Let $S = \{1, 2, 3, \ldots, 2006\}$. Prove that there are 14 subsets of $S$ which have the following property: if $s \in S$, then there are 7 of these 14 sets which have $s$ as their only common element.

Simple indeed.

Let $A$ be the set of positive integers satisfying the following condition: the binary representation of each element of $A$ consists of 14 digits, exactly 7 of which are equal to 1. (These numbers may start with one or more zeroes.) It is obvious that $A$ has ${14 \choose 7} = 3432 > 2006$ elements. Hence, we can choose some 2006 distinct numbers $a_1,a_2,\,\cdots,a_{2006}$ which belong to $A$. Let us now construct the desired subsets $S_1,S_2,\,\cdots,S_{14} \subset S$. For any number $s \in S$, let $s$ belong to $S_k$ if and only if the $k$-th digit of $a_s$ (in binary representation) is equal to 1. It is obvious that each $s \in S$ belongs to 7 of the sets $S_1,S_2,\,\cdots,S_{14}$ (since $a_s$ has exactly 7 digits equal to 1) and that each number $t \neq s$ cannot belong to the same 7 sets (since the numbers $a_s$ and $a_t$ differ in at least one digit in their binary representation). So, we're done. $\blacksquare$

Pretty much the exact same solution most of the UK squad got. I spent an hour on this question examining the word common trying to think whether I had misunderstood it.

I couldn't believe the question could be this easy.


[hide]The 10s digit of the sum of factorials from 1!...n! where n>9 will be the same digit as the tens digit in the sum 1!+2!+...9!. This is because after 9!, the unit and tens digits of n! are always 0 and thus will not affect the tens digit of the sum.

Assuming I memorized my factorials correctly and can add... is it 1? [/hide]

[hide]We can add up all the numbers, but to save some trouble, we should figure out which ones are actually going to matter.

So the question is, what is the least value of n such that n! has 2 zeroes at the end of it.

Now there's a formula for this, and it's basically n! has step(n/5) = step(n/25) + step(n/125) + ... zeroes.  The least value of n that'll give 2 is n=10, so starting with 10, we have a zero in the tens place.

1! = 01

2! = 02

3! = 06

4! = 24

5! = 120

6! = 720

7! = ..20*7 = ..140

8! = ..40*8 = ..320

9! = ..20*9 = ..180

1 + 3 + 6 + 4 = 14 so we need to carry a 1 into the tens place.

1 + 0 + 0 + 0 + 2 + 2 + 2 + 4 + 2 + 8 = 21 :equiv: 1 mod 10 so 1.[/hide]

P.S. I fixed your spoiler.


Triangle $ABC$ is inscribed in a circle.  The measure of the non-overlapping minor arcs $AB$, $BC$, and $CA$ are, respectively, $x+75^\circ$, $2x+25^\circ$, $3x-22^\circ$.  Then one interior angle of the triangle, in degrees, is:
\[ \text{(A)}\ 57\frac{1}{2}\ \ \ \ \text{(B)}\ 59\ \ \ \ \text{(C)}\ 60\ \ \ \ \text{(D)}\ 61\ \ \ \ \text{(E)}\ 122 \]

[hide]Adding all of the arcs together and setting it equal to $360$ we have $6x+78=360$.  Thus $x=47$ and the angles of the triangle are $61$, $59.5$, and $59.5$.  Thus the answer is $D$.[/hide]

[hide]
the sum of the measure of the arc lengths is 360, so:
x+75+2x+25+3x-22=6x+78=360
6x=282
x=47
plug x in x to find the measure of each arc. you will get AB=122 and the other two =119
the angle of an interior angle must be half of this based on some theorem or something that i forget, so:
[b]angle BCA=61 (D)[/b][/hide]


trigonometry
calculus
calculus computations

Integrate the following and set it equal to $ y$:

$ \frac{dy}{d\theta} \equal{} \frac{e^{y} \sin^2(\theta)}{y \sec{\theta}}$.

I assume that the purpose is to solve the differential equation:

$ ye^{\minus{}y}dy \equal{} \sin^2\theta \cos \theta d \theta \equal{}> \minus{}e^{\minus{}y}(y\plus{}1) \equal{} \frac{\sin^3\theta}{3} \plus{} C$.


function
modular arithmetic
induction
number theory
greatest common divisor
real analysis
floor function

Let $ m$ be the number of subsets $ A\subset \{1,2,\dots,n\}$ that satisfy $ \gcd(x\in A)\equal{}1$. Show that if $ n\geq 2$ then $ m$ is not a perfect square.

If $ A\equal{}\{a_1,a_2,\ldots,a_k\}$, do you mean $ \gcd(a_1,a_2,\ldots,a_k)\equal{}1$ or $ \gcd(a_1,a_2)\equal{}\gcd(a_1,a_3)\equal{}\ldots\equal{}\gcd(a_{k\minus{}1},a_k)\equal{}1$?

In my opinion ,this is a hard problem .I have found the general formulas but I failed to prove it was not a perfect square .

Well, it's not easy but you can post partial results or ideas  :)

Ok,I will display some ideas that I had thought . Let f(n) be the number of subsets that satisfy condition $ \gcd(x\in A) = 1$ . G(n) is the number of these subsets that contain '$ n$ as an element . It is so obvious that :
\[ f(n) = f(n - 1) + g(n)\]
[u]Claim 1 [/u]: The number of subsets that contain n as an element and $ \gcd(x\in A) = d$ is $ g(\frac {n}{d}) - 1$ for all d>1 (Notice that these sets contain n so $ \frac {n}{d}$ is an integer ) .
Proof :Consider a set that satisfies $ \gcd(x\in A) = d$. By the definition of  greatest common divisor function we have  $ \gcd(x\in A) = d$ iff  ${ \gcd(\frac {x}{d}|x\in A } = 1,|A| > 1$ . Because $ A\subset \{1,..,n\}$ so $ A'\subset \{1,..,\frac {n}{d}\}$ . The number of these sets is $ g(\frac {n}{d})$ , but we must notice that if A contains only $ n$ then it doesn't satisfy condition , that is why the number of these sets must be $ g(\frac {n}{d}) - 1$ .
[u]Claim 2 [/u]: $ \sum_{d|n} [g(d) - 1] = 2^{n - 1} - 1$ .
For each subset A of $ {1,..,n\}}$ and contains n as a term then the greatest common divisor of all elements in this set is unique . This means that all subset A and $ \gcd(x\in A) = d$ cover all subsets of $ {1,..,n - 1}$ when we get out  $ n$ from A . Therefore we have :
\[ g(n) + \sum_{d|n;d < n}[g(\frac {n}{d}) - 1 ] = 2^{n - 1}\]
This is equivalent to claim 2 . By mobious inverse function ,we have  :
\[ g(n)-1 = \sum_{d|n} (2^{d - 1} - 1).\mu(\frac {n}{d})\]
But we knew that $ \sum_{d|n}\mu(d) = 0$ so :
\[ g(n) = \sum_{d|n}2^{d - 1}.\mu(\frac {n}{d})+1\]
To get formulas of f(n) we only need to take sum of all g(i)

Nice, you are in fact very close to finishing. I suggest you compute some small values of $ f(n)$ and try to find a pattern  :lol:  :wink:

The number of subsets with $ \gcd(x\in A) \equal{} d$ is $ g(\frac{n}{d})$ not $ g(\frac{n}{d})\minus{}1$.
After you make the corrections, what is $ g(n)\pmod{3}$?

[quote="Albanian Eagle"]The number of subsets with $ \gcd(x\in A) \equal{} d$ is $ g(\frac {n}{d})$ not $ g(\frac {n}{d}) \minus{} 1$.
After you make the corrections, what is $ g(n)\pmod{3}$?[/quote]
I don't think so , it is true iff d=1 . If $ d>1$ then the number must be $ g(\frac{n}{d})\minus{}1$ because set $ \{\frac{n}{d}\}$  satisfies condition $ \gcd(x\in A)\equal{}1$ but  set $ n$ doesn't satisfy $ \gcd(x\in A)\equal{}d$ ,so the numbers must be $ g(\frac{n}{d})\minus{}1$ if $ d>1$ .For example ;$ g(2)\equal{}2,g(3)\equal{}4,g(4)\equal{}7$ ... I can prove that $ 3|g(n)\minus{}1$ by induction but that all I got .

how does the set $ \{\frac{n}{d}\}$ satisfy $ \gcd \equal{}1$  :huh:
For example. $ f(2)\equal{}2,f(3)\equal{}5$ so $ g(3)\equal{}3$...

[quote="Albanian Eagle"]how does the set $ \{\frac {n}{d}\}$ satisfy $ \gcd \equal{} 1$  :huh:
For example. $ f(2) \equal{} 2,f(3) \equal{} 5$ so $ g(3) \equal{} 3$...[/quote]
Excuse me ! When I saw this problem ,I didn't  know  if a set that contains only one element satisfies condition .But when saw second post , I think it also satisfies .

This is the definition of gcd I meant for this problem.
$ \gcd(a_1,a_2,\dots,a_n)\equal{}\prod_p p^{\min(\nu_p(a_i))_{1\le i\le n}}$

[quote="Albanian Eagle"]Let $ m$ be the number of subsets $ A\subset \{1,2,\dots,n\}$ that satisfy $ \gcd(x\in A) \equal{} 1$. Show that if $ n\geq 2$ then $ m$ is not a perfect square.[/quote]

We’ll use the following notation: $ \mathbb{N}: \equal{}\{0,1,2,\ldots\}$ and $ \mathbb{N}_0: \equal{}\mathbb{N} \setminus \{0\}$; $ f(n): \equal{}|\{A \subset \{1,2\ldots,n\} \text{ s.t. } \text{gcd}(x \in A)\equal{}1\}|$ for all $ n \in \mathbb{N}_0$; $ g(n): \equal{}|\{A \subset \{1,2\ldots,n\} \text{ s.t. } n \in A \text{ and }\text{gcd}(x \in A)\equal{}1\}|$ for all $ n \in \mathbb{N}_0$; $ *$ denotes the [url=http://mathworld.wolfram.com/Convolution.html]convolution product[/url] between two arithmetical function; $ a(n): \equal{}2^{n\minus{}1}$ for all $ n \in \mathbb{N}_0$; $ b(n): \equal{}(\minus{}1)^n$ for all $ n \in \mathbb{N}_0$; $ u(n): \equal{}1$ for all $ n \in \mathbb{N}_0$; $ I(n): \equal{}\lfloor n^{\minus{}1} \rfloor$ for all $ n \in \mathbb{N}_0$; $ \mu(\cdot)$ denotes the [url=http://en.wikipedia.org/wiki/M%C3%B6bius_function]Moebius function[/url] and $ (\frac{a}{p})$ denotes the usual [url=http://en.wikipedia.org/wiki/Legendre_symbol]Legendre symbol[/url].

Clearly $ \displaystyle f(n)\equal{}\sum_{i\equal{}1}^n{g(i)}$, and $ g(1)\equal{}g(2)\equal{}1$, and $ a(n)\equal{}(g*u)(n)$. By Moebius inversion and operating in $ \mathbb{F}_3$ we have $ g(n)\equal{}(a*\mu)(n)\equal{}\minus{}(b*\mu)(n)$: if $ n>2$ is odd then $ g(n)\equal{}\minus{}(u*\mu)(n)\equal{}\minus{}I(n)\equal{}0$; if $ n\equal{}2q$ for some odd $ q>1$ then $ g(n)\equal{}\minus{}(u*\mu)(q)\minus{}(u*\mu)(q)\equal{}\minus{}2I(q)\equal{}0$, otherwise $ n\equal{}2^tq$ for some $ t>1$ and $ q>1$ odd, then $ g(n)\equal{}\minus{}(u*\mu)(q)\minus{}\mu(2)(u*\mu)(q)\equal{}(u*\mu)(q)\minus{}(u*\mu)(q)\equal{}0$. We have concluded that $ f(n)\equal{}f(2)\equal{}2$ for all $ n>1$, but $ (\frac{2}{3})\equal{}\minus{}1$, so $ f(n)$ cannot be a perfect square.


The question is:

How many terms are in this series and find its sum.
2(0.1) + 2(0.1)^2 + .... +2(0.1)^10

My answer is:

there are 10 terms
the sum is:

(0.2(1-0.1^10))/(1-0.1) = 0.1

is this right??

Note that a sum which starts with $0.2$ and consists entirely of positive terms can't possibly sum to $0.1,$ a number smaller than that first term.

When you said $\frac{0.2(1-0.1^{10})}{1-0.1}$ that was correct. You just need to finish with the correct arithmetic.


1. Find the center and the radius of the circle
    x2 + y2 – 2x + 2y + 2 = 16

2. Find the equation of the inverse relation y = 2x - 8

Does anyone have a clue?

[hide="hint for 1"]

Complete the square.

[/hide][hide for 2]

Replace x with y and solve.[/hide]

I am still confused.  Is their anyway that you can show me the exact work for the problem.  This will help me to figure the steps out and truly grasp the material.


Let $ f: \mathbb R\to\mathbb R$ be a function such that $ f(0) \equal{} 0$ and $ \frac {f(t)}{t}$ is a monotonic function on $ \mathbb R/\{0\}$ .Prove that
\[ xf(y^2 \minus{} yz) \plus{} yf(z^2 \minus{} xy) \plus{} zf(x^2 \minus{} yz)\ge 0
\]

This is [color=red]T11/275 of M&Y[/color].

Sholud be locked.

As for the reason, every does know! :mad:


Fine...all right, I admit it, I do last minute studying for MC...life likes to stack stuff up...so any help on some of these babies from old State sprints would be very much appreciated. Once more. Yep.

1) State 2001 Sprint 13: Ugh. All right, I [i] know [/i] this is somewhere on the site, as I saw it before, but searching for "rectangle" is incredibly ineffective, so if anyone would bother to reiterate for me, I'd be very happy...

[i] What is the ratio of the width to the length of a folded sheet of paper if the rectangle formed when the sheet is folded in half as shown is similar to the original rectangular sheet? Express your answer as a common fraction in simplest radical form.

I'm sorry I can't use the drawing tools here effectively yet, but imagine a normal sheet of paper turned lengthwise, then folded in half that way like a greeting card, so if you had a rectangle with, say, length 3 and width 2, then it'd end up having a new width of half its length, 1.5, and its width being its new length, 2... [/i]

[hide] (2^(1/2))/2 [/hide]

2) State 2001 Target 5: No clue, really. I found the diameter and made a 9-12-15 triangle, and got 2...

[i] Point P is 9 units from the center of a circle of radius 15. How many different chords of the circle contain P and have integer lengths? [/i]

[hide] 12 [/hide]

3) State 2002 Sprint 24:
[i] A jet plane departing on time, flying between two airports at an average speed of 540 mph arrives eight minutes late. Departing on time and flying at an average speed of 480 mph it arrives fifty-three minutes later. What is the number of miles between the two airports? [/i]
I know this is a simple problem, but is my logic correct? It seems iffy...Should get d=3240:
d=rt
540(t+(8/60))=480(t+(53/60))
540t+72=480t+424
60t=352
If t is number of hours for a flight on time, 60t=minutes, so add 8 to that to get 360 minutes, and convert into 6 hours, then use to find d...is that right?

4) State 2002 Sprint 26: [i] What is the remainder when (2^87) +3 is divided by 7? [/i]
It's 4. But the problem is, it's the whole number, not just one of the ones that ask for the final two digits or so, and it's 7, not something like 5, so it throws the pattern out of whack...I found the 2 exponential pattern, added 3, and got something that ended in "eleven" (8+3=11), so I just took a remainder of 4 from 11/7 and got lucky. What's the right way to do this?

5) State 2002 Team 9: [i] A farmer has previously fenced a field in the shape of a trapezoid as shown. The farmer divides this field into exactly four congruent trapezoids. How many additional meters of fencing are needed? Express your answer to the nearest whole number.

Companion pic again...trapezoid is basically a 40mx40m square with a right triangle, both legs 40m, tacked onto it on the right. [/i]

Bah, my main problem is that I can't figure out how it's divided into the 4 trapezoids here...

[hide] 108 [/hide]

May have more later, but right now, thanks for any answers or help. *Nods* Tons of genii here, way higher than me...

Oh, and just out of curiosity, is there anyone here who can honestly look at (x+(1/x)=4) and figure out what to multiply to the x+(1/x) to find (x^3)+(1/(x^3)) in about three or four minutes on the sprint? Because I played all sorts of tricks, but couldn't get that...

[quote="HyperGFreak"]1)What is the ratio of the width to the length of a folded sheet of paper if the rectangle formed when the sheet is folded in half as shown is similar to the original rectangular sheet? Express your answer as a common fraction in simplest radical form.[/quote]If you set length as x and width as y, you'll see that if you fold the paper the new figure becomes a rectangle with length y and width x/2. So, since these are similar, you can set up a proportion: x/y=y/(x/2). Cross multiplying, you get (x^2)/2=y^2. Taking the square root of both sides, y=x/(sqrt2). So the ratio is 1 to sqrt2. After rationalizing the denominator, you get (sqrt2)/2.
(Sorry I can't do Latex...)

[quote="HyperGFreak"]Point P is 9 units from the center of a circle of radius 15. How many different chords of the circle contain P and have integer lengths?[/quote]Actually, you're on the right track. The chord of length 24 is the shortest one possible, while that of length 30 is the longest. There is only one possible segment more these two lengths. However, if you think about it, you will realize that though we don't know exactly how to draw them, other [i]possible[/i] lengths can range anywhere between these two numbers. The integers between 24 and 30? 25, 26, 27, 28, and 29, of course. Also, each of these lengths have duplicates, because they become different chords if you flip them over to the other side. So, 5*2+1+1=12.

[quote="HyperGFreak"]A jet plane departing on time, flying between two airports at an average speed of 540 mph arrives eight minutes late. Departing on time and flying at an average speed of 480 mph it arrives fifty-three minutes later. What is the number of miles between the two airports?[/quote]Yes, that's the answer. I didn't really look at how you did it though...
[quote="HyperGFreak"]What is the remainder when (2^87) +3 is divided by 7?[/quote]Well, if you wanted to do it the "finding patterns" way, you can do this:
(2^1+3)/7, R5
(2^2+3)/7, R0
(2^3+3)/7, R4
(2^4+3)/7, R5
...
As you can see, it repeats in cycles of 3. 87/3 has no remainder, so you take the third number in the pattern, or 4.
[quote="HyperGFreak"]A farmer has previously fenced a field in the shape of a trapezoid as shown. The farmer divides this field into exactly four congruent trapezoids. How many additional meters of fencing are needed? Express your answer to the nearest whole number.[/quote]I attached a pic to this post. That should give it away.

I don't have Latex... so I'll leave your last question to someone who can make it look good, unlike what my carats would do. :lol:    :wink:


inequalities
geometry
rectangle
inequalities proposed

Show  that  for any  interior  point  P  of a  rectangle  ABCD , the area  of the  rectangle  does  not exceed
 $PA\cdot PC+PB\cdot PD$
Thanks for your time .

See http://www.mathlinks.ro/Forum/viewtopic.php?t=20958 .

  Darij


I  really need to install the package fancyhdr for TeXNic, but when I do everything as the Wiki says, after I confirm the install, it says "fatal error", and it doesn't work.

[hide="This is everything from the log."]
This is pdfTeX, Version 3.141592-1.40.4 (MiKTeX 2.7) (preloaded format=pdflatex 2008.6.27)  5 AUG 2008 12:33
entering extended mode
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LaTeX2e <2005/12/01>
Babel <v3.8h> and hyphenation patterns for english, dumylang, nohyphenation, ge
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("C:\Program Files\MiKTeX 2.7\tex\latex\base\article.cls"
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
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File: size10.clo 2005/09/16 v1.4f Standard LaTeX file (size option)
)
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Package: amsmath 2000/07/18 v2.13 AMS math features
\@mathmargin=\skip43

For additional information on amsmath, use the `?' option.
("C:\Program Files\MiKTeX 2.7\tex\latex\amsmath\amstext.sty"
Package: amstext 2000/06/29 v2.01

("C:\Program Files\MiKTeX 2.7\tex\latex\amsmath\amsgen.sty"
File: amsgen.sty 1999/11/30 v2.0
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("C:\Program Files\MiKTeX 2.7\tex\latex\amsmath\amsbsy.sty"
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\pmbraise@=\dimen104
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("C:\Program Files\MiKTeX 2.7\tex\latex\amsmath\amsopn.sty"
Package: amsopn 1999/12/14 v2.01 operator names
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\inf@bad=\count87
LaTeX Info: Redefining \frac on input line 211.
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LaTeX Font Info:    Redeclaring font encoding OML on input line 567.
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LaTeX Info: Redefining \[ on input line 2666.
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("C:\Program Files\MiKTeX 2.7\tex\latex\amsfonts\amssymb.sty"
Package: amssymb 2002/01/22 v2.2d

("C:\Program Files\MiKTeX 2.7\tex\latex\amsfonts\amsfonts.sty"
Package: amsfonts 2001/10/25 v2.2f
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(Font)                  U/euf/m/n --> U/euf/b/n on input line 132.
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("C:\Program Files\MiKTeX 2.7\tex\latex\amscls\amsthm.sty"
Package: amsthm 2004/08/06 v2.20
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! LaTeX Error: File `fancyhdr.sty' not found.

Type X to quit or <RETURN> to proceed,
or enter new name. (Default extension: sty)

Enter file name:
! Emergency stop.
<read *>

l.6

*** (cannot \read from terminal in nonstop modes)

Here is how much of TeX's memory you used:
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 27i,0n,24p,280b,36s stack positions out of 5000i,500n,10000p,200000b,5000s
!  ==> Fatal error occurred, no output PDF file produced!
[/hide]

That's everything that it says inside my log.  I really need help, it's an assignment due tomorrow!  PLease help!!!!!!!!!!!!!!

It's not clear how you installed fancyhdr, but the first thing to do is to refresh the database. Start -> MiKTeX -> Settings -> Refresh FNDB and try again.

If that doesn't work then install it by going to Start -> MiKTeX -> Browse Packages, find fancyhdr then right-click and choose install.

If there's still a problem and you are in a hurry copy fancyhdr.sty into the same directory as your tex document and then LaTeX will find it. fancyhdr.sty should be in C:\Program Files\MiKTeX 2.7\tex\latex\fancyhdr but if it's not there you can download it from http://tug.ctan.org/tex-archive/macros/latex/contrib/fancyhdr/

This problem of finding fancyhdr.sty needs sorting out once you are not in a rush.

EDIT: Never mind.  Ignore this post please.

Oh, and it worked!  Thank you so much!

AIME15 u have a scary amount of post for someone who joined 5 months ago  :P


Solve in reals the following system of equations.

$x^2+y^2+z^2=125$
$x+y+z=25$

i don`t think it has solutions: $x^2+y^2+z^2\geq xy+xz+yz \rightarrow 125 \geq xy+xz+yz$ but $xy+xz+yz=\frac{(x+y+z)^2-x^2-y^2-z^2}{2}=250$ thus $125\geq 250$ contradiction!

At least one of $x,y,z$ must be positive.

1) If all $x,y,z$ are positive, then by QM-AM

$\sqrt{\frac{x^2+y^2+z^2}{3}}\geqslant\frac{x+y+z}{3}$

but $\sqrt{\frac{125}{3}}<\frac{25}{3}$, hence there's no solution in this case.

2) If two numbers are positive (WLOG we can take $x,y>0$) then

$x^2+y^2=125-z^2\land x+y=25-z$

Hence

\begin{eqnarray*}\sqrt{\frac{x^2+y^2}{2}}\geqslant \frac{x+y}{2} &\implies& \sqrt{\frac{125-z^2}{2}}\geqslant \frac{25-z}{2}\\ &\implies& 3z^2-50z+375\leqslant 0\end{eqnarray*}

As the discriminant of the last inequality is negative, it has no solutions.

3) If only one is positive, then two are negative, and similarly as in 2) we prove there's no solution.

What  if the  system  was : $\begin{cases}x^2+y^2+z^2=75\\x+y+z=15.\end{cases}$ ???

look back at my solution you get $x=y=z=5$

Or, more generally, what are the conditions on $m$ and $n$ ($m, n \in \mathbb{N}$ such that the system :
$\begin{cases}x^2+y^2+z^2=m\\x+y+z=n.\end{cases}$ has solutions! :)

$\\x^2+y^2+z^2\geq xy+xz+yz\\ \implies m\geq xy+xz+yz=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{n^2-m}{2}\\ 2m+m\geq n^2\\ m\geq \frac{n^2}{3}$
$\therefore$ if $m$ and $n$ staisfy $m\geq \frac{n^2}{3}$, then there are solutions to the system of equations.
Masoud Zargar


Is it possible to have more than one blog at once? I know you can contribute to someone's blog, but can you actually [b]own[/b] two or three blogs?

I think you can, by going to "my Profile" "blogs".

Someone should move this to the Questions, Suggestions, & Announcements subforum in the Site & General Information forum.

second question: how do i do that?

i don't see how to create another blog :maybe:


Find all ordered pairs of numbers $(a,b)$ such that $a\neq b$, and $a$ is divisible by $b$, where $a$ and $b$ are 7-digit numbers formed using each of the digits 1,2,3,4,5,6,7 exactly once each. Prove that the numbers you found are the only such numbers.

[hide="Hint"]$a \equiv b \equiv 1 \mod \mbox{?}$
Actually, come to think of it, there shouldn't be any solutions...  :? [/hide]

[hide]
Yeah, there aren't any solutions. And you appear to have already made the key step.[/hide]

a  :equiv: b :equiv: 1 (mod 3)
a = 4b because only 4 times 1 (mod 3) gives 1 (mod 3) again.
(7 times b will yield a number greater than 7654321)

Here's where I used my unique(?) method:

1  2  3  4  5  6  7

0  0  1  1  2  2  2
4  8  2  6  0  4  8

where the first row has the numbers that will form b.
and 2nd and 3row have the number that will form a.
1 times 4 is 4.
2 times 4 is 8.
3 times 4 is 12.
4 times 4 is 16, etc.
the 2nd row has the numbers that will be carried over to the next digit.
for example, 7 will yield 8 in its own digit but 2 will be carried over to the next digit.
Now 2 and 7 yield 8 in their digits, however, 8, 9, and 0 are not part of 7-digit numbers that has only 1,2,3,4,5,6,7 used exactly once. So there has to be at least a 3 that will carried over so 8+3 can at least yield 1, however it's impossible to have more than one 3 carry overs, so such a number can not be constructed.

or... you can have [hide="the one line proof"]$a \equiv b \equiv 1 \mod 9 \implies a \ge 10b \therefore \nexists (a,b)$[/hide] :D


limit
floor function
real analysis
real analysis theorems

Let $ \{a_n\}$ be a positive and decreasing sequence such that the series $ \sum a_n$ is convergent. Prove that $ \lim na_n\equal{}0$.

The condensation test kills this quickly.

The condensation test says that if $ a_n$ is positive and decreasing, then $ \sum_n a_n$ converges iff $ \sum_k2^ka_{2^k}$ converges.

(And this test is just a slight generalization of Oresme's 14th-century proof of the divergence of the harmonic series.)

So if $ a_n$ is decreasing and $ \sum_na_n<\infty,$ then $ \sum_k 2^ka_{2^k}<\infty.$

In particular, we have $ 2^ka_{2^k}$ tending to zero.

Now for a general $ n,$ there exists $ k$ such that $ 2^k\le n<2^{k\plus{}1}.$ We have

$ na_n\equal{}\frac{n}{2^k}2^ka_n<2\cdot 2^ka_n<2\cdot2^ka_{2^k}$

and that can be made as small as we please by making $ n$ large enough.

Let $ \varepsilon > 0$. Since $ \sum_{n = 1}^{\infty} a_{n}$ converges we have that $ s_{m} = a_{1} + \ldots + a_{m}$ is a Cauchy sequence. Hence, there exists $ M \in \mathbb{N}$ such that if $ m, n \geq M$ then $ \vert s_{m} - s_{n} \vert < \frac {\varepsilon}{2}$.

[b]Let[/b] now $ N = 2M$. If $ k > N = 2M$ it follows that

$ \left \lfloor \frac {k}{2} \right \rfloor + 1 > \frac {k}{2} \geq \left \lfloor \frac {k}{2} \right \rfloor \geq M.$

Therefore,

$ \\
\frac {\varepsilon}{2} > \left \vert s_{k} - s_{\left \lfloor \frac {k}{2} \right \rfloor} \right\vert \\
= \left \vert a_{k} + a_{k - 1} + \ldots + a_{\left \lfloor \frac {k}{2} \right \rfloor + 1} \right \vert \\
= a_{k} + a_{k - 1} + \ldots + a_{\left \lfloor \frac {k}{2} \right \rfloor + 1} \\
\geq \left(k - \left \lfloor \frac {k}{2} \right \rfloor \right) a_{k} \\
\geq \left(\frac {k}{2}\right)a_{k} \\
= \left \vert \frac {k \cdot a_{k}}{2} \right \vert$

and the proof terminates.

coquitao's choice to add up a "dyadic block" - that is, to add $ a_n$ for $ n$ ranging from $ \frac k2$ to $ k$ - shows that his proof has its roots in the same set of ideas as the Oresme argument and the condensation test. These ideas are all related.


inequalities
calculus
integration
inequalities unsolved

[quote="Pradhan P"]S  is greater than integral from 10  to 1000 of 1/x 2 and  S  is less than integral from 9  to 999 of 1/x 2. This gives the required bound.[/quote]
thnx,
i believe there is another solution without using integrals.. i found this problem as 'olympiade excercise' in a schoolbook for students who didn't learn integrals yet....

[quote="billzhao"]Note that
\[
\frac{1}{n} - \frac{1}{n+1} < \frac{1}{n^2} < \frac{1}{n-1} - \frac{1}{n}
\]
and sum the inequality for $n$ from 10 to 1000 we get
\[
\frac{1}{10} - \frac{1}{1001} < S < \frac{1}{9} - \frac{1}{1000}
\]
and the rest is easy.[/quote]
this is exactly what i meant. it's also useful to know how the other ways work.


[quote="guile"]Any recommendable site on modulo division?[/quote] Google would be your friend. I say that for I don't really know about the American schooling system, and when it says "High School Basics" my mind draws a blank!

You could check out [url]http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf[/url].

[quote="guile"]Im not from US. Anyway, high school basics in this forum are quite easy before but i think it got harder. Before, proofs are not allowed in the said forum.[/quote]guile is right and proofs should be very minimal in HSB, only the basic ones that should not involve too deeply.

Please read [url=http://www.mathlinks.ro/viewtopic.php?t=32837]High School Basics Difficulty - Read BEFORE Posting Problems[/url]


Can you help me with this question please?

A game involves choosing tiles from a bag. There are eight tiles in the bag. as

A A E E E I O U

Chris chooses two tiles at random from the bag. What is the probability that they have different letters? Please explain in your own words if possible.

Charikaar

Thanks for the promt reply.

That really helped. I have been working on probability for the last 10 days and could not get my head around them but I think I learnt something useful today.

Charikaar


through factor theorem (x+1) is not a factor of $(1+x^{18})$.

In general $(x+1)$ is a factor of $(x^{n}+1)$ iff n is odd.


Using each of the ten digits 0, 1, 2 ......9 just once, is it possible to form positive integers whose sum is exactly 100?

  At first this problem seem only solvable through a " guess- and- trial" approach, but in fact there is a very simple and elegant solution.

I don't know why, but I think the sum has to be divisible by 9.

Good observation, now can you expand on that?

That means that the sum can't be 100. But explaining it is my problem.

If the sum of the digits is divisible by 9, the number is divisible by 9. No matter what the combination of numbers you make, adding them all together will result in a number divisible by 9, so the sum of however many numbers are used will be a multiple of 9, which 100 is not.

Well, I can give examples... try it with any multiple of 9 that you want. For example, $27 \Rightarrow 2+7=9$. Okay, so let's try a higher number. $15183 \Rightarrow 1+5+1+8+3=18$. $4870296 \Rightarrow 4+8+7+0+2+9+6=36$. I'll try to prove it as well, but I'm not quite as sure about that...

Say you have a number in the form $A\cdot 10^n+B\cdot 10^{n-1}+C\cdot 10^{n-2}+...+X$, where $A,B,C...X$ are integers from 0 to 9. $10^n\equiv 1 \bmod{9}$, for any integer $n$, so essentially we are looking for $A\bmod{9}+B\bmod{9}+C\bmod{9}+...+X\bmod{9}=0$, and, using the digits from any multiple of $9$, this equation is equal to $0$. I'm not sure how to prove it any further than that.

[hide]
$N=10^{n-1} \cdot a_1 + 10^{n-2} \cdot a_2 + \cdots + a_n \equiv 0 \mod {9}$
$(1)^{n-1} \cdot a_1 + (1)^{n-2} \cdot a_2 + \cdots + a_n \equiv 0 \mod {9}$
$a_1 + a_2 + \cdots + a_n \equiv 0 \mod {9}$
$1+2+3+\cdots+9=45 \equiv 0 \mod {9}.$
Therefore, the sum of the numbers can never equal $100$.
[/hide]

That is indeed a nice solution, but a simple and more elegant one, is as follows:

 [hide]  The answer is no, it is not possible.

To prove this, let assume that several numbers are formed so that their sum is 100 and each of the digits 0, 1, 2 . . . . 9 is used just once.  Clearly these numbers can have one or two digits only.

  Let $t$ be the sum of the left digits of each of the two-digit numbers.  Then the sum of all remaining digits must be 45-$t$ as 0 + 1 + 2 + 3..........9= 45

  Hence 10$t$ +  (45-$t$) = 100

  Thus 9$t$ = 55 which is not possible, since $t$ is an integer.[/hide]

Nice, simple equation solving. I like that  :)


[quote="Dumel"]$ a,b,c,d > 0$ and
$ \frac {1}{1 \plus{} a^4} \plus{} \frac {1}{1 \plus{} b^4} \plus{} \frac {1}{1 \plus{} c^4} \plus{} \frac {1}{1 \plus{} d^4} \le 1$
prove that $ abcd \ge 3$
I tried Jensen for $ f(a) \equal{} \frac {1}{1 \plus{} e^{4a}}$ but it's not convex in $ (0, \infty)$    :|[/quote]
I think you should suppose $ abc\leq 3$ and prove that: $ \frac {1}{1 \plus{} a^4} \plus{} \frac {1}{1 \plus{} b^4} \plus{} \frac {1}{1 \plus{} c^4} \plus{} \frac {1}{1 \plus{} d^4} \ge 1$

from the condition we get :

$ \frac{d^4}{1\plus{}d^4} \ge \frac{1}{1\plus{}a^4}\plus{}\frac{1}{1\plus{}b^4}\plus{}\frac{1}{1\plus{}c^4} \ge \frac{3}{\sqrt[3]{(1\plus{}a^4)(1\plus{}b^4)(1\plus{}c^4)}}$

do the same with a,b,c and multiply... :maybe:


z is a complex number such that z + 1/z = 2cos(3 :^o: ). Find [z^2000 + 1/z^2000] + 1...

My solution:[hide]

Let z = e^(3i) = cos(3) + isin(3)

      1/z = e^(-3i) = cos(-3) + isin(-3) = cos(3) - isin(3)

z + 1/z = e^(3i) + e^(-3i) = 2cos(3)

so z^2000 + 1/z^2000 = e^(6000i) + e^(-6000i) = 2cos(6000)

2*cos(6000) + 1 = 2*(-1/2) + 1 = 0

I don't have the answers, so I don't know if it's correct, and also i'm not sure if 0 is a possible answer on the AIME... [/hide]

It's number 9 btw.

It's correct. When I did this problem for practice a few years ago, I was also shocked that the answer would be [hide]0[/hide]. But it's allowed.

I spend so long in Government class today trying to figure out if [hide]0[/hide] was an acceptable answer, Thanks.

Isn't there a radians-degrees issue there?

Well, actually, let me restate that.  The parts where you have e are actually wrong -- you should have converted to radians to make them true.  However, since you have to reconvert them back at the end, it turns out not to matter.  But just be aware of using the wrong kind of angle measure in the wrong place.

That's true. That's why I often write :ang::theta: rather than ei:theta:. Then I don't have to worry about radians and degrees.

I see what you mean... because [hide]e^3i implies 3 radians, while I have cis(3 :^o: )... [/hide] I see your point, I got lucky that it did not matter in the end... Thanks for pointing that out.


Find the least number of complete-bipartite graphs that partition the edges of $C_{15}\times C_{5}$.

What do you mean by graph multiplication?  What are the edges of $C_{15}\times C_{5}$?

Suppose $V(G)=\{x_{1},\dots, x_{n}\}, \; V(H)=\{y_{1},\dots,y_{m}\}$. Then$V(G\times H)=V(G)\times V(H)$ and $(x_{i},y_{j})$ is connected to $(x_{i'},y_{j'})$ iff $i=i'$ and $y_{j}$ is connected to $y_{j'}$ or $j=j'$ and $x_{i}$ is connected to $x_{i'}$.


factorial calculus integration function Euler geometry induction

I had a really cool prof last semester who was talking about factorial being defined for non-integers.  How is this possible?

It's the [tex]\Gamma[/tex] function. Unfortunately, it is defined so that [tex]\Gamma(s)=(s-1)![/tex] for positive integers s. Alternatively, Riemann used a function [tex]\Pi(s)=s![/tex] for nonnegative integers s, but for some reason, the [tex]\Gamma[/tex] function caught on instead. Anyway,

[tex]\displaystyle\Gamma(s)=\int_0^\infty e^{-t}t^{s-1}\; dt.[/tex]

See if you can prove that this definition matches factorials in the previously stated manner. (It isn't too hard.)

Give me like a week to answer this, we have <b> not</b>done the integrals w/the infinity sign yet in my calc class.

EDIT: I have not done them, typo.

While Julie's waiting for her calculus class to get to improper integrals (and integration by parts, which you can't live without), here's something from the pre-history of the subject. Before Riemann, before Euler, even before there was calculus. This is from the hand of John Wallis:

Step 1: Establish the following pattern, true for all nonnegative integers n:

[tex]\displaystyle\int_0^1 x^{n} (1-x)^{n}\; dx = (n!)^2/(2n+1)!.[/tex]

Step 2: Evaluate the integral in step 1 in the case n = 1/2.  Don't do this by integration; use that the integral is the area under a curve, and recognize the curve as something whose area you know.

Step 3: Make a completely unjustified leap of faith and just assume that the formula in step 1 makes sense when n is not an integer. Apply it when n = 1/2, and solve. Tell me what (1/2)! should be.

[Note: I wrote this before I saw Prof. Merryfield's post.]

Simon is right; that's the modern way we define factorials for non-integers.  Even though that is best, I would like to describe a method that early mathematicians could have used to define fractional factorials.  Note, I am not saying they did do it this way (I don't know the history -- was it Euler who did it first?); I'm just saying they could have.  I'll focus on the special case of (1/2)!.

Consider 9!, 10!, and 11!.  Note that 10! is 10 times 9!, and 11! is 11 times 10!.  That means that these three terms almost form a geometric sequence.  In a geometric sequence with three positive terms a, b, and c, it is known that the middle term b is the geometric mean of a and c, meaning b = :sqrt: (ac).  Thus 10! is approximately the geometric mean of 9! and 11!.

By analogy, we might expect 9!, 9.5! (whatever it is), and 10! to approximately form a geometric sequence.  In particular, 9.5! will approximately be the geometric mean of 9! and 10!, which is about 1,147,527.  Once we know 9.5!, we can figure out 8.5! by dividing by 9.5.  Then we can figure out 7.5!, 6.5!, ..., and so on until 0.5!.  By doing that, we get that 0.5! is approximately 0.897.  That's not too far off.

That is only approximate, though, because 9!, 9.5!, and 10! are not exactly a geometric sequence.  To do better, we could look at 99!, 99.5!, and 100! instead.  Following the steps in the previous paragraph, we would get an even better approximation for (1/2)!.  Of course, we can start this process using numbers as large as we want.  In that way, we will get a limiting value for (1/2)!.

It turns out that (1/2)! = :sqrt:(:pi:) / 2, which is about 0.886.  That's a surprising formula.  We could prove it using the process I described above and something called Wallis's infinite product for :pi: .  But that's not my main point.  My main point was that there is a method for coming up with fractional factorials that even precalculus students can understand.

Hi, thanks Ravi B, Kent Merryfield, and Simon for the replies.

Step 1:

Induction?

For n=1,

[tex]\int_{0}^1{x(1-x)}dx[/tex]

let u=x and v'=1-x, (so u'=1 and [tex]v=\int(1-x)=x-\frac{x^2}{2}+C[/tex])

[tex]\int_{0}^1{x(1-x)}dx=(x)(x-\frac{x^2}{2})-(\frac{x^2}{2}-\frac{x^3}{6})=\frac{x^2}{2}-\frac{x^3}{3}+C[/tex]

Evaluating this w/the fundamental theorem of calculus, you get 1/6.

For n=1:

[tex]\frac{(n!)^2}{(2n+1)!}=\frac{(1!)^2}{(2(1)+1)!}=\frac{1}{3!}=\frac{1}{6}[/tex]

So it works for n=1.

Assume that:

[tex]\int_{0}^1[x^k(1-x)^k]=\frac{(k!)^2}{(2k+1)!}[/tex]

Now, show that it is true for n=k+1.

Okay, here I am not sure what to do.  I have to find the indefinite integral of [tex]x^{k+1}(1-x)^{k+1}[/tex].  I tried a ton of manipulation using integration by parts and I could not get it.  I think my old prof (who mentioned this about factorial) also used factorial before when he proved that m^x would always dominate x^m or something like that.

Step 2:

[tex]x^{\frac{1}{2}}(1-x)^{\frac{1}{2}}=\sqrt{x-x^2}[/tex]

Let [tex]y=\sqrt{x-x^2}[/tex], thus [tex]y^2 + x^2 -x = 0[/tex].

[tex]y^2 + [x^2-x+(\frac{-1}{2})^2]=(\frac{-1}{2})^2[/tex]
[tex](x-\frac{1}{2})^2+y^2=\frac{1}{4}[/tex]

Okay, so this HALF (the positive half) of a circle w/a radius = 1/2, and an center at (1/2,0).

The only thing that is iffy about this, is that the curve never touches the x-axis.  But anyway, I will proceed.  The area of 1/2 the circle =  :pi: r^2= :pi: /4.  Okay, now you want to take the area of that rectange below the circle between it's diameter and the x-axis.  The height of this rectangle = 1/2, and the length = 2r = 2*(1/2)=1.  So the area of the rectangle is 1/2 *1=1/2.

[tex]\frac{\pi}{4}+\frac{2}{4}=\frac{\pi+2}{4}[/tex]

Am I correct so far?

On step 2: You've suffering from one of those momentary hallucinations that all mathematicians have whether they admit it or not: the center of the circle IS on the x-axis, and the curve does touch the x-axis. There's no extra rectangle.

On step 1: Depending on the mood I'm in, I might just settle for showing it works for n = 0, 1, and 2 and just guessing that it keeps working after that.

If you're serious about showing the validity of the formula, one approach might be to discover a general formula based on m and n for the expression

[tex]\displaystyle\int_0^1 x^{m} (1-x)^{n}\; dx[/tex]

by integrating by parts over and over again until the last integral either has nothing in it but a power of x or nothing in it but a power of (1-x). Keep to the definite integral for all of the integrations by parts, and use that the evaluations at 0 and 1 result in zero for everything except that last integral.

There's a recursion (induction) in there, but it's "horizontal", conecting the integral with (m, n) with the integral with (m-1, n+1) or with (m+1, n-1). There's no simple way to make the "vertical" step from (n, n) to (n+1, n+1).

[quote]On step 2: You've suffering from one of those momentary hallucinations that all mathematicians have whether they admit it or not: the center of the circle IS on the x-axis, and the curve does touch the x-axis. There's no extra rectangle. [/quote]

Yea, that was stupid.  What happened was that I graphed it on the TI-89 to make sure it was a circle before I posted, and that graph is above the x-axis for some odd reason.  Oh well, that will show me to trust calculators.  So the area is just pi/4.

<b>Step 3:</b>

So we have the equation [tex]\frac{(n!)^2}{(2n+1)!}=\frac{\pi}{4}[/tex].

If you leave the numerator alone, but substitute n=1/2 into the denominator, then isolate n!:

[tex]\frac{(n!)^2}{2}=\frac{\pi}{4}[/tex]
[tex](n!)^2=\frac{\pi}{2}[/tex]
[tex]n!=(\frac{1}{2})!=\sqrt{\frac{\pi}{2}}[/tex]

I must have gone wrong somewhere though b/c my answer is different from Ravi's.

Semicircle, not circle. The area is  :pi:/8, not  :pi:/4. Make that change and you'll be in agreement.

Now that you've done that, what do you suppose the right values are for (3/2)!, (5/2)! and (7/2)! (also known as Gamma(5/2), Gamma(7/2), and Gamma(9/2))? For that matter, how about (-1/2)! (also known as Gamma(1/2))?

I just happened to be studying the gamma function for the past few days since I was looking for the factorial function for non-integral values.  In a calculus book I found, they substituted t = u^2 the definition of gamma [tex]\Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} \, dt[/tex] to get

[tex]\displaystyle\Gamma(x) = 2\int_0^{\infty} u^{2x-1} e^{-u^2} \, du[/tex]

Then the let x = 1/2 and then evaluated

[tex]
\displaystyle 2\int_0^{\infty} e^{-u^2} \, du[/tex]

I don't really know how they did it though.  It showed how they did it, but it somehow had two integrals and more variables.

nr1337:

This is normally presented in a Calculus III or Vector Calculus course. You have to know about double integrals in both rectangular and polar coordinates. In particular, you have to know that when you change a double integral from rectangular to polar coordinates, [i]dx dy[/i] becomes [i]r dr d[/i]:theta:.

We're trying to compute
 [tex]  I = \int_{-\infty}^{\infty}  e^{-x^2}  dx[/tex]

Start with
[tex]\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2}e^{-y^2} dx dy[/tex]

If we do that as an iterated integral in rectangular coordinates, we get [i]I[/i] :^2:.

In polar coordinates, it becomes
[tex]\int_0^{\infty} \int_0^{2\pi} r e^{-r^2} d\theta dr = 2\pi\int_0^{\infty} r e^{-r^2} dr = \pi[/tex]

Hence, [i]I[/i] =  :sqrt::pi:

Now, a followup question. Let "1-measure" be length. Let "2-measure" be area. Let "3-measure" be volume. Now continue, so that there is such a thing in "4-measure" in 4-dimensional space, and so forth. Let the "unit ball" in R^n be the set of all points in n-dimensional space whose Euclidean distance from the origin is  :le: 1.

In 1 dimension the unit ball is the interval [-1, 1] and its 1-measure is 2.
In 2 dimensions the 2-measure of the unit ball is  :pi: .
In 3 dimensions, the 3-measure of the unit ball is 4:pi:/3.

What is the 4-measure of the unit ball in 4 dimensions?
More generally, what is the n-measure of the unit ball in n dimensions?
What happens to the n-measure of the unit ball in n dimensions as n tends to infinity?

[quote="Kent Merryfield"]What is the 4-measure of the unit ball in 4 dimensions?[/quote]

Having never been exposed to this type of stuff, I'll take a stab at it. In 4 dimensions, the unit ball is the region (x,y,z,w) that satisfies x^2+y^2+z^2+w^2 <= 1. I'll be integrating by cross sections perpendicular to the x axis, i.e. I'll be integrating

[tex]\displaystyle \int_{-1}^1 (\mbox{volume})\,dx[/tex]

where 'volume' stands for the volume of the cross section.

Notice that x^2+y^2+z^2+w^2<=1 is equivalent to y^2+z^2+w^2<=1-x^2, and for fixed x this represents a 3-dimensional ball with radius [tex]\sqrt{1-x^2}[/tex]. So the volume of this cross section is [tex]\frac{4\pi}{3}(\sqrt{1-x^2})^3[/tex]. Therefore, the 4-measure of the 4-dimensional unit ball is

[tex]\displaystyle \frac{4\pi}{3}\int_{-1}^1 \left(\sqrt{1-x^2}\right)^3\,dx[/tex],

which (after a trig sub) comes out to be [tex]\frac{\pi^2}{2}[/tex].

-------------------------

Awkward LaTeX occurrence: I was trying to write 'volume of cross section' instead of 'volume' in the equation above, but for some reason the LaTeX compiler won't accept the word 'section' in it:

[code]\text{section}[/code]
[tex]\text{section}[/tex]

[code]section[/code]
[tex]section[/tex]

any ideas why?

That's n=4. I know of at least three different methods; zabelman used one of the ones with the least connection to the rest of this thread. For n > 4?


I made up a game and then made it.  It's a fun game and easy to make so I decided to post it here.

Setup:
4 cups labeled "Penny, Nickels, Dimes, Quarters"
1 Penny labeled "5"
3 Nickels labeled "-15", "10", and "20"
4 Dimes labeled "0", "0", "10", and "10"
4 Quarters labeled "-10", "-10", "-10", and "50"
Place the coins in their corresponding cups.

Play:
2 - 4 players take turns picking a cup and pulling a random coind out of that cup.  They then look at the coin they got (but don't show it to anybody else) and put it in their pile.  If you don't want to take a coin, you may say "Pass", which removes you from the remainder of the round.  When either everybody has passed or all the coins are gone, everybody reveals the numbers on their coins and adds up their points for the round.  (Each coin's point value is the number labeled on it.)  The game ends when a player reaches a predetermined goal score.

Post thoughts, ideas, responses, and so forth in this thread.

Does the person set their own goal?

If so, couldn't a person make the goal 5, and then pick out of the penny cup?

No, the group decides on a number like 300.  Then, the first player to achieve that goal wins, much like many point games.

[quote="1=2"][hide="quack"]The expected value of your winnings if you draw out of any cup is 5.[/hide][/quote]Yes, I did that on purpose.  Otherwise, the person who went first would have an advantage.

Umm.... what does that have to do with anything?

it's a game...hmm everytime i upload an avatar it becomes blurry


A standard set of dominoes consists of 28 blocks of size 1 by 2.  Each block contains two numbers from the set $0, 1, 2, \cdots, 6$.  We can denote the block containing 2 and 3 by [2,3], which is the same block as [3,2].  The blocks $[0,0], [1,1], \cdots, [6,6]$ are in the set but there are no duplicate blocks.

a)  Show that it is possible to arrange the twenty-eight dominoes in a line, end-to-end, with adjacent ends matching, e.g. $\cdots [3,1][1,1][1.0][0,6] \cdots$.

b)  Consider the set of dominoes which do not contain 0.  Show that it is impossible to arrange this set in such a line.

c)  Generalize the problem and prove your generalization.

[hide="Helpful connection"]The seven bridges of Königsberg :wink:[/hide]


Find the smallest positive base 10 integer such that if the first digit is transfered to the very end of the number, the new number so formed is 3 times the original number.

[hide]Equation to solve is something like :

3(10^n*a+B) = 10B+a, with 1<=a<=9, B < 10^n, n >= 1

or

a*(3*10^n-1) = 7*B

So 7 | a (i.e. a=7) or 7 | (3*10^n-1).

a=7 is not correct (B = (3*10^n-1) > 10^n)) so 7 | (3*10^n-1). Using mod, smallest n is 5.

3*10^5-1 = 17*2521, so only possible value for a is 1 and smallest number looked for is 142857.[/hide]

In my youth, I had a book that called this number (142857) "The Ring Around The Rosie Number".  Does anybody know the name of this book?

There's a really cool alternative approach to this problem. Work on the problem a bit before you look at the hint.

Hint:[hide] Represent the number you are looking for as anan-1an-2...a1a0

Let N = 0.anan-1an-2...a1a0anan-1... (a repeating decimal number)

What can you do with this number N to solve the problem?[/hide]

For a cool extention of this problem, what happens if we change the 3 to some other number. For example, what if it were 3/2? What if it were 2?

What is the connection between my method and belenos's?

A professor here put that question on a problem sheet for the undergraduate math club but with 2 instead of 3. I realized that it was highly dependent on the base, so I wrote 142857 in base 100 as the answer. Later, I solved it for base 10, but I'll let other people do it here.

Just giving this thread a bump for those who have not tried it yet.

And by the way, there's a small problem with your solution Belenos. [hide]There's a typo in your factorization of 3*10^5 - 1, and a does not have to equal 1; though that will give you the smallest solution.[/hide].

If you take the leftmost digit of 714285 (the part that repeats in 1/14) and put it on the right, the new number is one-fifth of the old number.  And if you take 476190 (the part that repeats in 1/21) and take the [i]rightmost[/i] digit and put it on the left, you get a number that is one-tenth the old number.  I wonder if you can do something like this for all of the repeating parts of 1/(7 times some number)...

So what is the full solution to this problem?  I don't quite understand how you guys solved it.

Here is my solution.

Let's call the number that we're looking for N, and write it as [tex]\displaystyle {N = a_na_{n-1}a_{n-2} \ldots a_2a_1} [/tex] (written in base 10).

From this number N we create a new rational number x, where [tex]\displaystyle {x = 0.a_na_{n-1} \ldots a_2a_1a_n \ldots} [/tex] (a repeating decimal number).

Then by the condition asked of N, [tex]\displaystyle {3x = 0.a_{n-1} \ldots a_1a_n \ldots}[/tex] And of course [tex]\displaystyle {10x = a_n. a_{n-1} \ldots a_1a_n \ldots}[/tex].

So subtracting the first equation from the second we get, [tex]7x = a_n [/tex]. Since we were asked for the smallest number N that fits the condition we should let [tex]a_n = 1[/tex]. Therefore x = 1/7, and N is just the smallest repeating part of x, which is [tex]142857[/tex].

Notice that we could also have let [tex]a_n = 2[/tex], and the repeating part of 2/7 would have fit the condition. But the question asked for the smallest such number. Do you see why we couldn't let [tex]a_n = 3[/tex] though?

Once you get it, try this method out on the other problems.


Prove that 1/(1 + x + y) + 1/(1 + y + z) + 1/(1 + z + x) <= 1 for all positive reals x, y, z with product 1.

Put x=a^3, y=b^3 and z=c^3. Then 1+x+y=1+a^3+b^3>=1+ab(a+b)=(a+b+c)/c. Take the inverse and sum up.


geometry
geometric transformation
homothety
reflection
cyclic quadrilateral
power of a point
radical axis

Construct a tangency $ l$ of $ (O)$ and $ (O_1)$ then $ \angle GHD\equal{}\angle (l,DG)\equal{}\angle CED$ thus $ GH//CE$
Similarly, $ FI//DC$
But $ CG.CD\equal{}CF.CE$ then $ DGFE$ is cyclic
Therefore $ \angle GHD\equal{}\angle CED\equal{}\angle CGF\equal{}\angle CFI$, which follows that $ G,I,H,F$ are concyclic.

Inversion with center $ C$ and power $ CB \cdot CA$ takes $ \Gamma_1$ and $ \Gamma_2$ into themselves and $ \Gamma$ into a common external tangent $ \gamma$ to $ \Gamma_1,\Gamma_2$ $ \Longrightarrow$ $ F,G$ are the contact points of $ \gamma$ with $ \Gamma_1,\Gamma_2.$ Since $ \Gamma$ is internally/externally tangent to $ \Gamma_1$ and $ \Gamma_2$ (both cases hold), it follows that $ D,E$ are inverse points through a positive inversion $ (O,k^2)$ taking $ \Gamma_1$ into $\Gamma_2.$ Hence, $ H,I$ are inverse points and so are $ F,G$ obviously, thus $ OI \cdot OH \equal{} OG \cdot OF \equal{} k^2$ $ \Longrightarrow$ $ F,G,H,I$ are concyclic.

Dear Mathlinkers,
I can not resist to applied my favourite theorem..
1. Let (0) be the initial circle, (1) the one tangent at D, (2) the one tangent at E.
2. Let (3) the circle passing through D, G, E.
3. According to Monge or d'Alembert's theorem, it goes through F.
4. By Reim's theorem applied to (0) and (1), IF//DC
5. By a converse of the pivot theorem applied to (0), (1) and (3), GF is tangent to (1) at G
6. By a converse of Reim's theorem applied to (1) with IF//DG, F, G, H, I are cocyclics.
Sincerely
Jean-Louis

[quote="Concyclicboy"]\item Two circles $ \Gamma_1$ and $ \Gamma_2$ intersect at points $ A$ and $ B$. Consider a circle $ \Gamma$ contained in $ \Gamma_1$ and $ \Gamma_2$, which is tangent to both of them at $ D$ and $ E$ respectively. Let $ C$ be one of the intersection points of line $ AB$ with $ \Gamma$, $ F$ be the intersection of line $ EC$ with $ \Gamma_2$ and $ G$ be the intersection of line $ DC$ with $ \Gamma_1$. Let $ H$ and $ I$ be the intersection points of line $ ED$ with $ \Gamma_1$ and $ \Gamma_2$ respectively. Prove that $ F$, $ G$, $ H$ and $ I$ are on the same circle.[/quote]
It is enough to prove that $ \angle GHI\equal{}\angle GFI$.

$ DGFE$ is is a cyclic quadrilateral, as $ CG\cdot CD\equal{}CA \cdot CB\equal{}CF\cdot CE.$ So, $ \angle CFG\equal{}\angle D$
We know that a homothety with center $ D$ maps $ \Gamma_1 \to \Gamma$. So, $ GH \parallel EC$, and for the same reason $ FI \parallel FD$.
We have, $ \angle GHI\equal{}\angle E$, and $ \angle GFI\equal{}\pi\minus{}\angle CFG\minus{}\angle IFE\equal{}\pi\minus{}\angle D\minus{}\angle C\equal{}\angle E$.
Q.E.D.

[quote="diegu"][quote="diegu"][quote="livetolove212"]Construct a tangency $ l$ of $ (O)$ and $ (O_1)$ then $ \angle GHD \equal{} \angle (l,DG) \equal{} \angle CED$ thus $ GH//CE$
Similarly, $ FI//DC$
But $ CG.CD \equal{} CF.CE$ then $ DGFE$ is cyclic
Therefore $ \angle GHD \equal{} \angle CED \equal{} \angle CGF \equal{} \angle CFI$, which follows that $ G,I,H,F$ are concyclic.[/quote]

why if $ CG.CD \equal{} CF.CE$ then $ DGFE$ is cyclic?
and on the picture where is the tangency $ l$?
and why $ \angle GHD \equal{} \angle (l,DG) \equal{} \angle CED$?

thanks[/quote]

please...I don't understand[/quote]
The cyclicity of $ DGFE$ is a direct consequence of the Power of a Point Theorem's converse...
$ \angle GHD \equal{} \angle (l,DG) \equal{} \angle CED$ is a direct consequence of the Half-Inscribed Angle Theorem...
But I think $ \Gamma$ is inside $ \Gamma_1$ and $ \Gamma_2$, and not the opposite... Anyway, the difficulty is the same.
Here's my solution:

Let $ O$, $ O_1$ and $ O_2$ be the centers of $ \Gamma$, $ \Gamma_1$ and $ \Gamma_2$.
From the tangencies between the circles, we get the colinearities $ D \minus{} O \minus{} O_1$ and $ E \minus{} O \minus{} O_2$.
Since $ C$ lies on the radical axis of $ \Gamma_1$ and $ \Gamma_2$, the points $ D,E,F,G$ are concyclic.
With some angle chasing, it's easy to see it suffices to prove $ \angle DGH \equal{} \angle EFI$.
If we have the colinearity $ O_1 \minus{} E \minus{} O \minus{} D \minus{} O_2$, it's trivial, since $ \angle DGH \equal{} \angle EFI \equal{} 90^{\circ}$.
Otherwise it suffices to prove $ \angle DO_1H \equal{} \angle EO_2I$.
Let $ D'$ be the reflection of $ D$ w.r.t. $ O$.
Let $ D''$ be the reflection of $ D$ w.r.t. $ O_1$.
It's easy to see $ [DD'E]$ and $ [DD''H]$ are similar, so we have $ \frac {DE}{DH} \equal{} \frac {DD'}{DD''} \equal{} \frac {DO}{DO_1}$, therefore $ [DOE]$ and $ [DO_1H]$ are similar, so $ \angle DOE \equal{} \angle DO_1H$.
Analogously we prove $ \angle DOE \equal{} \angle EO_2I$, so $ \angle DO_1H \equal{} \angle EO_2I$, QED.


$ A\equal{}\begin{pmatrix}
    4&\minus{}3&\minus{}2\\
    2&\minus{}1&\minus{}2\\
    3&\minus{}3&\minus{}1
    \end{pmatrix}$
$ A^{2007}\equal{}?$

[quote="the_wonder"]$ A \equal{} \begin{pmatrix} 4 & \minus{} 3 & \minus{} 2 \\
2 & \minus{} 1 & \minus{} 2 \\
3 & \minus{} 3 & \minus{} 1 \end{pmatrix}$
$ A^{2007} \equal{} ?$[/quote]

The general method to do this is to find an invertible matrix $ B$ such that $ B^{ \minus{} 1}AB$ is a diagonal matrix $ D$

Now $ D \equal{} B^{ \minus{} 1}AB \Rightarrow D^{n} \equal{} (B^{ \minus{} 1}AB)^n \equal{} B^{ \minus{} 1}A^n B \Rightarrow A^n \equal{} BD^{n}B^{ \minus{} 1}$
This method is useful because raising a diagonal matrix to some power is a simple task.

So the only question that remains now is, how to diagonalize the matrix. Look [url=http://en.wikipedia.org/wiki/Diagonalizable_matrix]here[/url] for a general description and an example on how to do it.

Just diagonalize. (Note: in diagonalizing, I never worked out the full characteristic polynomial. I noticed that the rows all add to $ \minus{}1,$ so that $ \minus{}1$ is an eigenvalue. Together with the trace of $ 2$ and the determinant of $ \minus{}2,$ that's enough to get me the other two eigenvalues.)

$ A\equal{}\begin{bmatrix}1&1&1\\0&1&1\\1&0&1\end{bmatrix} \begin{bmatrix}2&0&0\\0&1&0\\0&0&\minus{}1\end{bmatrix} \begin{bmatrix}1&\minus{}1&0\\1&0&\minus{}1\\\minus{}1&1&1\end{bmatrix}$

$ A^{2007}\equal{} \begin{bmatrix}1&1&1\\0&1&1\\1&0&1\end{bmatrix} \begin{bmatrix}2^{2007}&0&0\\0&1&0\\0&0&\minus{}1\end{bmatrix} \begin{bmatrix}1&\minus{}1&0\\1&0&\minus{}1\\\minus{}1&1&1\end{bmatrix}$

$ \equal{}\begin{bmatrix}2^{2007}\plus{}2&\minus{}2^{2007}\minus{}1&\minus{}2\\2&\minus{}1&\minus{}2\\ 2^{2007}\plus{}1&\minus{}2^{2007}\minus{}1&\minus{}1\end{bmatrix}$

(Don't trust my matrix multiplication - check it for yourself.)


[quote="panos_lo"]Έστω ρ πρώτος χωρίς πολ/σια της μορφής (ν^2)-5. Ν.δ.ο. για κάθε ακέραιο α υπάρχουν άπειροι ακέραιοι μ ώστε ο (μ^5)+α να είναι διαιρετός διά ρ.[/quote]

re pano me ton tropo pou mou edeikses xtes, amfivalo pos tha tin lusei kaneis edw mesa ektos apo sena :P

Λοιπόν θα γράψω λίγο γρήγορα τη λύση μου γιατί είμαι στο σχολείο...Αν και νομίζω είναι λίγο ανορθόδοξη δεδομένου του ότι είναι για JBMO...Ίσως να υπάρχει και πιο απλή λύση δηλαδή...
Καταρχάς αρκεί να δείξουμε ότι υπάρχει τουλάχιστον ένας τέτοιος $ m$ γιατί τότε προφανώς (παίρνοντας τη σχέση $ modp$ ή απλά λόγω διωνυμικού) κάθε αριθμός της μορφής $ m\plus{}pk$ όπου $ k \in Z$ ικανοποιεί τα ζητούμενα.
Έχουμε από τη δεδομένη ότι η $ n^2 \equiv 5modp$ δεν έχει λύση, άρα $ (\frac{5}{p}) \equal{}\minus{}1$(όπου εννοούμε το σύμβολο του Legendre) όμως από τετραγωνική αντιστροφή προκύπτει και ότι $ (\frac{p}{5}) \equal{}\minus{}1$ άρα $ p \equiv 2mod5$ ή $ p \equiv 3mod5$
Τώρα ας υποθέσουμε ότι $ \minus{}a \equiv g^ymodp$ όπου (Σιλ μη με κράξεις :P) $ g$ γεννήτορας $ modp$ και έστω $ m \equiv g^xmodp$. Αρκεί λοιπόν η $ g^{5x} \equiv g^y modp$ να έχει πάντα λύση. Αυτό όμως ισοδυναμεί ότι η $ 5x \equiv y mod(p\minus{}1)$ να έχει λύση ως προς $ x$. Όμως λόγω των παραπάνω $ (5,p\minus{}1)\equal{}1$ άρα υπάρχει ο αντίστροφος του $ 5$ ως προς $ p\minus{}1$ και άρα έχουμε πάντα λύση, άρα τουλάχιστον ένα $ m$, άρα άπειρα τέτοια $ m$.

Ηλίας

Πολύ ωραία η άσκηση .
Αρκεί να βρούμε έναν $ m$ καθώς αν βρούμε έναν τέτοιο τότε $ (ps \plus{} m)^5 \plus{} a\equiv m^5 \plus{} a\equiv 0(\mod p)$  , οπότε με τον τρόπο αυτό βρίσκουμε άπειρους

Και για να βρούμε έναν αρκεί να τον αναζητήσουμε στην residue class mod p .
Θα δείξω ότι τα $ {0^5,\ 1^5,\ ..., \ (p \minus{} 1)^5}$ σχηματίζουν πλήρες σύστημα υπολοίπων mod p . Πράγματι έστω
ότι $ x^5\equiv y^5(\mod p)$ και έστω ότι $ x\neq y(\mod p)$ . Τότε $ x^4 \plus{} x^3y \plus{} x^2y^2 \plus{} yx^3 \plus{} y^4\equiv 0(\mod p)$ ή

$ (2(x^2 \plus{} y^2) \plus{} xy)^2 \minus{} 5x^2y^2\equiv 0(\mod p)$ , το οποίο εύκολα δίνει ότι το 5 είναι τετραγωνικό κατάλοιπο mod p  που είναι άτοπο .

Επομένως τα $ {0^5,\ 1^5,\ ..., \ (p \minus{} 1)^5}$ σχηματίζουν πλήρες σύστημα υπολοίπων mod p και επομένως τα
$ {0^5 \plus{} a,\ 1^5 \plus{} a,\ ..., \ (p \minus{} 1)^5 \plus{} a}$ σχηματίζουν πλήρες σύστημα υπολοίπων mod p  ΄και άρα ένα από αυτά διαιρείται με $ p$ .  :)

@panos : Διάβασε τις οδηγίες πάνω πάνω για να γράφεις σε latex . Είναι πολύ εύκολο . Απλά βάζεις ένα δολλάριο μπρος και πίσω από τη μαθηματική έκφραση
@ilia : ATIME ! Δεν τα γράφουμε αυτά ρεεεε . Δεν είδες πως το κουκουλωσα εγώ  :P

[quote="silouan"]

$ (2(x^2 \plus{} y^2) \plus{} xy)^2 \minus{} 5x^2y^2\equiv 0(\mod p)$ , το οποίο εύκολα δίνει ότι το 5 είναι τετραγωνικό κατάλοιπο mod p  που είναι άτοπο .

[/quote]

Ωραια ασκηση και ωραια λυση. Αν καταλαβα καλα εδω (επηδη δεν εχω διαβασει τετραγωνικα καταλοιπα ακομα), δειχνεις οτι το $ x^2y^2\neq 0 (modp)$ και πολλαπλασιαζεις με τον αντιστροφο $ q$ του $ x^2y^2(modp)$ και βγαζεις $ q^2x^2y^2(2(x^2 \plus{} y^2) \plus{} xy)^2 \minus{} 5\equiv 0(\mod p)$ το οποιο ειναι ατοπο απο την υποθεση?

Ναι ακριβώς αυτό !  :)


If you are interested post here. If there is enough interest, i will post the test.

I don't see the point when Mathcounts is already over.

yeah i have that too
~tinytim its for other people participating next year if they want.

Sure. I'll try. (I'm done w/ mathcounts but still interested)

[quote="lotsofmath"]Me!

(must not tie for 2nd like last contest...must tie for 1st  :rotfl: )[/quote]
ya, but this time I'm in the contest. First will be mine!
So ya sign me up.

ps. I really must take the credit for starting all these mock contests... lol

I will.

anyways jack, plenty of ppl have made mock contests b4 u.

ya I no, what I meant was the fact that a lot of them just popped up out of no-where. I was reffering to the present, rather the reffering to what has happened a while ago.

this is the sprint round. there may be some errors. post them here.

I can't read #24?

Is it $ \sqrt[3]{\sqrt{3^3 \cdot 9^9 \cdot 27^27}}$?

In that case I get $ 129140163$

You didn't give the side length of the equilateral triangle in problem 28. Do you want us to express in terms of $ s$?

problem 24 says $ \sqrt [3]{\sqrt [7]{3^3 \cdot 9^9 \cdot 27^{27}}}$

um.... can i still sign up??? Im sorry im late...

you don't need to sign up. Just do the problems and pm the answers to abacadaea.

triangle side length is 6.

yes you can still sign up

24 should be $ ^3\sqrt ^{17}{\sqrt{3^3*9^9*27^{27}}}$

[quote="BOGTRO"]@tinytim Yes, I caught that but I think it's one of those stupid

MC tricks...

@dynamo729: I think so.[/quote]

what makes you think so???

[quote="dynamo729"]I think that was supposed to be a trick. You may want to consider editing that.....

23.) is pair (a,b) different from (b,a)   ????[/quote]

The way the question is worded implies that (a,b) is different from (b,a). In other words (2,15) and (15,2) are different because it never said a>b.

[quote="BOGTRO"]There was one done rather recently.

Anyway, I'm interested.

Can you make a target round? The last one I did

didn't have a Target round  :([/quote]

Lol, BOGTRO, I did have a target round, but I guess you didn't notice it in your PM.

Btw, here was mine that was recently made: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=203522

I'll try this one as well.

[quote="dynamo729"]28.) There are no side lengths given in the problem!!!![/quote]

abacadaea said the side length was 6 earlier.

[quote="tinytim"]18) How many ways can you arrange 1 blue bead, 2 red bead, and 3 red beads in a line

you mean 5 red beads?[/quote]
i meant 2 white beads

CLARIFICATIONS!!!
#11-inclusive
#18-should read 1 blue bead, 2 white beads, 3 red.
#24 should be $ ^3\sqrt {^{17}\sqrt {3^3*9^9*27^{27}}}$
#28-the triangles side lenght is 6

my computer is being retarded and i cannot view #7. could someone post it please? i have an old answer key in front of me and i have changed some problems since then

i will admit this was a pretty poorly written test.

My computer won't download your test... :(

on mumber 29, on the left ( on the rectangle) there is like a little upside down 7 is it supposed to be 7?

erm...the height of the rectange is 4, the distance between the lower base and the x axis is 4. im not sure what you are saying

Is there a target round?

Also, can i get my score?

i will post the target round tomrow.

i will send you your score in a second


geometry
Euler
Gauss
projective geometry
geometry proposed

[quote="Leonhard Euler"]Let  $ \triangle{ABC}$ be a triangle and  $ I$ its incircle,that touch $ BC,CA,AB$ at $ D,E,F$.Denote $ F,G$ as  Feuerbach,Gergonne point of $ \triangle{ABC}$.Define the intersection $ X = AB\cap{DE}, Y = CA\cap{DF}, P = EF\cap{IG}, T = AP\cap{BC}$. Let $ M$ be a midpoint of $ XY$. prove that
a)$ TF$ is tangent to incircle of $ \triangle{ABC}$
b)$ P,A,M$ is collinear[/quote]

There is two $ F$. So I edited it and post here
Let  $ \triangle{ABC}$ be a triangle and  $ I$ its incircle,that touch $ BC,CA,AB$ at $ K,L,M$.Denote $ F,G$ as  Feuerbach,Gergonne point of $ \triangle{ABC}$.Define the intersection $ X = AB\cap{KL}, Y = CA\cap{KM}, P = IG\cap{LM}, T = AP\cap{BC}$. Let $ H$ be a midpoint of $ XY$. prove that
a)$ TF$ is tangent to incircle of $ \triangle{ABC}$
b)$ P,A,H$ is collinear

Dear Leonhard Euler
Your problem is very interesting and pretty difficult, could you post your solution?

[b]b)[/b]
[color=blue]lemma[/color]: http://www.mathlinks.ro/viewtopic.php?t=162991,
 $ IA$ bisect $ \angle{XAY}$ and $ M.L$ is projection of $ I$ on $ AX,AY$ and $ H$ is midpoint of $ XY$.so by lemma, we conclude that   $ ML,AH$ and line that through $ I$ and

perpendicular to $ XY$ are concur. Thereofore, It is suffice to show $ IG$ is perpendicular to $ XY$.  It is easy to see that $ XY$ is Pascal line of $ KKLLMM$ and $ G$ is Brianchon

point of $ KKLLMM$. Therefore $ IG$ is perpendicular to $ XY$

[b]a)[/b]
[color=blue]lemma[/color]: Let $ A,B,C,D$ are point on circle. Denote  $ X$ as  intersection of tangent line that pass $ A,B$,$ Y$ as intersection of $ AC,BD$ and $ Z$ as intersection of tangent line

that pass $ D,C$. Then $ X,Y,Z$ is collinear. This can be prove by Pascal theorem.

Let $ HAP$ meet $ MK$ at $ Q$. By Menelaus theorem in $ \triangle MXY$ and line $ QAH$, $ \frac{QM}{YQ}\frac{AX}{MA}\frac{HY}{XH}\equal{}\minus{}1$. Since $ HY \equal{} XH$,

$ \frac{QM}{YQ}\equal{}\minus{}\frac{MA}{AX}$ and since $ (BMAX)$ forms harmonic division, $ \frac{MA}{AX}\equal{}\minus{}\frac{BM}{XB}$. So $ \frac{QM}{YQ}\equal{}\frac{BM}{XB}$

which imply $ BQ$ is parallel to $ XY$. Let $ R \equal{} BQ\cap CA$.Since $ BR$ is parallel to $ XY$, midpoint of $ XY,A,BR$ are collinear. Since $ H$ is midpoint of $ XY$ and $ Q$ is

intersection of $ AH,BR$, we conclude $ Q$ is midpoint of $ BR$.

Let $ U,V$ be midpoint of $ BC,BA$. Since $ Q$ is midpoint of $ BR, Q$ lie on $ UV$. and since $ Q$ lie on $ MK$, $ Q$ is intersection of $ MK$ and $ UV$. Therefore by Fontene

theorem, $ Q$ lie on $ LF$. Let $ W$ be the intersection $ BC$ and line through $ F$ and tangent to incircle of $ \triangle ABC$. By lemma, $ W,Q,A$ are collinear. But $ Q,P,A$ is

collinear. Therefore $ W,P,A$ is collinear. So proven.

[quote="Leonhard Euler"]

[b]a)[/b]
[color=blue]lemma[/color]: Let $ A,B,C,D$ are point on circle. Denote  $ X$ as  intersection of tangent line that pass $ A,B$,$ Y$ as intersection of $ AC,BD$ and $ Z$ as intersection of tangent line

that pass $ D,C$. Then $ X,Y,Z$ is collinear. This can be prove by Pascal theorem.

Let $ HAP$ meet $ MK$ at $ Q$. By Menelaus theorem in $ \triangle MXY$ and line $ QAH$, $ \frac {QM}{YQ}\frac {AX}{MA}\frac {HY}{XH} \equal{} \minus{} 1$. Since $ HY \equal{} XH$,

$ \frac {QM}{YQ} \equal{} \minus{} \frac {MA}{AX}$ and since $ (BMAX)$ forms harmonic division, $ \frac {MA}{AX} \equal{} \minus{} \frac {BM}{XB}$. So $ \frac {QM}{YQ} \equal{} \frac {BM}{XB}$

which imply $ BQ$ is parallel to $ XY$. Let $ R \equal{} BQ\cap CA$.Since $ BR$ is parallel to $ XY$, midpoint of $ XY,A,BR$ are collinear. Since $ H$ is midpoint of $ XY$ and $ Q$ is

intersection of $ AH,BR$, we conclude $ Q$ is midpoint of $ BR$.

Let $ U,V$ be midpoint of $ BC,BA$. Since $ Q$ is midpoint of $ BR, Q$ lie on $ UV$. and since $ Q$ lie on $ MK$, $ Q$ is intersection of $ MK$ and $ UV$. Therefore by Fontene

theorem, $ Q$ lie on $ LF$. Let $ W$ be the intersection $ BC$ and line through $ F$ and tangent to incircle of $ \triangle ABC$. By lemma, $ W,Q,A$ are collinear. But $ Q,P,A$ is

collinear. Therefore $ W,P,A$ is collinear. So proven.[/quote]
This is Soo Hong Lee's(Little Gauss) solution


$ a_0\equal{} \frac 12 ,a_n \equal{} a_{n\minus{}1} \plus{} \frac {a_{n\minus{}1}^2}{n^2}, \Longrightarrow \frac {n\plus{}1}{n\plus{}2} <a_n <1 .(n \geq 1)$


vector
combinatorics proposed
combinatorics

We have a n*n grid filled with vectors - up, down, left, right. If we place a beetle in any of the squares, and it follows the vectors around it is known that it will eventually come back to the original position, and the beetle duznt go off the grid. Further, the vectors are arranged so the sum of the vectors in any column is parellel to the column (with exception the first and last column) and the sum of the vectors in any row is parallel to the row (with exception the first and last row). Prove that there is no such arrangement.

It is wrong!figure that I've attached shows such an arrangement for $n=4$

Well, the stated directions for the vectors are only $N$(orth), $S$(outh), $E$(ast) and $W$(est) - your model also has $NE, NW, SE, SW$ vectors.

However, the model
$E \ \ S$
$N \ \ W$
is OK for $n=2$.

So I didn't do the first principle every problem solver must consider:read the problem carefuly!and it costed me so much because I spent so much on this.
Well I think the problem is true for $n>2$.

for a set $A$ of squares we mean by $E(A),W(A),S(A),N(A)$ number of $E,W,S,N$s in its squares.
And By $R(i),C(i)$ We mean row(column) $i$.

By problems assumption the unit squares of our main square will be decomposed into cycles(this is not graph but I think you'll understand what I mean by cycles)

Now for every cycle $A$ and a number $1 \leq j <n $, It is not hard to see that $E(A\cap C(j))=W(A\cap C(j+1))$.summing on all $A$s we will deduce that $E(C(j))=W(C(j+1))$(*)
By the problem's assumption we have $E(C(j))=W(C(j))$ for $1<j<n$(**).
So by (*)&(**) there is a $k$ such that $E(C(1))=...=E(C(n-1))=W(C(2))=...=W(C(n))=k$

lets call set of squares of our main big square $B$.so $E(B)+W(B)=2k(n-1)$.By similar reasoning we deduce that $S(B)+N(B)=2k'(n-1)$ and so $n^{2}=2(k+k')(n-1)$ and so $n-1|n^{2}$ so $n-1|1$ so $n=2$.


algebra
polynomial
number theory proposed
number theory

Find all pairs of integers $x,y$ such that  $x^3+x^2y+xy^2+y^3=8(x^2+xy+y^2+1)$

Abdu i will explain you this later ...  :)

hallo,
i have an idea , but i don't really know if its right :)
what about, using symetric polynominals ?
and then perform the polynom division ?

hallo,
nice idea Davron. my idea was : (i hope its correct   :) )
to write the equation
$(x+y)^3=8((x+y)^2-xy+1)+2xy(x+y)$
now let $l=x+y$ and $k=xy$ we have
$l^3=8(l^2-k+1)+2kl$
$k=\frac{l^3-8l^2-8}{2l-8}$
from here $l$ is eaven so $l=2n$ and we have
$k=\frac{2n^3-8n^2-2}{n-2}$
By polynominal division we get:
$k=2n^2-4n-8-\frac{18}{n-2}$
I thik from here it is easy to see the solutions.


[hide="My solution, true or false"]
* It's easy to show 1505 vertices that there arent 4 vertices forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
* Now we will prove that it's OK for every choice of 1506 vertices. WLOG Let $A_{2007}$ is a vertex we don't choose. Divide other 2004 vertices  in to 501 sets
$X_{1}=\{A_{1},A_{2},A_{3},A_{4}\}, X_{2}=\{A_{5},A_{6},A_{7},A_{8}\}$
$.... X_{501}=\{A_{2001},A_{2002},A_{2003},A_{2004}\}$
with $A_{1},A_{2},...A_{2007}$ are the vertices of the polygon.
By Dirichlet theorem, there must be one set contains 4 chosen vertices. These are 4 vertices we need to find[/hide]

[quote="hien"][quote="hikaru123"][hide=My solution, true or false]
* It's easy to show 1505 vertices that there arent 4 vertices forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon. [/quote]

I think it's wrong because of if we choose consecutive three vertices, leave out next one following, then this event don't satify![/quote]
So that is the case I said ' easy to show'. what's wrong here?

[quote]So that is the case I said ' easy to show'. what's wrong here?[/quote]

How can you show that?

[/quote]

I think you are wrong, silver. The example I gave, can you see? There are 1505 vertices that are not satisfied, so the answer must be 1506.
@silver: did you take part in VMO?

oh!I'm wrong.1506 is true.
To hikaru:I don't take part in VMO
[/code]

[quote=hien]Given a regular 2007-gon. Find the minimal number $k$ such that: Among every $k$ vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.[/quote]

Sorry for the \bump,but why is the answer $1506$ not $1340?$

[quote=laikhanhhoang_3011][quote=Sprites][quote=hien]Given a regular 2007-gon. Find the minimal number $k$ such that: Among every $k$ vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.[/quote]

Sorry for the \bump,but why is the answer $1506$ not $1004?$[/quote]

are you Vietnamese ?[/quote]

Nah
Got it I will write a solution just in case:-
[hide=Storage]Let the vertices of the polygon be $A_1,A_2,........,A_{2007}$
We want any set of $k$ vertices to contain atleast $4$ consecutive ones and by the stronger form of pigeon hole principle we must need $4k>3 \cdot 2007 \implies k \ge 1506$ and showing $1506$ works is easy.[/hide]


http://www.urbandictionary.com/define.php?term=nope

Slang for "no".

[quote="K81o7"]... Are you kidding?[/quote]

Nope. He is not kidding.

[quote="conceive"]
Nope. He is not kidding.[/quote]

I was actually wondering if [i]you[/i] were kidding...

[quote="K81o7"][quote="conceive"]
Nope. He is not kidding.[/quote]

@K81o7: I was actually wondering if [i]you[/i] were kidding...[/quote]
Me? Kidding? Why? I am from Czech, not english speaking country.. you know I do not meet english speaking people every day..

Yeah, I figured...
But your English is pretty good, so I figured you'd probably know what "nope" meant.  Also because it was in the Games and Fun Factory, I thought you might be referring to something else.  I guess I just never thought of "nope" as slang, because it's used so much.

To me, it seemed to equivalent of asking what the function of "nai" is in Japanese (I study Japanese, so I know that much).  But if it is slang, I suppose it isn't so well known to most non-(English first language) people.

[quote="bubka"][quote="k81o7"]But if it is slang, I suppose it isn't so well known to most non-(English first language) people.[/quote]

What do you mean? My first language is Bengali (which should be yours as well). And I know what 'nope' means.

Okay, I guess at school English is my first language, but that is only theoretically. I am much more comfortable speaking Bengali, which has a beauty unmatched by English.

@conceive:
And how does 'nope' matter anyway? Why do you give that much importance to these things?

Finally, before the topic is locked, can somebody tell me what 'anyways' means? I am seeing it a lot thse days.[/quote]

Are you trying to blame me?  How could Bengali be my first language if I was never taught it and nobody I knew generally spoke it?  Notice that I live in the US...
If I had a choice, I would know Bengali, my mother's native language, by now (and would also know some Tamil, my father's native language).  I try to pick up and learn bits and pieces whenever I go to India, but it's not as easy as you might think...
Consider yourself lucky that you can understand Bengali.

Well clearly, you have used english in communcation more than Conceive has, and have picked up that slang.  What I meant was people who don't use any English in communication on a daily basis.

Don't try to teach me about languages.  I have heard enough Bengali (and Sanskrit, for that matter) to know that it is a softer language than English.

And don't reprimand Conceive either.  He probably heard the word a lot on these forums (likely) and wondered exatly what it meant.  This topic holds as much importance as any other in the G&FF: now it might become a discussion about what we actually mean by different english slang, so why should it be locked?

Anyways: as far as I can tell, it doesn't have meaning unless in conversation, but it is a word used to signify a change in topic or context, basically clearing the slate of what was said just before in the conversation, or summing up.  It's more of an interjection.

Nope means "no" It originated in 1913, when there were word contests, and one guy said the wrong word. The modrator said "nope" and everyone was confused. He explained.

[quote="K81o7"]But if it is slang, I suppose it isn't so well known to most non-(English first language) people.[/quote]
As you pointed out, the major reason is that most Japanese students have no chance to communicate in English. Sadly, teachers have to put an emphasis on reading, writing and grammer because Japanese college entrance exams are largely based on these three elements.

In fact, I had no idea what "nope" meant when I first saw that word, although I knew some of the difficult words in the SAT vocab book. (I learned English in Japan for about 3 years and came to the U.S. 3 years ago.) I believe that even if you learn English for 6 years in Japanese middle school/high school, you are not likely to learn what "yup" or "nope" means :(

When I offered a draw, my Fritz 6 (and later Fritz 7) chess engine often called out:
"NO! NEIN! NOT! NJET!" all four of which means the same.

But my point is what does it matter whether you are not likely to learn what 'yup' or 'nope' means? It is the same with most third world countries. If you frequent such sites as this one, you may be helped a bit by the knowledge of english slang, that's all. You should also consider the fact that a person in Europe or America trying to learn Japanese misses most of its finer points. Be proud of your own language! No country ever progresses by dumping its mother language for a foreign one.

And as for Krishanu, his mom must certainly be of the type which feels nothing for the language Bengali, otherwise she would make an effort to teach him Bengali. He himself cannot be blamed though ...

bubka that's not true.

Many parents worry over whether their child will speech problems due to 2 languages. My parents did not agree with that theory and hence I know their language and English

whoa...i thought nope was a type of Asian bear :o

[quote="PenguinIntegral"][quote="Treething"][quote="bubka"]No country ever progresses by dumping its mother language for a foreign one.
[/quote]

except for South Korea[/quote]
And the ghetto.

*PenguinIntegral high-fives Daesun*[/quote]

Ghettoes are in Europe? Or do you mean places like [url]http://en.wikipedia.org/wiki/New_Haven%2C_CT#Demographics[/url]

Actually if you live in England or America, it's easy to know these terms like nope yep, wuz up, uh-huh,


linear algebra
matrix
superior algebra
superior algebra unsolved

Let the set ${M=\{(\begin{array}{cc} a&b \\ c&d \end{array}) | a,b,c,d\in\mathbb{Z}, \ a+b=c+d}\}$. Find the inversible elements of the monoid $(M,\cdot)$.

If the matrix $\left( \begin{array}{cc} a &b\\ c& d\\ \end{array} \right)\in M$ is invertible then so is the matrix

$\left(\begin{array}{cc} c&d\\ a&b \end{array}\right)\in M$
so we can suppose $a>c$.
Denote by $k$ the sum $k: =a+b=c+d$.
Then we have to find the matrices of the form
$\left( \begin{array}{cc} a & k-a\\ c & k-c\\ \end{array} \right)$ with the determinant $\pm1$. That means $ak-ac-kc+ac=k(a-c)=\pm1$. But because $a>c$ we know that $c=a-1$.
For $k=1$ we get the matrices
$\left( \begin{array}{cc} a & 1-a\\ a-1& 2-a\\ \end{array} \right)$ with $a\in\mathbb Z$.
and for $k=-1$
$\left( \begin{array}{cc} a & -a-1\\ a-1 & -a\\ \end{array} \right)$ with $a\in\mathbb Z$.


look on the South Carolina Forum

Here's a link for the lazy people:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83953&sid=81af9e8288369cd5910e69acdb3ef56f


Let $\omega$ be a circle with center $O.$ Convex quadrilateral $AEDB$ inscribed in $\omega$ with segment $AB$ as a diameter of $\omega.$ Rays $ED$ and $AB$ meet at $C\,(CA>CB).$ Let $\omega_{1}$ denote the circumcircle of triangle $OBD,$ and let segment $OF$ be a diameter of $\omega_{1}.$ Ray $CF$ meet $\omega_{1}$ at $G.$ Prove that $A,\,O,\,G,$ and $E$ lie on a circle.

Dear April,

this is a very nice problem! Here's my solution:

Call $S$ the intersection point of $BD$ and $AE$.
We have that $FB \bot OB$ so $FB,FD$ are tangents of $\omega$ (Because $\left|FB\right| = \left|FD\right|$)

$S$ lies on the polar of $F$ wrt $\omega$, so it follows that $CF$ is the polar of $S$ wrt $\omega$.
Because $OG \bot CF$, we now know that $S$ lies on $OG$.

Now $\angle SED = \angle DBA = \angle DGS$ so $SDGE$ is cyclic.

Now we can finish our proof, because $\angle GOB = \angle GDS = \angle GEA$

[img]http://img517.imageshack.us/img517/9963/aprilmathlinkscyclicpolst8.jpg[/img]

[quote="April"][color=darkred]Let $AEDB$ be a convex quadrilateral which is inscribed in the circle $w=C(O)$ with the diameter $[AB]$.
Denote $C\in AB\cap ED$, where $CA>CB.$  Let $w_{1}$ be the circumcircle of the triangle $OBD$ and let $[OF]$ be a diameter of it.
Denote the second intersection $G$ between the circle $w_{1}$ and the line $CF$. Prove that the points $A,\ O,\ G,\ E$ are concyclically.[/color][/quote]
[color=darkblue][b]Very nice problem, April ! Thank you for it.[/b] Here is a proof with the pol-polar theory.

[b][u]Proof.[/u][/b] Denote the intersection $L\in AE\cap BD$ and the polar $x$ of the point $X$ w.r.t. the circle $\omega$.

Prove easily that the point $L\in c\cap f$. From the [b]Hire's theorem[/b] obtain $l\equiv CF$.

Thus, $\{\begin{array}{c}OL\perp CF\\\ OG\perp CF\end{array}$. Therefore, $G\in OL$, i.e. [b][u]the lines $AE$, $BD$, $OG$ are concurrently[/u][/b] (it is essentially !).

From the relations $\{\begin{array}{c}LE\cdot LA=LD\cdot LB\\\ LD\cdot LB=LG\cdot LO\end{array}$ obtain $LE\cdot LA=LG\cdot LO$, i.e. the quadrilateral $AOGE$ is cyclically.

[b]Remark.[/b] The orthocenter of the triangle $ABL$ belongs to the line $\overline{CFG}$.[/color]

[quote][color=darkred][b]Extension.[/b] The convex quadrilateral $AEDB$ is inscribed in the circle $w=C(O)$. Denote $C\in AB\cap ED$,
where $CA>CB$. Let $w_{1}$ be the circumcircle of triangle $OBD$ and let $[OF]$ be a diameter of it.
Denote the second intersection $G$ between the circle $w_{1}$ and the line $CF$. Prove that the points $A,\,O,\,G,\ E$ are concyclically.

[b]Remark.[/b] The lines $AD$, $BE$, $\overline{CFG}$ are concurrently.[/color][/quote]

Please April, what means MOSP ?

[b][color=blue]Thank you for your nice solutions, Virgil Nicula and Jan!  :lol:
MOSP means[/color] [color=red]Mathematical Olympiad Summer Program[/color]
[color=blue]And you can find more informations at here: http://www.unl.edu/amc/ [/color][/b]

I guess I have pretty much the same proof...drawing $BD\cap OG\cap AE$ is very natural for this problem...[working backwards $H$ is the radical center of $\omega$, $\omega_{1}$ and the circumcircle of $AOGE$...we just want to show that they are concurrent.]. I just want to make the logic a bit more clear:

Let $H=AE\cap BD$. Since $DB$ is the polar of $F$ wrt $\omega$ [$\angle ODF=\angle OBF=90^\circ$], $F$ lies on the polar of $H$, $h$, wrt $\omega$ by l'Hire's theorem. We know that $h\perp OH$ and $F\in h$. Since $OG\perp GF$, $G$ is the point where $h$ intersects $OH$. Hence $O$, $G$, and $H$ are collinear.

Let the circumcircle of $\triangle AOE$ intersect $w_{1}$ at $G'$.  We know that $AE$, $OG'$, and $DB$ are concurrent because they are the radical axes of $w$, $w_{1}$, and $C(AEO)$. i.e. $OG'H$ is a line. Hence $G$ and $G'$ are both points that lie on $OH$ and $w_{1}$ [distinct from $O$], hence $G\equiv G'$. By definition, $AOG'E\equiv AOGE$ is cyclic, as desired.

Comments:
We have that $CFG$ is the polar of $H$ [although this is not necessary]. We only need that $O$, $G$, and $H$ are collinear.


algebra
polynomial
vector
superior algebra
superior algebra unsolved

Let $f(x)\in\mathbb{Q}[x]$. Show that there is $g(x)\in\mathbb{Q}[x]$ so that $f(x)g(x)=a_{1}x^{2}+a_{2}x^{3}+a_{3}x^{5}+\cdots a_{n}x^{p_{n}}$ where $p_{n}$ is the $n$-th prime.

i guess $g(x)=x^{2}$,
but it would be too easy

That would not make the coefficients of all non-prime powers vanish, would it?

Of course your $g(x)$ depends on $f(x)$.

hi,
Let $R : = \mathbb{Q}[X] /f(X)$,now $R$ as a $Q$ vector space has dimension $n$ i.e degree of $f$,now consider the set ${( X^{p_{i}})}$ where $p_{i}$ is the $i$th prime,this set cannot be linearly independent in $R$ over $\mathbb{Q}$,this means there exist ${a_{i}}$ s.t $a_{i}$is zero except for finitely manny $i$ and $a_{1}X^{2}+.....$ belongs to the ideal $f\mathbb{Q}[X]$ and that gives the required result

i find the problem very ambigously stated and buddha's
proof beeing very sketchy.
either that or im pretty dumb  :blush:


Let N be a natural number that divides 171 with a remainder of 6. List all the two-digit numbers N can be.

[hide="solution"] If 171 divides it with remainder 6, then 165 must be divisible to N. The prime factorization of 165: 3, 5, 11. So N is $\boxed{33}$ or $\boxed{55}$. :D[/hide]

[quote="ch1n353ch3s54a1l"][hide]...What happened to 11 and 15?[/hide][/quote]

Oops.  :blush:

I hate saying this but vicseb, why do you post these 'lateral thinking' type of problems in Classroom Math? This is not how you use this forum... If you need help, please look at the guidelines before posting please. PLEASE. Thank you and have a good day.

[hide="do not click this if you do not want to see a solution"]$171-6=165$  Prime factor 165 $11*3*5$  So the two digit numbers are
11, 15, 33, and 55[/hide]

[hide]N divides 171-6=165.
We want all two-digit positive factors of 165.
165=3*5*11.
$(11,15,33,55)$.[/hide]


Jane is 25 years old.  Dick is older than Jane.  In $n$ years, where $n$ is a positive integer, Dick's age and Jane's age will both be two-digit number and will have the property that Jane's age is obtained by interchanging the digits of Dick's age.  Let $d$ be Dick's present age.  How many ordered pairs of positive integers $(d,n)$ are possible?

[hide="Answer"]We have $25+n=10a+b$, and $d+n=10b+a$. Subtracting them, we have $25-d=9a-9b=9(a-b)$. We know that $d>25$, so $b>a$. Also, $25-d$ is a multiple of $9$, so $d\equiv 7\bmod{9}$. Adding the equations, we have $25+d+2n=11a+11b=11(a+b)$, so $25+d+2n$ is a multiple of $11$. Also, $d+n<100$. From there, we find the possible starting pairs $(d,n)$ as $(34,9)$, $(43,10)$, $(52,11)$, $(61,1)$, $(70,2)$, $(79,3)$, and $(88,4)$, and from there we can add $11$ to $n$ as long as $d+n<100$. Therefore, our number of pairs is $6+5+4+4+3+2+1=\boxed{25}$.[/hide]

[b]EDIT:[/b] Fixed.

[hide="Solution"]
In $n$ years, Dick's age will be $d+n$. Jane's age will be $25+n$.
Let $D=d+n$, and $J=25+n$. Since $D$ and $N$ have two digits, let $D=10x+y$ and $J=10y+x$. From here we get

\begin{eqnarray*} 10x + y &=& d+n \\ x + 10y &=& 25+n \\ \end{eqnarray*}

Subtracting the second from the first, we get $9(x-y)=d-25$. $\mod 9$, we have $d \equiv 7 \pmod {9}$. Thus the possibilities for $d$ are 34, 43, 52, 61, 70, 79, 88, and 97.

If $d=34$, then $x-y=1$. The possibilities for $10x+y=34+n$ are 43, 54, 65, 76, 87, and 98. These yield $n=9, 20, 31, 42, 53$, or $64$, respectively. (6 possibilities)

If $d=43$, then $x-y=2$. The possibilities for $10x+y=43+n$ are 53, 64, 75, 86, and 97. These yield $n=10, 21, 32, 43,$ or $54$, respectively. (5 possibilities)

If $d=52$, then $x-y=3$. The possibilities for $10x+y=52+n$ are 63, 74, 85, and 96. These yield $n=11, 22, 33,$ or $44$, respectively. (4 possibilities)

If $d=61$, then $x-y=4$. The possibilities for $10x+y=61+n$ are 62, 73, 84, and 95. These yield $n=1, 12, 23,$ or $34$, respectively. (4 possibilities)

If $d=70$, then $x-y=5$. The possibilities for $10x+y=70+n$ are 72, 83, and 94. These yield $n=2, 13,$ or $24,$ respectively. (3 possibilities)

If $d=79$, then $x-y=6$. The possibilities for $10x+y=79+n$ are 82 and 93. These yield $n=3$ or $14$, respectively. (2 possibilities)

If $d=88$, then $x-y=7$. The only possibility for $10x+y=88+n$ is 92, and $n=4$. (1 possibility)

The sum is $6+5+4+4+3+2+1=\frac{6\cdot7}{2}+4=\boxed {25}$.
[/hide]

:huh:

If there are 25 pairs, can you please give all 25 of them and show that they all work?

I did show all of them...I'm not going to show that they all work, that would take a long time, but you can see Kalva's answer too: http://www.kalva.demon.co.uk/aime/soln/answ02.html

JesusFreak197 :

You forget that it can also be if d=34 ,

n=9,20,31,42,53,64

But not only n=9  :P

Other cases also the same ....

Oh, duh, I wasn't thinking... :oops: I'm not sure why I thought that I could add $11$ to some of those pairs and not to others... :huh:

Let $a > b$ be any two positive integers between 1 and 9. There exists an integer $n$ such that Jane's age + $n$ and Dick's age + $n$ are $\overline{ab}$ and $\overline{ba}$ in some order if and only if $\overline{ba} > 25$. That $\overline{ba} > 25$ is obvious; for the other direction, we note that we can set $n = \overline{ba} - 25$, whence Dick's age is $\overline{ab} - n = 25 + (\overline{ab} - \overline{ba}) > 25$.

We now count the number of integers $a,b$ with $0 \leq b < a \leq 9$ for which $\overline{ba} > 25$. Clearly, $b \geq 2$. If $b = 2$, then $a \in \{6,7,8,9\}$, so there are 4 pairs $(n,d)$ when $b = 2$. For $b > 2$, any pair $(a,b)$ with $3 \leq b < a \leq 9$ suffices, so there are $\binom{7}{2} = 21$ pairs $(n,d)$ in this case, giving an answer of $4+21 = \boxed{025}$.

Notice that if Jane's age + $n$ reversed is larger than Jane's original age, then we can choose $d$ to be Jane's age + $n$. This means that for every valid value for $d$ that we find, there exists an $n$ that can give us that value.

So, we basically have to find all two digit numbers greater than $25$ such that when reversed they are still greater than $25$. In the 20's we have 26...29, in the thirties we have 34...39, forties: 45...49, 56...59, 67...69, 78...79, and 89.

That gives $4+6+5+4+3+2+1=25$ possibilities.


geometry
incenter
inequalities
symmetry
circumcircle
geometry solved

Let ABC be an isosceles triangle so that BC=a, AB=AC=b. Take M \in (AC), N \in (AB) so that :
a 2 AM*AN=b 2BN*CM. Lines BM and CM intersect at P. Find the locus of P.

Let x=AN and y=AM. Then the given condition is met if a*x=b*(b-x) and a*y=b*(b-y). This equivalent to a/b=(b-x)/x=(b-y)/y or, BM and CN are angle bisectors of ABC. So the incentre I is one of points P possible.

Only a special case :-(

Eckard

It seems that I is the only point possible. Could it be? This would mean that there is some inequality which prevents other points. Let's have a closer look...

No, no, no. (I should not guess too much.)

The locus is an arc through B, C and I (excluding B and C). But why?

I finally remembered! I knew I had solved this before!

There's this thm: if AD and AE are isogonal cevians in the triangle ABC (meaning that angles BAD and EAC are equal) then (AD/DC)*(AE/EC)=AB^2/AC^2. The converse is also true. I forgot what it's called, but I can look it up in some books.

We rewrite the condition like this: a^2/b^2=(BN/AN)*(CM/AM). Now consider pt T on (AC) s.t. AT=AN. Then we will have (CT/AT)*(CM/AM)=a^2/b^2=BC^2/AC^2, so the cevians BT and BM are isogonal. This means that angle CBM= angle TBA and, for reasons of symmetry, angle TBA=angle NCA, so we get angle CBM=angle NCA= C-angle BCN, so angle BCP+angle CBP=C=B, so angle BPC=Pi-C=constant, so P is on the circumcircle of BIC, but P is in the interior of ABC, so the locus of P is the arc BC which contains I of the circumcircle of BIC. So Eckard was right.

Oh, yeah, now I remember. This problem was proposed by Dan Branzei, a tough-problem proposer (mainly geometry) and it was used for the second or third sel test (I can't remember exactly). I think Valentin participated in that one, didn't you Valentin?

[quote="Valentin Vornicu"]yeah I was the only guy in history to take the Seniors and the Juniors Tests (that year  I had the age for junior, but qualified for the Seniors too). That TST was really easy :)[/quote]

oh! you are great.
I can do this last year too. But those competition doesn't allow this.I 'm grade 10 last year, and i'm 13 last year. I can join a competition for grade7-grade9, but becoz of my grade, i can't join

yeah - at that moment exactly the selection tests for the JBMO were not in the same time as the TST for the IMO and I could take them both.


Ποιός είναι ο συντελεστής του $x^{23}$  του αναπτύγματος
$(1+x^5+x^9)^{100}$  ?

Πολύ ωραία και σωστά .Μπράβο Απ.....


calculus
integration
trigonometry
function
logarithms
real analysis
real analysis unsolved

Hello i'm studying antennas and need to solve integrals like

$\int{\frac{cos^{2}(a*cos(\theta))}{sin(\theta)}\;d(\theta)}$

where a is a constant

i know i must use fresnel sine and cosine integral ecuations because i have the solution but i don't know how to use them for this kind of integrals

this kind of integrals appear when you try to calculate the power radiated by a dipole and much others

thanks
Jose

Making substitution $a \cos x = v$

$I = \int \frac{\cos^{2}(a \cos x)}{\sin x}\cdot dx = a \int \frac{ \cos^{2}v}{v^{2}-a^{2}}\cdot dv = \frac{1}{2}\int \left( \frac{cos^{2}v}{v-a}-\frac{cos^{2}v}{v+a}\right) \cdot dv =$

$= \frac{1}{4}\int \left( \frac{1+\cos 2v}{v-a}-\frac{1+\cos 2v}{v+a}\right) \cdot dv = \frac{1}{4}\int \left( \frac{2a \cdot dv}{v^{2}-a^{2}}+\frac{\cos 2(v_{1}+a) \cdot dv_{1}}{v_{1}}-\frac{\cos 2(v_{2}-a) \cdot dv_{2}}{v_{2}}\right)$

with $v_{1}= v-a$ and $v_{2}= v+a$

The first integral is obvius and the last two ones can be integrated (expanding the trigonmetric functions) in terms of sine integral and cosine integral. Eventually

$I= C+\frac{1}{4}\ln \left| \frac{1-\cos x}{1+\cos x}\right|+$

$+\frac{\cos 2a}{4}\left[ \mathrm{Ci}(2a-2a \cos x)-\mathrm{Ci}(2a+2a \cos x) \right]+\frac{\sin 2a}{4}\left[ \mathrm{Si}(2a-2a \cos x)-\mathrm{Si}(2a+2a \cos x) \right]$

a lot of thanks greengrass it has been very useful

can you recomend me some document or book to study the use of sine and cosine integrals??

thanks again

Jose

Look here
Mira aquí

http://mathworld.wolfram.com/SineIntegral.html
http://mathworld.wolfram.com/CosineIntegral.html

http://foro.migui.com (spanish/español)


geometry
analytic geometry
graphing lines
slope
trapezoid

Let ABC be a right triangle with hypotenuse BC. Suppose that M is the midpoint of BC and H is the feet of the perpendicular dropped from A onto BC. A point P, distinct from A, is chosen on the opposite ray of ray AM. Let the line through H perpendicular to AB intersect PB at Q; and let the line through H perpendicular to AC meet PC at R. Prove that A is the orthocenter of triangle PQR.

I can see that you're new here, but please pay attention to the level of difficulty of the forums in which you post.  This problem is not appropriate for the Middle School Classroom Math forum.  I'm moving it.

[quote]A point P, distinct from A, is chosen on the opposite ray of ray AM.[/quote]
I don't understand this. What do you mean the opposite ray of ray AM? What is an opposite ray?

I think he means "choose $P$ on $AM$ such that $M$ lies on the segment $AP$".

My mother tongue is not English but I tried my best to translate the problem into English. I hope you can understand what I mean.

I think he means the point $A$ belongs to the segment $[PM]\ !.$

[hide="A easy analytical solution"] A analytical treatment of this proposed problem is recommendablely.
I will choose $A(0,0)$, $B(b,0)$, $C(c,0)$, $M\left(\frac b2,\frac c2\right)$,
$H\left( \frac{bc^2}{b^2+c^2}, \frac{b^2c}{b^2+c^2}\right)$, $P\left(d,\frac{cd}{b}\right)$, where $d<0<b<c$. We obtain:
$x_R=\frac{bc^2d}{(b^2+c^2)(b-d)}$, $y_R=\frac{b^2c}{b^2+c^2}$ $\Longrightarrow$ $m_{AR}=\frac{b}{cd(b-d)}.$
$x_Q=\frac{bc^2}{b^2+c^2}$, $y_Q=\frac{b^2cd}{(b^2+c^2)(b-d)}$, $\Longrightarrow$ $m_{AQ}=\frac{bd}{c(b-d)}.$
$m_{RQ}=\frac{y_Q-y_R}{x_Q-x_R}=-\frac bc$, $m_{PA}=m_{AM}=\frac cb$ $\Longrightarrow$ $AP\perp RQ.$
$m_{PQ}=m_{BQ}=\frac{y_Q-y_B}{x_Q-x_B}=\frac{cd}{b(d-b)}$ $\Longrightarrow$ $AR\perp PQ.$

Therefore, the point $A$ is the orthocentre of the triangle $PQR.$

[u]Remark.[/u] I note: the coordinates $x_P$ (abscissa), $y_P$ (ordinate) of the point $P$; the slope $m_d$ of the line $d$. I recall: that the slope of the line $d=AB$, where $AB\not \parallel Oy$ and $A(x_1,y_1)$, $B(x_2,y_2)$, is $m_d=\frac{y_1-y_2}{x_1-x_2}$; the lines $d_1$, $d_2$ are perpendicularly iff $m_1m_2=-1.$[/hide]

[hide="Here is a equivalent enunciation."] I like this problem and I am looking up a its syntetical solution.
" Let $ABCD$ be a orthodiagonal trapezoid ($AB\parallel CD$, $AC\perp BD$). I note the middlepoint $M$ of the side $[AB]$, the intersections $I\in AC\cap BD$, $J\in AD\cap BC$, the projection $S$ of the point $I$ on the line $AB$, the points $L\in AD$, $SL\perp AC$ and $K\in BC$, $SK\perp BD$. Then the point $A$ is the orthocentre of the triangle $JKL$."[/hide]