Let $\chi$ be a nonempty subset of the rational numbers which is bounded from above.  Prove that there exists $\alpha\in\mathbb{R}$ such that $\alpha =\sup\chi$.

The solution depends in a crucial way on the meaning of $\mathbb{R}$.

$\mathbb{R}$ means the set of all reals.

That misses the point. This could be the [i]definition[/i] of real numbers. With the Dedekind cut construction, it's very close. With some other definitions/constructions, it takes some work.

The definition we're using is that $\mathbb{R}$ is the set of equivalence classes of Cauchy sequences of rational numbers.  If $x=\left[\left\{x_{n}\right\}\right]$ and $y=\left[\left\{y_{n}\right\}\right]$, then $x\mathcal{R}y$ if and only if $\left\{x_{n}-y_{n}\right\}$ is a null sequence.

Claim: for any $n\ge 1$ there exists $a_{n}\in\chi$ such that $a_{n}+\frac{1}{n}$ is an upper bound for $\chi$.

Proof: if no such $a_{n}$ exists, then you can find an infinite sequence $b_{i}\in\chi$ such that $b_{i}\ge b_{i+1}+\frac{1}{n}$, which contradicts the boundedness assumption.

It remains to prove that $\alpha=[\{a_{n}\}]$ is the least upper bound from $\chi$. I will leave this to you (note that $\alpha=[\{a_{n}+1/n\}]$).

I know Rudin's proof using cuts, but I'd like to do something different.  Part of the following was given to me as a starting point for a different proof:

Take two sequences of rational numbers $\chi =\left\{x_{n}\right\}$ and $\beta = \left\{b_{n}\right\}$ such that $\beta$ is the set of upper bounds for $\chi$.  The sets $\chi$ and $\beta$ have the following properties:

[b]1.[/b] $x_{k}\leq x_{k+1}$
[b]2.[/b] $b_{k}>b_{k+1}$
[b]3.[/b] $2^{n-1}\left(b_{n}-x_{n}\right)\leq b_{1}-x_{1}$

Using this information, how might I prove that $\beta$ has a least element?

The least element of $\beta$ is the real number $[\{b_{n}\}]$. Indeed, $[\{b_{n}\}]\le b_{n}$ for any $n$, and for any $x\in\chi$ you have $x\le [\{b_{n}\}]$ because $x\le b_{n}$ for all $n$.


abstract algebra
group theory
superior algebra
superior algebra solved

a..Prove that Q is a torsion free Z-module but not free.
b..Prove that Q/Z is an infinite torsion Z-module. Describe it is
    n-torsion submodule.
c..Let A subset of Q be a finitely genetared nonzero subgroup.
    Show that A is isomorphic to Z.

a: obviously Q is torsion-free. each pair of rational numbers is linearly dependant over Z, so that Q cannot be free.
b: Q/Z has the rational intervall [0,1[ as a system of representatives, in particular its infinite. every element has the form a/b + Z, where a,b are coprime. it's easy to see, that b is the order of this element. from here, the rest is clear ...
c: show that the subgroup generated by a1/b1, ... , an/bn is already generated by a single element, namely kgV(a1,...,an)/ggT(b1,...,bn).

please think about it by yourself ...


Ok, so there was a list of stuff in the packet. But some stuff is unclear:

Are bicycles allowed?

[quote="not_trig"][quote="27r"]Why would you need a bike?[/quote]

walking is obviously superior[/quote]

false. biking is faster.

Now I just have to find someone to carry me... I got it! I'll pretend that I have a rare disease called "polioitis."  It is similar to polio, but causes me to not be able to be in a wheelchair because if I travel in a wheelchair, I'll start to bleed internally... yeah..

I mean... I HAVE POLIOITIS! I NEED A PERSONAL ASSISTANT TO CARRY ME EVERYWHERE!!

/me gets a miniature Barney to carry connaisance around.

Yay! its Barney! Tell me a story Barney!

Anyways, I really don't think that a bike is necessary... everyone else is going to be walking b/c they don't have a bike, and your going to miss the wonderful social opportunity to socialize w/ppl on the walk to lunch, dinner, classes, breakfast, etc.  And, pretend you have polioitis...

Don't bring a bike, if you haven't figured out already.

How would you bring a bike in the first place?
I mean, I don't think the airline company would let you (unless you live really close to UTD, as in driving distance)

ak909090 lives in driving distance from Santa Cruz :)

haha man there are a lot of useless replies here. I am not sure what they asked you to bring. Just a few other things to think of:

1. things to organize your class materials (i used a binder+punched paper+3 hole punch)...it actually payed off to organize things because now I still reference my amsp stuff occasionally

2. a Tupperware (you might be sneaky and take a snack for later :D)

If I think of more stuff, I'll post it... (I haven't exactly started packing yet haha)

my cargo shorts probably snuck out at least 50 cookies from the cafeteria last year. huge side pockets sure do come in handy every now and then.

[b][u][size=200]SUNSCREEN!!!!!!!!!!!!!!![/size][/u][/b]

if you dont want to get BURNEDDDD and turn CANDY APPLE RED after spending two hours in the pool after lunch....[/u][/b]

sooozyy. You are supposed to stay inside until after dinner. Then you go into the pool.

1=2
IT GETS WAYYY TOO CRWODEDDD
and all groooosss
andd
plus foood isnt that good so you feel all gross and nauseous afterwards
ahaha whatever
im not returning this year x)

but its not a bad idea to have sunscreen just in case.
i got burned a LOOOT

if you bring sunscreen, I'm assuming that you don't have to bring aloe vera gel... but considering how much people sweat in that heat (remember MC Nats? although that was in Florida...)... should we bring aloe vera gel?

Heh, I'll probably bring some, now that you mention it...

You have two options:

a) Be safe and bring it.
b) Don't have it and require it when you're there, purchase at store :D


vector
superior algebra
superior algebra solved

A,B \in M_n(R), such that there exist a,b,c \in R\(0)
aA+bB=cAB
Proove that AB=BA
---------------------------------------------------------------------
Dividing by a, A+xAB+yB=0 with x,y reals =/=0
(1) A(I+xB)+yB=0 or
(2) A(I+xB)+ y/x(I+xB) - y/xI =0
If I+xB is not invertible ker(I+xB)=/={0} we  have a non zero vector W
such that (I+xB)W=0 from (2) we get -y/xW= 0=>W=0 contradiction
So I+xB is invertible
(1)=> A=-yB(I+xB)^(-1) and  AB=BA

M.Omarjee

If you have a different solution i am interested, thank you.

Divide by c and get xA+yB=AB then add xyI and get 1/xy (A-xI)(B-yI)=I thus the matrices are invertible and are each others' inverse, thus the comute, and 1/xy (B-yI)(A-xI)=I which comes back to xA+yB=BA and we are done. :D it's my (obvious) generalisation of IMC 07/2003. :D

I see that you are translating the whole list moubi :D


BMO 1996

Determine all sets of non-negative integers x, y, and z which satisfy

$2^{x}+3^{y}= z^{2}$

BMO 1995

Find all triples (a, b, c) of positive integers such that

$(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) = 2$

BMO 2000
Given x, y, and z are positive reals and xyz = 32, find the minimum value of $x^{2}+4xy+4y^{2}+2z^{2}$

[hide="1"]
[b]Case 1:[/b] $y=0$. Then $2^{x}=(z+1)(z-1)$ so $z+1$ and $z-1$ are powers of 2 that differ by 2. This is only possible when the powers of 2 are 4 and 2, which means $z=3,\ x=3$.
[b]Case 2:[/b] $x\leq 1$. If $x=1$ there is no solution as the left hand side would be congruent to 3 or 5 (mod 8). If $x=0$ we get $3^{y}=(z+1)(z-1)$. The only powers of 3 that differ by 2 are 1 and 3, so $z=2,\ y=1$.
[b]Case 3:[/b] $x\geq 2,\ y\geq 1$. Working (mod 3), $x$ is even, and (mod 4), $y$ is even. So if $x=2j$ and $y=2k$, then $(2^{j},3^{k},z)$ is a primitive Pythagorean triple and there are integers $m$ and $n$ such that
\[mn=2^{j-1},\ m^{2}-n^{2}=3^{k}\]
From the first equation, $m$ and $n$ are powers of 2; let $m=2^{a}$ and $n=2^{b}$. Then $2^{a}+2^{b}$ and $2^{a}-2^{b}$ are powers of 3, but their difference is not a multiple of 3, so one of them is not a multiple of 3 and must equal 1. Then $2^{a}-2^{b}=1$, so $b=0$ and $a=1$. This leads to $x=4,\ y=2,\ z=5$.

This makes all of the solutions $\boxed{(0,1,2),(3,0,3),(4,2,5)}$.
[/hide]

The second problem (from BMO 1995) was solved here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110648[/url].


calculus
integration
real analysis
real analysis unsolved

Find $ \int\int_D \frac{dxdy}{1\plus{}(x\plus{}y)^4}$,

where $ D\equal{}\{(x,\ y)\in\mathbb{R}^2|x\geq 0,\ y\geq 0,\ x\plus{}y\leq 1\}.$


Find all continious functions $ f: (0,\infty)\rightarrow (0, \infty)$, such that for all positive reals $ x,y$:
          $ f(x)f(y)\equal{}f(xy)\plus{}f(\frac{x}{y})$.

$ y\equal{}1\to f(1)\equal{}2$.
$ x\equal{}y\to f(x^2)\equal{}f(x)^2\minus{}2$. Let $ f(x)\equal{}2chy(x)$, then $ y(x^2)\equal{}2y(x)$. By continiosly $ y(x)\equal{}c(lnx)$ or $ f(x)\equal{}x^c\plus{}x^{\minus{}c}$.

[quote="Rust"]$ y \equal{} 1\to f(1) \equal{} 2$.
$ x \equal{} y\to f(x^2) \equal{} f(x)^2 \minus{} 2$. Let $ f(x) \equal{} 2chy(x)$, then $ y(x^2) \equal{} 2y(x)$. By continiosly $ y(x) \equal{} c(lnx)$ or $ f(x) \equal{} x^c \plus{} x^{ \minus{} c}$.[/quote]
Could you explain each step in details?


A cylindrical log has diameter $ 12$ inches. A wedge is cut from the log by making two planar cuts that go entirely through the log. The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a $ 45^\circ$ angle with the plane of the first cut. The intersection of these two planes has exactly one point in common with the log. The number of cubic inches in the wedge can be expressed as $ n\pi,$ where $ n$ is a positive integer. Find $ n.$

[hide]
This is half the volume of the cylinder with height 12 inches and radius 12/2 = 6 inches:

$n=\frac{V}{\pi}=\frac{\frac12 \pi r^{2}h}{\pi}= \frac12(6^{2})(12)=6^{3}=\boxed{216}$

[/hide]

I'm really confused with this problem just because of the way it's worded.  First of all, why does it need to have a height of 12.  All the problem says is that the the planes of the two cuts form a 45 degree angle and the cuts go all the way through the log.  The cuts could be such that the first cut just goes through the log, parallel to the base, and the second cut goes through one point on the first cut on the edge of the cylinder and then all the way through the cylinder.  Why does the height need to be 12??  The height could just as well be 100 or 200 or 300, unless im getting confused with the meaning of "goes all the way through the cylinder".

No, you've understood it correctly, I think -- I'd just worded my solution awkwardly back in the day.  :roll:

The point I was trying to make is that two such wedges, joined together, form a cylinder of height 12 and diameter 12. We can easily calculate the volume of such a cylinder, and divide the result by 2, to get the volume of a single wedge.

The concept is similar to joining two congruent right triangles along their hypotenuses to make a rectangle, in order to calculate the area of one such right triangle.

How do you know the height is 12? I understand that the diameter would be 12, but how would you deduce that the height is 12 just from the fact that it forms a 45 degree cut?

because 45 degree make isosceles triangle! so sides are same size.


Where did 42 come from? Like I know it’s used in the Hitchhiker’s Guide to the Galaxy as the answer to everything but where did the author get that number? Sorry if this is a wrong place to ask this.

http://en.wikipedia.org/wiki/Answer_to_Life,_the_Universe,_and_Everything#The_number_42

In a nutshell, the author of that book says: "It had to be a number, an ordinary, smallish number, and I chose that one."

don't you see how it has the right twang in it?


induction
combinatorics unsolved
combinatorics

In a $ 2n\times 2n$ board $ (n\in N)$, the cells are all filled by $ 1\times 2$ dominoes, each domino lying either vertically or horizontally. Three kinds of moves are allowed:

(a) For any two horizontal dominoes, one lying directly above the other, the two can be replaced by two vertical dominoes.

(b) For any two vertical dominoes, lying directly next to each other, and there is a horizontal domino lying directly [b]above[/b] them, the two vertical dominoes can be moved [b]up[/b] a cell, and the horizontal domino moving [b]down[/b] two cells' length.

(c) For any two vertical dominoes, lying directly next to each other, and there is a horizontal domino lying directly [b]below[/b] them, the two vertical dominoes can be moved [b]down[/b] a cell, and the horizontal domino moving [b]up[/b] two cells' length.

Find all possible boards such that it is not possible to perform any more moves.

[quote="pbornsztein"]I do not understand the question.... :wink:
Are you asking for boards for which no move is possible?

Pierre.[/quote]

Yes  :D

Call the top left square $ (1,1)$ and the bottom right square $ (2n,2n)$. A horizontal domino covers $ (x,y), (x\plus{}1,y)$ and a vertical domino covers $ (x,y), (x,y\plus{}1)$.

Obviously there needs to be a domino covering $ (1,1)$. Suppose it is horizontal. Then we need a domino to cover square $ (1,2)$. If it is horizontal, move (a) is possible. So it is vertical. Then consider $ (2,2)$. Move (b) is possible if it is vertical, so it is horizontal. We can continue like this and show that for $ 1 \leq k \leq 2n\minus{}1$, $ (k,k\plus{}1)$ must always be vertical and $ (k,k)$ must always be horizontal. But we run into problems at $ (2n\minus{}1,2n)$, where only a horizontal domino fits, and we'll then get a move. Contradiction.

So the domino at $ (1,1)$ is vertical. There needs to be a domino covering $ (2,1)$. If it is horizontal, then by the arguments above, for $ 1 \leq k \leq 2n\minus{}1$, $ (k,k)$ has a vertical domino and $ (k\plus{}1,k)$ has a horizontal, and we run into problems at $ (2n,2n\minus{}1)$ when a vertical domino must be placed and move (b) is possible. So the domino at $ (2,1)$ is vertical. Then move (c) is possible if $ (1,3)$ is horizontal, so it is vertical. By the same laddering argument in the previous paragraph, $ (2,3)$ is also vertical. We can see that the entire first two columns must be covered with vertical dominoes.

Now an induction on $ m$ for a $ 2m \times 2n$ board will show that the entire board must be covered with vertical dominoes if no move is possible.


linear algebra
matrix
linear algebra unsolved

Prove, in general, that (AB)T = BTAT

where T indicates a transpose

Let $A$ be a $m\times n$-matrix and $B$ be a $n\times p$-matrix. Now for every $i,j$ we have
\begin{eqnarray*}(AB)^{T}(i,j)&=&(AB)(j,i)=\sum_{k=1}^{n}A(j,k)\cdot B(k,i)=\sum_{k=1}^{n}B(k,i)\cdot A(j,k)\\ &=&\sum_{k=1}^{n}B^{T}(i,k)\cdot A^{T}(k,j)=(B^{T}A^{T})(i,j) \end{eqnarray*}


function
trigonometry
calculus
calculus computations

This problem is for them who like to calculate and have much free time:
prove that
$ \frac {\sum_{k_{1} \plus{} k_{2} \plus{} k_{3} \equal{} n \minus{} 1}\frac {(2n \plus{} 1)!}{(2k_{1} \plus{} 1)!(2k_{2} \plus{} 1)!(2k_{3} \plus{} 1)!}}{3} \in N$
and
$ \frac {\sum_{k_{1} \plus{} k_{2} \plus{} k_{3} \equal{} n \minus{} 1}\frac {(2n \plus{} 1)!}{(2k_{1} \plus{} 1)!(2k_{2} \plus{} 1)!(2k_{3} \plus{} 1)!}}{3^{p}}$ is not natural number if $ p > 1$
I note that this is not interesting problem. it's only some particular case of some problem

It's not hard- just look at generating functions. The sum is $ \frac14(3^{2n\plus{}1}\minus{}3)$, which is clearly divisible by $ 3$ but not $ 9$ when $ n\ge 1$.

The proof: Write $ \left(x\plus{}\frac{x^3}{3!}\plus{}\frac{x^5}{5!}\plus{}\cdots\right)^3\equal{}\sum_{m\text{ odd}}\frac{a_m}{m!}$; we get
$ a_m\equal{}\sum_{\substack{i\plus{}j\plus{}k\equal{}m\\1,j,k\text{ odd}}}\frac{m!}{i!\cdot j!\cdot k!}.$
As functions, the first expression is $ (\sinh x)^3\equal{} \frac{\left(e^x\minus{}e^{\minus{}x}\right)^3}{8}\equal{} \frac{e^{3x}\minus{}e^{\minus{}3x}}{8}\minus{}3\frac{e^x\minus{}e^{\minus{}x}}{8}\equal{}\frac14(\sinh 3x\minus{}3\sinh x).$ The formula for the sum follows easily.

Yes, here is another way ( like your way) :
$ \sin 3a \equal{} 3\sin a \minus{} 4 \sin^{3}a$
or
$ \sum ( \minus{} 1)^{i}\frac {(3a)^{2i \plus{} 1}}{(2i \plus{} 1)!} \equal{} 3\sum( \minus{} 1)^{i}\frac {x^{2i \plus{} 1}}{(2i \plus{} 1)!} \minus{} 4(\sum( \minus{} 1)^{i}\frac {x^{2i \plus{} 1}}{(2i \plus{} 1)!})^{3}$
or
...thinking...
and we have
$ \frac {3^{2p \plus{} 1}}{(2p \plus{} 1)!}( \minus{} 1)^{p}x^{2p \plus{} 1} \equal{} 3( \minus{} 1)^{p}\frac {1}{(2p \plus{} 1)!} \minus{} 4(\sum_{k_{1} \plus{} k_{2} \plus{} k_{3} \equal{} p \minus{} 1}( \minus{} 1)^{k_{1} \plus{} k_{2} \plus{} k_{3}}x^{2k_{1} \plus{} 2k_{2} \plus{} 2k_{3}}\frac {1}{(2k_{1} \plus{} 1)!(2k_{2} \plus{} 1)!(2k_{3} \plus{} 1)!})$
and we have done
Byt this problem we can say much problem like this by using trigonometric equation


The rectangular stage at Radio City Music Hall in New York
City measures 144 feet wide and 60 feet deep. What is the
number of square yards in its area?

Note: Three feet in a yard, thus we divide the number of feet by 3 to find how long it is in yards.

[quote="Brut3Forc3"]In yards, it is 36 by 20, so it is 720 square yards area.[/quote]

144/3=48...

48*20=960

answer : 960 square yards

Can anyone explain it more clearly and precisely?


Let $ a,b,c>0$ and $ a\plus{}b\plus{}c \leq 3$
$ \frac{a}{\sqrt{b}}\plus{}\frac{b}{\sqrt{c}}\plus{}\frac{c}{\sqrt{a}}\geq 3$

[quote="nguyensilva"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \leq 3$
$ \frac {a}{\sqrt {b}} \plus{} \frac {b}{\sqrt {c}} \plus{} \frac {c}{\sqrt {a}}\geq 3$[/quote]

The inequality does not hold

 a counter-example  is
$ a\equal{}\frac{1}{2},b\equal{}\frac{1}{14},c\equal{}\frac{1}{12}$

then

$ \frac {a}{\sqrt {b}} \plus{} \frac {b}{\sqrt {c}} \plus{} \frac {c}{\sqrt {a}}\minus{}3\equal{}\minus{}0.76388434676257.....$

so...............................

It's false for $ a\equal{}b\equal{}c\equal{}.5$.

[quote="nayel"]It's false for $ a \equal{} b \equal{} c \equal{} .5$.[/quote]

I'm sorry, $ a\plus{}b\plus{}c \geq 3$
easy... sorry...

[quote="nguyensilva"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \geq 3$
Prove that: $ \frac {a}{\sqrt {b}} \plus{} \frac {b}{\sqrt {c}} \plus{} \frac {c}{\sqrt {a}}\geq 3$[/quote]
Let $ a\equal{}x^2,$ $ b\equal{}y^2$ and $ c\equal{}z^2,$ where $ x,$ $ y$ and $ z$ are positive numbers.
Hence, we need to prove that $ \frac{x^2}{y}\plus{}\frac{y^2}{z}\plus{}\frac{z^2}{x}\geq\sqrt{3(x^2\plus{}y^2\plus{}z^2)}.$ :wink:

[quote="nguyensilva"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \leq 3$
$ \frac {a}{\sqrt {b}} \plus{} \frac {b}{\sqrt {c}} \plus{} \frac {c}{\sqrt {a}}\geq 3$[/quote]
This is my solution:
$ LHS \geq \frac {2a}{b \plus{} 1} \plus{} \frac {2b}{c \plus{} 1} \plus{} \frac {2c}{a \plus{} 1} \geq \frac {2(a \plus{} b \plus{} c)^2}{ab \plus{} ac \plus{} bc \plus{} a \plus{} b \plus{} c}$(By Schwarz)
$ a \plus{} b \plus{} c \geq 3$ so we have:$ ab \plus{} ac \plus{} bc \leq \frac {(a \plus{} b \plus{} c)^2}{3},a \plus{} b \plus{} c \leq \frac {(a \plus{} b \plus{} c)^2}{3}$


Explain and/or give the reason for why your avatar was chosen.
This was made kind of as a sister to "What's your name mean" or whatever its callled.

 :D

P.S. I just chose mine. Because I'm from Texas. Texas pride!  :yankchain:

Mine's fairly striaghtforward.  My username is B-tone_party, therefore, my avatar is a baritone.  I don't even play the baritone (at least, not regularly.)

I first played fifa 2005 with Cambridge. This is their logo.

it has no correlation i just chose it randomly

I personally feel I really dont need an avatar....Besides, if I have an avatar, all my posts take that extra bit of time to load...........mainly because of my rather slow connection...

because i look great

lol, j/k,
simply because its my picture :)

[quote="stupidkid"]Because Linux is awsome.[/quote]

Your avatar is awsome as well........where on earth did you find that stuff ?? :D

'Tis my little sisters and my computer, all of which I really like.

I don't really like art, and the only type of art I CAN draw is abstract art! :)

because its prettiful and its got bobbing smilies :lol:

I don't know exactly how to load an  image as an avatar  so I asked  our dog, who  is quite handy with such things ... so he uploaded  his own favorite image . I guess :)[img]http://www.artofproblemsolving.com/Forum/images/avatars/45961890041e4a6671adbd.jpg[/img]

(Yes that's a keyboard and a computer ... and using keyboard is sometimes difficult with paws, ... but he is very handy with mouse.)

I was inspired to draw my avatar after listening to John Cage's [i]4'33"[/i].

i change my avatar like every 2 weeks, i won t bother to check what it is now because in a couple minutes youll probably see a new avatar.

I choose mine because the nuke silo is the most powerfull weapon in Red Alert 2 (my favorite game).

It's a scene from my favourite movie: Requiem for a dream

My avatar is ... now mmm ... true  :wacko:

it's rage against the machine, the best band ever, and it also kinda fits with my screenname  :lol:

The nicest fractal I invented ever (but shrinking it to this size has badly influenzed it).

my is my work i will announce soon and i would like to hear what you will say

Because my signature, name, and avatar are all correlated in a certain way...

Mine is a high resolution pic of where I live  ;)

Well my name is PARTIALLY derived from eminem(a.k.a slim shady) n plus he's looking cool in it.

Mine is a weird game of Weiqi that looks even more horrible than it did at 96x96.


Suppose that 6$ \%$ of the eighth graders and 3$ \%$ of the seventh
graders at Washington Junior High participate in
MATHCOUNTS. There are 1.5 times as many 8th graders as
7th graders at the school. What percentage of the 7th and 8th
graders, taken together, participates in MATHCOUNTS?
Express your answer to the nearest tenth.

Suppose there are $ x$ 7th graders and $ 1.5x$ 8th graders.  Then $ 0.06(1.5x)$ 8th graders and $ 0.03(x)$ 7th graders participate in MATHCOUNTS.  This simplifies to $ 0.09x\plus{}0.03x\equal{}0.12x$, out of $ 2.5x$ total students, for $ \frac{0.12x}{2.5x}\equal{}\frac{12}{250}\equal{}\boxed{4.8 \%}$.


Yeah, because he said "equation." :oops:

you add 1 line to the plus sign to make 545+5+5=555


Is there any quick comparison of the MathCounts and Math Olympiad programs? Obviously one can look at the websites, but does anyone have any experience with both such that they can offer some idea about the advantages and disadvantages of both for a 6th grader? TIA.

MathCounts and MOEMS are somewhat different competitions.

MathCounts gives you problems that most people can do, but its very fast paced at 30 problems, 40 minutes for the sprint, 6 minutes for 2 problems on the Target, and Countdown is even quicker (most of the time).

MOEMS on the other hand, gives you problems that most people can do, but gives you much more time to solve it (I think the average is 4-5 minutes, as opposed to 10-80 seconds on MC).

Therefore, I'd suggest doing Mathcounts for a 6th grader, assuming you want a challenge.

However, if you are just getting started, and are looking for a competition to get some fairly easy problems that you are given 4,5 minutes for, to get an idea of Math Competitions, try MOEMS, and then move on to MathCounts in 7th or 8th grade.

I personally would join mathcpounts this year and study. If you don't do well this year, then you have all that more practice in Mathcounts for next year. I did Mathcounts in 6th grade and 7th grade and now i'm doing it this year. Both of those years' experiences are going to help me do well this year. I wish I had done AMC * last year but, what the heck, I'm doing AMC 10 this year and after a few praactice tests I realized that what I learned in Mathcounts will help me in AMC 10 too.

Try the AMC 8 also.  The problems are a bit easier than MathCounts, and give more time.


For all positive numbers $ a$, $ b$, $ c$ and $ d$ prove that

$ \frac{1}{cd(a+b)}+\frac{1}{ad(b+c)}+\frac{1}{ab(c+d)}+\frac{1}{bc(d+a)}\leq\frac{a+b+c+d}{2abcd}$

 :)

by $ AM-HM$

$ \sum_{cyc}\frac{2ab}{a+b}\leq \sum_{cyc}\frac{a+b}{2}=\sum_{cyc}a$
this gives after dividing throught out by $ 2abcd$

$ \sum_{cyc}\frac{1}{cd(a+b)}\leq \frac{ \sum_{cyc}a}{2abcd}$


Inside rectangle $ABCD$ a point $M$ is given such that $\angle{AMB}+\angle{CMD}=180^0$
Find $\angle{ABM}+\angle{CDM}$

 :)

Really nice !!!

 As Shobber pointed out , if $N$  is the image of  $M$  , that is $ADMN$ is a parallelogram , then $ANBM$ is inscribed and so $\angle MBA = \angle MNA = \angle ADM$. Thus the desired sum is equal to  $\angle D = 90$.

 [u]Babis[/u]

Translate $\triangle CMD$ on top of $\triangle ABM$, such that you create $\triangle ABE$.
$\angle AMB+\angle AEB=180^{\circ}$.  Quadrilateral $AMBE$ is thus cyclic.
$\angle MDC=\angle EAB$.
$\angle CDM+\angle ABM=\angle ABM+\angle EAB=\frac{\widehat{EB} + \widehat{AM}}{2}=\angle EPB=90^{\circ}$
where P is the intersection of $AB$ and $EM$.

EDIT: Made the $\text{\LaTeX}$ better.
Masoud Zargar


Pe forumul in lb. engleza apare topicul:
-----------------------------------------------------
bertrand's assumption n < p <2n for n>2
-----------------------------------------------------
Care afirma ca: "intre n si 2n putem gasi intotdeauna un numar prin".
Se cere o demonstratie a acestei conjuncturi enuntate de catre Bertrand.
Conjunctura a fost demonstrata de catre Cebasev si demonstratia ei,
cred ca poate fi gasita(n-am verificat, dar trebuie sa fie acolo!) in cartea lui
Sierpinski: Elementary theory of numbers
care apare la pozitia 42 in pagina web de pe siteul(de unde se poate descarca gratuit):
http://matwbn.icm.edu.pl/ksspis.php?wyd=10&jez=
sau in cartea lui: Hardy&Wright: An Introduction to the Theory of Numbers
(care poate fi descarcata in aceleasi conditii de pe undeva de pe net!,
eu o am tot de acolo, pe undeva prin biblioteca personala)

Topicul anterior se dorea a fi un raspuns la o intrebare care a aparut pe FORUMUL in lb. engleza, si vroiam
sa-l postez pe subforiumul in lb. engleza. Dintr-o greseala!, a mea am postat-o aici.
Din aceasta cauza am citat in interventia mea anterioara niste carti in lb. engleza. Poate aceste carti
le vor fi insa de folos si cititorilor romani(care nu inteleg engleza).

-Iata insa 2 carti aparute la noi(in lb. romana), in care gasiti demonstratia teoremei lui Cebasev:
1) Laurentiu Panaitopol, Alexandru Gica:
O INTRODUCERE IN ARITMETICA SI TEORIA NUMERELOR, Ed.Univ.Bucuresti(2001)9o gasiti la pag.76-79);
2). Victor Alexandru, Niculaie Marin Gosoniu: ELEMENTE DE TEORIA NUMERELOR, Ed.Univ.Bucuresti(1999)
(gasiti demonstratia la pag.40 s.u.)
Mihai Miculita-ORADEA.

Era sa uit, o alta carte(in lb.romana) in care gasiti dem. teoremei Bertrand-Cebasev,
este cartea lui:

Dumitru Busneag, Florentina Boboc, Dana Piciu:
ARITMETICA SI TEORIA NUMERELOR, Ed.Univ. Craiova-1999.
(v. paginile: 78 s.u.)

Aceasta carte (si nu numai), o puteti descarca (GRATUIT!) de pe pagina
web personala a primului autor:
http://www.inf.ucv.ro/~busneag/books/
(eu am cartea cumparata de la autor!)


modular arithmetic
combinatorics proposed
combinatorics

two player  play a game on the board .first there exists number $n$ in positive integer  on  board .after that first player subtract $x^2$ such that x^2<n .(x is  positive integer)then second player keep on this rule .each person who reach 0 win. prove there exist infinitive many $n$ such that
second player has strategy to win. :?:

I think you can actually determine when the first player wins and when the second does, but for a quick proof of what you want here, assume $B$ (the second player) only wins when $n=n_1,\ldots,n_k$. Now, by the Chinese Remainder Theorem, you can choose some large $n$ (larger than all $n_i$) such that $n-n_i\equiv p_i\pmod{p_i^2},\ \forall i\in\overline{1,k}$, where $p_i$ are distinct primes. There's no perfect square $u^2$ such that $n-u^2\in\{n_i\ |\ i=\in{1,k}\}$, so no matter what $A$ (the first player) does, after his first move they end up in a position favorable to the first player that moves, and that player is now $B$. We have a contradiction.


vector
geometry
geometric transformation
combinatorics unsolved
combinatorics

Show that there is a finite set of points in the plane such that for any point P in the set we can find 1993 points in the set a distance 1 from P.

Start from any two points in the plane, say $ A_1,A_2$, such that $ A_1A_2 \equal{} 1$.
Now, assume that for some $ k$, we have a finite set $ P_k$ of points in the plane such that for each point $ A$ from $ P_k$ there are exactly $ k$ other points $ M$ from $ P_k$ such that $ AM\equal{}1$.
Then, choose any unit vector $ \overrightarrow{u}$ such that for any $ A \in P_k$ we have $ \overrightarrow{u}\equal{}\overrightarrow{AM}$ where $ M \notin P_k$ and $ BM \neq 1$ for each $ B \in P_k\minus{}\{A\}$ (which can be done since $ P_k$ is finite, so it forbid a finite number of unit vectors, but we have uncountably many).
Then defined $ P_{k\plus{}1}$ as the union of $ P_k$ and $ t(P_k)$ where $ t$ is the translation with vector $ \overrightarrow{u}$. It os easy to verify that for each $ A \in P_{k\plus{}1}$ there are exactly $ k\plus{}1$ other points $ M$ in $ P_{k\plus{}1}$ such that $ AM\equal{}1$.

The desired result now follows by induction.

Pierre.

[quote="pbornsztein"]Start from any two points in the plane, say $ A_1,A_2$, such that $ A_1A_2 \equal{} 1$.
Now, assume that for some $ k$, we have a finite set $ P_k$ of points in the plane such that for each point $ A$ from $ P_k$ there are exactly $ k$ other points $ M$ from $ P_k$ such that $ AM \equal{} 1$.
Then, choose any unit vector $ \overrightarrow{u}$ such that for any $ A \in P_k$ we have $ \overrightarrow{u} \equal{} \overrightarrow{AM}$ where $ M \notin P_k$ and $ BM \neq 1$ for each $ B \in P_k \minus{} \{A\}$ (which can be done since $ P_k$ is finite, so it forbid a finite number of unit vectors, but we have uncountably many).
Then defined $ P_{k \plus{} 1}$ as the union of $ P_k$ and $ t(P_k)$ where $ t$ is the translation with vector $ \overrightarrow{u}$. It os easy to verify that for each $ A \in P_{k \plus{} 1}$ there are exactly $ k \plus{} 1$ other points $ M$ in $ P_{k \plus{} 1}$ such that $ AM \equal{} 1$.

The desired result now follows by induction.

Pierre.[/quote]
Can you explain about $ k\plus{}1$ points $ C$ such $ MC\equal{}1$($ M$ in  $ t(P_k)$)
How can we easily found them?

Notice that $ t(P_k)$ has the same shape as $ P_k$, thus for any point C in $ t(P_k)$ there are k other points in $ t(P_k)$ that are 1 away from C. Additionally, according to the definition of $ t(P_k)$, there is a corresponding point to C in $ P_k$ that are 1 away from C. Thus gives a total of k+1 points 1 away from C.


limit
real analysis
real analysis solved

1.Let (a_n)n>=0 be a sequence satisfying the relation
(n+1)a_(n+2)-(n+2)a_(n+1)+a_n = 0, a_0=1, a_1 = 3.
a)Determine the general term of the sequence a_n and proove the convergence of (a_n)n.
b)Compute the limit lim{n->oo}x_n, where x_n=(2+2e-a_n)^n!.

a)This one is simple just see that we can write the sequence like so:

(n+1)(a_(n+2)-a_(n+1))=a_(n+1)-a_n
and denote x_n=a_(n+1)-a_n

then we have x_(i+1)=x_i/(i+1); now multyply all relations of this kind to get x_n and then you get a_(k+1)-a_k=2/k! (I think)

now sum all relations of this kind from k=1 to n to get a_n and then
(a_n)n is divergent because we have lim(a_n)=2e+1 and so ct*n=infinity
I think u meant (a_n)n>=1 like so? right? so it is lim(a_n)=2e+1

b)if I am not mistaking is 1
it's just a case of 1^oo so you treat it like so, it's not too hard! I hope i didn't make any computations errors! :D :D

no it is not 1
it's
e^2 :D... and this one was easy too :(

eh well.. I guess that in my oppinion e^(2/1)-->1 shish.. :( sorry!

:D
i made smth like this in the contest
look how i'm making the following computation:
a+2a+2+2=3a+2...
i spent 45 minutes to figure it out why does the result is not good :)
finally i found it.

Lagrangia and Kuba I found
 b) with Landau technic
a(n)=1+2 \sum _{k=0 to n-1)1/k! = 1+2e - r(n) where
r(n)= \sum _(k \geq n) 1/k!

2+2e-a(n)=1+2r(n)=1+2/n! + r(n+1)

(2+2e-a(n))^n! = exp(n!.ln(1 + 2/n! + r(n+1))
= exp( 2 + n!r(n+1) + o(n!r(n+1)))
but n!.r(n+1)=  1/(n+1).t(n) where t(n) is positive and bounded
 \lim n! r(n+1) = 0

and  \lim (2+2e-a(n))^n! = e^2

this is the first way how i tried to do it but i couldn't finish it...
i don't know what's Landau technic
maybe u can give me some reference moubi
thx

Some material for Landau technic in french " dveloppements limits"

http://perso.wanadoo.fr/lavau/mpsi2003/DLTAYLOR.PDF


$m$ and $n$ are both three digit integers under 500.  Let $\frac{m}{n}$ be the most accurate estimate of $\cos \theta$, where $\sin \theta = \frac{1}{2}$.  Find $n-m$.

Edit: Can be solved with pencil and paper, although may not seem like it...

[hide="Solution Maybe"]
$\sin{\theta}=\frac{1}{2}\implies\cos{\theta}=\frac{\sqrt{3}}{2}$

So we must find the closest fraction to $\sqrt{3}$..one way is to do it by newton's method of finding squareroots (not sure if this is the best way)..

The steps:

1) We guess a number that we think may be evenly divisible by 3, so call it $x$
2) We take the number and divide it from 3 (e.g. $\frac{3}{x}$)
3) We then take the average of 1) and 2) (e.g. $\frac{\frac{3}{x}+{x}}{2}$)

We repeat this process until we get a number when divided by three almost gives us the exact number..however the next time we repeat it our $x$ is the new average found in 3)

So for instance if you guess $2$

Step 2: $\frac{3}{2}$
Step 3: $\frac{\frac{3}{2}+2}{2}=\frac{7}{4}$

Now $\frac{7}{4}$ is our new guess

Step 3 reveals: $\frac{97}{56}$

When we take this number and put it through the process we see:

Step 2:$\frac{168}{97}$
Step 3 $\frac{\frac{168}{97}+\frac{97}{56}}{2}$

The numerator in this case is greater then 500..so our closest value is either $\frac{97}{56}$ or $\frac{168}{97}$

$\frac{168}{97}$ seems to be closer to $\sqrt{3}$ therefore:
(I used a calculator but you can crunch it out..(i hope i read it right..))

$\Rightarrow\frac{m}{n}=\frac{\frac{168}{97}}{2}=\frac{168}{194}$ Thus:

$(n-m)=194-168=\boxed{26}$
[/hide]

[quote="SM4RT"]
The numerator in this case is greater then 500..so our closest value is either $\frac{97}{56}$ or $\frac{168}{97}$
[/quote]

Remember that you're dividing the final fraction by 2 as well; you can go a bit further before it goes over 500.  But you're close!

[hide]Or you could just remember that $\sqrt{3}\approx 1.732$ so $\frac{1}{2}\sqrt{3}\approx 0.866$ and then it's trivial[/hide]

[hide=" I got"]$\frac{181}{209}$ with continued fractions. The error is about 3E-6. Then m-n=28.[/hide]

That's it.  :wink:

It's useful to remember that continued fractions give the most accurate representation of a square root per "amount" of digit.  (Note that this is not neccesarily base 10, in forms of 1 digit, 2 digit, 3 digit, so that's why it's limited to 500, as there are more accurate representations higher up.)  I just used Pell's method to finish it off.

@potatotheory:

Chess64 meant this

$\frac{\sqrt{3}}{2}\approx 0.866=\frac{866}{1000}=\frac{433}{500}$

Ya I was planning on doing something like this but I knew that it may not reveal the correct answer

EDIT: hmm i still cant seem to quite get that answer (following my method)..

Actually that doesn't work since $m,n<500$. But it's basically the same thing.

So how would you solve it?  The more methods, the better.

What is the continued fraction form of $\sqrt{3}$


I read the syllabus of WOOT, and on the assignments list,
it says there are 3 problem sets...
What are they?
Are they just a group of problems and generally how many are there?
Thanks,

The problem sets are groups of challenging problems at a variety of difficulty levels designed to reinforce the lessons in the classes.  The problems will be grouped by difficulty level and the students will be able to choose the set that's most appropriate for them (or mix-and-match, choosing problems of varying difficulty).  Students will receive solutions to all the problems, even those they don't choose to do.

We have not yet determined the length of the problem sets.


How many ways can the letters in the word "MIXED" be rearranged such that no letter maintains it original position?

I looked at the link and calculated the derangements possible to be 44.
I just wanted to make sure if this is correct...


inequalities
function
inequalities solved

[quote="Diogene"]Manifestly F(x,y,z) = 2*(x^3+y^3+z^3) - y*(x^2) - z*(y^2) - x*(z^2)  is a symetrical fonction .
Suppose for ex. x =< y =< z ===> y*(x^2) + z*(y^2) + x*(z^2) >= 2*(x^3) + y^3  ===> F(x,y,z) =<  y^3 + 2*(z^3) =< 3[/quote]
=========
Hi,
F(x,y,z) is not symetric and you can not suppose that x =< y =< z.

Namdung


Which book is more commonly used on Amc10/12 and aime, the topics covered in intro to number theory or intro to counting/prob

I've already ordered the classics and I'm on a limited budget. I've been reading the forums and it seems that vol 1 covers geom and alg enough to skip their intro books but that there is little coverage on the others. My goal is to get to intermediate level of understanding (that is, will reading vol alone prepare me for interm geom, alg, and possibly vol 2 and would I need the intro series in nt and counting to prepare for the int seris in the topics).

Thanks

I skipped NT, but have Intro to C/P. I think it really depends on your level of understanding for each topic if your goal is to get to the intermediate level. To check what you know, you can take the pre-tests and post-tests for the corresponding classes:

[url=http://www.artofproblemsolving.com/Classes/IntroCounting/PreTest.pdf]Intro. C/P Pre-test[/url]
[url=http://www.artofproblemsolving.com/Classes/IntroCounting/PostTest.pdf]Intro. C/P Post-test[/url]
[url=http://www.artofproblemsolving.com/Classes/IntroNumbers/PreTest.pdf]Intro. NT Pre-test[/url]
[url=http://www.artofproblemsolving.com/Classes/IntroNumbers/PostTest.pdf]Intro. NT Post-test[/url]

For frequency on the AMCs, see the chart in [url=http://www.unl.edu/amc/f-miscellaneous/faq.shtml#amc1012]FAQ #2 for AMC 10/12[/url].

The pretests were easy but I got maybe 2/3 of the count/prob and very few of the nt on the post tests. Based on the chart the 3 main principles on the amc12 is nt/geom and then a little algebra. Thanks I think I'll order the nt book

edit: not including calculus


so basically use good punctuation and capitalization to create a story
you cannot post twice and try to copy the post above yours and put your word in bold....
EX.
[b]Julie[/b]
Julie [b]went[/b]
Julie went [b]to[/b]
Julie went to [b]a[/b]
Julie went to a [b]park.[/b]

i'll start:

Summer

Ok ignore that post.

Summer is really the [b]best[/b]

It's a person?
Well in that case..

Summer is really the best person [b]at[/b]

Summer is really the best person at [b]Summer[/b]

Summer's a person?

Summer is really the best person at Summer [b]vacation[/b]

Summer is really the best person at Summer Vacation [b]School.[/b]

Summer is really the best person at Summer Vacation School [b]in[/b]

Summer is really the best berson at Summer Vacation School in [b]the[/b]

Summer is really the best person at Summer Vacation School in the [b]USA[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT).  westiepaw

Find some way to spam me out.  :P

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw [b]atesometurkey,[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, [b]stuffing,[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, [b]whichwasofcourse disgusting[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, whichwasofcourse disgusting [b]sohethrewup.[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, whichwasofcourse disgusting sohethrewup. [b]Then,[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, whichwasofcourse disgusting sohethrewup. Then, [b]k00lperson[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, whichwasofcourse disgusting sohethrewup. Then, k00lperson [b]smiledatakitty.[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, whichwasofcourse disgusting sohethrewup. Then, k00lperson smiledatakitty. [b]However,[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, whichwasofcourse disgusting sohethrewup. Then, k00lperson smiledatakitty. However, [b]15!!![/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, whichwasofcourse disgusting sohethrewup. Then, k00lperson smiledatakitty. However, 15!!! [b]didn't[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, whichwasofcourse disgusting sohethrewup. Then, k00lperson smiledatakitty. However, 15!!! didn't [b]liveanddied[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, whichwasofcourse disgusting sohethrewup. Then, k00lperson smiledatakitty. However, 15!!! didn't liveanddied [b]"YAY!' spaceguycried[/b]

Summer is the best person at Summer Vacation School in the USA rain. She ate some macaroni and cheese with extra pickles. She loved the artificial flavors and tang with a dill pickle. It stung her beloved tongue, which is red. Then she lost her beloved computer monitor. After college was jail. Then later, someone asked her to get a glass case filled with money. Bob Meyers accepted many suitors for his interesting clients. He threw parties at palaces in Detroit!!!!! He lost a piece of his favorite tomato pie. He then ate pasta.Then something ate Bob Meyers and forgot its lunchbox. Connecticut was stunned.They were in deep pie. However, Billy started to climb very mostly high to mount everest without help.After he went home to fight the tomato monster and then glared at Summer. "Today I want mushroom with eggs under my sandwich. k00lperson is a person.hjoon went to die in the cliff NOT. Monkeygirl13 and Taylor Swift went swimming after they were left to the alligators, HAHA!Ernie thought. Kanye West went to save Beyonce and died." Dead, King! jumped, along with AceOfDiamonds and levans. They sat at a table, and picked their next target as cottontail. They killed each other harshly twice an d went to jail. The next fatty jumped off a fatty fan and died. Meewhee noticed the wonderful but judged orange pizza which sat perpendicular to spaceguy524, the great looser. Pyropower3, k00lperson, spaceguy524 monkeygirl13, and the meewheeofawesomness died (jk) (not). k00lperson stepped onto lizards whichthenkilledMagicBard whoisindestructibleanddidn'tdie. MagicBardwishedthathewasindestructiblebutwasn'tsohedied. Meewhee and k00lperson didnt die.
MagicBard died (not jking). Then, k00lperson lived. Spaceguykilledk00lpersonthen. (NOT) (NOT). westiepaw atesometurkey, stuffing, whichwasofcourse disgusting sohethrewup. Then, k00lperson smiledatakitty. However, 15!!! didn't liveanddied "YAY!' spaceguycried
"MarhcepalloozaIamawizardfearme!!!!"[b] lalalalalal.[/b]


LaTeX inequalities parameterization percent probability blogs function

I hope this is the right forum for this, I wanted to discuss the following:

Background: The MMPC, or Michigan Mathematics Prize Competition, is a contest (in Michigan...) with two parts - Part I is a 40 multiple choice test and Part II is a 5 proof question test, where the parts are weighted 40% to 60%.

Scenario:  Question #33 on this year's Part I was quite obviously wrong.  Here is a link to the test and answers:
http://www.svsu.edu/mmpc/registration/2009_Part_I.html

For those who do not wish to go to the link, the problem is as follows:
A triangle has vertices A,B and C, and the respective opposite sides have
lengths a, b and c. This triangle is inscribed in a circle of radius R. If b = c = 1
and the altitude from A to side BC has length $ \sqrt{\frac{3}{2}}$ (no I did not LaTeX it wrong) then R equals:
A: 1/rt(3), B: 2/rt(3), C: rt(3)/2, D: 1, E: none of the others.

Problem:  The tests were scored as though the correct answer were E.

I would like to discuss our thoughts about this - if it matters, last year, the top scores differed by .2 points, and the top ranked individual gets ~3000 dollars, if you think money issues are at hand.

Well, then what did you get?

I got R=[hide="answer"]$ \sqrt {\frac {2}{3}} \equal{} \frac {2}{\sqrt {6}} \equal{} \frac {\sqrt {6}}{3}$[/hide]

Is there really a problem?

[hide="solution"]
Draw the diagram.  the distance from the center, O, to A is R and to B is also R.
Now, the length AD is given as sqrt(3/2) so BD=CD=1/sqrt(2)  (where D is the point where the altitude from A hits BC.
Use Pythagorean Theorem on triangle BOD or COD.

[/hide]

This is the correct forum if...
you wanted to talk about the justification of messing up an answer on the answer key for a contest with such high stakes.
whether it would be ok if they got it wrong and they did not/did change the answer key, regraded everything, and redistributed

This is still the correct forum if...
this is about ethics.

This is not the right forum if...
You're posting a problem that you want the answer to and you want input on whether it is correct...

For now, you made the right choice.  But the fact that (in my opinion) the answer is E, possibly makes this more suitable somewhere else.

Note that this violates the Triangle Inequality  :roll:

Many people, including myself and kyyuanmathcount, assumed that there was a typo and the altitude was supposed to be $ \frac{\sqrt{3}}{2}$ and thus got B. I can't guarantee what other people were thinking, but I assumed that this question would be thrown out anyway, so I marked the answer as B, which, as the original poster pointed out, was marked incorrect.

At this point, it doesn't really matter for me, but there is a chance that the top two scores overall will be the same after Part II has been graded. I happen to know that one of those two put down E for this question, while the other put down B; in other words, this question more or less determines the winner.

I personally don't think this is fair at all; is it our fault for interpreting an incorrect problem in a way that isn't being accepted by the director?

[quote="xpmath"]Note that this violates the Triangle Inequality[/quote]

lol, I didn't realize that.

Did you guys bring it up during or immediately after the competition? Is this a live competition, or one where testing occurs at a variety of locations?

Have you guys tried contacting the people who run this competition.

In my opinion, now that I see the problem, maybe they should throw the problem out and maybe give some tiebreakers.  Whether they will do that is up to them...but money here is a big thing.

But you obviously can't take away the money from someone who already got it and possibly spent it or invested it already.  I'm not sure if they'd be willing to give a second $ \$$ 3000 to the person who should have tied/won.

Right now, Part II has ended, but the grading for Part II and the announcement of winners doesn't happen for a while.

Our school's head math teacher asked the director of the problem, and the response was just that "the correct answer was (E)."

I just think there should be no other choice than to throw it out?

Well, if changing their minds doesn't happen, well...
you can just mess around with students when you get older and host math competitions  :P

(jk, don't do that  :( ,  as you see they'll get pretty angry)

Someone needs to be contacted [i]by phone[/i], so that the issue can't be blown off with a one-liner response e-mail.

This is a sticky situation, since the following are both true:

(1) Having an incorrect question is completely unfair, because some contestants may have spent a lot of time on the problem (for example, running into a contradiction and looking for arithmetic errors), which is less time available for other problems. There's no entirely fair way to compensate each student to the damage caused by the incorrect question.

(2) There is no [i]good[/i] reason to assume that a particular parameter of the problem should have been a different particular number. If the answer to the problem, by any mathematical process, does not come out to one of the four numerical choices, the only reasonable answer is E.

A friend of mine gave the following reasoning:

[quote]First, the format is "if {FALSE STATEMENT}, then {answer choices}." Therefore, the statement is vacuously true. Thus, if an answer choice of "all of the above" were provided, it would be the best, though not very fair.

As it is, it seems like the best idea would be to throw out the question. Choices A,B,C, and D complete the statement as a true implication, so accepting all of these would be fair. Choice E, if anything, is possibly the worst choice, because, even though it still yields a true statement, it seems to deny the truth of the other four choices. Still, this is not reasonable grounds to disallow choice E. Thus, the only reasonable options are essentially the same: throw out the question, or accept all answers. [/quote]

Do you think this is valid?

We had a similar poblem recently with qualifier exams: one of the problems was unsolvable. There was a long discussion as to what to do in a couple of cases where it mattered and the points of view were radically different. The one that finally prevailed was that if a student tried the standard method and arrived at some answer, he should get full credit (I was strongly opposed to that maintaining that if a student cannot see that there is a problem with the standard technique, the grade should be 0). Why am I saying this? Just to show that there may be many contradictory viewpoints on this with some valid idea behind each of them.

If you want my opinion, in this situation, I would just drop the problem altogether when determinning the winner. On the other hand, I wouldn't bother to correct the grades of other people at all: they won't gain or lose much from this grading error.

If you want to make sure that the exam committee is aware of the issue, just send them an e-mail. In any case you'll have to abide by their decision. Breaking close ties is always unfair: a few percent of your score are determined just by whether you slept well, had a good breakfast, etc., so it is always some sort of probability game.

Formally speaking, your friend is correct: all answers and all their combinations are equally valid here: "$ R\equal{}1$ AND $ R\equal{}2$ AND $ R\ne 1,2,3$" is as good a logical conclusion from a false assumption as any other. As long as the conclusion is a statement or a combination of statements, the implication holds (even if the statements in the combination clearly contradict each other). On the other hand, I doubt very much that the choice of the winner will be determined by this observation about formal logic.

It's an unfortunate situation.  I would recommend throwing the problem out.  The friend of kyyuanmathcount makes a good argument.

Tanya Khovanova wrote a blog post about a similar error in a US standardized test:
[url]http://blog.tanyakhovanova.com/?p=131[/url].

The only honorable response from the organizers is to admit that a mistake was made, throw the question out, and apologize profusely.  Of course, that doesn't mean that this will actually happen.  I can only assume that kyyuanmathcount's teacher already explained why the problem is nonsense; perhaps the director of the competition just doesn't care.

By the way, this is what I hate about NOTA.  It allows test writers to just be lazy and think, "Well, if we screw the question up, that just means that NOTA is the right answer, so no harm done."  And that would make perfect sense, if it were actually true.

This is why multiple choice exams with NOTA are completely pointless. It's of course designed to make people second-guess themselves if they get a different answer, and make sure not to make stupid mistakes. However, imagine if the problem Ravi B linked to was on an exam and you realized it was impossible. Do you put the "obvious" answer of 30? Or do you put NOTA and hope the person who wrote the test did it on purpose? The only fair thing to do after something like this comes up is to either throw out the problem, or accept all answers that could possibly be correct.

As for something like what happened here, in this situation I am always that annoying little kid who goes up to the test proctors and asks if there was a typo. Most of the time they'll ignore you, but its worth a shot.

[quote="Demosthenes"]i think the director said that the problem was intentionally worded that way, and that is why 'E' is the only correct answer
[/quote]
Then he is obviously wrong (mathematically, not ethically). You'd better check whether this information is correct.

[quote="Demosthenes"]i think the director said that the problem was intentionally worded that way, and that is why 'E' is the only correct answer
However, it is still unfair to not give credit to those who put B[/i][/quote]

I'm assuming you are from Michigan and you have contact with Dr. Al-Halees?  You mean to say he said he meant to write rt(3/2)?

I don't think this is correct, as I am family friends with one of the members of the problem writing committee.  When we told her of the situation, she, too, was surprised, confirming the error, and later told us that she discussed the situation with the writing committee.

Also, just if it is relevant, the contest has had such an error, as two years ago on the Part II, one of the five problems was unsolvable.  I think they handled the error to the best of their ability in that situation, though.

This whole discussion is reminding me of why I try, if I can, to stay away from "none of the above" or similar formulations in the contest I'm responsible for. But I went back to look, and wouldn't you know, I found a question from a year ago that I'm a little queasy about. Here it is:

9. For all real numbers $ t$ we have that $ \left|\overrightarrow{BA}\minus{}t\overrightarrow{BC}\right|\geq \left|\overrightarrow{AC}\right|.$ Then $ \triangle ABC$ must be

(A) an acute triangle (B) an obtuse triangle (C) a right triangle
(D) degenerate (E) none of the above

All right, we probably should have had something in there about that vertical bar notation being the (Euclidean) length of a vector. But the stipulation that we're really missing is that we should have also stated that all three of $ A,B,$ and $ C$ were different points. And I'm not even sure how "degenerate" got in there - we needed something to fill out five multiple choice slots.

As it turns out, the question had very strong discriminating power. Only 18 out of 189 competitors got it right, and that did include 11 of the top 17 students overall. More people (28) answered (B) than (C), and yes, that is what we had in mind in making this a question - that (C) is right and (A) and (B) are wrong.

But 31 people, including 5 of the top 15, answered (D), degenerate - and that's really not what we had in mind. And, if we didn't specify up front that $ A,B,$ and $ C$ were all different, I suppose there's an argument that we could have $ B\equal{}C.$ Now, you can't say that the triangle [i]must[/i] be degenerate, so I'll still argue that (D) can't be the right answer - but ... aargh! We try not to do this. (And it was nearly a year ago, and no one complained at the time, and I'm not changing anything now.) The hazards of multiple choice.

We did resort to the "other" answer in one other place, although it was a bit more specific - the following exploration of the meaning of "if ... then":

15. The function $ f$ has these properties: if $ n$ is odd, then $ f(n)$ is divisible by $ 3$; and if $ n$ is divisible by $ 3$, then $ f(n)$ is even. Suppose $ f(n) \equal{} 2009$. Which of the following is a possible value of $ n$?

(A) 0 (B) 1 (C) 2 (D) 3 (E) this is impossible

That question worked just fine, and 86 out of 189 people got it right, with 49 leaving it blank and 54 getting it wrong. OK, the single most common wrong answer, with 32, was (E), impossible. But of course this is possible. But as a medium-difficulty question, its discriminating powers weren't as strong as the much harder #9.

It could be reasoned that no such triangle exists, so R doesn't exist. Thus, the answer is 'it is undefined', which isn't one of A, B, C, or D. The only correct answer would then be E.

I would agree that the above reasoning is quite unfair and the problem should not be counted.

People do not like to admit mistakes, so I wouldn't be surprised if what Demosthenes is saying is true.  NOTA means never having to say you're sorry.

fedja makes the point that even if the wording of the question was completely intentional, the grading of the question is still mathematically wrong.  But I believe that this is beside the point.  Every test has unwritten rules, and one of those unwritten rules is that the question is not vacuous.  If you do not believe me, ask yourself:  In every math test you've ever taken, do you always check to make sure that every problem is well-formulated?

@Professor Merryfield, unfortunately, that question 9 is lousy.  While you cannot say that the triangle *must* be degenerate, you also cannot say that the triangle *must* be a right triangle, so logically, (C) is not any more correct than (D).  To make matters worse, since the "triangle ABC" is referred to in the problem, the reader can (and should) reasonably assume that the points A, B, and C are distinct, BUT FOR the fact that the "degenerate" situation is explicitly included in the answer choices.  Oy!  By including choice (D), you are explicitly telling the reader to take into account the degenerate possibility.  I would argue that the only reasonable way to deal with this question is to admit that a mistake was made, accept all of C, D, and E (as well as non-answers if there was a guessing penalty) as correct answers and, apologize profusely.  While I'm guessing that this is not what you did, I'm sure that you tried to do what was you thought was right.

yenlee: Thanks for the thoughts, but I didn't even know there was anything wrong with it until I looked just a few days ago. It was graded with (C) as the only correct answer, and that all happened long enough ago that no redress is possible.

In retrospect, we created the problem by feeling the need to pad out the responses to make it a 5-answer multiple-choice question, so we threw in "degenerate" as a late addition to the question without really thinking it through. We'd have been better off by just saying for both (D) and (E) "This is not the answer", just making it a three-answer problem.

We spent most of our time on post-mortem discussing a question that really didn't work. It was a short-answer free-response question for a geometric figure that gave specific numerical values for $ AD\equal{}3$ and $ EC\equal{}2$ and asked for $ DE,$ which turned out to be $ \sqrt{13}.$ Only 3 out of 189 got it, and none of those three were in the top 20 overall - so in effect it was a wasted questions. Had it been a proof question - draw the same figure, and ask for a proof that $ DE^2\equal{}AD^2\plus{}EC^2$ - it would have been much easier, despite being a proof question.

I emailed him on January 3rd and I have just finally gotten his response, which is as follows after the stars - my email is below.

[color=green]Moderator note: I've edited to remove e-mail addresses, as well as the response you received.  (It's very bad form to publicly post personal e-mails without the sender's consent.)[/color]

**********

----- Original Message -----
From: "Allen Yuan"
To: "Dr. Al-Halees"
Sent: Sunday, January 3, 2010 11:03:26 PM GMT -05:00 US/Canada Eastern
Subject: MMPC: A Concern

Dear Dr. Al-Halees,

        I am Allen Yuan, a participant of the 2009-2010 MMPC, and I am writing to you because of a concern I have with the grading of this year’s Part I exam.  Number 33 on Part I asked the students to consider an isosceles triangle of leg length 1 that had an altitude from the vertex with length equal to the square-root of 3/2.  In a Euclidean geometry, this does not exist, so the condition of the problem itself is a false statement (see reason 1).  It has come to my attention recently that only an answer of (E) none of the above was accepted.  I believe that the question should be thrown out (in that either all participants are given a point for this question, or none are given the point), and my reasons are as follows:

1.    By formal logic:  the question format is "if {FALSE STATEMENT}, then {answer choices}." Therefore, the statement is vacuously true. Thus, if an answer choice of "all of the above" were provided, it would be the best, though not very fair.  Choices A,B,C, and D complete the statement as a true implication, so accepting all of these would be fair. Choice E, if anything, is possibly the worst choice, because, even though it still yields a true statement, it seems to deny the truth of the other four choices. Still, this is not reasonable grounds to disallow choice E. Thus, the only reasonable options are essentially the same: throw out the question, or accept all answers.
        I provide this reasoning to show that accepting only E would be mathematically false.  Obviously, students will not be expected to come up with a formal logical deduction as such, but I believe that this argument unambiguously shows that the current answer is, mathematically, incorrect.

2.  The principles of this competition:  The Part II (I know, not Part I) instructions read “If you have difficulty understanding what is required in a given problem, note this on your answer
sheet and attempt to make a nontrivial restatement of the problem. Then try to solve the restated
problem.”  I understand that this is not the Part I instruction, but it seems within the ideals of the contest that the contestants try to make such a “nontrivial restatement.”  Knowing this, it is natural for a participant to consider the altitude as equal to one-half times the square-root of three.  This would yield a perfectly solvable problem with the correct answer of (B)  (Furthermore, familiarity with math-typesetting programs lets one acknowledge the ease of accidentally making the square-root over the whole fraction).  It is certainly unfair to accept only (B) on the grounds that a student should guess a correct version of the problem, but it is similarly unjust, I believe, to penalize the many who solved the problem correctly with the restatement.  The purpose of the exam, in a perspective, is to find who can solve more problems correctly, rather than who knows how to deal with an error in the contest.  Again, this thought leaves the only choice as discounting the question.

3.  It is simply unfair to ask students to deal with a question that is not fairly proposed mathematically.  This is quite similar to point number 2, but I think that just in the integrity of the contest, it is desired to have the question not be counted against or for anybody.  Suppose a test-taker knows of the error, and the answer in the case that the problem used sqrt(3)/2 instead of sqrt(3/2) – it is a dilemma that, for the test-taker, will waste time and thought.  Furthermore, extra time will inevitably be spent checking this problem, and this time cannot be recovered.  It is only fair that the problem not affect the scores of the participants even more by being counted incorrect.

I hope that these points have shown that the most desirable way to deal with the situation is to have all the choices counted or discounted for the problem.  However, I understand that in reality, it is clearly impractical to re-grade so many tests, and that such a little error is not something for which much ado should be made.  Nevertheless, the rankings, especially near the top, are, as experience shows, quite close (if we take last year, the top two ranks were different by 0.2 of a point).  Therefore, I propose an alternative that I believe is certainly both practical and fair:

The alternative is that when the top 100 are decided after Grading Day and the Part II tests of the top 100 are to be re-graded (I believe this is done from my knowledge, but my alternative proposal does not rely on it), that the Part I scores be adjusted with the new scoring.  This procedure is quick, because only 100 scores need to be adjusted (or if you desire, maybe only top 50 or top 20, because the rankings are most important and serious near the top).  I think that this simple procedure can guarantee the fairness of the contest, and I am sure that nearly all participants will favor this alternative.

Please let me know of any concerns you have, and I look forward to your favorable and just consideration of my writing,
Happy New Year to you,
-Allen Yuan

kyyuanmathcount, I've edited your post to remove e-mail addresses as well as the response  you received: it's very bad form to post either without the owner's consent.  I suggest you edit your post to include a summary or paraphrase of the response you received.  (If you can't because of forum constraints, feel free to PM me and I can edit it for you.)

I'm very sorry about that, thanks for the info and I'll work on the edit soon.


Does anyone know a place (preferably internet) where I could find a good deal of elementary proofs? (My geometric proving ability is good enough, I'd like to focus more on algebraic proofs) Also, is there a site that could help me with writing algebraic proofs?

Thank you very much

Yes. I am looking mainly for easy algebraic proofs that could be done without using anything above algebra 2. (Although geometric proofs are ok too, but my main purpose is algebra).


Is there a forum that you can just post random problems that you don't know how to solve?

There are many such forums, depending on the level and subject area of the problem.

I think the problems would be for middle schoolers about 2 times better at math than a normal math student.  I don't know what grade though.

You could post it in the high school basics forum, as long as it's not too easy...


Prove that $ \binom{ap}{bp} \equiv \binom{a}{b} \mod p^{3}$ for $ p > 3$ where p is prime.

$ ap! \equal{} a!p^a((p \minus{} 1)!)^aC(a)$, were $ C(a) \equal{} \prod_{k \equal{} 0}^{a \minus{} 1}B(k), \ B(k) \equal{} \frac {\prod_{i \equal{} 1}^{p \minus{} 1}(i \plus{} kp)}{(p \minus{} 1)!}\equal{}\binom{kp\plus{}p\minus{}1}{p\minus{}1}.$
Therefore $ \binom{ap}{bp} \equal{} \binom{a}{b}\frac {C(a)}{C(a \minus{} b)C(b)}$.
$ B(k) \equal{} 1 \plus{} kpS_1 \plus{} k^2p^2T \plus{} O(p^3)$ were
$ S_1 \equal{} \sum_{j \equal{} 1}^{p \minus{} 1} \frac 1j , \ T \equal{} \sum_{1\le i < j\le p \minus{} 1}\frac {1}{ij} \equal{} \frac 12 (S_1^2 \minus{} S_2).$
Because for $ p > 3$ $ S_1 \equal{} O(p^2),S_2 \equal{} \sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{i^2} \equal{} O(p)$ we get $ B(k) \equal{} 1 \plus{} O(p^3)$.
It give $ \frac {C(a)}{C(a \minus{} b)C(b)} \equal{} 1 \plus{} O(p^3)$ or $ \binom{ap}{bp} \equal{} \binom{a}{b}(1 \plus{} O(p^3)).$


Find function $ f: \mathbb{R}\rightarrow(0,+\infty),$ such that $ f(x)=F(x)F(-x),$ where F is a primitive function.

$ f(x) = a^{2}\text{sech}^{2}(ax)$ and $ F(x) = a(1+\tanh (ax))$ where $ a$ is a nonzero constant.

My solution:

Since $ f$ is even, $ y(x) = F(x)-F(0)$ is odd and if we let $ a = F(0)$, then the given equation becomes $ y'+y^{2}= a^{2}$ with the initial condition $ y(0) = 0$. Now all you have to do is to solve it.