A radio station in central Pennsylvania is having a birthday contest.  One of the 366 days of the year is randomly selected with equal probability.  If the month of a contestant's birthday matches the month selected - or the day of the month of the contestant's birthday matches the day of the month selected - the contestant wins a compact disc.  Otherwise, the contestant wins nothing.  What is the probability that a contestant with a December 31 birthday will win?  Express your answer as a common fraction.

[hide]
$\frac{1}{12} \text { is the prob. that the month matches, and }$
30 days in Sept., April, June, Nov. -4 months
31 days in Jan, March, May, July, Aug., Oct. Dec. -7 mos.
29 days in Feb. - 1 mos.
The probability that the month selected has 31 days and the 31st day is picked is $\frac{7}{12} \cdot \frac{1}{31}=\frac{7}{372}$.

The sum is $\frac{1}{12}+\frac{7}{372}=\frac{38}{372}=\frac{19}{186}$. Is that right? I have a feeling it isn't... If it is, then good luck to him  :lol:
[/hide]

[hide]He has 31 (matching dec) + 7 (matching day 31) ways to win minus the one for December 31. That is $\frac{37}{366}$ which is the final answer.

However you could consider winning nothing still winning making the answer $1$[/hide]

[hide][color=indigo]there are of course 366 total possibilities. to match the month there are 31 possibilities, and to match the day (besides december 31 because i already counted that) there are 6 possibilites. 6 + 31 = 37, so the chances of him wining is 37/366. you really don't want to call in if your birthday's on the 31st... if it was on the 1st or something it would be better.[/color][/hide]


I've been reading around some of the other topics in this section of AoPS and keep reading that for quant jobs it is almost essential to have good coding skills.
My question is, what kind of coding abilities are you talking about? Like being able to just write a small method to perform a simple task or being able to write a "full blown program"? Also what language(s) would you recommend I know? Currently I know C++ and Java, and am learning Python.

Full-blown program -- that gives you the most flexibility.

What language will be most important will almost certainly change by the time you get there, and varies from place to place.  Showing some mastery of serious languages (C++, Java, and Python count, HTML does not) is most important.

Thanks for the reply. Are there any places where I could gain experience writing programs outside of class?

[quote="gauss202"]You could try these problems from [url=http://projecteuler.net/]The Euler Project[/url]

You might also find [url=http://www.acsl.org/samples.htm]these problems[/url] from the American Computer Science League contest interesting enough to try.[/quote]

Thank for the sites, but those are the kinds of programs I am currently doing and would classify them as "small methods". I suppose am looking for a good way to get involved with an open source project but something I'll be able to contribute to instead of being lost.

Think of programs of your own that you'd like to write, and write those.

[quote="Kalle"]You should learn Haskell, so that you can know why every other language is vastly inferior.[/quote]
Spam...

You should also learn how to build layers of abstraction while writing big programs so you don't have to write the same stuff over and over again. For example, if you consistently have to certain extract parts of input, make a class/method/function to do it and you can then just call that function.


function
real analysis
real analysis unsolved

$P \in \mathbb{C}[X]$. prove that $P$ maps every open subset of $\mathbb{C}$ into an open subset of  $\mathbb{C}$

This is simply a special case of the well-known theorem that every analytic function is open.

Let $V$ be an open beighbourhood around a point $x_0$.

Take $\varepsilon>0$ so small that $|p(x)-p(x_0)|>2\varepsilon$ for all $x$ on a circle $\mathcal C$ which bounds a disk $\mathcal D$ inside $V$.

Take any $y$ s.t. $|p(x_0)-y|<\varepsilon$. We have $|p(x)-y|>\varepsilon,\ \forall x\in\mathcal C$. If $p(x)-y$ has no zeros inside $\mathcal D$, then we contradict the minimum modulus principle, since $|p(x)-y|$ does not reach its minimum on $\mathcal C$, the boundary of $\mathcal D$, so there must be an $x$ inside $\mathcal D$ s.t. $p(x)=y$.


The set S contains all the points that are less than 1 away from any point of a square length 4. What is the area of S. (S includes the points in the square also)

[hide]
After a diagram is drawn, it's obvious that the total area is just the area of the square + a circle with radius 1 + 4 1 by 4 rectangles.

So for calculation

$\displaystyle 4(4) + 4(1(4)) + 1^2\pi=16+16+\pi=32+\pi$[/hide]

I'm pretty sure he meant to phrase the problem this way - at least the way I've seen it.

The set S contains all points less than or equal to 1 unit away from the square.

A square is not defined by its interior, so he means any point 1 unit or less away from the perimeter I think.

So the answer might be: [hide="Answer"] $ 4^2 - 2^2 + 4(4) + \pi = 28 + \pi$[/hide]

Doing this problem enough times prompted me to find an expression (and I hope I remembered it right) to solve this given some side length s and maximum distance r.  Working out the problem like before, we can say the area is
$s^2+4sr+\pi r^2$
Plugging in 4 for s and 1 for r here, we get $32 + \pi$

White horse king, the area of the square is $4^2$, not $2^2$  :)

i think i did this wrong, since everyone else has $\displaystyle\pi $ but here goes, i drew a square, and because it had to be one or less away from the edge, i started at the edges and extended 4 segments sqrt(2) units inwardly because if it were one unit to both the adjacent sides, that would give a square with diagonal sqrt(2), i thought that because it had to be one unit, and a circle would come closer to the center at the edges, i made 4 congruent isosceles trapezoids on the exterior with side lengths of sqrt(2), bases of 4 and 6, and with height 1, i did the same, but for the inside and got trapezoids with bases 2 and 4, so add together the areas, and they give you $32$ units, you could also see that it's 2 similar squares with sides 6 and 2, so 62 -22 =32...wont be surprised if im wrong

[hide="original problem"]A circle represents all the points a certain distance away.  So if we have a circle with radius 1, and drag it along the edges of the square, we find four strips, which are extensions of the edges, and 4 quarter circles are added.  The area of those is equal to $(4)((4\times1)+\frac{\pi}{4})$, plus the area of the square itself, which is 16.  So our final answer is $32+\pi$.[/hide]

[hide="whitehorseking wording"]See answer above, but instead of adding 16 we must add $4(1\times1)+4(2\times1)=12$, to account for the 4 edges added, which gives us an answer $28+\pi$.[/hide]

Never mind - my understanding is different.

Look at my post again to see what I thought.

hmmm....I got 32+pi, here's how I did it. In this question, you have a square and an area around the square, I don't have a picture, sorry. But the area around the square is equal to the square itself plus 1 whole circle (there are 4 quarter circles). Therefore the area of the square (16) + the area around the square (16+pi)=[b]32+pi[/b]

The way the question is worded, every point in the square is included, and the only points not in the square that are included are those up to 1 unit away from any point on the perimeter of the square.

Therefore, the area will be, as some others have already said:

$s^2 + 4s + \pi = 32 + \pi $ when $s=4$


Fifty light switches, labeled $ 1$ to $ 50$, are all initially set to off. Consider the $ 50$-step process which proceeds as follows. On step $ k$, if $ k$ is prime, then every light switch labeled with a multiple of $ k$ is switched from on to off, or vice versa. If $ k$ is not prime, do nothing. Compute the number of light switches that are on at the end of the process.

[hide="answer"]
We can clearly say that a number from 1 to 50 can have a maximum of 3 prime number divisors. They can be 2,3,5 or 2,3,7
Note that if a number has exactly 2 prime divisors it will be turned off while if it has exactly 1 or 3 prime divisors, it will be on.

Therefore the numbers which will be on at the end of the process include all the primes + all the perfect powers of the primes (like $ p^2 , p^3 \cdots$ if p is a prime).
Answer is 15 (primes) + (4 (powers of 2) + 2 (powers of 3) + 1 (powers of 5) + 1 (powers of 7) ) + 2 (30 and 42)

Answer = $ \boxed{25}$
Sorry if there are calculation mistakes  :)
[/hide]

@madness: I got the same answer, so I don't think you made any errors.


Is anyone here going to the Michigan Math and Science Scholars Program this summer at the University of Michigan?

Yep!  Find me there.  Which classes are you taking?

Explorations of a Field Biologist and Geometry and the Imagination. I went last year and took Fibonacci Numbers, so I can't do that one again. I'm hoping to get a scholarship as the prices are a bit high for me.

Oh yeah....what are you taking? and is there anyone else here going?

Is the Michigan Math and Science program for MI residents only, or can anyone apply? I ask, because a parent on an email list I'm on just asked me about Midwest summer programs. That sounds like a cool program.

Anyone can apply. It's an international program. In fact, my roommate last year was from Peurto Rico. It's a great program and the proffessors that teach it are great. the website is  http://www.math.lsa.umich.edu/mmss if you want that to. It sould be the right one.

Thanks for sharing the link. Those math courses look very interesting--I haven't seen anything quite like them in other summer programs.

It's no problem really. I'm happy to help.

I am taking Fibonacci numbers and decoding. :(  Will I get to meet you?

I don't know.....we would still see each other around though.....anyways, how would I know if I saw you?

[quote="Ragingg"]We'll arrange something, or I could just wear a big sign that says Ragingg or... "Looking for Mr. Hikari" :)[/quote]

Isn't hikari a girl?

yes I am a girl. I'm wondering how you thought I was a boy.

you could always use ragging's name. its one of these:
Zheng Nan   OR  Daniel  Overbeek  OR  Brandon Long   OR Edward  Weng . Wear a big sign with all four of these names on it. :)

I can't really tell if you're a boy or a girl unless I can see you.  I guess I'm just a fool.

no, I'm pretty sure you aren't a fool. You know if I do see you there, I'll need to know your name. Maybe you could be kind enough to pm it to me sometime.


When asked to multiply

$(x^{2a+1}-y^{3b})(x^{a}-y^{2b-1})$

the answer is:

$x^{3a+1}-x^{2a+1}-y^{2b-1}-x^{a}y^{3b}+y^{5b-1}$

I don't understand the $-$ sign before the $y^{2b-1}$ term.

[quote="AstroPhys"]It looks like the answer is wrong it should be $-x^{2a+1}y^{2b-1}$.[/quote]

I agree with you.


conics
parabola
geometry
circumcircle
analytic geometry
ratio
complex numbers

We are given a parabola with its focus $F$. Three tangent lines at points $A,B,C$ form a triangle. Prove, that point $F$ lies on the circumcircle of that triangle.

Putting the coordinate origin in the focus F, the parabola equation is

$y+\frac{1}{4k}= kx^{2}$

Equations of the tangents $t_{A},\ t_{B},\ t_{C}$ at $A = (x_{a}, y_{a}),$ $B = (x_{b}, y_{b}),$ $C = (x_{c}, y_{c})$ are then

$y = 2kx_{a}(x-x_{a})+y_{a}= 2kx_{a}x-kx_{a}^{2}-\frac{1}{4k}$

$y = 2kx_{b}(x-x_{b})+y_{b}= 2kx_{b}x-kx_{b}^{2}-\frac{1}{4k}$

$y = 2kx_{c}(x-x_{c})+y_{c}= 2kx_{c}x-kx_{c}^{2}-\frac{1}{4k}$

Let $A' \equiv t_{B}\cap t_{C},$ $B' \equiv t_{C}\cap t_{A},$ $C' \equiv t_{A}\cap t_{B}$ be pairwise intersections of the 3 parabola tangents. Solving the pairwise equations of the tangents and expressing their pairwise intersections A', B', C' by complex numbers $\alpha,\ \beta,\ \gamma,$

$\alpha = \frac{x_{b}+x_{c}}{2}+\text{i}\left(kx_{b}x_{c}-\frac{1}{4k}\right)$

$\beta = \frac{x_{c}+x_{a}}{2}+\text{i}\left(kx_{c}x_{a}-\frac{1}{4k}\right)$

$\gamma = \frac{x_{a}+x_{b}}{2}+\text{i}\left(kx_{a}x_{b}-\frac{1}{4k}\right)$

The points A', B', C', F are concyclic iff the cross ratio $\frac{\alpha-0}{\alpha-\gamma}\cdot \frac{\beta-\gamma}{\beta-0}= \frac{\beta-\gamma}{\alpha-\gamma}\cdot \frac{\alpha}{\beta}$ is real. Calculating the 1st fraction,

$\alpha-\gamma = \frac{x_{c}-x_{a}}{2}+\text{i}kx_{b}(x_{c}-x_{a}) = \frac{x_{c}-x_{a}}{2}(1+2\text{i}kx_{b})$

$\beta-\gamma = \frac{x_{c}-x_{b}}{2}+\text{i}kx_{a}(x_{c}-x_{b}) = \frac{x_{c}-x_{b}}{2}(1+2\text{i}kx_{a})$

$\frac{\beta-\gamma}{\alpha-\gamma}\sim \frac{1+2\text{i}kx_{a}}{1+2\text{i}kx_{b}}\sim 1+4k^{2}x_{a}x_{b}+2\text{i}k(x_{a}-x_{b})$

where any real proportionality coefficients have been omitted. Calculating the 2nd fraction,

$\frac{\alpha}{\beta}= \frac{2k(x_{b}+x_{c})+\text{i}(4k^{2}x_{b}x_{c}-1)}{2k(x_{c}+x_{a})+\text{i}(4k^{2}x_{c}x_{a}-1)}\sim$

$\sim 4k^{2}(x_{b}+x_{c})(x_{c}+x_{a})+(4k^{2}x_{b}x_{c}-1)(4k^{2}x_{c}x_{a}-1)$ $+$

$+\ 2\text{i}k[(x_{c}+x_{a})(4k^{2}x_{b}x_{c}-1)-(x_{b}+x_{c})(4k^{2}x_{c}x_{a}-1)] =$

$= (4k^{2}x_{a}x_{b}+1) (4k^{2}x_{c}^{2}+1)-2\text{i}k (x_{a}-x_{b})(4k^{2}x_{c}^{2}+1) \sim$

$\sim 1+4k^{2}x_{a}x_{b}-2\text{i}k (x_{a}-x_{b})$

where again, any real proportionality coefficients have been omitted. In conclusion, the 2 fractions $\frac{\beta-\gamma}{\alpha-\gamma}$ and $\frac{\alpha}{\beta}$ are proportional (with real proportionality coefficients) to complex conjugate numbers, respectively, which means that their product is real and the points A', B', C', F are concyclic.

Let $t_{A},\ t_{B},\ t_{C}$ be the parabola tangents at A, B, C, forming a triangle $\triangle A'B'C',$ where $A' \equiv t_{B}\cap t_{C},$ $B' \equiv t_{C}\cap t_{A},$ $C' \equiv t_{A}\cap t_{B}.$ Locus of the feet of normals from the parabola focus F to the parabola tangents $t_{A},\ t_{B},\ t_{C},$ (the sides lines of the triangle $\triangle A'B'C'$), i.e., the parabola pedal curve with respect to the focus F, is the parabola tangent $t_{V}$ at its vertex V, i.e., it is a line. Thus the pedal triangle of the triangle $\triangle A'B'C'$ with respect to the point F degenerates to the line $t_{V}$, the Simson line of the triangle $\triangle A'B'C'$ with the pole F, hence F lies on the circumcircle of this triangle.


So the problem asks me to find the Fourier series $ F_{f}$ of the function $ f$ given by

$ f(x) = \left\{ \begin{array}{cc}{c 1}
0 & x\in [-1,0) \
x & x\in [0,1) \
\end{array} \right$

on [-1,1) and then extend the periodicity to the whole real line. Is  $ F_{f} = f(x)$ for all real numbers x? Explain.

My question is: what does the question mean by extending the periodicity to the whole real line? There are a few formulas available for $ F_{f}$ so that is not my main concern, but when I get $ F_{f}$, what do I do with it, to answer the last part?

I finally figured it out, the problem just wanted exactly what it had asked, whether the resulting $ F_{f} \equal{} f$ or not.


let $f(x)=\sum_{n=1}^{+\infty}\frac{\{nx\}}{n^{2}}$, where $\{ x \}$ denotes the fractional part of the real number,x. the problem is to find all discontinuites of f and infact show that they form a countable dense set.

The function $f$ is discontinuous at every rational number and continuous at every irrational number.

First, suppose $x$ is irrational. Let $\epsilon>0$ be arbitrary. Chose $N$ such that $\sum_{n=N+1}^{\infty}\frac1{n^{2}}<\frac{\epsilon}3.$

Let $\delta_{1}=\max\{|x-k/n|: k,\in\mathbb{Z},1\le n\le N\}.$

Then $g_{N}(y)=\sum_{n=1}^{N}\frac{\{ny\}}{n^{2}}$ is a finite sum of functions, each of which is continuous for $|y-x|<\delta_{1}.$ Hence $g_{n}$ is continuous on that interval, and $\exists\,\delta\le\delta_{1}$ such that if $|y-x|<\delta$ then $|g_{N}(y)-g_{N}(x)|<\frac{\epsilon}3.$

We also have that for all $x,\ |g_{N}(x)-f(x)|\le\sum_{n=N+1}^{\infty}\frac1{n^{2}}<\frac{\epsilon}3.$

Then for $|y-x|<\delta,$

$|f(y)-f(x)|<|f(y)-g_{N}(y)|+|g_{N}(y)-g_{N}(x)|+|g_{N}(x)-f(x)|< \epsilon.$

Now suppose $x=\frac{p}{q}$ in lowest terms is rational. Write

$f(y)=\sum_{q|n}\frac{\{ny\}}{n^{2}}+\sum_{q\not{|}n}\frac{\{ny\}}{n^{2}}=f_{1}(y)+f_{2}(y)$

Then $f_{2}$ is continuous at $x$ by essentially the same argument given above, while each term of the sum for $f_{1}$ has a jump discontinuity at $x$ and each one of these jumps is in the same direction (downwards). Hence, $f_{1}$ and thus $f$ has a jump discontinuity at $x.$

Since the convergence is uniform, $f$ is continuous at each point where all the terms are continuous. Thus, $f$ is continuous at irrational numbers. For a rational number $a$ let $B=\{n\ge 1\colon na\in\mathbb Z\}$ and $G=\{n\ge 1\colon na\notin\mathbb Z\}$. The sum $\sum_{n\in G}\frac{\{nx\}}{n^{2}}$ is continuous at $a$ for the above reason. The sum over $B$ has a jump discontinuity at $a$: it jumps down by $\sum_{n\in B}\frac{1}{n^{2}}$.

(Edit: this is basically a condensed version of the reply by Kent Merryfield.)


calculus
integration
function
trigonometry
real analysis
real analysis unsolved

Let $f: R\to R_{+}^{*}$, a continuous function with the following property: $f(x)\cdot f(-x)=1,\ \forall{x\in R}$. Show that $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\sqrt{\cos{x}}}{1+f(x)}}dx\leq\sqrt{\frac{\pi}{2}}$.


geometry
geometric transformation
reflection
perpendicular bisector
angle bisector
geometry solved

A convex quadrilateral  $ABCD$ has an incircle. In each corner a circle is inscribed that also externally touches the two circles inscribed in the adjacent  corners. Show that at least two circles have the same size.

Denote $ABCD$ the tangential quadrilateral and $r_A, r_B, r_C, r_D$ radii of the circles $a, b, c, d$ inscribed in the corners $A, B, C, D$. Let $t_A, t_B, t_C, t_D$ be their tangent lengths from the vertices $A, B, C, D$ and  $t_{AB}, t_{BC}, t_{CD}, t_{DA}$ the segment lengths of the common external tangents of the circle pairs $(a, b), (b, c), (c, d), (d, a)$:

$AB = t_A + t_{AB} + t_B$
$BC = t_B + t_{BC} + t_C$
$CD = t_C + t_{CD} + t_D$
$DA = t_D + t_{DA} + t_A$

Since the quadrilateral is tangential, $AB + CD = BC + DA$ and

$t_A + t_{AB} + t_B + t_C + t_{CD} + t_D = t_B + t_{BC} + t_C + t_D + t_{DA} + t_A$
$t_{AB} + t_{CD} = t_{BC} + t_{DA}$

Since the circle pairs $(a, b), (b, c), (c, d), (d, a)$ pairwise touch,

$t_{AB} = 2 \sqrt{r_Ar_B}$
$t_{BC} = 2 \sqrt{r_Br_C}$
$t_{CD} = 2 \sqrt{r_Cr_D}$
$t_{DA} = 2 \sqrt{r_Dr_A}$

$\sqrt{r_Ar_B} + \sqrt{r_Cr_D} = \sqrt{r_Br_C} + \sqrt{r_Dr_A}$

Squaring this equation:

$r_Ar_B + 2 \sqrt{r_Ar_Br_Cr_D} + r_Cr_D = r_Br_C + 2 \sqrt{r_Br_Cr_Dr_A} + r_Dr_A$
$r_Ar_B + r_Cr_D = r_Br_C + r_Dr_A$
$r_Ar_B -  r_Br_C  = r_Dr_A - r_Cr_D$

$r_B(r_A - r_C)  = r_D(r_A - r_C)$

Either $r_A -  r_C = 0$, i.e., $r_A = r_C$, or $r_A - r_C \neq 0$, but then $r_B = r_D$.

It does not seem possible that such circles exist, unless the diagonals $AC, BD$ of the tangential quadrilateral $ABCD$ are perpendicular to each other, i.e., unless $ABCD$ is at least a kite, or unless at least two of the circles ($a, c$ or $b, d$) coincide with the quadrilateral incircle.

Yetti wrote :
........
It does not seem possible that such circles exist, unless the diagonals AC, BD of the tangential quadrilateral ABCD are perpendicular to each other, i.e., unless ABCD is at least a kite ...
......................

   Why not start the solution here as a physicist (what you claim you are) and not as a
   mathematician (what you always wanted to be), without being carried away
   by algebra?

   Maj. Pestich

[quote="pestich"]Why not start the solution here as a physicist (what you claim you are) and not as a
   mathematician (what you always wanted to be), without being carried away
   by algebra?

   Maj. Pestich[/quote]

You call this being carried away with algebra, major?

The following is going to be a combination of Yetti's solution of the original problem (slightly simplified) with some combinatorial-geometrical ideas to answer positively Yetti's question.

[color=blue][b]Theorem 1.[/b] A convex quadrilateral ABCD has an incircle. Four circles x, y, z, w are given such that the circle x touches the segments AD and AB, and so on cyclically, and such that the circles x and y are externally tangent to each other, and so on cyclically.

Then, either the quadrilateral ABCD is a kite, or the circles x and z coincide, or the circles y and w coincide.[/color]

[i]Proof.[/i] We will require two lemmata:

[color=blue][b]Lemma 2.[/b] If two circles with radii u and v touch each other externally, and one of the two external common tangents of these two circles touches them at the points P and Q, respectively, then $PQ=2\sqrt{uv}$.[/color]

Lemma 2 was proven at http://www.mathlinks.ro/Forum/viewtopic.php?t=104489 .

[color=blue][b]Lemma 3.[/b] Given a line g through a point A, some positive number r, and a circle k touching the line g at a point L. Then, there exists at most one circle t which is externally tangent to the circle k, has radius r, lies in the same halfplane with respect to the line g as the circle k and touches the line g at a point lying on the extension of the segment AL beyound L.[/color]

We will call this circle t (when it exists) the [b](A, g, r, k)-circle[/b]. The main purpose of Lemma 3 is to show that this notation is well-defined, i. e. the (A, g, r, k)-circle is (if it exists) uniquely determined from A, g, r, k. (By the way, it always exists, but we don't need to know that.)

[i]Proof of Lemma 3.[/i] We prove Lemma 3 by contradiction: Assume that there are two different circles $t_{1}$ and $t_{2}$ which are both externally tangent to the circle k, have radius r, lie in the same halfplane with respect to the line g as the circle k and touch the line g at two points both lying on the extension of the segment AL beyound L. Let $T_{1}$ and $T_{2}$ be the centers of these two circles $t_{1}$ and $t_{2}$, and let $M_{1}$ and $M_{2}$ be the points where these two circles touch the line g. Both of these points $M_{1}$ and $M_{2}$ lie on the extension of the segment AL beyound L.

The points $T_{1}$ and $T_{2}$ must be distinct (since, if they were equal, the circles $t_{1}$ and $t_{2}$ would be equal, having the same center $T_{1}=T_{2}$ and the same radius r).

Let K be the center and $r_{k}$ the radius of the circle k.

Since the circle $t_{1}$ with center $T_{1}$ and radius r is externally tangent to the circle k with center K and radius $r_{k}$, we have $KT_{1}=r_{k}+r$. Similarly, $KT_{2}=r_{k}+r$. Thus, $KT_{1}=KT_{2}$.

Since the circles $t_{1}$, $t_{2}$, k have centers $T_{1}$, $T_{2}$, K and touch the line g at the points $M_{1}$, $M_{2}$, L, respectively, the lines $T_{1}M_{1}$, $T_{2}M_{2}$ and KL are all perpendicular to g, and since r is the radius of the circles $t_{1}$ and $t_{2}$ which pass through the points $M_{1}$ and $M_{2}$, respectively, we have $T_{1}M_{1}=r$ and $T_{2}M_{2}=r$.

Since the circle $t_{1}$ with center $T_{1}$ and radius r touches the line g, the distance from the point $T_{1}$ to the line g equals r. Similarly, the distance from the point $T_{2}$ to the line g equals r. Thus, the points $T_{1}$ and $T_{2}$ are equidistant from the line g. Also, the points $T_{1}$ and $T_{2}$ lie in the same halfplane with respect to the line g (in fact, the circles $t_{1}$ and $t_{2}$ must lie in the same halfplane with respect to the line g as the circle k; hence, the centers $T_{1}$ and $T_{2}$ of these two circles must also lie in this halfplane). Thus, $T_{1}T_{2}\parallel g$. Hence, the perpendicular bisector of the segment $T_{1}T_{2}$ must be perpendicular to the line g. On the other hand, this perpendicular bisector of the segment $T_{1}T_{2}$ passes through the point K (since $KT_{1}=KT_{2}$). Hence, this perpendicular bisector is the perpendicular to the line g through the point K. Since $KL\perp g$, this perpendicular also passes through the point L. Thus, we have shown that the perpendicular bisector of the segment $T_{1}T_{2}$ passes through the point L, and this entails $LT_{1}=LT_{2}$. Together with $T_{1}M_{1}=T_{2}M_{2}$ (because $T_{1}M_{1}=r$ and $T_{2}M_{2}=r$), this yields the congruence of the right-angled triangles $LT_{1}M_{1}$ and $LT_{2}M_{2}$; thus, $LM_{1}=LM_{2}$. Consequently, either the points $M_{1}$ and $M_{2}$ coincide, or the point L is the midpoint of the segment $M_{1}M_{2}$. But the points $M_{1}$ and $M_{2}$ cannot coincide because this would yield $T_{1}T_{2}\perp g$ (since both points $T_{1}$ and $T_{2}$ would lie on the perpendicular to the line g at the point $M_{1}=M_{2}$), contradicting $T_{1}T_{2}\parallel g$; and the point L cannot be the midpoint of the segment $M_{1}M_{2}$, since both points $M_{1}$ and $M_{2}$ lie on the extension of the segment AL beyound L. Hence, we get a contradiction, and our assumption that there exist two different circles $t_{1}$ and $t_{2}$ both satisfying the condition of Lemma 3 was wrong. Thus, we have shown that there exists at most one such circle, and Lemma 3 is proven.

Let's solve the problem now. Let the circle x have center X and radius $r_{x}$ and touch the sides AD and AB at the points $X_{d}$ and $X_{b}$, respectively. Similarly define the objects Y, $r_{y}$, $Y_{a}$, $Y_{c}$ for the circle y, the objects Z, $r_{z}$, $Z_{b}$, $Z_{d}$ for the circle z, and the objects W, $r_{w}$, $W_{c}$, $W_{a}$ for the circle w.

Since the points $X_{b}$ and $Y_{a}$ both lie on the segment AB, two cases are possible: Either the points A, $X_{b}$, $Y_{a}$, B lie on the line AB in this order, or the points A, $Y_{b}$, $X_{b}$, B lie on the line AB in this order. But the second of these two cases, the case when the points A, $Y_{b}$, $X_{b}$, B lie on the line AB in this order, is actually impossible: if it was the case, the segments $X_{b}X_{d}$ and $Y_{a}Y_{c}$ would intersect at a common interior point; this point would then be an interior point of both circles x and y, but these two circles x and y are externally tangent to each other and therefore cannot have common interior points. Hence, the points A, $X_{b}$, $Y_{a}$, B lie on the line AB in this order. Similarly, the points B, $Y_{c}$, $Z_{b}$, C lie on the line BC in this order; the points C, $Z_{d}$, $W_{c}$, D lie on the line CD in this order; the points D, $W_{a}$, $X_{d}$, A lie on the line DA in this order.

Since the quadrilateral ABCD has an incircle, AB + CD = BC + DA. In other words, $\left(AX_{b}+X_{b}Y_{a}+Y_{a}B\right)+\left(CZ_{d}+Z_{d}W_{c}+W_{c}D\right)=\left(BY_{c}+Y_{c}Z_{b}+Z_{b}C\right)+\left(DW_{a}+W_{a}X_{d}+X_{d}A\right)$. But since the two tangent segments from a point to a circle are equal, we have $AX_{b}=X_{d}A$, $BY_{c}=Y_{a}B$, $CZ_{d}=Z_{b}C$, $DW_{a}=W_{c}D$, so this equation simplifies to $X_{b}Y_{a}+Z_{d}W_{c}=Y_{c}Z_{b}+W_{a}X_{d}$. Now, since the circles x and y with radii $r_{x}$ and $r_{y}$ touch each other externally, and the line AB, which is one of their external common tangents, touches them at the points $X_{b}$ and $Y_{a}$, by Lemma 2 we have $X_{b}Y_{a}=2\sqrt{r_{x}r_{y}}$. Similarly, $Y_{c}Z_{b}=2\sqrt{r_{y}r_{z}}$, $Z_{d}W_{c}=2\sqrt{r_{z}r_{w}}$ and $W_{a}X_{d}=2\sqrt{r_{w}r_{x}}$. Hence, the equation $X_{b}Y_{a}+Z_{d}W_{c}=Y_{c}Z_{b}+W_{a}X_{d}$ becomes

$2\sqrt{r_{x}r_{y}}+2\sqrt{r_{z}r_{w}}=2\sqrt{r_{y}r_{z}}+2\sqrt{r_{w}r_{x}}$
$\Longleftrightarrow\ \ \ \ \ \sqrt{r_{x}r_{y}}+\sqrt{r_{z}r_{w}}=\sqrt{r_{y}r_{z}}+\sqrt{r_{w}r_{x}}$
$\Longleftrightarrow\ \ \ \ \ \sqrt{r_{x}r_{y}}+\sqrt{r_{z}r_{w}}-\sqrt{r_{y}r_{z}}-\sqrt{r_{w}r_{x}}=0$
$\Longleftrightarrow\ \ \ \ \ \left(\sqrt{r_{x}}-\sqrt{r_{z}}\right)\left(\sqrt{r_{y}}-\sqrt{r_{w}}\right)=0$
$\Longleftrightarrow\ \ \ \ \ \sqrt{r_{x}}-\sqrt{r_{z}}=0\text{ or }\sqrt{r_{y}}-\sqrt{r_{w}}=0$
$\Longleftrightarrow\ \ \ \ \ r_{x}=r_{z}\text{ or }r_{y}=r_{w}$.

We can WLOG assume that $r_{x}=r_{z}$. Denote $R=r_{x}=r_{z}$.

Now, consider the line AB through the point B, the positive number R, and the circle y touching the line AB at the point $Y_{a}$. The circle x is externally tangent to the circle y, has radius $r_{x}=R$, lies in the same halfplane with respect to the line AB as the circle y (in fact, both circles x and y lie in the halfplane with respect to the line AB which contains the quadrilateral ABCD, because the circle x touches the segment AD and the circle y touches the segment BC) and touches the line AB at the point $X_{b}$ lying on the extension of the segment $BY_{a}$ beyound $Y_{a}$. Thus, the circle x is the (B, AB, R, y)-circle. Similarly, the circle z is the (B, BC, R, y)-circle.

Now, the reflection with respect to the angle bisector of the angle ABC maps the line AB to the line BC, the point B to itself, and the circle y to itself (since the center of the circle y lies on the angle bisector of the angle ABC, because the circle y touches the sides BA and BC of this angle). Hence, this reflection maps the (B, AB, R, y)-circle to the (B, BC, R, y)-circle. In other words, it maps the circle x to the circle z. Hence, this reflection maps the center X of the circle x to the center Z of the circle z.

Now, if X = Z, then the circles x and z coincide (because they have the same center X = Z and the same radius $r_{x}=r_{z}$) and Theorem 1 is proven. So it remains only to consider the case when $X\neq Z$. In this case, the fact that the reflection with respect to the angle bisector of the angle ABC maps the point X to the point Z can be rewritten as follows: The angle bisector of the angle ABC is the perpendicular bisector of the segment XZ. Similarly, the angle bisector of the angle ADC is the perpendicular bisector of the segment XZ. Hence, the angle bisectors of the angles ABC and ADC are one and the same line; this line must be the line BD. Hence, $\measuredangle ABD=\measuredangle CBD$ and $\measuredangle ADB=\measuredangle CDB$. Together with BD = BD, this yields that the triangles ABD and CBD are congruent, so the quadrilateral ABCD is a kite. Theorem 1 is thus proven.

  Darij


There is at least one two-digit number such that when it is added to the two-digit number having the same digits in reverse order, the sum is a perfect square. Find the sum of all such two digit numbers.

[hide]If one of the numbers is 10x+y, the other is 10y+x, and if you add them up, you get 11(x+y).  Therefore, the sum must be a multiple of 11, and the only perfect square that can possibly be a sum of two two-digit numbers is 121.  The numbers that fit this condition are 29, 38, 47, 56, 65, 74, 83, 92, whcih sum up to 121(4)=[size=150]484[/size][/hide]

[hide] Let x be the tens digit and y be the units digit of the original number.

10x + y + 10y + x = perfect square
11x + 11 y = perfect square
11(x+y) = perfect square.

Since 11 is prime, and 11*(x+y) must be a perfect square, x+y must be equal to 11. 29, 38, 47, 56, 65, 74, 83, and 92 all work. Add those up and you get... well, you add them up  ;) [/hide]

[hide]This two digit number can be represented as 10x+y. With the digits reversed, it's 10y+x. Adding these two numbers gets 11x+11y, or 11(x+y). This is a representation of the perfect square, so that square must be a multiple of 11. Also, the maximum value of x+y is 18, so the squares must be less than 11*18=198. There is only one perfect square, 121, that is a multiple of 11 less than 198. x+y must equal 11. The numbers that work are 29,38,47,56,65,74,83,and 92. Their sum is [b]484[/b].[/hide]


Find all Posititve integer $m$ such that:
$a_1a_1...a_k=-1$(mod$ m$)
With $a_i$ are all positive in $[1,m]$ and $(a_i,m)=1$.

This is the third time i saw this problem ! Try find it in this section...

Because I don't know,sorry.

Here is two other problem,I hope It's not post before.

1,Find all $(x,y,z)=1$ is odd integer and:
$\frac{1}{x^2}+\frac{1}{y^8}=\frac{1}{z^2}$

2,Find all integer roots of:
$y^2=x^3+(x+4)^2$

Please, post them into new and separated threads...

Pierre.


Hi!

Let $a_{1},\cdots, a_{k}$ be integers $\geq 1$ such that $a_{1}+\cdots+a_{k}=n$ where $n$ is a given integer.

Compute
\[\max\prod_{i=1}^{k}a_{i}. \]

[quote="arkhammedos"]$Max =1$ :?:  :rotfl:[/quote]
Nope. Try to solve the case $k=2,n=3$.

we do devision with reaminder of n to 3.
if $n=3p$ =>  $Max=3^{p}$
if $n=3p+1$ => $Max=23^{p}$.
if $n=3p-1$ => $Max=43^{p-1}$  :wink:


what 's the proof of Zsigmondy's theorem? which is  http://mathworld.wolfram.com/ZsigmondyTheorem.html

See http://www.mathlinks.ro/Forum/viewtopic.php?p=739526 . That's $ b\equal{}1$ only, but it easily generalizes.


geometry
circumcircle
incenter
symmetry
Euler
projective geometry
cyclic quadrilateral

Let $ ABCD$ be the inscribed quadrilateral with the circumcircle $ \omega$.Let $ \zeta$ be another circle that internally tangent to
$ \omega$ and to the lines $ BC$ and $ AD$ at points $ M,N$ respectively.Let $ I_1,I_2$ be the incenters of the $ \triangle ABC$ and $ \triangle ABD$.Prove that $ M,I_1,I_2,N$ are collinear.

[img]http://img231.imageshack.us/img231/1443/asan2lt5.png[/img]
I'm sorry for my not perfect picture,it is look like $ A,I_2,C$ are collinear :blush: ,but it isn't true.Thank you. :)

I think you know that there is exits a circle, which is tangent to $ AC$, $ BD$ and $ (O)$ at $ E$, $ F$ and $ P$, respectively and the line $ EF$ is passes through the incenters $ I_1$, $ I_2$ of triangles $ ABC$, $ ABD$. Denote by $ X$, $ Y$, $ Z$, $ T$ be the middle points of arcs $ AC$, $ AD$, $ DB$, $ CB$. Applying [b]Pascal's theorem[/b] for the cyclic hexagon $ PYBDAZ$, we have $ N \equal{} PY\cap DA$, $ I_2 \equal{} YB\cap AZ$ and $ F \equal{} BD\cap ZP$ are collinear. Similarly, $ M$, $ I_1$ and $ E$ are collinear. Hence we conclude that six points $ M$, $ N$, $ E$, $ F$, $ I_1$ and $ I_2$ are collinear.

[b]P.S.[/b] About the result $ EF$ passes through $ I_1$, $ I_2$, I think it's well-known (maybe it's called as [b]Thebault[/b] theorem)

Here is an image for your nice solution:
[img]http://img87.imageshack.us/img87/205/i1i2gp6.png[/img]

P.S:
Yes,I know this solution,which use your facts above.
I'll give a proof for them later. :wink:

[quote="April"]I think you know that there is exits a circle, which is tangent to $ AC$, $ BD$ and $ (O)$ at $ E$, $ F$ and $ P$, respectively[/quote]

I would like to see a proof of this!

Oops!  :ewpu:  Thank you very much dear [b]nsato[/b]! This is exactly what I am wavering. Please give me a little time to check and try to prove it. And for now, I'll give another solution for this problem, to atone for one's mistake.

Denote by $ X$ the intersection of $ PN$ with the circle $ \omega$ and $ I_2$ the intersection of $ BX$ with $ MN$. We have $ X$ is the middle point of the arc $ AD$ of the circle $ \omega$.
Consider the common tangent $ Pt$ at $ P$ of two circles $ \omega$ and $ \zeta$.
We have $ \angle I_2MP\equal{}\angle NPt\equal{}\angle I_2BP$, therefore the quadrilateral $ BPI_2M$ is cyclic. It implies that $ \angle PI_2B\equal{}\angle BMP\equal{}\angle MNP$, which follows that $ \triangle XI_2P\sim\triangle XPI_2$. Hence we conclude that $ XI_2^2\equal{}XN\cdot XP$ $ (1)$

On the other hand, we can easily to see that $ \triangle XDN\sim\triangle XPD$, so $ XD^2\equal{}XN\cdot XP$ $ (2)$

Combining $ (1)$ and $ (2)$, we get $ XI_2\equal{}XD$. And notice that $ X$ is the middle point of arc $ AD$ of the circumcircle of triangle $ ABD$ and $ I_2$ is a point lies on $ BX$, so we conclude that $ I_2$ is the incenter of triangle $ ABD$.

Similarly, $ MN$ also passes through the incenter $ I_1$ of riangle $ ABC$. And our proof is completed.

$ \bullet$ Hope that I didn't get any mistake in this proof  :read: .

$ \bullet$ I believe that if $ ABCD$ is a given cyclic quadrilateral, then there always exits two circles $ \omega$ and $ \Omega$, where $ \omega$ is tangent to $ (O)$ - the circumcircle of quadrilateral $ ABCD$, at $ P$ and also to $ AC$, $ BD$, whereas the circle $ \Omega$ is also tangent to $ (O)$ at $ P$ and to $ AD$, $ BC$.

[quote="April"]
$ \bullet$ I believe that if $ ABCD$ is a given cyclic quadrilateral, then there always exits two circles $ \omega$ and $ \Omega$, where $ \omega$ is tangent to $ (O)$ - the circumcircle of quadrilateral $ ABCD$, at $ P$ and also to $ AC$, $ BD$, whereas the circle $ \Omega$ is also tangent to $ (O)$ at $ P$ and to $ AD$, $ BC$.[/quote]
Yes,of course,it is obviously that there exist such circle $ \omega$.But we must prove that $ \omega$ is tangent to diagonals  at points $ E,F$.
Here is another lemma to this problem:
Let $ ABCD$ be a cyclic quadrilateral and $ I_1,I_2$ are incenters of $ ABC$ and $ ABD$ respectively.Denote $ O\in AC\cap BD$,$ E\equal{}\in AC\cap I_1I_2$,$ F\in BD\cap I_1I_2$.Prove that $ OE\equal{}OF$.
This problem is from All-Russian olympiad,i'll post a solution to it later. :wink:

Here is an image to your nice proof:
[img]http://img458.imageshack.us/img458/2408/apriljl1.png[/img]

To April:
I think that it must be $ \triangle XI_2P\sim \triangle XNI_2$ in your proof,instead of $ \triangle XI_2P\sim \triangle XPI_2$.
Thank you.Great proof.

Let $ R$ be the point of intersection $ AC$ and $ MN$.Prove that $ PR$ is an angle bisector of $ \angle APC$.
[img]http://img211.imageshack.us/img211/1372/erkenslemmatoasnasproblri0.png[/img]

$ \frac {AN}{AP} \equal{} \frac {CP}{CM} \equal{}$constant$ (\sqrt{\frac{R\minus{}r}{R}}$ where $ R,r$ is radius of $ w,\zeta$). So $ \frac {PC}{AP} \equal{} \frac {CM}{AN}$. Let $ BC$ and $ AD$ meet $ X$. Since $ XM$,$ XN$is tangent line of $ \zeta$,$ XN \equal{} XM$. Hence $ sin\angle ANR \equal{} sin\angle CMR$ and $ sin\angle ARN \equal{} sin\angle CRM$. So by law of sin in triangle $ ANR,CRM,\frac {CM}{AN} \equal{} \frac {RC}{AR}$. Henc $ \frac {PC}{AP} \equal{} \frac {RC}{AR}$ Therefore, $ RP$ bisect $ \angle APC$

Wow it is a nice solution!
but i don't know why  AN/AP=CP/CM
could you solve this more detail?
 :lol:

Partial Solution:

Let $ PM$ and $ PN$ intersect $ \omega$ again at $ M'$ and $ N'$. Let $ P'$ be the midpoint of small arc $ AB$.

Lemma 1: $ PP'$, $ AB$, and $ MN$ are concurrent.
Proof: I don't have this yet, but I believe that it should be easy with an inversion.

It is well known that $ M'$ and $ N'$ are the midpoints of small arcs $ BC$ and $ DA$, respectively.

Apply Pascal on $ AM'PP'CB$ to get that $ I_1\equal{}AM'\cap P'C$, $ M\equal{}M'P\cap CB$ and $ X\equal{}PP'\cap BA$ are collinear.

By symmetry of argument,  $ I_2$, $ N$, and $ X$ are collinear.

By lemma 1, $ X$, $ M$, and $ N$ are collinear, so $ I_2$ and $ I_1$ lie on $ XMN$, so $ M,N,I_1,I_2$ are collinear.

I hope to come back with a proof for lemma 1.

In my post above, lemma 1 is easy to prove.

If we look in triangle $ PAB$, $ PP'$ is the external angle bisector of the angle at $ P$. So if $ PP'\cap AB \equal{} X_1$, then $ BX_1: X_1A \equal{} PB: PA$ is easy to calculate.

If we connect $ AD$ and $ BC$ to meet at $ Z$, then in triangle $ XAB$, we get that $ MN$ intersects $ AB$ at $ X_2$, we get $ BX_2: X_2A \equal{} BM: AN$ by menelaus.

It is a pretty easy lemma to show that $ PB: PA \equal{} BM: AN$ (it is the same fact that [b]Leonhard Euler[/b] uses in his solution to the second problem in this topic).

Comment: This is a very interesting and (in my opinion) difficult problem. The reason is this: If we consider the quadrilateral $ ABCD$ fixed, there are two circles that are tangent to $ \omega$, $ BC$, and $ AD$. Point $ P$ can either lie on small arc $ AB$ or $ CD$. If $ P$ is on small arc $ CD$, then the problem is true. Another way to state this is, if $ PBCDA$ is convex, the problem is true, if $ PDABC$ is convex, the problem is not true.

So when you use the methods that you use, they must differentiate between these two cases. Obviously, Pascal's theorem is projective, so it does not differentiate between these cases.

The reason that my solution here works is because it implicitly invokes a sort of betweenness argument. If you look at point $ N$, it could potentially have two positions based on how I used it: in particular, there are  two tangents from point $ A$ to the smaller circle. When I used menelaus, it differentiated between these two cases (because $ X_2$ would be on segment $ AB$ otherwise).

I have not read April's post, but I'm sure the solution can be interpreted as I did above.

Proof of [b]April's[/b] lemma is also here http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf .

Here is my solution
First, we construct the common targent Tx of two circles
Hence, we have BMI1T is a cyclic quadrilatetal
Let M is intersection of CI1 with (w). Easy to proof that MI1=MA=MX, so I1 be the incenter of triangle ABC.

Here is my solution
First, we construct the common targent $ Tx$ of two circles
Hence, we have $ BMI_{1}T$ is a cyclic quadrilatetal
Let $ M$ is intersection of $ CI_{1}$ with $ (w)$. Easy to proof that $ MI_1\equal{}MA\equal{}MX$, so $ I_1$ be the incenter of triangle $ ABC$.

This problem isn't Chinese TST 2007 6th quiz P2. maybe orl have some mistakes.

Is there somebody who knows where this question comes from??? :help:


$ACEG$ is a quadrilateral circumscribed about a circle and tangent to the circle at points $B,D,F,H$
If $AB=3,CD=4,EF=5,GH=6$, find the radius of the circle

Denote $a = AB = 3, c = CD = 4, e = EF = 5, g = GH = 6$. The formula for radius of the incircle is

$r = \sqrt{\frac{ace + aeg + ecg + acg}{a + c + e + g}} = \sqrt{\frac{60 + 90 + 120 + 72}{18}} = \sqrt{\frac{342}{18}} = \sqrt{19} \doteq 4.36$

For different ways to obtain this formula, see the thread [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28864]Looks so easy[/url] in the Olympiad Section.


Let $ BD$ and $ CE$ are bisectors of  $ \Delta ABC$ with $ \measuredangle BAC\ge60^{\circ}.$ Prove that $ AD\plus{}AE\le BC.$

$ AD\plus{}AE\le BC\Leftrightarrow \frac{bc}{c\plus{}a}\plus{}\frac{bc}{b\plus{}a}\le a$
$ \Leftrightarrow a(a\plus{}b)(a\plus{}c)\geq (a\plus{}b)bc\plus{}(a\plus{}c)bc \Leftrightarrow (a^2\minus{}bc)(a\plus{}b\plus{}c)\geq 0$
It is OK because $ \angle{A}\geq 60^0\Leftrightarrow a^2\geq b^2\plus{}c^2\minus{}bc \Rightarrow a^2\geq bc$


[quote="elvenchamp777"]My favorite quotes are:
Don't ask a question and appear dumb then ask and remove all doubt.
                                      - Mark Twain
There are 3 types of lies: lies, damed lies, and statistics.
                                      - Mark Twain
I was waiting for like 3 hours from 7:00 to 8:00!
                                      - One of my friends

what are your favorite quotes?

elvenchamp777[/quote]

wasnt this discussed before?

[quote="elvenchamp777"]My favorite quotes are:
Don't ask a question and appear dumb then ask and remove all doubt.
                                      - Mark Twain
There are 3 types of lies: lies, damed lies, and statistics.
                                      - Mark Twain
I was waiting for like 3 hours from 7:00 to 8:00!
                                      - One of my friends

what are your favorite quotes?

elvenchamp777[/quote]

another version of number one is one of Abraham Lincoln's Code of Conduct rules...i should kno, i had to write them 10 times cuz i got caught chewing gum in math class last year. :) anyways, his version is:

"better to be silent and thought a fool than to speak out and remove all doubt."

my fave quote tho, which is really cool is in my sig...i think...is:

"Great spirits have always encountered violent opposition from mediocre minds." ~Albert Einstein

gobraves, this probably has been discussed b4, topics seem to recycle after a while. this is partially due to ppl not looking back at old threads and reopening them. but that's ok.

Rediscussion is good because every time, we are older and wiser, and can share new knowledge.

older yes but wiser is questionable  :D

elvenchamp777

"It ain't illegal if the police don't see it"

My friend's motto

O yeah that reminds me of another one.

No one saw me do it! You can't prove anything!
                            - Bart Simpson

elvenchamp777

"Quitters never win, winners never quit, but those who never quit AND never win are idiots."

--not sure

this topic was discussed very recently. but anywho:
President George W. Bush said "He is a man of good experiences."

"Education is my top priority. But education is not my top priority"
"they misunderestimate me"-both george bush

"think about it; potatoe, add an extra e on the end, good!" dan quayle, "correcting" a students correct spelling of "potato"
"I admit i made a mistake there (in reference to the previous quote) but as mark twain said, you should never trust a man who has only one way to spell things" Dan quayle (mark twain did not say that, andrew jackson did)
"Andrew jackson got his name "stonewall" because of his policy of vetoing almost every bill" Dan quayle again

[quote="MathFiend"]gobraves, this probably has been discussed b4, topics seem to recycle after a while. this is partially due to ppl not looking back at old threads and reopening them. but that's ok.[/quote]

I thought reopening old threads was bad, or do those rules not apply in G&FF?

"Advertising is legalized lying." ~Someone

[quote="RC-7th"][quote="MathFiend"]gobraves, this probably has been discussed b4, topics seem to recycle after a while. this is partially due to ppl not looking back at old threads and reopening them. but that's ok.[/quote]

I thought reopening old threads was bad, or do those rules not apply in G&FF?[/quote]

well, rather than to have 3 or 4 threads on the same topic, having just one with everyone's thoughts on it would be the best way to do it.

"Genius can't win effort; effort can't win the fun."
- don't remember.. but I think it was some Korean (supposedly famous) guy who said it.. my translation sucks, though.

A cute one I like: "Happiness is a warm puppy" -Charles Schultz.

"Attention to health is life's greatest hindrance."
- Plato (427-347 B.C.)

"Plato was a bore."
- Friedrich Nietzsche (1844-1900)

"Nietzsche was stupid and abnormal."
- Leo Tolstoy (1828-1910)

"I'm not going to get into the ring with Tolstoy."
- Ernest Hemingway (1899-1961)

"Hemingway was a jerk."
- Harold Robbins

hehehe...

Dreaming of the person you want to be is a waste of the person you are.
- Kurt D. Cobain

I dont have a fav.. heres a good one:

"Underachieve. You never know what u might not achieve"

btw that one about friends and clovers.. i think its Sarah Jessica Parker.

[quote="addicted2sharpiz"]My friend said once: "wow, there are a lot of asians in china."[/quote]
LOL

Here's a good one:
If you are one in a million, there are at least 1200 people in China just like you.

hmmm... my favorite would have to be a Bushism.

JamesD
[quote]mine is "Luke, I am your father." -Darth Vader[/quote]

YOU MISQUOTED STAR WARS!!! it's "NO, I am your father." that's the correct quote.

:starwars:

Mine is: "I wish I could take the pain away. If you can make it through the night there's a brighter day. Everything'll be allright if you hold on. It's a struggle everyday gotta roll on"

Kinda long but it's by Tupac.

"With great power, comes great responsibilty."-i don't know who said that, but it's in Spiderman

[quote="Iversonfan2005"][quote="mathbomberII"][quote="GoBraves"]"Indians are smarter"
- Me[/quote] i think so to[/quote]

Ofcourse they are!! :D[/quote]
i second that! :D

[quote="xdl615x"]"Liar Liar Pants on fire" has to be my favorite.[/quote]
that one is funny but old

bubalas sig is a good one
"is our children learning"


Never really thought of that.  I never thought I was doing math to prepare for the AMCs...

Funny.  You get one ha for that.  Ha.

However, some people voted "nothing", so that would tell you something...

Wait a sec. It's NOT possible to do ALL of the above. Otherwise, you're doing NOTHING, and you're doing all the others (SOMETHING!).


It's equivalent to prove : $ \sum_{cyclic}\frac{\sqrt{c}}{\sqrt{c\plus{}a}} \leq \frac{3\sqrt{2}}{2}$
See here : http://www.mathlinks.ro/viewtopic.php?t=158729

[quote="Erken"]This problem was at CWMO too, so it isn't new[/quote]
What is "CWMO"  :?:

[quote="mai quoc thang"]
\[ \ xyz = 1;x,y,z > 0
\]
Prove that:
\[{ \frac {3}{\sqrt2}} \geq \frac {1}{\sqrt {1 + x^2}} + \frac {1}{\sqrt {1 + y^2}} + \frac {1}{\sqrt {1 + z^2}}.
\]
:rotfl:  :rotfl:[/quote]
Using Cauchy-Schwarz Inequality,we have

$ (\sum {\sqrt {\frac{a}{{a + b}}} } )^2  = (\sum {\sqrt {a(a + c)} .\sqrt {\frac{1}{{(a + b)(a + c)}}} } )^2  \le (\sum {a^2 }  + \sum {ab} )(\sum {\frac{1}{{(a + b)(a + c)}}} )$
then applying p,q,r . :)


[quote="gemath"]Given $x,y,z>0$ prove that
$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\frac{9(x^{2}+y^{2}+z^{2})}{(x+y+z)^{2}}$
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120859


induction parameterization geometry puzzles

Many people have heard of the unexpected examination (or unexpected hanging-same idea) paradox. I have been thinking about it for quite a while and I still can't come up with a good explanation. Maybe someone else can!

The paradox goes like this: At the beginning of the week, a teacher tells the class about an upcoming exam. He tells them that the exam is that week; however, he doesn't give a specific day. In addition, he says that they will not know the day of the exam until the actual exam. In other words, the exam will be a surprise for them. Suddenly, the teacher is interrupted by one of his students, who tells his classmates not to worry because according to what the teacher told them, the exam can never take place. He explains this in the following way: "Suppose it's Friday, and we're still awaiting the exam. Since the exam is this week, it will have to be on Friday. But, in that case, we will know the day of the exam before the time of the exam, which clearly contradicts the teacher. Therefore, the exam cannot be on Friday. Well, suppose its Thursday and we're still awaiting the exam. We already know that it can't be on Friday, and using the same reasoning we can rule out Thursday as well. In fact, using this reasoning, we can rule out every day of the week." The teacher looks at the student and smiles. That week, they have the exam and everything the teacher told them is true. How can this be? What's wrong with the student's logic?

there is already a forum discussing this on,

 http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml

The logic doesn't work because you are assuming things that aren't given. If after each day, the teacher says that it won't be on that day, then the logic is fine. However, he/she doesn't so if it is on wednesday, you wouldn't know unless the teacher has said it isn't on friday and it isn't on thursday

But it appears that according to what the teacher said, you can deduce that it can't be on Friday, and then (by induction) conclude that it can't be any of those days

It is [i]because[/i] the student deducts that the test cannot be given. In this way, he/she will assume the test will not be given on any day of the following week, in which case the teacher may give the test on [i]any[/i] day of the following week, and the student will not be expecting it.

That was my first reaction, but it's not a good explanation. The students would indeed be surprised, but they would be surprised "on a different level", kinda like with the sentece "This sentence is not true" -Is it true or false? Well, it's true on one level and false on another level (a self-contradictory sentence, unlike other sentences, does not have to be either true or false). The paradox is asking about the flaw in the student's logic. In other words, suppose that in addition to his two original statements, the teacher also said "and anything that you logically deduce will occur". Then the explanation would not suffice.

You have to wait until Thursday to conclude that it is not on Friday, AND THEN you would go back to Wednesday and conclude that it is not on Thursday.

And so on, and so on. That's impossible.

So if the exam is on Wednesday, you don't know that. On Monday you don't know, on Tuesday you don't know, and on Wednesday: SURPRISE!!! Hope we studied. On Thursday it's too late.

Hope that helps.

I saw this in a book and couldn't figure it out either.

Are you saying that in the process of concluding that the exam cannot be on a given day (except Friday), circular logic is involved? It seems that if the exam can't be on day $n$, then it can't be on day $n-1$. Since you know it can't be on Friday, you can use this "backward induction" to prove that it can't be on the very first day. In fact, you can do this before the first class of the week starts. However, if you take a different approach, there is no reason why the test can't be on the first day, on the second day, etc. It is this I don't understand.

I looked this up in an encyclopedia and it gave some weird explanation I didn't understand... Something about some people knowing when the event will happen before it happens and others not knowing when the event will happen until after it happens. I don't see what that has to do with it.

I can give you very easy explanation.
[hide="explanation with one line"]The term "unexpected" is not well defined. That is why we get a paradox.[/hide]

Suppose you think of it this way.  If the test is on Friday, then the students will not know that until Thursday.  However, the teacher said that they won't KNOW which day the test is on until it comes.  So, assuming that they assumed that there was a test after all (otherwise the teacher can just put it any day and surprise them), let's say that it's Wednesday.  They "know" that it can't be on Friday, because then they would know by Thursday.  So they know it's on Thursday.  But that is assuming that they didn't "know" which day the test was on when it was Wednesday.  Therefore, if it's Wednesday and they "know" which day it's on, then they can't rule out the possibility of its being Friday yet.  Hope that works.  You can set it back a few days to make it easier if that helps.  But if they just stick to their logic, then the teacher can put it on any day, they won't know, and they'll be surprised.

[quote="lightrhee"][quote="Bob"]Can someone give a clear explanation of what exactly is wrong with the student's inductive logic?[/quote]
I already gave an explanation. Since we do not clearly define what "unexpected" means, the students' logic sadly fail. When he(or she) thinks that
[i]"If the exam is on Friday, it will not be an unexpected thing"[/i],
this step is already logically very weak since "unexpected" is not defined.[/quote]

But the problem defined "unexpected" as "something you don't know until the day it happens".

My encyclopedia: "The paradox is a subtle one. The crucial point lies in the fact that a statement can be known to be a true prediction by one person but not to another person until after the event happens."

What?! :huh:

pf) His(or her) first statement is wrong.
[hide]
Suppose that the exam was not written until Thursday. The student would think on Thursday that

[i]"This is not unexpected! The exam has to be on Friday now!"[/i]

If the student is not expecting the exam on Friday, based on his(or her) thinking(if the exam is on Friday, there is a contradiction. Therefore, no exam on Friday), he(or she) will face an unexpected exam on Friday.

If the student is expecting the exam on Friday, this proves that his(or her) first step is wrong.

Therefore, the first step is already wrong.[/hide]

We can also consider 1 day, instead of 5 days. You will find out that his(or her) thinking does not work.

i think the problem is that the student's logic works from friday to monday while the time passes in exactly the opposite direction. if the week would start with friday and end with monday it would be quite right what he said. :)

Which is what I said in my previous post. But you can still use induction in that direction.

For example, let's say you know an event will occur on a certain day. You've been tipped that the event will not occur on the 24th, and furthermore, if it will not occur tomorrow, it will not occur today either. What can you conclude? It will not occur on any day on or before the 24th.

This problem is similar. Except that you know the exam will occur next week but it can't occur on Friday or any day before a given day.

Isn't the simplest solultion, as stated above,

The students concluded that no exam could occur; therefore, if an exam occured, it would be unexpected and the teacher would be right (always one step ahead?).

Like I said before, theres a forum discussing this, heres a direct link: http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1028031934

Its called the " Pop Quiz Riddle" ,   unexpected examination sounds kinda lame...........

Yes, but then the students know that the teacher will try to surprise them and give them an exam, so they will expect it, and the teacher will expect them to expect it and won't give an exam, but then the students know that the teacher will try to surprise them and give them an exam, so they will expect it, and the teacher will expect them to expect it and won't give an exam, but then the students know that the teacher will try to surprise them and give them an exam, so they will expect it, and the teacher will expect them to expect it and won't give an exam, but then the students know that the teacher will try to surprise them and give them an exam, so they will expect it, and the teacher will expect them to expect it and won't give an exam, but then the students know that the teacher will try to surprise them and give them an exam, so they will expect it, and the teacher will expect them to expect it and won't give an exam, but then the students know that the teacher will try to surprise them and give them an exam, so they will expect it, and the teacher will expect them to expect it and won't give an exam, but then the students know that the teacher will try to surprise them and give them an exam, so they will expect it, and the teacher will expect them to expect it and won't give an exam, ad infinitum.

And besides, if you use my definition of unexpected, you come up with a totally different problem that you can't say is wrong any more than you can say that Euclidean geometry is the only correct type of geometry. I just make different assumptions that I then reason from and make a different cool problem.

[quote="Bob"]"Things should be made as simple as possible, but not simpler" -Albert Einstein[/quote]

Umm... are you implying that our solution is simpler than it should be? Why?

mathnerd, you can go ahead and make up your own problem all you want, but I can't understand why you won't accept that [b]we aren't dealing with your definition of "unexpected."[/b] I don't think that problem is solvable, anyway.

By the student's reasoning, they should expect an exam everyday. The problem comes up because what is required to expect a pop quiz is never given, therefore the student is applying his own definition of expect to the statement, whereas the teacher may have an entirely different definition of expect. Expect needs to be defined, because the two definitions are whether it is possible (student's definition) or an event will occur because it must occur (teacher's definition).

The students argument is based on the fact that if it is possible that there is a pop quiz on a given day then the student's will expect it, while the teacher's argument is that they can only expect the pop quiz if it is know that it must take place.

The teacher gave the pop quiz on (choosing random day) Thursday and surprised the students because the student didn't know that his/her argument had them expect the quiz on all the days......

Hope that made sense.

[quote="JesusFreak197"][quote="Bob"]"Things should be made as simple as possible, but not simpler" -Albert Einstein[/quote]

Umm... are you implying that our solution is simpler than it should be? Why?

[/quote]

Your solution doesn't really answer the question: What's wrong with the student's inductive logic? Imagining myself as one of the students, I can see why before the start of the week, I have no way of knowing for sure whether the exam is on Monday, Tuesday, etc. However, that still doens't answer the question.

Yes it does completely answer the question! The students deduce that it cannot be held on any day of the week, so the exam will be unexpected! Seriously, how can you not understand that? Nothing whatsoever is wrong with the student's deductive logic, but he then fails to realize that he has made the teacher's statement possible, because he will not expect it no matter what day it's on.

OK, let's stay within this definition of unexpected and discuss your question. After the student deduces that the teacher cannot give the exam, the teacher gives it, expecting the students to make such a conclusion, and then the students know the teacher will do that and will expect the exam. It goes back in a circle onto itself.

Okay then, so we've found the problem with the student's logic: he stops when he decides that the exam cannot be given.

[quote="fdump"]so.....let's consider this
let's suppose that the teacher doesn't give them any exam till friday and friday comes.  the students will know that if there will be a test must it must be that day, cause it must be that week and friday is the last day of the week. but since the teacher said it will be a total surprise it means that the test cannot be on friday cause they expect it to be on friday, so the students will know that there won't be any test on friday. therefor, if the teacher DOES give them a test on friday it will be a total surprise for them. :)[/quote]

The teacher said that the exam would be this week, and that the students would not expect it. If the students do expect the exam, though, this does not necessarily mean that the exam will not be given. The teacher made two different statements, and one of them could be false with the other one still true. Similarly, if I said "Pi is a number, and it is rational," you could not conclude that pi is not a number just because it isn't rational.

But the problem said that everything the teacher said was true. Therefore the problem was wrong.

Which was exactly what I said.

By the way, what if the exam truly was unexpected? What if unexpected fits my definition? The problem was about the student's inductive logic, not about whether the teacher was wrong. If we run the induction from Friday to Monday, the exam cannot happen. But if we run the induction from Monday to Friday, there is no problem with having the exam. And we have proven that it doesn't matter which way it goes.


Solve the equation $x^4-5x^3+5x^2-5x+1=0$.

Since $x=0$ isn't a root of the given equation, dividing by $x^2,$ we can rewrite the given equation as follows.

\[ x^2+\frac{1}{x^2}-5\left(x+\frac{1}{x}\right)+3=0.  \]

I don't see how this brings us any closer to the solution :huh: . Or is it something really obvious that I don't see :? ?

Let $a=x+\frac{1}{x}$ and see what happens ;)

Oh, I see now :oops:  . That was something really obvious that I'd overlooked :blush: . I'll let somebody else post the solution though, since you practically led me to it.


So I was doing this one problem that dealt with two blocks, one on top of the other, the bottom one lying on a frictionless plane. We want to find the maximum force that can be applied to the lower block (say in the rightward direction) so that both blocks move together.
A breakdown of the forces acting on the systems is as follows: We have the static friction force acting between the two blocks--it acts on the upper block in the right ward direction as well. Because this is an action reaction pair, there is a force of equal magnitude acting on the lower block in the leftward direction. This force opposes the force we are applying to the lower block. Through calculation, we find that the maximum force that can be applied to the lower block  to maintain our condition is achieved when the static friction force between the two blocks is at its max.
Now, we know from everyday experience, that if we push the bottom block with enough force, the upper block will move in the leftward direction. But if we retrace our steps (by the direct proportionality existing between the two forces), this corresponds to increasing the static friction force beyond its max. I'm confused as to why that would translate to the block moving to the left. I understand that the static friction force acting to the right serves to maintain its stationary state, but how can the top block possibly move to the left.
Sorry if the above description was convoluted. Thanks!

[quote="kevmo314"]Since the top block is sliding on the bottom block, it's kinetic friction that's causing the top block to move, not static friction.  In either case, the static friction between the blocks is causing the top block to stay stationary relative to the bottom block, not to the ground, so it's trying to force the top block to keep up with the bottom one.  With enough force, that friction isn't strong enough, thus it switches to kinetic friction, which produces the same effect but fails to keep the top block accelerating at the same rate as the bottom.[/quote]

Thanks for the reply.
I'm still a little confused as to how kinetic friction is causing the top block to move. Could you explain that a little more?

Remember this always:
[b]The role of friction is to prevent [u]relative motion[/u] between the two surfaces in contact which are experiencing the frictional force.[/b]


The graph of $ x^2 \plus{} y \equal{} 10$ and the graph of $ x \plus{} y \equal{} 10$ meet in two points. The distance between these two points is:

$ \textbf{(A)}\ \text{less than 1} \qquad
\textbf{(B)}\ 1\qquad
\textbf{(C)}\ \sqrt{2}\qquad
\textbf{(D)}\ 2\qquad
\textbf{(E)}\ \text{more than 2}$

[hide="My Solution"]
Solving both equations for $ y$ and equating, we obtain $ \minus{} x^2 \plus{} 10 \equal{} \minus{} x \plus{} 10\implies x^2 \minus{} x \equal{} 0\implies x(x \minus{} 1) \equal{} 0$.

Substituting each value of $ x$ back into the linear equation, we find that the order pairs satisfying the system are $ (x,y) \equal{} (0,10),\,(1,9)$. The distance between the two points is $ \sqrt {(0 \minus{} 1)^2 \plus{} (10 \minus{} 9)^2} \equal{} \boxed{\sqrt {2}\text{ (C)}}$.

[/hide]


Hi guys,
I was trying to solve these problem and I was wondering if I am right or not. I got two solution too

In How many ways can the elements of n be permuted if 1 is to precede
2 and 3 is to precede 4?  I got (n+1! / n!)

and

In how many ways can the elements of [n] be permuted if 1 is to precede
both 2 and 3? I got (n-1!)

thanks in advance!

Both answers are wrong, as you could easily have discovered by checking some small value of $ n$ (say, $ n \equal{} 4$ in both cases).  In both cases, there is a natural way to collect all the permutations of $ [n]$ into groups, each group having the same size and containing one "good" permutation.  In both cases, the grouping should depend on the relative positions of the numbers mentioned in the problem (1, 2, 3 and 4 in the first problem; 1, 2 and 3 in the second).

Equivalently, you need to argue what fraction of all permutations of $ [n]$ are good in each case -- since this involves the relative position of a fixed collection of elements, this fraction actually turns out to be independent of $ n$.


geometry
rectangle
search
number theory unsolved
number theory

We have a square table of $ k^2\minus{}k\plus{}1$ rows and $ k^2\minus{}k\plus{}1$ columngs, where $ k\equal{}p\plus{}1$ and $ p$ is a prime number.  For each prime $ p$, give a method of distributing the numbers $ 0$ and $ 1$, one number in each square of the table, such that in each row and column there are exactly $ k$ zeroes, and moreover no recatngle with sides parallel to the sidse of the table has zeroes at all four corners.

The solution I see is using the projective plane of order p, but I don't understand what this means.

Could someone explain this and hence be able to solve the problem.

Have a look here
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=518076053&t=16699

Pierre.


Thanks !
 As I ahve seen on the links , all solutions are on the same way as the official solutions.

 BAbis

[quote="stergiu"]See the figure! You will enjoy this nice problem.[/quote]


Demonstrate that for any 5 numbers not necesarily different at least the addition of three of them would be a multiple of 3.

Please explain it step by step and clear :)

[hide="Solution"]Obviously, there can be at most two numbers divisible by any given number, because if three numbers share the same remainder modulo $ 3$, then we add those three numbers together, which is equivalent to $ 3k$ modulo $ 3$, or $ 0$ modulo $ 3$.

Therefore, there are two cases: either there exist three numbers with the same remainder modulo $ 3$, or not. The first case is handled as above. We now consider the second case.

Since we have to have $ 5$ numbers, we must have at least one number equivalent to $ 0$ modulo $ 3$, one number equivalent to $ 1$ modulo $ 3$, and one number equivalent to $ 2$ modulo $ 3$ by the above logic. Add those three numbers together; they are equivalent to $ 0$ modulo $ 3$.

Therefore, within a set of $ 5$ arbitrary integers, one can always select $ 3$ integers such that their sum is divisible by $ 3$.[/hide]

[quote="Truffles"]@dark: The numbers don't have to be distinct.[/quote]

It doesn't matter. We just need to look at the remainder mod 3, of which there are only 3 possibilities: 0,1,2. Then, either there are 3 of one remainder, or by pigeonhole, one of each, so we win.


good morning
i'm looking for a good dictionary for maths from english to greek!

Maybe members of the [url=http://www.mathlinks.ro/Forum/index.php?f=371]Greek Community[/url] on AoPS-MathLinks might help you.


number theory
prime factorization
number theory unsolved

Let $ m_1$, $ m_2$, $ \cdots$, $ m_r$ (may not distinct) and $ n_1$, $ n_2$ $ \cdots$, $ n_s$ (may not distinct) be two groups of positive integers such that for any positive integer $ d$ larger than $ 1$, the numbers of which can be divided by $ d$ in group $ m_1$, $ m_2$, $ \cdots$, $ m_r$ (including repeated numbers) are no less than that in group $ n_1$, $ n_2$ $ \cdots$, $ n_s$  (including repeated numbers).

Prove that $ \displaystyle \frac{m_1 \cdot m_2 \cdots m_r}{n_1 \cdot n_2 \cdots n_s}$ is integer.

[hide]
By taking d = 1, we get $ r \ge s.$ Let $ a_i, b_i$ be the exponent of the greatest power of an arbitrary prime p that divides $ m_i,n_i,$ respectively. We order these numbers, so $ a_1 \ge a_2 \ge \ldots \ge a_r$ and $ b_1 \ge b_2 \ge \ldots \ge b_s.$ Then, we prove that if $ 1 \le i \le s,$ then $ a_i \ge b_i.$ Assume that $ a_i < b_i.$ Let $ d \equal{} p^b_i.$ Then, we get a contradiction, because since the $ a_i,b_i$ are ordered, at least i $ n_k$s are divisible by d, but at most i-1 $ m_k$s are divisible by d, a contradiction, so the power of p in $ m_1m_2\ldots m_r$ is clearly greater than or equal to the power of p in $ n_1n_2\ldots n_s.$ This is true for every prime p, so QED.[/hide]

[quote="pythag011"]
By taking d = 1, we get $ r \ge s.$[/quote]

Just to point out that this isn't valid, since $d>1$, but pythag didn't use it anyways.

Let $v_p(a)$ be the exponent of p in prime factorization of a. For each integer d let $M_d, N_d$ be the number of $m_i, n_i$ divisible by d respectively. Then clearly $v_p(m_1m_2...m_r)=M_p+M_{p^2}+...+M_{p^n}+...$ and $v_p(n_1n_2...n_s)=N_p+N_{p^2}+...+N_{p^n}+...$. But since $M_d\ge N_d$ for all d, $v_p(m_1m_2...m_r)\ge v_p(m_1m_2...m_r)$ and we're done.


pigeonhole principle
combinatorics unsolved
combinatorics

We have $ n\times n$ board.Find the maximum number of knights on the board,such that for every knight there exist non-occupied square,which knight could attack.Solve the problem for $ n\equal{}6,7,8,9$.

Is there any solution?I think that it is not hard to solve this problem for $ n \equal{} 6$?Thank you. :)

I can't believe that nobody solved this famous problem.Maybe somebody has any ideas...

If we put 4 knigths in a 4-squares square leaving a 32-squares cross whicch are either occupied or "attacked", that will work for n=6, by leaving the 4 squares in the corner alone. We can put 4 knights so there are 8 knights. Let's assume we can have an arrangement with 7 knights. These 7 knights must cover 29 squares. By pigeonhole principle, some knight must attack 5 squares but we cant have a knight attacking only 5 squares but 6. Then one knight attacks at least 6 squares. Then 6 knights must attack 23 squares, and some knight attacks less than 4 squares.

Case 1.- Suppose it attacks 3 squares. then we have that 5 knigths attack 20 squares, and some knight attacks 4 squares. but no knight can attack only 4 squares, so these 5 knights will attack either 3,6 or 8 squares. if so, there must be an even number of knigths that attac only 3 squares. If we have them, the number of knights covering 3 squares must be more than two, so we have 4 knigths that attack 3 squares and one attacking 8 squares. There are only 8 squares (the squares adjacent to the corners) that can cover 3 other squares, there are 5 knights in these squares, by pigeonhole principle, the two squares adjacent to some corner are occupied, but there is one square that both knights attack. So it's not all covered.
-Now suppose that we dont have knights that cover 3 squares, so we must have one attacking 8, one attacking 6 and three attacking 2. These three are in the corner. Consider the knight that attacks 8 squares. It can be placed in only 4 squares, the ones in the center, and given the 3 corners, we can only put it in one place, the further square to the missing corner. Then one corner is not covered. There are only 2 places from where we can attack these corner and also these places have 6 squares to attack, so we must put the 6-square knight in one of these places, but these places are attaaced by the 8 squares knight. So we cover one square twice, so we dont cover all.

Case 2.- Suppose it attacks 2 squares. Then we have 5 knights attacking 21 squares. By pigeonhole principle, one knight attacks 6 squares, so 4 knights attack 15 squares. By pigeonhole one knight attacks less than 4 squares so we have knights that attack either 2,3,6 or 8 squares. We must have at least one 3-squares knight, so we haver 12 squares for 3 knights. There may be 1 8-knight, 2 2-knights or 1 6-knight and 3 2-knights. In first case, we leave one corner without covering, and there is an 8-knight so we can conclude as case 1.  Now suppose the second distribution, we have the 4 corners with knight, 3 6-knights and 1 3-knight. SInce 6-knights attack corners, there are three squares twice covered. So it's not all covered.

Then it cant be done with 7 pieces and 8 is the minimum for n=6.

We can make a cross as described in my earlier post (a 32-cross with 4 knights in a 4-squares square) and we'll notice that such a cross can be put ((n/6)(n/36)(n/216)...)((n/6)(n/36)(n/216)...) times in a nXn board. then we may take cases (a lot of cases) where the knight dont cover squares...[/quote]


prove that given a triangle ABC where AB>=AC>=BC, then given any point P inside triangle ABC, AP+BP+CP<=AC+BC is always true iff ABC is equilateral

edit: fixed sign

This doesn't appear to be true...

[hide]Suppose $P$ is the intersection of the medians (which are also angle bisectors and altitudes) of an equilateral triangle with side length $2\sqrt{3}$.

Extending the medians to the opposite sides splits each into parts of length $\sqrt{3}$. This is the length of a side opposite a 60 degree angle in a 30-60-90 triangle. Here $AP=BP=CP=$ the length of the hypotenuse, 2.

But if the if statement were true, then $6>4\sqrt{3}$ would have to be true. And it isn't...[/hide]


geometry
3D geometry
sphere
trigonometry

A large sphere is ona  horizontal filed on a sunny day.  At a certain time the shadow of the sphere reaches out a distance of 10 m from the point where the sphere touches the ground.    At the same instant a meter stick (held vertically with one end on the ground) casts a shadow of length 2 m.  What i the radius of the sphere in meters?  (Assume the sun's rays are parallel and the meter stick is a line segment.

(a) 5/2

(b) 9-$ 4\sqrt{5}$

(c) $ 8\sqrt{10}$-23

(d) 6-$ \sqrt{15}$

(e) $ 10\sqrt{5}$-20

Thanks in advance!

[hide][asy]size(200);pathpen=linewidth(0.7);pair A=(0,0),B=(10,0),C=(4*5^.5,2*5^.5),O=(10,10*5^.5-20);D(A--O--B--A--C--foot(C,A,B));D(CP(D(O),B));MP("10",(A+B)/2);D(anglemark(B,A,O,10));MP("\theta/2",A,(10,1),fontsize(8));MP("r",(O.x,O.y/2),E);[/asy]
So $ \tan \frac{\theta}{2} \equal{} \frac r{10}$, where $ \theta$ is the smallest angle of a $ 1\minus{}2\minus{}\sqrt{5}$ right triangle. Hence $ r \equal{} 10 \tan \frac {\theta}2 \equal{} \frac{10\sin \theta}{1 \plus{} \cos \theta} \equal{} \frac{10}{5\plus{}\sqrt{2}} \equal{} 10\sqrt{5} \minus{} 20$.
[/hide]


2002 AMC 12B #14:
Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?
A) 8 B) 9 C) 10 D) 12 E) 16
The solutions packet says that the answer is A. However, it then proceeds to explain that the answer is D) 12 (which is what I got). Is the 'A' a typo or am I missing something?

The answer is D, as confirmed by me, the book, and you.  It must have been some sort of typo (A is sort of close to D on the keyboard? :D).

Right. Now a question I've been curious about...is the maximum C(n,2)*2 for n circles always possible, even for really large n?

Yes.  You can actually show this visually, but I'm too lazy to mess with paint and get a diagram up.  Intersect two circles so that they almost coincide.  Then keep going like that.  You might be able to see what I mean.

Well, there are $\binom{n}{2}$ pairs of circles and each pair has a maximum of $2$ intersections, so then the total is $2\binom{n}{2}$? I think it is achievable with some careful drawing (though it may not be very easy or fun).

In September, I sent an e-mail to the person in charge of AMC publications pointing out this mistake, but I never got a response.


Find all pairs $ (m$, $ n)$ of integers which satisfy the equation
\[ (m \plus{} n)^4 \equal{} m^2n^2 \plus{} m^2 \plus{} n^2 \plus{} 6mn.\]

see, by checking divisibility of squares by $ 3$ we obtain remainder as $ 1$ or none.

checking the above equation, either both $ 4m$ and $ n$ need to be multiples of $ 3$ or then w.l.g. $ m\equiv 1mod3$ and $ n\equiv 2mod3$.

in the first case the power 4 in LHS is high, compulsing $ m$ and $ n$ to be multiples of $ 9$ and so on, ultimately solving only at values $ >N$.

the second case is possible...and needs to be worked..
ll be back soon

My solution:

Denote $ (m \plus{} n)^2 \equal{} x$ and $ mn \equal{} y$. Then the original equation is equivalent to $ x^2 \minus{} x \equal{} y^2 \plus{} 4y$. Solving this equation wrt to $ y$, one gets $ y \equal{} \minus{} 2 \pm \sqrt {x^2 \minus{} x \plus{} 4}$. Hence $ x^2 \minus{} x \plus{} 4$ has to be a square. But $ (x \minus{} 1)^2 < x^2 \minus{} x \plus{} 4 < x^2$ for $ x\ge 5$. Hence there are just 5 easy cases left.


trigonometry function calculus integration conics hyperbola

how can hyperbolic sin and cosine functions be used in problem solving?
(probably not HSB its from AoPS volume 2 chapter 9)

They're inticately connected with exponential functions.  In calculus, many integrals result in the hyperbolic functions.

And there are also used to represent hyperbolas in parametric form (I'm not really sure if this is helpful though).


Assuming the expression converges, determine the largest integer n with $ n \leq 9, 000, 000$ for  which

$ \sqrt{n\plus{}\sqrt{n\plus{}\sqrt{n...}}}$ is rational.

My attempt to solve it:

[hide]Let $ N\equal{}\sqrt{n\plus{}\sqrt{n\plus{}\sqrt{n...}}}$, then

$ N\equal{}\sqrt{n\plus{}N}$
$ N^2\equal{}n\plus{}N$
$ N^2\minus{}N\minus{}n\equal{}0$

$ N\equal{}\frac{1 \pm \sqrt{1\plus{}4n}}{2}$

For N to be rational, all we need is that $ 4n\plus{}1$ is a perfect square, but I dont know what to do from there. Maybe you do $ 4n\plus{}1\equal{}x^2$ but I still wouldn't know what to do from there...[/hide]

[hide="In particular..."]We have $ N \equal{} \frac {1 \plus{} \sqrt {1 \plus{} 4n}}{2}$ and so $ N$ is integral iff $ 1 \plus{} 4n$ is an odd square.  :rotfl: Note that all odd squares are congruent to $ 1$ mod $ 4$ anyhow, so we can have for the maximum $ 1 \plus{} 4n \equal{} 9000000\cdot4 \minus{} 11999\Rightarrow n \equal{} 8997000$.[/hide]

[quote="Vworldv"]maximum $ 1 \plus{} 4n \equal{} 9000000*4 \minus{} 11999\Rightarrow n \equal{} 8997000$.[/quote]

What did yo do there?

[hide="In Detail"]$ \sqrt{4n\plus{}1} \equal{} m$ for some odd $ m$.
The upper bound on $ m$ happens when $ n \equal{} 9000000$.
We find $ \sqrt{4 \cdot 9000000\plus{}1} \approx \sqrt{4 \cdot 9000000}\equal{}6000 \geq m$
So the highest odd $ m$ we can make is $ 5999$.

We have to solve $ \sqrt{4n\plus{}1} \equal{} 5999 \implies n \equal{} 8997000$

What he did there was just to say that $ 5999^2 \equal{} 6000^2 \minus{} 6000\minus{}5999$[/hide]

Well, the largest value for $ n$ is $ 9000000$ by the given, but unfortunately that will not yield a perfect square... the number that it does yield is $ 1\plus{}4\cdot9000000$. The greatest square below that is $ 4\cdot9000000$, but this is not odd. The greatest square below that is $ \left(\sqrt{4\cdot9000000}\minus{}1\right)^2\equal{}4\cdot9000000\minus{}11999$.  :)


I don't think this is the fastest way to do this
[hide]
First, find the least common denominator of 3, 5, and 13, which is 195. B is at 130, and you want A to be between 91 and 169. Since A is a multiple of 15, just find the number of multiples between 91 and 169 which is 5. Since there are 14 numbers to pick, your answer is $\frac{5}{14}$.
[/hide]

[quote="ch1n353ch3s54a1l"][quote="mr. math"]how would u do this problem?

illian will randomly choose a value for P from the integers 0 through 13, inclusive, and will make P/13 the coordinate of point A. The coordinate of point B is 2/3. Points A and B will be on the same number line. What is the probility that point A and point B will be less than 1/5 of a unit form each other. Answer using a fraction in simplest form.[/quote]
[hide]
"A" must be between 2/3-1/5=7/15 and 2/3+1/5=13/15. The maximum integer for P is 6 that doesnt make P/13 exceed 7/15 (7 integers) and the miniumum integer for P is 11 that doesnt make P/13 less than 13/15 (13-11+1=3 integers). The probability is (7+3)/14=$\boxed{\frac{5}{7}}$.[/hide][/quote]
How did you find $\frac{6}{13}$ and $\frac{11}{13}$ as the maximum and minimums though? Did you use a calculator? Also, you counted each number twice.

[hide]Ignite is right.

The answer is $\frac{5}{14}$ because the minimum value for P is 7 and the max value is 11 :gleam: [/hide]

[hide]Min is 7, Max is 11, so the answer is 5/14.[/hide]


Here is a question.

Let $C$ be circle and $p,q$ lines. Let $t$ be tangent of a point $P$ on a circle $C$.
Let $X = t \cap p$ and $Y = t\cap q$. Is map $X\mapsto Y$ projectivity from $p$ to $q$,
when we move $P$ on circle?

That's not even a well-defined map: take a point on the line $p$ which lies outside the circle. There are two tangents through this point to the circle, so the point we chose has two distinct images through this map.


Find all $(x,y,t,u) \in (\mathbb{N}^*)^3$ such that $\frac{x^2+y^2 \pm 8}{t^2+u^2}=5$ and $x+t=2u$.

I am pretty sure that something is wrong with this problem, since we know that $t^2+u^2$ represents an arbitrary natural number, which doesn't contain prime factors of the form $4k+3$ in odd power. The same for $x^2+y^2$. So the problem reduces to $A\pm 8=5B$, where $A$ and $B$ contains prime factors $4k+3$ in even power. Do you think we can analyze it? ;)

Oops! :blush: I have forgotten this hypothesis: $x+t=2u$.

Then the problem is trivial :), since it reduces to $(2t+u)^2=y^2\pm 8$. So $2t+u=1$ and $y=3$ OR $2t+u=3$ and $y=1$.

I think that the (only) condition  $y-x=t+u$  leads to an interesting problem...
[i]Collins[/i]: What is the source of your question?

I tried to build a puzzle...But I failed :(
Perhaps blang's condition is interesting?


The desert temperature, $ A$, oscillates daily between $ 30$ F at 5 am and $ 70$ F at 5 pm. Write a possible formula for $ A$ in terms of $ t$, measured in hours from 5 am.

If $ t$ is the number of hours after $ 5$ A.M., $ A(t) \equal{} 30 \plus{} \frac {10}3t$ works.

Edit: Sorry, I was being stupid. :blush:

[hide="Using the properties of the sine function"]Define $ t=0$ as 5:00 AM, and $ t=12$ as 5:00 PM, so that $ t$ is measured in hours. Let $ A=a\sin(bt+c)+d$

First off, let's consider that $ A$ must oscillate between $ 30$ and $ 70$. We have $ -a\leqq\sin t\leq a$, so $ a$ must be $ 20$.

We're shifted up slightly. We want the minima of $ A$ to be $ 30$, and so far we have $ -20\leq\sin t$, so $ d=20+30=50$.

Now the only issue is that $ A(0)=a\sin0+d=d=50$, not $ 30$. For this reason, we have to introduce a constant $ c$ so that $ A(0)=20\sin c+50=30\implies\sin c=-1\implies c=-\frac{\pi}{2}$.

Now $ A=20\sin\left(bt-\frac{\pi}{2}\right)+50$. Our concern now is whether or not $ A=70$ when $ t=12$.

So far, $ A=20\sin t+50$. Evaluated at $ t=12$ we have $ A=20\sin\left(12-\frac{\pi}{2}\right)+50$, which is not quite equal to $ 70$. We need to change the coefficient of $ t$ until $ 20\sin\left(12b-\frac{\pi}{2}\right)+50=70$.

Solving for $ b$, we have $ 12b-\frac{\pi}{2}=\arcsin 1=\frac{\pi}{2}\implies 12b=\pi\implies b=\frac{\pi}{12}$.

If you graph this, you'll see that it works fine, though $ A$ doesn't really oscillate between $ t=0$ and $ t=12$. So, instead, we can say that $ b=\frac{\pi}{12}+2\pi n$ for any integer $ n$. The larger $ n$ is, the more oscillations you will have between $ t=0$ and $ t=12$.

Finally, we can use the identity $ \sin\left(\theta-\frac{\pi}{2}\right)=-\cos\theta$ to clean it up a bit.

Our answer is $ \boxed{A\left( x \right)=-20\cos \left( \frac{\pi x}{12}+2\pi x \right)+50}$[/hide]

What about when the values are $ 40 F$ and $ 80 F$ rather than $ 30 F$ and $ 70 F$?

You'd have to recalculate $ a$, $ b$, $ c$, and $ d$, but it shouldn't be too hard to do so.
I'm very lazy, so I'll leave you to try to find a general formula for an oscillating function between $ t\equal{}t_1$ and $ t\equal{}t_2$ ($ t$ being time in hours, of course), for temperatures between, say, $ T_\mathrm{low}$ and $ T_\mathrm{high}$.

I might play around with it and post the solution I get, or write it up in TeX if I'm really feeling up to it  :wink:

In any case, it's a good exercise; you'll be more familiar with properties of sine and cosine graphs than you thought possible after coming up with a function like that one.

These problems sound like homework to me. Maybe we should give hints instead of answers.

Good call.

Edit: Wow this post is spam.


Will we have to cancel tournies and this thread becuase of loss of chat. Its still possible to do tournies without chat. It just will be a little harder

Mm nah, it'll still work. There's PMs, G-mail chat, AIM, and so forth if communication is really needed. The most important part is that we can still play countdowns.

theres no need. chat will be back tomorrow.

and we would be able to do it via PM, etc., so it wouldn't be a concern, regardless.


The price of gasoline has increased from $ \$1.05$ per gallon to $ \$1.41$ per gallon. By how many more cents will the price have to rise to achieve a total increase of $ 40\%$.

If it's currently \$1.05, then it will be a 40% increase when it reaches $ 1.05 \times 1.40 \equal{} 1.47$. Since it increased to \$1.41, there are still $ \boxed{6}$ more cents to go.


A super ball is dropped from 100 feet and rebounds half the distance it falls each time it bounces. How many feet will the ball have traveled when it hits the ground the fourth time?

100
hits ground
50+50
hits ground
25+25
hits ground
12.5+12.5
hits ground

After 4 hits, it has traveled 100+100+50+25=275 feet.


[quote="marko avila"]if my calculations are correct then this problem is correct :

Let $A$ , $B$ , $C$  be angles of a triangle prove that  one of :

$sin(2C-B)$ , $sin(4B-C)$ , $sin(2B-C)$

is the sum of the other two.

i hope its right!! :([/quote]

Afraid it's not true; take $A = 30^\circ$, $B = 60^\circ$, and $C = 90^\circ$.  Then the three quantities are $\sqrt{3}/2$, $1/2$, and $1/2$.


Let a;b;c >0
We have
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b}\plus{}\sqrt{\frac{ab\plus{}ac\plus{}cb}{a^2\plus{}b^2\plus{}c^2}}\ge \frac{5}{2}$

[quote="tranvanluan"]Let a;b;c >0
We have
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \sqrt {\frac {ab \plus{} ac \plus{} cb}{a^2 \plus{} b^2 \plus{} c^2}}\ge \frac {5}{2}$[/quote]
by cauchy we have
\[ \sum\frac{a}{b\plus{}c}\geq\frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}\]
so we must prove that
\[ \frac{a^2\plus{}b^2\plus{}c^2}{2(ab\plus{}bc\plus{}ca)}\plus{}\sqrt {\frac {ab \plus{} ac \plus{} cb}{a^2 \plus{} b^2 \plus{} c^2}}\geq\frac{3}{2}\]
but by AM-GM we have
\[ \frac{1}{2}(\sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{(ab\plus{}bc\plus{}ca)}}\plus{}\sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{(ab\plus{}bc\plus{}ca)}}\plus{}
\frac{a^2\plus{}b^2\plus{}c^2}{(ab\plus{}bc\plus{}ca)}\geq\ 3)\]
Nice inequality :wink:

Your solution has a little mistake. But it's right and nice.

[quote="M.A"]by cauchy we have
\[ \sum\frac {a}{b \plus{} c}\geq\frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)}
\]
[/quote]
You mean by Cauchy-Schwarz's Inequality? How did you apply it to prove this?

It is known as the Engels form of Cauchy-Schwarz:
$ \sum_{i\equal{}1}^{n}\frac{a_{i}^{2}}{b_{i}}\geq \frac{(\sum_{i\equal{}1}^{n}a_{i})^{2}}{\sum_{i\equal{}1}^{n}b_{i}}$
Here:
$ \sum_{cyc}\frac{a}{b\plus{}c}\equal{}\sum_{cyc}\frac{a^{2}}{ab\plus{}ac}\geq \frac{(a\plus{}b\plus{}c)^{2}}{2(ab\plus{}bc\plus{}ca)}$ :wink:


calculus
derivative
integration
algebra
system of equations

Hey,

I'm a newbie to all this math competition stuff. I'm a 9th grader, currently taking and going to be competing in Algebra 2.

A friend of mine recently told me that for one of the things we are going to be competing in [mu alpha theta], for the individual part, you cannot use a graphing calculator. I only own a graphing calculator, so I'm wondering if anyone can tell me what is the best/most powerful/superior scientific calculator? I am just going to be using it for Mu alpha theta individual tests.

:) Thanks. And when you tell me which one is good, can you also give me an explanation why? I'd like to know why things are the way they are.

Oh yeah,  a friend of mine recently told me that this one is the best: http://www.officedepot.com/ddSKU.do?level=SK&id=397554 and that a plethora of people in comps use that scientific calculator. Is his speculation true? Is that the best?

Thanks in advanced.

My personal preference is the TI-36x Solar, but taht was 6 years ago.  But that's just me, because I know where all the buttons are.  The one that you have a link to looks like it should suffice, though.

Well, the best calculator that you can get is a well-trained brain.  On most math competitions the calculator should only help for checking the adding, subtracting, dividing etc.  I, however, understand you wanting the best calculator.  I don't really know much about scientific calculators but the link you posted seems to be very good for a scientific one.  I wouldn't worry too much about the calculator.  That time could go to practicing for the competition more.

[quote="kool_dudy"]That would have been a nice answer for a math competition but he needs it for school.

On another note, the scientific calculator I've been using right now is a fx-991ms.  It's been working great for me but I'm not too sure about the "best" scientific calculator.[/quote]

From what I gathered he will only be using it for the Mu alpha theta competitions.  Plus he already has a graphing calculator for school.

I also live in FL, and compete in Mu Alpha Theta..

The bad thing about this competition is that so much of it calculator dependant. So much number crunching, rounding, memorization, and not much problem solving.

If you want a good MAO calculator, Casio's new one: fx 115 MS

I didn't notice the link you gave was the one I mentioned..

yes, that calculator will get you through Calculus. It has a lot of stuff..

[quote="joml88"][quote="kool_dudy"]That would have been a nice answer for a math competition but he needs it for school.

From what I gathered he will only be using it for the Mu alpha theta competitions.  Plus he already has a graphing calculator for school.[/quote][/quote]

I will be using it both for school and for Mu Alpha Theta competitions. All the Mu Alpha Theta tests that I have practiced on are usually calculator-dependent. Plus, I'm going to need a calculator to check my work, because with it being timed and all, I won't have time to hand-do the problems and see if I got them right or if I got them wrong.

Mathkid: Yeah, My friend recommended me that one because it does everything a standard scientific calculator does, plus it does factoring and system of equations [I'm not sure what system of equations is, I'm guessing its equations with 3 variables?]

So.. It's between fx-115ms and the fx-991MS? Hrrmmm, here's a comparison of them I found: http://www.casio.co.jp/edu_e/product/eu/scientific/eu_fx_2line_a.html (Not that I know what it means or anything)

Yes, 3 var. systems.. no factoring. You'll find it useful in pre-calc when you find that it can do complex #s, and also numerical differentiation and integration for calculus.

Well, I'm gonna get my fx-115ms tomorrow then, unless anyone objects tonight or tomororw mroning [ill check it one last time before i buy it]

Thanks for the help.

seriously guys? the so called "power" on a scientific calculator won't help you at all for math competitions. Don't succumb to noob tactics. Your time is much better spent doing actual real math, not this calculator nonsense.

Why revive an 8 year old thread, especially when the policy isn't even the same for Mu Alpha Theta tests anymore? It was changed nearly 6 years ago to no calculators allowed for nearly all tests.


this is a totally random topic and i dun care if the mods deletit it, but....who lykes the simpsons? HOMER IS SOOO COOL.  so is bart

I think the Simpsons are ok. I haven't watched them in like.., well, since school started. What are your favorite episodes??

[quote="Rep123max"]Simpsons were funny like 3 years ago, but now it is all about South Park and Family Guy. Those are the funniest cartoons and I can't wait till Friday to see the new movie by the South Park creators. Also, besides the great news that Family Guy is coming back next year, the creators have a new show that will be on Fox called American Dad I think.[/quote]

Amen.

yea but theyre still playing now and um...its in the genius book of world records for sumthing. 4got wat. and my favorite episode is when lisa got framed. so much stuff happened in it and it was so funny.

I'm not allowed to, nor do I [i]want[/i] to, watch the Simpsons.

ive never watched south park. but wats wrong with the simpsons?  the "innapropriate" stuff make it funny.  and all the characters are so cool and the  picture is really good cause its nonrealistic. um...yes...fits with the plot.

I don't like the references it makes, and such. I just have no desire to watch it. It seems as if it has no moral values, and is a completely pointless show. Thus, I don't watch it.

Likewise, I'm not much for either the Simpsons or South Park.  Too much lowbrow humor for my tastes.  Not saying its bad, or trying to offend those who watch either show, I just don't find it to my liking.

well yea moral values are...important, but sum ppl just like comedy and moral values ruin it for sum shows.  and sum episodes do have morals lyke the one where lisa gets 2 see her future with this british guy. but mostly its just a way of relaxing ppl from the daily "MORALS" of everyday life and not be such a sensible person. otherwise thats just pointless. what? nonsensiblilty is important too!!

[quote="sunnyray"]OMG!!! i HATE the simpsons, i think South Park is JUST as DISTURBING, and ive never heard of Family guy[/quote]

You've never heard of Family Guy? It's so hilarious. Same thing with South Park.

[quote="dinoman"]simpsons is funny but i dont watch it everday, maybe once or twice a week or so
favorite episode? hhmm.....that zorro one is good, or that star wars one
the halloween ones are cool too
homer is hilarious
"DOH!!!!!"
 :D[/quote]

FINALLY someone who actually appreciates the finer points of the simpsons. and homer  "mmmmmmm donuts"

[quote="RobinHood3000"]When on earth did I bring up morals?  Morals plus comedy gives you 7th Heaven.  I was talking about shows like Spin City.[/quote]

Hey, there are still a bunch of great sit-coms out there.

[quote]drink beer everyday![/quote]

My dad drinks a beer every day during the summer. Got an issue with that?

[quote="Scrambled"]its just a fault and i dont think gameworld thinks mhes a good role model for one of the most widely watched shows[/quote]

But then again, we must remember we live in an age where people pay money to watch a movie about people humiliating and hurting themselves. (Jackass)

The Famous Last Words of a Jackass:

"Hey!  Watch this!"

[quote]its just a fault and i dont think gameworld thinks mhes a good role model for one of the most widely watched shows[/quote]

the most watched shows rarerely star role models. i hope no one watches the simpsons (or for that matter, almost every show on tv) to find a role model.

[quote="childoftheflam"]you people need lives, i mean come on, the simpsons? that is a total waste of time, but who am i to judge waste your own brains[/quote]

That was totally uncalled for. What do you waste your time on?

[quote="childoftheflam"]you people need lives, i mean come on, the simpsons? that is a total waste of time, but who am i to judge waste your own brains[/quote]

sum ppl wouldnt call the simpsons a "waste" i do have other things 2 do besides watch tv geez arent sum ppl entitled to do unhelpful time consuming things? but since this thread is about the simpsons everyone has a right to offend it i gess. but it wont come from me.

and for gameworld7's post (if it hasnt been deleted): *hint hint* maybe thares a reason. even tho im not a mod.

I love the early Simpsons episodes (I have the seasons 1-5 on DVD), but the writers are running out of ideas (well obviously after over 300 episodes). The jokes are still funny but predictable :?
My favourite character is definitely Homer:
[list]Lisa, vampires are make-believe, like elves, gremlins, and eskimos

Marriage is like a coffin and each kid is like another nail.

I don't have to be careful, I've got a gun[/list]
More quotes on [url=http://en.thinkexist.com/]www.thinkexist.com[/url] (Dan Castellaneta quotes).
I hope the 17th season is going to be an improvement. :D

[quote="xxreddevilzxx"]u shouldnt revive old threads.[/quote]

and why not??

[quote=RedFireTruck][quote=ythomashu]because it is not good for the forums[/quote]

i know right. people should stop bumping. it clogs stuff.[/quote]

YOU HYPOCRITE