
\]
[i]Robert Szasz[/i][/quote]
$ F(a,b,c)\equal{}a^{2}b^{2}c^{2}\minus{}(3\minus{}a)(3\minus{}b)(3\minus{}c)$
$ F(\frac{a\plus{}b}{2},\frac{a\plus{}b}{2},c)\equal{}\frac{(a\plus{}b)^{4}}{16}c^{2}\minus{}(3\minus{}(a\plus{}b))^{2}(3\minus{}2c)$
$ F(\frac{a\plus{}b}{2},\frac{a\plus{}b}{2},c)\minus{}F(a,b,c)\equal{}(a\minus{}b)^{2}[\frac{c^{2}}{16}(a^{2}\plus{}b^{2}\plus{}6ab)\plus{}2c\minus{}3]$
We have $ \frac{c^{2}}{16}(a^{2}\plus{}b^{2}\plus{}6ab)\plus{}2c\minus{}3\leq 2\frac{c^{2}}{16}(3\minus{}c)^{2}\plus{}2c\minus{}3$
we suppose c=min{a,b,c}
$ \Rightarrow F(a,b,c)\geq F(x,x,c)$ (x=a+b/2)
we need prove :
$ F(c)\equal{}\frac{1}{16}c^{2}(3\minus{}c)^{4}\minus{}c^{2}(3\minus{}2c)\geq 0$
$ G(c)\equal{}\frac{1}{16}(3\minus{}c)^{4}\minus{}(3\minus{}2c)$
G'(c)<0 $ \Rightarrow G(c)\geq G(1)\equal{}0$
Done !

Use sin<A+sin<B+sin<C≤3√/2 in triangle. It can be the easy inequality if se put sin<B+A=sin<C... This problem is the direct consequence of that statement.

[quote=ThAzN1]This is equivalent to

$$(a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) \leq 27a^2b^2c^2$$

[hide=*]Let $x = -a + b + c, y = a-b+c, z = a+b-c$ and note that at most one of $x,y,z$ can be negative (since the sum of any two is positive).  If $x < 0$ then $(a+b+c)^3xyz \leq 0 < 27a^2b^2c^2$.  If $x,y,z > 0$, note $x+y+z = a+b+c, x+y = 2c$ etc so our inequality becomes

$64xyz(x+y+z)^3 \leq 27(x+y)^2(y+z)^2(z+x)^2$

Note that $9(x+y)(y+z)(z+x) \geq 8(x+y+z)(xy+yz+zx)$ and $(xy+yz+zx)^2 \geq 3xyz(x+y+z)$.  Combining these completes our proof[/hide]![/quote]
$$(a+b)(b+c)(c+a)(b+c-a)(c+a-b)(a+b-c) \leq 8a^2b^2c^2$$https://artofproblemsolving.com/community/c6h1785721p11793688

Let $x = a+b-c$,$y = b+c-a$ and $z = c+a-b$ Note that  we need to prove $xyz \le ((\frac{x+z}{2})(\frac{y+x}{2})(\frac{z+y}{2}))^2$ Note that if one of $x,y,z$ is negative then inequality is obvious so assume $x,y,z \ge 0$ and also Note that $x+y+z = 3$. Note that we need to prove $64xyz \le ((x+z)+(y+x)+(z+y))^2$ or $81xyz \le ((x+y+z)(xy+yz+zx))^2$ or $9xyz \le (xy+yz+zx)^2$ which is true since $(xy+yz+zx)^2 \ge 3(x+y+z)xyz$ and $x+y+z = 3$.


Let $f: \mathbb{N}\rightarrow \mathbb{N}$ such that $f(n+1) > f(n)$ and $f(f(n))=3n$ for all $n \in \mathbb{N}.$ Determine $f(1992).$

[Note: If you know the solution, please don't write it. Just provide a hint or two to those who wish to attempt to solve it.]

[quote="ashwinrk_jain"]$f(n) = \sqrt{3}n$
hence $f(1992) = 1992\sqrt{3}$[/quote]

the range of $f(x)$ is integers greater than 0, so that does not work

EDIT: oh and a hint:

[hide="..."]consider $f(f(f(n)))$[/hide]

$f^{2k}(n) = 3^{k}n, f^{2k+1}(n) = 3^{k}f(3n)$

Also, you're basically trying to find what x is s.t. $f(x) = 5976$..

i don't have the time to think about this but in the 10 min i looked over it i got the following:
let $3*3^{k}>n \ge 3^{k}$ then we have $f(n)=n+3^{k}$ if $2*3^{k}\ge n$ and
$n \ge 3^{k}$.else we have $f(n)=3n-3*3^{k}$.i guess we can prove this by induction :wink: .if i'll have the time i will post my entire solution later.so $f(1992)=3789$.but i did not have the time to check this so it might be wrong.

[hide="here is my progress:"]Using the equation, we get: \[f(f(f(n)))=f(3n)=3f(n)\] If for natural numbers $n$ and $m$, $n>m$, we can show that $f(n)>f(m)$ through induction. Therefore, the fucntion is one-to-one and has an inverse.

Also I have been looking at the fact that $f(f(n))$ is divisibly by 3, but I am not sure if that helps. [/hide]

i have checked what i wrote above and it is correct.if anyone wants i'll post my solution or some more hints


Four positive integers $ a,b,c,$ and $ d$ have a product of 8! and satisfy\begin{align*}ab \plus{} a \plus{} b &\equal{} 524\\
bc \plus{} b \plus{} c &\equal{} 146\\
cd \plus{} c \plus{} d &\equal{} 104.\end{align*} What is $ a \minus{} d$?

$ \textbf{(A)} \ 4 \qquad \textbf{(B)} \ 6 \qquad \textbf{(C)} \ 8 \qquad \textbf{(D)} \ 10 \qquad \textbf{(E)} \ 12$

[hide]add $1$ to both sides of each equation and factor to get:
$(a+1)(b+1)=525=3*5^2*7$
$(b+1)(c+1)=147=3*7^2$
$(c+1)(d+1)=105=3*5*7$
Divide the first by the second
$\frac{a+1}{c+1}=\frac{25}{7}$
thus $a+1$ is a multiple of $25$, and $c+1$ is a multiple of $7$
trying out $25$ and $7$ for these, we get $b+1 = 21$, and $d+1=15$
in this case, $abcd=24*20*6*14=8*3*5*4*6*7*2=8!$, so $24-14=\boxed{10 (D)}$[/hide]

[hide="solution"]
This solution's a little messy, but here it goes:
Note that we can factor these equations as
$(a + 1)(b + 1) = 525$
$(b+1)(c+1) = 147$
$(c + 1)(d + 1) = 105$

Now let $a + 1 = a'$, and similarly for the other variables.  We know that because $a,b,c,d$ are positive integers, $a', b', c', d'$ must also be positive integers greater than one such that

$a' b' = (3)(5^2)(7)$
$b' c' = (3)(7^2)$
$c' d' = (3)(5)(7)$

Looking at the equations quickly shows that there are two possible solutions, depending on how we distribute the 3s to the variables: either $a' = (3)(5^2), b' = 7, c' = (3)(7), d' = 5$, or $a' = 5^2, b' = (3)(7), c' = 7, d' = (3)(5)$.  Calculating $abcd$ from these solutions shows that the second set is the correct one, and we have $a - d = 24 - 14 = 10$, or $(D)$.
[/hide]

Edit: Never mind, I thought the equations were symmetric. :blush:

[hide="A Number Theoretic Approach"]So start off with:

$(a+1)(b+1) = 5^{2}\cdot 3 \cdot 7$
$(b+1)(c+1) = 7^{2}\cdot 3$
$(c+1)(d+1) = 3 \cdot 5 \cdot 7$

Note that we have $3$ possible values for $b+1$ by observing the first two equations.

[b]Case 1: b + 1 = 3[/b]
A contradiction arises in the third equation, as this implies that $c+1 = 7^{2}.$

[b]Case 2: b + 1 = 7[/b]
Solving, we get $a = 74, b = 6, c = 20, d = 4.$ However, $37|a$ and $37$ is a prime that doesn't go into $8!.$ Contradiction.

[b]Case 3: b + 1 = 7 \cdot 3[/b]
This case works out well. Solving, we get $a = 24, b = 20, c = 6, d = 14.$ Subtracting yields $a-d = 10.$[/hide]

I understand the solution, but what I don't understand is how it occurred to yall to separate 525 into 3(5^2)(7), 147 into 3(7^2), and 105 into 3(5)(7). What I'm really asking is if there is a series of steps that ran through your minds that allowed you to see that factoring those out would get you further in the problem. Thanks for your help!

Mary Catherine

I think you're saying that you wonder how they thought of prime factorizations?  The whole idea of products sharing similar factors.  If it's not obvious to you at first, and you want to be able to see it quicker, it's the same as learning when to apply other theorems, tips, and tricks.  Practicing those types of problems.

Whenever you're trying to solve problems which are restricted to integers you should always try to factor as much as possible.

Wait this problem is really troll because on the pdf it shows everything on one line, so I misread it as $ab+a+b = 524bc+b+c =146cd+c+d = 104$ because that's what it showed up as... smh. Why does the formatting always mess up in the pdfs?!?!?!


Functional Analysis
calculus
calculus computations

Consider the Banach space $ l^1$. Let $ S\equal{}\{x\in l^1| \parallel{}x\parallel{}\leq1\}$. Is $ S$ a compact subset of $ l^1$

If I understand the problem correctly, no; it is easy to exhibit sequences with no convergent subsequences, like the sequence $ s_k$ where each entry is zero except for a one in the $ k^{th}$ position.  Every element of the sequence is at a distance $ 2$ from every other element.

Inappropriate to this forum [b]and[/b] very recently posted in another forum. countdownmath, please stop posting such things. I'm about to delete several of your other posts.


probability
calculus
integration
expected value

Here's a problem I came up with. For some reason I haven't solved it yet (at least not well enough to be certain of my answer).

I have an unfair coin. p of the time it lands heads and 1-p of the time it lands tails. I flip it n times, and I get all tails except that on the last flip I get heads. What is the expected value of p?

Well I have no idea if this is how expected values are usually found out over continuous ranges but here we go anyway..
First switch p and 1-p around to make it easier :).
For a given p the probability of that happening is y=(1-p)p^(n-1).
So to find expected value of p (or 1-p), we want:
integral from 1 to 0 of ((1-p)^2 p^(n-1)) dp) over the integral from 1 to 0 of ((1-p)p^(n-1)). Or basically, the sum of all py over the sum of all p.

After a bit of algebra this is (1/n+2 - 2/n+1 + 1/n)/(1/n(n+1))
= 2/n+2.

So the expected value of p as far as I can work is out is n/n+2. Interestingly this isn't equal to n-1/n which you would probably expect.. unless I just made a mistake!

umm... you could try using Bernoulii's Formula. It finds the probability of an event happening a specific number of times out of a total number of trials, i.e. flipping a coin 10 times and getting exactly 7 heads. The formula is:
nCr * p^r * q^n-r
where n is the total number of trials, r is the desired number of specific events, p is the probability of the event happening, and q is the probability of the event not happening (usually 1-p.) If you put the values of your problem into the formuloa, you get:
nCn-1 * p^n-1 * r^n-n-1
nCr for any number where r is one less than n simplifies to n.
Therefore, you can simplify it to:
n * p^n-1 * 1/r
Don't forget, r is p-1.
Therefore, as far as I can simplify, it comes to:

[u]n * p^n-1[/u]
   p-1

This finds out the probability of geting n-1 tails and 1 head in any order. Since we know the head is last, we drop the nC1. Also watch out when you are subtracting n and n-1, their difference is 1 not -1.

I used this formula when working out the expected value of p.. see my other post. (in which I think I mixed up a 1-p and p somewhere but the answer still comes to n/n+2.)


calculus
integration
SFFT
blogs
special factorizations

How many distinct positive integral solutions of the equation $ a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ac \equal{}89$ are possible?

[hide="In its most specific case, this"] "Simon's Favorite Factoring Trick," as it is often called:

$ ab\plus{}a\plus{}b \equal{} (a\plus{}1)(b\plus{}1)\minus{}1$ [/hide]

[quote="t0rajir0u"][hide="Hint"] A variant of SFFT might be handy. [/hide][/quote]

Wouldn't an $ abc$ term be needed for that here?  :maybe:

Hm. I've got a solution involving SFFT, but it is not satisfying at all.

[hide="Major casework bash"]

WLOG $ a\leq b\leq c$. It is easy to see that $ a\leq4$.

$ a \equal{} 1$:

$ bc\plus{}2b\plus{}2c\plus{}1 \equal{} 89$
$ (b\plus{}2)(c\plus{}2) \equal{} 92$

$ (1,2,21)$

$ a \equal{} 2$:

$ bc\plus{}3b\plus{}3c\plus{}2 \equal{} 89$
$ (b\plus{}3)(c\plus{}3) \equal{} 96$

$ (2,3,13)$
$ (2,5,9)$

$ a \equal{} 3$:

$ bc\plus{}4b\plus{}4c\plus{}3 \equal{} 89$
$ (b\plus{}4)(c\plus{}4) \equal{} 102$

No solutions.

$ a \equal{} 4$:

$ bc\plus{}5b\plus{}5c\plus{}4 \equal{} 89$
$ (b\plus{}5)(c\plus{}5) \equal{} 110$

$ (4,5,6)$

So we have four unique solutions:

$ (1,2,21)$
$ (2,3,13)$
$ (2,5,9)$
$ (4,5,6)$
[/hide]

I don't know how much this will help; just throwing out an idea?

[hide]
$ \sum_{cyc}ab\plus{}a \equal{}\sum_{cyc}ab\plus{}\frac{a}{2}\plus{}\frac{b}{2}\equal{}\sum_{cyc}\left( a\plus{}\frac{1}{2}\right)\left( b\plus{}\frac{1}{2}\right)\minus{}\frac{1}{4}\equal{}\frac{1}{4}\left(\sum_{cyc}(2a\plus{}1)(2b\plus{}1)\minus{}1\right)$

So substituting $ 2a\plus{}1 \equal{} A$, etc, it becomes:
$ AB\plus{}BC\plus{}CA \equal{} 359$, solve for triple of odd integers (A,B,C).
The thing is, though, I don't know if this is any simpler  :(

I suppose you could then be like, $ (A\plus{}C)(B\plus{}C) \equal{} 359\plus{}C^{2}$, but I don't know if that'll help either.

Somehow, I have a feeling that no matter what we do, it'll boil down to casework of the same nature as that presented above.
[/hide]

This looked promising at first but then I'm stuck...
[hide]Major step: add 1.
$ a\plus{}b\plus{}c\plus{}ab\plus{}ac\plus{}bc\plus{}1 \equal{} 90$
Now we can do
$ (b\plus{}1)(c\plus{}1)\plus{}a(b\plus{}c\plus{}1) \equal{} 90$ and so on...
then... um....
maybe $ (a\plus{}1)(b\plus{}c\plus{}1)\plus{}ab\equal{}90$

[/hide]

If that's your approach, t0rajir0u's hint suggests adding $ abc$ to both sides as well. This gets you $ (a\plus{}1)(b\plus{}1)(c\plus{}1) \equal{} 90\plus{}abc$. Looks all nice and clean, but I failed to do anything with it. Maybe it would make the case bash easier though.

[quote="MellowMelon"]If that's your approach, t0rajir0u's hint suggests adding $ abc$ to both sides[/quote]

Actually, I was thinking about what The Zuton Force did.  If nothing else, it makes the casework slightly more straightforward.

Why is it called "Simon's Favorite Factoring Trick"? And who is "Simon"?

I believe Simon is one of the older members of this forum, and apparently this trick is Simon's favorite factoring trick.

[[Simon's Favorite Factoring Trick]] includes a link to the thread on which the name is based.


It is given a trapezium ABCD it is known that AD=BC. In the triangles ACD and ABC we know that the radii of incircles are r1 and r2 respectively. It is known that we can inscribe a circle in ABCD. Find the radii (r) of inscribed circle in ABCD in therms of r1 and r2.

The answer given is $ r\equal{}\frac{r_1\plus{}r_2\plus{}\sqrt{r_1^2\plus{}r_2^2}}{2}$
I can express r_1, r_2 and r in therms of a, and b (bases of the trapezoid). But there are long expressions.
Is there a better approach?
I also considered the tangents formed by k1 and k2 and the bases of the trapezium but i still don't have a good results.

How to attack this problem?

The problem is solved by me and two more people on a Bulgarian math forum. My question above is not relevant.


With [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=29491]ElChapin[/url] and his team :)

Our Bangladesh team. From left to right: Alba (our guide), me, Ishfaq, Moon, Nazia.

Well, me with two Croatian contestants :P

The Great Exam Hall  :)

(From the IMO official website)

Ishfaq and me with [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=32134]bambaman[/url] :)

With two contestants from Cyprus and Greece

Us with the Indonesian team, before going to the closing


MATHCOUNTS blogs algebra polynomial Vieta calculus conics

OK, now that everyone has submitted their answers, it's time to release the final results!! :D

[hide="In fourth place, with 34 points, is..."]
[size=100]PI-Dimension[/size]!
[/hide]
[hide="In third place, with 35 points, is..."]
:winner_third: [size=134]Bachukas[/size]!
[/hide]
[hide="In second place, with 37 points, is..."]
:winner_second: [size=167]andersonw[/size]!!
[/hide]
[hide="And the champion of the 2008 Yongyi National MATHCOUNTS Competition, with 39 points, is..."]
:winner_first: [size=200]tinytim[/size]!!!
[/hide]

Congratulations to all who participated... you did an amazing job on my mock test!

Special thanks to tinytim, who donated his time to proofread my test and fix semantical errors!

A full analysis of individual contestants, as well as the total number of people who got a certain problem correct, as well as the problems in order of difficulty, can be found in my Excel file.

The answer key can also be found below.

Note: Isabella2296 requested that her score not be shown, therefore you will not find a row containing her score.

At this point, feel free to discuss problems, such as your favorite (and least favorite) problems!! :lol:

[hide="Interesting information"]
[list]
[*]Nobody correctly answered #25 on the sprint (Answer: [b]299[/b]) or #8 on the target (Answer: [b]209.06[/b]).
[*]Andersonw and Bachukas were the only two people to correctly answer #6 on the sprint (Answer: [b]34[/b]).
[*]Andersonw, math154, and xpmath were the only three people to correctly answer #11 on the sprint (Answer: [b]128[/b]).
[*]Tinytim, 27r, and isabella2296 were the only three people to correctly answer #27 on the sprint (Answer: $ \frac 43$)[/list]
[/hide]

The test was even harder than nats so i desperately failed :(
im gonna go to my blog crying why do i have to be so stupid :(

Don't worry. There's a reason I didn't want my score to be shown.  :roll:

Ha!!! I predicted that I would lose to tinytim.  :lol:

Anyone know how to do #25?

I thought #24 would have been a better question if the polynomial wasn't factorable (unless if vieta's formulas aren't mathcounts level, I'm not sure).
#2 on the target was pretty cool, and #3 on the target was very tricky.
I still don't understand #8 on the target :P

It is very wrong to be dead-last when your score is substantially lower than the person second-dead-last.

#29 on the sprint was fun.

Argh, stupid target. Well, more like stupid me, since I only turned in 3 target answers.

i liked #20.

idk if theres an other way to do it, but i used calculus to solve it  :rotfl:

For #20, I assumed the center of the bottom of the tunnel was at (0,0).
Then, the equation of the archway would be $ f(x)\equal{}25\minus{}\frac{1}{4}x^2$ since it's a parabola with a y-intercept of 25 and x-intercepts of 10 and -10.
Then, plug in 4 because the train is 8 feet wide, so $ 25\minus{}\frac{16}{4}\equal{}\boxed{21}$

The target really pwned me so I only got 4 problems right :(

*gasp* YAY 1000th POST!!!!!!!!! :D

I just decided to pick random numbers for #2 on the target and on my 4th try, I got it!! :rotfl:

[quote="andersonw"]For #20, I assumed the center of the bottom of the tunnel was at (0,0).
Then, the equation of the archway would be $ f(x) \equal{} 25 \minus{} \frac {1}{4}x^2$ since it's a parabola with a y-intercept of 25 and x-intercepts of 10 and -10.
Then, plug in 4 because the train is 8 feet wide, so $ 25 \minus{} \frac {16}{4} \equal{} \boxed{21}$[/quote]

heres how i solved:

i assumed the lower left corner of the tunnel was at (0,0), and since the parabola is passing the origin, it must have the form of $ f(x) \equal{} ax^2 \plus{} bx$.

$ f(20) \equal{} 400a \plus{} 20b \equal{} 0$
$ f(10) \equal{} 100a \plus{} 10b \equal{} 25 \implies 400a \plus{} 40b \equal{} 100 \implies 20b \equal{} 100 \implies b \equal{} 5, a \equal{} \minus{} \frac {1}{4}$

so $ f(x) \equal{} \minus{} \frac {1}{4}x^2 \plus{} 5x$

were looking for $ f(6)$ and its $ \minus{} 9 \plus{} 30 \equal{} 21$

oh hey i didnt use calculus after all  :rotfl:

That's exactly what I did...thanks to conic sections :D

What two crazy people put "as easy as A-B-C"?  :maybe:

I put easy as A-B-C just to mess up the survey. :D  :rotfl:  :rotfl:

[quote=alpha_2]how is #29 solved on sprint?[/quote]

pells equation

it's sprint
is there a faster way?
it was literally ranked the 6th easiest on the excel stats spreadsheet

[quote=alpha_2]it's sprint
is there a faster way?[/quote]

it shouldn’t take too long but idk any other way
also the excel sheet says it’s the 5th hardest problem

Sad to see almost all the people who posted above are now inactive :(

lol this test was taken like 2 years before i was born XD

also, @5above   these people are probably all of a job and a life, and probably dont even know that they still have an Aops account

[quote=Jndd]bru i was like 4 months old when this thread started[/quote]

Lol! I wasn't even born!!

[quote=alpha_2][quote=kn07][quote=alpha_2]no that's the wrong thing
you're just reading the number of people who got it right
but i still need a solution tho[/quote]

i think thats the ranking literally, 6 was the second hardest because only 2 people got it right, and no one got 25 right, hence why its the hardest problem. for example, 2 is the easiest, 1 is the second easiest, and so on.[/quote]

6 was not the 2nd hardest
it was the 15th hardest
refer to the order of difficulty

2 is the easiest, then 1 is the second easiest, then 3, 13, 18, and then 29

just use the literal ranking[/quote]

"Nobody correctly answered #25 on the sprint (Answer: 299) or #8 on the target (Answer: 209.06).
Andersonw and Bachukas were the only two people to correctly answer #6 on the sprint (Answer: 34).
Andersonw, math154, and xpmath were the only three people to correctly answer #11 on the sprint (Answer: 128).
Tinytim, 27r, and isabella2296 were the only three people to correctly answer #27 on the sprint (Answer: $ \frac 43$)"-yoongyi

Hmmmm... it's almost as if my evaluation of the excel sheet is correct, these 4 problems are all in the top 4 hardest problems. The ranking describes each problem listed from easiest to hardest. 29 is the 5th hardest problem.


parameterization
algebra
polynomial
conics
parabola
hyperbola
system of equations

Find the values of the parameter $ a$ such that the system of equations:

$ ax^2 \minus{} 2x \plus{} y^2 \minus{} 2 \equal{} 0$
$ \minus{} 3x^2 \plus{} x \minus{} 2y \plus{} 2 \equal{} 0$
has [b]two[/b] solutions in real numbers.
I tried to substitute y from second to first equations but this leads to a really nasty 4-degree equation in x-- not even applying it the Rolle sequence did't give me much...

You aren't solving for $ x$; you're finding the cases where there are only two solutions for $ x$.  When does a fourth-degree polynomial have two solutions?

Alternately, a geometric interpretation:  when do the graphs of a circle and a parabola intersect exactly twice?  Generically, they intersect in four points.  (Actually, the first graph can be a circle, parabola, or hyperbola depending on the sign of $ a$.)


Given a set of coins in the plane, all with different diameters. Show that one of them is tangent to at most 5 of the others.

[hide]

Consider the smallest coin. Suppose this coin is tangent to more than five other coins. If it is tangent to more htan six coins, than one of those coins is smaller than our given coin (trivial to show by angles etc), and we have a contradiction. Suppose six coins are tangent to it; than either they are all equal to our coin, or one of them is smaller to our coin.. the first case violates the fact that they are all different diameters, and the second contradicts our assumption that the given coin is minimal. Thus, our coin can be tangent to at most five other coins.

[/hide]

Thanks, I was trying to prove with maximality instead of minimality... I don't know how i couldn't see it.. :blush:


An organic acid A on heating with $ AlPO_4$ at 700 degree forms B. B also reacts with A to give C. C on reaction with 1,3,5 trimethyl benzene in presence of $ AlCl_3$ gives a ketone D and $ CH_3COOH$. The compound D on treatment with $ Na(Hg)/HCl$ gives an aromatic hydrocarbon A. Give structures of A to E with proper reasoning.

Haven't looked up on organic chemistry for a really long time.... so dont believe my answers straightaway

Okay, $ AlPO_4$ on heating to high temperatures gives $ P_2O_5$, if thats what you wanted, Da Vinci.

So, A is acetic acid.
B=ketene  :D
C=acetic anhydride
D=Just attach a $ CH_3CO\minus{}$ group anywhere on the benzene ring of mesitylene.
E=Reduce the above carbonyl.

[quote="hell_ever"]
Okay, $ AlPO_4$ on heating to high temperatures gives $ P_2O_5$, if thats what you wanted, Da Vinci.

[/quote]

just what i needed,...thanks....... :)


There is a graph $G$. The number of vertices is even and equals to $N$.
The number of edges is $M$.
The degree of each vertex is between $1$ and $3$.
We want to paint the vertices in the following way:
1) We paint the vertices in black or white.
2) There is no $2$ vertices such that they connected with an edge and are both white.
3) The number of white vertices is bigger that the number of black vertices.
You need to find the easiest way for constructing a group of vertices A such that for all accessible paintings of the graph the number of black vertices in group A will be more that the number of white vertices in group A.

I had a mistake in the problem I have written here.
About the group - you need to find a group such that the number of black vertices will be more than the number of white vertices in all accessible paintings.


Hello,

I'm having a problem with how to prove that if $T \in L(V)$ is diagonalizable then $V=im(T) \oplus ker(T)$.
Could anyone help? ^^

Use a basis of eigenvectors. The eigenvectors of the eigenvalue zero are a basis for the null space, and the remaining eigenvectors are a basis for the image.


hi every body
Given two triangle with same area, is there a manner to cut one of them into pieces and form the other one.(MIT TECH REVIEW, january 2009)

Thanks a lot.
But i would like to see at least one manner;and to have an idea about the minimum cuts.


How many numbers greater than ten thousand can be formed with the digits 0, 1, 2, 2, 3 without repetition?? (Note that the digit 2 appears exactly twice in each number formed)

WOW!!! This is SO CONFUSING!!!

We can arrange these digits in $ \frac{5!}{2}$ ways.

There are $ \frac{4!}{2}$ combinations which start with a digit 0.

So the answer is $ \frac{5!}{2}\minus{}\frac{4!}{2}\equal{}60\minus{}12\equal{}48$ numbers  :)

I don't quite get it...

Why is it over 2??

When you are trying to find the number of ways to arrange the letter in the word MOON, there are not exactly 24 ways to arrange them, because MONO is the same word as the word you get, when you switch the O's. There are a total of two ways to arrange the O's in every word when there are ever two spaces left. So instead of having 24, we dvide by 2 to get 12.

Similarly, we do the same with numbers. The trick we use now is called complementary counting. Instead of counting the number of numbers that aren't having 0 as the ten-thousands digit, we find the number of five digit numbers, and then we subtract the number of numbers where 0 is the ten-thousands digit. There are a total of 5! ways to arrange the five digits. That is wrong, because there are two 2's. We divide 5! by 2 to get; 60. Then we subtract the number of ways to get a number with 0 as the ten-thousands digit. Once zero is the ten-thousands digit, there are 12 ways to arrange the remaining four digits because there are two 2's left. So the answer is 60 - 12. Which is 48.

If you still don't understand, then I suggest you get the Intro to Counting and Probability. It helps.

Also, instead of complementary counting:

There are $ 4$ choices for the first digit.  This is because we cannot use $ 0$ as a first digit.

There are $ 4$ choices for the second digit.  This is because we cannot use the first digit as the second.

There are $ 3$ choices for the third digit.  This is because we cannot use the first nor second digit as the third.

There are $ 2$ choices for the fourth digit.  This is because we cannot use the first, second, nor third digit as the fourth.

There is $ 1$ choice for the fifth digit.  This is because we cannot use the first, second, third, nor fourth as the fifth.

Thus, we have $ 4 \cdot 4 \cdot 3 \cdot 2 \cdot 1\equal{}96$ numbers we can make, right?

Wrong.

Let one of the $ 2$'s be denoted as $ 2_1$, and the other as $ 2_2$.  Then $ 32_112_20$ and $ 32_212_10$ are two different numbers, right?  No.  $ 32120\equal{}32120$, so each of our numbers has been double-counted!  To resolve this, we simply divide our total by $ 2$, to get our final answer of $ \frac{96}{2}\equal{}\boxed{48}$.

I hope I helped.

[quote="Bachukas"]Wow, nice teaching Wicke and AIME15  :thumbup:[/quote]

AIME15's solution was better. I think AIME15's solution is better for harder problems, but it can still work in this one.

I think I should clarify one of Wickedestjr's points.  We're not really dividing by $ 2$, we're dividing by $ 2! \equal{} 2$ (sorry if that sounds weird).  Here's the generalization: Suppose we have a collection of $ n$ digits.  Now suppose that digit $ \hat{p}$ repeats $ p$ times, digit $ \hat{q}$ repeats $ q$ times, and so on up to digit $ \hat{t}$ which repeats $ t$ times.  Then the total number of words that can be formed is
$ \frac {n!}{p!\cdot q!\cdots t!}$

Here's a practice problem from this year's VTRMC:
How many sequences of $ 1$'s and $ 3$'s can be made such that the sum of the digits is $ 16$?  For example, $ \{3,3,3,3,3,1\}$ and $ \{3,1,3,1,3,1,3,1\}$ both work.
[hide="Answer"]The answer is 277.  I'll leave you to provide the solution.  It's not hard.[/hide]

[quote="JRav"] We're not really dividing by $ 2$, we're dividing by $ 2! \equal{} 2$  quote]

I know, that is pretty straightforward, because if the two digits are u1 and u2, and they are the last two digits there are 2 for the tens digit, and 1 remaining for the ones digit.

This product is 2 x 1 which equals 2.

Obviously.  Anyone have a solution for my practice problem?  It's not too hard.

Alright, we know that $ 16 \equal{} 3a \plus{} b$, and $ a$ and $ b$ are natural numbers.

So if $ a \equal{} 1$, then we have $ (1,13)$.
If $ a \equal{} 2$, then we have $ (2,10)$.
If $ a \equal{} 3$, then we have $ (3,7)$.
If $ a \equal{} 4$, then we have $ (4,4)$.
If $ a \equal{} 5$, then we have $ (5,1)$.

If we have one three and thirteen ones, then there are $ 13$ ways to arrange these digits.
If we have two three's and ten ones, then there are $ 12!/10!2!$, which equals $ 66$.
If we have three three's and seven ones, then there are $ 12!/3!7!$ which equals $ 15840$.
If we have four three's and four ones, then there are $ 12!/[(4!)^2]$ which equals $ 831600$.
If we have five three's and one one, then there are $ 12!/5!1!$ which equals $ 3991680$.

Adding all these numbers together, yields the sum; $ 4839186$.

You really messed up your algebra.  That's not even close.

Actually, it was a typo.  The answer is 277, which I've corrected.  Either way, 4839186 is way off.

I can't figure out what I did wrong. Is there a really easy way of doing it, or is it solved by casework?

[quote="AIME15"]You have a good approach, however, your later cases were evaluated incorrectly.

For example, in your last case, with five three's and one one, how do you get 12!/5!1!?  Shuldn't it be 6!/5!1!?[/quote]

Oh, I think I accidentally did $ 12!$ for all the cases except the first. I'm an addicted to the number $ 12$.  :lol:

Alright, lets do this long strenuous problem once again.

We know that $ 16 \equal{} 3a \plus{} b$, and $ a$ and $ b$ are natural numbers.

So if $ a \equal{} 1$ then we have $ (1,13)$.
If $ a \equal{} 2$, then we have $ (2,10)$.
If $ a \equal{} 3$, then we have $ (3,7)$.
We also have $ (4,4)$ and $ (5,1)$.
Oh, wait a minute. I forgot $ a \equal{} 0$. Aww, I'm so stupid. If $ a \equal{} 0$, then we have $ (0,16)$.

For $ (0,16)$, there is one way to arrange sixteen ones. $ 1$
For $ (1,13)$, there are fourteen ways to arrange the thirteen ones and the three. $ 14$
For $ (2,10)$, there are sixty six ways. $ 66$
For $ (3,7)$, there are $ 120$ ways.
For $ (4,4)$, there are $ 70$ ways.
For $ (5,1)$, there are $ 6$ ways.
Add these numbers up;

$ 1 \plus{} 14 \plus{} 66 \plus{} 120 \plus{} 70 \plus{} 6$
$ 15 \plus{} 186 \plus{} 76$
$ 201 \plus{} 76$
$ 277$

Yay, I solved it.

@JRav, your formula is just like choosing 16 objects from 16 objects, 1 object from 14 objects, ... 5 objects from 6 objects. etc.

Is that the easiest way of solving it?

That's how I solved it on the contest.  The one additional thing I did was prove that no such sequence could contain an odd number of elements.  Can you do it?  It's a simple proof by contradiction.

Here's the "official" solution:
For each positive integer $ n$, let $ f(n)$ denote the number of sequences of $ 1$'s and $ 3$'s.  That sum to $ n$.  Then $ f(n\plus{}3)\equal{}f(n\plus{}2)\plus{}f(n)$, and we have $ f(1)$, $ f(2)\equal{}1$, and $ f(3)\equal{}2$.  Thus $ f(4)\equal{}f(3)\plus{}f(1)\equal{}3$, $ f(5)\equal{}f(4)\plus{}f(2)\equal{}4$, $ f(6)\equal{}6,\ldots,f(15)\equal{}189$, $ f(16)\equal{}277$.  Thus the number of sequences required is $ 277$.

[quote="JRav"]That's how I solved it on the contest.  The one additional thing I did was prove that no such sequence could contain an odd number of elements.  Can you do it?  It's a simple proof by contradiction.

Here's the "official" solution:
For each positive integer $ n$, let $ f(n)$ denote the number of sequences of $ 1$'s and $ 3$'s.  That sum to $ n$.  Then $ f(n \plus{} 3) \equal{} f(n \plus{} 2) \plus{} f(n)$, and we have $ f(1)$, $ f(2) \equal{} 1$, and $ f(3) \equal{} 2$.  Thus $ f(4) \equal{} f(3) \plus{} f(1) \equal{} 3$, $ f(5) \equal{} f(4) \plus{} f(2) \equal{} 4$, $ f(6) \equal{} 6,\ldots,f(15) \equal{} 189$, $ f(16) \equal{} 277$.  Thus the number of sequences required is $ 277$.[/quote]

Well let $ a$ be the number of threes, and let $ b$ be the number of ones. If $ 16 \equal{} 3a \plus{} b$, then that means that $ b \equal{} 16 \minus{} 3a$. The number of elements is $ a \plus{} b$.

We can rewrite $ a \plus{} b$ by substituting to get $ 16 \minus{} 2a$. $ 16 \minus{} 2a$ is divisible by $ 2$, so it is therefore even.

In the solution to that question, why does $ f(n \plus{} 3) \equal{} f(n \plus{} 2) \plus{} f(n)$ ?


Let [b]a, b, c[/b] be non-negative numbers such that a^2+b^2+c^2=3. Then,
[b]1/(5-2ab)+1/(5-2bc)+1/(5-2ca)<=1[/b]

[quote="Inequalities Master"]Let $ a, b, c$ be non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}3.$ Then,
$ \frac{1}{5\minus{}2ab}\plus{}\frac{1}{5\minus{}2bc}\plus{}\frac{1}{5\minus{}2ca}\leq1$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=109029

Sorry,arqady.but in that forun is only a proof for the following inequality:
If $ a^2\plus{}b^2\plus{}c^2\equal{}1$ then $ \dfrac{1}{5\minus{}6ab}\plus{}\dfrac{1}{5\minus{}6bc}\plus{}\dfrac{1}{5\minus{}6ca}\leq1$.
But the original question is the next:
If $ a^2\plus{}b^2\plus{}c^2\equal{}3$ then $ \dfrac{1}{5\minus{}2ab}\plus{}\dfrac{1}{5\minus{}2bc}\plus{}\dfrac{1}{5\minus{}2ca}\leq1$.
Do they have the same proof? :?:

[quote="Inequalities Master"]Sorry,arqady.but in that forun is only a proof for the following inequality:
If $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ then $ \dfrac{1}{5 \minus{} 6ab} \plus{} \dfrac{1}{5 \minus{} 6bc} \plus{} \dfrac{1}{5 \minus{} 6ca}\leq1$.
But the original question is the next:
If $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ then $ \dfrac{1}{5 \minus{} 2ab} \plus{} \dfrac{1}{5 \minus{} 2bc} \plus{} \dfrac{1}{5 \minus{} 2ca}\leq1$.
Do they have the same proof? :?:[/quote]
Yes! If in your inequality $ a\equal{}\sqrt3A,$ $ b\equal{}\sqrt3B$ and $ c\equal{}\sqrt3C$ then $ A^2\plus{}B^2\plus{}C^2\equal{}1$ and
we need to prove that $ \frac{1}{5\minus{}6AB}\plus{}\dfrac{1}{5\minus{}6BC}\plus{}\dfrac{1}{5\minus{}6CA}\leq1,$ which is a Vasc's inequality.

Thanks a lot,arqady! You helped me one more time. :thumbup:


Let $P(x) = (x-1)(x-2)(x-3)$. For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x) \cdot R(x)$?

\[ \text{(A) } 19 \qquad \text{(B) } 22 \qquad \text{(C) } 24 \qquad \text{(D) } 27 \qquad \text{(E) } 32 \]

What is wrong with my attempt at this problem?

[hide]$\deg R(x)=3\Rightarrow \deg P(Q(x))=6\Rightarrow \deg Q(x)=2$. Write $Q(x)=ax^2+bx+c$, $a\neq 0$. $P(x)|P(Q(x))$ iff 1, 2, and 3 are all roots of $P(Q(x))=\left(ax^2+bx+c-1\right)\left(ax^2+bx+c-2\right)\left(ax^2+bx+c-3\right)$. Then, if $x$ is 1, 2, or 3, then $ax^2+bx+c$ must be 1, 2, or 3 in any order. But doesn't that only produce 6 possible systems of 3 equations for a, b, and c, and therefore at most 6 possible polynomials $Q(x)$? This is far less than any of the options.[/hide]

Never mind, I found my mistake.

[hide="Finishing my solution"]$Q(x)$ could also be 1,2, or 3 twice and another of the numbers once, for $x\in \{1,2,3\}$. In this case there are 3 options for doubled number, 2 options for singled number, 3 places to put singled number, for 18 more cases. That gives at most 24. However, $Q$ cannot be linear. This occurs when $Q(\{1,2,3\})$ is equal to $\{1,2,3\}$ or $\{3,2,1\}$. Discard these 2 solutions to find 22 options for $Q(x)$. [b](B)[/b][/hide]


Our basketball team has 10 players, including Steve and Danny.  We need to divide into two teams of 5 for an intra-squad scrimmage.  In how many ways can we do this if Steve and Danny insist on playing on the same team?

This problem is quite easy if you do it the way of the solution. I kept on trying to to find a new way to do it, but I couldn't.

If Steve and Danny are on the same team, there are 8 players to choose from for the other 3 spots on the team they're on, so there are 8C3=56 ways total.

Wouldnt it be $\binom{9}{4} = 126$ because you can treat Steve and Danny as one super person, then you just have to choose 4 players (which is 5 because of Steve and Danny) out of 9?

You can't treat Steve and Danny like only one person.  You may treat them like a block, but they must be a block equivalent to two players.  If Steve and Danny were one person, then you would have a 6 person team and a 4 person team.

That's why you choose a 4-person team, like in my solution. Is that valid?

Eh, good thought!  I didn't completely understand your solution the first time  :oops:

The problem with your solution is that once you take Steve and Danny away from 10 people, you have 8 people.  You may treat them as one person, but then there would only be 9 people in total; we need 10 players on two teams.

Can someone explain why the answer would not be $112$? Isn't there two ways? If Steve and Danny are on Team 1, or if Steve and Danny are on Team 2?

The teams are indistinguishable, so we overcount, and hence we must divide $112$ by 2 to get $\boxed{56}$.

Wait yeah they are indistinguishable, so its $\dbinom{8}{3}=56$ but doesn't [i]that[/i] overcount by a factor of 2 so 28? .-.

No, we already dealt with the overcounting by 2.
So it is $\dfrac{\dbinom{9}{2}}{2}$

Nice solution, but please hide it :)
also, I think you counted the opposite

[hide = My Solution]Once you put Danny and Steve on one team, there's $8$ teammates left. We need to fill in the remaining $3$ spots on Danny and Steve's team (the remaining automatically go on the other team) so this is just $\binom{8}{3} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2} = \boxed{56}$.

This is the same as $\binom{8}{5}$ if you wanted to choose the people on the opposing team.[/hide]
I'm pretty sure this is right.

:oops_sign: solution of wlm2 is also correct

[quote=RunarEC][hide=solution]If Steve and Danny are on opposite teams, there are 8 other players to choose from for the other 4 spots on Steve's team, so there are $\binom{8}{4} = \boxed{70}$ choices.[/hide][/quote]
This is the actual answer Alcumus gave which means it's right.

[quote=Williamz1483]say how its correct not it is correct[/quote]

well, [hide=here's how] Since there are $10$ players, and steve and danny are on the same team, we can put them on a team beforehand. Since steve and danny don't need to be chosen, there are $10-2=8$ other people left. There are only $3$ other spots on steve and danny's team so you pick $3$ out of $8$ people which is $\binom{8}{3} = \boxed{56}$[/hide]


Let ABC be an acute-angled triangle whose side lengths satisfy the inequalities AB < AC < BC. If point I is the center of the inscribed circle of triangle ABC and point O is the center of the circumscribed circle, prove that line IO intersects segments AB and BC.


logarithms
probability
number theory theorems
number theory

Type "prime number theorem" in Google. Yeah, it's hard. :)

[quote="grobber"] Yeah, it's hard. :)[/quote]  :( ;) I fetch on Google... Thank you.

[quote="bénabar"]Hello, a few of probability...
How prove that: \[P( n \in \mathbb{P}) = \frac{1}{\ln{n}}\]
Is it hard?  :?  :blush:[/quote]

What exactly does this mean? Since $\mathbb N$ is not a finite set, there is no canonical probability distribution. I presume you want Dirichlet density or natural density, in which case the statement is correct.

Andrei


Five cards are to be selected at random from 1 to 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median ?

[hide="Solution"]The number of ways to select $ 5$ numbers in the set $ \{1,2,3,4,5,6,7,8,9,10\}$ such that the mean equals the median is $ 24$ by some case analysis.

Now observe that there is a 1-1 correspondence between the number of ways to select $ 5$ numbers with median greater than the mean and the number of ways to select $ 5$ numbers with median less than the mean. This is evident by the fact that if $ a,b,c,d,e$ are the 5 selected numbers with either property, then $ 11\minus{}a,11\minus{}b,11\minus{}c,11\minus{}d,11\minus{}e$ has the other property.

Let $ A$ be the number of ways to select the $ 5$ numbers with mean $ >$ median, $ B$ be the number of ways with mean $ \equal{}$ median, and $ C$ be the numbe of ways  with mean $ <$ median. Then
$ A \plus{} B \plus{} C \equal{} {10 \choose 5}$
But we know that $ B \equal{} 24$ and $ A \equal{} C$ from the previous paragraphs. So
$ 2A \plus{} 24 \equal{} {10 \choose 5} \equal{} 252$
$ 2A \equal{} 228$
$ \boxed{A \equal{} 114}$

If this is wrong, it's probably because I screwed up the calculation of $ B$ in the first paragraph; I kinda rushed.[/hide]


is there any reason it is called alcumus? (just wondering)


1. $2 +\sqrt{u}$=$\sqrt{2u+7}$

2. $\sqrt{X-5}$$-\sqrt{X + 3}=4$

I keep getting different answer then whats in the back of the book  :|

And my teacher didnt even bother explaining this :(

hello, for (1) we get after squaring both sides of the given equation
$ 4\plus{}u\plus{}4\sqrt{u}\equal{}2u\plus{}7$
$ 4\sqrt{u}\equal{}3$
squaring again and isolating $ u$ we get $ u\equal{}\frac{9}{16}$ as the solution.
Sonnhard.

hello, your second equation has no real solution
Sonnhard.

For (2), add sqrt(x+3) to both sides, square both sides, simplify and square again to get x=6.

hello, $ x\equal{}6$ is no solution of your second equation, to prove this you can insert $ x\equal{}6$ .
Sonnhard.

You're right, of course, Sonnhard.  6 is an extraneous solution which doesn't work: as you rightly point out there are no real solutions. 6 arises when squaring sqrt(x+3) = -3 which could just as well have been squaring -sqrt(x+3) = -3. So 6 is the (only real) solution of sqrt(x-5) = 4 - sqrt(x+3).

[quote="FlaME_HiRo"]1. $ 2 \plus{} \sqrt {u}$=$ \sqrt {2u \plus{} 7}$[/quote]

$ 2 \plus{} \sqrt {u}$=$ \sqrt {2u \plus{} 7}$ $ \iff$ $ u\geq 0$ and $ (2 \plus{} \sqrt {u})^2$=$ (\sqrt {2u \plus{} 7})^2$

$ \iff$ $ u\geq 0$ and $ 4\plus{}4\sqrt u \plus{} u\equal{}2u\plus{}7$

$ \iff$ $ u\geq 0$ and $ 4\sqrt u\equal{}u\plus{}3$

$ \iff$ $ u\geq 0$ and $ (4\sqrt u)^2\equal{}(u\plus{}3)^2$

$ \iff$ $ u\geq 0$ and $ 16u\equal{}u^2\plus{}6u\plus{}9$

$ \iff$ $ u\geq 0$ and $ u^2\minus{}10u\plus{}9\equal{}0$

$ \iff$ $ u\geq 0$ and $ (u\minus{}1)(u\minus{}9)\equal{}0$

$ \iff$ $ u\in\{1,9\}$

[quote="FlaME_HiRo"]2. $ \sqrt {X \minus{} 5}$ $ \minus{} \sqrt {X \plus{} 3} \equal{} 4$[/quote]

$ \sqrt {X \minus{} 5}$$ \minus{} \sqrt {X \plus{} 3} \equal{} 4$ $ \iff$  $ \sqrt {X \minus{} 5}\equal{}4\plus{} \sqrt {X \plus{} 3}$

$ \iff$ $ X\geq 5$ and $ (\sqrt {X \minus{} 5})^2\equal{}(4\plus{} \sqrt {X \plus{} 3})^2$

$ \iff$ $ X\geq 5$ and $ X\minus{}5\equal{}16\plus{}8\sqrt{X\plus{}3}\plus{}X\plus{}3$

$ \iff$ $ X\geq 5$ and $ 24\plus{}8\sqrt{X\plus{}3}\equal{}0$

So no solution

hello, oh, thank you my solution for the first equation is also wrong.
Sonnhard.

$ \minus{} 5 < 3\Rightarrow x \minus{} 5 < x \plus{} 3\Rightarrow \sqrt {x \minus{} 5} < \sqrt {x \plus{} 3}\Rightarrow\sqrt {x \minus{} 5} \minus{} \sqrt {x \plus{} 3} < 0<4\Rightarrow \\
\Rightarrow \sqrt {x \minus{} 5} \minus{} \sqrt {x \plus{} 3}\neq 4; (\forall)x\ge 5.$

1. $ \left\{\begin{array}{ll} \sqrt {2u \plus{} 7} \minus{} \sqrt {u} \equal{} 2 & \quad \\
\sqrt {2u \plus{} 7} \plus{} \sqrt {u} \equal{} \frac {u \plus{} 7}{2} & \quad \end{array} \right.\Longleftrightarrow \left\{\begin{array}{ll} \sqrt {u} \equal{} \frac {u \plus{} 3}{4} & \quad \\
\sqrt {2u \plus{} 7} \equal{} \frac {u \plus{} 11}{4} & \quad \end{array} \right.$

$ \Longleftrightarrow \boxed{u \equal{} 1,\ 9}$

2. $ \left\{\begin{array}{ll} \sqrt {X \minus{} 5} \minus{} \sqrt {X \plus{} 3} \equal{} 4 & \quad \\
\sqrt {X \minus{} 5} \plus{} \sqrt {X \plus{} 3} \equal{} \minus{} 2 & \quad \end{array} \right.$ , yielding no real solution.


algebra
polynomial
quadratics
Rational Root Theorem

I have no idea what the "factor theorem" is but...

The first thing you're going to want to do is use the rational root theorem mentally,
$\frac{\text {Factors of lowest degree coefficient}}{\text{Factors of highest degree coefficient}}$,

the only real roots possible are $\pm{5}$ and $\pm{1}$. You an easily see that $\pm{1}$ won't work, so synthetically divide $\pm{5}$ and you can see that $5$ works (since the remainder is zero).

So now factor out the $(x-5)$ to $\boxed{(x-5)(x^2+x+1)}$. Notice that there is no sign change in $(x^2+x+1)$ so you don't have to even try $-5$

Remember that when you're dealing with a polynomial that has the largest degree coefficient of one... the roots of the polymonial are factored in the form $(x-a)$, where $a$ is the root

Hope that helps.

[quote="king_23"]Can I get some help with factor theorem, it's kind of new to me. I need to use factor theorem to factor x^3- 6x^2 + 6x - 5.[/quote]

The factor theorem is basically the same as rational root theorem (I think) which is basically test numbers that make the polynomial $=0$.

The only possible values are $\pm 5, \pm 1$ (I don't really know how to explain why, but it has to do with the last term and the first term).

Test $5$. It works.  So we know that $x-5$ is a factor.

Now do whatever division you want to find the other factor (which will be a quadratic).  The factor the quadratic if possible (in this case it's not) and there's the answer.


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra unsolved

I have a matrix A=
1  0
-1 1

a 2 by 2 matrix. I want to show that this has infinite order.

I have that A^n =
1  0
-n 1
is it enough to say that there is no n such that A^= identity.

Yes; clearly the only $ n$ such that $ A^n \equal{} I$ occurs when $ n \equal{} 0$.  Alternately, you could compute its characteristic polynomial.

I have found char poly then what to do with it?

What must the minimal polynomial of a matrix satisfying $ A^n \equal{} I$ look like?


I'm not sure how to do the entire thing right now, but i would imagine it would have something to do with the relationships between the sin values. They follow a circular pattern, you are really only dealing with two values.

[quote="mysmartmouth"]Can someone help me with number 30 on the [url=http://www.math.sc.edu/contest/1997/exam/PDF/exam1997.pdf]1997 USC Math Contest[/url]?  It's a trig problem, and it looks rather hard.  Any solutions appreciated![/quote]

What are you doing? You still need to learn your identities and you expect to do a problem like that? Learn your idenitities first! And then learn complex numbers and Demoivres in order to solve it! There's really no point to post a solution because you won't understand any of it! ;)  ;)

[quote="arkmastermind"]well...i dont think you use complex numbers to solve it.

i think the hint is to use the double angle formula for sine...i'm sick though, and won't work it all out now. theres a smiliar problem in a later usc test with cosines, look for the solution they give. it is really similar.[/quote]

hmm...if i remember correctly, the usc solution to this problem involved demoivres :?

that might be true. i cant seem to remember what i did when i did this for practice...im pretty sure i didnt use demoivres though...

oh well.

Well the solutions are here http://www.math.sc.edu/contest/1997/solutions/PDF/solutions1997.pdf

and it looks like they did use demoivres. I'm not a big fan of the USC solutions though, it seems they make problems more complicated than they need to be.

[quote="Iversonfan2005"]and it looks like they did use demoivres. I'm not a big fan of the USC solutions though, it seems they make problems more complicated than they need to be.[/quote]

That problem [i]is[/i] complicated...  Show me an easier way to do it than complex numbers....


No.  The football team brings a substantial amount of money to my city and to my school.  In addition, I feel the funding for the various academic programs is sufficient.  You could throw more money into academics, but until you find a more efficient use for it, I feel it wouldn't help enough for the action to be justified.

Also, given the bad state of American nutrition and exercise, I feel that athletic programs should be encouraged...especially at younger ages, but high school programs can be just as beneficial.

[quote="elston"]Post stuff..[/quote]

To tell the truth, my school needs to spend [b]more[/b] money on sports. We were playing Ultimate Frisbee, and half the frisbees are those really cheap and flexible ones that you can fold up and that don't fly well.

I heard that my friend's school has to [b]share sports jerseys![/b]  :o

My school doesn't spend too much money on sports.

[hide="Then again..."]I am homeschooled... :D [/hide]

wow..where i live, like everything is football football football baseball baseball soccer soccer track track track football football....

sports are big at my school and spending the money is justifable as they bring in money. if you can figure out a way to get people to buy tickets to watch a math contest, i'd love to hear it.

More money on sports would be nice, so we don't have to depend on donations to keep our sports up.

[quote="hwenterprise"]Our school's athletic program is too cheap to even afford to give us 2 busses for our 56 member cross country team to meets.

Our school doesn't spend much money.[/quote]

while another school near us has those super luxury buses (they only have like 20). Not fair  :mad:

depends on the sport, football we hire outside coaches, nice buses for every game, nice pads and equipment, but other sports are dependant on teachers who are willing to drive. Fortunately, we have begun to spend some money on academics, even thouhg i still dont have my economics textbook becuase we dont have any and its been almost 3 weeks. At one time, our school was using outdated textbooks, sharing desks and more, while the football teams funds never seemed to have a limit. Other ECs are the sole responsibility of students/coaches to finance like our reach team, where we paid $ \$ $ 170 without any school assistance, but we were fortunate that our coach was a resourceful fundraiser.

Yeah actually they took funds out of our already depleted funds for football and basketball.  But I guess since our athletic teams own our academics like crazy, maybe it's justified? I don't know.

No.  If anything, they should spend more money on sports.

Well, sometime the lack of money can hurt, especially this [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=2695]time we got lost[/url].


algebra
polynomial
number theory
prime factorization
algebra unsolved

How many factors are there in $ x^{2007}\minus{}1$?

[quote="exoeragonius"]How many factors are there in $ x^{2007} \minus{} 1$?[/quote]
 hdo you mean by factors $ card(A)$ that :
$ A \equal{} (x1.x2...........xn)$ and  $ p(x) \equal{} \prod (x \minus{} xi)$ ?
 and $ x \in C$  ???????? :wink:

[quote="exoeragonius"]I mean if you factor it completely, how many factors will there be?[/quote]
$ p(x) \equal{} x^2007 \minus{} 1$
$ p(x) \equal{} 0$ ==> $ x^2007 \equal{} 1$ so $ x_k \equal{} e^(2ki \pi /2007)$ and $ k \in (0.1.2.3............2006)$
so the number of factor is 2007
nb = $ x \in C$

[quote="Akashnil"]I think he didn't mean factors with complex coefficients.
Number of irreducible factors of $ x^{2007} \minus{} 1$ is equal to the total number of divisors of $ 2007$ which is $ \boxed{6}$[/quote]So, you mean the factors are $ x^d\minus{}1$, where $ d|2007$? I don't think these factors are necessarily irreducible. Take, for instance, $ d\equal{}3$. You'll get $ x^3\minus{}1$ which is reducible to $ (x\minus{}1)(x^2\plus{}x\plus{}1)$. Besides, is $ x^{669}\minus{}1$ also a factor? ;)

I guess there are five irreducible factos: $ (x\minus{}1)(x\plus{}1)(x^2\minus{}x\plus{}1)(x^2\plus{}x\plus{}1)(x^{223}\plus{}x^{214}\plus{}x^{205}\plus{}\ldots\plus{}x^7)$. So, we have $ 2^5\equal{}32$ factors, not necessarily irreducible. :)

There are in fact six.  $ x^{2007} \minus{} 1 \equal{} \prod_{d | 2007} \Phi_d(x)$ where $ \Phi_d(x)$ is the $ d^{th}$ [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]cyclotomic polynomial[/url].

I don't understand what cyclotomic polynomial is. Anyway, what's wrong with my solution?

Well, cyclotomic polynomials give the answer.  It's pretty easy to verify that your factorization is wrong; $ x + 1$ is clearly not a factor and the polynomial you gave has degree $ 1 + 1 + 2 + 2 + 223 = 229$.  (You don't seem to have the prime factorization of $ 2007$ correct.  Actually, I don't understand how you got that last factor at all.)

Edit:  The correct factorization is

\begin{eqnarray*} x^{2007} - 1 & = & \Phi_1(x) \Phi_3(x) \Phi_9(x) \Phi_{223} (x) \Phi_{669} (x) \Phi_{2007}(x) \\
& = & (x - 1)(x^2 + x + 1)(x^6 + x^3 + 1)(x^{222} + x^{221} + ... + x + 1) \Phi_{669}(x) \Phi_{2007}(x) \\
\end{eqnarray*}

where $ \Phi_{669}(x) = \frac {x^{446} + x^{223} + 1}{x^2 + x + 1}$ and $ \Phi_{2007}(x) = \frac {x^{1338} + x^{669} + 1}{x^6 + x^3 + 1} = \Phi_{669}(x^3)$.

Edit 2:  The essential pattern here is that $ 2007 = 3^2 \cdot 223$; a square of a prime times a prime.  The smallest such number is $ 12 = 2^2 \cdot 3$, and here

\begin{eqnarray*} x^{12} - 1 & = & \Phi_1(x) \Phi_2(x) \Phi_4(x) \Phi_3(x) \Phi_6(x) \Phi_{12}(x) \\
& = & (x - 1)(x + 1)(x^2 + 1)(x^2 + x + 1)(x^2 - x + 1)(x^4 - x^2 + 1) \\
\end{eqnarray*}.


This is actually HMMT problem but instead of giving numbers, I put variables to make it an algebraic question.

A man, standing on a lawn, is wearing a circular sombrero of radius $ n$ feet. Unfortunately, the hat blocks the sunlight so effectively that the grass directly under it dies instantly. If the man walks in a circle of radius $ k$ feet where $ k > n > 0$, find the area of dead grass.

$ \text{(A)} \ \pi nk \ \qquad \text{(B)} \ 2 \pi nk \ \qquad \text{(C)} \ 4 \pi nk \ \qquad \text{(D)} \ \pi \sqrt{17} nk \ \qquad \text{(E)} \ \text{none of given}$

[hide]
The path will be an annulus of inner radius $ k\minus{}n$ and outer radius $ k\plus{}n$. Hence the area is

$ \pi (k\plus{}n)^2 \minus{} \pi (k\minus{}n)^2 \equal{} 4kn\pi$, or $ C$.
[/hide]


So, up to now which one of you is sure of about going to MOP 2005? :) Please [u][b]post only once[/b][/u] so that we can count easily :D
Good luck to the TSTs to the Winners and HM members taking it.

I'll be going as well, although I didn't exactly expect to do this well on my first USAMO.  Anyways, I can't really complain.  :)

I'm going, but only because I happened to see number 5 in [u]Math Olympiad Challenges[/u] (which I am now recommending to everyone :) ).

I'm going. So much for "post only once" and "post if you're sure you're going," but the list can still probably be sorted through.

-Adam Hesterberg

I'm going, which is pretty awesome! This is my first year taking the USAMO, just barely qualifying with a 9 (as a sophomore). Somehow, I managed to get a 21 on USAMO, and I think I just barely qualified for MOSP too. I can't wait!

LOL.

Red MOP here.

Looks like my first post was deleted.


Solve equation:
    $ 3^{cos^2x}\plus{}2^{sin^2x}\equal{}2^{(\frac{x}{\pi})^2\minus{}3}\plus{}2^{9\minus{}\frac{4x}{\pi}}$

Prove that $ f(x) \equal{}2^{(\frac{x}{\pi})^2\minus{}3}\plus{}2^{9\minus{}\frac{4x}{\pi}} \minus{} 3^{cos^2x}\minus{}2^{sin^2x} \geq 0 \forall x$. The equality occurs if and only if $ x \equal{} 2\pi$.


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

[quote="Fanatic"][quote="13375P34K43V312"]Don't worry about countdowns, there's really no point in preparing for those. However, it's a good trick to know nonetheless, since it is a time-saver.[/quote]

I beg to differ ;)... i don't know all the elites but I know I was practicing countdown with Nathan Benjamin last year and he sure didn't think there was "no point" of this.

Of course, I guess if you don't think you can make top 12 nationals... but then again if you make nationals you should always prepare for a) the worst -- emotional trauma is the worst that can happen. and b) the best. U get in countdown and bam u could be the national champion under this system!!

I know i'll be practicing countdown if I make nationals this year.[/quote]
I kinda figured that Nathan Benjamin would be really fast since that would be all that he needed to practuce, but Treething roasted him...

any given set of 5 problems, anything could happen.

It's so luck based. I think Nathan was actually doing better at first: He answered first on questions but made dumb mistakes. Treething finished him off.

Exampel: 2002 nats champ (albert ni) was down 0-2, he had answered first twice but made two dumb mistakes (similar to what happened to nathan). Albert missed the 3rd question but the dude he was facing couldn't finish him off (difference between what happened to nathan). albert then "counter-roasted" the dude he was facing and took 3 questions easily and became nats champ one round later.

That one question could make the difference. I remember thinking that nathan, down 2-0, could possibly win anyways -- it'd be ironic deja vu for me after i watched the albert do it. but nope, didn't happen.

Edited: changed names to "albert  ni" to make less confusing

[quote="Fanatic"]any given set of 5 problems, anything could happen.

It's so luck based. I think Nathan was actually doing better at first: He answered first on questions but made dumb mistakes. Treething finished him off.

Exampel: 2002 nats champ was down 0-2, he had answered first twice but made two dumb mistakes (similar to what happened to nathan). He missed the 3rd question but the dude he was facing couldn't finish him off (difference between what happened to nathan). The eventual champion then "counter-roasted" the dude he was facing and took 3 questions easily and became nats champ one round later.

That one question could make the difference. I remember thinking that nathan, down 2-0, could possibly win anyways -- it'd be ironic deja vu for me after i watched the 2002 nats champ do it. but nope, didn't happen.[/quote]
You mean Albert Ni?

[quote="13375P34K43V312"]Countdown really doesn't teach you a whole lot. You don't see olympiad winners sitting in their rooms practicing doing obvious math problems in 3 seconds or less.

Sergei Bernstein, who owned everyone in the written in 2005, didn't give a care about the countdown. With all due respect to him, he was terrible at it. He got 'roasted' by Neal 3-0 in his first countdown matchup. Later, he beat everyone in the Masters Round by a wide margin. He nailed it. He knew that there are much more important things then doing mindless problems in 3 seconds. This year, he got one of the highest freshman USAMO scores.

Some people are just naturally fast and very good at the other stuff as well (Neal). However, I believe that spending a whole lot of time practicing for countdown is pointless. I"m not saying that at the end of a practice you shouldn't do a mini-countdown for fun. I happen to be pretty decent at countdown and enjoy it myself, but I know that it's not worth dying for.[/quote]
You know that Neal got a 13 on the USAMO this year?

Um, so you try so hard and u get so far and you're at the point where you're up on the stage, the lights are pointing at you, there's 8,000, a trip, and national recognition at stake and your attitude is "there's no point doing mindless buzzer blah blah blah"?

I'm not saying practice countdown over written, I'm saying practicing countdown is basically essential for top competitors. I mean why are you even DOING mathcounts if you know u can a) make the top 12 and b) aren't going to try countdown???

IF sergei believed that countdown was worthless, fine by me, but logically if you're going to try that hard and go so far then you might as well try to win it all in countdown too.

[quote="jhredsox"]is a 13 good?

isn't it out of 14?
-jorian[/quote]
Well, It is out of 42, but a 13 is really really good (especially since he was an 8th grader). That is like a 4th grader doing mathcounts and making nationals. You know that at some point he is going to win written at nationals.

[quote="jhredsox"]lol.......he did win written :o

but good point...they must be hard for eith graders
-jorian[/quote]
He won countdown too...

yes i know!, but not in the same year (please don't try to be offended nebula)

-jorian

p.s. i think we've gotten the point of this question, so can a mod lock this up?

I agree with 13375P34K43V312. The countdown round don't teach you a lot. Knowing how to do problems is a lot more important.

My point is why the heck are u even bothering working on mathcounts problems AT ALL if you aren't interested in countdown? I point you at AMC 8 --> AMC 10 --> AIME --> USAMO and other math contests. You don't see olympiad winners hunched over 40 minute timed tests, either, do you?

Sure, the obvious and "best" answer would be "we should only learn math to be better at it, and since countdown doesn't teach math, then screw it".

But countdown is a skill nonetheless. Not necessarily direct math ability, but reactions and timing too. and I DO NOT FIND IT A COINCIDENCE that of the people that I know that are good at math, all the ones that are good at countdown are gregarious people. The other ones remind me of furniture more often than not, or are just less lively in general. I know of course this is people I know and this cannot be generalized, but as I type this I realize I can really separate people into two groups: good at cd (countdown) and sociable / not good at cd - antisocial. Wow. I'm not even joking, not one bit. But that's just people I know.

And if you argue that my point is void, because the only point of mathcounts should be pure math, then I gotta say mathcounts isn't the right contest for you. Might as well start training AIME now rather than later. Don't waste your time on mathcounts. I'm sure a lot of elites, if they could go back in time, would never touch mathcounts and work directly on harder stuff (AMC 8 to compare to regional level - easy state level, AMC 10 to compare to upperstate -->nationals level) I'm more of a countdown guru than any other round but I've been training AIME material about... 100 times more than mathcounts. Hours spent and effort dedicated.

Albert ni (2002 nats champ) once said: "Mathcounts isn't about who's good at math, it's about who's seen more problems".

Yes, I know about Neal's beast score on USAMO. I doubt he practiced countdown day in and day out, he probably practice it a few times, but I'd be willing to bet it's not what he focused on.

I'm just saying, some people are naturally fast, it's not worth working on that. But it's pretty clear that Neal spent a lot of time working on Olympiad math (or at least 20 hours total, say)

[quote="Fanatic"]who said anything about practicing countdown day in and day out?[/quote]
Anyone who does that is stupid, I'm sorry if I offended anyone.
Perhaps the week before the competition it is okay.
[quote="Fanatic"]if they could go back in time, would never touch mathcounts and work directly on harder stuff (AMC 8 to compare to regional level - easy state level, AMC 10 to compare to upperstate -->nationals level)[/quote]
You mean like MysticTerminator?

The thing about countdown is that it is fun. It makes mathcounts different, and I enjoy a good round of countdown even if it costs me first place (which it has...)... In countdown, everyone makes stupid mistakes. They are often costly. In countdown, you have to be in like hypermode. You have to react faster than you can think. Finally, you have to be smart :D.

[quote]
Anyone who does that is stupid, I'm sorry if I offended anyone.
Perhaps the week before the competition it is okay.
[/quote]
Just in case you couldn't tell (I'm not sure), my post was meant to reply to leetspeakfever's (13375p34k43v312) post. It meant "wait, you said 'I doubt he practiced countdown and day in and day out' as a counter to my argument, but I (nor anyone else) every mentioned anything about practicing day in and day out".

Just making sure you could tell that I wasn't actually advocating practicing countdown day in and day out. I personally won't until I find out whether or not I make nationals.

[quote="bpms"]if they could go back in time, would never touch mathcounts and work directly on harder stuff (AMC 8 to compare to regional level - easy state level, AMC 10 to compare to upperstate -->nationals level)
You mean like MysticTerminator?[/quote]

Who's MysticTerminator?

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=614892&h#614892
He got fifth at mc If I recall correctly... He also is an IMO gold medalist.

Meh... u can take his opinion or leave it.

Personally mathcounts is my "style" of math competition -- I feel as if I have a shot to place high nationally in it but not the harder math competitions later on (i'm good at speed but not endurance, for example), so that's why I emphasize mathcounts.

Then again I've done like 20 hours of mathcounts in the last 3 months and like hundreds of harder math...

but my attitude is still "mathcounts is my chance to DO SOMETHING in math"... man I need to start working on mathcounts :(.


on the 2001 state #24

It asks the number of partitions for 7

Is there a formula or sumthin??

Yes, I remember there was this other formula, but it involved lots of calc and other things. I think it would be better to just list.

Uh, it takes like 1-2 minutes (less for some people) to just list them out.
Not that bad for a state #24.

ehh, it has to do with triangular numbers i think.

EDIT: scratch that.  there's a partitioning function.  it looks awfully complicated though T_T

Hm, I looked on wikipedia and they have it as the reciprocal of Euler's Function... o.0

i think you have to take the closest integer function of that?

Well, there's a recursive function from which you can find the number of partitions of $ n$ based on the number of partitions of each of the numbers less than it...

I bet it's something that's kinda pointless to remember

I know i sound really dumb but what is a partition??? i think ive heard of them but i have no idea what they are!

[quote="darkmatter47"]A partition is a way to express an integer as a sum of one or more positive integers.
Just list them, the only real strategies are for how. Grouping as abacadea suggested would work perfectly. I mean, they're not gonna ask for the number of partitions of 129. If they do, good luck buddy.[/quote]

ah thank you so much!!!


$\text{NO WAY! THIS IS AWESOME!}$

I am so pumped that we have biology on the forums now. I want to moderate this thing! Okay all you biology lovers out there from Ecology/Evolutionary Biology to Molecular/Cellular/Developmental Biology and Genetics, it's time to represent our science! There's a bunch of math in biology and that is really awesome too. I am stoked.

Hopefully [b]you[/b] guys will actually start using this forum, since you won't be getting any bio input from any of us the admins :P :D

Ha ha! Actually Valentin, I think this forum will draw you into it...and then you will be so intrigues by this enigmatic science that you will dump math altogether and do biology. Or maybe not...

I like math too, hopefully there will be discussions here on population growth and ecology and other stuff where mathematics are highly applicable (note my signature on the bottom, it's the formula for logistic population growth lol)

[quote="adidasty"]Ha ha! Actually Valentin, I think this forum will draw you into it...and then you will be so intrigues by this enigmatic science that you will dump math altogether and do biology. Or maybe not ...[/quote]Indeed, maybe not :D I used to like biology, I can remember most of my classes in high-school, but I was never capable of understanding how a IBO question would look like (I imagined something like: given two alien species :alien: predict how will the kids look like given their DNA strings :D).

[quote="adidasty"]ha ha, yeah sometimes biology questions, especially IBO questions are worded pretty awkwardly.[/quote]
there is also IBO!?

[quote="GoBraves"]I dont know a lot of biology....I just know the stuff we learned in school (8th grade) I hope this forum won't do just complicated stuff only but a balance of easy stuff and hard stuff. Otherwise its useless to me.[/quote]I'm with GoBraves. All I have is what I've learned up through 8th grade biology. (Did we even do biology this year?)

Hmmm... next year though I'm taking biology D, so that will be nice.  :)

yeah IBO is definitely everyones goal who loves biology. if im not mistaken, there may be an AoPSer who went to IBO by the name of Kay Aull. she got 2nd overall in the IBO (pretty much astoundingly amazing) and i think she roamed these boards too...shes at MIT now.

yeah, Kay went to IBO twice and got a silver medal in Belarus and 3rd in the world last year... she is my personal goddess. =P

yeah she is definitely awesome. no doubts about that.


a quick rough sketch on a reply to show 6 points of intersection between a triangle and a circle


I had two number $a,b$ , $a,b>1$ And $a+b<100$ . I have two friend $A\&B$ ; A knew accumulate of $a,b$ And B knew
Value worth of a+b . After that , i asked A: Can you read name of two number? . A said : I can't . B said :Ofcouse you unable
 divine they . A said : Oh i know they . B said : So do i .
To everybody : What are they ? ($a\&b$)

Answer: 4 and 13

A's number is 52. A cannot determine whether the numbers are 2 and 26, or 4 and 13.

B's number is 17. B sees that the possible numbers are 2 and 15, 3 and 14, 4 and 13, 5 and 12, 6 and 11, 7 and 10, or 8 and 9. Then A's number would respectively be 30, 42, 52, 60, 66, 70, 72. All of these numbers can be resolved into two factors >1 in at least two ways. Therefore B knows that A cannot determine the numbers.

A assumes for the sake of contradiction that the numbers are 2 and 26. Then B would hold 28. One possibility that B would then consider is that the numbers are 5 and 23. Since these are primes, no other numbers >1 share their product 115. A would be able to determine the numbers in this case, so B would not know that A could not determine the numbers. Since B reported to the contrary, A now knows that the numbers are 4 and 13.

To facilitate the verification of B's last statement, I first note that in general A cannot determine the numbers at the outset unless both are primes, or one is a prime and the other its square. Let us call a number under 100, but over 3, which is the sum of two primes, or of a prime and its square, a [i]D-number.[/i] Then all even numbers, as well as numbers 2 more than a prime, are D-numbers.

B's first statement shows that the sum of this problem's two numbers is a non-D-number. Since this is all that A knows (aside from A's own number), A can find the numbers if and only if exactly one of the factor pairs of A's number has a non-D sum. B now tries each possibility in turn, and all but one give two factor pairs with non-D sums:

30: 2,15=17; 5,6=11
42: 3,14=17; 2,21=23
52: 4,13=17 only
60: 5,12=17; 3,20=23
66: 6,11=17; 2,33=35
70: 7,10=17; 2,35=37
72: 8,9=17; 3,24=27

Thus B can conclude that the numbers are 4 and 13.

To find this solution, I used trial and error mixed with some eliminative mathematics. Since B's number is non-D, I paralleled B's last calculation using the smallest non-D-number 11:

18: 2,9=11 only
24: 3,8=11 only
28: 4,7=11 only
30: 5,6=11; 2,15=17

B cannot now determine the numbers, but is instead faced with three possibilities. Taking the next non-D-number 17, I arrived at the solution (4,13). I have not looked at larger cases but the way in which the problem is phrased implies a unique solution.

Thank you
In may class , No one can kill it :D


[quote="dingzhou"]Given a random string of 33 bits (0 or 1), how many (they can overlap) occurence of two consecutive 0's would you expect?[/quote]
Hold on, what about more than 2 in a row? how many do they count as? nC2?[/hide]

whoa.. this is really similar to the 1989 AHSME #30 (thanks to the future for pointing it out a few months ago).. which i never understood  :blush:

[quote="serialk11r"][quote="IamSleepy"][hide](1/2)*(1/2)=1/4 chance to get two consecutive 0's.
(1/4)*32=8
32 because there are 32 pairs of numbers total. Is 8 correct?[/hide][/quote]
Uh....no...
And also, there's a good reason why I made the above post.
Consider 000. How many pairs of zeros are there? well, first 2 zeros are a pair, and second 2 are as well.[/quote]Really? I thought that I was considering that... It was a 1/4 chance to get a pair of 00 and there were 32 pairs (out of 33 numbers), not 16 or whatever, just like there are 2 pairs in 000.

[hide="Solution"]
Let $ f(n)$ be the average number of 00s in a bit string of length n.

Clearly, $ f(2) = \frac{1}{4}$.

\begin{eqnarray*}f(n)&=&\frac{1}{2^n} ( \mbox{(no. of 00s in strings starting with 1) + (no. of 00s in strings starting with 0)}\\ &=& \frac{1}{2^n} ( (2^{n-1}\cdot f(n-1)) + (2^{n-2} + 2^{n-1}\cdot f(n-1))\\ &=& \frac{1}{4} + f(n-1)\end{eqnarray*}
$ \implies f(33) = 32\cdot \frac{1}{4} = 8$
[/hide]