algebra
polynomial
vector
linear algebra
matrix
calculus
derivative

Define $ V_{n,m} \equal{} \text{span}\{ (a_1 x_1 \plus{} ... \plus{} a_n x_n)^m | (a_1, ... a_n) \in \mathbb{C}^n \}$, regarded as a subspace of $ \mathbb{C}[x_1, ... x_n]$.  Compute $ \dim V_{n, m}$ and write down a basis for it.

[hide="Remark"] This dimension is at most $ {m \plus{} n \minus{} 1 \choose n\minus{}1}$, the dimension of the span of the monomial terms that could possibly appear, but I'm having trouble proving that this is the case.  For $ n \equal{} 2$ the resulting coefficients, with the basis $ v_i \equal{} {m \choose i} x_1^i x_2^{m\minus{}i}$, form a Vandermonde matrix, so we can set $ a_2 \equal{} 1$ and let $ a_1$ take on $ m \plus{} 1$ distinct values.  It seems like this should generalize. [/hide]

You have a nice scalar product $ \langle P,Q\rangle\equal{}[P(D)Q](0)$ on polynomials. Now just notice that if you have a homogeneous polynomial of degree $ m$ and take its scalar product with the $ m$-th power of a linear form, you get the $ m$-th directional derivative at $ 0$ in the direction given by the coefficients of that form. If it is $ 0$, the polynomial is $ 0$ along the line. If a polynomial is $ 0$ along each line, it is identically $ 0$. The rest should be clear.

Sorry, could you explain more?  It's not clear to me what $ [P(D)Q]$ means.

$ D$ is the diffrerentiation, so $ x_i^k$ is replaced by $ \frac{\partial^k}{\partial x_j^k}$. For example, if $ P(x)\equal{}5x_1x_2^3x_3^2$, then $ P(D)\equal{}5\frac{\partial^6}{\partial x_1\partial x_2^3\partial x_3^2}$. Thus you turn a polynomial $ P$ into a differential operator $ P(D)$, apply it to $ Q$, and take the value of the resulting polynomial at $ 0$. The monomials $ x^\alpha\equal{}x_1^{\alpha_1}\dots x_n^{\alpha_n}$ form an orthogonal basis with respect to this scalar product with $ \|x^\alpha\|^2\equal{}\alpha!\equal{}\alpha_1!\dots\alpha_n!$.

fedja's approach is wonderful, but the problem can be attacked in a more straightforward way:

The $ m$-th degree component of $ \mathbb{C}\left[x_1,x_2,...,x_n\right]$ is a vector space of dimension $ \binom{m \plus{} n \minus{} 1}{n \minus{} 1}$. We claim that

$ \left\{\left(a_1x_1 \plus{} a_2x_2 \plus{} ... \plus{} a_nx_n\right)^m\mid\ a_1,a_2,...,a_n\text{ are nonnegative integers with }a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} m\right\}$

is a basis of this vector space. Because of the equality of dimensions, it is enough to either prove that it is a linearly independent subset or that it spans the whole vector space. The latter easily follows from the identity

$ \sum_{T\subseteq\left\{1,2,...,m\right\}}\left( \minus{} 1\right)^{\left|T\right|}\left(\sum_{i\in T}u_i\right)^m \equal{} \left( \minus{} 1\right)^mm!u_1u_2...u_m$

for any $ u_1$, $ u_2$, ..., $ u_m$ (this identity is proven using multivariate finite differences, or by assuming WLOG that $ u_1$, $ u_2$, ..., $ u_m$ are nonnegative integers, interpreting combinatorially and applying the Sylvester sieve formula - is there a more direct proof?). The former assertion may turn out more difficult (Vandermonde tensors? this would certainly be worth studying, given the usefulness of the Vandermonde determinant), but we don't need it anymore.

Captain Obvious speaking: Of course, the only thing we used about $ \mathbb{C}$ here is that $ 1$, $ 2$, ..., $ m$ are invertible in $ \mathbb{C}$. And we need this condition, since the assertion epically fails if the characteristic of the field is $ m$.

  darij

[quote="darij grinberg"](this identity is proven using multivariate finite differences, or by assuming WLOG that $ u_1$, $ u_2$, ..., $ u_m$ are nonnegative integers, interpreting combinatorially and applying the Sylvester sieve formula - is there a more direct proof?). [/quote]
It looks like a consequence of Mobius inversion (in the general form); thanks!  This is pretty much the nicest result I could hope for.

An exercise: find the mistake in my previous post (#5).
Hint: the identity
$ \sum_{T\subseteq\left\{1,2,...,m\right\}}\left( \minus{} 1\right)^{\left|T\right|}\left(\sum_{i\in T}u_i\right)^m \equal{} \left( \minus{} 1\right)^mm!u_1u_2...u_m$
is correct.

It seems that every second proof of mine which I don't formalize beyound readability is bound to have errors...

  darij

[quote="fedja"]If a polynomial is $ 0$ along each line, it is identically $ 0$. The rest should be clear.[/quote]
I think I understand what you're saying; the orthogonal complement of $ V_{n,m}$ in the ideal of homogeneous polynomials of degree $ m$ is the zero vector, right?  But I'll be more specific:  when I said "write down a basis," what I really mean is "what's a good way of writing monomials as a linear combination of terms of the given form"?

In hindsight, one easy way to do this is to use, for example, the $ (m\plus{}1)^{st}$ roots of unity filter on each of the $ x_i$, but this requires a lot of terms.  Ideally I'd like something like darij's solution (is the error that $ T$ must be a set and not a multiset?) that works over arbitrary fields of characteristic zero.

(In any case, perhaps I should mention that what I'm really looking at is the $ G$-symmetrization of such a vector space with respect to some group $ G$, and $ G$-symmetrization doesn't commute with the roots of unity filter, so this problem might be more subtle.)

[quote="t0rajir0u"]In hindsight, one easy way to do this is to use, for example, the $ (m \plus{} 1)^{st}$ roots of unity filter on each of the $ x_i$, but this requires a lot of terms.  Ideally I'd like something like darij's solution (is the error that $ T$ must be a set and not a multiset?) that works over arbitrary fields of characteristic zero.[/quote]

Nope, it is totally irrelevant whether $ T$ is a set or a multiset (it is just an index set). The mistake is that the $ \left(\sum_{i\in T}u_i\right)^m$ don't belong to what I claimed to be the basis. What I actually proved is that

$ \left\{\left(a_1x_1 \plus{} a_2x_2 \plus{} ... \plus{} a_nx_n\right)^m\mid\ a_1,a_2,...,a_n\text{ are nonnegative integers with }a_1 \plus{} a_2 \plus{} ... \plus{} a_n \leq m\right\}$

spans the vector space $ V_{n,m}$, but whether the terms with $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} m$ already suffice is not clear.

[quote="t0rjarir0u"](In any case, perhaps I should mention that what I'm really looking at is the $ G$-symmetrization of such a vector space with respect to some group $ G$, and $ G$-symmetrization doesn't commute with the roots of unity filter, so this problem might be more subtle.)[/quote]

Out of curiosity, a $ G$-symmetrization is the subspace of all elements fixed by $ G$ ? So you are basically doing invariant theory?

  darij

[quote="t0rajir0u"]Essentially.  So my real question:  does $ V_{n,m}^G$ equal $ I_{n,m}^G$, where $ I_{n,m}$ is the span of the monomials of degree $ m$, and if so, what is a good choice of $ a_i$ in this case?  (So far I've been acting under the assumption that the answer is the same, but I'm starting to question that assumption.)  Here $ G$ is a finite subgroup of $ GL_n(\mathbb{C})$.[/quote]

I don't get this. Take $ G\equal{}S_n$ (well, $ S_n$ embedded into $ \mathrm{GL}_n\left(\mathbb{C}\right)$ in the usual way): The symmetric polynomials are not the span of the symmetric monomials (of which there is at most $ 1$, but more often $ 0$ in every degree) and not multiples of $ \left(x_1\plus{}x_2\plus{}...\plus{}x_n\right)^m$ either. Maybe I don't understand how $ G$ is supposed to act on the polynomials?

  darij

I guess I should clarify.  Let $ a_1 x_1 \plus{} ... \plus{} a_n x_n \equal{} \mathbf{a} \cdot \mathbf{x}$; the vector space I'm looking at is the span of terms of the form $ \sum_{g \in G} (\mathbf{a} \cdot (g \mathbf{x}))^m$ and I want to compare it to the span of terms of the form $ \sum_{g \in G} i(g \mathbf{x})$ where $ i$ is a monomial of degree $ m$.

So for example, when $ n \equal{} m \equal{} 2, G \equal{} S_2, V_{n,m}^G$ contains $ (x_1 \plus{} x_2)^2 \equal{} x_1^2 \plus{} 2x_1 x_2 \plus{} x_2^2$, but it also contains $ (x_1 \plus{} 2x_2)^2 \plus{} (2x_1 \plus{} x_2)^2 \equal{} 5x_1^2 \plus{} 8x_1 x_2 \plus{} 5x_2^2$, hence it contains $ x_1^2 \plus{} x_2^2$ and $ x_1 x_2$.

- We saw above that every polynomial $ p\in V_{n,m}$ is a linear combination of some $ \left(\mathbf{a}\cdot\mathbf{x}\right)^m$ with $ \mathbf{a}$ nonnegative and satisfying $ \left|\mathbf{a}\right|\leq m$. (Here, $ \left|\mathbf{a}\right|$ means $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n$, and $ \mathbf{a}$ is called nonnegative if all $ a_i$ are nonnegative.)
- Every $ G$-invariant polynomial $ p\in V_{n,m}$ satisfies $ p\left(\mathbf{x}\right) \equal{} \frac {1}{\left|G\right|}\sum_{g\in G}p\left(g\mathbf{x}\right)$ (because $ p\left(\mathbf{x}\right) \equal{} p\left(g\mathbf{x}\right)$ for every $ g\in G$).
Thus, every $ G$-invariant polynomial $ p\in V_{n,m}$ is a linear combination of some $ \sum_{g\in G}\left(\mathbf{a}\cdot\left(g\mathbf{x}\right)\right)^m$ with $ \mathbf{a}$ nonnegative and satisfying $ \left|\mathbf{a}\right|\leq m$.
This works, at least, for fields of characteristic not dividing $ \left|G\right|$.

  darij


MATHCOUNTS
rotation
probability
geometry
3D geometry
perimeter
ratio

I need somebody to confirm that these answers are correct. (They are from the [b]2000 Nationals Mathcounts Team Round)[/b]

1. Right triangle ABC with legs AB = 3 cm and CB = 4 cm is rotated about one of its legs. What is the greatest possible number of cubic centimeters in the volume of the resulting solid? Express your answer to the nearest whole number.

1. 30

2. Erica flips a fair two-sided coin. Nate then flips the coin and wins if it matches Erica's flip. If it doesn't match, then Noah flips the coin and wins if it matches Nate's flip. If Noah doesn't win, the Erica flips and wins if it matches Noah's flip. The game continues until there is a winner. What is the probability that Nate wins the game? Express your answer as a common fraction.

2. 1/5

3. What is the sum of the digits needed to write all the whole numbers from 0 to 10,000, inclusive?

3. 49995

6. Four number cubes, with the numbers 1 to 6, one per face, are rolled. What is the probability that two numbers are the same even number and that the other two numbers are the same odd number? Express your answer as a common fraction.

6. 1/24

7. Five positive integers have a sum of 92. What is the greates possible product of these five numbers?

7. 2046528

8. All possible three-digit positive integers are formed using the digits                                 0, 1, 3, 6, 7, 8 and 9 with no digit repeated in any number. If one of these numbers is selected at random, what is the probability that it will be greater than 300 and divisible by 4? Express your answer as a common fraction.

8. 3/20

10. The perimeter of a right triangle is 60 inches. When the lenghts of the sides are squared and added, the sum is 1352. Determine the number of inches in the positive difference between the lengths of the two legs.

10. 14

If I am wrong, please post your solution and answer.
I didn't give 4, 5, or 9 because they required a diagram.

I'm to busy to solve them but i'll tell if you're right.
You got #6 and #10 right.

[hide="number 1"]
Its either a cone with radius 3 height 4 or radius 4 height 3. Since the radius is squared, we obviously want the bigger one to be squares.
The base is thus $16\pi$ and the height $3$. The answer is thus $16\pi\approx50$[/hide]
[hide="number 2"]
This requires an infinite series.
The probability that Nate wins first is $\frac 12$.
THen there is a $\frac\cdot\frac 12$ chance that Noah wins next, then a $\frac 14 \cdot \frac 12=\frac 18$ chance that Erica wins next. You keep multiplying by $\frac 12$.
Since you have to multiply by $\frac 12$ 3 times to get back to Nate, the common ratio is $\frac 18$ The original term is $\frac 12$
$\frac{\frac 12}{\frac 78}=\frac 47$[/hide]
[hide="number 3"]
For the first 9999, you clearly use each digit the same number of times. If you do a lot of counting, you will get that it is 4000 (i think). Thus you have to find the sum of 1-9 4000 times. That is $4000\cdot45=180000$. Add 1 for 10000, to get $180001$[/hide]
[hide="number 6"]
There are $\binom 42=6$ ways to arrange which are the same even and which are the same odd.
For any specific case, there is a $\frac 12 \cdot \frac 16 \cdot \frac 12 \cdot 16=\frac{1}{144}$. Multiply by 6 to get $\frac{1}{24}$[/hide]
[hide="number 7"]
You want them to be as close together as possible. 92/5=18.4
So make 3 = 18 and 2 = 19.
2,105,352[/hide]
[hide="number 8"]
For it to be divisible by 4, the last two digits have to as well.
The last two digits thus can be:
08
16
36
60
68
76
96
80
For 16, the first digit can be 3,7,8,9
For 36, it can be 7,8,9
For 60, it can be 3,7,8,9,
For 68, it can be 7,3,9
For 76, it can be 3,8,9
For 96, it can be 3,6,7,8
For 08, it can be 3,6,7,9
For 80, it can be 3,6,7,9
Summing the total working, we get 4+4+4+4+4+3+3+3=29
For the total number of 3 digit numbers, the first digit can't be 0.
There are 6 choices for the first digit.
Then 6,5.
Thus 180 choices total.
$\frac{25}{180}=\frac{29}{180}$[/hide]
Ill do 10 later, im getting kicked off the computer
EDIT:
[hide="number 10"]
The perimeter is 60, thus the sides are integers.
Might it be a 10,24,26 triangle?
yes. The squares of the sides are 100,576,676.
Summing gets 1352.
So the legs are 24 and 10
The difference is 14[/hide]

For number 1, can't you use the hypotenuse to rotate the triangle?

Oh...That makes sense.
So how is 16 pi the answer, since they need the volume?

oh nvm :blush:
I'm so sorry.
I forgot that you already divided by three.
I was confused since that wasn't in your work.
[size=75]
[color=orange][mod solafidefarms edit: Double posts are bad. Edit your previous post, don't make 2 in a row.][/color][/size]

Either way, the problem says to rotate it on one of its legs.

[hide="10"]$a+b+c=60$
$a^2+b^2+c^2=1352$
$a^2+b^2=676$
$a+b=60-\sqrt{a^2+b^2}$
$a+b=34$
$c=26$
$a=10, b=24$
$24-10=\boxed{14}$[/hide]

Edit: I'm using a half broken touchpad so all the messages here are practically useless. Sorry.

Looks like no one can get numbers 4,5,8, and 9.

Yeah, I have the answers 2000 up.Those are the only ones I can get.

number 8 is
[hide]

since the number can end in
08
16
36
60
68
76
80
96
so then you have 5*6*6= 180
and you work out when you have 3 how any you can make.... or just the ones above without 3 in it since you cant repeat. so you do that with 3,6,7,8,9
and you get 28/180 or 7/45[/hide]

I have the answers too, but I don't won't my mom to give them to me.
This is my Algebra 2/MathTeam hw.
I have to be able to show how to do it in two ways to my math coach/teacher.

Oh and I think runescapesux is right. I accidently had one extra, 868. So it is 28/180 which is 7/45.

wait

i forgot one number... so its actually 29/180

[quote="goldendomer"]Where did you get the 2000 mathcounts problems?[/quote]
My nats coach had them because she went that year and the years after so she let us have a copy.

#3 is wrong, you do the 9000 4-digit integers times 4(# of digits), the 900 3-digit integers times 3, the 90 2-digit integers times 2, the 9 1-digit integers times 1, plus 5 for 10,000.


inequalities
algebra
polynomial
calculus
derivative
induction
inequalities solved

Here's a generalisation I've made from two vietnamese inequalities :
Let n  \geq  3 be an integer and x_1,...,x_n be non-negative real numbers, such that :
(n-2) \sum {i<j} x_i*x_j +  \sum {i<j<k} x_i*x_j*x_k = 2n(n-1)(n-2)/3.
Prove that :
 \sum {1 \leq i \leq n} x_i  \geq  (2/(n-1))* \sum {i<j} x_i*x_j
and determine when equality occurs.

For n= 3,4 you will have Vietnam 1996.

Pierre.

I stayed a lot to solve that problem from Vietnam and unfortunately my method doesn't work with this one. Well, I said I don't like generalizations, but it seems I will change my mind. This one definitely won't take 4 minutes.

:D  :D  It makes me feel more confident, since I spent some time on it....

Nice flag, Harazi, did you become a romanian?

Pierre.

Thank you very much. I was always romanian. Please, don't ask me the whole story. Valentin knows it and believe me it's not interesting.

Could anyone please post the condition for a,b,c,d such that the equation ax^3+bx^2+cx+d=0 has all roots real?

well you put y=x+b/(3a) then the ecuation becomes:
y^3+3py+2q=0 where 3p=(3ac-b^2)/(3a^3) and 2q=2b^3/(27a^3)-bc/(3a^2)+d/a

if q^2+p^3>0 then the ecuation has a real root and 2 complex ones
<0 it has all roots real and distinct at the same time
if =0 and p=q=0 then the equation has all roots =0
if =0 and p^3=-q^2<>0 then it has all the roots real and 2 are equal.

I hope this is what you wanted! :D

Thank you very much. Now I realised that my poor idea has died. I was thinking to consider the polynomial p(x)=(x-x_1)...(x-x_n). Then its n-3-th derivative has all its roots positive and the equation is:
   n(n-1)(n-2)/6-(n-1)(n-2)/3*(sum x_i) x^2+(n-2)/2*sum x_i*x_j*x-2n(n-1)(n-2)/3+(n-2)sum x_i*x_j=0. That is why I asked for a realtion between coefficients. But I really don't want to substitute these values in the relations you posted.
  So, the problem remains unsolved. Too bad!

I would like to see a solution to this problem. I really don't have any idea that doesn't kill me with computations.

We prove the result by induction on n  \geq  3.

- For n = 3 :
Let x,y,z  \geq  0 such that xy + yz + zx + xyz = 4. (1)
With no loss of generality, we may assume that x \leq  y  \leq  z. Thus, from (1), we deduce that 3x 2  + x 3   \leq  4, that leads to x  \leq  1. Similarly, we have z  \geq 1.
Moreover y = (4 - xz)/(x+z+xz).
The desired inequality is then equivalent to :
f(x,z)  \geq  0 where f(x,z) = x(x+z+xz)+4-xz+z(x+z+xz)-x(4-xz)-z(4-xz)-zx(x+z+xz).
But it is easy to see that f(x,z) = (x+z-2) 2  + xz(z-)(1-x), hence result and equality occurs iff x=y=z=1 or (x,y,z) is a permutation of (0,2,2).

- Now suppose that the inequality holds for n  \geq  3.
Let x_1,...,x_(n+1)  \geq 0.
Let S_1 = sum{1 \leq i \leq n+1} x_i, S_2 = sum{1 \leq i < j \leq n+1} x_i*x_j,
and S_3 = sum{1 \leq i<j<k \leq n+1} x_i*x_j*x_k.
And suppose that (n-1)S_2 + S_3 = 2n(n+1)(n-1)/3.  (2)
The desired inequality is S_1  \geq 2*S_2/n.  (3)

Let P(x) = (x - x_1)(x - x_2)...(x - x_(n+1)) = x n+1  - S_1*x n  + S_2*x n-1  - S_3*x n-2  + ...
Then P'(x) = (n+1)(x n - (n*S_1/(n+1))x n-1  + ((n-1)*S_2/(n+1))x n-2  - ((n-2)S_3/(n+1))x n-3  + ...)

Since all roots of P are non-negative real numbers, it follows from Rolle's theorem that all the roots of P' are non-negative real numbers, say y_1,...,y_n. Thus P'(x) = (x-y_1)...(x-y_n).
Let S'_1, S'_2 and S'_3 be defined as above but for the y_i's.
Identifying the coefficients of P', we deduce that :
S'_1 = (n/(n+1))S_1, S'_2 = ((n-1)/(n+1))S_2 and S'_3 = ((n-2)/(n+1))S_3.

From (2), we then have (n-2)S'_2 + S'_3 = 2n(n-1)(n-2)/3.
Thus, we may use the induction hypothesis for the y_i's and deduce that S'1  \geq  (2/(n-1))S'_2, from which we easily (3), which ends the induction steps.

it is not difficult now to see that equality occurs iff :
n = 3 and (x,y,z) is a permutation of (0,2,2), or
n > 3 and x_1 = .... = x_n = 1.

Pierre.

VERY NICE and VERY HARD solution for an very very difficult inequality.

That really is a difficult and beautiful solution. Congratulations!


ratio
conics
ellipse
hyperbola
geometric sequence
calculus
calculus computations

Tangent is drawn at the point $ (x_i , y_i)$ on the curve $ y \equal{} f(x)$ , which intersects the $ x$ axis at $ (x_{i \plus{} 1} , 0)$. Now again a tangent is drawn at $ (x_{i \plus{} 1} , y_{i \plus{} 1})$ on the curve which intersects the $ x$ axis at $ (x_{i \plus{} 2},0)$ and the process is repeated $ n$ times i.e. ( $ i \equal{} 1,2,....n$)

1)If $ x_1 , x_2 ,..x_n$ form an AP with common difference equal to $ log_2(e)$ and curve passes through $ (0,2)$, the curve is

a) y = $ 2e^x$
b) y = $ 2e^{ \minus{} x}$
c) y = $ 2^{1 \minus{} x}$
d) y = $ 2^{1 \plus{} x}$

2) If $ x_1, x_2 ,....x_n$ form a GP with common ratio equal to $ 2$, then curve is

a)parabola
b) ellipse
c) rectangular hyperbola
d) none of these

The answer is [b](d)[/b].
If we are in the point $ (x_{i}, y_{i})$ and we want to find the intersection of the tangent to the curve with the $ x$-axis, we just write the equation of the tangent and express $ x_{i\plus{}1}$ from it. So $ (y\minus{}y_{i})\equal{} (y_{i})^{\prime}(x\minus{}x_{i})$. Putting $ y\equal{}0$ into it gives $ x_{i\plus{}1}\equal{}\minus{}\frac{y_{i}}{(y_{i})^{\prime}}\plus{}x_{i}$.

In case of $ (d)$ you get $ y_{i}\equal{}2^{1\plus{}x_{i}}, (y_{i})^{\prime}\equal{}2^{1\plus{}x}ln(2)$. So $ x_{i\plus{}1}\equal{}\minus{}\frac{1}{ln(2)}\plus{}x_{i}$, which is $ x_{i\plus{}1}\equal{}ln(2)\plus{}x_{i}$.

Thnx but you did a little computational mistake at the end.

Thank you, Conjurer.

In $ (d)$ $ x_{i\plus{}1}$ in fact comes $ x_{i\plus{}1}\equal{}\minus{}log_{2}(e)\plus{}x_{i}$. In case of $ (c)$ $ x_{i\plus{}1}\equal{}log_{2}(e)\plus{}x_{i}$. So $ (c)$ is a correct answer.

$ (2)$ If we take a rectangular hyperbola with equation $ xy\equal{}1$, we get $ y^{\prime}\equal{}\minus{}\frac{1}{x^{2}}$, so $ x_{i\plus{}1}\equal{}\minus{}\frac{\frac{1}{x_{i}}}{\minus{}\frac{1}{x_{i}^{2}}}\plus{}x_{i}\equal{}2x_{i}$. So the points $ x_{1}, x_{2}, x_{3},...$ are in the geometric progression with the common difference $ 2$.


Balance the equation
$CuCO_3+HNO_3=======CuNO_3+CO_2+H_2O$
If you think it is wrong please tell. :?

[quote="inteluser"]That should be $\text{Cu(NO}_3\text{)}_2$

You would probably want to write this net ionically too:
$\text{CuCO}_3 + \text{H}^+ \rightarrow \text{Cu}^{2+} + \text{CO}_2 + \text{H}_2\text{O}$

You can try balancing that.[/quote]
So my question is wrong????
Im not argueing or anything
$\text{CuCO}_3 + 2\text{H}^+ \rightarrow \text{Cu}^{2+} + \text{CO}_2 + \text{H}_2\text{O}$

[hide="Complete Equation"]
\[\text{CuCO}_3(s)+2\text{HNO}_3(aq)\longrightarrow \text{Cu(NO}_3)_2(aq)+\text{CO}_2(g)+\text{H}_2\text{O}(l)\][/hide]

[hide="Net Ionic"]
\[\text{CuCO}_3(s)+2\text{H}^+\longrightarrow \text{Cu}^{2+}+\text{CO}_2(s)+\text{H}_2\text{O}(l)\][/hide]

Thank you for all your replies, but is my question wrong?
Should it be
$\text{CuCO}_3(s)+\text{HNO}_3(aq)\longrightarrow \text{Cu(NO}_3)_2(aq)+\text{CO}_2(g)+\text{H}_2\text{O}(l)$????


trigonometry
calculus
calculus computations

Assume that $ \phi(x)\equal{}\left\{\begin{array}{ll}
x^3\sin\frac{1}{x} & x\neq0\\
0 & x\equal{}0
\end{array}
\right.$

$ f(x)$ is derivable when $ x\equal{}0$.

Find $ F'(x)$, here $ F(x)\equal{}f[\phi(x)]$

I think you mean $ F'(0)$. So, using the chain rule, $ F'(0)\equal{}f'(\phi(0))\phi'(0)$. Figure out $ \phi'(0)$ and you're done.


Find all f:N to N such that:
    f(f(n)) + f(n) =2n for all n in N

It is obviosly $f(1)=1$. Let $f(k)=k$ for all $k<n$, then if $f(n)<n$ we have contradition ($f(f(n))+f(n)=2f(n)<2n$). If $f(n)=m>n$, then $f(m)=2n-m<n\Longrightarrow 2m=f(f(m))+f(m)=2f(m)$ give contradition. Therefore $f(n)=n \ \forall n\in N.$


You have a $2003$ digit number. Now re-arrange the digits to make another $2003$ digit number. Call the first number $A$ and second number $B$. Find all values of $A$ and $B$ such that $A+B$ has $9$ for every digit.

Is there even a solution to this question?

Suppose there is;  you want us to give you a $2003$-digit number?

I think he would want you to explain the ordering of the digits rather than writing the entire number down.

Oh yeah, that $9$ in the middle messed it up.  Nevermind.

It really smells as there's no solution: Since the sum of two digits can never be $19$, the only way they can give a niner in the sum is that they add up to $9$. Hence every digit $d$ in $A$ must have its pair $9-d$ within $A$, but as there's an odd number of digits, that can't be accomplished. (Of course, no digit can be its own pair since $9$ is an odd number.)

Yeah. That makes sense. Good job Farenhajt! :clap2:

Yes

I was about to say any permutation of 181818...1, but then I realized it wouldn't work.

So I think maybe it is easy since there is no solution.

That's right, no solution. It's actually really easy once you know the trick. :)

Okay, that is cool.

If it was 2004 digits though...we would have a lot of solutions.

Is there a reason why you labeled this 'easy problem'?

[quote="lingomaniac88"]If it [i]was[/i] 2004 digits, it would be $a_{1}a_{2}a_{3}\dots a_{2004}$ so that for all $i$, $a_{i}+a_{2005-i}=9$.  Of course, $a_{1}\ne0$ and $a_{2004}\ne0$.  So that allows for $8\cdot10^{1001}$ solutions, I believe.[/quote]

You limited yourself to a single rearrangement of the digits (reversing them), but the original problem allows for any permutation.  That leads to the following question:

How many numbers $N$ are there with $2n$ digits such that the digits of $N$ can be rearranged to give a new number $M$ and $N+M$ has only the digit 9?

Also, consider smaller bases: what if $N$ is in base 2 (and 9 is replaced with 1)?

If we use an odd base, we no longer have to limit ourselves to an even number of digits.  So, how many numbers $N$ are there with $n$ base 3 digits whose digits can be rearranged to give a number $M$ (perhaps $M = N$) such that $M+N$ has only the digit 2?

Finally, just to pile on a little, naming a topic "Easy problem" is obnoxious because (a) it might not really be easy and (b) it gives no information whatsoever about the actual content of the question.


function
real analysis
real analysis unsolved

Hi dears,

It is said that there exists an ONTO continuous function $ f : (0,1)\rightarrow [0,1]$ ( i.e, f is continuous and onto ). Can you guys give me an example?
Many thanks .

[quote="Vodanhvn"]
It is said that there exists an ONTO continuous function $ f : (0,1)\rightarrow [0,1]$ ( i.e, f is continuous and onto ). Can you guys give me an example?[/quote]

Define $ f(x)\equal{}0$ for all $ 0<x\le 1/10$.
Define $ f(x)\equal{}1$ for all $ 9/10\le x<1$.
You should be able to come up with the rest yourself.

[quote]Define  for all .
Define for all .
You should be able to come up with the rest yourself.
[/quote]

Thanks, but what we need is f is onto . i.e $ f(0,1)\equal{}[0,1]$

Note that the function has only been partially defined. The hard part is figuring out how to include the end points. test20 has taken care of that aspect. Now just fill in the middle with your favorite continuous function such that $ f(1/10)\equal{}0$ and $ f(9/10)\equal{}1$...(I'd probably do it linearly).


calculus
integration
conics
ellipse
function
geometry
calculus computations

Use a triple integral to find the volume of the given solid:

The solid bounded by the elliptic cylinder $4x^2+z^2=4$ and the planes $y=0$ and $y=z+2$.

I did an integration from the y direction first and said that we must go from $y=z+2$ to $y= 0$, and then the projection of the figure into the xz plane gives an ellipse $4x^2+z^2=4$. I solved for z to get the bounds and said x went from -1 to 1. However, I always get stuck while doing the z partial integral because it is not an elementary function, yet I have trouble representing the ellipse in polar coordinates. The answer is not in the solution manual (even numbered question). Could someone please help? Thanks.

Two things will make this problem much easier. One is odd-even symmetry. The other is this:

What is $\int_{-r}^r\sqrt{r^2-u^2}\,du$ ?

Instead of trying to make substitutions and compute antiderivatives, visualize that as the area of a semicircle. Hence,

$\int_{-r}^r\sqrt{r^2-u^2}\,du=\frac{\pi r^2}2.$

We have:

$\int\int_{4x^2+z^2\le4}z+2\,dx\,dz = \int_{-1}^1\int_{-\sqrt{4-4x^2}}^{\sqrt{4-4x^2}} z+2\,dz\,dx.$

Since $z$ is an odd function of $z,$ it integrates to zero over this symmetric set. That leaves us with the integral of the constant, $2.$

$=\int_{-1}^12\cdot2\cdot\sqrt{4-4x^2}\,dx=8\int_{-1}^1\sqrt{1-x^2}\,dx = 8\cdot\frac{\pi}2=4\pi.$

Note: the "z partial integral" [b]is[/b] an elemetary function, even though I chose to avoid it.

Yes, if the z-partial integral is an elementary function, how would you integrate it?

The function was something like $\sqrt{1-z^2}$, but how do you integrate that?


quadratics
modular arithmetic
Diophantine equation
special factorizations
number theory solved
number theory

I think this works: rewrite the equation as \[ (6n - 1)^2 - 3\left[2(m - 2n)\right]^2 = 1.  \] This is a Pell equation of the form $x^2 - 3y^2 = 1$. If we can prove that no solution $(x,y)$ to such an equation satisfies $x \equiv -1\ (\textrm{mod}\ 6)$, we will be done. The smallest solution is $(2,1)$ and the other solutions $(x_n,y_n)$ satisfy the recurrence relations $x_{n + 1} = 2x_n + 3y_n$ and $y_{n + 1} = x_n + 2y_n$. Computing the sequence of solutions $(x_n,y_n)$ modulo 6 gives $(2,1),\ (1,4),\ (2,3),\ (1,2),\ (2,5),\ (1,0),\ (2,1),\ \dots$ and the sequence is periodic. So, all solutions satisfy $x_n \equiv 1$ or $x_n \equiv 2$ (mod 6) and we are done.

I didn't want to post this in the post in which I wrote my solution, since this is for the resources project, but - does anyone have a better way to do it? I don't like mine and it's a bit complicated...  :huh:

This doesn't use Pell, though it's probably just as complicated :lol:  :  rewrite it as
\[ \left(\dfrac{2m^2}{n}-4m+1\right)^2=(2m-1)(6m-1) \]
and note that as $2m-1$ and $6m-1$ are coprime $6m-1$ should be a square, however it is congruent to 2 mod 3.

[quote="Rushil"]Show that there do not exist positive integers $m$ and $n$ such that \[ \frac{m}{n} + \frac{n+1}{m} = 4 .  \][/quote]

We can do as : $m^2+n^2+n=4mn$ so $m^2+n^2=n(4m-1)$. Cause $4m-1$ must have an odd prime divisor $p$ in form $4t-1$ so $p\mid m^2+n^2$. But $p\mid 4m-1$ so $(p,m)=1$ and it leads us to contraction cause in fact that $p\mid m^2+n^2\rightarrow p\mid m,n$

Not really worthwhile now, but to amplify on my previous solution:  treating the equation as a quadratic in $\frac{m}{n}$ we can write
\[ \left(\frac{m}{n}\right)^2-\left(4-\frac{1}{m}\right)\left(\frac{m}{n}\right)+1=0 \]
and hence
\begin{eqnarray*} \frac{m}{n}&=&\frac{4-\frac{1}{m}\pm\sqrt{\left(4-\frac{1}{m}\right)^2-4}}{2}\\ &=&\frac{4-\frac{1}{m}\pm\sqrt{\left(2-\frac{1}{m}\right)\left(6-\frac{1}{m}\right)}}{2} \end{eqnarray*}
so $(2m-1)(6m-1)$ must be a square.

[quote="pluricomplex"]We can do as : $m^2+n^2+n=4mn$ so $m^2+n^2=n(4m-1)$. Cause $4m-1$ must have an odd prime divisor $p$ in form $4t-1$ so $p\mid m^2+n^2$. But $p\mid 4m-1$ so $(p,m)=1$ and it leads us to contraction cause in fact that $p\mid m^2+n^2\rightarrow p\mid m,n$[/quote]

That's so neat! :o :) Thanks, I'll remember this one.

[quote="pluricomplex"][quote="Rushil"]Show that there do not exist positive integers $m$ and $n$ such that \[ \frac{m}{n} + \frac{n+1}{m} = 4 .  \][/quote]

Cause $4m-1$ must have an odd prime divisor $p$ in form $4t-1$ so $p\mid m^2+n^2$. [/quote]

Could someone tell me why $p\mid m^2+n^2$?

Also,Can we write the R.H.S as $1+3$ or $2+2$ or $3+1$ and then equate the values in L.H.S and thus prove that each value is not possible,thus m,n integers do not exist?

Please do clear my doubt above.

[quote="shadysaysurspammed"][quote="pluricomplex"][quote="Rushil"]Show that there do not exist positive integers $m$ and $n$ such that \[ \frac{m}{n} + \frac{n+1}{m} = 4 .  \][/quote]

Cause $4m-1$ must have an odd prime divisor $p$ in form $4t-1$ so $p\mid m^2+n^2$. [/quote]

Could someone tell me why $p\mid m^2+n^2$?[/quote]
Because:
[quote="pluricomplex"]We can do as : $m^2+n^2+n=4mn$ so $m^2+n^2=n(4m-1)$[/quote]

[quote="shadysaysurspammed"]
Also,Can we write the R.H.S as $1+3$ or $2+2$ or $3+1$ and then equate the values in L.H.S and thus prove that each value is not possible,thus m,n integers do not exist?

Please do clear my doubt above.[/quote]
Why have the summands to be integers¿

[quote="ZetaX"]

Because:
[quote="pluricomplex"]We can do as : $m^2+n^2+n=4mn$ so $m^2+n^2=n(4m-1)$[/quote][/quote]

Umm,ZetaX,That's the thing i don't get.Why :? If you could please kiondly explain......

[quote="ZetaX"]The given equation is $\frac{m}{n} + \frac{n+1}{m} = 4$, or after multiplying with $mn$: $m^2+n^2+n = 4mn$.
But $m^2+n^2+n = 4mn \iff m^2+n^2=4mn-n = n(4m-1)$.[/quote]

 :lol: hehe,sorry i had quoted the wrong thing previously .I undrstand the sol now.. :D

i solved it here [url]http://www.mathlinks.ro/weblog.php?w=1052[/url] xD!  :D  :lol:

why

p|m^2 +n^2 -> p|m,n

Is m,n the same as gcd(m,n)

We have
$ m^2 \minus{} 4nm \plus{} n^2 \plus{} n \equal{} 0$, so the discriminant has to be a perfect square.
So $ 12n^2 \minus{} 4n$ is a perfect square, or $ 3n^2 \minus{} n \equal{} n(3n \minus{} 1)$ is a perfect square. But $ gcd(n,3n \minus{} 1) \equal{} 1$, so $ 3n \minus{} 1 \equal{} b^2$, contradiction mod 3.

Also, can someone answer linH's question? I think the statement is wrong, as $ 5|6^2 \plus{} 7^2$ but 5 does not divide 6 or 7.

[quote="dgreenb801"]Also, can someone answer linH's question? I think the statement is wrong, as $ 5|6^2 \plus{} 7^2$ but 5 does not divide 6 or 7.[/quote]

It is true because in that solution p was taken to be of the form $ 4t\minus{}1$.
The proof in that case is fairly simple: When $ p\equiv3\pmod4$, if $ n\not\equiv0\pmod p$ and $ m^2\plus{}n^2\equiv0$, $ (mn^{\minus{}1})^2\equiv \minus{}1\pmod p$, but of course -1 is not a quadratic residue modulo such primes p, contradiction.

[hide="Try using completing the square"]$ \& \frac{m}
{n} + \frac{{n + 1}}
{m} = 4  \\
  \&  \implies {m^2} + {n^2} + n = 4nm  \\
  \&  \implies {\left( {m + n} \right)^2} = 6mn - n = n\left( {6m - 1} \right)  \\
  \&  \implies {\left( {m + n} \right)^2} = n\left( {6m - 1} \right)$

let $ g = \gcd \left( {n,6m - 1} \right)$

$ \&  \implies g|n,g|\left( {6m - 1} \right)  \\ \\
  \&  \implies {g^2}|{\left( {m + n} \right)^2}  \\ \\
  \&  \implies g|m + n  \\ \\
  \& {\text{But }}g|n \implies g|m  \\ \\
  \& {\text{But }}g|6m - 1 \implies g|1 \implies g = 1$

this means that $ n$ and $ 6m-1$ are coprime.
this means that they both have to be squares individually.

but $ 6m - 1 \equiv 5\left( {\bmod 6} \right)$. hence it can never be a perfect square.
hence ther are no integers satisfying the above equation[/hide]

$ m^2\plus{}n^2\plus{}n\equal{}4mn$
$ m^2\plus{}4n^2\minus{}4mn\equal{}3n^2\minus{}n$     which means $ 3n^2\minus{}n$ must be a perfect square.

Because of $ (3n\minus{}1,n)\equal{}1$   3n-1 and n must be perfect squares which is not impossible since
$ 3n\minus{}1\equiv\minus{}1(mod 3)$

Does anyone has a proof involving infinite descent?
I have but its too complicated.

[quote="Rushil"]Show that there do not exist positive integers $m$ and $n$ such that \[ \dfrac{m}{n} + \dfrac{n+1}{m} = 4 . \][/quote]
Very nice Problem
Ideas very easy
$\dfrac{m}{n} + \dfrac{n+1}{m} = 4\Leftrightarrow m^2+n^2=4mn-4n$
$\Leftrightarrow m^2+n^2=n(4m-1)$
$4m-1|m^2+n^2$
Imposible

Why would $g|n_1$ be a contradiction? For example, $gcd(42,36)=6$, and $6|6(=36/6)$.

no.mehmetcantu.
3n-1=-1(mod 3) it is right.But beside n must be a square.So if (n,3)=1 then 3n-1 can't be a square.And if n=3k^2 then also 3n-1 cant be a square.
So no intiger solutions

There is prime divisor form $p|4m-1, p=-1\mod 4$.
Because $p|m^2+n^2$ must be $p|n, p|m$, It is imposible, because $p|4m-1, p|m\to p|1$.

Я вообще удивляюсь. Как можно судить о решениях не решая само уравнение?
Надо сперва решить уравнение, а потом смотреть на их вид.
Это понятно, что решения определяются решениями уравнения Пелля, но их надо же записать!
Ладно я это сделаю.
$\frac{m}{n}+\frac{n+1}{m}=4$

Если мы возьмём уравнение : $p^2-12s^2=1$ и воспользуемся его решениями:
Тогда:
$n=\frac{p^2}{2}+2ps+2s^2$
$m=\frac{p^2}{2}-2s^2$

Если же воспользуемся решениями уравнения: $p^2-3s^2=1$

$n=-s^2$
$m=-s(p+2s)$

Форма решений или будет дробная или же отрицательная. Значит целых положительных решений нет. Что и требовалось доказать!

P.S. Видно, что уравнение $\frac{m}{n}+\frac{n+2}{m}=4$  имеет решения в целых положительных числах.

Попросили решить уравнение:    $\frac{m+1}{n}+\frac{n+1}{m}=a$
Если воспользуемся решениями уравнения Пелля:
$p^2-(a^2-4)s^2=1$
Тогда решения имеют вид:
$n=2(p-(a+2))s$

$m=-2(p+(a+2))s$
И ещё.
$n=\frac{2p(p+(a-2)s)}{a-2}$

$m=\frac{2p(p-(a-2)s)}{a-2}$

Если у уравнения $\frac{m+1}{n}+\frac{n+1}{m}=a$
Такой коэффициент, что
 воспользовавшись решениями уравнения: $p^2-(a^2-4)s^2=4$  , тогда:

$n=\frac{p-(a-2)s+2}{2(a-2)}$

$m=\frac{p+(a-2)s+2}{2(a-2)}$

Stop using russian(¿) in an english forum!

Another solution: rewrite as $m^2+n^2+n=4mn$. Set $d=(m,n),m=d\mu,n=d\nu$, then $d(\mu^2+\nu^2)+\nu=
4d\mu\nu $. We see $d|\nu$ and with $\nu=de$ we find $\mu^2+\nu^2+e=4\mu\nu$.
Since $e|\nu$ it follows $e|\mu^2,\nu$ but $\mu,\nu$ are coprime. Hence $e=1$ and $\mu^2+\nu^2+1=4\mu\nu$.
But squares of integers are either $0,1$ modulo 4 and we find $\{0,1\}+\{0,1\}+1=0$ modulo 4, contradiction.

This can be very easily solved using vieta root jumping.


$A'\in BC$, $B'\in AC$ and $C'\in AB$ are points in $\triangle ABC$ such that $AA',BB'$ and $CC'$ intersect in a point. Prove that if two of the quadrangels $ABA'B', BCB'C', CAC'A'$ are inscribed in a circle then so is the third.

WLOG we may assume that quadrilaterals $ABA'B'$ and $BCB'C'$ are
cyclic, and $AA', BB', CC'$ have a point of intersection $X$. From
that $\angle ABB'=\angle AA'B'$ (from quadrilateral $ABA'B'$) and
$\angle ABB'=\angle C'BB'=\angle C'CB'$ (from $BCB'C'$). Now we see
that $\angle XA'B'=\angle XCB'$ and from that quadrilateral $XA'CB'$
is cyclic so $\angle A'CC'=\angle A'B'B=\angle A'AB$ which implies
that quadrilateral $A'CAC'$ is also cyclic.

[quote="M4RI0"]Just use trig Ceva :D .[/quote]

:D Niiiiiice


Hi everyone! I need to prepare m-nitrobenzenesulfonic acid or metanilic acid but I can't find the techniques to do it. I can prepare m-nitrobenzenesulfonic acid by reaction of benzenesulfonic acid and nitric and sulfuric acid but I don't the techniques. I need to know experimetal conditions of reaction and of isolation. The same with metanilic acid, which I can get by reduction of the latter.

Can anyone help me? please send me a mail, I will really appreciate it

Thank you so much

Leandro from La Plata, Argentina

[quote="Carcul"]Welcome to this forum Leandrods. Have you considered to go to a library and consult an experimental organic chemistry book?[/quote]

thank you for the answer. Yes I have considered it, but I didn't find anything

Well, to get m-nitrobenzenesulfonic acid you have to do two things: first, sulfonate benzene with fuming sulphuric acid ($ \ce{H2SO4}$ with $ \ce{SO3}$ dissolved in it) to get benzenesulfonic acid; second, you have to nitrate benzenesulfonic acid - there are several reagents to make this, the most common being simply a mixture of concentrated nitric and sulphuric acids. Does this help?


Find all quadruples $(a,b,m,n)$ of nonnegative integers that satisfy
\[a^{n}b^{m}=a^{m}b^{n}+a^{m+n}+b^{n}\]

If $abmn=0$ then trivial.
If $abmn>0$ and all great than $1$ then easy see that set all prime divisor of $a$ same as of $b$. Write $a=p_{1}^{x}\cdot...$ and $b=p_{1}^{y}\cdot...$, then $nx+my=\min (mx+ny,(m+n)x,ny)$ , contradiction!
If at least $a$ or $b$ are equal to $1$ then easy , too.


algebra
polynomial
Vieta
Rational Root Theorem

Please help:
Solve $ 3x^3\plus{}4x^2\plus{}11x\plus{}2 \equal{} 0$

You want the [url=http://mathworld.wolfram.com/CubicFormula.html]cubic formula[/url]. The solutions are:

$ x \equal{} \frac {1}{9} \left( \minus{} 4 \minus{} \frac {83}{\sqrt [3]{287 \plus{} 18 \sqrt {2019}}} \plus{} \sqrt [3]{287 \plus{} 18 \sqrt {2019}}\right)$

$ x \equal{} \frac {1}{18}\left( \minus{} 8 \plus{} \frac {83 \left(1 \plus{} i \sqrt {3}\right)}{\sqrt [3]{287 \plus{} 18 \sqrt {2019}}} \minus{} \left(1 \minus{} i \sqrt {3}\right) \sqrt [3]{287 \plus{} 18 \sqrt {2019}}\right)$

$ x \equal{} \frac {1}{18}\left( \minus{} 8 \plus{} \frac {83 \left(1 \minus{} i \sqrt {3}\right)}{\sqrt [3]{287 \plus{} 18 \sqrt {2019}}} \minus{} \left(1 \plus{} i \sqrt {3}\right) \sqrt [3]{287 \plus{} 18 \sqrt {2019}}\right)$

you cant unless its in a special form (like one of the root is 1 or -1)

cubic formula or the formula for higher degreed equations' roots are very complicated.

[quote="Poincare"]what i meant was how do you solve like any polynomial of any degree?[/quote]
The rather surprising answer is (in short), "you can't."  Polynomials of degree five and higher cannot, in general, be solved by root extraction.  See [url=http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem]Abel-Ruffini theorem[/url].

Instead, one usually looks for special features of a given polynomial - symmetries, rational roots - and works from there.

In general, you can use Vieta's formulas (in special cases) and the Rational Root Theorem (but this ONLY applies for integer coefficients). A lot of that is discussed in Chapters 6-7 in the Intermediate Algebra book.

For higher level polynomials, I would use the graph on your handy TI-calc. Then you can use the Zero tool to find the solutions.


I have learned that $ \L^{\infty}[0,1]$ is not separable.
How about $ \L^{\infty}(E)$?
$ E$ is a subset in $ R^{n}$ with positive Lebesgue measure.
Thanks for any help.
Happy new year~

You've, probably, learnt it with proof. The same proof works. All you need is to split $ E$ into countably many disjoint sets of positive measure and put $ \pm 1$ on them in all possible ways to get an uncountable separated family of functions.

I think I can split $ E$ into countably many disjoint sets of positive measure when $ E$ is a subset in $ R^{n}$ with positive Lebesgue measure(by partition of $ R^n$).
how about  $ E$ is a subset in arbitrary measure space with positive  measure ?
Is it still not separable?

Thanks fedja!
I will try to think about this after my final exam. :)

As long as you can carry out the same construction, the answer is "yes". The only alternative is that the space is finitely dimensional, which happens if and only if the measure restricted to $ E$ is a finite sum of atoms. The proof of this requires some work. See if you can find it :).


AMC
USA(J)MO
USAMO
probability
summer program
PROMYS
algebra

Try to be as honest as possible :). I seem to have predicted my score quite a bit lower than it was...

Way higher. I can't talk right now. The tears are falling on the keyboard and it's hard to see. :noo:

I predicted around a 10 (i think it fell under 10-14 in the other poll), and I got a 9.  So -1 (although hopefully USAMO qualifiers can do that kind of math  :P )

Just out of curiosity probability what did you get/predict as you seem happy about your actual compared to your predicted  :)

Well...he did make winner as a 10th grader!

[quote="sirorange"]How is +10 bad? 10 more than one's expected score is incredible![/quote]

Yeah but that's pretty unlikely, and it means that the original guess was unrealistically conservative.

-2

ARRGH still don't know where I lost points but eh

my guess was a range, with mean 15.5. Since I got a 15, I guess that means -0.5? It's not an option, so I'll round. Since I don't like to  brag, I'll say -1, but everyone knows it rounds to I'M AWESOME AT GUESSING

I was pretty close... I was expecting 14-15 and got 14, so -1 I guess.

And I thought I was being conservative with my guess of 28... if only I had been right... actual = 26. difference = -2 (possible reason: notice I never said 26-28 = -2)

[quote="nr1337"][quote="Phelpedo"][quote="probability1.01"]I thought I would get (-1 or -2) on #1 for sucky proof-writing (and being lazy occasionally), plus (-2 or -3) on #3 for not actually solving the problem ;) plus (-1) on a bit of lax proof-writing on #5, plus possibly (-1) on #6 for saying "symmetry." Turns out I only lost one of those points... so I guess that's +4ish (maybe not the +10ish some of you expected? ;)).[/quote]

I'm just curious...to what would you attribute your success this year? I realize that you've always been on the radar when it comes to USAMO, having achieved a great score as an 8th grader, but jumping from 13 last year to 41 this year (as only a 10th grader!) is simply amazing! I'm sure all prospective USAMOers here would be eager to hear anything you had to say about how you practiced, maybe any really good books you read, etc.[/quote]

Seeing that he has not responded yet, one of his parents said he took WOOT after going to freshman MOSP, so that's quite a bit of good training...[/quote]
Nah, he was amazing at freshman MOP to begin with. My guess is that his 13 was a really bad day.

Heh, all I can say is do a lot of problems, and do them in a variety of subjects. Learn theorems you don't know, but you won't know how to use them unless you have very good intuition or you read several solutions that employ them.

I had complete solutions to #1 and #2 (half of each was identical to the one in the answer key, and the other half was a bit awkward but still correct), a complete solution to #4 except the case n=7 (most of my proof was identical to the answer key), and a small part of a solution to #5.  I predicted 20.  I got 13.  I can think of only four possibilities:
1) The answer sheets to day 2 were never sent.
2) The graders couldn't read my writing (I thought it was fairly neat, though I used pencil).
3) I received no points for my nearly-complete solution on #4 (for any reason or lack thereof) and partial credit for my complete solutions on #1 and/or #2 (too unclear????).
4) Two or more of the above occurred.
Any comments are welcome.

It's common to overestimate your score on the USAMO. About partial credit though: 1 crap point=1 crap point. 1 crap point + 1 crap point = 1 crap point. 1 complete response + 1 technical error = 5-6 points. 1 technical correction = 0 points. So basically

1 = 1. 1+1 = 1. 7=7. 7-1=5. -(-1)=0.

What I feel would be best for those who are doubting there scores would be to post their solutions in the problem threads, and have other people respond to them. But that's just me, and I'm likely mistaken  :P

[quote="Ravi12346"][quote="randomdragoon"]1 = 1. 1+1 = 1. 7=7. 7-1=5. -(-1)=0.[/quote]
Are these also true?
7 = 6
6 = 0
7 + 7 + 0 + 6 + 2 + 0 = 7 + 7 + 0 + 4 + 0 + 0 = 13

Also, there is one more possibility that I overlooked.  They may have deducted (six or more) points for bad proofwriting (aka the last half of each of the first two problems and a bit of #4).
P.S. I'm in 8th grade.[/quote]

My guess is that you were wrong and the USAMO graders are right ;). First of all, you may have lost several points on #4 for missing the 7 case, and when people say they got a "small part" of a problem, it usually means they get 0 points on it. Then, I'm guessing that you lost a significant number of points on #2, based on the large number of people who claimed #2 but ended up losing significant points on it. You could call it "bad proof-writing skills," but more likely it was "didn't solve the problem rigorously." I would guess that your score distribution was something like 6 + 1 + 0 + 6 + 0 + 0. If you don't believe me, post your proof of #2 :).

Edit--BTW, 13 is still a good score for an 8th grader, so please don't get discouraged by what I've written. However, please do realize that the score you got was almost certainly fair (a probability close to 1.01).

Yeah you seem to know your stuff probability so see if my score analysis seems reasonable:

At first I predicted around a 14 for myself, thinking I got 1 and 2 completely right.  Then I noticed a pretty key flaw in number 1, which made my "if" proof completely invalid.  So I started thinking closer to 10 because I thought I'd get half-credit for 1 and full credit for 2, and since all the other problems constituted insignificant, unconsequential ideas I'd figure I'd get at most 2 points out of all those.

As most first-timers do, I overestimated.  I really got a 9.  So i'm guessing I got 2 7 0 0 0 0.  Would getting the "only if" of an "iff" proof give you 2 points?  And for number two my proof didn't really seem like a proof.... I more just found the minimum value of N by slowly manipulating the inequalities they gave you and the fact that if the terms are written in order from least to greatest, consecutive terms must have a difference greater than or equal to 1.  Would that get me a full 7?

eh, you should analyze my score, zhai (STILL haven't figured it out . . . )

[quote="Ravi12346"]I had complete solutions to #1 and #2 (half of each was identical to the one in the answer key, and the other half was a bit awkward but still correct), a complete solution to #4 except the case n=7 (most of my proof was identical to the answer key), and a small part of a solution to #5.  I predicted 20.  I got 13.  I can think of only four possibilities:
...
Any comments are welcome.[/quote]
Maybe the other half of #1 and #2 were actually incorrect, or at least not rigorous. Keep in mind that solving "half" of the problem isn't always equivalent to getting half the points. For example, take #1. Half of it is to prove "if" and half to prove "only if." Proving the latter and not the former would probably only earn 1-2 points.

More USAMO math!

1/2 (7) = 1 for most values of 1/2
1/2 (7) = 2 for large values of 1/2
7-(non-rigor) = 1 for most values of non-rigor
7-(non-rigor) = 6 for trivial values of non-rigor

Problem 1: (7/2)*good + (7/2)*bad = 4 + 2sin(x) for some x
Problem 2: ditto
Problem 3: 0
Problem 4: 6 - oops = 4.5 + sin(x) for some x
Problem 5: 1*epsilon = 0
Problem 6: 0
Total: 13
Wow, I feel stupid.

Yes, SEVERE over prediction for me.

Problem 1: 7=7
Problem 2: 7-(last part+answer)=3 for very small values of last part+answer and 0 for large values of last part + answer
Problem 3: 7-(all except for correct ideas)=0 in most cases
Problem 4: 7-(7 case)=4, or 3 for nonrigorous case 5
Problem 5: 7+misreading question=0 :oops:
Problem 6: 7-most cases of quadrilaterals except for trapezoids=0

Total=14  :(  :(  :(  :(  :(

[quote="jrshoch"]And for number two my proof didn't really seem like a proof.... I more just found the minimum value of N by slowly manipulating the inequalities they gave you and the fact that if the terms are written in order from least to greatest, consecutive terms must have a difference greater than or equal to 1.  Would that get me a full 7?[/quote]

It depends...if you literally did [i]exactly[/i] what you just said, then you would most likely get a 6. This is because none of what you said included actually showing that the minimum attainable value is actually possible to attain (of course, this is trivial once you prove it as the minimum, but it still makes your proof imperfect). If you actually did do that in your proof, then you would get a 7 (I know this because this is exactly what I did and I am 99% sure that I got a 7 on that proof based on my USAMO score...if I got a 6 on that proof then my score would require unreasonably high amounts of partial credit).

Also in my opinion that is probably the best and most rigorous way to prove the required statement. I think many people tried to make the argument that elements should always be as close together as possible, but it really requires either an extremely thorough argument or handwaving to make that claim. The latter method is obviously not rigorous and so prone to losing points (especially since it trivializes the problem).


Show that if the root of the algebraic equation [tex]f(x)=0[/tex] is of the form [tex]a+bi[/tex] and [tex]a,b\in\mathbb{R}[/tex]
 [tex]f(x)\in\mathbb{R}[x][/tex], then the number [tex]a-bi[/tex] is another root of the equation.

Was this really proved by Viete?  I thought Viete did the solutions to the cubic in terms of trigonemtric functions and that was pretty much it?

I don't know, by who has it been proved, but I know that this can be proved using the theorem of Viete (the one, that connects the roots of a polynomial with it's coeficients).


[quote="Tokyo"]How come there are way more male than female in this forum?[/quote]
?... Shouldn't this be in the Fun & Games forum?

Well, in general there are always more males than females who do math.  Its said that females tend to be more well rounded, and males will more likely specialize in something, like math.

it's probably untrue for most people, but

in my school, all the asian guys are good at math but suck at english, while the asian girls are better all around but not as good at math.

For example: My neighbor who's a 7th grade girl got about 600 in eng. and 500 on math in the SATs, while my classmate who's a guy and was in 7th grade got like a 480 in eng. and 690 in math

Hello,

[quote="Tokyo"]so sorry. can someone help me move this topic there?[/quote]

Tokyo, I believe the only ones that can do that are the administrators (I don't know if the moderators), you can ask them directly to do it. The thread would be great in round table.

[quote="sky9073"]
Generally, women are scarce in math/science areas.
It's not long ago that women escaped from household chores (some still haven't)..
And discrimination still exists.. we just haven't had as much opportunity to pursue these interests.
[/quote]

sky9073 is totally right. There are great brains in mathematics that have been placed in women's heads, and even in competitive mathematics (I believe that two or three years ago it was a woman, Ana Cariani, the highest score of the Romanian Team on the IMO, I didn't check, but I believe it is right), but any way, there are something that are written very deep in the collective conscience of every culture, and those paradigms are hard to break. Starting in the fact that Aristotelic Philosophy was the basis of the Western culture for more than two millenia, and I tend to think that Aristoteles is the greates misoginistic  and chobinistic of the 'great' philosophers.

There are tons of misconseptions that makes people take maybe not the most accurate path. In most of the cultures I dare to guess, the spectancy for a guy is to become something "productive" for the society, an engineer, a lawyer, a manager, a businessman, and almost anything else means that he is a failure. On the other hand, if you are a woman, it is spected for you to, either be a housewife and raise the children and have the meal warm when your husband comes in the evening; or otherwise, you can be a school teacher, a social worker, a waitress, or something similar. The problem is that so many people settle for that, specially women with few access to education.

I really wouldn't like to be a 'househusband', but I believe that it is OK if a man (in a common agreement with his wife) decides that is his duty to raise the children and have the house clean and the meal warm when his wife comes in the evening after a hard day of work, and it is perfectly OK for a woman to be an engineer, a manager or whatever is she choose to be.

Finally, there are other two issues that are important in this respect. Unfortunatelly a lot of men consider offensive to have a woman as a chief. During two years I had a job where my superiors in three levels (including the highest) were women, and I consider that if a woman is qualified to have and authority charge, good bless her, if is my duty to have her as my boss, it is perfectly OK. Second, unfortunatelly if you take a man and a woman (even in the same company) who have the same qualification, the same experience and the same possition, you are going to find most likely that the man earns considerably more than the woman (and I mean, if the woman earns $x$, probably the man earns at least $1.5x$!)

Of course, there are unjustices commited for women against men, specially in the recent times, but that is another story, and probably something that is rated R, as Valentin says!

Best,


1) Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \sqrt {4a^2 \plus{} bc} \plus{} \sqrt {4b^2 \plus{} ca} \plus{} \sqrt {4c^2 \plus{} ab} \le \frac {5}{2}.
\]
2) Let $ a,b,c$ be the side lengths of a triangle such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \sqrt {9a^2 \plus{} 16bc} \plus{} \sqrt {9b^2 \plus{} 16ca} \plus{} \sqrt {9c^2 \plus{} 16ab} \ge 5.
\]
3) If $ a,b,c$ are nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}1$ then
\[ \sqrt {a^2 \plus{} bc} \plus{} \sqrt {b^2 \plus{} ca} \plus{} \sqrt {c^2 \plus{} ab} \le \frac {3}{2}.
\]
Both of them have nice solutions with just using classical Inequalities. Hope you all like it, especially Toan and Loc. :)
PS: 3) is a problem of Hungkhtn.

It's hard for me to prove these ineq by Am-Gm and chebysev . Do you want to kill me ?:D
Can you tell me , the problem below is true or false ?
if $ a,b,c \geq 0$ then
$ \sum \frac {2a^2 \plus{} ab \plus{} ac \minus{} bc}{2a \plus{} b \plus{} c} \geq \frac {3(a \plus{} b \plus{} c)}{4}$
:D

[quote="can_hang2007"]1) Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \sqrt {4a^2 \plus{} bc} \plus{} \sqrt {4b^2 \plus{} ca} \plus{} \sqrt {4c^2 \plus{} ab} \le \frac {5}{2}.
\]
[/quote]
easy this one by using Jensen (i think):

$ \sqrt {4a^2 \plus{} bc} \plus{} \sqrt {4b^2 \plus{} ca} \plus{} \sqrt {4c^2 \plus{} ab} \le 3\sqrt{\frac{4(a^2\plus{}b^2\plus{}c^2)\plus{}ac\plus{}bc\plus{}ca}{3}}\le 3\frac{2(a\plus{}b\plus{}c)}{\sqrt{3}}\equal{}2\sqrt{3}$
Done  :D

[quote="mehdi cherif"][quote="can_hang2007"]1) Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \sqrt {4a^2 \plus{} bc} \plus{} \sqrt {4b^2 \plus{} ca} \plus{} \sqrt {4c^2 \plus{} ab} \le \frac {5}{2}.
\]
[/quote]
easy this one by using Jensen (i think):

$ \sqrt {4a^2 \plus{} bc} \plus{} \sqrt {4b^2 \plus{} ca} \plus{} \sqrt {4c^2 \plus{} ab} \le 3\sqrt {\frac {4(a^2 \plus{} b^2 \plus{} c^2) \plus{} ac \plus{} bc \plus{} ca}{3}}\le 3\frac {2(a \plus{} b \plus{} c)}{\sqrt {3}} \equal{} 2\sqrt {3}$
Done  :D[/quote]
You have mistake, my friend. Since
\[ 4(a^2 \plus{} b^2 \plus{} c^2) \plus{} ac \plus{} bc \plus{} ca \ge \frac{5(a\plus{}b\plus{}c)^2}{3}\]

[quote="mehdi cherif"]yess sorry  :maybe:
but i don't see where i'm wrong in my demo  :([/quote]
Ok, I am sorry since I haven't checked your solution carefully, but it is still... wrong. Since If you use
\[ 4(a^2\plus{}b^2\plus{}c^2) \plus{}ab\plus{}bc\plus{}ca \le 4(a\plus{}b\plus{}c)^2\]
then
\[ LHS \le 2\sqrt{3}\]
But $ 2\sqrt{3}>\frac{5}{2}$, it is not what we need.

[quote="can_hang2007"]1) Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \sqrt {4a^2 \plus{} bc} \plus{} \sqrt {4b^2 \plus{} ca} \plus{} \sqrt {4c^2 \plus{} ab} \le \frac {5}{2}.
\]
2) Let $ a,b,c$ be the side lengths of a triangle such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \sqrt {9a^2 \plus{} 16bc} \plus{} \sqrt {9b^2 \plus{} 16ca} \plus{} \sqrt {9c^2 \plus{} 16ab} \ge 5.
\]
3) If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$ then
\[ \sqrt {a^2 \plus{} bc} \plus{} \sqrt {b^2 \plus{} ca} \plus{} \sqrt {c^2 \plus{} ab} \le \frac {3}{2}.
\]
Both of them have nice solutions with just using classical Inequalities. Hope you all like it, especially Toan and Loc. :)
PS: 3) is a problem of Hungkhtn.[/quote]
Here is my solution for the first and the third. :)
For the second, I think VasC will like it. :) (Because he is the one who use Chebyshev solve this type of inequality :) )

Very nice can_hang2007! :blush:
I like your AM-GM solution. :roll:


Pascal\u0027s Triangle
number theory
relatively prime

[i]For any prime numbered row, or row where the first element is a prime number, all the numbers in that row (excluding the 1’s) are divisible by the prime[/i]

Why is this so? I got a vague explanation about why. Let the prime number be n. nC2 (n choose 2) is what you use to solve it. Let's say n is 13. 13!/(11!2!)=78. That's divisible by 13 because you can't cancel out the 13. This will work for all prime numbers. But where does the 2 in nC2 come from?

suppose n is prime
the nth row consists of (excluding the 1's)
$ \binom{n}{1}, \binom{n}{2}, \binom{n}{3}, \ldots \binom{n}{n\minus{}3}, \binom{n}{n\minus{}2}, \binom{n}{n\minus{}1}$
suppose you have a n by n grid of lattice points (n^2 squares, (n+1)^2 vertices)
you start from the upper left corner and you walk down or right turning only
we are interested in the points on the line going from the upper right and lower left vertices
we could number the triangle like the pascals triangle.
the reason this is significant is because every path to a certain point either comes from the corner above it or to the left (and you add these, this is almost exactly what happens in the pascal's triangle)
so we have 1 representation of the numbers
it is quite clear that to get to any point on the diagonal mentioned, you need to make n steps
and say you want to get to the $ i^{th}$ point on the diagonal counting from the lower left with 0 at the vertex (this represents the $ i^{th}$ or $ i\plus{}1^{th}$ point on the $ n^{th}$ row of the pascals triangle, i forgot which)
you need to move down n-i times and right i timesand there are $ \binom{n}{i}$ ways to do this
so we have found 2 different representations for the numbers in the pascals triangle, so therefore the representations are the same

but after that, youre explanation is pretty good

[hide="meh..."]
Let $ n$ be a prime. Thus, all numbers less than n are relatively prime to it.
$ \binom{n} {k}\equal{}\frac{n!} {k!(n\minus{}k)!}$. All numbers in the denominator are relatively prime to n, so n must be a factor of $ \binom{n} {k}$.
[/hide]


A nail of mass M is driven into a board against a constant resistive force F by a hammer of mass m, which is allowed to fall freely at each stroke through a height h. The hammer does not rebound after striking the nail. Find the distance the nail is driven in at each blow. Show that the total energy expended in raising the hammer during the operation of driving the nail fully into a depth of d is independent of the value of h. and can be decreased by making the hammer more massive.

I get something that sort of contradicts the "prove that.." statement (which obviously means I went wrong somewhere).

When the hammer falls through $ h$, (let mass of nail be $ m'$).
$ v\equal{}\sqrt{2gh} \\ \\
mv \equal{} (m\plus{}m')v_f \;\;\;\;... \text{   since there's no rebound, by COM} \\ \\
\frac{1}{2}(m\plus{}m')v_f^2 \equal{} Fd$

Here we assume $ d<<<h$.

$ d \equal{} \frac{g\color{red}h\color{black}m^2}{F(m\plus{}m')}$

No replies for this yet? Is my analysis correct or not?

I think it's correct, but would be more appropriate if we didn't assume d>>h.

Without that assumption, I got:

d= (m^2) gh/[(m+M)(F-(m+M)g)]

Solution:
[quote]Find the distance the nail is driven in at each blow.[/quote]

The momentum of the hammer just before impact = $ m\sqrt {2gh}$. Since the hammer does not rebound, the velocity of the hammer and the nail are equal. If this velocity is $ v$ , then, using conservation of momentum at impact:

$ (m \plus{} M)V \equal{} m\sqrt {2gh}$
$ \Rightarrow V \equal{} \frac {m\sqrt {2gh}}{m \plus{} M}$

KE of system immediately after impact = $ \frac {1}{2}(m \plus{} M)v^2 \equal{} \frac {m^2gh}{m \plus{} M}$

If the nail is driven in a distance $ x$, the work done against resistance = $ (F \minus{} (m \plus{} M)g)x$ = loss in KE
$ \Rightarrow \boxed {x \equal{} \frac {m^2gh}{(m \plus{} M)\{F \minus{} (m \plus{} M)g\}}}$

[quote]Show that the total energy expended in raising the hammer during the operation of driving the nail fully into a depth of d is independent of the value of h.[/quote]

So if the total distance driven is $ d$, the number of blows required = $ \frac {d}{x}$ = $ \frac {d(m \plus{} M)\{F \minus{} (m \plus{} M)g\}}{m^2gh}$

Each of these blows requires an energy expenditure of $ mgh$ in order to raise the hammer. So the total enery expended is $ \frac {d}{x}$ = $ \frac {d(m \plus{} M)\{F \minus{} (m \plus{} M)g\}}{m^2gh} \times mgh$ = $ \boxed {\frac {d(m \plus{} M)\{F \minus{} (m \plus{} M)g\}}{m}}$, which is independent of $ h$.

[quote]... and can be decreased by making the hammer more massive.[/quote]

$ \frac {m \plus{} M}{M} \equal{} 1 \plus{} \frac {M}{m}$, which decreases as $ m$ increases. Also, the term $ F \minus{} (m \plus{} M)g$ decreases as $ m$ increases. So the total energy expended is decreased by making the hammer more massive.


inequalities
geometry
inequalities proposed

actually I am not good in algebraic inequality ! is it wrong ? or are you intersted in ? :blush:

It is like this one
http://www.mathlinks.ro/Forum/viewtopic.php?t=52376&start=20

Langrange Multipliers or geometric interpretation.

Actually I haven't try with Langrange Multipliers but It think it can't kill this problem , I only try it in computer and I see it true,I don't have any solution  :D .I don't know geometric method help us to solve it or it help us to solve the problem in geometry!


linear algebra
matrix
linear algebra unsolved

Give $ A \equal{} [a_{ij}]_{n.p} , rank(A) \equal{} r > 0$ . Prove that exist $ A_1,..,A_r \in M_{n.p}(R)$ such that
$ 1)  A \equal{} A_1 \plus{} A_2 \plus{} ... \plus{} A_r$
$ 2)  rank(A_i) \equal{} 1, i \equal{} 1,2,..,r$

Use the following :  http://www.mathlinks.ro/viewtopic.php?t=187409


I propose a new game, in which everyone starts out as a human and tries to defeat (kill) the person who posted last in a creative way. Once you defeat someone, you can turn them into whatever you want and next time they play, they have to defeat the user above them as whatever they were turned into last.

I also reserve the right to create new rules as the game progresses.

yeesh
i shove you down a toilet
and you drown
 :)

so what am I?
assuming you decided not to change me, I drop an ocean liner on you.  :lol:
and I transform you into a slug.

This is a slight variation of King of the Hill, and should be locked.


[quote="orangefronted"]A clock loses 3 minutes every hour. It is set correctly at 1pm. What will the clock read when the correct time is 10am the next day?[/quote]

$ 21$ real hours pass, so the clock loses $ 1: 03$.

Thus, it reads $ 9: 57$.

10 am is 21 hours away, so we've lost 21*3=63 min.
So instead of reading 10:00, it reads one hour and 3 minutes before that, which is 8:57.

(BOGTRO: 9:57 is 3 minutes before 10:00, not 63 min  :wink: )


Fina all $z$ satisfying $| \dfrac{z+6}{z+4}| = \dfrac{5}{3}$ and $| \dfrac{z+2}{z+4} | = 1$

Since we have $|z+2|=|z+4|$ . Letting $z=x+yi$ , we obtain $x=-3$

Also we have $3|z+6|=5|z+4|$ which is $3|3+yi|=5|1+yi|$ . Solving this we get
$y=+-\dfrac{\sqrt{14}}{2}$

Hence $z=-3+\dfrac{\sqrt{14}}{2}i$ and $z=-3-\dfrac{\sqrt{14}}{2}i$


Construct a triangle with three given altitudes.

[hide="Solution for an acute triangle"]
Call the altitude from point $A$ $a$, from point $B$ $b$, and from point $C$ $c$.  The point where they intersect is $D$.  The intersection of three altitudes divides the altitudes in the ratio $2 : 1$, so $AD = \frac{2a}{3}, BD = \frac{2b}{3}, CD = \frac{2c}{3}$.

Call the intersection from altitude $a$ to $BC$ $E$, the intersection of altitude $b$ $F$, and the intersection of altitude $c$ $G$.  Then from above we know $DE = \frac{a}{3}, DF = \frac{b}{3}, DG = \frac{c}{3}$.

Consider the triangles formed by these lines.  We have, for example, $\triangle ADG$ with $\angle AGD = 90^{\circ}$.  Using Pythagoras on all six of these, we get:

$AB = \sqrt{ \left( \frac{2a}{3} \right)^2 - \left( \frac{c}{3} \right)^2} + \sqrt{ \left( \frac{2b}{3} \right)^2 - \left( \frac{c}{3} \right)^2 }$
$BC = \sqrt{ \left( \frac{2b}{3} \right)^2 - \left( \frac{a}{3} \right)^2} + \sqrt{ \left( \frac{2c}{3} \right)^2 - \left( \frac{a}{3} \right)^2 }$
$BC = \sqrt{ \left( \frac{2c}{3} \right)^2 - \left( \frac{b}{3} \right)^2} + \sqrt{ \left( \frac{2a}{3} \right)^2 - \left( \frac{b}{3} \right)^2 }$

Or do you mean something more specific by "construct"?
[/hide]

[quote="t0rajir0u"][hide="Solution for an acute triangle"]
Call the altitude from point $A$ $a$, from point $B$ $b$, and from point $C$ $c$.  The point where they intersect is $D$.  The intersection of three altitudes divides the altitudes in the ratio $2 : 1$, so $AD = \frac{2a}{3}, BD = \frac{2b}{3}, CD = \frac{2c}{3}$.
[/hide][/quote]
what if you have a right triangle?

[quote="t0rajir0u"]The intersection of three altitudes divides the altitudes in the ratio $2 : 1$[/quote]

This is wrong - what you said holds for medians, not altitudes.

[hide]Sides are inversely proportional to altitudes ($ah_a=bh_b=ch_c$). Therefore, if we construct segments ${1 \over h_a},{1 \over h_b},{1 \over h_c}$, the triangle made of those segments would be similar to the desired triangle. And it's easy to construct the reciprocal of a segment. Let $Op$ and $Oq$ be any two rays from arbitrary point $O$. On $Op$ find $M$ and $N$ such that $OM=1$ and $ON=h_a$. On $Oq$ find $S$ such that $OS=1$. Connect $N$ and $S$. Through $M$ draw the paralel to $NS$ until it meets $Oq$ in $R$. Then $OR={1 \over h_a}$.

When we construct the similar triangle, it's easy to change its proportions according to the given values of altitudes.[/hide]

[hide="Much simpler solution"]If area of the desired triangle is $P$, then $h_a={2P \over a}, h_b={2P \over b}, h_c={2P \over c}$. Let's construct the triangle with sides $h_a, h_b, h_c$ and let its area be $P'$. Then its altitudes are $h_a'={2P' \over h_a}={2P' \over {2P \over a}}={P' \over P}a, h_b'={P' \over P}b, h_c'={P' \over P}c$. Therefore, the altitudes of the auxilliary triangle are proportional to the sides of the desired triangle.

Construction is now clear: Construct the triangle with sides $h_a, h_b, h_c$. Take its altitudes $h_a', h_b', h_c'$ and construct the triangle with those as sides. The second triangle is proportional to the desired triangle, so now scale it in order for its altitudes to equal the given ones.[/hide]

[quote="shobber"][quote="t0rajir0u"] The intersection of three altitudes divides the altitudes in the ratio $2 : 1$, so $AD = \frac{2a}{3}, BD = \frac{2b}{3}, CD = \frac{2c}{3}$.
[/quote]
Why? I think it is only true when the Euler's line paralel to the base.[/quote]

Oh, whoops, I was thinking of medians  :blush:


What is the greatest number of interior right angles a convex octagon can have?

Oh dear, this is one I got wrong.

[hide]
Pretend you're drawing the octagon with your pencil. You start by drawing a line pointing down, and every time you add a right angle, you add $ 90^\circ$ to the total angle of your pencil. This angle cannot exceed $ 360^\circ$ if you want it to be convex. Therefore, the maximum number of right angles you can have is $ \boxed{3}$.[/hide]


analytic geometry
trigonometry
quadratics
ratio
trig identities
Law of Cosines
geometry

Given the right triangle $ ABC$ with legs $ BC \equal{} 3$, $ AC \equal{} 4$. Find the length of the shorter angle trisector from $ C$ to the hypotenuse:

$ \textbf{(A)}\ \frac {32\sqrt {3} \minus{} 24}{13}\qquad \textbf{(B)}\ \frac {12\sqrt {3} \minus{} 9}{13}\qquad \textbf{(C)}\ 6\sqrt {3} \minus{} 8\qquad \textbf{(D)}\ \frac {5\sqrt {10}}{6}\qquad \textbf{(E)}\ \frac {25}{12}$

[hide="Solution"]Yay, coordinate bash time!
Let $ C\equal{}(0,0),A\equal{}(4,0),B\equal{}(0,3)$.
Let the two trisectors intersect $ AB$ at $ D$ and $ E$, with $ D$ on the side closer to $ A$. We have $ \angle CED\equal{}\angle B\plus{}30, \angle CDE\equal{}\angle A\plus{}30$, so $ \angle CED > \angle CDE$. Thus, $ CE$ is the shorter bisector.

Now, line $ AB$ is just $ 3x\plus{}4y\equal{}12$, and line $ CE$ is just $ y\equal{}\sqrt{3}x$. Solving this system, we have $ x(3\plus{}4\sqrt{3}\equal{}12$, which reduces to $ x\equal{}\frac{16\sqrt{3}\minus{}12}{13}$. We have $ CE^2\equal{}x^2\plus{}(\sqrt{3}x)^2 \iff CE\equal{}2x$, so the answer is $ \boxed{A}$.[/hide]

Can anyone see the flaw in my solution?

[hide="My Solution"]
We have $ AB \equal{} 5$. Let $ E$ be the point on $ \overline{AB}$ such that $ BE \equal{} \frac{1}{3}AB \equal{} \frac{5}{3}$ and let the length of the shorter trisector be $ x$. Applying Law of Cosines on $ \triangle CBE$, we have

$ \left(\frac{5}{3}\right)^2 \equal{} 3^2 \plus{} x^2 \minus{} 2(3)(x)\cos{30^\circ}\\
\frac{25}{9} \equal{} 9 \plus{} x^2 \minus{} 6x\cdot\frac{\sqrt{3}}{2}\\
25 \equal{} 81 \plus{} 9x^2 \minus{} 27\sqrt{3}x\\
9x^2 \minus{} 27\sqrt{3}x \plus{} 56 \equal{} 0.$

Now it turns out that the roots of this quadratic don't even match up with any of the choices. So what did I do wrong?
[/hide]

Thanks chris, I just realized my gaping flaw. It was that I assumed that $ E$ was a point of trisection.

[quote="gfour84"]Where did this question come from?[/quote]Read the first original post.... it says AHSME 1960 OK?

[hide]That was kind of rude.  But, yea, that was pretty dumb.

I got the same anwer as Burt3Forc3 except, I used similar triangles without any coordinates.

As Burt3Forc3 deduced, $ \overline{CE}$ is shorter.  If we label the trisection points $ D$ and $ E$ ($ D$ closer to $ A$) and then draw a perpendicular to $ \overline{CA}$ from $ E$ and label their intersection we $ F$, we have $ \overline{CF} \equal{} a$ which gives $ \overline{EF} \equal{} a\sqrt {3}$ and $ \overline{CE} \equal{} 2a$.  Then, we have the ratios
\[ \frac {3}{4} \equal{} \frac {a\sqrt {3}}{4 \minus{} a}.\]
It follows that $ \overline{CE} \equal{} \frac {32\sqrt {3} \plus{} 24}{13}$. [/hide]


What time do you wake up in the morning, and what time do you go to bed at night? Oh, and post your grade as well. :)

During summer, I wake up at 11AM-1PM in the morning, go to bed at 2AM-3AM.

During school, I wake up at 6AM the morning, go to bed at 10PM-12AM.

Uhh let's see
Sleep normally from 12 AM - 1 AM
Wake up: school is 8 AM, holiday is around 10 AM

I think there was another topic like this a while back.

Holydays, Wake up around 12 and go to bed around 2am
On school wake up around 8 am and go to bed around 12

Holidays, wake up at 8-9 and go to sleep at 11 (parents.. :dry: )
School, wake up 6 go to bed 10-11

Wow you have good habits math92, getting lots of sleep I see.  :roll:

Over the summer...go to bed around 11:30-12, wake up at 6.

SUMMER: Sleep:11 PM
              WAKE: 7 AM

SCHOOL: SLEEP: 10:30 PM
             WAKE: 5 AM

weird as it is... it's true

school wake: 7:00
school sleep: 9:30

holiday wake: 10:30
holiday sleep: 11:00

School: sleep from 10 or 10:30 to 6:00

Summer: sleep from 11 or 12 to around 9

:D

[quote="math92"]summer: bedtime: 9-10pm
              wake up: 7-8am
fall: bedtime 8-9pm
           wake up-6am
winter: bedtime 8-9pm
           wake up-6am
spring: bedtime 8-9pm
           wake up-6am
 :lol:[/quote]

LIARRRRRR!! Haha, you were awake for parties at like 4 in the morning and you were talking to me. :P

all of the time:

wake up: 6AM-9AM

go to sleep: 7PM-11:30PM

6 am?

r u forced or do u do it by ur self ?

Usually I wake up at 4AM, sleep from 6AM-8AM, and then sleep again at about 9 or 10.

I wake up 7:30 summer, 6-6:30 school year. But next year I'll have too many afterschool activities so my training runs will have to be done in the morning, so probably 5:30 school year this upcoming year.

I go to bed around 10-11. Yearround. I probably should go to bed earlier school year though.

I've been getting up at 8-9 in the past couple of days.

And sleeping at 11:30-12.

But it feels a lot better waking up at 8 than at 11.  And no matter what, I'm tired.  But waking up early is the good kind of tired, and waking up late isn't.

summer:  sleep anywhere from 10-2PM, wake up between 5am-1pm

school:  sleep 9-12pm, wake up 6-7am (depending on the day)

school and summer:

wake up- whenever
bed time - whenever

 :D

just kidding.

uhh.

summer :
wake up - 7-9 am
bed time - 10:30 pm -12am

school :
bed time - as early as possible  :P
wake up- 6-7:30 am

year round - go to sleep about 1 AM

school - wake up at 7:30AM
not school - wake up at noon

sleep in summer:
anywhere from 10 to 2, though I did sleep at like 8 once and I did sleep at 4 once...

wake in summer:

As early as 7ish, usually about 9-11... though I did wake at 2 pm the day I slept at 4...

During school:
Sleep: um... something like 10-12:30ish...
Wake: 6-6:30

I go to bed at 11pm and wake up at 8am during summer and holidays, and during school I go to bed at 10:30pm-11pm and wake up at 7am. :sleep2:

ah, this is fun  :D

let's see, summer and weekends I go to bed whenevers and wake up whenevers, but usually sleep like 2 AM to 10 AM.

school my parents force me into bed at 10 PM but I stay up and do my homework like at latest 2 AM and wake up at 7:37 AM (yes that precise) since I wanna get as much sleep as possible but I need to get to school by 7:53 AM since teachers expect you to be in class 7 minutes before class starts  :(


quadratics
Vieta
AMC
AIME
algebra
polynomial
number theory

The equation $2000x^{6}+100x^{5}+10x^{3}+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}{r}$, where m, n and r are integers, m and r are relatively prime, and $r>0$. Find $m+n+r$.

[quote="zanttrang"][quote="matt276eagles"][hide]Let $x=\frac{y}{\sqrt{10}}$.  Substituting, we obtain

$2y^{6}+\frac{y^{5}}{\sqrt{10}}+\frac{y^{3}}{\sqrt{10}}+\frac{y}{\sqrt{10}}-2=0$

Divide by $y^{3}$ to get

$2y^{3}+\frac{y^{2}}{\sqrt{10}}+\frac{1}{\sqrt{10}}+\frac{1}{y^{2}\sqrt{10}}-\frac{2}{y^{3}}=0$
$2\left(y^{3}-\frac{1}{y^{3}}\right)+\frac{1}{\sqrt{10}}\left(y^{2}+\frac{1}{y^{2}}\right)+\frac{1}{\sqrt{10}}=0$

Now let $z=y-\frac{1}{y}$.  Notice that $y^{3}-\frac{1}{y^{3}}=z^{3}+3z$ and $y^{2}+\frac{1}{y^{2}}=z^{2}+2$.  Substituting, we obtain

$2(z^{3}+3z)+\frac{1}{\sqrt{10}}\left(z^{2}+2\right)+\frac{1}{\sqrt{10}}=0$
$2z^{3}+\frac{z^{2}}{\sqrt{10}}+6z+\frac{3}{\sqrt{10}}=0$
$(z^{2}+3)\left(2z+\frac{1}{\sqrt{10}}\right)=0$

If $x$ is real, then $y$ and $z$ are real also.  Therefore, we ignore imaginary solutions for $z$.  Therefore,

$2z+\frac{1}{\sqrt{10}}=0$
$z=-\frac{1}{2\sqrt{10}}$
$y-\frac{1}{y}=-\frac{1}{2\sqrt{10}}$

Since $x=\frac{y}{\sqrt{10}}$, we have $y=x\sqrt{10}$.  Substituting, we obtain

$x\sqrt{10}-\frac{1}{x\sqrt{10}}=-\frac{1}{2\sqrt{10}}$

Multiplying by $2x\sqrt{10}$, we get

$20x^{2}-2=-x$
$20x^{2}+x-2=0$

Using the quadratic formula, we obtain $x=\frac{-1\pm\sqrt{161}}{40}$, and the answer is $-1+161+40=\boxed{200}$.[/hide][/quote]
You just basically did a full Vieta's substitution on that problem.  Is there any sneakier way to attack it?[/quote]
Or a more standard way?

[quote="matt276eagles"]Another solution idea:

The two real roots form a quadratic which is a factor of the given polynomial.  Find the factor, then use the quadratic formula to find the root we're looking for.[/quote]
Can you show how that would work?

[quote="ShoeFactory"][quote="kimby_102"][hide="solution"]
we are given $2000x^{6}+100x^{5}+10x^{3}+x-2=0$.
notice that the middle three terms form a geometric sequence.
$100x^{5}+10x^{3}+x=x\frac{1000x^{6}-1}{10x^{2}-1}$.
So we have:
$\frac{x(1000x^{6}-1)}{10x^{2}-1}=-2(1000x^{6}-1)$.
by multiplying $10x^{2}-1$ on both sides and factoring out we get:
$(1000x^{6}-1)(20x^{2}+x-2)=0$.
now we just find the roots of $20x^{2}+x-2=0$ which is
$\frac{-1\pm\sqrt{161}}{40}$.
[/hide][/quote]

nice soln.[/quote]
That is probably the solution AIME writers had in mind when they wrote the problem.

[quote="Phelpedo"]Why does the problem say that there are only 2 real roots? Doesn't $1000x^{6}-1$ yield another root?[/quote]

the equation would be undefined because of division by zero

In other words, that term tells us that $\frac{1}{\sqrt{10}}$ times the sixth roots of unity except for $1,-1$ are the other four complex solutions :)

[quote="zanttrang"][quote="matt276eagles"][hide]Let $x=\frac{y}{\sqrt{10}}$.  Substituting, we obtain

$2y^{6}+\frac{y^{5}}{\sqrt{10}}+\frac{y^{3}}{\sqrt{10}}+\frac{y}{\sqrt{10}}-2=0$

Divide by $y^{3}$ to get

$2y^{3}+\frac{y^{2}}{\sqrt{10}}+\frac{1}{\sqrt{10}}+\frac{1}{y^{2}\sqrt{10}}-\frac{2}{y^{3}}=0$
$2\left(y^{3}-\frac{1}{y^{3}}\right)+\frac{1}{\sqrt{10}}\left(y^{2}+\frac{1}{y^{2}}\right)+\frac{1}{\sqrt{10}}=0$

Now let $z=y-\frac{1}{y}$.  Notice that $y^{3}-\frac{1}{y^{3}}=z^{3}+3z$ and $y^{2}+\frac{1}{y^{2}}=z^{2}+2$.  Substituting, we obtain

$2(z^{3}+3z)+\frac{1}{\sqrt{10}}\left(z^{2}+2\right)+\frac{1}{\sqrt{10}}=0$
$2z^{3}+\frac{z^{2}}{\sqrt{10}}+6z+\frac{3}{\sqrt{10}}=0$
$(z^{2}+3)\left(2z+\frac{1}{\sqrt{10}}\right)=0$

If $x$ is real, then $y$ and $z$ are real also.  Therefore, we ignore imaginary solutions for $z$.  Therefore,

$2z+\frac{1}{\sqrt{10}}=0$
$z=-\frac{1}{2\sqrt{10}}$
$y-\frac{1}{y}=-\frac{1}{2\sqrt{10}}$

Since $x=\frac{y}{\sqrt{10}}$, we have $y=x\sqrt{10}$.  Substituting, we obtain

$x\sqrt{10}-\frac{1}{x\sqrt{10}}=-\frac{1}{2\sqrt{10}}$

Multiplying by $2x\sqrt{10}$, we get

$20x^{2}-2=-x$
$20x^{2}+x-2=0$

Using the quadratic formula, we obtain $x=\frac{-1\pm\sqrt{161}}{40}$, and the answer is $-1+161+40=\boxed{200}$.[/hide][/quote]
You just basically did a full Vieta's substitution on that problem.  Is there any sneakier way to attack it?[/quote]
What is a vieta substitution?

I'm not sure why they're not solutions, but by Descartes' Sign Rule, there is one positive and one negative real root.

[quote="Phelpedo"]@SplashD, you seem to like to hide the quote button, so I'll have to do this manually.

[quote]
the equation would be undefined because of division by zero[/quote]

Where does she divide by $1000x^{6}-1$?[/quote]

It's not division by zero. It is when both sides are multiplied by $10x^{2}-1$, which has roots $\frac{1}{\sqrt{10}}$, the same real roots as $1000x^{6}-1$. So those roots are added in. The complex roots of $1000x^{6}-1$ do work.


linear algebra
matrix
probability
expected value
conditional probability
advanced fields
advanced fields unsolved

a monkey is randomly typing on a keyboard that has $ 27$ keys ($ 26$ letters and the space key). What is the expectation of the number of letters the monkey will type before the word "mathlinks" appear ?

Except very very rough upper bounds, I have not been able to say anything clever about this one. Basically, this is a random walk on a directed graph with $ N \equal{} 27^9$ vertices, and this graph is $ 27$-regular (with some possible loop). One are trying to compute one hitting time. (the eigenvalues of the adjency matrix of the graph are $ 0$ and $ 27$)

This doesn't look too hard, especially because "mathlinks" doesn't repeat any letters.  Let $ E_0$ be the expected number you asked about, let $ E_1$ be the expected number of letters given a free "m" at the start, let $ E_2$ be the expected number of letters given a free "ma" at the start, and so on up to $ E_8$ being the expected number of letters given a free "mathlink" at the start.  Then we get equations like
\begin{align*} E_0 & = 1 + \frac {26}{27} E_0 + \frac {1}{27}E_1 \\
E_1 & = 1 + \frac {25}{27} E_0 + \frac {1}{27} E_1 + \frac {1}{27} E_2 \\
E_2 & = 1 + \frac {25}{27} E_0 + \frac {1}{27} E_1 + \frac {1}{27} E_3. \end{align*}
We can express all the variables in terms of $ E_0$ and $ E_1$, and then solve for those two.

I've got the expected values; it's not terribly complicated if we look at it right.

Version 1: we don't care about spaces, so something like "akhbmathlinksd apq" contains "mathlinks".
Here, the "m" is letter number $ 27^9 \minus{} 8$ on average.
[hide="Argument"]Consider a very long string of typing. On average, our 9-letter string appears as $ \frac1{27^9}$ of all 9-letter substrings. These appearances are independent, except that no two can appear less than nine spaces apart; "mathlinksmathlinks" is as close as two can get. Now, replace each occurrence of "mathlinks" with "*". The * are $ \frac1{27^9 \minus{} 8}$ of all letters.[/hide]
A model for the distribution here: If we reached letter $ n \minus{} 9$ without starting the "mathlinks", it starts at letter $ n$ with conditional probability $ \frac1{27^9}$. I've confirmed the expected value on some small cases.

Version 2: We want "mathlinks" to appear as a space-delimited word, as in "vejnl mathlinks ikmc" or as the first nine letters followed by a space.
In this case, the "m" appears as letter $ 27^{11} \minus{} 9$ on average.
[hide="Argument"]Here, we look at "words", delimited by spaces. If we include the terminal space, the length of a word is a geometric distribution with mean 27. $ \frac {26^{9}}{27^{10}}$ of those are 10 letters, and $ \frac1{26^9}$ of all 10-letter words are "mathlinks ". It costs $ 27^{10} \minus{} 1$ preceding words on average to reach "mathlinks " (distributed geometrically-1), and the non-mathlinks words preceding it average $ \frac {27 \minus{} 10\cdot 27^{ \minus{} 10}}{1 \minus{} 27^{ \minus{} 10}}$ letters. That's $ 27^{11} \minus{} 10$ letters before the "m" on average.[/hide]

[Edit: reverted a mistaken change]

Nice argument, jmerry.

[quote="Mr.Doe"]Can someone explain who did Ravi B get these equations?[/quote]
Well, consider the first equation $ E_0 \equal{} 1 \plus{} \frac{26}{27} E_0 \plus{} \frac{1}{27} E_1$.  Let's look at the first letter typed.  That letter costs 1 unit, which explains the "1" in the equation.  That first letter is either an "m" or not.  If it's an "m", which happens with probability $ \frac{1}{27}$, then we are in the state represented by $ E_1$.  If it's not an "m", which happens with probability $ \frac{26}{27}$, then we are back to the starting state.

I usually think of these problems by drawing a state diagram (in this case, with 9 or 10 states), and then assigning a variable $ E_0$, $ E_1$, etc., to each state.  But I am too lazy to draw the diagram here.

Here's another version of my first calculation, which actually proves the expected value $ 27^9\minus{}8$:
[hide="Calculation"]We consider "words" delimited by each appearance of the letter "s". We are looking for the first appearance of a word ending in "mathlink". As in the space-delimited version, words average 26 letters plus the terminal "s". The probability that a word ends in "mathlink" is $ \frac1{27^8}$, and these words average an additional 26 letters before the "m". We have a mean of $ 27^8\minus{}1$ preceding words, distributed geometrically-1, and these average $ \frac{27\minus{}(26\plus{}9)\cdot 27^{\minus{}8}}{1\minus{}27^{\minus{}8}}$ letters, for a total average cost of $ 27^9\minus{}35$. Add another $ 26$ letters on average in the last word, and we have $ 27^9\minus{}9$ average letters before the "m".[/hide]

Here's the general result.

Given a pattern $ w \equal{} X_1X_2X_3\cdots X_m$ over an alphabet of size $ d$, we calculate the auto-correlation string of $ w$. This is the string over $ \{0,1\}$ whose $ i$th position indicates whether the overlap of $ w$ with itself-shifted-$ i$-places to the right is identical. This will be more clear with an example.

Suppose $ w \equal{} {\tt ABACABA}$ (a pattern over an alphabet of size $ 3$). If we shift it 0 places (that is, no shift at all), we're comparing it with itself, and of course it's identical:
[code]
ABACABA
ABACABA
[/code]
So we write "1". Now shift it once to the right and compare the overlap:
[code]
ABACABA
 ABACABA
[/code]
the overlapping segment is not identical so we add a 0 to the overlap string which is now "10". Pushing another step to the right:
[code]
ABACABA
  ABACABA
[/code]
still no match, so it's another 0: "100". Another push:
[code]
ABACABA
   ABACABA
[/code]
still no match, overlap is now "1000". On the next push things change:
[code]
ABACABA
    ABACABA
[/code]
The overlapping segment ABA is now the same, so we attach a 1 to our overlap string which is now "10001". The final two steps are:
[code]
ABACABA
     ABACABA
[/code]
(no match)
[code]
ABACABA
      ABACABA
[/code]
(match of a single letter A). So the final auto-correlation string is

1000101

We interpret this as a base-$ d$ number and call this $ \mbox{Aut}(w,d)$. In our case, $ \mbox{Aut}({\tt ABACABA}, 3) \equal{} 1000101_3 \equal{} 1 \plus{} 3^2 \plus{} 3^6 \equal{} 739$.

[b]Theorem[/b]: The expected number of (uniform, independent, random) characters we need to generate in order to obtain $ w$ as a consecutive word is $ d \mbox{Aut}(w,d)$.

In our example, the waiting time is $ 3 \times 739 \equal{} 2217$.

In the original "mathlinks" problem, the overlap string is the trivial $ 100000000$ which, interpreted in base $ 27$ is just $ 27^8$. Thus the waiting time is $ 27^9$. This is identical to the result obtained by jmerry (note that if the word was finally generated after $ 27^9$ steps, the "m" showed up at step $ 27^9 \minus{} 8$. It turns out that it's a little bit cleaner to look at the time to reach the final letter).

The proof of the theorem is not hard, and uses essentially the state-diagram method described by Ravi B. One defines $ x_n$ to be the number of words of length $ n$ without any occurrence of $ w$, and $ y_n$ to be the number of words with a single occurrence of $ w$ at the end. The recursion relations for $ x_n$ and $ y_n$ can be written down and they depend on the auto-correlation string (because when you attach $ w$ as a suffix to a $ w$-free string, you may get multiple overlapping occurrences of $ w$ at the end).

I can reconstruct all the details if someone's interested. I first read about this nice result in a wonderful paper of Apostolico, some years ago. The same method also leads to a beautiful formula for the [url=http://www.boingboing.net/2006/03/27/nontransitive-short-.html]non-transitive coin-tossing game[/url], but this is a subject to a separate post.


Let $p$ be an odd prime number. How many $p$-element subsets $A$ of $\{1,2,\ldots \ 2p\}$ are there, the sum of whose elements is divisible by $p$?

Here's my solution from MathLinks:

Define $X: =\{ 1,2,...,p\},Y: =\{p+1,p+2,...,2p\}$.
See $X$ and $Y$ as a representant system for $\mathbb{Z}/ p \mathbb{Z}$, since not more is necessary (so all identities concerning $X$ or $Y$ must be seen $\mod p$).
For a subset $A \subset X$ and $z \in \mathbb{Z}/ p \mathbb{Z}$ define $A+z: =\{ a+z \in X| a \in A \}$ and similar for a subset $B \subset Y$.
Call $A \subset X$ trivial iff $A=\emptyset$ or $A=X$, similar for $B \subset Y$ again.
Now there are only $4$ subsets $P \subset (X \cup Y)$ such that both $P \cap X$ and $P \cap Y$ are trivial, so call these subsets trival from now on too.
Then define for a nontrivial subset $P \subset (X \cup Y)$ a 'translation':
if $P \cap X$ is nontrivial define $P+z : = ((P\cap X) +z) \cup (P \cap Y)$ and $P+z : = (P \cap X) \cup ((P\cap Y) +z)$ otherwise.
Now it's easy to see that when $z$ goes through all possible residue classes, that also the sum of the elements of $P+z$ goes through all of them, so there is exactly one sum that is divisible by $p$.
So the 'translation' divides the nontrivial subsets into equivalence classes of $p$ sets each, all sets of the same class having same number of elements.

Since there are exactly $\binom{2p}{p}$ subsets of the desired order $p$ and $2$ of them are trivial, we get that there are $\frac{\binom{2p}{p}-2}{p}+2$ such subsets.

(note that this technique trivially generalises to other types of subsets and all sets of type $\{1,2,...,kp\}$)


Can the number obtained by writing the numbers from 1 to n in order (n>1) be the same when read left-to-right and right-to-left?


What's your attitude when solving problems? As in, what is your motto or advice for solving problems?

Me: *when I can solve it: "Eh, whatever."
*when I can't: "Heck, might as well try.*dive in*"

There's no such thing as overposting, but this thread sounds slightly like spam.

General problem solving strategies don't belong in this forum either..


Prove bellow inequalitiies where $a,b,c$ be possitive real numbers and $n$ is a natural number except 0
$\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b}\ge \frac{\sqrt[n]{3(a^{n}+b^{n}+c^{n})^{n-1}}}{2}\ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}$

\[\sum \frac{a^{n}}{b+c}\geq\frac{\left(\sum a^{n}\right)\left(\sum \frac{1}{b+c}\right)}{3}\geq\frac{3(a^{n}+b^{n}+c^{n})}{2(a+b+c)}\geq\frac{3(a^{n}+b^{n}+c^{n})}{2\sqrt[n]{a^{n}+b^{n}+c^{n}}}\] and the second inequality is obvious.


calculus logarithms function trigonometry

[quote="skimnc"][quote="Cardinnus"]Hi everyone,
I've become stuck on a problem (which I gave myself so I have no-one else to blame :roll:)
Basically, how would I go about solving something of the form 5^x=3+x or 4^(x+1)=7+x etc.  :?:
I have graphed it and stuff and found approximations but how would one solve this problem using algebra? I'm not an expert by any means so please be gentle and detailed :blush:
Again, any help would be very much appreciated,
Thanks![/quote]

$5^{x}=3+x$ or $4^{x+1}=7+x$

I don't think you can solve either of them algebraically, I think you must use those approximations.  There is a method called "Newton's Method" in calculus that you can use to figure them out, but obviously you're a bit away from that.[/quote]

hmm...you could use logarithms.

$\log_{5}(3+x)=x$?

Logarithms seems like a better approach.

[quote="b-flat"]$\log_{5}(3+x)=x$

I don't think this can be solved without approximating, or advanced methods, either.[/quote]

 :D  at first I thought you could do $\log_{5}(3+x)= \log_{5}3+\log_{5}x$ but that doesn't work

[quote="DiscreetFourierTransform"]Actually, you can. You need to use something called Lambert's W Function.
[url]http://mathworld.wolfram.com/LambertW-Function.html[/url]

[color=red][size=59]Preview your posts.
b-flat[/size][/color][/quote]

Could you possibly show me an example of a problem that uses this (or just one of the ones posted here) and a solution using it?  I see that you use a function $W$ but I don't see any formal equation for this function.. is it just the series posted?  Thanks.

Formal equation (From maclaurin series stuffs)
$W(x) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}(n)^{n-2}}{(n-1)!}\cdot x^{n}$
They give a sample solution [url=http://en.wikipedia.org/wiki/Talk:Lambert_W_function]Here[/url]

Is that helpful?
Most generally, it's the inverse function of $f(x) = x \cdot e^{x}$ where $x \in \mathbb{C}$.

Yeah, thanks.  But isn't that just another approximation method because you can only really approximate series like that?  Just asking because i'm truly curious about it.

Well, yes and no.
The function is a series expansion, so yes, it's an approximation.
However, functions like $\sin x$ have  a series expansion, and those are NOT approximations.
If you leave it in terms of the W function, then you have an explicit value (like $\frac{W(\ln12)}{2\ln6^{1.5}}$)...
That's not an approximation, but evaluating it is.

[quote="Cardinnus"]Thanks for the helpful replies, that clears up most of my question nicely   :)
Is there a website where I can find a (relatively) simple derivation of that Lambert W Function thing. Just so I can see how it was arrived at?
Thanks[/quote]

Try [url]http://mathworld.wolfram.com/LambertW-Function.html[/url].


Hi every body, I am a newbee in Measure theory, I don't understand much of notation used in this discipline. Could some one explain me this notation, please

$E(X 1_{Y\in B})$= $\sum{E(X|Y=r)P\left\{Y=r\right\}}$

Thing I really don't understand is the left term
$E(X 1_{Y\in B})$


trigonometry
geometry
trig identities
Law of Sines
angle bisector

A circle centered at $ O$ has radius $ 1$ and contains the point $ A$. Segment $ AB$ is tangent to the circle at $ A$ and $ \angle{AOB} \equal{} \theta$. If point $ C$ lies on $ \overline{OA}$ and $ \overline{BC}$ bisects $ \angle{ABO}$, then $ OC \equal{}$

[asy]import olympiad;
unitsize(2cm);
defaultpen(fontsize(8pt)+linewidth(.8pt));
labelmargin=0.2;
dotfactor=3;

pair O=(0,0);
pair A=(1,0);
pair B=(1,1.5);
pair D=bisectorpoint(A,B,O);
pair C=extension(B,D,O,A);

draw(Circle(O,1));
draw(O--A--B--cycle);
draw(B--C);

label("$O$",O,SW);
dot(O);
label("$\theta$",(0.1,0.05),ENE);

dot(C);
label("$C$",C,S);

dot(A);
label("$A$",A,E);

dot(B);
label("$B$",B,E);[/asy]

$ \textbf{(A)}\ \sec^2\theta \minus{} \tan\theta \qquad \textbf{(B)}\ \frac {1}{2} \qquad \textbf{(C)}\ \frac {\cos^2\theta}{1 \plus{} \sin\theta} \qquad \textbf{(D)}\ \frac {1}{1 \plus{} \sin\theta} \qquad \textbf{(E)}\ \frac {\sin\theta}{\cos^2\theta}$

[hide="Contest Style Solution"]
We first conveniently consider the angle $ \theta \equal{} 30 ^{\circ}$, and find that $ OC \equal{} 2/3$. Looking at our choices, this leaves D and E standing.

We also know that $ OC < 1$ for $ 0^{\circ} < \theta < 90^{\circ}$. Choice E blows up as we approach $ \theta \equal{} 90^{\circ}$, whereas choice D stays bounded, so the answer is $ \boxed{D}$.[/hide]
[hide="Direct Solution"]
Let $ \phi$ be the measure of angles $ \angle OBC$ and $ \angle CBA$. From $ \triangle OBC$ we use the Law of Sines to find that $ \frac {OC}{\sin \phi} \equal{} \frac {BC}{\sin \theta} \implies OC \equal{} \frac {sin \phi}{\sin \theta} BC$.

We must now find $ \sin \phi$, which comes up quite conveniently in $ \triangle CBA$ as $ \sin \phi \equal{} \frac {CA}{BC} \equal{} \frac {1 \minus{} OC}{BC}$.

Substituting this back into the formula for $ OC$ gives that $ OC \equal{} \frac {1 \minus{} OC}{\sin \theta} \implies OC \sin \theta \equal{} 1 \minus{} OC \implies OC (1 \plus{} \sin \theta) \equal{} 1$.

From here, it is clear that the answer is $ \boxed{D. \; \frac {1}{1 \plus{} \sin \theta}}$[/hide]

Ob=sec y, AB=tan y, by angle bisector theorem on angle B, sec y/OC=tan y/(1-OC)$ \implies$ OC=sec y/(secy+tany)=1/((secy+tany)/sec y)=1/(1+sin y).

[quote="ghjk"][hide]Quick approach :)
Applying the angle bisector theorem we have: OC/OB=CA/AB=1/OB+AB.Then we get:
OC=OB/(OB+AB)=1/(1+AB/OB)=1/1+sin(alpha)(since triangle OAB is right at A).So D is correct(Q.E.D)[/hide][/quote]

CA/AB=1/OB+AB

How did you achieve this?

If $ \frac {a}{b} \equal{} \frac {c}{d}$, then they both also equal $ \frac {a \plus{} c}{b \plus{} d}$. In fact, for any m and n, they equal $ \frac{am\plus{}cn}{bm\plus{}dn}$.

Sorry for the bump, but I have a question about this problem.

In the AoPS wiki there is a solution (solution 4) that says $\frac{h}{h+\sin{\theta}} = \frac{1}{1+\sin{\theta}}$, is this wrong? Are you able to factor out the denominator like this? Attached is a screenshot

Ok... so you can't factor things like that, there's no such property in trig like that right? Sorry if this is dumb...

Yeah, in general you should not be able to do that. They're saying that $\frac{h}{h+\sin \theta}=\frac{h}{h(1+\sin \theta)}$, so $h + \sin \theta = h(1+\sin \theta)$. That would mean $\sin \theta = h \sin \theta$, which would only hold if $h=1$.

Also one more question (double posting so that everything isn't crammed together)

In solution 1 of the problem, I get that we have to use the angle bisector theorem, but how do you get that $AB = \tan \theta$ and $OB = \sec \theta?$ Isn't $\tan \theta$ opposite/adjacent?

Indeed it is - we have $\tan \theta = \frac{AB}{AO} = AB$ since the radius, which is $AO$, is equal to $1$.

Hmm ok, so what I missed is that the radius is $1$. Thanks so much for the help!

[quote=OlympusHero][hide=Solution]From the Angle Bisector Theorem, we have $\frac{AC}{OC}=\frac{AB}{OB} = \sin \theta$. The radius is $1$, so $AO=1$ and hence $AC=1-OC$. We now have $\frac{1-OC}{OC}=\sin \theta$, and solving for $OC$ gives $\frac{1}{\sin \theta + 1} \implies \boxed{\text{(D)}}$.[/hide][/quote]

Nice! It is very similar to Solution 1 on the AoPS wiki, but it's more intuitive this way. Thanks!