geometry
algebra
difference of squares
special factorizations

If the lengths of each of the sides of a square are increased by 2 centimeters, then the area of the square increases by 12 square centimeters. Find the number of square centimeters in the area of the original square.

Let the first square have side length x. Thus, the larger square will have side lengths of x+2.

We are given

$ (x \plus{} 2)^2 \minus{} x^2 \equal{} 12$.

Factoring using difference of squares gives

$ (x \plus{} 2 \minus{} x)(x \plus{} 2 \plus{} x) \equal{} 2(2 \plus{} 2x) \equal{} 12$.

Dividing by 4 gives

$ x \plus{} 1 \equal{} 3$.

Thus,

$ x \equal{} \boxed{2}$, and the area is $ \boxed{\boxed{4}}$.

Checking, $ 4^2 \minus{} 2^2 \equal{} 16 \minus{} 4 \equal{} 12 \checkmark$.


I am looking for a neat solution to a^4 + b^4 +c^4 +2abc(a+b+c) >= (ab+bc+ca)(a^2+b^2+c^2) for a, b, c>0. Any ideas please?

After some algebraic passages you get the following statement

$\sum_{cyc}a^{4}+\sum_{cyc}a^{2}bc\geq \sum_{sym}a^{3}b$

Which can equivalently be rewritten as

$\sum_{cyc}a^{2}(a-b)(a-c)\geq 0$

Which is exactly Schur's inequality

You can find many similar inequalities by schur inequality and muiheard theorem


Prove that: $ \displaystyle \left (1\plus{}\frac{1}{n} \right )^n <3$, $ \forall n \in N^*$.

[quote="thanhnam2902"]Prove that: $ \displaystyle \left (1 \plus{} \frac {1}{n} \right )^n < 3$, $ \forall n \in N^*$.[/quote]

$ \left(1 \plus{} \frac{1}{n}\right)^n \equal{} \sum_{k\equal{}0}^{k\equal{}n}{\binom{n}{k}}\frac1{n^k} \equal{} 1\plus{}\sum_{k\equal{}1}^{k\equal{}n}\frac1{k!}\left (1 \minus{} \frac {1}{n} \right )\left (1 \minus{} \frac {2}{n} \right )\left (1 \minus{} \frac {3}{n} \right )\cdots\left (1 \minus{} \frac {k\minus{}1}{n} \right)$

Thus

$ \lim_{n \to \infty}\left(1 \plus{} \frac{1}{n}\right)^n \equal{} 2 \plus{} \sum_{k\equal{}2}^{k\equal{}n}\frac1{k!}$

So if we prove that $ \sum_{k\equal{}2}^{k\equal{}n}\frac1{k!} < 1$, we are through. To prove this we observe that $ \sum_{k\equal{}2}^{k\equal{}n}\frac1{k!} < \sum_{k\equal{}2}^{k\equal{}n} \frac1{2^{k\minus{}1}} \equal{} 1$

Wouldn't it be sufficient to prove that if

$ f(n) \equal{} \left(1 \plus{} \frac {1}{n}\right)^n$

then, since

$ f'(n) > 0$

$ \lim_{n \rightarrow 0} f(n) \equal{} 1$

$ \lim_{n \rightarrow \infty} f(n) \equal{} e \approx 2.71828$

we have

$ f(n) < 3$

Yes, but I think we cannot base our proof  on the existence of the limit, because the proof of existence is based upon finding a restriction :wink:
(plus I don't think we can have $ f'(n)$)


analytic geometry
geometry proposed
geometry

In $ \triangle ABC$, where $ \angle C \equal{} 90$, CH is an altitude, M is the midpoint of AB P is a point inside $ \triangle ABC$ such that AP=AC. Show that PM bisects $ \angle HPB$ if and only if $ \angle A \equal{} 60$.
I have seen two proofs of this in Olympiad Problems from Around the World, but I would like to know if there is a nicer one. By the way, why can't I post in solved problems?

Hello!

Let $ PM'$ be the bisector of the angle $ \angle HPB$, $ M'\in AB\Rightarrow\angle HPM' \equal{} \angle BPM'$     (1)
$ \frac {AH}{AP} \equal{} \frac {AC\cdot cos\angle A}{AC} \equal{} cos\angle A \equal{} \frac {AC}{AB} \equal{} \frac {AP}{AB}\Rightarrow$

$ \Rightarrow \frac {AH}{AP} \equal{} \frac {AP}{AB}\Rightarrow$ the triangles $ AHP$ and $ APB$ are similar $ \Rightarrow \angle APH \equal{} \angle ABP$    (2)
(1), (2) $ \Rightarrow\angle APM' \equal{} \angle AM'P\Rightarrow AM' \equal{} AP\Rightarrow AM' \equal{} AC$

$ \angle BAC \equal{} 60^{\circ}\Leftrightarrow \angle ABC \equal{} 30^{\circ}\Leftrightarrow AC \equal{} \frac {AB}{2}\Leftrightarrow AM' \equal{} \frac {AB}{2}\Leftrightarrow M' \equal{} M$ :)

Thank you for the proof, it's simpler than both of the two in the book. One used coordinates, the other had a circle center A radius AC.

[quote="dgreenb801"] By the way, why can't I post in solved problems?[/quote]

I think the "solved problems" are only for problems that are solved here in this forum. So only moderators can post there.

I don't exactly know what your answers are. Any way I send a small way to you.
Let BA Intersects  circle  w(A,AC) in R and Q(BR<BQ). It's clear that BC is tangent to w.So (BRHQ)=-1
Note RPQ=90 so PR bisects BPH.So R=M.So AM=AC.So ^A=60.


Prove or disprove that there is a power of 2 that ends in 1000 1's and/or 2's.

I don't really understand the 1000 1's part, since a power of 2 will always be an even number, it can never end in a 1 right?
For the 1000 2's, I can't think of a really good solution, but if you start listing the last two digits of powers of 2, you see that 2^22 ends in 04 and will start repeating and since none of the other powers end in 22, no power of 2 can end in 1000 2's.   :P

I think he means it can end with a mixture of 1s and 2s.

I claim that for every $ k$ there exists a power of $ 2$ ending in at least $ k$ digits which are all either $ 1$s or $ 2$s.  What technique does this generalization suggest?

[hide="If you guessed..."] induction, that works great :) [/hide]
[hide="You could also try..."] a counting argument.  How many such strings are there?  How many residues $ \bmod 2^k$ are there?  Can you prove uniqueness? [/hide]

Can you work from the rightmost digit to the left like this?:

-the last digit must be 2, so it is even
-the next to last must be 1, so it ends in 12 so is divisible by 4.
-the 3rd to last must be 1, so it ends in 112 so is divisible by 8.
-the 4th to last must be 2
-5th to last is 2
6th to last is 1...etc. Can a pattern be found and generalized/applied/proven?


Let $ T_{F}$ be a temperature in degrees Fahrenheit, and $ T_{C}$ be the same temperature in degrees Celsius. For the following problems, use $ T_{C}\equal{}\frac{5}{9}\left(T_{F}\minus{}32\right)$.

1. Determine the temperature at which $ T_{F}\equal{} T_{C}$.

2. Determine the smallest positive values of $ a$ and $ b$ if $ T_{F}\equiv a\bmod{b}$ that make $ T_{F}$ and $ T_{C}$ integers for all temperatures.

3. Determine the smallest positive integer $ T_{C}$ for which $ T_{F}$ is a power of $ 2$.

[hide="1"]Since $ T_{F}\equal{}T_{C}$, substitute $ T_{C}$ to represent both...

$ T_{C}\equal{}\frac{5}{9}(T_{C}\minus{}32)\Leftrightarrow$
$ \left(\frac{4}{9}\right)T_{C}\equal{}\frac{\minus{}160}{9}\Leftrightarrow$
$ T_{C}\equal{}\minus{}40$[/hide]

I'll leave the rest for someone smarter than me.

[hide="2"]
$ T_{F}\minus{}32$ must be divisible by nine, i.e. $ \equiv 0\bmod{9}$.
$ 32\equiv 5\bmod{9}$, so $ T_{F}\equiv 5\bmod{9}$.
[/hide]

[hide="3"]
The lowest power of two that produces a positive integer $ T_{C}$ is $ 2^{11}\equal{} 2048$.
That gives $ T_{C}\equal{} 1120$.
[/hide]

Did I miss anything?  That seemed a little too easy.

EDIT: Time to edit! Mwahaha!
Thanks i_like_pie for pointing out *howdidimissthat* flaw in #2

[quote="CircleSquared"][hide="2"]
$ T_{F}\minus{}32$ must be divisible by nine, i.e. $ \equiv 0\bmod{9}$.
$ 32\equiv 5\bmod{9}$, so $ T_{F}\equiv 4\bmod{9}$.
[/hide]

[hide="3"]
The lowest power of two that produces a positive $ T_{C}$ is $ 2^{6}\equal{} 64$.
That gives $ T_{C}\equal{}\frac{160}{9}$.
[/hide]

Did I miss anything?  That seemed a little too easy.[/quote]

Hey, you're back :).

[quote="CircleSquared"][hide="2"]
$ T_{F}\minus{}32$ must be divisible by nine, i.e. $ \equiv 0\bmod{9}$.
$ 32\equiv 5\bmod{9}$, so $ T_{F}\equiv 4\bmod{9}$.[/hide][/quote]
It's not $ 32\minus{}T_{F}$, but $ T_{F}\minus{}32$, so you shouldn't do $ 9\minus{}T_{F}\equiv5\bmod9$, but instead $ T_{F}\minus{}9\equiv5\bmod9$.

[quote="CircleSquared"][hide="3"]
The lowest power of two that produces a positive $ T_{C}$ is $ 2^{6}\equal{} 64$.
That gives $ T_{C}\equal{}\frac{160}{9}$.[/hide][/quote]
Sorry, I forgot to mention $ T_{C}$ should be an integer.


Let $ a, b, c$ be real numbers satisfying $ a < b < c, a \plus{} b \plus{} c \equal{} 6$ and $ ab \plus{} bc \plus{} ca \equal{} 9$.

Prove that $ 0 < a < 1 < b < 3 < c < 4$.

[quote="AndrewTom"]Let $ a, b, c$ be real numbers satisfying $ a < b < c, a \plus{} b \plus{} c \equal{} 6$ and $ ab \plus{} ac \plus{} ca \equal{} 9$.

Prove that $ 0 < a < 1 < b < 3 < c < 4$.[/quote]
Let  $ f(x)\equal{}x^3\minus{}6x^2\plus{}9x$  :wink:


i had to do a lab on DNA extraction, and i did evything except testing it for acidity with a blue litmus paper.... does any one knwo how acid or base the DNA of an onion is ???  thank you

-titan

I believe that DNA is acidic. I'm not sure how much so though.

because of nucleic acid? thats why its called DNA ^^


Let a,b,c,d be positive real nunbers.
Prove that A=a/(a+b+c) + b/(b+c+c) + c/(c+d+a) +d/(d+a+b)
CAN NOT be a positive integer.
I can't prove that.

[quote="Nguyen Ngoc Linh"]
Prove that A=a/(a+b+c) + b/(b+c+c) + c/(c+d+a) +d/(d+a+b)
[/quote]

Don't you this that should be :

[code]A=a/(a+b+c) + b/(b+c+d) + c/(c+d+a) +d/(d+a+b)[/code]

[quote="Euclid_great"][quote="Nguyen Ngoc Linh"]
Prove that A=a/(a+b+c) + b/(b+c+c) + c/(c+d+a) +d/(d+a+b)
[/quote]

[code]A=a/(a+b+c) + b/(b+c+d) + c/(c+d+a) +d/(d+a+b)[/code][/quote]

I think that, too!

Or [code]A=a/(d+b+c) + b/(a+c+d) + c/(b+d+a) +d/(c+a+b)[/code]

HUH? I'm sorry.
Yes, I was wrong. b/(b+c+d) not b/(b+c+c).
thank you, Euclid and Euler.

We can prove that $ 1 < \frac{a}{{b \plus{} c \plus{} d}} \plus{} \frac{b}{{a \plus{} c \plus{} d}} \plus{} \frac{c}{{b \plus{} a \plus{} d}} \plus{} \frac{d}{{b \plus{} c \plus{} a}} < 2$.

If a=6 and b=c=d=1, the sum is greater than 2. Your ineq is true when sum of the 2 smallest numbers is greater than the greatest number.

[quote="L_Euler"]We can prove that $ 1 < \frac {a}{{b \plus{} c \plus{} d}} \plus{} \frac {b}{{a \plus{} c \plus{} d}} \plus{} \frac {c}{{b \plus{} a \plus{} d}} \plus{} \frac {d}{{b \plus{} c \plus{} a}} < 2$.[/quote]

Is that wrong? :?:

[quote="L_Euler"]We can prove that $ 1 < \frac {a}{{b \plus{} c \plus{} d}} \plus{} \frac {b}{{a \plus{} c \plus{} d}} \plus{} \frac {c}{{b \plus{} a \plus{} d}} \plus{} \frac {d}{{b \plus{} c \plus{} a}} < 2$.[/quote]

Oh, I'm sorry, I think that is $ 1 < \frac{a}{{a \plus{} b \plus{} c}} \plus{} \frac{b}{{b \plus{} c \plus{} d}} \plus{} \frac{c}{{c \plus{} a \plus{} b}} \plus{} \frac{d}{{d \plus{} a \plus{} b}} < 2$ :)

Hello everyone!
[hide="My solution"]Assume $ a<b<c<d$ $ \Rightarrow \frac{b}{b\plus{}c\plus{}d}<\frac{b}{b\plus{}c\plus{}a}$ and $ \frac{c}{c\plus{}d\plus{}a}<\frac{c}{c\plus{}b\plus{}a}$
So we have $ A<\frac{a}{a\plus{}b\plus{}c}\plus{}\frac{b}{a\plus{}b\plus{}c}\plus{}\frac{c}{a\plus{}b\plus{}c}\plus{}\frac{d}{d\plus{}a\plus{}b}<2$  because $ \frac{d}{d\plus{}a\plus{}b}<1$
And $ A>1$ is too easy!! :) [/hide]


integration
logarithms
function
calculus
derivative
parameterization
probability and stats

Say $ X$ has a density $ f(x) \equal{} 3x^{ \minus{} 4}$ if $ x > 1$, and 0 otherwise. Now say $ X_1,...,X_{16}$ are independent with density $ f$. Let $ Y \equal{} (X_1X_2...X_{16})^{1/16}$. Find $ E(Y)$ and $ Var(Y)$.

Some equations that may help are

$ Var(Z) \equal{} E(Z^2) \minus{} [E(Z)]^2$
$ E(Z) \equal{} \displaystyle\int^\infty_{ \minus{} \infty} zf(z)\,dz$

I found $ E(X) \equal{} \frac {3}{2}$ and $ Var(X) \equal{} \frac {3}{4}$, using the above formulas. However, I still do not even know how to approach the actual question of the problem.

Let $ X'_i\equal{}\ln X_i$ and $ Y'\equal{}\ln Y$. Now $ Y'$ is the mean of the $ X'_i$, and you should know how to handle that.
The next step of the calculation is to find the density of the transformed distribution $ X'$.

[quote="jmerry"]Let $ X'_i \equal{} \ln X_i$ and $ Y' \equal{} \ln Y$. Now $ Y'$ is the mean of the $ X'_i$, and you should know how to handle that.
The next step of the calculation is to find the density of the transformed distribution $ X'$.[/quote]

I can see how $ Y'$ is the mean of the $ X'_i$ by the transformation, but I don't think I am certain as to how to calculate the new transformed distribution.  I would say that $ X' \equal{} \ln (3x^{\minus{}4})$, but this would be too easy, and would make the transformation useless. I feel as though I am missing something obvious.

Could you offer some more insight? This and another problem have been bugging me for days now.  :(

Cumulative distribution functions transform by direct substitution $ f(x)\mapsto y$ (at least for increasing $ f$). Densities are derivatives, and transform by the chain rule; there's a factor of the derivative of $ f$ involved.
It's also easier to look at the inverse of the transformation.

In this case, the cumulative distribution function $ F_X(x)\equal{}P(X\ge x)$ is $ \begin{cases}1&x<1\\ x^{\minus{}3}&x\ge 1\end{cases}$. We substitute $ t\equal{}\ln x,x\equal{}e^t$, and get $ F_{X'}(t)\equal{}P(X'\ge t)\equal{}\begin{cases}1&t<0\\ e^{\minus{}3t}&t\ge 0\end{cases}$.
The density of $ X'$ is the negative of the derivative, $ 3e^{\minus{}3t}$ for $ t\ge 0$.

This is an exponential random variable with parameter $ 3$.


geometry
calculus
probability
ratio
combinatorics solved
combinatorics

Alright, the answer for this question is actually needed for a marketing report.  I tried going to a local high school with the following equation and 4 algebra, geometry, and calculus teachers tried looking at it, and none of them could seem to figure it out, or tell me what additional information they needed TO figure it out, but one of them did give me this website, and said if anyone could figure it out, it would be someone on here.

So here was the problem:

~ 1000 gardeners have insect infestations every year.
~ One in ten gardeners is an organic gardener.
~ One third of all insect infestations happen to organic gardeners.
~ How many times more likely is it for an organic gardener to have an infestation then a non-organic gardener?

Since organic gardeners are far more rare then inorganic gardeners and yet a third of all infestations that occur happen to them, I would assume that organic gardeners are far more likely to get infestations, but I can’t seem to lock down exactly what the number is or what the formula would be to figure this out.

I'm looking for an answer like, "Organic gardeners are 30 times more likely to have an infestation then an inorganic gardener".

Can anyone figure this out OR tell me what information they need me to dig up so they can figure it out?  :?

It's really pretty simple. The 1000 infestations is irrelevant, and doesn't tell you anything extra unless you also know the number of gardeners.

$N_g=$ number of gardeners, $N_i=$ number of infestations.
$P_o=$ probability that an organic gardener is infested $=\frac{\frac13 N_i}{\frac1{10} N_g}=\frac{10}3\cdot\frac{N_i}{N_g}$
$P_n=$ probability that a non-organic gardener is infested $=\frac{\frac23 N_i}{\frac9{10} N_g}=\frac{20}{27}\cdot\frac{N_i}{N_g}$
The ratio is $\frac{P_o}{P_n}=\frac{\frac{10}{3}\cdot\frac{N_i}{N_g}} {\frac{20}{27}\cdot\frac{N_i}{N_g}}=\frac92$.

i do not know if this is a correct solution , but i hope it is  ,  :P
ok let "G" be the number of all gardners ( either healthy or not )
"O" is the number of organic gardners , "NO" is the number of non organic gardners
so 1/10G are O                  &                   9/10G are NO
let I be the number of infected people
so 1/3I are O                      &                   2/3I  are NO
the rate of infection of organic people is 10I/3G & for non or ganic ones is 20I/27G

so %O/%NO = (10I/3G)/(20I/27G)=9/2
that means for every 2  non organic gardners infected there is 9 organic ones infetcted
i hope that this will help  :)

Wow, I never realized it was such a complicated process to figure this out...   :huh: at least it's complex to me I guess, ha.  I guess I've been out of school and math classes too long.

Thank you VERY much guys for the help.  So to put it in the correct format, I can also phrase this by saying, "Organic gardeners are 4 and a half times more likely to get an infestation then an inorganic one", correct?

Thanks again!


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be the group of integer under addition, let $H \subset G$ be the subgroup of  $G$ which different from $G$. Prove that $H = (x)$, that is $H$ consits of all multiple of fixed integer $x$.

When the subgroup is $\neq \{0\}$, it must contain a smallest positive element. All others are mutliples of it then by division with remainder through it.


Prove that if $x$, $y$, $z$ are nonnegative reals such that $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$, then

$\left(1-x\right)^{2}\cdot\left(1-y\right)^{2}\cdot\left(1-z\right)^{2}\geq 2^{15}\cdot xyz\cdot\left(y+z\right)\left(z+x\right)\left(x+y\right)$.

Very nice inequality. When do you post your solution?

Thank you very much.

Well, the problem is that I have "discovered" this inequality two or three years ago and now I don't remember the idea. So, you will have to wait a little bit. Sorry

Hi all,

I suggest my solution. Put a = sqrt(x), b = sqrt(y), c = sqrt(z) then
1 - x = 1 - a^2 = (a+b+c)^2 - a^2 = (b+c)(2a+b+c).
We have to show that ((a+b)(b+c)(c+a)(2a+b+c)(a+2b+c)(a+b+2c))^2 >= 2^15*a^2b^2c^2(a^2+b^2)(b^2+c^2)(c^2+a^2).
Using inequality ab(a^2+b^2) <= (a+b)^4/8  (this equivalent to (a-b)^4)
and  (2a+b+c)(a+2b+c)(a+b+2c) >= 8(b+c)(c+a)(a+b) (AM-GM) we are done.

Have joy,

Namdung


There are 3 concentric conducting shells A,B and C of radii a,2a and 4a respy.
A is earthed and consists of a key K1 connecting A to earth.
B and C are connected to each other with a key K2.
Initially both the keys are open. Now a charge Q is given to B.
K1 is closed so that A gets connected to the earthed.
Now answer the following questions:
[color=red](1)[/color] If K2 is closed, how much charge flows from the earth to A?
[color=red](2)[/color] After K2 is closed what is $ V^{C}\minus{}V^{A}$ ? (where $ V^{C}$ and $ V^{A}$ represent the potentials of C and A)
[color=red](3)[/color] If U1 and U2 represent the electrostatic potential energy of the system before and after closing K2, find $ \frac{U2}{U1}$?

Thanks for any hints.

When a charge is given to concentric shells or parallel plates, half the charge surfaces on the outermost faces of the arrangement. However, charge is conserved. Hence the inner surface of A and outer surface of C have $ \frac {Q}{2}$, outer surface of A, $ \frac { \minus{} Q}{2}$. Both surfaces of B get $ \frac {Q}{2}$ each and inner surface of C gets $ \frac { \minus{} Q}{2}$.

If A is earthed, the corresponding surface loses all charge but the charge on other surface is retained. Assuming inner surface of A is earthed, $ \frac { \minus{} Q}{2}$ flows from earth to the shell.

The charge Q will go to the outermost surface, C, and no electric field exists inside the system . So, no charge flows from earth to A, there is no potential difference and no energy variation.

(1) As   B and C  are short-circuited, we can consider them as a solo shell with inner and outer radius b and c, respy( aha!!!).

So the superposition of effects of charge  $ Q_A$ that flows from the earth to A, on the surface of radius $ a , - Q_A$ on the surface of radius $ b$ and $ Q + Q_A$ on the outermost surface must lead to a null potential of $ A$.
  That for
\[ Q_A = - Q {{ab}\over{ba + bc - ac}}
\]
(2) $ V_C - V_A = V_C = {{Q + Q_A}\over{4\pi\epsilon_0c}}$, so
\[ V_C - V_A = {{Q(b - a)}\over{4\pi\epsilon_0(ba + bc - ac)}}
\]
(3) Before closing $ K_2$ (with $ K_1$ already closed) , there will be a charge $ Q_A'$ with radius $ a$ and $ Q$  on radius $ b$ in order to make null $ A$s  potential. That for $ Q_a' = - Q{a\over b}$.

 It's easy to calculate $ U_1 = {1\over2}\sum{Q_i'V_i'} = {{Q^2(b - a)(b + c)}\over {8\pi\epsilon_0b^2c}}$, and $ U_2 = {1\over2} \sum{Q_iV_i} = {{Q^2(b - a)\over {8\pi\epsilon_0(ba + bc - ac)}}}$, and divide.

Where is the difference in energy?

There are nice problems of this kind in a Brazilian book: Problemas de Física - Eletricidade - Eletrostática , by Eduardo Wilner.

[quote="Immanuel"]
Where is the difference in energy?

[/quote]

I didn't understand that!

[quote="Immanuel"]
There are nice problems of this kind in a Brazilian book: Problemas de Física - Eletricidade - Eletrostática , by Eduardo Wilner.

[/quote]

Unfortunately, I can't understand the language.

My question about the vanishing energy is mor or less the same as in the subject Capacitors in Advanced Physics, from about march, this year. Btw I still didn't see a better answer for that one...


linear algebra
matrix
Putnam
linear algebra unsolved

Two players in turn replace asterisks in the $ 4\times4$ matrix $ \left(\begin{smallmatrix}* &* & * & *\\ * &* & * & *\\ * &* & * & *\\ * &* & * & *\\ \end{smallmatrix}\right)$ by any real numbers (at each turn one may replace any asterisk). If they form a singular matrix then the first player wins, else the second player wins. Has anybody of players a winning strategy? If somebody has, then who?

The singular player wants to make the determinant zero, and the non-singular player wants to make the determinant non-zero. I think if the singular player used all of his turns entering zero he would win. It would turn into tic-tac-toe, the invertible player would have to keep the singular player from filling an entire row or column with zeros.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146653]Putnam 2002 A4[/url] was similar, with three differences:

1. In that other problem, the matrix was $ 3\times 3$.
2. In that other problem, the player trying to make the matrix singular moved second.
3. In that other problem, only zeros and ones were used, not arbitrary real numbers.

The solution was a matter of tedious casework, which leads me to suspect that the answer to this problem, the $ 4\times 4,$ will be a matter of vastly more tedious casework. In examining my proof on the other thread, point #3 doesn't seem to matter - I don't think the non-singular player could win even if you let him choose arbitrary values. But I'm not 100% sure of that.

In this case, the second player can easily force it to be singular, and no zeros are required at all, unless the first player also uses some. In fact, the second player can easily force a rank 2 matrix.

It's a lot harder as written.


Let a,b,c be positive real.Prove that:
3(a+ :sqrt: ab+(abc) 1/3 )<=(8+2 :sqrt: ab/(a+b))(a(a+b)/2(a+b+c)/3) 1/3
Do you generalize this ineq? ;)

This is the LaTeX version of the problem:
[quote]Let $a,b,c$ be positive reals. Prove that:
\[ 3(a+\sqrt{a b}+\sqrt[3]{a b c}) \leq \left(8+\frac{2\sqrt{ab}}{a+b}\right)\left(a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}\right)^{\frac{1}{3}}.\]
[/quote]

By Holder, we have:
\begin{eqnarray*}
a+\sqrt{a b}+\sqrt[3]{a b c}
&=& a^{\frac{1}{3}}a^{\frac{1}{ 3}}a^{\frac{1}{3}} + a^{\frac{1}{3}}(\sqrt{a b})^{\frac{1}{3}}b^{\frac{1}{3}}
	+ a^{\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}} \\
&\leq & (a+a+ a)^{\frac{1}{3}} (a+ \sqrt{a b} + b)^{\frac{1}{3}} (a+b+c)^{\frac{1}{3}} \\
&= & 3^{\frac{2}{3}}  \left(a+ \sqrt{a b} + b\right)^{\frac{1}{3}} \left(\frac{2}{a+b}\right)^{\frac{1}{3}}     \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.
\end{eqnarray*}

It suffices to show:
\[3^{\frac{5}{3}}  \left(a+ \sqrt{a b} + b\right)^{\frac{1}{3}} \left(\frac{2}{a+b}\right)^{\frac{1}{3}} \leq 8+\frac{2\sqrt{ab}}{a+b}.\]
Or equivalently,
\[3^5  \left(2+\frac{2\sqrt{ab}}{a+b}\right)  \leq \left(8+\frac{2\sqrt{ab}}{a+b}\right)^3.\]

Let $\displaystyle x = \frac{2\sqrt{ab}}{a+b}$, then by AM-GM:
\[3^5(2+x)  = 3^3[3\cdot3\cdot(2+x)] \leq 3^3 \left(\frac{3+3+(2+x)}{3}\right)^3= (8+x)^3. \]

Q.E.D.

This is a slight improvement of Kiran Kedlaya's inequality:

\[  \frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}. \]

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6474

Cheers!  :D  :D


Hey all
I've been trying to warn as many people as I could...
If any of u r on Orkut, please leave it as soon as possible!!
It's sad that I have to say this about a GOOGLE organisation but Orkut is simply vile... :mad:
I have documentary proof(which sadly is too large to upload here but I can show it to u if u pm me ur email id) of how photos are being taken from orkut and are being modified...I won't say more...
Also, the community called Hackers there had revealed some time back that they use the orkut database to hack people's email ids....and they can read anyone's mail if they are on orkut...!!!
My OWN profile had been nodified when I was there and some extremely Smart person has written a bunch of rubbish including that I speak Zulu and...a load of other things that I won't put here...!!
The same thing is happening on fropper,hi5 and all those other "friendship" networks
So please leave orkut  as soon as you can coz it's not safe and please DON'T post ur pic or ur details on any network
Hope this warning comes in handy... :)

Not only that .u should encourage a more and more people to [color=indigo][u][b]join Aops [/b][/u][/color].....this site is just amazing...u get to know so many things...

I've been doing that...literally shouting my head off...sadly not many of the people I know capable of showing interest in this amazing subject...the rest are already on their way here!!  :D  I still keep trying though...

The problem is that many of my friends think that this site is meant is meant for "SCHOLARS" (with huge spectacles!)  who do nothing but solve loads and loads of problems....a trait typical of an overworked IITian....and hence they feel this site is not meant for them.....This perception needs to be removed..
they fail to understand that we guys here are much more fun loving than them....the thing is that we derive our fdun in a slightly different and refined way...not like those idiots at ORKUT...who get fun out of abusing others..

[quote="ashwath.rabindranath"]Absolutely. Viva la AoPS!! I live here more than at home, man!
@Padmavathi: Thanks, I've quit the stupid thing. It was starting to get bugging anyway. I'll tell everyone I can. The problem is people are addicted and really don't care. Still we must try.. :|[/quote]
Yeah! I live here so much that these guys can get really rich if they start collecting rent  :wink:
Just make sure that you've quit completely...Coz I quit but then people could still see my account for a week...to make sure, you can go to google, personalise it, check out the My Account page and it may still be there! I could delete mine only from that page...
Yeah..it gets really bugging...
Even though at one point I hadn't logged on for a week or two, One of my cousins received a BAD scrap...from ME!! It's so easy to manipulate profiles and stuff there...and then I got a mail from one of my best friends saying that my profile said I was a BI(ok no, I won't say it here)!!!!!
I just HATE the place now...
@Ashwin- I USED TO know so many IITians on the site...who would vent out their frustration on users...call them names..send them spam..there was one "brilliant" IITian who wrote in his profile that he tried to commit suicide after his professor came to the hostel and banned him from logging on to orkut for a day... [NO FURTHER COMMENT]
Even though there ARE so many good communtities like Maths, Physics, I Love Chemistry, the High IQ Club and others, the bad part is that there are DOUBLE the number of bad ones.
Addiction is quite right!!  I just get the feeling that people get this false identity, try to become someone else on Orkut...they portray themselves as "uber-cool" when in reality they are like the plain-Janes of our class...it's just so silly...And when you tell them, they are too thick-skulled to listen...shouting from the rooftops won't help..you have to hire a blimp and a set of megaphones...
Check out what happened sometime back-
Me-"Hey why did you skip your lunch? that isn't good for your health."
Friend-"But i HAD to log on to Orkut...in case I got a scrap from...."
Me-"But then Orkut is so dumb...especially all the communities you've joined...why don't you do something USEFUL? Plus it's dangerous..people can hack your email id and stuff."
Friend-"Look, don't try to make me quit, I can't live without orkut."
Me-"I know this really cool site...the best in the world" (And then I told her about AoPS)
Friend-"So dumb, that site's only for nerds like you..I WON'T QUIT ORKUT.It's fun,safe and you can meet so many people. it's changed my life..."
A week later she comes saying her id has been hacked,her profile has been modified and her worst enemy has got hold of all her photos through orkut... :D
AoPS Ki Jai!! I rarely log on to anything other than this and wiki...it's helped so much!! This site is the best....
Still, I'm trying to spread the word...hope  many people listen yaar!!

Yeah..that site is EVIL...
Last year, after a slew of crank calls,an intruder was caught in one of my friends' houses because she had put her details on Orkut. He later confessed that he had got her phone number from Orkut and had got the address by hacking her Yahoo! id which was in the orkut database. The police(in Pakistan) were clueless about what he was saying. They had to call an engineer and get him to explain!!
It wasn't even MENTIONED in the papers... :mad:
The bad thing is that we don't have enough laws and they are not well phrased enough to combat cyber crime... :mad:

[quote="Cyber-Shreyas"][quote="sonny"]What is Orkut? is it like some myspace of India?[/quote]
http://orkut.com - Just another Socialnetworking site. But much more popular than Myspace/Facebook in India and Brazil.

Anyways, the thing about Orkut is that, though it's not really secure and all that, it helps in keeping me connected with a lot of friends who I had lost contact with..[/quote]

ah okay.....thanks for the info..........

wel
orkut is not bad if u r smart enough
who tlls u to put ANY of ur authentic details

:rotfl:  :rotfl:  :rotfl: cool thats another part os the story
just like a coin has two sides!

i won't mind flipping that coin :rotfl:  :rotfl:

and i wont mind losing the toss :rotfl:  :rotfl:  :rotfl:

I won't mind hitting both of u with that da. don't use this for nonsense talk all the time.

sorry but i think u r also engaged in all this!


-5^-2 = - 1/25

How?

My book didn't really go into detail about things such as

-x^-n

By the way, how do you guys do that cool formula stuff where it's bold and looks nice?

To answer your second question, it's called Latex... go to the testing forum to try it out.
:edit:
you can just go to the latex link on the left side of the site

Now to answer your first question...

$-5^{-2} = \frac{-1}{25}$ -5^-2 really equals -(5^-2)
:-D showing off my latex skills

so anyway, for all numbers, $x^{-y} = \frac{1}{x^{y}}$

so... we have:
$ \frac{1}{-5^{-2}}$

we simplify the denominator to -25 and we have $\frac{-1}{25}$as our answer.

When you have fraction such as

$\frac {p^k}{p^m}$

That's equal to $p^{k - m}$

So $x^{-n} = \frac 1{x^n}$ because ${\frac {x^0}{x^n} = x^{0 - n} = x^{-n}}$

That's correct. Exponents and Logarithms.
(I enjoyed rationalizing denominators but that's not really applicable to this post)


[quote="luiseduardo"](OMRN) [i]n [/i] is a positive real [i]k [/i] a positive rational. Suppose that (2.N)^k = 1944 and n^k = 486.√[2].
 Determine the number of units n³.

Thanks.[/quote]

This is a visually flawed presented problem.  You must be using two variables
(instead of three, so one case would be used for each letter), please don't use
a period for multiplication, and I do not know what you mean by "determine the
number of units n^3 (my version)."  I don't know why you used square brackets
in the radicand, when parentheses could have managed instead, unless you had
additional intentions for them.

If you had instead typed something along the lines of, but not limited to, one of these:

Suppose that (2n)^k = 1944 and n^k = 486*sqrt(2), determine n^3. . . . . Or

Suppose that $ {(2n)^k = 1944}$ and $ n^k = 486\sqrt {2}$, determine $ n^3.$

OK, I'm sorry.

Determine the number of units n³.  = Find the number of units of n³. Or Find the algarism of n³.

Thanks.

Dividing the first equation by the second we have $ 2^k\equal{}2\sqrt2 \implies k\equal{}\frac32$

Thus, $ n^{\frac32}\equal{}486\sqrt2 \implies n^3\equal{}\boxed{486^2 \cdot 2}$ which I am too lazy to calculate  :P

Thanks, yes, my answer was 472392. :D

n³ = ...2

Language note: that's the [i]units digit[/i] of that number.  None of the other terms you used for this are standard English (and in fact I wasn't able to figure out what you were asking for until you wrote your last line).


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

Show that $\int^{a}_{0}f(x) dx = \int^{a}_{0}f(a-x) dx$ for all $x$and use this result to find $\int^\frac\pi{2}_{0}ln(sinx) dx$

i have shown the first part but i can't connect it to the second part.

i'll just give you a hint. to prove your statement, make a change of variable such as $x= a-y$, plug in the appropriate limits, relate the differentials $dx$ and $dy$

So we have $I = \int_{0}^{\frac{\pi}{2}}\ln{(\sin{x})}dx = \int_{0}^{\frac{\pi}{2}}\ln{\left(\sin{\left(\frac{\pi}{2}-x\right)}\right)}dx = \int_{0}^{\frac{\pi}{2}}\ln{(\cos{x})}dx$.

Adding the two together, we get

$2I = \int_{0}^{\frac{\pi}{2}}\ln{(\sin{x})}dx+\int_{0}^{\frac{\pi}{2}}\ln{(\cos{x})}dx = \int_{0}^{\frac{\pi}{2}}\ln{\left(\frac{1}{2}\sin{2x}\right)}dx$.

But this is just

$2I =-\frac{\pi}{2}\ln{2}+\frac{1}{2}\int_{0}^{\pi}\ln{(\sin{u})}du$,

after making the substitution $u = 2x$. Fortunately, since $\sin{x}= \sin{(\pi-x)}$, we know

$\int_{0}^{\pi}\ln{(\sin{u})}du = 2 \int_{0}^{\frac{\pi}{2}}\ln{(\sin{u})}du = 2I$,

so our equation becomes

$2I =-\frac{\pi}{2}\ln{2}+I \Rightarrow I =-\frac{\pi}{2}\ln{2}$.

Maybe there's a shorter way?

[quote="math_music"][quote]...Fortunately, since $\sin{x}= \sin{(\pi-x)}$, we know

$\int_{0}^{\pi}\ln{(\sin{u})}du = 2 \int_{0}^{\frac{\pi}{2}}\ln{(\sin{u})}du = 2I$,

...[/quote]
I don't get it, could you explain it a little more? Thanks in advance.[/quote]
\[\int_{0}^{\frac{\pi}{2}}\ln{(\sin{u})}du =\int^{\pi}_{\frac{\pi}{2}}\ln{(\sin{u})}du\]

I got it now, i don't was thinking geometrically (i was thinking in a substitution). Thanks


function
vector
real analysis
algebra proposed
algebra

is there some resource or[b] counter example[/b] to prove:
when f is not a continuous function,the conclusion following is WRONG:
$ f: \mathbb{R}\rightarrow \mathbb{R}$,such that $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)<\equal{}>f(x)\equal{}ax(a\in\mathbb{R})$

[quote="wxydx00"]is there some resource or[b] counter example[/b] to prove:
when f is not a continuous function,the conclusion following is WRONG:
$ f: \mathbb{R}\rightarrow \mathbb{R}$,such that $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) < \equal{} > f(x) \equal{} ax(a\in\mathbb{R})$[/quote]

Yes, consider $ \mathbb R$ as a $ \mathbb Q$-vector space and $ A$ and $ B$ two non trivial (neither $ \{0\}$, neither $ \mathbb R$) supplementary sub vector spaces  of $ \mathbb R$ (whose existence may be demonstrated thru axiom of choice and Zorn lemma). Let $ a(x)$ and $ b(x)$ the projections of $ x$ in $ A$ and $ B$ (such that $ x \equal{} a(x) \plus{} b(x)$)

Then $ f(x) \equal{} u\cdot a(x) \plus{} v\cdot b(x)$ is one of the numerous counter-examples.

(notice that trivial sub vector spaces  give continuous solution)

I agree with most of pco's answer except that it is not the choice of $ A$ and $ B$ that requires the Axiom of Choice.
For example I can just take $ A$ to be $ \mathbb{Q}$ and $ B$ to be $ \sqrt{2}\mathbb{Q}$.

It is when you define the projection operators that you need the AoC. Projection operators onto a vector sub space are not unique
They depend on a basis or (something else like an inner product)

$ \mathbb{R}$ as a vector space over $ \mathbb{Q}$ is infinite dimensional so you have to choose an infinite number of independent
numbers as a basis. This requires Zorn's Lemma which is equaivalent to the AoC. e.g. see
[url]http://myyn.org/m/article/every-vector-space-has-a-basis/[/url]

[quote="PhilG"]I agree with most of pco's answer except that it is not the choice of $ A$ and $ B$ that requires the Axiom of Choice.
For example I can just take $ A$ to be $ \mathbb{Q}$ and $ B$ to be $ \sqrt {2}\mathbb{Q}$. [/quote]

Hemm, I said "supplementary" subvector spaces of the $ \mathbb Q$ vector space $ \mathbb R$. $ \mathbb{Q}$ and $ \sqrt {2}\mathbb{Q}$ are not supplementary subvector spaces of the $ \mathbb Q$ vector space $ \mathbb R$.

And the existence of non trivial supplementary $ \mathbb Q$-subvector spaces of the $ \mathbb Q$ vector space $ \mathbb R$ may be demonstrated with AC.

Oh I see what you mean now. I was just taking the word supplementary in its English sense rather than a mathematical sense, sorry.

[quote="wxydx00"]is there some resource or[b] counter example[/b] to prove:
when f is not a continuous function,the conclusion following is WRONG:
$ f: \mathbb{R}\rightarrow \mathbb{R}$,such that $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) < \equal{} > f(x) \equal{} ax(a\in\mathbb{R})$[/quote]

Cauchy's equaltion.

Can you construct an example without using the Axiom of Choice?

No.  This is related to the fact that you can't construct non-Lebesgue measurable functions without the axiom of choice.


geometry
rhombus
symmetry
incenter
angle bisector
geometry proposed

Prove that the quadrilateral formed by the adiacent quadrisectors of the angles of a rombus is a square.

This is nice! I rewrite your problem to make it more suitable for solving:

[i]Let ABCD be a rhombus. Let the quadrisectors of the angles DAB and ABC nearest to the side AB intersect each other at a point X. Similarly define three points Y, Z and W. Then prove that the quadrilateral XYZW is a square.[/i]

In fact, from the symmetry of the rhombus with respect to its diagonals, it is clear that the points X and W are symmetric to each other with respect to the line AC. Thus, $XW\perp AC$. Similarly, $XY\perp BD$. But since the quadrilateral ABCD is a rhombus, we have $AC\perp BD$. Hence, $XW\perp XY$. In other words, < WXY = 90; similarly, < XYZ = 90, < YZW = 90 and < ZWX = 90, and consequently, the quadrilateral XYZW is a rectangle. In order to prove that it is a square, it will now suffice to show that XW = XY.

Let the diagonals AC and BD of the rhombus ABCD meet at a point P. From the symmetry of the rhombus, it is clear that < ABP = < ABC / 2; on the other hand, by construction, we have < ABX = < ABC / 4, and thus < ABX = < ABP / 2. Hence, the point X lies on the angle bisector of the angle ABP. Similarly, the point X lies on the angle bisector of the angle BAP. Hence, the point X is the incenter of triangle ABP. This implies that the point X has equal distances from the sides AP and BP of this triangle. In other words, the point X has equal distances from the lines AC and BD.

Since the points X and W are symmetric to each other with respect to the line AC, the midpoint U of the segment XW lies on the line AC, and we have $XW\perp AC$. Thus, the segment XU equals the distance from the point X to the line AC. Similarly, if V is the midpoint of the segment XY, then the segment XV equals the distance from the point X to the line BD. But since the point X has equal distances from the lines AC and BD, we thus have XU = XV. Since the points U and V are the midpoints of the segments XW and XY, we have XW = 2 XU and XY = 2 XV, and thus XU = XV implies XW = XY, and we are done.

  Darij


Aflati numerele naturale $x$ si $y$ astfel incat:

$x^3+y^3+1=x^2+y^2+x+y+xy$

Daca in dreapta e -xy si nu +xy atunci se descompune
 $(x^2+y^2-xy-1)(x+y-1)=0$

[quote="andyciup"]Solutia pt "$+xy$" este "clasica",luam $s=x+y$ si $p=xy$ si devine o ecuatie in necunoscutele $s$ si $p$,care se rezolva exprimand $s$ in functie de $p$ sau invers si dupa aia se merge pe divizibilitate.[/quote]
     Te complici.Ecuatia e echivalenta cu (x+y)(x+1-x^2+xy-y^2)=(1-y)(1+y), si de aici mergem pe mana divizibilitatii. :lol: E usoara problema.


What is the least number of minutes after 8:00 that the lesser angle between vertical and the hour hand equals the lesser angle between vertical and the minute hand?

this is what I got
[hide]
1min. = .5 deg. on the hour hand
1 hour = 6 deg. on the min. hand
the angle between the hour hand at 8:00 and vertical is 120 deg., and each min. it becomes .5 deg. smaller.  the min. hand is 6 deg. times the number of min., so
120-.5X = 6X
120=6.5X
X=18.4 deg.
[/hide]


modular arithmetic
number theory unsolved
number theory

Find all $ a\in N$ so that$ :$ exist $ n\in Z^\plus{}$ such that $ (a^n\plus{}1)\vdots n^2$

[quote="DCTPKTCSPHN"]Find all $ a\in N$ so that$ :$ exist $ n\in Z^ \plus{}$ such that $ (a^n \plus{} 1)\vdots n^2$[/quote]

Hemmm, $ \forall a\in\mathbb N$, we have $ 1^2|a^1\plus{}1$

[quote="pco"][quote="DCTPKTCSPHN"]Find all $ a\in N$ so that$ :$ exist $ n\in Z^ \plus{}$ such that $ (a^n \plus{} 1)\vdots n^2$[/quote]

Hemmm, $ \forall a\in\mathbb N$, we have $ 1^2|a^1 \plus{} 1$[/quote]
Sorry pco.I forgot conditions $ n>1$

[quote="DCTPKTCSPHN"][quote="DCTPKTCSPHN"]Find all $ a\in N$ so that$ :$ exist $ n\in Z^ \plus{}$ such that $ (a^n \plus{} 1)\vdots n^2$[/quote]
No one idea :)[/quote]

It's easy to see that if $ (a \plus{} 1)$ has an odd divisor $ p > 1$, then $ a \plus{} 1 \equal{} 0\pmod p$ and $ a^{p \minus{} 1} \minus{} a^{p \minus{} 2} \plus{} a^{p \minus{} 3}... \plus{} a^2 \minus{} a \plus{} 1 \equal{} 0\pmod p$ and so $ a^p \plus{} 1 \equal{} 0\pmod {p^2}$

So it remains to study the case where $ (a \plus{} 1)$ only has even divisors, so $ a \equal{} 2^m \minus{} 1$

To be continued ...  :wink:

the case where a+1 isnt a power of 2 is settled as pco said

Consider a+1 is a power of 2.
Assume such an n exists. We see that n is odd. ( If n is even then $ a^n\plus{}1\equal{}2(mod4)$ whereas 4 divides $ n^2$ )
let p be the smallest prime divisor of n. then we have $ a^2\equal{}1(modp)$ and hence $ a\equal{}1 or \minus{}1 (modp)$ in the first case, we have p=2 which is false since n is odd. in the second, we have p divides a+1 and since it is a power of two, we again have p=2 which is impossible.

Hence, for integers a of the form $ 2^n\minus{}1$, such an n doesn't exist

[quote="lajanugen"]the case where a+1 isnt a power of 2 is settled as pco said

Consider a+1 is a power of 2.
Assume such an n exists. We see that n is odd. ( If n is even then $ a^n \plus{} 1 \equal{} 2(mod4)$ whereas 4 divides $ n^2$ )
let p be the smallest prime divisor of n. then we have $ a^2 \equal{} 1(modp)$ and hence $ a \equal{} 1 or \minus{} 1 (modp)$ in the first case, we have p=2 which is false since n is odd. in the second, we have p divides a+1 and since it is a power of two, we again have p=2 which is impossible.

Hence, for integers a of the form $ 2^n \minus{} 1$, such an n doesn't exist[/quote]
Nice  :) .Thank you

[quote="lajanugen"]the case where a+1 isnt a power of 2 is settled as pco said

Consider a+1 is a power of 2.
Assume such an n exists. We see that n is odd. ( If n is even then $ a^n \plus{} 1 \equal{} 2(mod4)$ whereas 4 divides $ n^2$ )
let p be the smallest prime divisor of n. then we have $ a^2 \equal{} 1(modp)$ and hence $ a \equal{} 1 or \minus{} 1 (modp)$ in the first case, we have p=2 which is false since n is odd. in the second, we have p divides a+1 and since it is a power of two, we again have p=2 which is impossible.

Hence, for integers a of the form $ 2^n \minus{} 1$, such an n doesn't exist[/quote]

Hemmm, sorry but ....

why $ a^2 \equal{} 1\pmod p$ ?

$ a^{2n}\equal{}1(modp)$ and $ a^{p\minus{}1}\equal{}1(modp)$ imply $ a^{gcd(2n,p\minus{}1)}\equal{}1(modp)$ implies $ a^2\equal{}1(modp)$ ( n is odd )

[quote="lajanugen"]$ a^{2n} \equal{} 1(modp)$ and $ a^{p \minus{} 1} \equal{} 1(modp)$ imply $ a^{gcd(2n,p \minus{} 1)} \equal{} 1(modp)$ implies $ a^2 \equal{} 1(modp)$ ( n is odd )[/quote]

Quite clear, thanks  :)


quadratics
number theory unsolved
number theory

Hey all;
Let a,b be positive integers s.t.
a^2 + b^2 - ab | (ab)^2
(i) Prove that for any fixed a, finitely b exist satisfying the condition.
(ii) Find all pairs (a,b) satisfying the condition.
I thought part (i) was quite trivial, but i cant get much on (ii) .. obviously all a=b works but there are also some other random sols
lyk (3,6). so yeh .. thanks in advance.

The problem with (ii) is that there are too many solutions. Let $a=pr$, $b=qr$ with $p$ and $q$ relatively prime. Then $a,b$ works if and only if $p^2-pq+q^2|r^2$.

oh ok .. sorry my bad, i should have tried the r=(a,b),p=a/r,q=b/r fing .. i totally forget abt that and only tried quadratic stuf.
so yah .. (ii) is really stupid. mayb this tread should b deleted.


In which grade did you take your first AP test/class?

You're kdding, right?  You have options for 3 years of middle school but only 2 years of high school?  I've added new options to your poll.  (Yay moderator powers!)

I don't get it...I've heard people complain as freshman that they're taking onnnnllly 4 AP classes...I'm trying to figure it out...is it really that important to take so many APs...I mean, I take three right now, but it's my junior year and I haven't taken any before this...

I took 4 AP courses in all of high school if that makes you feel any better.

Also, unrelated to this thread topic -- Stan, I owe you an e-mail.  It will come this weekend.  Sorry for the long delay, but it completely slipped my mind.

which school you go to often affects when and how many aps you take.  smaller schools often don't offer as many ap classes.  my school offers 22 ap classes, and the most anyone has taken in my school has been 15.  the average honors student though doesn't begin taking aps until junior year and probably takes about 7 by the end of senior year.  i am current taking calc bc and physics c as a freshman, but it's certainly not ordinary.  colleges understand what environment you're coming from and as long as you take the most rigorous (or more) schedule you can, you're fine.

Oh...ok...thanx Joel...
                                                                Stan

well, some high schools have more strict requirements than others, and some states as well (like for NY you have to finish a whole bunch of required classes before even being able to think of taking an AP)
for my school, we have to pick a "major" area to concentrate in and that takes away much of the elective/AP room and people end up with much less APs than at other places. The maximum number of APs i know anyone graduating with is like 7.

yea, who does take an ap exam in middle school.
i took ap us as a sophmore.

At my school, we're very limited in the courses we can take (the downside to having, oh, 250-300 students in the whole school), so I didn't take any AP courses before this year, and only one this year, although I would have liked to take more, specifically Computer Science (is it true that it's Java now?) and Physics.

Actually, I just took my first AP exam today (AP Calc AB)

Oh yeah, I almost forgot, I'm a tenth-grader.

[quote]yea, who does take an ap exam in middle school. [/quote]

I did last year (BC)

Also: it seems that others on this site have.  Look at ur poll results.

I took AP Calculus as a sophomore <shrugs>

I guess you could plow ahead in high school math courses, but keep in mind that there is really no race to understand all the material. Just dont strain yourself.

That said, it is not infeasible to do calculus in middle school if you start early enough and have a non-public school math education. I mean, I could have started (non-superior) algebra in elementary school, seeing as I spent all of third grade doing 5th grade busy work (I hate my teacher)

[quote="Danbert"](is it true that it's Java now?) [/quote]
yes, this is the second year for the comp sci test to be in java

[quote="paladin8"][quote="stupidkid"]just took my ap world history test today. failed essay.[/quote]

same. except maybe without the failed essay part :D i thought it was pretty easy compared to the practice ones i've taken; if luck is on my side, i'll pull a 5.

i took 1 ap last year as a freshman, ap comp sci AB in its first year as java -- wish i could've done C++. oh well.[/quote]
-_-
too many smart ppl in this world *sigh*

[quote="Mariag"] The maximum number of APs i know anyone graduating with is like 7.[/quote]

Yes I'm graduating with 7 :)
BC Calculus / Chemistry / Physics B / Java A / Biology / Macroeconomics / Microeconomics

What Maria said is very true tho.
I was the only junior in my BC class last year.
you have to take a lot of math classes, as well as regents for each math.
Regents suck.

actually to take an AP in middle school u only need a flexible school district :)

of course ur right about there not being a race and such

if there were, I wouldn't do AMC stuff of anything, I'd just plow ahead w/ more courses

but what would be the point in that?

[quote="Mariag"] The maximum number of APs i know anyone graduating with is like 7.[/quote]

Well, I know one crazy person that took 6 soph year, is taking 3 this year, and will take at least 3 senior year!

Sophomore: Calc BC, Macro and Micro Econ, Comp Sci, Chem, Bio
Junior: Stat, World History, Physics (I think C)
Senior: US History, Gov. Pol, French Language

I seriously do not get trying to finish calculus before high school, but that's just me. I'm probably just weird.

As to the topic- probably taking 3, maybe 2 next year. Hoping for 4-6 as a junior, so its a fairly large amount. But APs don't matter that much. They're just tests. I'd never want to try to study super hard in order to be able to take 15 APs.

[quote="stupidkid"][quote="Derek"]This year is my first taking AP. AP World and CS AB.

You will never guess what my grade is.[/quote]
8th grade?[/quote]

or senior...

[quote="Geehoon"]What Maria said is very true tho.
I was the only junior in my BC class last year.
you have to take a lot of math classes, as well as regents for each math.
Regents suck.[/quote]

We have a freshman and a sophomore (me) in our calc BC class... but that's probably because my school promotes AP almost religiously ;)

[quote="xxreddevilzxx"]What is a regent?[/quote]New York State has a number of statewide [url=http://www.nysedregents.org/testing/hsregents.html]Regents Exams[/url], in all of the core high school subjects (English, social studies, math, science, and foreign languages).  NY has fairly strict graduation requirements, and one has to pass a certain subset of Regents Exams in order to graduate high school. (I grew up in NY, so I know these exams pretty well.  If I remember correctly, every time you take one, you have to sign this absurd statement that you didn't cheat.)

To specifically answer your question, a "regent" is a member of the [url=http://www.regents.nysed.gov]Board of Regents[/url], which is in charge of all public education in New York.

I have read a lot of times about AP Classes, but, what are they exactly?

[quote="stupidkid"]They are simply advanced classes and at the end of each school year you can take a test and if you do well, you can get college credit.[/quote]Thanks for the information.

[quote="mcalderbank"][quote="themonster"]Is it true that one is not allowed to use a Computer for the AP Computer Science Exam?[/quote]

Yes, it is all purely theoretical.[/quote]

Yeah, you have to write code by hand... not fun especially if you missed a line or something.

how many hours do I have to work each day after school if I am taking 4 AP class next year(I am a freshman) :?

that depends on a LOT of factors. what classes are they? what are your core classes like? do you generally do well on tests without studying? what score would you like to make- enough to get you credit at most universities, or enough to make your chest puff out in pride?

personally, i took three this year. cal bc, biology, and us history. im going to end up graduating with eight, i think. i thought about taking english, but we are required to take high school english as well, and that just seems absurd. dont take aps just for the point of taking them; taking art history or music theory when you intend to major in economics is probably not that hot of an idea. for those of you who are taking cal in middle school- just remember, you are still a kid. please dont get offended because i said that. when you are in eighth grade, you are still a kid- ENJOY IT. dont become pretentious because you are ahead of all of your friends; dont miss out on middle school things to do. and DO NOT ignore your other classes to get ahead in math (this is for all of you poor spellers and grammar-challenged individuals, you know who you are)- colleges will NOT be impressed. last point- everything is not a race to get into "the best college" the fastest. enjoy your life.

From what I hear, AP Chem is quite difficult and has a lot of hw... but it does vary from school to school... Bio is unpredictable... Calc BC (if you haven't taken AB) can be a lot of work because it really is quite a bit of material... Stats should be no problem.

i dont think AP chem is hard, i'm taking it this year. it is actually one of my easiest classes, (just a side note, maybe it is because of me--i took the National Chemistry olympiad last year in Iran, i was ranked 12th in the country....) but anyways my total notes taken in Chem is like 10 pages total,  :P
in BC my total notes is not even a page... i never took a single note in class--but i'm quite certain i'll get a five on the test.
now the first time i took an AP class, was the last year, my first year in USA, i took AP calc AB as a sophomore. i should've taken BC, cause i knew absoloutly 100% of all they teach in the AB or BC class, before i came to USA.... :lol:

[quote="amirhtlusa"]in BC my total notes is not even a page... i never took a single note in class--but i'm quite certain i'll get a five on the test.[/quote]

Same here! ;)

If you have a good start on chem you will most likely do good in ap chem. Probably.

[size=34]I do not take any responsibility for what I've just said!  :D [/size]

yeah in our school, for Chem, above $\%80$ is an $A$.  ;)
my percent was $\%102$ till last week, but i'm $\%98.9$ now :(

[quote="IPERFERPI"]k, they are AP chem , AP bio, Calc BC, and AP stat, i generally get Straight As  :D[/quote]
Considering your classes are all science/math, I don't think you would have any probablems with them (sine AoPSers are generally math geniuses).  If you took more social studies/english APs then you might have to work a bit harder.

Alright. I'm in 8th grade, and I took trig at the local university this semester. I didn't study at all out of class (maybe an hour before the final), and as far as I know I got a 99% in the class. If you go to this site and AP classes are no more difficult then college classes, you'll do fine! No matter how little you study (study a little, though). Especially if you do a lot of math contests.

Oh yeah, and I'm taking my first AP class as a freshman (calc AB).

SAS (Shanghai American School) has a policy of not letting underclassmen take AP's though I don't know why.

I took 2 AP's as a freshman this year (Calc AB, Music Theory), and had to get a waiver from the admistration. Next year i'll be taking Calc BC, US History, Statistics, and Comp Gov. Then i'll go to the IB program in junior year.

AP Chem in 8th grade was my first AP test.  :)


Hi,

Dividing Polynomial $ P(x)$ by $ x\plus{}1$ we have $ 3$ as the residue ,and dividing by $ x\minus{}1$ the residue is $ 5$ .Find the residue when $ P(x)$ is dived by $ x^2 \minus{}1$.

This should be moved to Intermediate (or even HSB).

[hide]$ P(\minus{}1)\equal{}3, P(1)\equal{}5$

$ P(x)\equal{}(x^2\minus{}1)Q(x)\plus{}Ax\plus{}B$. Plugging in the above conditions, we get

$ \minus{}A\plus{}B\equal{}3\land A\plus{}B\equal{}5\iff A\equal{}1\land B\equal{}4$, hence $ R(x)\equal{}x\plus{}4$[/hide]


LaTeX
number theory proposed
number theory

Find the last three digits of $ 2007^{{2008}^{2009}}$

i dont how to get it right in latex but it is 2007 to the power 2008 to the power 2009

For your $ \text{\LaTeX}$ question, take a look at your post now.
For the mathematical question, it was posted before (I think). Otherwise, you need to find the exponent $ e=2008^{2009} \mod 400$ and then $ 7^e \mod 1000$.

hey how did u do that in latex... and kindly send me a link of the solutions.... i cant understand whats u said :( :(


Four cubes of volumes 1,8,27 and 125 are glued together at their faces. What is the smallest surface area possible in the resulting solid?

Each cube should touch every other cube.

Since the side lengths are $ 1,2,3,5$, the preliminary surface area is $ 6(1^2 \plus{} 2^2 \plus{} 3^2 \plus{} 5^2) \equal{} 234$.

However, we have overcounted each of the following two times: three of the unit cube's faces, two of the $ 2$-cube's faces, and one of the $ 3$-cube's faces, which means we have overcounted by $ 2(3\cdot1^2 \plus{} 2\cdot2^2 \plus{} 1\cdot3^2) \equal{} 40$.

Thus, our final answer is $ 234 \minus{} 40 \equal{} \boxed{194}$.

Edit: Thanks for the correction, ocha. :)

I agree with Math154, but we have actually counted each of those faces twice since there are two faces touching eachother.
So i believe the answer should be $ 234\minus{}2.20\equal{}194$


AMC
USA(J)MO
USAMO
AIME
AMC 12
AMC 10
ratio

[quote="calc rulz"]Due to the recent controversy over the current USAMO qualification policy and the current general indecision of the best solution to the current, I have come up with a new qualification procedure that I think will solve the current issues. It goes as follows:

         1. The goal is to select about 375 of the top scorers from the prior AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.
         2. Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12.
         3. The first selection will be the approximately 240 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.
         4. The second selection will be the approximately 120 highest AIME scores of students in grades 7 through 10 who have not already qualified by the first selection. The lowest AIME score among those in the second selection will be dubbed the "floor" value.
         5. The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.
         6. To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests.
         7. The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests.

First of all, this system allows for 7th through 10th graders to qualify for the USAMO solely via their AIME score without running into the same problems that occured during the 2006 qualification. This "floor" will not select too many students, which would force the CAMC to use AMC scores to determine USAMO qualification for those in the second selection, causing possible discrimination between AMC 10 and AMC 12 participants. On the other hand, it will preserve the original intention of the floor system: to allow 7th through 10th graders to qualify solely via their AIME score.

Second of all, the AMC 12 is favored in this process. It allows students to qualify for USAMO without getting as high AIME scores. In our new era, where the floor defined above may be 1 or even 2 (which would have happened had this system been used in 2006) higher than the minimum possible AIME to qualify by the second selection, students in 7th through 10th grade have a reasonable chance of qualifying in the first selection but not getting an AIME score high enough for the second selection. This way, the AMC 12 is indeed favored over the AMC 10.

Finally, upon careful reading of the process, one can see that only those who take the AMC 12 may qualify by step 5, giving an extra advantage.

Thankyou for reading my proposal, and I hope you take this into consideration. Feel free to post comments, criticisms, or suggestions below  ;)[/quote]

Interesting proposal...
But make the following hypothetical situation, there are 80 people who scored 9 or higher on aime, and 40 scored 8, how would you select 120(or anywhere near 120) then?

[quote="Sly Si"]pieterminate: In that case, you set the floor at 8 and get exactly 120 qualifiers.

What I think you meant to ask is, what if we got 80 floor qualifiers with 9 or better, and 200 (for example) with 8.  80 is arguably too few, and 280 is completely implausible.  There would need to be a tiebreaker.  My ideas:

1. Set an AMC12 cutoff for those with 8s AND an AMC10 cutoff.  Those achieving either cutoff on the appropriate test would qualify.  I don't think this is a good idea, since the cutoffs--and the ratio of students with 8s qualifying from the 12 and 10--would come off as arbitrary.  Indeed, they probably would need to be somewhat arbitrary.
2. Take all the 8s who took the AMC12.  Thus the floor is 9 for AMC10 takers, and 8 for AMC12 takers.  I don't like this one, either, because it (a) gives a VERY large advantage to AMC12 takers, and (b) when AIME score is all that matters, why discriminate between 12 and 10 takers?
3. Break the tie by grade level: the floor becomes 9 for 10th graders and 8 for 9th and below.  I would favor this one.  Actually, something like this was already done in 2000: they selected all 10th graders who took AMC12 and got 8 on AIME, and all 9th graders (and below) who took AMC12 and got 7 on AIME.  This tiebreaker would be essentially the same, but dropping the requirement that students take the AMC12 to make the floor.

Another slight problem: Generally, any state with an AIME qualifier is guaranteed a USAMO qualifier (although that may have changed this year; four states have AIME qualifiers but no USAMO qualifier).  Under your method, in the unlikely event that all AIME qualifiers from a state go through AMC10, that state would get no USAMO qualifier.  Fortunately, I don't think this is likely to happen.

Overall, I really like your idea!  This is exactly what I would have suggested.[/quote]

Yeah, that's what I meant, and I definitely favor 2 :lol: .  AMC 10 won't exist then :)

I crunched some numbers from the AIME statistics and got the following data:

Students (10th grade and below) with 10 or more on AIME: 93

9th graders and below with a 9: 37
10th graders with a 9: 68

9th graders and below with an 8: 55
10th graders with an 8: 81

Total 10th graders and below with a 9 or more: 198
Total 10th graders and below with an 8 or more: 334

Using tiebreaker #3 (the only one we can figure this out for), the floor would be 9 for 9th graders and below.  Depending on how many floor-eligible students qualify through the index, and also on whether AMC preferred to err on the side of inviting too few or too many students, the floor for 10th graders would be either 9 or 10.

Just thought some people might be curious.

Also, the last bit of data (334 potential floor qualifiers!) makes clear why the AMC had to set a second cutoff, and probably explains why it was so close to the main cutoff, as well.

[quote="randomdragoon"]I think the teacher manual needs a clause

"Students may be put at a large disadvantage if they do not take both the A and B AMC tests. It is also recommended that they do one 10 and one 12 because who knows what kind of distribution will come up this year"

OR:

Don't take anyone from the lower floor value at all. Round up all floor values, and save the funds for next year. Every few years or so there will be enough funds left over for a massive expansion  :P[/quote]

No, that won't work because less good schools like ours only offer amc B :mad: .

But the 2nd suggestion seems nice :D

3 does sound like it would be a good idea. That keeps it to the AIME, which eliminates the controversial "What?! I wasn't expecting my AMC score (or choice of test) to count for USAMO qualification!" No one is going to feel cheated if they miss qualification simply because of their grade.

The CAMC could give even have some leeway in its policy though, such as "if a tie-breaker is required, the CAMC will decide which of the following three tie-breakers to use:"

At any rate, how often would a tie-breaker be needed anyway? The CAMC has been able to find floors that give the correct number of students for the past few years.

Note that a tie-breaker would only apply to those who made the floor. Anyone scoring above the floor would automatically get in.

[quote="beta"]Eh I think we should eliminate AMC10 as part of USAMO qualification process completely.[/quote]

Agreed, the AMC 10 is pretty stupid. But I don't think the CAMC wants to do that... :?

[quote="calc rulz"]

[quote="beta"]Eh I think we should eliminate AMC10 as part of USAMO qualification process completely.[/quote]

Agreed, the AMC 10 is pretty stupid. But I don't think the CAMC wants to do that... :?[/quote]

And not possible within 3 years either :|

I kind of like the old way they did it.(With only the AHSME).  Can someone please explain me why they changed the system?

[quote="calc rulz"]At any rate, how often would a tie-breaker be needed anyway? The CAMC has been able to find floors that give the correct number of students for the past few years.[/quote]

Take a look at the data I posted: It's somewhat likely that a tiebreaker would be required this year, though a lot depends on how many floor-eligible students make the index cutoff instead.

Furthermore, until yesterday, we could have said exactly the same thing about paring down the floor qualifiers: "How often would more than 120 people make the floor, anyway?  The floors the last few years have given suitable numbers..."

[quote="beta"]THey can make AMC10 a separate competition(like AMC 8), I don't see why they have to use it in USAMO selection.[/quote]

If this were done, I think it would be absolutely necessary to move the 10 to a different date.  Junior high kids don't have to choose between an exam intended specifically for them (AMC8) and one that carries the possibility of advancement (AMC10/12), and I think it would be bad policy to force 9th and 10th graders to do so.

[quote="pieterminate"][quote="beta"]THey can make AMC10 a separate competition(like AMC 8), I don't see why they have to use it in USAMO selection.[/quote]

Because there are theoretically 9-10th graders with excellent problem solving skills, but haven't taken more advanced classes.  Of course, these 9-10th graders who qualify for usamo have a good chance of already being in advanced classes, but....[/quote]

Show me a topic that appears on AMC12 but not on  AIME...

I'm just writing this post to clear up a little confusion.

My formula suggestion is designed [b]only for tiebreakers[/b]. I don't think it's at all fair to award someone a spot on USAMO for only solving two or three really hard problems. Or perhaps they only study one subject and thus get all four of those problems. That's not how USAMO qualification works. We would still have the complete index and floor system, but the only change would be as a tiebreaker between those <11th graders who got the AIME floor.

For anyone that's confused about my formula, here's another attempt to explain it:

You just took the AIME and fall into the tiebreaker category (<11th grade, AIME floor). You solve problems $\{a_1,a_2,\ldots,a_k\}$, where $k$ is the AIME floor. Suppose $n_i$ people solve problem $i$. You should get more points for solving harder problems, where we define harder as "fewer people solved them." Since the function $f(x)=1/x$ is strictly decreasing, I would use this to flip them. Therefore problem $i$ is worth $1/n_i$ points if you get it and 0 if you don't. Thus, your total tiebreaker score (ATI) is $\displaystyle\sum_{i=1}^k\frac1{n_{a_i}}$. This formula would work if there was only one AIME.

In order to accomodate for having two AIME's, we define the $m_i$ to be the portion of correct answers that were submitted for problem $i$, out of the total number of correct answers given on that entire AIME. In this way, the problems are ranked against each other, so we use statements like "10% of the correct answers on the AIME-I were for problem 1" (and $m_1=.1$ in that case). If everyone taking a particular AIME added their scores together, let that value be $N$. We have $m_i=\frac{n_i}N$, so instead we have $ATI=\displaystyle\sum_{i=1}^k\frac1{m_{a_i}}=N\displaystyle\sum_{i=1}^k\frac1{n_{a_i}}$.

[quote="Phelpedo"]Well I think that the difficulty of the problem is slightly exaggerated because it is at the end (i.e. if it were given MathCounts Target style, more people would get it). [/quote] I agree. However, if there was some value (if only for a tiebreaker) to solving harder problems, more people would try them out. This is a stated goal of AMC (I believe), and a good policy, since the later problems have the better math.

By the way, $N$ (see my previous post) was 49549 for the AIME-I and 12857 for the AIME-II this year. Assuming that on the AIME-I, 10000 solved 1-3, 3000 solved 4-6, and 1000 solved 7-9, a student getting 1-7 and 9 would have an ATV in the range of 164. This is just to give a sense of the range of the function, somewhat near the range for USAMO indices. (Of course, if one solves a problem that only 100 people get, they recieve 500 points toward this tiebreaker index!). Notice too that 100 of those points came from two problems. It's extreme, but it's just the tiebreaker.

[quote="JSteinhardt"]

I think, though, that we are really arguing without any common ground. In my opinion all that extra time should be used to safeguard against careless errors and the like. In your opinion, at least from what I can gather, it should be used to explore problems you couldn't get the first time through, and hopefully up your AIME score by a few points. Perhaps you are a more accurate problem solver than I (I wouldn't be surprised!) or perhaps you are just more daring. I concede that this really comes down mostly to personal philosophy and experience, and at least any argument I form will be based a lot on opinion. Perhaps then we should just agree to disagree in this case.[/quote]

Actually, I think you're quite right about that. I finished 1-7 in the first hour or so, and spent the rest of the time checking those problems in addition to four more. I ended up only solving up to #12, but in retrospect I think I could have solved #14 and 15.

I think that a lot of people decide to check their work instead of attempting the later problems because there is some sort of impressions that #15 is impossibly difficult (for the average AIME taker obviously),  so it's better to go for the easy point than the hard point. Obviously, this is a great strategy, and I saved myself a couple of points by checking instead of moving on.

But my point is that the number of people who solved the problem isn't really an accurate measure of difficulty because of the time constraint as well. In my case, I think I could have done 15 (it was relatively easy this year) if I decided to forgo checking.

[quote="bookaholic"]Hmm, interesting ideas. I was going to propose subtracting 20 points from any AMC-10 score when calculating the index for tiebreakers, since that's how AIME qualification is set up, but I think this weighting by difficulty is a pretty good idea. Even though it is very, very, very bad for me.[/quote]

See that isn't exactly fair either. I don't see why we can't just make AMC10 on a separate date and separate it from USAMO qualification.

[quote="Phelpedo"][quote="MathChauffeur"]:) Joining in for the fun of it:
1. Get rid of AMC10, just AMC12A and 12B
2. No guessing penalty, but 6 choices instead of 5
3. Assign AMC pts to 5 sets of 5 problems each based on statistics:
------ 1.0  for the easiest 5
------ 1.2  for the next 5
------ 1.4 for the next 5
------ 1.7 for the next 5
------ 2.0 for the hardest 5
4. Qualify top 10,000 +/- for AIME by higher of 12A/B scores
5. Assign AIME pts to 3 sets of 5 problems each based on statistics:
------ 1.0  for the easiest 5
------ 1.5  for the next 5
------ 2.0 for the hardest 5
6. Qualify top 375 +/- for USAMO by AIME x 10 + AMC[/quote]

I think we should leave the AMC as it is, or in any case there should definitely be a penalty for guessing, as it is a multiple choice exam.[/quote]

Ummmmm, there won't be a "penalty" for guessing by next year anyways...

[quote="pieterminate"]
Ummmmm, there won't be a "penalty" for guessing by next year anyways...[/quote]

There is a penalty, but a smaller penalty.

[hide="This is wrong!!"]Actualy, the expected value for guessing on next year's AMC and the expected value for leaving things blank will be the same, instead of giving a 1 point advantage for leaving it blank as in previous years.

My point is, not really a "penalty", is it.[/hide]

Whoops, I can't do math while thinking about the impending SATs at the same time  :blush:

[quote="randomdragoon"]Actualy, the expected value for guessing on next year's AMC and the expected value for leaving things blank will be the same, instead of giving a 1 point advantage for leaving it blank as in previous years.

My point is, not really a "penalty", is it.[/quote]

The expected value of guessing 5 questions is 6 pt. and leaving 5 blank is 7.5 pt... :?

Your mathematical expectation blind guessing 5 questions is lower than leaving the 5 blank (1.5 lower, so 0.3 penalty for every 5 questions)

My suggestions to USAMO Qualifications (all assuming 360 people, which is fair considering states without representation and other things). The amounts of people (240, 120, or 360) may not be exact, however, should be as close as possible and have a rough sum of 360 (excluding those chosen from states specifically for representation)

1) Abolish the floor and completely separate the AMC10 and the AMC12. 240 people are selected via AMC12+AIME*10 and 120 people are selected via AMC10+AIME*10. This system still favors the AMC12 takers, as the ratio of people qualifying as a result of taking the AMC12 compared to AMC10 is 2:1. I'm not sure how ties would work.

2) Abolish the AMC10. 240 people selected via AMC12+AIME*10, and the remaining 120 be selected via floor (the lowest AIME score of the original 240). Ties in 120 people will be broken via index.

3) Abolish both the floor and AMC10. All 360 people must qualify by AMC12+AIME*10. No complications, although a bit extreme of a solution.

Comments?

[quote="pieterminate"]3 IS too extreme and would be like going back in time(i.e. the AHSME) ! :lol:  2 sounds just pointless if there isn't an amc 10.... 1 would be good, but for the 7-10th graders taking the amc 12...[/quote]

#1 - Well, they would be informed about this and be aware about the advantages (more chance for qualification) and disadvantages (difficultly) of taking the AMC12 at their age if they visited the website or something. However, that brings up the issue: What if their schools don't give them the choice? So meh.

#2 - I don't understand your comment.

#3 - Yeah, I agree, but it is the most full proof one either way.

On #2, he's saying that if you determine the ties on the AIME floor with the index, then those that took the AMC 10 simply cannot qualify. Or, if you have enough vacancies, then there's a huge tie because all of them have an index of 10 times the floor (having not taken the AMC 12). On the other hand, if you abolish AMC 10 for [i]AIME[/i] qualification, then you don't meet their goal of getting more young people involved early. If the 10 and 12 are offered at the same time, EVERYONE will take the 12, since the 10 is a dead end.


hey everyone. i will be hosting a countdown, ftw tournament. basically, i will wait for either 8 or 16 signups. then, using random.org, seeding will be determined. a bracket will be created. each set will be 5 games each.

sign up by posting here or pming me. i will post an updated list as people sign up.

lets hope this doesnt get locked...

Just because you won't join, doesn't mean someone else won't, AIME15 :wink:
I am rarely on FTW these days, so won't join.

[quote="AIME15"]We don't need more tourneys in the middle of the school year.  Izzy/vallon/jjx, PLEASE lock.[/quote]

ehh? why would they lock it, its a tournament. and if we have time for all the middle school tournaments, why not TWO ftw tourneys?

signups:
1. theprodigy
2. ernie
3. SuperPawn
4. akhilupneja
5. SonyWii
6. nt811581147

ill edit as people join.

COME ON, PEOPLE.

I would like to demonstrate my beastliness! I join!

[quote="jjx1"]I think I'll let this go on and see what happens. izzy, if you think you should lok this, lock it. Also, people before you sign up, please just make sure you'll be on basically everyday or so.[/quote]

thanks! please dont lock it, if its starts getting where its just prolonged, ill tell you.

I'm studying for mathcounts.  Are there any good websites I can see for it, because my regional competition is this Saturday?

Wrong forum, it should go [url=http://www.artofproblemsolving.com/Forum/index.php?f=132]here[/url].

I suggest [url=http://www.artofproblemsolving.com/Edutainment/g1/index.php]here[/url].

ALL CDS MUST BE COMPLETED BY FEBRUARY 20TH. THERE WILL BE STRICTER DEADLINES IN UPCOMING ROUNDS. PLEASE PM YOUR OPPONENT AND SET UP A TIME.

highlight??
[color=white] I LOST! [/color]

[hide=" "]GAH I LOST I HATE YOU NIKE[/hide]

So, AIME what did you have in your highlight?

What's up with the blank posts?

SuperPawn (2) vs CA Math (3)
Done.phew.

[quote="AIME15"]Last year, I know someone won a region with a score of 25, so regional champ isn't something to be proud of unless you have a tough region.
[/quote]

Way to be positive  :wink:

yeah, that would be me.

:) not my fault.

[quote="AIME15"]Last year, I know someone won a region with a score of 25, so regional champ isn't something to be proud of unless you have a tough region.[/quote]

I have a pretty tough region, but i'm not that proud of being regional champ.  I was just trying to make a point.  I got an 87 on AMC 10, which is pretty bad.

lol i got a 144...CAN YOU PEOPLE PLEASE PLAY YOUR GAMES?!

double post :P
and yes we'll play our games

[quote="ernie"][quote="nikeballa96"]wow, ernie and super both gone. to be honest, i had money and ernie, nt, ca, and super going in...how did two of them end up eliminated in fisrt round?

(beats me.)

FINISH EVERYONE ELSEEEE[/quote]

thats bcuz [b][size=200]SOMEONE!!!!![/size][/b] use random.org for seeding...[/quote]

lol ernie
and i get the easy side too :P

oh and by the way what chapter were u in akhilupneja?

yeah i got beat by the prodigy 3-0.... but hey i'm new i'll learn sometime.  idk the name of our chapter but it was in pa.


This is from the 91 exam...
Consider the equation x = √19 + 91/(√19 + 91/(√19 + 91/(√19 + 91/(√19 + 91/x)))). Let k be the sum of the absolute values of the roots. Find k2.

Kalva's explanation says that althgouh one can find the end quadratic that determines x through a LOT of annoying algebra, it is better to just write x =  :sqrt: 19   + 91/x   = x. Why does this work? I would understand it working if it was infinite, but being that its finite, i dont understand why it works...

[hide]

Now that you know there are only two values for x, if you actually found those two, you'd be done. Then if you happen to see that if you assumed x= √19 + 91/x, then the RHS equals:

√19 + 91/(√19 + 91/(√19 + 91/(√19 + 91/(√19 + 91/x))))

= √19 + 91/(√19 + 91/(√19 + 91/(√19 + 91/x)))

= √19 + 91/(√19 + 91/(√19 + 91/x))

= √19 + 91/(√19 + 91/x)

= √19 + 91/x

= x.

So the two values from the easier quadratic will also work in the bigger one, and there you go.

[/hide]

Isnt (what MMM did) pretty much just like applying circular logic to make what u want to make work? Ur assuming x = "blah", and then u plug in blah for x, and what do u know, it coems out to equal x. And to Yannik, if it isnt infinite, then why can u think of it like that? Thats like saying that the summation of n=1 to 70bajillion of 1/2^n = 1, but it doesnt.

If you want, you can restate the important step of what MMM did like this:

Let z be a number which satisfies z = √19 + 91/z.  Then we can repeatedly substitue the right side into the right side for z to derive that z also satisfies the given problem.

Ok thats not fair. That summary by JBL really encapsulated the whole question. Hmm....


vector
algebra
polynomial
linear algebra
linear algebra solved

Find the maximal number of nxn matrices A such that any two commute and any one of them verifies $A^k=I$, where k is a natural number given.

just a question: what is the field of the vector space. If it is C it seems easy ....

The field is C. Could you please show an easy solution?

We are in $M_n(C)$

Here what you can say with the statement:

The condition $A^k =I_n$ means $A$ vanishes $X^k-1$ wich split and has simple roots, so all $A$ are diagonalisable.(This is a theorem in chapter of diagonalisation)

You have also a exercice in MPSI or MP*  in reduction chapter diagonalisation
wich say

If a family of matrices is diagonalisable and any two pair commute, then exist
a common basis such that all the matrices are diagonalisable.

The eigenvalues satisfy $z^k=1$ so $z=e^{\frac{2iq\pi}{k}}$ with $0\leq q \leq k-1$

Yes, Moubi, this is the solution I also found, but I do not find it so easy.

Moubinool didn't explicitly say it, but it's right there in his answer: the maximum cardinality for this set of matrices is $k^n$. After all, if the matrices have a common diagonalization, you might as well assume they are all diagonal, and then any of the $n$ diagonal elements can be any of the $k$ roots of unity.

This is an upper bound for the cardinality in any field, since $x^k-1$ can have at most $k$ roots. For any particular field, the question will be how many roots of unity will be in the field; that's why alekk asked about the field.

oh,I'm sorry. I've just said that because the fact that the matrices were diagonalizable in the same basis was mentioned explicitly by our teacher in my course. I thought it was a well known fact.

I think that with this result it is easy to prove that GL(n,C) and GL(m,C) are not isomorphic. Is this that result you were referring to ?

Well, I think it is well known. The bigger problem for me was to show that if $A^k=I$ then A is diagonalisable. I know it is not a very big problem, but for me it was. Anyway, this was the reason for which I posted the problem: from this problem it follows that GL_n(C) and GL_m(C) are not isomorphic. I tried a lot to prove this when you first posted the problem, but I did not manage. And here it is a very nice solution for it.

[quote="harazi"] ...show that if $A^k=I$ then A is diagonalisable.[/quote]

You can also used "le thorme de dcomposition des noyaux"

If $P=A.B$ where $P,A,B$ are polynomial with $gcd(A,B)=1$
and $f:E\rightarrow E$ a linear map on $E$ a R or C vector space finite dimension
Then $ker(P(f))=ker(A(f))\oplus ker(B(f))$

See chapter reduction-diagonalisation MPSI or MP* for the proof

For our problem $P(x)=x^k-1=\prod_{q=0}^{k-1}(x-e^{2iq\pi}{k})$
each factor is prime to each other, $A$ vanishes $P(x)$, then "thorme de
dcomposition des noyaux" gives
$ker(P(A))=\oplus_{q=0}^{k-1}ker(A-e^{2iq\pi}{k}I_n)$
but LHS is $E$
$E=\oplus_{q=0}^{k-1}ker(A-e^{2iq\pi}{k}I_n)$ now you take an appropriate
basis with the eigenspace, you have a basis in wich the linaer map wich
represent $A$ is diagonalisable

Thanks a lot, Moubinool, very neat proof.


Problem:
The base imidazole has a Kb of 9.8 * 10 ^-8 .
In what amounts should 0.02 M HCL and 0.02 M imidazole be mixed to make 100 ml of a buffer at ph 7?

Im confused about the reaction which forms the salt. The doubt i have is , would the base react with acid to form equal number of moles of salt or do we consider the dissociation of base, then find the moles of cation of base and react this with HCL?

If my question is rather confusing, a detailed solution to this problem would be appreciated.

Thanks

Let $ Im$ be the imidazole. Then in the buffer we will have the equilibrium

$ Im \plus{} H_2O \rightleftharpoons ImH^\plus{} \plus{} HO^\minus{}$

and $ K_b \equal{} \frac{[ImH^\plus{}][HO^\minus{}]}{[Im]} \equal{}> \frac{K_b}{[HO^\minus{}]} \equal{} \frac{[ImH^\plus{}]}{[Im]} \equal{}> pK_b \minus{} pOH \equal{} \log \frac{[Im]}{[ImH^\plus{}]}$
$ \equal{}> pH \equal{} 14 \minus{} pK_b \plus{} \log \frac{[Im]}{[ImH^\plus{}]}$

We know from the data the pH of the final solution and pKb, so from the equation above you get [Im]/[ImH+]. The reaction between Im and HCl is

$ Im \plus{} HCl \longrightarrow ImH^\plus{} \plus{} Cl^\minus{}$

I think you can now proceed from here.

heres another one...
0.1M NaoH is added to 20 ml of CH3COOH solution. Ka=10^-5.
What is the ph when 0.1M (2ml) HCL is added after 1/4th neutralisation point?

What does the question mean by 1/4th neutralisation point?

i think 1/4th neutralisation pt means 1/4th of the CH3COOH is consumed by NaOH  :maybe:

well, i dont seem to be getting the right answer that way. I might have gone wrong elsewhere then. Would u mind posting the solution?


modular arithmetic
number theory
number theory proposed

Show that for all odd natural numbers $ n$ we have $ n|2^{n!}\minus{}1$

It is really easy to see that if $ a$ and $ b$ are coprime and the condition is fullfilled for $ n\equal{}a$ and $ n\equal{}b$ then it is also true for $ n\equal{}ab$ (It's just easy congruences)
Then we only have to prove the condition for $ n\equal{}p^x$ ($ p$ is a prime)
 Since $ (p^x)!$ is divisible  by $ p^{x\minus{}1}(p\minus{}1)$ we can now use fermat's little theorem and lifting the exponent to prove the problem claim

http://en.wikipedia.org/wiki/Fermat's_little_theorem
http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf

Daniel

I don't think you need Lifting the Exponent. It follows from what you said by Euler's Theorem - $ \phi(p^k)|(p^k)!$, and $ (p^k,2)=1$, thus ${{ 2^{(p^k)!}=(2^{\phi(p^k)})^\frac{(p^k)!}{\phi(p^k)}} \equiv 1 ^\frac{(p^k)!}{\phi(p^k)}} \equiv 1 \pmod {p^k}$

You only need that $ \varphi(n) \leq n$, which is really trivial.


Forty-five minutes is what fractional part of a day? Express your answer as a common fraction.

$ \frac {45\text{ minutes}}{1\text{ day}} \equal{} \frac {\frac34\text{ hours}}{24\text{ hours}} \equal{} \boxed{\frac1{32}}$


function
trigonometry
calculus
analytic geometry
real analysis
real analysis unsolved

I have created an equation that expresses the position of a particular point on an object as a function of variable time and other known quantities.  The equation is as follows:

a+bt+cos(c+dt)e=f+gt+cos(h+it)j   equation (1)

where [a] is the initial position of the center of mass of object [1], [b] is the translational velocity of the center of mass of object [1], [c] is the initial angle in radians of a point (on object [1]) of interest with respect to the center of mass of object [1], [d] is the rotational velocity of object [1] in radians, and [e] is the distance between the point of interest and the center of mass of the object (may also be considered the radius of the circle that the point of interest makes as it moves through its trajectory around the center of mass).  From the form of the equation, it can be seen that [f], [g], [h], [i] and [j] represent the same values as [a], [b], [c], [d] and [e] but these values pertain to object [2] rather than object [1]. It should also be noted that [a],[b],[c],[d],[e],[f],[g],[h],[i] and [j] are known.  This equation is derived from the equation:

x(t)=a+bt+cos(c+dt)e   equation(2)

which is the position function of a point on object [1] with respect to time.  So I simply created another position function representing a point on object [2].  And so, what I'm trying to get at is during a collision, there exists a point that is shared by both object [1] and object [2] (the point of impact).
  And so when I derived equation (1) I realized that if I solve for t, or the time, I can essentially calculate the exact moment in time when two objects collide.  And with this value, I can more importantly, calculate the exact position at which the two objects collide using either of the two position functions (the one for object [1] or object [2]).
  I am writing the program in Java.  As you may or may not know, such a program runs on a loop which increments every set amount of time.  Currently I have it set to loop every 33 milliseconds so it comes to around 30 frames a second.  The reason I mention this is the reason for the problem.  At any given frame, by the time a collision has been detected between two objects, I have no idea of exactly where the two objects collided.  Although it may be possible to try to make some abstract geometric assumptions to try and figure out an approximation, I would like to solve for [t] to have a discreet method of finding the point of impact.

I guess the difficulty can be found in the fact that there is a variable both inside and outside the cosine expression.  Temporarily removing the constants, the equation can be simplified as follows:

y=x+cos(x)   equation(3)

I have tried every possible means of solving for x, but to no avail.  I have even asked my Calculus I  professor, and short on time, he just said that I might need to use polar points to solve the equation.  So I went online to try to teach my self the basics of polar points, but it only seemed to make the problem more complex then any simpler.   I guess the reason for this is that there are essentially two poles that collectively express the position of a point on the object (one being the origin of the coordinate system, and the other being the center of mass of the object).  Anyway, I derived the following equations:

rsin(theta)=rcos(theta)+cos(rcos(theta))   equation(4)
r(sin(theta)-cos(theta))=cos(rcos(theta))   equation(5)

I could go on and on, but as you can see, all this did is add more sine and cosine terms, both of which I want to limit.  I guess in this situation I want to get cosine of theta by it self as to aid me in solving for x.

So could someone please show me how to solve the function y=x+cos(x) for x, or better yet solve the original equation (1) for t.

Thanks in Advance

Wow, so this question is that impossible huh?  Who knew, if you simply add an x outside the cosine function, then it apparently becomes completely impossible to solve for x ;) .  Even if no one knows how to do this problem, atleast some feedback would be appreciated, I am interested to know what others may think of this problem.  I'm only taking Calc I right now, so I'm no math genius, but I never figured that people who have taken higher level math wouldn't know how to solve such a "simple" problem (My Calc teacher provided no help), but I guess looks can be decieving.  If someone sees a blatant impossiblity of finding a solution, please point out that fact so I can move on to some other way of tackling this problem rather than to linger on an impossiblity :D

As you've said yourself, there's no general closed-form solution. There are many numerical methods you can use to find approximate solutions with high accuracy- Newton's method should work well in this case, since you have a formula you can differentiate.


algebra
polynomial
function
algebra unsolved

[quote="chessfreak"]Do there exist four real polynomials such that the sum of any three of them has a real zero while the sum of no two polynomials has a real zero?[/quote]

No, it does not :

$ \forall i,j\in\{1,2,3,4\}$, we have either $ P_{i}\plus{}P_{j}> 0$ $ \forall x$, either $ P_{i}\plus{}P_{j}< 0$ $ \forall x$

But $ P_{1}\plus{}P_{2}$, $ P_{1}\plus{}P_{3}$ and $ P_{2}\plus{}P_{3}$ can't have same sign, else their sum would have constant sign and $ P_{1}\plus{}P_{2}\plus{}P_{3}$ would never be 0.

So, WLOG (choice or order + replacement $ P_{i}$ by $ \minus{}P_{i}$) say :
$ P_{1}\plus{}P_{2}> 0$
$ P_{2}\plus{}P_{3}> 0$
$ P_{1}\plus{}P_{3}< 0$

Since $ P_{1}\plus{}P_{3}< 0$, and since $ P_{1}\plus{}P_{3}$, $ P_{1}\plus{}P_{4}$ and $ P_{3}\plus{}P_{4}$ can't have same sign, $ P_{1}\plus{}P_{4}$ or $ P_{3}\plus{}P_{4}$ is $ > 0$. WLOG say :
$ P_{1}\plus{}P_{4}> 0$

Since $ P_{1}\plus{}P_{2}> 0$ and $ P_{1}\plus{}P_{4}> 0$, and since $ P_{1}\plus{}P_{2}$, $ P_{1}\plus{}P_{4}$ and $ P_{2}\plus{}P_{4}$ can't have same sign :
$ P_{2}\plus{}P_{4}< 0$

Then since $ P_{2}\plus{}P_{3}> 0$ and $ P_{1}\plus{}P_{4}> 0$, then $ P_{1}\plus{}P_{2}\plus{}P_{3}\plus{}P_{4}> 0$
But since $ P_{2}\plus{}P_{4}< 0$ and $ P_{1}\plus{}P_{3}< 0$, then $ P_{1}\plus{}P_{2}\plus{}P_{3}\plus{}P_{4}< 0$

And so contradiction.
(and this property is available for any set of continuous functions)

i dunno how many times u need to be told that IMOTC postals are not to be posted at Mathlinks.
trust me, ppl who have easy access to the net have a gr8 advantage over those who dont.
in due recognition of the fact that u r not from IMOTC, i request u to kindly not post ANY OF THE IMOTC postal probs.

[quote="borislav_mirchev"]Internet is great place for people to share their knowledge. You should post all problems that you like! I dislike monopolism in math!!![/quote]
sure but I think that venkata meant that these problems are in a contest which is still open,so the students are supposed to solve the questions by them selves untill the given time,so it is not a good idea to post it here,its kind of cheating...

Yes, you are maybe right. But I think the problems are took from somewhere maybe some other competition or some textbook. What if the student use same textbook? I think, at this time when the world is a "globbal village" some competition without presence of the students if it is not time limited or aren't there some other limitations may be a warranty only that students have the desire and they want to solve proposed problems. They also may ask for help their friends parents, or other web sites not so good but, different from mathlinks... I think after I-st round there should be some communication with the students...maybe interview or round where they solve problems. Other idea I have is electronic competition with fixed time limit, but you again cannot be sure who is in front of the PC...

[b]chessfreak[/b] i suppose u r relly intesrested in seeing the Postal problems solved  :P
How is it that u r not a IMOTC member(as u had claimed  :wink: ) and yet u r the first each time to post the postal problems.
well we have given u enough warnings i think it's time u better stp posting them otherwise we have to egt each of ur posts scrutinized and deleted if they are have postal problems .and if u still continue this u know what lies next.
now no more requests just
[color=red][size=200]STOP POSTING THEM BECAUSE OF U EVERYONE ELSE IS DOING SO [/size][/color]

Dear borislav_mirchev
As Bhabak Ghalebi mentions, this is an ongoing competition , and the rules clrly state: NO EXTERNAL HELP TO BE TAKEN.
as far as "global villaging" goes, no 1 minds the posting of problems ONCE THE DEADLINE IS OVER.
also, in all these competitions, a lot hinges on TRUST and ETHICS. so things like asking frnds, parents etc usually dont happen at this level(so i bliv) :maybe:
btw, ill surely suggest ur examination methods to our profs :wink: