function
algebra
functional equation
algebra unsolved

Find all functions $ f : \mathbb{R}  \ \to \ \mathbb{R}$ such that
\[f((x\plus{}1)f(y)) \ \equal{} \ y (f(x) \plus{}1) , \quad\forall x,y \in \mathbb{R}.\]

[quote="nguyenvuthanhha"][i]Find all function $ f : \mathbb{R} \ \to \ \mathbb{R}$ such that :

   $ f((x \plus{} 1)f(y)) \ \equal{} \ y (f(x) \plus{} 1) \ \forall \ \ x ; y \ \in \ \mathbb{R}$

 [hide]I think the problem will become very easy if we can prove $ f(1) \equal{} 1$ but i can't reach that arguement [/hide][/i][/quote]
Thanks for the gift :)

[hide="My solution"]
Let $ P(x,y)$ be the assertion $ f((x \plus{} 1)f(y)) \equal{} y(f(x) \plus{} 1)$
Let $ a \equal{} f(1)$
$ f(x) \equal{} 0$ $ \forall x$ is not a solution. So $ \exists c$ such that $ f(c)\neq 0$

If $ f(0)\neq 0$, $ P(\frac {c}{f(0)} \minus{} 1,0)$ $ \implies$ $ f(c) \equal{} 0$, which is wrong. So $ f(0) \equal{} 0$
$ P(0,x)$ $ \implies$ $ f(f(x)) \equal{} x$ and $ f(x)$ is an involution, and so is bijective.
$ P( \minus{} 1,1)$ $ \implies$ $ f( \minus{} 1) \equal{} \minus{} 1$
Since $ f( \minus{} 1) \equal{} \minus{} 1$ and $ f(0) \equal{} 0$ and $ f(x)$ is bijective, $ a \equal{} f(1)\notin\{ \minus{} 1,0\}$

Since $ f(f(x)) \equal{} x$, we get $ f(a) \equal{} 1$
$ P(x \minus{} 1,a)$ $ \implies$ $ f(x) \equal{} a(f(x \minus{} 1) \plus{} 1)$ and so $ f(x \minus{} 1) \plus{} 1 \equal{} \frac {f(x)}a$
$ P(x \minus{} 1,f(y))$ $ \implies$ $ f(xy) \equal{} f(y)(f(x \minus{} 1) \plus{} 1)$ $ \equal{} \frac {f(x)f(y)}{a}$

We can compute then $ f(4)$ in two ways :
$ P(1,a)$ $ \implies$ $ f(2) \equal{} a^2 \plus{} a$
$ P(2,a)$ $ \implies$ $ f(3) \equal{} a^3 \plus{} a^2 \plus{} a$
$ P(3,a)$ $ \implies$ $ f(4) \equal{} a^4 \plus{} a^3 \plus{} a^2 \plus{} a$

But, using $ x \equal{} y \equal{} 2$ in $ f(xy) \equal{} \frac {f(x)f(y)}{a}$ implies $ f(4) \equal{} \frac {(a^2 \plus{} a)^2}a \equal{} a^3 \plus{} 2a^2 \plus{} a$

So $ a^4 \plus{} a^3 \plus{} a^2 \plus{} a \equal{} a^3 \plus{} 2a^2 \plus{} a$ and so $ a^4 \equal{} a^2$ and so $ a\in\{ \minus{} 1,0,1\}$ but we already know that $ a\notin\{ \minus{} 1,0\}$ and so $ a \equal{} 1$

So $ f(xy) \equal{} f(x)f(y)$
Then $ P(x,f(y))$ $ \implies$ $ f((x \plus{} 1)y) \equal{} f(y)(f(x) \plus{} 1) \equal{} f(x)f(y) \plus{} f(y)$ $ \equal{} f(xy) \plus{} f(y)$ and $ f(xy \plus{} y) \equal{} f(xy) \plus{} f(y)$ and so $ f(u \plus{} v) \equal{} f(u) \plus{} f(v)$

Since we also have $ f(xy) \equal{} f(x)f(y)$ $ \implies$ $ f(x^2) \equal{} f(x)^2$, we get that $ f(x)$ and $ x$ have same signs and so $ f(x)$ is increasing.

Hence $ f(x) \equal{} x$ (increasing solution of Cauchy equation with $ f(1)\equal{}1$) and it's immediate to verify that this mandatory result indeed is a solution.
[/hide]

[quote="mehdi cherif"][quote="nguyenvuthanhha"][i]Find all function $ f : \mathbb{R} \ \to \ \mathbb{R}$ such that :

   $ f((x \plus{} 1)f(y)) \ \equal{} \ y (f(x) \plus{} 1) \ \forall \ \ x ; y \ \in \ \mathbb{R}$

 I think the problem will become very easy if we can prove $ f(1) \equal{} 1$ but i can't reach that arguement [/i][/quote]

let $ P(x,y)$ be the assertion $ f((x \plus{} 1)f(y)) \equal{} y(f(x) \plus{} 1) \forall x,y$.

$ P( \minus{} 1,0)\implies f(0) \equal{} 0$

$ P( \minus{} 1, 1)\implies f( \minus{} 1) \equal{} \minus{} 1$

$ P(0,y)\implies f(f(y)) \equal{} y$

$ P( \minus{} 2, \minus{} 1)\implies f( \minus{} f( \minus{} 1)) \equal{} f(1) \equal{} \minus{} (f( \minus{} 2) \plus{} 1)$ so $ f( \minus{} 2) \equal{} \minus{} 1 \minus{} f(1)$ $ (1)$

$ P( \minus{} 2,f(1))\implies f(( \minus{} 2 \plus{} 1)f(f(1))) \equal{} f( \minus{} 1) \equal{} \minus{} 1 \equal{} f(1)f( \minus{} 2) \plus{} f(1)$  $ (2)$

from $ (1)$ and $ (2)$ we have $ f(1)^2 \equal{} 1$

and since $ f(f(1)) \equal{} 1$ , $ f(1) \equal{} 1$

.....

 :wink:[/quote]

Yess, nicer and quicker than my own path to $ f(1)\equal{}1$
Congrats :)

[i]Yes , after prove $ f(1) \equal{} 1$

   We have $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$ ( put $ y \equal{} 1$ into the initial equation )

 Consider $ P( x \minus{} 1; f(y) ) \rightarrow f(xy) \equal{} f(x) f(y)$

  $ f(x \plus{} y) \equal{} f \left(y \left( 1 \plus{} \frac {x}{y} \right) \right) \equal{} f \left( \left( 1 \plus{} \frac {x}{y} \right) f(f(y))\right) \equal{} f(y)\left( f \left( \frac {x}{y} \right) \plus{} 1 \right) \equal{} f(x) \plus{} f(y) \forall x ; y \ \in \ \mathbb{R} ; y \ \neq \ 0$

But $ f(0) \equal{} 0$
 So we have $ f(xy) \equal{} f(x) f(y) ; f(x \plus{} y) \equal{} f(x) \plus{} f(y) \ \forall x ; y \ \in \ \mathbb{R}$

  By a well - known lemma , we have known that $ f(x) \equal{} x \forall x \ \in \ \mathbb{R}$
or $ f(x) \equal{} 0 \forall x \ \in \ \mathbb{R}$

  After an easy check , we can eliminate the case $ f(x) \equal{} 0 \forall x \ \in \ \mathbb{R}$

 and $ f(x) \equal{} x \forall x \ \in \ \mathbb{R}$ is the unique root of the functional equation :lol: [/i]


Assume that [tex]\displaystyle{{CN\over NA}={CM\over MB}=r}[/tex] instead of assuming N and M are the midpoints of AC and CB respectively. Prove that [tex]a^2+b^2=kc^2[/tex] for some constant [tex]k[/tex]. Also find the value of [tex]k[/tex] in term of [tex]r[/tex].

Note: [hide]it's not necessary to find the ratio of NP/PB and MP/PA as in the solutions.[/hide]

Let AP be x, BP be y. [tex]c^2=x^2+y^2[/tex].
AN=[tex]\frac{a}{r+1} [/tex]  BM=[tex]\frac{b}{r+1}[/tex]
NC=[tex]\frac{ar}{r+1}, [/tex]CM = [tex]\frac{br}{r+1}[/tex]
Since, [tex]\Delta CMN \sim \Delta CBA[/tex]
NP=[tex]\frac{yr}{r+1}[/tex]  PM=[tex]\frac{xr}{r+1}[/tex]
[tex]b^2=(r+1)^2(({\frac{yr}{r+1})}^2+x^2)[/tex]
[tex]a^2=(r+1)^2(({\frac{xr}{r+1})}^2+y^2)[/tex]

add them together we have

[tex]x^2(2r^2+2r+1))+y^2(2r^2+2r+1)=(2r^2+r+1)(x^2+y^2)[/tex]=[tex](2r^2+r+1)(c^2)[/tex]. Therefore[tex] k=2r^2+2r+1)[/tex]

edit--I fixed the mistake

beta, I think there is a calculation error somewhere since k=5 when r=1 according to the solution on the AoPS site. My answer is:

[hide]k = r^2+(r+1)^2 = 2r^2+2r+1[/hide]

I will give a hint here without finding NP and MP in terms of AP and BP resp.

Hint: [hide]It's well known that the diagonals of a convex quadrilateral are perpendicular to each other if and only if the sums of squares of the correponding opposite sides are equal. Consider the (convex) quadrilateral ANMB, we have AN^2+MB^2 = NM^2+AB^2. The rest is obvious.[/hide]

I don't know about how "well known" that fact is, but:

[hide]It helps a lot to note that MN is parallel to AB, and that MN and AB are both part of right triangles...[/hide]


group theory
abstract algebra
superior algebra
superior algebra theorems

1) Let $ a$ be an element of $ Z_n$. State and prove something of interest concerning the cyclic groups<a> and <-a>.

My conjecture: $ <a> \equal{} <\minus{}a>$

Im not really good at this, so please comment on my proof. I don't have any idea if what im doing is correct.

[hide="My Proof"]
Let {$ a, a^2,a^3, ..., a^n$} be elements of $ Z_n$. The group generated by $ <a>$ will be {$ a^m, a^{2m},a^{3m}, ..., a^{km}, e, a^{\minus{}m}, a^{\minus{}2m},a^{\minus{}3m}, ...$}, $ m,k$ belongs to $ Z$ and where the terms negative exponents are inverses of those with postive exponents, which are elements of $ Z$.

Now, the group generated by <-a> which is the inverse of $ a$ will be {$ ...,a^{\minus{}m}, a^{\minus{}2m},a^{\minus{}3m}$} its inverse elements,{$ a^m, a^{2m},a^{3m}, ..., a^{km},...$} and the identity $ e$.

Therefore, $ <a> \equal{} <\minus{}a>$  [/hide]

2.) Let $ <G,\plus{}>$ be an abelian group. Let $ H$ and $ K$ be subgroups of $ G$. Define $ H\plus{}K \equal{} \{h\plus{}k|h \in H, k \in K \}$. Prove that $ H\plus{}K$ is a subgroup of $ G$.

[hide="My Proof"]Since $ H$ and $ K$ are subgroups of $ G$, they contain both the identity element $ e$. But $ e \plus{} e \equal{} e$, therefore, $ e \in H\plus{}K$

(This is the part of the proof which might have  a flaw.)
$ h$ is a unique element in $ H$ and $ k$ is a unique element in $ k$. Hence, $ h \plus{} k$ is unique in $ H \plus{} K$ which still belongs to the group $ G$ since both $ H$ and $ K$ are subgroups of $ G$. Therefore, $ H \plus{} K$ is closed under $ \plus{}$.

For each $ h \in H$ and $ k \in K$, there is a corresponding inverses $ \minus{}h \in H$ and $ \minus{}k \in K$. Therefore, each element $ h\plus{}k \in H \plus{} K$ has a corresponding inverse $ \minus{}h \minus{} k$.

Hence, $ H \plus{} K$ is a subgroup of $ <G>$[/hide]

For the second one, you didn't show closure. It means showing two elements in $ H\plus{}K$ have a sum that is also in $ H\plus{}K$.

To do this, say two elements are in $ H\plus{}K$, the first element can be written (not necessarily uniquely) as $ h_1\plus{}k_1$, and the second element can be written as $ h_2\plus{}k_2$.

Then find $ h,k$ such that the sum of $ h_1\plus{}k_1$ and $ h_2\plus{}k_2$ is $ h\plus{}k$.

(Here, $ h_1,h_2\in H, k_1,k_2\in K$, and you need $ h\in H,k\in K$, of course.)

The first one looks right but there's a simpler way, using: $ <x>$ is a subgroup of a group $ G$ if and only if $ x\in G$.

[quote="scorpius119"]For the second one, you didn't show closure. It means showing two elements in $ H \plus{} K$ have a sum that is also in $ H \plus{} K$.

To do this, say two elements are in $ H \plus{} K$, the first element can be written (not necessarily uniquely) as $ h_1 \plus{} k_1$, and the second element can be written as $ h_2 \plus{} k_2$.

Then find $ h,k$ such that the sum of $ h_1 \plus{} k_1$ and $ h_2 \plus{} k_2$ is $ h \plus{} k$.

(Here, $ h_1,h_2\in H, k_1,k_2\in K$, and you need $ h\in H,k\in K$, of course.)

The first one looks right but there's a simpler way, using: $ < x >$ is a subgroup of a group $ G$ if and only if $ x\in G$.[/quote]

Thanks.  You know, the problem, is that the books here in our school is very limited, and the examples in most of the books are more or less the same, so you don't really know, if your proof is correct. That is the reason why you can't practice.


symmetry
geometry
geometric transformation
reflection

Prove that a line tangent to a circle is perpendicular to the radius at the point of tangency.

Suppose that line $ \ell$ is tangent to circle $ \Gamma$ with center $ O$ at the point of tangency $ T$.

Choose an arbitrary point $ X\in\ell$. Observe that point $ X$ is outside of the circle since $ \ell$ is tangent to $ \Gamma$.

Draw $ XO$ and let $ C$ be the point of intersection of $ XO$ and $ \Gamma$. This means that point $ C$ is between $ X$ and $ O$ and so $ XO\equal{}XC\plus{}CO$.

$ T\in\Gamma$ and $ C\in\Gamma$, and so, by definition of a radius, $ TO$ and $ CO$ are radii so that $ TO\equal{}CO$.

$ XC>0$ (because it is outside of the circle and not on $ \Gamma$), then $ XO>CO\equal{}TO$.

Thus, $ TO$ is the shortest segment from line $ \ell$ to the center $ O$.

By definition of distance from a point to a line, $ TO\perp\ell$ at $ T$.

[b]Symmetry:[/b] You can use the fact that the entire circle O is reflectivly symmetric across the line containing the radius. From here you can argue that the normal at any point on the circle, say C, must be its own reflection over OC, and must thus be in the direction of OC. Then, the tangent line is clearly perpendicular to OC.


Suppose $ P$ is the point $ (5,3)$ and $ Q$ is the point $ (-3,6)$. Find point $ T$ such that $ Q$ is the midpoint of $ \overline{PT}$.

There are two equations you must solve:
\[ \frac{5\plus{}x}{2}\equal{}\minus{}3,\frac{3\plus{}y}{2}\equal{}6
\]

This implies $ (\minus{}11,9)$


function
combinatorics proposed
combinatorics

I'll post some nice combinatorics problems here, taken from the wonderful training book "Les olympiades de mathmatiques" (in French) written by Tarik Belhaj Soulami.
Here goes the first one:

Let $\mathbb{I}$ be a non-empty subset of $\mathbb{Z}$ and let $f$ and $g$ be two functions defined on $\mathbb{I}$. Let $m$ be the number of pairs $(x,\;y)$ for which $f(x) = g(y)$, let $n$ be the number of pairs $(x,\;y)$ for which $f(x) = f(y)$ and let $k$ be the number of pairs $(x,\;y)$ for which $g(x) = g(y)$. Show that \[2m \leq n + k.\]

I've the book.... No interest for me to answer :(

Pierre.

The book does not give any specific source for this problem.

This is problem 10.6 from the Russian olympiad in 1994.

For each $t$, let $n_f(t)$ be the number of values of $x$ for which $f(x)=t$, and define $n_g(t)$ similarly, except that this time we use $g$ instead of $f$. We have $m=\sum_t n_f(t)n_g(t),\ n=\sum_t n_f^2(t),\ k=\sum_t n_g^2(t)$, and the result follows (the sums are taken after all those $t$ for which at $n_f(t)n_g(t)>0$).


Let $ f(x)\equal{}5x^{13}\plus{}13x^5\plus{}9ax$. Find the least positive integer $ a$ such that $ 65$ divides $ f(x)$ for every integer $ x$.

$ 65\equal{}5\cdot13$ so we just need it to be divisible by both.
For divisiblity by $ 5$ let's  assume $ 5$ doesn't divide $ x$, then by Fermat's little theorem we need $ 5|13x\plus{}9ax \Longrightarrow 5|1\plus{}3a$
Similarly we need $ 13|5\plus{}9a$. Now we just use Chinese Remainder theorem and find the least positive $ a$.

    We note that by the Chinese Remainder Theorem, it suffices to find $a$ such that $f(x) \equiv 0$ modulo 5 and modulo 13.

    First, modulo 5 we have
    \begin{align*}
        f(x) = 5x^{13} + 13x^5 + 9ax \equiv 13x^5 - ax \pmod{5}.
    \end{align*}
    We note that if $5 \mid x$, then $f(x) \equiv 0$, so we can assume that $5 \nmid x$ and apply Fermat's little theorem to obtain
    \begin{align*}
        f(x) \equiv 13x - ax \equiv (3 - a)x \pmod{5}
    \end{align*}
    so if $f(x) \equiv 0$ for all $x$, we need $3 - a \equiv 0 \pmod{5}$, that is, $a = 3 \pmod{5}$.

    Now modulo 13, we again ignore the case of $13 \mid x$ as $f(x) \equiv 0$ if that is true, then we apply Fermat's little theorem as before.
    $$
    f(x) = 5x^13 + 13x^5 + 9ax \equiv 5x + 9ax \equiv (5 + 9a)x \equiv 0 \pmod{13},
    $$
    so we need $9a \equiv 8 \pmod{13}$.

    Thus we solve the system of congruences
    \begin{align*}
        a &\equiv 3 \pmod{5}\\
        9a &\equiv 8 \pmod{13},
    \end{align*}
    and applying the Chinese remainder theorem, we have $a$ is unique modulo 65, thus $a \equiv 63 \pmod{65}$, and $a = 63$ is the smallest solution.

Another solution:

$f(x)=5x^{13}-5x+5x+13x^{5}-13x+13x+9ax=5(x^{13}-x)+13(x^{5}-x)+9(a+2)x$

By Fermat's little theorem, we have $x^{13}-x\equiv 0 \pmod {13}$, so $5(x^{13}-x)\equiv 0 \pmod{65}$ and $x^{5}-x\equiv 0 \pmod{5}$, so $13(x^{5}-x)\equiv 0 \pmod{65}$. $65 \nmid 9$, then $65 \mid a+2$ and its easy to see that $a=63$ is the solution.


number theory
combinatorics proposed
combinatorics

Given 2 piles of stones. 2 players alternately make moves. In a move player can take any number of stones from 1 pile or equally from 2 piles. Player that takes the last stone wins. Find, in which initial positions first player wins. Find his winning strategy.

This has been discussed before. Look for the topic "A golden game" (started by Mindflyer, I think) in the Number Theory section. It's called [url=http://mathworld.wolfram.com/WythoffsGame.html]Wythoff's game[/url].

OK. What about if we have n piles and player can take equally from k piles (1<=k<=n)?

[quote="Sasha"]Is my generalization too crazy?[/quote]

No. This game has been analyzed completely by Fraenkel and Zusman:

  Aviezri Fraenkel and Dmitri Zusman
  A new heap game.
  Theoretical Computer Science 252 (2001),  pp 5-12.


I was looking, and I discovered basically the answer to the first problem on Wikipedia. It says that there are 288 ways to fill a $2 \times 2$ Sudoku thing, and well it's pretty easy to see that one divides that by $4!$ to get the answer to problem 1.

[url=http://en.wikipedia.org/wiki/Mathematics_of_Sudoku#Sudoku_with_rectangular_regions]Wikipedia Hint at #1[/url]

I guess it's like how last year one of the problems was actually on mathworld...

NOTE that I found this after I had submitted my solutions

After I finished my solution, I googled 2x2 sudoku and found the same thing.


geometry
3D geometry
tetrahedron
sphere
geometric transformation
reflection
geometry unsolved

Could anyone help me with some English terms in solid geometry?

1.) In the plane geometry, we have the term median to refer to the line segm
ent joining a vertex to the midpoint of the opposite side. In solid geometry, can we have the term median plane of a tetrahedron: through an edge and the midpoint of the opposite edge?

2.) Similarly, can we have the term bisector plane of a tetrahedra? or other term?

3.) In a sphere, the center of the sphere is the intersection of what planes?

4.) A plane that is the reflection of median plane about the bisector plane ... how do we call this plane?

5.) We call $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ a sphere, but how we call the set of points such that $(x-a)^2+(y-b)^2+(z-c)^2\leq r^2$

In my math (our country) the [u]bimedian[/u] is a segment which unites the midpoints of the opposite edges (there are three medians) and the [u]median[/u] is the segment which unites a vertex with the centroid of the opposite face (there are four medians). The common point of all seven above segments is the [u]centroid[/u] of the tetraedron. Such is here, in Romania.

Let me help where I can:

[quote="pvthuan"]3.) In a sphere, the center of the sphere is the intersection of what planes?[/quote]

For any two points A and B, the locus of all points P in space such that PA = PB is called the [i]perpendicular bisecting plane[/i] of the segment AB.

The center of a sphere lies on the perpendicular bisecting plane of every chord of the sphere.

[quote="pvthuan"]5.) We call $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ a sphere, but how we call the set of points such that $(x-a)^2+(y-b)^2+(z-c)^2\leq r^2$[/quote]

This set is called a [i]ball[/i] (more exactly, a [i]closed ball[/i]; an [i]open ball[/i] would be the set of all points such that $\left(x-a\right)^2+\left(y-b\right)^2+\left(z-c\right)^2<r^2$).

  Darij

Thank you, Nicula and Darij. I want to buy the book [i]Modern Pure Solid Geometry [/i]
by Nathan Ashiller Court. It is out of print, I am willing to buy used book or just a copy.


inequalities
function
inequalities unsolved

$a+b+c = 1, a,b,c>0$

Prove $\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\geq \sqrt{\frac{3}{2}}$

[hide]

since $a+b+c=1$, the inequality becomes

$\frac{a}{\sqrt{1-a}}+\frac{b}{\sqrt{1-b}}+\frac{c}{\sqrt{1-c}}\geq \sqrt{\frac{3}{2}}$

let $f(a)=\frac{a}{\sqrt{1-a}}$. This function is convex, cause $f''(a)=\frac{3}{4}*\frac{1}{(1-a)^{\frac{5}{2}}}$.

So, by Jensen, $f\left (\frac{a+b+c}{3} \right)\leq \frac{1}{3}(f(a)+f(b)+f(c))$, and we have the proof

[/hide]


MATHCOUNTS
probability
blogs
number theory
relatively prime
prime numbers

Hmm I got this question: Of the set  {20,21,22,23,24,25,26,27,28,29,30}, exactly 2 members are selected. What is the probability that the two numbers are relatively prime?

Its confusing to me!

Can someone help?

Well, in total there's $ \binom{11}{2} \equal{} 55$

Now I would proceed with casework....blah. I don't have a better method. You do know what releatively prime means...correct?

[hide="Here goes...."] Ok, so with 20, we have 21, 23, 27, 29 who are relatively prime...

21: 22, 23, 25, 26, 29 who are relatively prime.

22: 23, 25, 27, 29.

23: 24, 25, 26, 27, 28, 29, 30

24: 25, 29.

25: 26, 27, 28, 29

26: 27, 29,

27: 28, 29

28: 29

29: 30

Add them up, perhaps? $ \boxed{32}$ in total I believe....[/hide]

well....i do know what relatively prime is...but is there an easier way to do this w/o casework (i hate casework  :P )

First of all, the number of ways to choose two numbers, as has been noted, is $ \binom{11}{2}\equal{}55$. We next proceed by complementary counting. We thus consider all combinations of multiples of $ 2$, multiples of $ 3$, multiples of $ 5$, etc. (we can limit our case work to prime numbers since all composite numbers can be expressed as the product of prime numbers). So first, we consider even numbers. There are $ 6$ even numbers, and hence the number of ways to choose $ 2$ of them is $ \binom{6}{2}\equal{}15$. Next we consider multiples of $ 3$. There are $ 4$ multiples of $ 3$, and hence in order to choose two of them we have $ \binom{4}{2}\equal{}6$. However, we must neglect the case of $ (24,30)$, since we have already counted it. In the same way, there are $ 3$ multiples of $ 5$, and thus $ \binom{3}{2}\equal{}3$. This time we neglect the case of $ (20,30)$, since it was already counted. We only have one case for $ 7$: namely, $ (21,28)$. We don;t have any more cases to consider, so we have

$ \frac{55\minus{}15\minus{}5\minus{}2\minus{}1}{55}\equal{}\boxed{\frac{32}{55}}$.

Thanks!

I like complementary counting - but for this one I think casework is better (I think i wouldve overcounted)

I sent but I do not know what happens beyond

[quote="CRICKET229"]Erm that link .... sends me to pinkys blog[/quote]

You are probably clicking the wrong link.


symmetry algebra function domain conics ellipse

I have come across a very nice problem, that can be solved using high school physics but is quite challenging, so I thought you'd like it:

We have an infinite superconductive plate in empty space and a poing charge Q in distance R from this plate. There is a very small hole in this plate exactly in the closest point to the charge. If we let the charge move freely, it is attracted to the plate and starts to oscillate back and forth (it can pass thorugh the hole). Calculate the frequency (or period, whatever) of this movement.

... still nothing?

[hide="hint"]The infinite conductive plate guarantees a zero potential everywhere on it. That means all the electric field must be of the kind that there is zero potential in that plane, and also the field is perpendicular to the plane ... There is another simple situation (with no such ugly thing as an infinite plane) where this condition si satisfied. The field around the charge is the same, so the force must also be the same ...[/hide]

this is the most basic image problem (though I do not consider it a high school level), you place a charge -Q a distance R on the other side of the plane and then pretend the plane is gone. This satisfies the boundary conditions. The existance and uniqueness theorem guarantees this is a unique solution. The force on the charge can be calculated to be $\frac{-q^{2}}{16\pi \epsilon R^{2}}$. This does not however apply to moving charges, so I don't see where you are going with this problem.

I have encountered this problem in a high-school competition, it does involve some tricks, especially if you're a high-school student, but I think it's a nice one.

And I admit, that the non-stationary effects may affect the result, but let's assume that they won't. For example, let's set the mass of the charged thing to be sufficiently large so that it doesn't move too fast. Can you somehow quantify those effects? I don't see through electromagnetism that much, (I am still in high school;-) but on a first glance, it seems like there should be no magnetic force, from the symmetry of the whole situation .. or is there?

Umm, well if it's superconducting it will repel the induced magnetic charge. Just making a conducting plane. Anyway, if you want to ignore the physics (mass has nothing to do with it in E&M), then your motion equation is:

$F=m\frac{d^2x}{dt^2}=\frac{-q^2}{16\pi \epsilon x^2}$

I do not know of any elementary solution to this equation solved in the time domain. I suppose if you only want the period, then you change it to a radial domain and perhaps the bounds could be obtained.

What do you mean by "it will repel the induced magnetic charge."? If you can explain, I would like to know more about that - what happens when the charge starts moving.

I said the mass is large enough so that the force on a static charge is small and it starts to move slowly and the magnetic effects are not too big. Do you think that the other effects are there regardless of this?!

what are you unclear about, the superconducting plate replelling? or the induced charge in the first place. Unfortunately, a moving point charge is much more complex than can be explained at the high school level. A whole world of things happen when the charge starts moving, but all of the physics is very mathematically rigorous, as it deals with relativistic effects and electromagnetic waves.

I am curious about the original problem though, let's assume for a second that no magnetic charge is induced, and the moving charge is treated like the static case, the equation I gave above represents the necessary equation for motion. As you should note, it is analogous to a mass traveling in a gravitational field with a different proportional constant. In general, this differential equation cannot be solved. I am very curious how such a problem could be put on a high school exam.

... well, i admit, that the problem was one of the "stoppers" - the last problems that ensure that noone solves everything in the time limit ...

This was my solution - the situation is analogous to two masses that start 2R apart and attract each other. To find out the movement of one of them, instead of the second you can put a fixed mass in between them, with proper mass. Now solve the system. It is like if it was was orbiting the central fixed mass, but with a very eccentric orbit. According to the Kepler's ?th law, its period is the same as if the orbit was circular with radius R (because the main axis of the quasi-ellipse is 2R). This is an ordinary satellite-motion problem. I know my explanation is a bit fuzzy, but I can give you figures if it's too bad.

And, if you can explain, I'm really curious about what happens when the charge starts to move, at least qualitatively (yes, the repelling and effects of the motion of the charge) . You don't have to keep at high school level, the worst thing that can happen is that I don't get it;-)

Your solution seems like it'd be the only way to solve the problem, using kepler's laws and making the analogy to a planet in orbit.

As far as what happens, I would suggest reading Griffiths, Introduction to Electrodynamics for the answers you seek.

I do see what you're saying with a very large mass now though, it would cause the object to move slower and thus better approximate a point charge. I suppose that in the limit where R is extremely small compared to M, the problem might be ok.

Check this out, it shows the field of a moving point charge. Trust me, the math behind them is intense.

http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html

Will that magnetic field not influence the motion of charge? I'm not sure about that.

i think that pretty much any problem done with high school physics is flawed.. but since thats all i know im going to try it with that: (i am going to assume that it was meant to be an infinite plane of charge)
[hide="is this right?"]say it is a charged plane with surface charge density $-\sigma$ instead of a superconductor..

so there is a point charge of mass $m$ and charge $q$.

$E=\frac{\sigma}{2\epsilon_o}$ (twoards the plane)

from that, $F=\frac{q\sigma}{2\epsilon_o}=ma$

therefore, $a = \frac{q\sigma}{2m\epsilon_o}$

now, since the electric field is constant wrt $x$ (the distance from the plate), $U=\frac{qx\sigma}{2\epsilon_o}$

now to find the velocity at the center, $v_o$

$\frac{qx\sigma}{2\epsilon_o}=\frac{1}{2}mv_o^2$

from that, we have $v_o = \sqrt{\frac{qx\sigma}{m\epsilon_o}}$

now we know the acceleration is constant, so we can use $\Delta x = v_ot+\frac{1}{2}at^2$

after the charge goes out and comes back, $\Delta x = 0$

therefore, $0 = \frac{1}{2}at^2+v_ot=t(\frac{1}{2}at+v_o)$

ignoring the trivial $t=0$, we have $t = \frac{-2v_o}{a}$

because $a$ is twoards the plane, and $v_o$ is away from the plane, the negative sign goes away, and the magnatudes of both can be used...

therefore..

$t=2\sqrt{\frac{qx\sigma}{m\epsilon_o}}\frac{2m\epsilon_o}{q\sigma} = 4\sqrt{\frac{qx\sigma m^2 \epsilon_o^2}{m \epsilon_o q^2 \sigma^2}} = 4\sqrt{\frac{x m \epsilon_o}{q \sigma}}$

because this is only half, then period, $T$ is given by $T=2t$

therefore, $T=8\sqrt{\frac{x m \epsilon_o}{q \sigma}}$, or $f=\frac{1}{8}\sqrt{\frac{q \sigma}{x m \epsilon_o}}$

[/hide]

Sky, unfortunately that is not how it works. You are neglecting the induced surface charge density from the point charge onto the plane. This induced surface charge is not constant, it's a function of the radial distance away from the point as well as the distance from the point to the plane.

You typically solve this problem using what's called "The method of images". Basically, you seek to find a configuration such that the potential on the plane is zero everywhere. There is a methamtical theorem known as the Existance and Uniqueness Theorem that guarantees that if you find such a configuration such that the boundary conditions are satisfied, then you have a unique solution. This means that once you find a configuration such that the potential on the plane is zero, and the potential at infinity (in the direction perpendicular to the plane) is 0, you are done.

Consider the having a point charge +Q a distance d away from a plane in the +z direction, and you then have have a point charge -Q a distance d away from the plan in the -z direction, and pretend the plane is gone. This artificial -Q is known as the "image charge".

Calculate the potential when z=0. Calculate the potential at z = $\pm \infty$. Are the conditions satisfied? If so, then this situation is equivalent to the original. You can then calculate the force and you will find that the force does indeed depend on the distance of the point from the plane, thus making stuff very complicated.

ohh. lol. :(  now i understand what people were talking about before..

Sorry to interrupt... what category of Arts and Sciencesdoes engineering fit into?


I'm not sure if this is really the right place to ask but I'm stuck on a problem, and I'm not that good with math.. I'm working on a mod for a game called Tribes, and I need help figuring out the time to impact for a projectile. The things I have to work with are Initial Velocity, Acceleration (per second), Maximum Velocity and the distance to the target.

The projectile starts off at Initial Velocity, and increases speed by Acceleration every second untill it reaches Maximum Velocity. I hope that's enough info, any help would be greatly appreciated.

Hm, I program myself, but am not too well with making algorithms as fast as possible for computing.  And I don't know what Tribes's scripting is like, so I hope you understand basic C++/Java notation.  If not, I can tell you what it's doing.

[code]
//function returns the amount of frames it will take for projectile modeled by inital velocity, maximum velocity,
//and acceleratoin will take to reach a certain distance
public int calculateTime(double initialVelocity, double maximumVelocity, double acceleration, double distance)
{
     double framesUntilMax, tempDistance = 0;
     int totalFrames = 0;
     if( (framesUntilMax = (maximumVelocity - initialVelocity) / acceleration) < 0)
          framesUntilMax = 0;

     while(tempDistance < distance)
     {
          dist += (totalFrames >= framesUntilMax) ? maximumVelocity :
               totalFrames * acceleration + initialVelocity;

          totalFrames += 1;
     }

     return totalFrames;
}
[/code]

Now I know that isn't very elegant, and maybe not very fast, but I think it should work, returning the number of frames that must pass before the projectile reaches the distance that you want.  I didn't test it, though, so I'm not completely sure.

There's actually a whole forum for Computer Science... it's towards the bottom of the main page. It's times like these I wish I were a mod  :(

No kidding?  I never traveled down that far on the page  :blush:   I'll have to go there some time... I know this algorithm can be done faster, and I'm sure a lot of mine can too.

That's plenty good enough, thanks [b]alot[/b] for the help.  :)


Hi Guyz,
                I got a new project to create some latex documents for publications for my professor, as I am new to this language, I tried to gather all the information from the net and thru tutorials and almost am done with the project. I am now stuck with inserting pictures into the latex documents. I see many figures in the source soft copy of the file( which is to be converted into a latex document) which are graphs and mathematical related diagrams. All the source document is in word document. I am using the Winedit software to edit the latex documents. I saw different ways to include the pictures in the latex documents. The diagrams should be in eps or pdf forms. Please let me know how i can produce the figures which are in the word document into the eps or pdf forms. I tried to paste the document in the photoshop and later changed it to the .eps form then I used the following code:

--------------EPS form-----------

\documentclass[8pt]{article}
\usepackage{natbib}
\usepackage{graphicx}
\usepackage{graphics}
\begin{figure}[htp]
\includegraphics[height=60mm, width=45mm]{fig1}
\end{figure}

                    Then I tried to copy the figure into another word document and convert it directly into pdf by clicking the icon on the word document; "convert to pdf". well I got the fig1.pdf file now with me. Later I tried using the following code

------------PDF form ---------------

\documentclass[8pt]{article}
\usepackage{natbib}
\usepackage[pdftex]{graphicx}
\begin{figure}[htp]
\includegraphics*[viewport=100 200 500 650]{fg1}
\end{figure}

                  In either case I am getting the follwing error:

Cannot determine the size of the grapic in fig1 <no BoundingBox>

                Please help me with this....please guyz

See the solutions in [url=http://groups.google.com/group/comp.text.tex/browse_thread/thread/6e048135606fb4a2/116bb8364e02312d#116bb8364e02312d]comp.text.tex newsgroup[/url] on using EPS generated by Photoshop

Personally I use the option to save as EPS in OpenOffice's drawing program and that seems to work well with LaTeX.


algebra
polynomial
search
number theory proposed
number theory

find all polinomials P with real coefficientes such that iff P(r) is an integer, then r is an integer.

It was already discussedand anyway it is allmost trivial.

Here is the other topic where you will find a solution.
http://www.mathlinks.ro/viewtopic.php?t=6122&highlight=iran

Pierre.

[quote="Leonardo"]Where can i get the Iran Math Olympiad???[/quote] search after Iran in the problem source field. You will find tons of Iran problems.


calculus
advanced fields
advanced fields unsolved

I am in a lower math class and am being tutored by a friend of mine who is in higher math (calculus).  She has done an amazing job helping me out, but I feel it has caused her problems with her own class.  She was given a take home quiz and she started crying because she is not so clear on this section (derivatives).  I would love to help her as a way of returning the favor.  Is there anyone that will work out a few problems for me?  It would mean the world to her and I would feel better about her falling behind.  She told me the answers she has gotten just do not make sense.  Thank you for the consideration!

Welcome to AoPS!

1) There is a separate forum for calculus. The sort of problems it sounds like you want to post will go in the Calculus Computations and Tutorials forum.

2) We on this forum prefer not to do other people's homework for them. We're happy to help people who are having trouble with math (at any level), but most of us are more inclined to explain and stuff rather than do someone's homework. If you post some problems, you will receive (from me at least, if I answer) an explanation of how to do that sort of problem, not just free answers.


Two players play a game on a $2000 * 2001$ board. Each has one piece and the players move their pieces alternately.
 A short move is one square in any direction (including diagonally) or no move at all.
 On his first turn each player makes a short move.
On subsequent turns a player must make the same move as on his previous turn followed by a short move.
This is treated as a single move.
The board is assumed to wrap in both directions so a player on the edge of the board can move to the opposite edge.
The first player wins if he can move his piece onto the same square as his opponent's piece.
For example, suppose we label the squares from $(0, 0)$ to $(1999, 2000)$, and the first player's piece is initially at $(0, 0)$ and the second player's at $(1996, 3)$.
The first player could move to $(1999, 2000)$, then the second player to $(1996, 2)$.
Then the first player could move to $(1998, 1998)$, then the second player to $(1995, 1)$.
 Can the first player always win irrespective of the initial positions of the two pieces?


LaTeX
Functional Analysis
real analysis
real analysis unsolved

In my Hilbert space course, i came across the question "write down an explicit isomorhism from $ l^2(\mathbb{Z})$ to $ l^2$ ". I cannot find such an isomorphism. Is it something very obvious?
Thanks.

Can you find a bijection between $ \mathbb Z$ and $ \mathbb N$?

I think that a bijection from $ \mathbb{N}\to \mathbb{Z}$ is $ n\to k$ if $ n\equal{}2k$, and $ n\to \minus{}k$ if $ n\equal{}2k\plus{}1$. Is this correct?

That's only correct if your version of $ \mathbb{N}$ includes $ 0.$ I would normally define $ \mathbb{N}$ so as not to include zero.

But anyway - use the same idea, tweak it slightly, you'll make it work. Do you see how do get the desired isomorphism of Hilbert spaces from this?

$ \text{\LaTeX}$ tip: for these Hilbert spaces, I like \ell better than l. That is, $ \ell^2$ rather than $ l^2.$ It's just a preference about appearances.


geometry
trapezoid
circumcircle
geometry proposed

In an hexagon $ ABCDEF$ the oposite sides are parallel and the diagonals $ AD,BE,CZ$ are equal.Prove that the hexagon is inscribed in a circle.

 Babis.

Consider ABDE: it is a trapezium, as AB//DE(they are opposite sides). But its diagonals AD and BE are equal; that means, ABDE is an isosceles trapezium. Similarly, one can obtain that AFDC and FBCE are both isosceles trapeziums. But every isosceles trapezium can be inscribed in a circle. We notice that the circumcircles of the three trapeziums share, one by one, a string: AD,BE,CF. Thus, these strings are the root axis of the three circles, and so are concurrent. But that common point is a point belonging also to three perpendicular bisectors: the common of AB and DE, the common of AF and CD, and the common of FE and BC(common because of the isosceles trapeziums). That is, the hexagon is inscribed in a circle  whose center is the root center of the three circles described above(that now appear to be the same circle).


How much would it help, if I took it this year, and the SAT two times this year?

Pretty much, is this enough to get 600-700s on the SAT verbal and math:
Some algebra 1
Intro to Geomatry
Intro to problem solving

I don't see how it would help on the verbal...

Otherwise, yes, though I'd say most people who really understand them should get 800, or very close.

Heh. Meant to say math and that I was also doing verbal :X


MATHCOUNTS AMC USA(J)MO USAMO trigonometry ARML

How many of last year's top 8 finishers were eighth graders?

I don't think this really matters, does it? But probably 5-6 of them. Just because you're in 6th or 7th doesn't mean you can't be better than 8th graders... and just because you're an 8th grader doesn't mean you can't beat seniors on USAMO (Evan O'Dorney... who also won the national spelling bee...)

dude that was a pointless comment

eighth graders obviously beat seniors on the USAMO every year

by virtue of the fact that eighth graders often make USAMO

WHILE MOST SENIORS DO NOT IN FACT TAKE USAMO GASP

also it was pointless since you referenced that other competition which we shall not deign to mention here

Anyway, to answer the question,

Places 1-5 were 8th grade for certain.  There were probably some more 8th graders 6-8, though not_trig and MysticTerminator are correct (I did my best at state mathcounts in 6th grade).

me and calvin (7th and 6th graders last year respectively) got 7th and 8th. I think bryce got 5th and jenny C. got 2nd and Grey Symon got 1st and this homeschool kid and 'insert name that I know but don't remember' got 3rd and 4th. Arjun got 6th.

I think the homeschool dude wasn't in 8th, (he might of been), so probably 5, but maybe 6...

I still am mad of how i didn't get a trophy thing (7th ind. and 6+ got trophies, 5th team and 4+ got trophies!  :furious: )

[hide]
btw, WHY AM I NOT ON THE CENTRAL NC MATHCOUNTS PAGE!! >_> i should be... I got first...

heres the website (unfortunately without my picture) http://www.penc.org/Student-Youth/MATHCOUNTS/MCCentralCarolina.aspx
[/hide]

I suspect this "homeschool dude" was Brendan Fletcher? Or David Lucia (who went to nats but may or may not be homeschooled?) Also Gray Symon is not that good... well maybe at MC... w/e.

When you get out of middle school, you see how silly MathCounts was.

1.  Gray Symon - 41???
2.  Jenny Chen - 39?
3.  Brendan Fletcher - 38?
4.  David Lucia - 37
5.  Bryce Taylor - 37

Also, david goes to Providence Day, I think.

[quote="not_trig"]why do you know this? and why does it matter? it was last year's state mathcounts... which has no bearing whatsoever on this year because they were all 8th graders. Oh, well I guess that's the relationship? No one in the top 5 last year is gonna be there this year? ... ok[/quote]

Memory.  Yes, comparing the results of old 8th graders is rather pointless, but someone asked for it, so I gave them the information.

[quote="PI-Dimension"]I think david goes to hanes middle?

so brendan must be the homeschool dude[/quote]

Umm, no.  David goes to Providence Day High (in Charlotte) and went to some middle school in Charlotte when he was in middle school.  Hanes Middle School is in Winston-Salem (I went there for middle school).

However, Brendan is indeed home schooled.

[quote="dnkywin"]So, in all, the only people who were in top 8 last year in NC who was not in 8th grade was a 6th grader (me) and a seventh grader (PI-Dimension)?[/quote]

Pretty much, yeah.

[quote="dnkywin"]Wait...
What grade was Christine Hong in?[/quote]

She is a freshman this year (i.e. 8th last year).

[quote="b-flat"][quote="dnkywin"]Wow. NC lost a [i][b]lot[/b][/i] talent this year :!: Increases my chances of getting to nats :lol: .[/quote]

The talent was not lost (with the exception of Jenny); it's just been reallocated to use for different purposes (i.e. arml, AMCs, etc)  :)[/quote]

ha, ha

so 8 seniors graduating from ARML A team don't count? (5 or 6 of which were USAMO qualifiers? and arnav/pardon/yasha/misha included?)

No. Didn't you get the memo? This is now the NC middle school forum.

Well, if it's the NC [i]Middle School[/i] forum, then NC [i]did[/i] lose a lot of talent. :D  :lol:

It's nice to see the next generation emerging... I feel old
I'd better do well my final year :/

[quote="dnkywin"]Wow. NC lost a [i][b]lot[/b][/i] talent this year :!: Increases my chances of getting to nats :lol: .[/quote]

you $ \ne$ grammar

Well, math $ \neq$ gramm[i]e[/i]r  :lol:

[quote="dnkywin"]Well, math $ \neq$ gramm[i]e[/i]r  :lol:[/quote]

I assume also that $ \text{math} \neq \text{spelling}$?

How very hilarious. I assume this thread has served its purpose. Locked.


Let $ a,b,c > 0 $ and $ ab+ac +bc =3 $ .Prove that :
 $ a^2+b^2+c^2+abc \geq a+b+c+1 $

Multiply both sides by two and add $2(ab+bc+ca)$ gives
$(a+b+c)^2+a^2+b^2+c^2+2abc-6 \ge 2(a+b+c)+2$
Recall that $a^2+b^2+c^2+2abc + 1 \ge 2(ab+bc+ca)=6$,
$L.H.S. \ge (a+b+c)^2 -1$
It remains to prove $(a+b+c)^2-2(a+b+c)-3 \ge 0$ or $(a+b+c-3)(a+b+c+1) \ge 0$
But this simple, since $a+b+c \ge \sqrt{3(ab+bc+ca)} = 3$

Nice solution ! but you used the ineq $ a^2+b^2+c^2+2abc + 1 \geq 2(ab+bc+ca) $ which is more difficult to prove (of course mixing variables ;) ) than the original ineq

   This problem seemed to be posted a long time ago .
  Of course , this problem has a short solution but try thinking about my solution :
 $ a^2+b^2+c^2+abc \geq a+b+c+1 \iff (a+b+c-3)^2 +5(a+b+c) +abc -16 \geq 0 $
 It's enough to prove : $ 5(a+b+c) +abc -16 \geq 0 $

  Let's post solutions for the last ineq  :)

[quote="siuhochung"][quote="nttu"]Nice solution ! but you used the ineq $ a^2+b^2+c^2+2abc + 1 \geq 2(ab+bc+ca) $ which is more difficult to prove (of course mixing variables ;) ) than the original ineq [/quote]

Nono, we only need to use $2abc+1 \ge 3\sqrt[3]{a^2b^2c^2}$ and then schur! It has become famous after apmo 04 no.5  :lol:[/quote]

   Really nice , Siuhochung ! I saw this ineq many times on ML , but i forgot its nice solution  :(  till you remind me  ;)


probability
geometry
3D geometry
modular arithmetic

1. The sum of three binomials each of the form ma+nb, where m and n are natural numberds, is 4a +8b. how many distinct sets of such three binomials exist?

2. Emily randomly selected a diagonal of a regular octagon. Tim randomly selected one of the remaining diagonals. Find the probability that the diagonals they selected have the same length. Expres answer as comon fraction.

3. Exactly one ordered pair of positive integers (x,y) satisfies the equation 37x + 73y = 2016. What is the sum of x+y ? (Is there any other way than brute force)

4. What is the least two-digit integer whose cube has the same tens and units digits as the integer?

5. What is the smallest positive integer that cannot be expressed as the sum of eight or fewer cubes of positive integers?

6. A rectangular floor is covered with square tiles. The floor is 81 tiles long and 63 tiles wide. If a diagonal is drawn across the floor, how many tiles will it cross?

Answers:

[hide]1. 12

2. 31/95

3. 36

4. -99

5. 23

6. 135[/hide]

[hide="1"]

so find total ways to get 4, total ways to get 8, and multiply.

so for 4, there is 2,1,1

so for 8, there is 6,1,1 5,2,1 4,2,2 4,3,1 3,3,2

1*5=5
(I bet I missed something
[/hide]

[hide="2"]
so there are 20 diagonals, by n(n-3)/2

so, use casework, there are 4 long diagonals, so 1/5*3/19 chance of that happening
there are 8 semi long diagonals, so 2/5*7/19 chance of that
There are 8 short diagonals, so 2/5*7/19 chance of that
so 3/95+14/95+14/95=31/95

[/hide]

[hide="3"]

so you have 37x+73y=2016
(The only way I can find out how is brute force)

[/hide]

[hide="4"]

remember negatives. So basically, first, the last digit cubed has to equal itself.
so the choices are 0,1,4,5,6,9. In addition, 9times the units digit times the first digit squared has the units digit as the tens digit. (sorry if this is unclear) So if this is a target, try them out, starting at the nineties. so, 9*x*9^2=9 (or something with a units digit of 9. so taking only units digits,9x=9. So take the highest possible one out of the units digits I stated above (since they are negative) to get the lowest answer. It turns out to be   -99.

[/hide]

[hide="5"]
so list out the cubes, 1,8,27,64.....
so first, you know that it can't be under 8. So try over it. Since 24, wouldn't work, try numbers under that. You can quickly find that 23 is the answer.
[/hide]

[hide="6"]
The equation for these is m+n-GCF(m,n) where m and n are the dimensions of the floor.
81+63-9=135.[/hide]

[quote="kl2836"]oops, sorry, #1 is 5[/quote]

#1 is 12. Keep the answer key as it is. It's Ihatepie who messed up. :D

argh this is getting me dizzy. oh well, its either 5 or 12 but mostly likely 12 :roll:

[quote="l Bob l"]1. The sum of three binomials each of the form ma+nb, where m and n are natural numberds, is 4a +8b. how many distinct sets of such three binomials exist?

#1 is 12, not 5. It even says on the answers.

4 can be expressed as a)$ 1 \plus{} 1 \plus{} 2$
8 can be experssed as b)$ 1 \plus{} 1 \plus{} 6$,___c)$ 1 \plus{} 2 \plus{} 5$,___d)$ ,1 \plus{} 3 \plus{} 4$,___ e)$ ,2 \plus{} 2 \plus{} 4$,___ f)$ 2 \plus{} 3 \plus{} 3$. the 2 in a) can match up with either $ 1$ or $ 6$ in b). The other 1's are forced to pair up with  the two left in b). That's $ 2$ ways. For c), 2 in a) can pair with 1, 2, or 5, and again, 1's are forced to pair with the rest. That's $ 3$ possibilities. For d), there are 3, 2 for e, 2 for f. $ 2 \plus{} 3 \plus{} 3 \plus{} 2 \plus{} 2 \equal{} [12]$[/quote]

o, I see.

[hide="3"]#3 - diophantine equations

$ x\equal{}\frac{2016\minus{}73y}{37}$

Thus, the top must be divisible by 37, so take that mod 37

$ 2016\minus{}73y \equiv 18\minus{}y \mod 37$

So the smallest, and only $ y$ that will yield a positive pair is $ 18$.

$ x\equal{}\frac{2016\minus{}18*73}{37}$.

Which should give the desired answer.[/hide]

HA HA, I'm going to make you all look stupid (actually, Mr. Frost is :P)

Notice that the coefficients are all multiples of 36 plus 1 and that 2016 is divisible by 36. So take both sides mod 36...

$ x \plus{} y \equiv 0 \pmod{36}$

and that makes the problem much simpler! :D

Slight Brute force method but easier for #3 (doesn't require mods and other stuff):

[hide]The least multiple of 73 less than 2016 is 73*27=1971.
The two numbers have a difference of 45.
Now, what happens if you add 37 3 times and then subtract 1 73?
You get +38.
Add 2 more 37's and subtract 1 more 73's.
You get +39.
Continue this on and on until 45 find that there are
17 37's and 27-8=19 73's

17*37+19*73=2016
17+19=36[/hide]

nice yongyi... I should've seen that

[hide="number 3"]$ 37x \plus{} 73y \equal{} 2016$
$ 37x \plus{} 73y \equal{} 36(56)$
$ 36x \plus{} x \plus{} 72y \plus{} y \equal{} 36(56)$
$ 36(x \plus{} 2y) \plus{} x \plus{} y \equal{} 36(56)$

Thus x+y is  a multiple of 36.
If $ x \plus{} y$ is greater than or equal to 72, $ x \plus{} 2y > 72$.
But $ x \plus{} 2y < 56$ since $ 36(x \plus{} 2y) \plus{} x \plus{} y \equal{} 36(56)$ and x,y are positive integers.
Since $ x \plus{} y$ cannot be 0, it is 36.
Please correct me if i'm wrong  :)  [/hide]


We consider a prism with 6 faces, 5 of which are circumscriptible quadrilaterals. Prove that all the faces of the prism are circumscriptible quadrilaterals.

[hide="hint"]just use Pithot's lemma (did i spelled corectly?)[/hide]

What's that lemma? I googled it, but no results...

$ABCD$ is a circumscriptibile quadrilateral $\Longleftrightarrow AB+CD = AD+BC$.

using it you will get the result faster then using the official solution.
is this problem  a geometry or combinatorics one?


Counting with Multiplication Part 1

We visualize how we can use multiplication in basic counting problems with a tree.

I thought that the material in the lecture was taught clearly.

i like the way he used a very simple topic to make his point. I have seen people teach this with like 15 of each thing making it very complicated.

I agree with Gamebot. Using trees to solve combinatorics problems are quite useful.

hey guys ... i know i'm a bigginer ...

but i really need to watch the vedio but i have a stupid question :D

where is the link ...???

best wishes

On the top, where there's all the links, click Resources.  Then, among the subtitles (like where you see Profile and Members and all that) click Videos.  Then, click the first link, "A collection of over 70 videos".  Then, scroll down into Chapter 1 Section and find Counting with Multiplication Part 1.  Now, just watch!  :D

if ur lazy here is link [url]http://www.artofproblemsolving.com/Resources/video.php?video_id=18[/url]


calculus
integration
algebra
polynomial
modular arithmetic
quadratics
superior algebra

Let $ \mathbb{Q}(\sqrt{n})$ be an extension of field $ \mathbb{Q}$. Prove that if $ 4$ does not divide $ n\minus{}1$, then $ \mathbb{Z}(\sqrt{n})$ is the integral closure of $ \mathbb{Z}$ in $ \mathbb{Q}(\sqrt{n})$. In other words, if a polynomial over $ \mathbb{Z}$ admits a root  the form $ a\plus{}b\sqrt{n}, a,b \in \mathbb{Q}$, then actually $ a,b \in \mathbb{Z}$, What if $ 4$ divides $ n\minus{}1$?

If $ 4$ divides $ n\minus{}1$, the integral closure (if I understand it correctly) is $ \mathbb{Z}\left[ \frac{1 \plus{} \sqrt{n}}{2} \right]$.

if $ D\equiv 1\pmod{4}$, $ D$ square-free, the algebraic integers in the quadratic field $ \mathbb{Q}(\sqrt {D})$ are those numbers of the form $ \frac {m \plus{} n\sqrt {D}}{2}$, where $ m$ and $ n$ are of the same parity. recall that an algebraic integer is a number whose minimal polynomial has integer coefficients. so yes.


Let $ a,b,c$ be the sidelengths of a triangle. Prove that
\[ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \plus{}\frac{3(ab\plus{}bc\plus{}ca)^2}{a^3b\plus{}b^3c\plus{}c^3a\minus{}abc(a\plus{}b\plus{}c)} \ge 5.\]
When do we have equality?

[quote="can_hang2007"]Let $ a,b,c$ be the sidelengths of a triangle. Prove that
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {3(ab \plus{} bc \plus{} ca)^2}{a^3b \plus{} b^3c \plus{} c^3a \minus{} abc(a \plus{} b \plus{} c)} \ge 5.
\]
When do we have equality?[/quote]

We assume that $ a \ge b \ge c$ then
\[ a^3 b \plus{} b^3 c \plus{} c^3 a \minus{} abc\left( {a \plus{} b \plus{} c} \right) \equal{} a^2 b\left( {a \minus{} c} \right) \plus{} b^2 c\left( {b \minus{} a} \right) \plus{} c^2 a\left( {c \minus{} b} \right) \le a^2 b\left( {a \minus{} c} \right) \le a^2 b^2
\]

It suffices to prove
\[ \frac{{a^2  \plus{} b^2  \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}} \plus{} \frac{{3\left( {ab \plus{} bc \plus{} ca} \right)^2 }}{{a^2 b^2 }} \ge 5
\]
with $ a \plus{} b \equal{} 1$ and $ x \equal{} ab$ then the inequality can rewrite as
\[ \frac{{c^2  \plus{} 1 \minus{} 2x}}{{c \plus{} x}} \plus{} \frac{{3\left( {c \plus{} x} \right)^2 }}{{x^2 }} \ge 5
\]
\[ \frac{{1 \minus{} \left( {c \plus{} 2} \right)x}}{{c \plus{} x}} \plus{} \frac{{3c^2 }}{{x^2 }} \plus{} \frac{{6c}}{x} \ge 2 \minus{} c
\]
Hence it's enough to prove in the case $ x \equal{} \frac{1}{4}$ which is equivalent with

\[ \frac{{2 \minus{} c}}{{4c \plus{} 1}} \plus{} 48c^2  \plus{} 26c \minus{} 2 \ge 0
\]

\[ \Leftrightarrow c\left( {48c \plus{} 26 \minus{} \frac{9}{{c \plus{} 1}}} \right) \ge 0
\]

Obvious true. The inequality is proved.
Equality holds if and only if $ a\equal{}b,c\equal{}0$.


What is the pH of the solution when

(a) $0.2$ mol/l $H_{2}SO_{4}$ mix with $0.1$ mol/l $Na_{2}CO_{3}$, in $1$l solution.

(b) $0.2$ mol/l $H_{2}SO_{4}$ mix with $0.3$ mol/l $Na_{2}CO_{3}$, in $1$l solution.

(a) The concentration of hydronium ion from sulphuric acid is given by

$[H_{3}O^{+}] = \frac{\sqrt{C^{2}+6KC+K^{2}}+C-K}{2}$

where C is the concentration of $H_{2}SO_{4}$ and K is the ionization constant of hydrogensulphate ion (K = 0.012). In this case, C = 0.2 and substituting gives $[H_{3}O^{+}] = 0.21 mol/L$. This concentration of acid is enough to convert all the carbonate ion in carbonic acid, so in the "end" we will have a solution of $H_{2}CO_{3}$ with concentration 0.20 mol/L (this is not correct, since a reasonable amount of carbon dioxide will bubble out of the solution, but for the present case that doesn't matter) in which $[H_{3}O^{+}] = 0.01 mol/L$. Then we have the following equilibrium in solution

$H_{2}CO_{3}+H_{2}O \rightleftharpoons HCO_{3}^{-}+H_{3}O^{+}$

and the equilibrium concentrations will be

$[H_{2}CO_{3}] = 0.2-x \approx 0.2$ [since $K_{a}(H_{2}CO_{3})$ is very small, $4.3 \times 10^{-7}$
$[HCO_{3}^{-}] = x$, $[H_{3}O^{+}] = 0.01+x$.

Solving the resultant quadratic equation we get $x = 8.6 \times 10^{-6}$ and so $[H_{3}O^{+}] \approx 0.01 \Rightarrow  pH = 2$.


geometry
ratio
calculus
derivative
probability
expected value

The spectroscopic methods of structure determination – infrared, NMR, and UV spectroscopies and mass spectrometry – are very useful tools available to the organic chemist for the complete determination of the structure of an unknown compound. The study of the basic aspects of these techniques is an important part of any standard Organic Chemistry course, and some spectroscopy problems also sometimes appear in Chemistry Olympiads. In this thread, several problems regarding those four spectroscopic methods are presented in order to provide suitable training for those interested. The problems are roughly organised by increasing difficulty: however, a more challenging exercise may appear in the first problems, and an easy one can come up in the last problems. Most problems include a molecular formula, but for some this information is not provided: instead, data from combustion/elemental analysis and mass spectral information is provided. The first 19 problems address basic knowledge of NMR and IR spectroscopy. The next 37 problems are mainly structure determination problems, where the task is to deduce the structure of one or more compounds from the spectral data provided (sometimes, some chemical reactivity information is also given). Most of these problems include NMR information. Problems 57 to 66 provide specific training in IR spectroscopy, Mass spectrometry, and UV spectroscopy. From problem 67 onwards, additional problems and more advanced problems are provided. Whenever not specifically stated, NMR data is given as ppm units, IR data in cm-1, and MS data in m/z values. For the description of the intensity and shape of the IR absorption bands, the following abbreviations were adopted: w = weak; m = medium; s = strong; vs = very strong; vbr = very broadened band (like an –OH band from alcohols, phenols, and carboxylic acids); br = medium broadened band (like the C=O absorption bands), sbr = small broadened band (like the N-H, C-O, and some aromatic bands); sh = sharp. Above all, don’t forget: try to have some fun with these problems!

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

[size=150][b]Problem 1[/b][/size]

How many types of non-equivalent protons are there in each of the following molecules?

(a) $ CH_{3}CH_{2}Br$.
(b) $ CH_{3}CH_{2}OH$.
(c) $ CH_{3}\minus{}O\minus{}CH_{3}$.
(d) The isomers of formula $ C_{2}H_{4}Cl_{2}$.
(e) $ CH_{3}CH_{2}\minus{}O\minus{}CH_{2}CH_{3}$.
(f) $ CH_{3}\minus{}O\minus{}CH_{2}CH_{2}CH_{3}$.
(g) $ CH_{3}\minus{}O\minus{}CH(CH_{3})_{2}$.
(h) $ CH_{3}CH_{2}CH_{2}CH_{2}OH$.
(i) The four isomers of formula $ C_{3}H_{6}Br_{2}$.
(j) $ CH_{3}CH_{2}CH_{2}NO_{2}$.
(k) $ CH_{3}\minus{}O\minus{}CH_{2}CH(CH_{3})_{2}$.
(l) 2-Methylbut-1-ene.
(m) Acetone.
(n) $ CH_{3}CH_{2}CHO$.
(o) Toluene.
(p) $ CH_{2}\equal{}CHCH_{2}OH$.
(q) 2-Chlorobutan-2-ol.
(r) $ C_{6}H_{5}CH_{2}CH_{3}$.
(s) 1,2-Dichloropropane.
(t) Methylcyclopropane.
(u) cis-3-Hexene.
(v) p-Methoxyaniline.
(w) Styrene ($ C_{6}H_{5}CH\equal{}CH_{2}$).
(x) N,N-Dimethylacetamide ($ CH_{3}CON(CH_{3})_{2}$).
(y) 3-Methylpen-1-yne.
(z) $ CH_{2}\equal{}CH\minus{}CO_{2}\minus{}CH_{2}CH_{3}$.

[size=150][b]Problem 2[/b][/size]

How many 1H- and 13C-NMR signals are expected in the spectrum of each of the following compounds?

(a) Ethane.
(b) Propane.
(c) $ CH_{3}\minus{}O\minus{}C(CH_{3})_{3}$.
(d) 2,3-Dimethylbut-2-ene.
(e) Z-But-2-ene.
(f) E-But-2-ene.
(g) n-Butane.
(h) Ethanol.
(i) Propene.
(j) 1,2-Dibromopropane.
(k) 1,1-Dimethylcyclopropane.
(l) trans-1,2-Dimethylcyclopropane.
(m) cis-Dimethylcyclopropane.
(n) Pen-1-ene.
(o) 1-Chloropropan-2-ol.
(p) n-Propylbenzene.
(q) 1,1-Dimethylcyclohexane.
(r) Naphthalene.
(s) 1-Methylcyclohexene.
(t) p-Diethylbenzene.
(u) $ (CH_{3})_{3}C\minus{}CO_{2}\minus{}CH_{3}$.
(v) 3-Mehylpentanal.
(w) 1,1,4-Tribromobutane.
(x) 2-Chloropropene.
(y) $ p\minus{}CH_{3}O\minus{}C_{6}H_{4}\minus{}CH\equal{}CHCH_{2}CH_{3}$.
(z) 1,1-Dichlorocyclopropane.

[size=150][b]Problem 3[/b][/size]

Identify the compound(s) that meet each of the following descriptions:

(a) $ C_{2}H_{6}O$; only one singlet in its 1H-NMR spectrum.
(b) $ C_{5}H_{8}$; IR absorptions at  3300 and 2150m cm-1.
(c) An hydrocarbon with 7 signals in its 13C-NMR spectrum.
(d) $ C_{5}H_{12}$; only one signal in its 1H-NMR spectrum.
(e) $ C_{4}H_{8}O$; strong IR absorption at 3400 cm-1.
(f) $ C_{3}H_{7}Cl$; one doublet and one septet (1H-NMR).
(g) A compound with six carbons and only 5 signals in its 13C-NMR spectrum.
(h) $ C_{4}H_{8}O$; strong IR absorption at 1715 cm-1.
(i) $ C_{4}H_{8}Cl_{2}$; two triplets (1H-NMR).
(j) $ C_{5}H_{10}$; only one signal in its 1H-NMR spectrum.
(k) A compound with 4 carbons that shows 3 signals in its 13C-NMR spectrum.
(l) $ C_{8}H_{10}$; IR absorptions at 1600 and 1500 cm-1.
(m) $ C_{4}H_{8}O_{2}$; one singlet, one triplet, and one quartet (1H-NMR).
(n) $ C_{4}H_{8}O_{2}$; only one signal in its 1H-NMR spectrum.

[size=150][b]Problem 4[/b][/size]

How would you use infrared spectroscopy to distinguish between these pairs of isomers?

(a) $ HC\equiv CCH_{2}NH_{2}$ and $ CH_{3}CH_{2}C\equiv N$
(b) $ CH_{3}COCH_{3}$ and $ CH_{3}CH_{2}CHO$
(c) $ CH_{3}C\equiv CCH_{3}$ and $ CH_{3}CH_{2}C\equiv CH$
(d) $ CH_{3}COCH\equal{}CHCH_{3}$ and $ CH_{3}COCH_{2}CH\equal{}CH_{2}$
(e) $ H_{2}C\equal{}CH\minus{}O\minus{}CH_{3}$ and $ CH_{3}CH_{2}CHO$

[size=150][b]Problem 5[/b][/size]

Three isomeric dimethylcyclopropanes give, respectively, 2, 3, and 4 signals in their 1H-NMR spectra. Identify the isomer that gives each of those numbers of signals.

[size=150][b]Problem 6[/b][/size]

Each of the following compounds have an 1H-NMR spectrum characterized by having only one signal, which appear at the indicated chemical shift. Identify each compound.

(a) $ C_{8}H_{18}$; 0.9 ppm.
(b) $ C_{5}H_{10}$; 1.5 ppm.
(c) $ C_{8}H_{8}$; 5.8 ppm.
(d) $ C_{4}H_{9}Br$; 1.8 ppm.
(e) $ C_{2}H_{4}Cl_{2}$; 3.7 ppm.
(f) $ C_{2}H_{3}Cl_{3}$; 2.7 ppm.
(g) $ C_{5}H_{8}Cl_{4}$; 3.7 ppm.
(h) $ C_{12}H_{18}$; 2.2 ppm.
(i) $ C_{3}H_{6}Br_{2}$; 2.6 ppm.

[size=150][b]Problem 7[/b][/size]

For each of the following compounds, predict the approximate positions, shapes, and intensities of as many bands as you can that are expected to be observed in their infrared spectra:

(a) Butanone.
(b) 4-Methylpen-1-ene.
(c) 4-Methylpen-1-yne.
(d) Methyl butanoate.
(e) 2-Methyloctan-2-ol.
(f) m-Diethylbenzene.
(g) 4-Methylbenzaldehyde.
(h) Acetophenone.
(i) cis-Pen-2-ene.
(j) Styrene.
(k) 4-Methylphenol.
(l) Anisole (methoxibenzene).
(m) Isobutyric acid.
(n) Vinyl acetate.
(o) Propionamide.
(p) N-methylaniline
(q) Methyl benzoate.
(r) Phenyl acetonitrile.
(s) 4-Nitrotoluene.

[size=150][b]Problem 8[/b][/size]

Distinguish the stereoisomers of 1,3-dibromo-1,3-dimethylcyclobutane on basis on their 1H-NMR spectra:

- isomer X: 2.13 ppm (singlet, 6H), 3.21 ppm (singlet, 4H).

- isomer Y: 1.88 ppm (singlet, 6H), 2.84 ppm (doublet, 2H), 3.54 ppm (doublet, 2H) (both doublets have equal spacings).

[size=150][b]Problem 9[/b][/size]

In the 1H-NMR spectrum of each of the following compounds there are only two signals (both singlets), having the indicated chemical shifts (in ppm). Identify each compound.

(a) $ C_{6}H_{8}$; 2.7 (4H) and 5.6 (4H).
(b) $ C_{5}H_{11}Br$; 1.1 (9H) and 3.3 (2H).
(c) $ C_{6}H_{12}O$; 1.1 (9H) and 2.1 (3H).
(d) $ C_{6}H_{10}O_{2}$; 2.2 (6H) and 2.7 (4H).

[size=150][b]Problem 10[/b][/size]

How many 1H-NMR signals are expected in the spectrum of each of the following compounds?

(a) 1-Chloro-2-metilpropane.
(b) 1-Bromobutane.
(c) 1-Butanol.
(d) Butane.
(e) 1,4-Dibromobutane.
(f) 2,2-Dibromobutane.
(g) 2,2,3,3-Tetrabromobutane.
(h) 1,1,4-Tribromobutane.
(i) 1,1,1-Tribromobutane.
(j) Vinyl bromide ($ CH_{2}\equal{} CH\minus{}Br$).
(k) 1,1-Dibromoethene.
(l) cis-1,2-Dibromoethene.
(m) trans-1,2-Dibromoethene.
(n) Allyl bromide ($ CH_{2}\equal{} CHCH_{2}Br$).
(o) 2-Methylbut-2-ene.
(p) 2-Bromopropene.
(q) 1-Butyne.

[size=150][b]Problem 11[/b][/size]

One of the following compounds has only one peak in its 1H-NMR spectrum and only two peaks in its 13C-NMR spectrum. Which one?

(a) Cyclohexane.
(b) Cyclopropane.
(c) Ethane.
(d) 2,3-Dichlorobutane.
(e) 2,2-Difluoro-1,3-Dioxacyclopentane.

[size=150][b]Problem 12[/b][/size]

Each of the following molecules shows only one peak in their 1H-NMR spectra. Order them according to increasing chemical shift: methane, benzene, acetylene, acetone, dichloromethane, cyclohexane, and ethylene.

[size=150][b]Problem 13[/b][/size]

The infrared spectrum of cis-cyclopentan-1,2-diol shows an –OH band at a lower frequency relative to the position of a free –OH group band. It is also observed that the band doesn’t disappear even at high dilution. On the other hand, trans-cyclopentan-1,2-diol doesn’t show that band. Can you suggest an explanation for this observation?

[size=150][b]Problem 14[/b][/size]

What kind of 1H-NMR spectrum is expected for the compound $ (Cl_{2}CH)_{3}CH$? Make a simple drawing of the spectrum, showing the relative position of each signal and its splitting pattern.

[size=150][b]Problem 15[/b][/size]

A compound of molecular formula $ C_{3}H_{6}O$ contains a carbonyl group. How could we used 1H-NMR spectroscopy to ascertain if the compound is an aldehyde or a ketone?

[size=150][b]Problem 16[/b][/size]

Consider the isomers with molecular formula $ C_{4}H_{9}Cl$, and select the one that has in its 1H-NMR spectrum:

(a) just one peak.
(b) several peaks, including one doublet at 3.4 ppm.
(c) several peaks, including one triplet at 3.5 ppm.
(d) several peaks, including two distinct peaks (each one integrating for 3H): a triplet at 1.0 ppm and a doublet at 1.5 ppm.

[size=150][b]Problem 17[/b][/size]

The 1H-NMR spectra of formic acid ($ HCO_{2}H$), maleic acid ($ cis\minus{}HO_{2}C\minus{}CH\equal{}CH\minus{}CO_{2}H$), and malonic acid ($ HO_{2}C\minus{}CH_{2}\minus{}CO_{2}H$) are very similar because they all are made up of just two singlets with equal intensity. Make the correspondence between compounds A, B, C and the appropriate acid:

Compound A: peaks at 3.2 and 12.1 ppm.
Compound B: peaks at 6.3 and 12.4 ppm.
Compound C: peaks at 8.0 and 11.4 ppm.

[size=150][b]Problem 18[/b][/size]

Show how one could use 1) 1H-NMR spectroscopy, and 2) IR spectroscopy to distinguish the following pairs of isomers:

(a) Pent-2-ene and ethylcyclopropane.
(b) Diethyl ether and methyl n-propyl ether.
(c) Ethyl acetate and methyl propanoate.
(d) 3-Methylbut-3-en-2-one and pent-3-en-2-one.
(e) 3-Methylcyclohex-2-enone and 4-acetylcyclopentene.
(f) p-Dimethylbenzene and ethylbenzene.

[size=150][b]Problem 19[/b][/size]

How many peaks are expected in the 1H- and 13C-NMR spectra of the following brominated cyclopropanes?

(a) Bromocyclopropane.
(b) 1,1-Dibromocyclopropane.
(c) cis-1,2-Dibromocyclopropane.
(d) trans-1,2-Dibromocyclopropane.
(e) 1,1,2-Tribromocyclopropane.

[size=150][b]Problem 21[/b][/size]

Compound A, $ C_{8}H_{10}$, yields three monobrominated substitution products, $ C_{8}H_{9}Br$, by reaction with bromine. The 1H-NMR spectrum of A shows a complex multiplet at 7.0 ppm (4H) and a singlet at 2.30 ppm (6H). What is the structure of A?

[size=150][b]Problem 22[/b][/size]

The red fox ([i]Vulpes vulpes[/i]) uses a chemical communication system based on scent markers in the urine. A recent investigation showed that one of the components of the fox’s urine, compound B, is an organic sulphide. Spectral analysis of pure compound B yielded the following results: MS: molecular ion at m/z = 116; IR: 890 cm-1 (strong band); 1H-NMR: 1.74 (singlet, 3H), 2.11 (singlet, 3H), 2.27 (triplet, 2H, J = 4.2 Hz), 2.57 (triplet, 2H, J = 4.2 Hz), 4.73 (broad peak, 2H). Propose a structure for compound B that is consistent with all these spectral properties.

[size=150][b]Problem 23[/b][/size]

Treating butanone (1 eq.) with molecular bromine (2 eqs.) in aqueous HBr gives a compound with molecular formula $ C_{4}H_{6}OBr_{2}$. The 1H-NMR spectrum of this compound shows the following peaks: 1.9 (doublet, 3H), 4.6 (singlet, 2H), e 5.2 (quartet, 1H). Identify the product.

[size=150][b]Problem 24[/b][/size]

Compounds C and D are isomeric diketones of molecular formula $ C_{6}H_{10}O_{2}$. The 1H-NMR spectrum of C shows two peaks, both singlets, at 2.2 ppm (6H) and 2.8 ppm (4H). The 1H-NMR spectrum of D also shows two peaks: 1.3 ppm (triplet, 6H) and 2.8 ppm (quartet, 4H). What are the structures of C and D?

[size=150][b]Problem 25[/b][/size]

The 1H-NMR spectrum of compound E ($ C_{8}H_{8}O$) consists in two singlets of equal area at 5.1 ppm (sharp) and 7.2 ppm (broad). By treatment with excess HBr, compound E yields a single dibromide ($ C_{8}H_{8}Br_{2}$). The 1H-NMR spectrum of this dibromide is similar to the spectrum of E, as it also shows two singlets of equal area at 4.7 ppm (sharp) and 7.3 ppm (broad). Suggest reasonable structures for E and its dibromide.

[size=150][b]Problem 26[/b][/size]

Compound F has molecular formula $ C_{10}H_{14}O_{3}$. Its 1H-NMR spectrum shows the following peaks: 3.25 (singlet, 6H), 4.38 (singlet, 1H), two doublets at ~ 4.6 (1H each), 7.29 (multiplet, 5H). By hydrolysis in diluted HCl, F yields compound G (m. p. 164-165ºC), which has molecular formula $ C_{16}H_{16}O_{4}$. In the infra-red spectrum of G there is a band at 3550 cm-1, but there is no band in the C=O region. What are the structures of F and G?

[size=150][b]Problem 27[/b][/size]

The acid catalysed dehydration of 1-methylcyclohexanol yields a mixture of two alkenes as products. Who are them? After you separated them by chromatography, how would you use 1) 1H-NMR and 2) infrared spectroscopy to distinguish them?

[size=150][b]Problem 28[/b][/size]

By bromination of 3-methylbutanone two isomeric compounds of molecular formula $ C_{5}H_{9}OBr$ are obtained in a ratio 95:5. The 1H-NMR spectrum of the major isomer H is the following: 1.2 (doublet, 6H), 3.0 (septet, 1H), 4.1 (singlet, 2H). The 1H-NMR spectrum of the minor isomer I shows the following peaks: 1.9 (singlet), 2.5 (singlet), with the peak at lower field having half the area of the peak at higher field. Suggest reasonable structures for isomers H and I.

[size=150][b]Problem 29[/b][/size]

Compounds J and K are isomers of molecular formula $ C_{4}H_{8}O_{3}$. Identify them through their 1H-NMR spectra:

Compound J: 1.3 (triplet, 3H), 3.6 (quartet, 2H), 4.1 (singlet, 2H), 11.1 (broad singlet, 1H).
Compound K: 2.6 (triplet, 2H), 3.4 (singlet, 3H), 3.7 (triplet, 2H), 11.3 (broad singlet, 1H).

[size=150][b]Problem 30[/b][/size]

Explain why 3-methylbutan-2-ol shows [i]five[/i] peaks in its 13C-NMR spectrum, at 17.90, 18.15, 20.00, 35.05, and 72.75 ppm, and make the correspondence between each signal and the appropriate carbon.

[size=150][b]Problem 31[/b][/size]

The 13C-NMR spectrum of cyclooctatetraene shows only one peak, but different 13C-NMR spectra are observed for 1,2-dimethylcyclooctatetraene and 2,3-dimethylcyclooctatetraene, which indicates that these molecules are isomers and not resonance structures. Examine the structures of these three compounds (models may help) and explain 1) why cyclooctatetraene shows only one peak, and 2) the expected number of 13C peaks of each dimethyl cyclooctatetraene.

[size=150][b]Problem 32[/b][/size]

Compound L ($ C_{10}H_{14}O$) is soluble in aqueous NaHO but insoluble in aqueous sodium bicarbonate. This compound reacts with bromine in water to yield a dibrominated derivative, $ C_{10}H_{12}OBr_{2}$. The infrared spectrum of L shows a broad band centred at 3250 cm-1 and a strong band at 830 cm-1. The 1H-NMR spectrum of L is the following: 1.3 (singlet, 9H), 4.9 (singlet, 1H), 7.0 (multiplet, 4H). What is the structure of L?

[size=150][b]Problem 33[/b][/size]

The 13C-NMR spectrum of a commercial sample of 2,4-pentanediol shows not three but [i]five[/i] peaks, at 23.3, 23.9, 46.5, 64.8, and 68.1 ppm. Explain why.

[size=150][b]Problem 34[/b][/size]

Compound M ($ C_{9}H_{10}$) reacts with bromine in carbon tetrachloride. Some spectral data of M follows:

IR: 3035 (m), 3020 (m), 2925 (m), 2853 (s), 1640 (m), 990 (s), 915 (s), 740 (s), 695 (s).

1H-NMR: 3.1 (doublet, 2H), 4.8 (multiplet), 5.1 (multiplet), 5.8 (multiplet), 7.1 (multiplet, 5H).

The UV spectrum of M shows a maximum at 255 nm. Propose a structure for M.

[size=150][b]Problem 35[/b][/size]

Compound N ($ C_{9}H_{10}O$) gives a negative iodoform test, and have the following spectral properties: IR: 1690 (s); 1H-NMR: 1.2 (triplet, 3H), 3.0 (quartet, 2H), 7.7 (multiplet, 5H). What is the structure of N?

[size=150][b]Problem 36[/b][/size]

Compound O is an isomer of compound N from the previous problem, but gives a positive iodoform test. Spectral data for O: IR: 1705 (s); 1H-NMR: 2.0 (singlet, 3H), 3.5 (singlet, 2H), 7.1 (multiplet, 5H). What is the structure of O?

[size=150][b]Problem 37[/b][/size]

Identify each of the following two isomers of molecular formula $ C_{20}H_{18}O$ from the 1H-NMR data given:

Isomer P: 2.23 (singlet, 1H), 3.92 (doublet, 1H, J = 7 Hz), 4.98 (doublet, 1H, J = 7 Hz), 6.81 (singlet, 10H), 6.99 (singlet, 5H).

Isomer Q: 2.14 (singlet, 1H), 3.55 (singlet, 2H), 7.25 (broad peak, 15H).

These two isomers could be distinguished by a single chemical test. What is it?

[size=150][b]Problem 38[/b][/size]

By treatment of methallyl chloride, $ H_{2}C\equal{}C(CH_{3})CH_{2}Cl$, with a solution of sodium amide in THF, a hydrocarbon with molecular formula $ C_{4}H_{6}$ is obtained. The 1H-NMR spectrum of this hydrocarbon shows the following peaks: 0.83 (doublet, 2H, J = 2 Hz), 2.13 (doublet, 3H, J = 1 Hz), 6.40 (multiplet, 1H). What is the most likely structure for this hydrocarbon, and what is the mechanism of its formation?

[size=150][b]Problem 39[/b][/size]

Compound R have molecular formula $ C_{8}H_{18}O_{2}$ and does not react with periodic acid. Its 1H-NMR spectrum shows three singlets at 1.2 (12H), 1.6 (4H), and 2.0 (2H). What is the structure of R?

[size=150][b]Problem 40[/b][/size]

By treatment of 2,4,6-tri-tert-butylphenol with bromine in cold acetic acid a compound with molecular formula $ C_{18}H_{29}OBr$ is obtained quantitatively. The infrared spectrum of this compound shows bands at 1630 and 1655 cm-1. Its 1H-NMR spectrum shows only three singlets at 1.2, 1.3, and 6.9 ppm, with peak areas of 9:18:2, respectively. What is the structure of this compound?

[size=150][b]Problem 41[/b][/size]

Compound S, with molecular formula $ C_{9}H_{8}O_{2}$, absorbs two moles of hydrogen (with a deactivated palladium catalyst) to form compound T. By hydrogenation over an active platinum catalyst, compound T absorbs another two moles of hydrogen to give compound U, which also forms directly from S by hydrogenation with the platinum catalyst. All three compounds react with hydroxylamine to form oximes. Compound T, by ozonolysis, yields two moles of a substance and one mol of another substance. The 1H-NMR spectrum of S shows only two singlets at 2.1 and 2.6 ppm, with peak areas of 3:1, respectively. Identify compounds S, T, and U.

[size=150][b]Problem 42[/b][/size]

One of the many products that can be obtained from the chlorination of propane is a pentachloropropane with the following 1H-NMR spectrum: 4.5 (triplet), 6.1 (doublet), with peak areas 1:2, respectively. Identify this compound.

[size=150][b]Problem 43[/b][/size]

Compound V has molecular formula $ C_{9}H_{11}NO$ and the following spectral properties: IR: 1680 (vs); 1H-NMR: 2.8 (singlet, 3H), 2.9 (singlet, 3H), 6.7-7.7 (multiplet, 4H), 9.7 (singlet, 1H). When $ D_{2}O$ is added to the sample, the peak at 9.7 ppm disappears. Identify compound V.

[size=150][b]Problem 44[/b][/size]

Compound W is an isomer of compound V from the previous problem. In its IR spectrum the most prominent bands appear at 1652 (s), 1610 (vs), 1511 (vs), 1246 (vs), 1029 (s), and 829 (s). A weak band appears at 2760 cm-1, but this is due to a contamination from a precursor of compound W. When $ D_{2}O$ is added to a sample of W, there is no change in its 1H-NMR spectrum. Identify compound W.

[size=150][b]Problem 45[/b][/size]

By treatment of neopentyl chloride with sodium amide, an hydrocarbon with molecular formula $ C_{5}H_{10}$ was obtained. This hydrocarbon is easily dissolved in concentrated sulphuric acid, but is not oxidized by cold, neutral, diluted solutions of potassium permanganate. The 1H-NMR spectrum of this hydrocarbon shows peaks at 0.20 and 1.05 ppm, with relative areas of 2:3, respectively. When the isotopically labelled alkyl halide $ (CH_{3})_{3}CCD_{2}Cl$ was allowed to react under the same conditions, the product obtained had its molecular ion at m/z = 71. Identify the hydrocarbon and propose a mechanism for its formation.

[size=150][b]Problem 46[/b][/size]

A compound with molecular formula $ C_{4}H_{6}$ gives a 13C-NMR spectrum with two peaks of nearly equal intensity: one, at 30.2 ppm, corresponds to a CH2 carbon; the other, at 136 ppm, is of a CH carbon. Identify this compound.

[size=150][b]Problem 47[/b][/size]

A compound with molecular formula $ C_{10}H_{16}O_{2}$ absorbs one mol of hydrogen  by catalytic hydrogenation. In the UV region it shows an absorption maximum at 236 nm ($ \epsilon \equal{} 2\times 10^{4}$), and its IR spectrum shows characteristic absorptions at 1667, 1620, and 1100 cm-1 (the last band is strong and broad). Its 1H-NMR spectrum shows the following peaks: 1.0 (triplet, 3H, J = 7 Hz), 1.0-2.0 (multiplet, 6H), 2.4 (singlet, 3H), 3.2 (quartet, 2H, J = 7 Hz), 4.8 (triplet, 1H), 6.2 (triplet, 1H). Propose a structure for this compound.

[size=150][b]Problem 48[/b][/size]

Compound X reacts with magnesium metal to form a product which, when added to acetone, produces an alcohol. Elemental analysis shows it to be 15.40% C and 3.23% H. IR spectrum: 2973 (s), 2910 (m), 1440 (m), 1378 (m), 1202 (vs, sbr), 951 (s). Mass spectrum: m/z = 27, 29 (base peak), 127, 156 (molecular ion). Identify compound X.

[size=150][b]Problem 49[/b][/size]

Compound Y has the following elemental composition: 71.4% C, 9.5% H, 19.0% O. Spectral data for Y:

Mass spectrum: m/z = 43, 69 (base peak), 83, 84 (molecular ion, vanishingly small peak, almost undetectable).

IR spectrum: 3400 (s, vbr), 3300 (sh, vs), 2987 (sh, s), 2200 (w), 1460 (m), near 1378 (two sharp bands – close enough to each other that they almost give a single broadened band -, the low frequency one stronger than the other), 1170 (s, sbr), 640 (s, sbr).

Identify compound Y.

[size=150][b]Problem 50[/b][/size]

Compound Z has the following elemental composition: 62.1% C, 10.3% H, 27.6% O. Spectral data for Z:

Mass spectrum: m/z = 26, 27, 28, 29 (base peak), 57 (weak), 58 (molecular ion).

IR spectrum: 2989 (m), 2813 (m), 2716 (m), 1739 (vs), 1461 (m), 1415 (m), 1391 (m).

What is the structure of Z?

[size=150][b]Problem 51[/b][/size]

Identify the following isomers of molecular formula $ C_{3}H_{5}Br$ from the spectral information provided:

(a) Isomer A’ (1H-NMR): 2.31 (singlet, 3H), 5.38 (singlet, 1H), 5.59 (singlet, 1H).

(b) Isomer B’ (13C-NMR): 32.6 (CH2), 118.8 (CH2), 134.2 (CH).

(c) Isomer C’ (13C-NMR): 12.0 (CH2), 16.8 (CH). The low field peak has half the intensity of the other peak.

[size=150][b]Problem 52[/b][/size]

Two isomeric lactones D’ and E’, with molecular formula $ C_{5}H_{8}O_{2}$, show a characteristic IR absorption at 1810 cm-1. In their 1H-NMR spectra, isomer D’ shows two singlets at 1.1 and 2.2 ppm (peak areas 3:1, respectively), while isomer E’ also shows two singlets but at 1.2 and 4.0 ppm (peak areas 3:1, respectively). Suggest reasonable structures for D’ and E’.

[size=150][b]Problem 53[/b][/size]

(a) By catalytic hydrogenation, compound F’ ($ C_{5}H_{8}$) gave cis-1,2-dimethylcyclopropane. On the basis of this result, three possible isomeric structures were proposed for F’. Which were they?

(b) The lack of an absorption band near 890 cm-1 in the IR spectrum of F’ turned one of the proposed structures unlikely. Which one?

(c) The 1H-NMR spectrum of F’ showed peaks at 1.04 and 2.22 ppm with relative areas 1:3, respectively. Which of the three structures from (a) is compatible with this data?

(d) The base peak in the mass spectrum of F’ appears at m/z = 67. Which ion is very probably responsible for this peak? How do you explain its abundance?

(e) Compound F’ can be prepared from open chain compound(s) in a single step. How?

[size=150][b]Problem 54[/b][/size]

Two isomers G’ and H’, with molecular formula $ C_{8}H_{12}O_{4}$, are easily converted in compound I’ by catalytic hydrogenation. Spectral data (1H-NMR) is given below for all three compounds. Identify them.

G’: 1.30 (triplet, 3H, J = 7 Hz), 4.28 (quartet, 2H, J = 7 Hz), 6.28 (singlet, 1H).
H’: 1.32 (triplet, 3H, J = 7 Hz), 4.27 (quartet, 2H, J = 7 Hz), 6.83 (singlet, 1H).
I’: 1.25 (triplet, 3H, J = 7 Hz), 2.62 (singlet, 2H), 4.15 (quartet, 2H, J = 7 Hz).

[size=150][b]Problem 55[/b][/size]

Elemental analysis of compound J’ showed only the presence of C, H, and O. By combustion analysis, the following composition was obtained: 79.37% C, 8.88% H. The spectral properties of J’ are the following:

IR: 3360 (s, vbr); three sharp, weak-medium bands between 3100 and 3000 cm-1; three strong and sharp bands between 3000 and 2850 cm-1; 1600 (m); 1500 (s); 760 (m); 700 (vs).

1H-NMR: 0.91 (triplet, 3H), 1.75 (quartet, 2H), 2.0 (singlet, 1H, broadened peak), 4.5 (triplet, 1H), 7.3 (singlet, 5H).

13C-NMR: 10, 31, 70, 125, 127, 128, 146 (short peak).

Identify compound J’.

[size=150][b]Problem 56[/b][/size]

Hydrocarbon K’, with molecular formula $ C_{6}H_{6}$, has an 1H-NMR spectrum with only two signals, at 3.84 ppm and 6.55 ppm, with peak areas 1:2, respectively. By heating in pyridine for 3 hours, K’ is quantitatively converted in benzene. By moderate hydrogenation K’ gives L’, whose spectra give the following information: MS – molecular ion at m/z = 82; IR – there are no double bonds; 1H-NMR – broad peak at 2.34 ppm.

(a) Propose structures for K’ and L’.
(b) In the 1H-NMR spectrum of K’, the high field peak is a quintet, while the low field peak is a triplet. How can these splittings be explained?


Here is #18 from 1998 MMPC.

A long strip of paper 0.01 cm thick and 10 cm wide is wound tightly about a wooden cylinder 5 cm in diameter.  The resulting reel of paper is 10 cm in diameter.  Which of the following numbers is closest to the length of the strip of paper (in cm)?

(A)4700          (B)5100          (C)5500          (D)5900          (E)6300

Could you please give a detailed solution and answer?

Thanks

If I give detailed solution, you won't be able to learn to develop your skills. Instead, I'll give you steps on how to solve it.

[hide="Steps"]Always draw pictures. It gives you an idea what it looks like.
1.- Get the collective thickness of the paper rolled around the cylinder.
2.- Now determine the number of times the paper is rolled around the cylinder.
3.- Use the formula for the length of circumference, find the length of the paper. Remember each circumference of paper around the cylinder increases a little for every roll around the cylinder due to the thickness of the paper.[/hide]

[hide]
Find area of combined paper and cylinder - the cylinder which is $\frac{75\pi}{4}$. Then, since thickness if 0.01cm, multiply by 100 to get $1875\pi$. Finally, multiply by $\pi$ and round to the nearest hundred to get $\boxed{D}$.[/hide]

[hide="Slighty longer solution"] The radius must increase from 2.5 cm to 5 cm at 0.01 cm per layer. Therefore, there must be 250 layers. The layers have a length, or circumference, of $5\pi$, $5.02\pi$, $5.04\pi$,..., $10\pi$. Using the formula for arithmetic sequence, we get $\frac{250}{2}\cdot15\pi=1875\pi\approx5890.5$. So the answer is $\boxed{(D) 5900}$. [/hide]


I'm looking for information and uses for something called Mollweide's Formula, $\frac{a+b}{c}=\frac{\cos\frac{1}{2}(\alpha-\beta)}{\sin\frac{1}{2}\gamma}$.

Derivation, general uses, and contest problems would be appreciated. Thank you. :)

this is a random formula, this is not useful, this is much like the law of tangents

The expression on the right was used with the same expression in some other cyclic permutation of the variables on some old mandelbrot problem...it's derivation is probably just LOC and LOS.


The trisection points of the sides of an equilateral triangle are connected to each other by drawing segments parallel to the sides of the triangle.  Many triangles are formed when this is done.  What is the ratio, expressed as a common fraction, of the area of the original triangle?  Express your answer as a common fraction.

Ratio of what?

My guess:
[hide]1/9[/hide]

I must have mistyped the problem.  I don't know if I'll get back to this.


I have suggested Mr.  Goebel set up a math forum under his current web page.  An example is as follows:

http://forum.texasmath.org/index.php

Mr. Goebel thinks it was a good idea but he does not know how to do it.

Do you know what is involved to set on of these up?  If you know something about this and like to help, please contact Mr. Goebel at  goebel@ncssm.edu

Thanks.

Just run a quick google search. [url]http://www.createforum.net/[/url] There are plenty of other options too.

Or you could ask the folks at the Texas forum how they made theirs. I'm sure they would be glad to share.

Here is one that I just created for test purpose only. It is free. Mr. Goebel may need to pay to buy better one ( I really do not know).

http://ncmathcontest.informe.com/


If one side of a regualr hexagon has a side of length $r$, then what is the area of the hexagon?

There should be two ways to do this problem.

[hide="HINT"] Notice something special about a regular hexagon. Divide it into triangles. Now, for those special triangles, what is the area for each one? Times it by 6. This is one way.

You could also use trigonometry, or you could put the hexagon inside a rectangle. I think that works also. [/hide]

[quote="shinwoo"][hide]3 ( r ^ 2 )[/hide][/quote]

Thats not right. Remember you should always post how you got the solution as well, not just the final answer - then other people can tell you where you went wrong.

Oh I'm sorry, I made a mistake :blush: ...

I will do it again..

[hide] you first have to find out the regular triangle's height...

the height is  :sqrt: -r/2+r^2

3(r *   :sqrt: -r/2+r^2)

did I get it right?

Could I simplyfy it little farther?
[/hide]

I'm pretty sure that you have the height wrong shinwoo.

[hide]split it into 6 equilateral triangles.  Area of an equilateral triangle is s^2*sqrt(3)/4 (use the 30-60-90 triangle created by dropping the altitude) then just do times 6 or 3s^2*sqrt(3)/2.[/hide]

elvenchamp777

Please post in spoiler until everyone who wants to try the problem has already tried it. Thanks.

sry i forgot. :blush:   i'll remember next time.

elvenchamp

There's still another way if you want to try it.

[hide="HINT"]Try putting the hexagon in a triangle[/hide]

[quote]Try putting the hexagon in a triangle [/quote]

That's what I did

and I had to find the height of the triangle in order to get the area, but I screwed up that part


Support
MIT
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Harvard
Yale
geometry
geometric transformation

Hi folks, I am currently an undergraduate student.

I have some queries pertaining to Spring admission as well as Financial Assistance in Spring Semester.

Query 1)

How many schools in the U.S. admit International students in the Spring Semester (for the Computer Science program, can anyone suggest some good/above average universities that admit international students in Spring semester)?

Query 2)

I would also like to know how many schools ( notably those who admit students in the Spring semester) proffer financial assistance to International Students in Spring Semester? Can anyone suggest any such schools that offer financial assistance as well as an admit to an International student in Spring?

Query 3)

How much "weightage" do the standardized tests (GRE,TOEFL) have while applying for financial assistance?

I currently don't have an extraordinary GPA, assuming If I manage to secure an outstanding GRE/TOEFL score, can I be guaranteed financial aid while applying for graduate program in Spring semester?

I had already perused sites of some universities, but I would like to seek someone's advice (people  in this forum).

I ASK PEOPLE IN THIS FORUM TO SUGGEST SOME UNIVERSITIES THAT MEET THE ABOVE CRITERION.

I thank you all in advance!

[quote="Gen8"]
Query 1)

How many schools in the U.S. admit International students in the Spring Semester (for the Computer Science program, can anyone suggest some good/above average universities that admit international students in Spring semester)?
[/quote]

[url=http://grad-schools.usnews.rankingsandreviews.com/grad/com/search]Here[/url] is a (not necessarily definitive) list of the best graduate programs in CS.  [url=http://www.google.com/]Here[/url] is a tool which will allow you to find the webpages of the graduate programs on that list and see whether they offer spring admission.

[quote="Gen8"]
Query 2)

I would also like to know how many schools ( notably those who admit students in the Spring semester) proffer financial assistance to International Students in Spring Semester? Can anyone suggest any such schools that offer financial assistance as well as an admit to an International student in Spring?
[/quote]

Use the above tool again!  The answer may be different depending on whether you are pursuing a Master's or a Ph.D.

[quote="Gen8"]
Query 3)

How much "weightage" do the standardized tests (GRE,TOEFL) have while applying for financial assistance?

I currently don't have an extraordinary GPA, assuming If I manage to secure an outstanding GRE/TOEFL score, can I be guaranteed financial aid while applying for graduate program in Spring semester?
[/quote]

Pretty much none.  At a reputable program, if you are getting in, there should be some sort of financial package attached (some combination of TAship + RAship + stipend + tuition waivers).  Some M.Sc programs won't offer financial support though.  These will be given to everyone admitted and are not dependent on GRE or TOEFL, although the admissions decision itself may be.  It's pretty absurd to think that you would get any sort of benefit from having a good TOEFL score though -- a bad one might disqualify you, but even a perfect one basically says that your english level may almost be as good as a native speaker, and I don't think that's a qualification that graduate programs get particularly excited over!

Hmm, Then I would like to know the importance of these standardized tests. I had earlier perused through the site of a university ( a pretty nice one at that!) which proffered a fellowship for a good GRE score/ a good GPA.

I think securing a good GRE/TOEFL score does have some "weightage".

Well, I was talking about some less putative schools (in relative to MIT,  Harvard, Yale etc), which I feel proffer some financial assistance based on the GRE/TOEFL score.

I think a TA is awarded based solely on the TOEFL score, ain't it?

How do you convert % into GPA? Is there any generalized formula?

As blahblahblah said, GRE (the general one) and TOEFL scores only matter in the sense that very poor scores might hurt your chances.  Getting a score above some minimum to show that you are capable of communicating in English is all that's really needed, and anything more than that is nice but ultimately irrelevant to the admission decision.  I'm not sure what you mean by TAships being awarded based 'solely on TOEFL' scores, but I don't believe that's the case (except that, again, if you score very poorly it might preclude you from getting a TAship because the committee might conclude that you will not be able to communicate with students).

The standardized test that [i]might[/i] be of somewhat more importance is the GRE subject test in Computer Science or Mathematics (some programs don't require you to take subject tests, but 'strongly encourage it').  Admission and fellowship committees do seem to put weight on these, though not as much as on the GPA.  If your GPA is poor then scoring very well on the GRE subject test might help in making the case that for whatever reason your grades don't accurately reflect your ability to learn.  Similarly, if your GRE subject test score is poor, it may nullify a solid GPA by creating the impression that grades notwithstanding you simply don't know the basics.

Perhaps even more important than GPA, and almost certainly more important than any standardized test scores, are the recommendations that you get, and the interest you show in research.  Ultimately admission committees are looking for self-motivated people who are going to succeed at research - not at getting grades, and not at scoring high on tests.  If you can demonstrate a genuine interest in the field by engaging yourself in undergraduate research, you'll be doing far more to improve your chances of getting into the school you want than by cramming for GREs.

[quote="Gen8"]Okay, Then what are the criterion that are evaluated before granting a Teaching Assistantship?[/quote]

um, basically none at a lot of (most?) schools.  some will require some sort of course to be taken before one can actually teach a class, but TAships are given to everyone at the schools I have gone to.

I wonder what schools you had attended! Then, on what basis are R.A.'s awarded? Can a person be awarded with a T.A. as well as an R.A. simultaneously?

I had earlier perused through the websites of some schools in which they were some minimum thresholds regarding GRE and TOEFL score, GPA , so provided one has surpassed these thresholds is he/she likely to get admitted at that particular school or do they also look for something more?

[quote="Gen8"]I wonder what schools you had attended! Then, on what basis are R.A.'s awarded? Can a person be awarded with a T.A. as well as an R.A. simultaneously?
[/quote]

RAs are generally paid out of individual faculty member's (usually your supervisor!) research grants, whereas TAs are paid out of the department's budget.  At my school people who do not have external funding usually get a full TAship during September to April (12 hours/week) and then some sort of RAship from their supervisor during the summer.

[quote="Gen8"]I had earlier perused through the websites of some schools in which they were some minimum thresholds regarding GRE and TOEFL score, GPA , so provided one has surpassed these thresholds is he/she likely to get admitted at that particular school or do they also look for something more?[/quote]

No, in general surpassing the thresholds is not enough  This is why they are called 'minimum thresholds' and not 'necessary and sufficient requirements' for admission.  References are important, as are research experience and various similar things.  This has already been said by others in this thread!

Hiii  blahblah..., I attempted the GRE some time ago, since I was busy, I couldn't access this forum. I had perused some random posts relating to GRE, and also some posts pertaining to IIT-JEE, I observed some posts made by you in which you claimed (or could have) solved some IIT-JEE problems in minutes ( You said a particular intricate (that was someone else's opinion) JEE problem took you 2 minutes to solve!).

Here's the link regarding the problem and your response.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42989&search_id=889566661&start=20

Secondly, I also read a post of yours on GRE-Verbal in which you mention that-- the GRE Verbal ability section is extremely tough (I think you also mentioned that you managed to secure 730 (mostly through some guessing as you have said)).

I think you are very adept at problem solving, then, I find it very weird that you found GRE Verbal as extremely tough, the GRE Verbal and JEE problems aren't comparable at all!

So, may I know, How can a person like you who had managed to solve JEE problems in minutes find the GRE-Verbal ability section as being extremely hard (The GRE Verbal is all about memorizing some words)?

[b]I am not trying to offend you in any sense, I just wanted to know how come an adept problem solver like you find GRE Verbal extremely hard? [/b]

Between,I didn't manage a stellar score on the Verbal, I just got only a 700.

Thanks!

Well, if you are doing well on the verbal section of the GRE, the problems get harder very quickly, to the point where I think I ended up having to compare two words that I had never seen before.  I am saying that the verbal section of the GRE is hard in the sense that it must be very difficult to score 800.

It's also a test in which problem-solving skills aren't that useful -- certainly you can use logic to increase your chances of getting the correct answer, but it mostly boils down to how many words you know and how many definitions you know for the words you know.  Any mathematical problem-solving skills I may or may not have don't translate particularly well to that setting.

In any event any score in the 700s is quite good though -- I think that 690 and above are 99th percentile, or something similar.  It's certainly not worth retaking.

Haha, You are smart! You have viewed my previous post (might be the case!) in which I mentioned my GRE score, and have been able to discern that I might be contemplating of re-taking the GRE (mostly in order to bolster my Quants score), that's pretty good!

But I would like to ask the graduate program(also the school) you are pursuing (That is if you like to tell, I observed that you are very proficient in Math) and I also realized a possible mistake of mine,in a certain PM I might have offended you through the usage of certain words that might not actually depict your personality in reality. But one should accept that every person makes (or is prone to making mistakes), it might have been the case that at that point of time I might have reacted in a strong manner (or might have been touchy).


This is a simple game based on the prisoners' dilemma.

Objective: You are a prisoner, among others, incarcerated by an evil master. To observe the spirit of democracy, the master holds a voting for the prisoners: each prisoner can either (a) vote to save himself and kill others or (b) vote to kill himself and save others. Then each prisoner will be killed or saved according to the voting (that is, a person is killed if more people vote to kill him than to save him and saved otherwise). Communication between prisoners are allowed. Your goal is simply survival.

Scoring (Cumulative): # survival / # total rounds played.

Round 1: PM me your choice (a) or (b) by Friday, Aug. 7. The results will be posted afterwards.

The rules are subject to future adjustment.

Round 1 Results:
SuperNerd123 (a) killed
maybach (a) killed
zmesdr (a) killed

Cumulative Ranking
1.
SuperNerd123 0/1
maybach 0/1
zmesdr 0/1

Round 2: PM me your choice (a) or (b) by Wed, Aug. 12. The results will be posted afterwards.

NOTE: you are allowed to communicate with each other. Feel free to post in this thread for others to know your name.

[quote="funtwo"]i'm not in the game, but if everyone votes b, then won't everyone survive?[/quote]

The point is that not everybody will do that.

In the case of 2 players, this game is akin to the Prisoner's Dilemma.

[quote="maybach"]I don't get the point of voting b.   You just forfeit points.[/quote]

Umm look at what happened the first round when no one voted b.