trigonometry
floor function
AMC
USA(J)MO
USAMO
number theory
prime factorization

how many perfect squares are divisors of the value 1!*2!*3!*4!*5!*6!*7!*8!*9!  ?

Ok, you do the prime part factorization (heh got that term from zeitz...)

1! = 1
2! = 2
3! = $2\times3$
4! = $2^{3}\times3$
5! = $2^{3}\times3\times5$
6! = $2^{4}\times3^{2}\times5$
7! = $2^{4}\times3^{2}\times5\times7$
8! = $2^{7}\times3^{2}\times5\times7$
9! = $2^{7}\times3^{4}\times5\times7$
10! = $2^{8}\times3^{4}\times5^{2}\times7$

now you multiply to get
\[2^{(1+1+3+3+4+4+7+7+8)}\times3^{(1+1+1+2+2+2+4+4)}\times5^{(1+1+1+1+1+2)}\times7^{(1+1+1+1)}= 2^{38}\times3^{17}\times5^{7}\times7^{4}\]

now you do cool combo stuff:
$2^{38}$ gives 20 options for squares ($2^{0\times2}\rightarrow 2^{19\times2}$); similarly 3 gives 9, 5 gives 4, and 7 gives 3. Then you multiply to get $20\times9\times4\times3 = 2160$.

(hmm may be an arithmetic mistake but I hope there isn't)

I think that your method only found perfect squares that have a prime factorization of one number. It excluded numbers like 100 because you did each part separately and forgot to count the combinations. You could add them back in, but that would get very tedious.

EDIT: I misunderstood not_trig's solution. He is absolutely correct. Good job.  :)

Assuming the factorization is correct, not_trig's method does work. It doesn't consider each prime factor separately; that would be $20+9+4+3$. Since he's multiplying $20 \cdot 9 \cdot 4 \cdot 3$ he is accounting for combinations of primes. For example, if we had $100 = 2^{4}\cdot 5^{2}$, we would see that there are $3 \cdot 2 = 6$ square divisors because any combination of $\{2^{0}, 2^{2}, 2^{4}\}\times \{5^{0}, 5^{2}\}$ is a distinct square.

okay...so what's the answer?...its not in the 2000s...its between 800 and 1000 something....

dude...there's no 10!
not_trig's prime factorization is incorrect

[hide="a continuation of not_trig"]Then it is $2^{30}\cdot 3^{13}\cdot 5^{5}\cdot 7^{3}$
$2$ gives $16$ options for squares.
$3$ gives $7$.
$5$ gives $3$.
$7$ gives $2$.
$16\cdot 7\cdot 3\cdot 2=672$[/hide]

This isnt an old amc 10 question.  You must have misread the answer of something mihail. (only directed towards him)

[quote="mihail911"]wait..boo...how did you get that like 7^3 gives you three squares....and the other ones too.?....no..i distinctly remember this from some amc 10 or some amc 10 practice?[/quote]
$7^{2(0)}, 7^{2(1)}$
There are two squares.

In a little more detail...

[hide] A number is a perfect square iff the exponents in its prime factorization are all even. Given a prime factorization such as $2^{30}\cdot 3^{13}\cdot 5^{5}\cdot 7^{3}$ we want to find all possible combinations of primes raised to even exponents. If the maximum exponent is $k$, then we can choose $0, 2, 4, ...$ for a total of $\left\lfloor \frac{k}{2}\right\rfloor$ possible exponent values [/hide]

still no comprendo... :(
so then wouldn't k/2 for 7^3 be 3/2 which would be 1.5...?

Hrm... that should read $\left\lfloor \frac{k}{2}\right\rfloor+1$ instead. $7^{3}$ has two possibilities for perfect squares as mentioned above: $7^{0}$ and $7^{2}$.

Try not to get so stuck on the formulas; learning the concepts behind any method to solving a problem is both quicker and will help you solve a larger number of other problems. Later at USAMO level, formulas are useless without understanding them.


trigonometry
LaTeX
function
algebra
polynomial
calculus
calculus computations

(1) Solve $ y''(t) \plus{} 3y'(t) \plus{} 2y(t) \equal{} f(t)$, with $ y(0) \equal{} \minus{} y'(0) \equal{} 1$, where f(t) is defined as follows: $ f(t) \equal{} 0$, for $ 0 \leq t < \frac {\pi}{2}$, and $ f(t) \equal{} \sin t$, for $ t \geq \frac {\pi}{2}$.

Here's how I did it, using Laplace Transform.

$ y''(t) \plus{} 3y'(t) \plus{} 2y(t) \equal{} \theta(t\minus{}\frac{\pi}{2})\sin t$.

LT gives

$ 2 Y \plus{} z^2 Y \plus{} 3 (z Y \minus{} 1) \minus{} z \plus{} 1 \equal{} \frac{e^{\minus{}\frac{\pi z}{2}}z}{1 \plus{} z^2}$, where $ Y$ denotes LT of $ y$.

Now, $ Y\equal{}\frac{1}{1\plus{}z}\plus{}e^{\minus{}\frac{\pi z}{2}}\frac{z}{2 \plus{} 3 z \plus{} 3 z^2 \plus{} 3 z^3 \plus{} z^4}$

or, simplified: $ Y\equal{}\frac{1}{1\plus{}z}\plus{}e^{\minus{}\frac{\pi z}{2}}\frac{z}{(1 \plus{} z) (2 \plus{} z) (1 \plus{} z^2)}$

It is not hard to make the Inverse Laplace Transform, as all of these fractions give LT's from the tables, so

$ y\equal{}e^{\minus{}t}\plus{}\frac{1}{10}(\theta\left(t\minus{}\frac{\pi }{2}\right) \left(4 e^{\pi }\minus{}5 e^{\frac{\pi }{2}\plus{}t}\plus{}e^{2 t} (\minus{}3 \cos t\plus{}\sin t)\right))$

which gives

$ y\equal{}e^{\minus{}t}$ for $ t<\frac{\pi}{2}$
$ y\equal{}e^{\minus{}t}\plus{}\frac{1}{10}(\left(4 e^{\pi }\minus{}5 e^{\frac{\pi }{2}\plus{}t}\plus{}e^{2 t} (\minus{}3 \cos t\plus{}\sin t)\right))$ for $ t>\frac{\pi}{2}$

Maybe there's a mistake somewhere... Some of the calculations were tedious.

I see it now ;) Where did I get those $ \pi$s in the exponents? Hmmm... However, The solutions should be:

$ y \equal{} \frac {1}{10} \left(e^{ \minus{} 2 t} \left( \minus{} 2 \plus{} 15 e^t\right) \minus{} 3 \cos t \plus{} \sin t\right)$ for $ t\geq\frac {\pi}{2}$
$ y \equal{} e^{ \minus{} t}$ for $ t < \frac {\pi}{2}$

Hmmm, maybe meddling with thetas and LTs wasn't really necessary-just casework for two intervals: first one gives a homogeneous equation, and the second one gives the non-homogeneous variant of it in which we can seek for the particular solution in the for of $ a \sin t \plus{} b \cos t$. It is easier and less erroneous.

Still, there's something I do not understand. I used Mathematica for checking my solution. The first version, using LT and thetas:

[code]\frac{1}{10} \left(e^{-2 t} \left(-2+15 e^t\right)-3 \text{Cos}[t]+\text{Sin}[t]\right)

{{y[t] -> -1/
      10 \[ExponentialE]^(-2 t) (-4 \[ExponentialE]^\[Pi] -
       10 \[ExponentialE]^t + 5 \[ExponentialE]^(\[Pi]/2 + t) +
       3 \[ExponentialE]^(2 t) Cos[t] - \[ExponentialE]^(2 t)
         Sin[t]) + (\[ExponentialE]^-t +
       1/10 \[ExponentialE]^(-2 t) (-4 \[ExponentialE]^\[Pi] -
          10 \[ExponentialE]^t + 5 \[ExponentialE]^(\[Pi]/2 + t) +
          3 \[ExponentialE]^(2 t) Cos[t] - \[ExponentialE]^(2 t)
            Sin[t])) UnitStep[\[Pi]/2 - t]}}

[/code]

I know it is hard to get through this code, so here's $ \LaTeX$ form:

$ \text{DSolve}[\{y\text{''}[t]\plus{}3y'[t]\plus{}2y[t]\text{\equal{}\equal{}}\text{HeavisideTheta}[t\minus{}\text{Pi}/2]\text{Sin}[t],y[0]\text{\equal{}\equal{}}1,y'[0]\text{\equal{}\equal{}}\minus{}1\},y[t],t]$
$ \left\{\left\{y[t]\to \begin{cases}
 e^{\minus{}t} & 2 t\leq \pi  \\
 \frac{1}{10} \left(e^{\minus{}2 t} \left(4 e^{\pi }\plus{}10 e^t\minus{}5 e^{\frac{\pi }{2}\plus{}t}\right)\minus{}3 \text{Cos}[t]\plus{}\text{Sin}[t]\right) & \text{True}
\end{cases}
\right\}\right\}$

But then, after some time, I decided to solve the equation without theta:

$ \text{DSolve}[\{y\text{''}[t]\plus{}3y'[t]\plus{}2y[t]\equal{}\equal{}\text{Sin}[t],y[0]\equal{}\equal{}1,y'[0]\equal{}\equal{}\minus{}1\},y[t],t]$
$ y[t]\to \frac{1}{10} \left(e^{\minus{}2 t} \left(\minus{}2\plus{}15 e^t\right)\minus{}3 \text{Cos}[t]\plus{}\text{Sin}[t]\right)$

What is the lapsus mentis I'm making?

Yes, that's the correct answer. Why did you stop using the unit step function? In your first post the error is when you pass from the expression for Y to the expression for y(t).

Exactly. I just had to use the exp member for the shift and ILT the fractions multiplied by it. Both approaches are valid, though.

(2) Solve, in $ [0, \plus{} \infty)$, the equation $ y''(t) \plus{} 2y'(t) \plus{} y(t) \equal{} f(t)$, $ y(0) \equal{} y'(0) \equal{} 0$, where $ f(t) \equal{} t^2$ for $ 0 \leq t < 2$, and $ f(t) \equal{} 2$ for $ t \geq 2$.

This can be also done in the two ways described above.

[hide="Solution"]First way: Solving $ y''(t) + 2y'(t) + y(t) = t^2$ is simple: the homogeneous equivalent is straightforward, and the particular solution is a 2nd degree polynomial (and we can use the initial conditions). Same stands for the second equation: the particular solution is just the constant $ 2$.

Second way: Using unit step functions:

$ y''(t) + 2y'(t) + y(t) = 2\theta(t - 2) - (\theta(t - 2) - 1)t^2$ and applying the Laplace transform:

$ Y (1 + z)^2 = \frac {2 e^{ - 2 z} (e^{2 z} - (1 + z)^2)}{z^3}$

Now, $ Y = - \frac {2 e^{ - 2 z} (1 - e^{2 z} + 2 z + z^2)}{z^3 (1 + z)^2}$

ILT gives $ y = - 2 e^{ - t} (t+3) - (6 - 4 t + t^2) + (t - 2)^2 \theta(t-2)$

i.e.

$ y = \begin{cases} 2 - 2 e^{ - t} (t+3) & t\geq 2 \\
6 + ( t - 4 ) t - 2 e^{ - t} (t+3 ) & 0\leq t < 2 \end{cases}$

I hope this is less erroneous then the first one :)[/hide]


ratio
quadratics
AMC
AIME
arithmetic sequence
geometric sequence

Three numbers form an arithmetic sequence, the common difference being 11. If the first number is decreased by 6, the second is decreased by 1 and the third is doubled, the resulting numbers are in geometric sequence. Determine the numbers which form the arithmetic sequence.

[hide]Let x, x+11, and x+22 be the numbers, x-6, x+10, 2x+44 be the numbers after manipulation.

In a geometric sequence, there is a common ratio, so $\frac{x+10}{x-6}=\frac{2x+44}{x+10}$, $x^2+12x-364=0$, $x=14, -26$. -26 doesn't work, so the numbers in the arithmetic sequence are 14, 25, and 36. [/hide]

And why doesn't $-26$ work?? Is that some general and authomatic discarding of negative solutions of quadratics?

If $x=-26$, then the arithmetic progression is $-26, -15, -4$ and the geometric one is $-32, -16, -8$ - all good.

Great point. There are TWO solutions.  :)

[hide="Answer"]$a+11=b$, $b+11=c$
$(a-6)(2c)=(b-1)^2=(a+10)^2=(a-6)(2a+44)\Rightarrow a=-26,14$
So we have either $(-26,-15,-4)$ or $(14,25,36)$.[/hide]

My guess is that they were looking for the positive solution.

[b][color=#00f000]J[/color][color=#0ce80c]e[/color][color=#18e018]s[/color][color=#24d824]u[/color][color=#30d030]s[/color][color=#3cc83c]F[/color][color=#48c048]r[/color][color=#54b854]e[/color][color=#60b060]a[/color][color=#6ca86c]k[/color][color=#78a078]1[/color][color=#849884]9[/color][color=#909090]7[/color][/b]

If they were looking for the positive solution, they would say "positive numbers". As they said just "numbers", all solutions are valid and hence required.

Since your assumption is no better than mine, we should take the text at the face value.

[quote="JesusFreak197"]That's assuming the question was copied exactly and assuming that there's no restriction on this contest like there is on the AIME for positive numbers only (although I rather doubt that).

[b][color=#00f000]J[/color][color=#0ce80c]e[/color][color=#18e018]s[/color][color=#24d824]u[/color][color=#30d030]s[/color][color=#3cc83c]F[/color][color=#48c048]r[/color][color=#54b854]e[/color][color=#60b060]a[/color][color=#6ca86c]k[/color][color=#78a078]1[/color][color=#849884]9[/color][color=#909090]7[/color][/b][/quote]

All questions are usually copied word by word (not all cases; check one of the problems in Part A of this year in COMC) but in this case, it is posted with exact wording. So, Farenhajt's argument is correct.

However, I understand how you feel toward the answer. In United States, the questions usually take the "positive" value and ignore the negative value. So, it's natural that you believed that the problem was looking for positive solution.

Silverfalcon

Well, I'm assuming that they were looking for only one value. In that case, it's more natural to take the positive value than the negative one.

[b][color=#00f000]J[/color][color=#0ce80c]e[/color][color=#18e018]s[/color][color=#24d824]u[/color][color=#30d030]s[/color][color=#3cc83c]F[/color][color=#48c048]r[/color][color=#54b854]e[/color][color=#60b060]a[/color][color=#6ca86c]k[/color][color=#78a078]1[/color][color=#849884]9[/color][color=#909090]7[/color][/b]

let's stay positive guys!

....

lolk, that was a really corny joke.


[b]1[/b]     The ratio of the interior angles of two regular polygons is 3:2.  How many such pairs are there?

[b]2[/b]     Given that $ \frac{2}{x}\equal{}\frac{y}{3}\equal{}\frac{x}{y}$, find $ x^{3}$.

Solution to number 2

[hide]Cross multiply to give $ x^2\equal{}2y, 3x\equal{}y^2, xy\equal{}6$.  First, realize that multiplying the first two gives us $ x^3\equal{}\dfrac{2y^3}{3}$.  Then, the third equation gives us $ x\equal{} \dfrac{6}{y}$.  Substitute into the second to get $ y\equal{}\sqrt[3]{18}$.  Another substitution yields $ x^3\equal{}12$.[/hide]

[hide]1. The size of an interior angle of a regular polygon is given by $ \frac {(s \minus{} 2)180}{s}$ where $ s$ is the number of sides of the polygon.

Our equation is then $ 2\cdot\frac {(n \minus{} 2)180}{n} \equal{} 3\cdot\frac {(m \minus{} 2)180}{m}$

or $ 2\cdot\frac {n \minus{} 2}{n} \equal{} 3\cdot\frac {m \minus{} 2}{m}$

Multiplying by $ mn$ we have $ 2mn \minus{} 4m \equal{} 3mn \minus{} 6n$ or $ mn \minus{} 6n \plus{} 4m \equal{} 0$

Subtracting 24 from both sides we can factor the LHS to get $ (m \minus{} 6)(n \plus{} 4) \equal{} \minus{} 24$

Since $ m,n > 0$ and we want non-degenerate polygons we have 3 such pairs.
[/hide]

Edit: Fixed

wait a sec 4 pairs?

if u mean using
m-6 and n+4 as
-4 and 6
-3 and 8
-2 and 12
-1 and 24

the first case gives m=n=2 which is extraneous

^You're right, there are actually 3 pairs.


group theory
abstract algebra
analytic geometry
superior algebra
superior algebra unsolved

$G$ is a finite simple group. How many normal subgroups does $G \times G$ have ?

( I found $5$ of them ).

The answer depends on what $G$ is. If $G$ is the cyclic group of order $p$, there are $p+3$ normal subgroups.

Hi,

I precised $G$ must be simple (which makes it possible to prove there are only $5$ normal subgroups for $G \times G$). My proof was kinda tedious to write though.

Well, a cyclic group of order $p$ is simple ... so this doesn't contradict your assertion. I'm going to check.

I think the answer should be $5$ if the group is non-abelian though.

Suppose, for instance, that a normal subgroup contains an element $(x_1,x_2)$ with non-trivial $x_1$. Take some $a\in G$ not belonging to the centralizer of $x_1$. Our subgroup will contain the element $(ax_1a^{-1},x_2)$, and thus the element $([a,x_1],1)$, where $[a,x_1]\ne 1$. This means that the intersection of our subgroup with $G\times\{1\}$ is a non-trivial normal subgroup of $G\times G$ which is also a subgroup of $G\times\{1\}$. This means that our subgroup actually contains $G\times\{1\}$. It's clear how to proceed, I think.

Thanks for the correction jmerry and for the confirmation Grobber.
After verification I found that I overlooked the case where $p$ is prime.
Eventually I proved that the number of normal subgroups must be $5$ if $p$ is not prime, $p+3$ otherwise. (It's a little stronger than what Grobber proved, but maybe his proof is adaptable to the general abelian case; I will read later since I need to get ready to go to school).

So... In the nonabelian case, what's the fifth normal subgroup? I see $\{1\}\times\{1\}$, $\{1\}\times G$, $G\times \{1\}$, and $G\times G$, with a convincing proof that there can't be any others. If the normal subgroup contains an element with both coordinates nontrivial, grobber's proof shows that it contains both coordinate axes and thus all of $G\times G$.

(by the way, I don't think "commutator" means what you want it to; try "centralizer" or just write it out in words)

I was influenced by Julien's $5$ :). Indeed, I meant $4$.

And yes, I meant "centralizer" :).


inequalities
algebra
polynomial
inequalities proposed

prove that
$2(a^3+b^3+c^3+3abc-a^2b-a^2c-b^2a-b^2c-c^2a-c^2b)\leq \frac{ab}{c}(a-b)^2+\frac{ac}{b}(a-c)^2+\frac{bc}{a}(b-c)^2$

where a,b,c are positive reals.
quite a nice inequality and i've got a BEAUTIFUL proof for it...

Peter

I didn't know this one and it's really nice.Then you must refer to the following proof:
 Write the inequality as $ ( ab(a-b)+bc(b-c)+ca(c-a))^2\geq 0 $. And its done. Its really not difficult to see this decomposition: just take $ x=ab(a-b), y=bc(b-c), z=ca(c-a) $.

it is really easy, but i just liked the following argument: multiply by $abc$ and subtract the right side to get something of the form $P(a,b,c)\geq 0$, where obviously P is a symmetric homogeneous polynomial of degree at most 6. now it is very easy to verify that we have equality if $a=b$. thus we can factor $(a-b)$ out of $P(a,b,c)$, and since it is homogenous, we can even factor $(a-b)^2$, analogeous $(a-c)^2$ and $(b-c)^2$. thus $P(a,b,c)=k(a-b)^2(a-c)^2(b-c)^2$ where k is a constant. it is easy to see that k=2(looking at the coefficient of $a^4b^2$ or something), and we're done.
i like that proof because it involves nearly no computations and it is more algebra than inequality :)

Peter

I think you two noticed your solutions are absolutely the same.
This inequality also can be solved by putting a=b+y=c+x+y; x,y>0.
You don't need to multiply by abc

i know our solutions are the same. but my argument avoids nasty computations to get that it is equivalent to $(a-b)^2(a-c)^2(b-c)^2\geq 0$.

Peter, I think you can factor $ (b-a)^2 $ as it is symmetric and not as it is

homogeneous.

If I am in error, can you please explain it to me.

Thank you very much.

yes that's what i meant, i wanted to write it but somehow i must have been thinking of something else :)

Peter

Peter 's solution is one of the methods to decomposite a polynomial. :D But it 's nice.


Let $ AB$ and $ CD$ be the chords of circle $ (O)$ ($ AB$ and $ CD$ don't intersect each other). Let $ I$ be a point on $ CD$. Construct point $ M$ on $ (O)$ so that $ AM$ and $ BM$ intersect $ CD$ at $ E$ and $ F$ and $ I$ is the midpoint of $ EF$.

Anyone have any idea of this problem?

It is on my textbook and I'm surprised about its difficulty.

It 's not a difficult problem. I will give you the construction.

Construct B' as the symetric point of B about I.

Create an arc $ \overarc{B'mA}$ of an angle $ \pi - \frac {\angle {AOB}}{2}$

Then E is the intersection of the arc and CD.

From E, M is followed.

I 've thought a new  problem. I haven't found the answer.

It 's:

Given a circle (C) and a point I. Suppose CD and AB are chord of (C) and CD does not intersect AB. Construct the chord EF of (C) such that $ I \in EF$ and AE, BE, CD are concurrent.

Thanks mathangel for your reply
BTW, I don't understand your problem if $ AE, BE, CD$ are concurrent then $ E$ must coincide $ C$ or $ D$.

I 'm sorry for late response. I have a lot of problem in my university study.

It 's a typo error. it must be AE, BF and CD not AE, BE and CD.


[quote="bpms"]Can I print out my solutions and give them to others who are talented in math for proofreading if they don't give me help, just if they tell me where my proof wasn't concise enough?[/quote]
I wouldn't recommend it - for example, what if your proof is wrong, or has some gap or flaw in it?  They'd tell you, and that would be a violation of independent work on the problems.  I'd recommend learning from the grading process and from showing your solutions to others after you turn them in; that way you'll learn more quickly anyway than if you always had others improve your solutions beforehand.  Also, there's really no such thing as "not concise enough."  There is of course "awkwardly wordy," which means that you're no longer clearly stating what you mean to say; but a solution is never incorrect just because you're very thorough (more than you might need to be be).

[quote="bpms"]Can I print out my solutions and give them to others who are talented in math for proofreading if they don't give me help, just if they tell me where my proof wasn't concise enough?[/quote]No.


In 2 hours, X can run 1 km further than Y. With a 2 minute head start, Y could tie X in a 4 km race. Assume that all speeds are constant. What is Y's rate, in km per hour?

Let us convert all hours to minutes while doing this problem.

Let x be X's speed
Let y be Y's speed

120x=120y+1
120x=122y
thus
122y=120y+1
2y=1
y=1/2 km per minute
  = 30 kmph

I'm new, so tell me if my format is wrong.


This is again taken from Arthur Engel's Problem Solving Strategies...

[Prove that] $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$ divides $ x^{44} \plus{} x^{33} \plus{} x^{22} \plus{} x^{11} \plus{} 1$

The solution says...
"Let $ w^5 \equal{} 1$ Then $ w^{44} \plus{} w^{33} \plus{} w^{22} \plus{} w^{11} \plus{} 1 \equal{} w^4 \plus{} w^3 \plus{} w^2 \plus{} w^1 \plus{} 1$ All roots of the left side are also roots of the right side. This is implies the stated divisibility."

My question is: how can you simply substitute any value you want for x and then draw a conclusion from it. Aren't you only considering a special case - that $ w^{44} \plus{} w^{33} \plus{} w^{22} \plus{} w^{11} \plus{} 1 \equal{} w^4 \plus{} w^3 \plus{} w^2 \plus{} w^1 \plus{} 1$ Don't you want to prove it for ALL X?????

If for any reason this question isn't clear...

"Factorize over Z: (a) $ x^2 \plus{} x \plus{} 1$..."

The solution says...
"Inserting the third root of unity w for x, we get $ w \plus{} w^2 \plus{} 1 \equal{} 0$ Thus, $ x^{10} \plus{} x^5 \plus{} 1$ has factor $ x^2 \plus{} x \plus{} 1$..."

My question here is: When you substitute w, a root of unity, for x, won't you merely get a special case (but you want to factorize the expression for all x)?

Maybe I'm not getting something here... can someone explain it to me?????

A polynomial $ P(x)$ divides a polynomial $ Q(x)$ if and only if each of the roots of $ P$ is also a root of $ Q$ with at least the same multiplicity.  The roots of $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$ are the primitive fifth roots of unity.

The philosophical point here is that a polynomial's behavior is determined by a finite amount of data: specifying its constant term and its roots will do, as will specifying its values at more points than its degree.

Well, realizing that $ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1\equal{}\frac{x^5\minus{}1}{x\minus{}1}$, all four of its roots are roots of unity such that $ w^5\equal{}1$ and $ w\neq1$.

As long as these 4 roots are ALSO roots of $ x^44\plus{}x^33\plus{}x^22\plus{}x^11\plus{}1$ then it will be a factor of it.

thanks!!!!!!
you guys were very helpful!  :D

Just one question: so the roots of unity are basically roots of the equation and not merely a substitution of variables, right?

Thanks!

The nth roots of unity are the n roots to the equation $ x^n\equal{}1$

For example, the 4th roots of unity are the solutions to $ x^4\equal{}1$ so they are $ 1,\minus{}1,i,\minus{}i$

Thanks!

The the answer to my question is yes, right?

Thanks again!


[quote="南盘江"]I am from China.I  want to publish my paper on how to use complex number to solve plane geometry.Are there any suitable magazines for me?Help me,please. Thanks a lot.[/quote]
You mean you want to publish your paper on a foriegn magzine?? Why not have a look at [b]Mathmetical Journals[/b] at the High school section.

[quote="南盘江"]I am from China.I  want to publish my paper on how to use complex number to solve plane geometry.Are there any suitable magazines for me?Help me,please. Thanks a lot.[/quote]
Hi ,

[b]try to send your paper [/b] as a .TEX file /together with the corresponding .PDF -flie/  to one
of the following journals :
1)

[b] GAZETA MATEMATICA -Seria A
 Str. Academiei  nr. 14
 70109 BUCURESTI , ROMANIA[/b]
E-mail: [i] office@rms.unibuc.ro[/i]

or to

2)
[b] Revista de Matematica din Timisoara
  Birchi - Damian Ion , Bd. Sudului nr. 2 , ap.8 ,
  300686- TIMISOARA , ROMANIA[/b]
E-mail: [i] dbirchi@xnet.ro [/i]

[quote="shobber"]Romania magzines?? Do they publish in English?[/quote]
Yes, these are Romanian magazines.
 In ,,Gazeta Matematica" -Seria A, there are published papers  also in English. /Alex


1) Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$ Prove that
\[ \sqrt{\frac{a^4\plus{}2b^2c^2}{a^2\plus{}2bc}}\plus{}\sqrt{\frac{b^4\plus{}2a^2c^2}{b^2\plus{}2ac}}\plus{}\sqrt{\frac{c^4\plus{}2a^2b^2}{c^2\plus{}2ab}}\geq a\plus{}b\plus{}c\]

2) Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ \sqrt{a^3\plus{}b^2c\plus{}bc^2}\plus{}\sqrt{b^3\plus{}a^2c\plus{}ac^2}\plus{}\sqrt{c^3\plus{}a^2b\plus{}ab^2}\geq\sqrt{(a\plus{}b\plus{}c)^3}\]

[quote="arqady"]1) Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that
\[ \sqrt {\frac {a^4 \plus{} 2b^2c^2}{a^2 \plus{} 2bc}} \plus{} \sqrt {\frac {b^4 \plus{} 2a^2c^2}{b^2 \plus{} 2ac}} \plus{} \sqrt {\frac {c^4 \plus{} 2a^2b^2}{c^2 \plus{} 2ab}}\geq a \plus{} b \plus{} c
\]
2) Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ \sqrt {a^3 \plus{} b^2c \plus{} bc^2} \plus{} \sqrt {b^3 \plus{} a^2c \plus{} ac^2} \plus{} \sqrt {c^3 \plus{} a^2b \plus{} ab^2}\geq\sqrt {(a \plus{} b \plus{} c)^3}
\]
[/quote]

1) is posted before.
see here : http://www.mathlinks.ro/viewtopic.php?t=260824

(2)squaring and Cauchy works
    $ \sum \sqrt {a^3 \plus{} b^2c \plus{} bc^2} \sqrt {ac^2 \plus{} a^2c \plus{} b^3} \ge \sum a^2c \plus{} 3abc \plus{} \sum b^2c$

[quote="arqady"]1) Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that
\[ \sqrt {\frac {a^4 \plus{} 2b^2c^2}{a^2 \plus{} 2bc}} \plus{} \sqrt {\frac {b^4 \plus{} 2a^2c^2}{b^2 \plus{} 2ac}} \plus{} \sqrt {\frac {c^4 \plus{} 2a^2b^2}{c^2 \plus{} 2ab}}\geq a \plus{} b \plus{} c
\]
2) Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ \sqrt {a^3 \plus{} b^2c \plus{} bc^2} \plus{} \sqrt {b^3 \plus{} a^2c \plus{} ac^2} \plus{} \sqrt {c^3 \plus{} a^2b \plus{} ab^2}\geq\sqrt {(a \plus{} b \plus{} c)^3}
\]
[/quote]

Another proof of 2)
By cauchy,
$ (a^3\plus{}b^2c\plus{}bc^2)(a\plus{}c\plus{}b)\ge(a^2\plus{}2bc)^2\Leftrightarrow \sqrt{a^3\plus{}b^2c\plus{}bc^2}\ge \frac{a^2\plus{}2bc}{\sqrt{a\plus{}b\plus{}c}}$.
so we have$ \sum\sqrt{a^3\plus{}b^2c\plus{}bc^2}\ge \sum\frac{a^2\plus{}2bc}{\sqrt{a\plus{}b\plus{}c}}\equal{}\sqrt{(a\plus{}b\plus{}c)^3}$.


Here is a puzzle I am trying to solve.  I think I have the answer, but I wanted to check with someone here to make sure that I have thought about the problem correctly.

This is the puzzle:

Suppose you have, say, a stone tablet of some sort. The tablet has 16 pieces missing. But lucky you, you have the pieces! You can tell by looking at the pieces which ones are the four corners, which ones are the eight side pieces, and which ones are the inside pieces. Otherwise, you can't tell where on the tablet they should go or how they should be oriented.

[img]http://images.neopets.com/games/conundrum/134_tablet.gif[/img]

If it takes 90 seconds to place all of the pieces, how many days will it take to try every possible combination? Please round to the nearest full day.

This is what I think the answer is:

Assuming all pieces are different, the TOTAL number of possible combinations for the placement of the 16 pieces (with the individual different possible combinations for the corner, side and center pieces) is solved as follows (I think):

4! * 8! * 4! =  24 * 40,320 * 24 = 23,224,320

Then the number of possible combinations in a day would be solved as follows (I think):

Number of seconds per day = 86,400  (60 * 60 * 24)

86,400 seconds per day / 90 seconds per combination = 960 combinations per day

23,224,320 total possible combinations / 960 combinations per day =  24,192

So the answer to the puzzle would be: 24,192 days to try all possible combinations of the 16 pieces

UNLESS....... when the problem says "or how they should be oriented", I am not sure what they mean, and if it does mean something particular (could it mean that each piece has four possible placements on each of its possible spots?), then have no idea what the solution might be.   Can anyone help me with this?  Thank you ahead of time!

From what you wrote, the answer looks right. I'm sure the problem is asking what you thought.

thats probably the right answer, but the picture does have some pieces that look the same shape.

if there are two alike lookin pics? :?:  :D


Find all polynomials F(x) satisfying the equation

x.F(x-1) =(x-26).F(x)

General problem:

Let a,b are nonnegative real . Find all P(x) in R[x] such that:

xP(x-a) = (x-b)P(x)

$P(ka)=0$,were $k\in Z_{+}, ka\not =b$
It give polinomial solution if and only if $\frac{b}{a}\in N$.

if we put a=1, b=26 we have the original problem  :D


floor function
factorial
complex numbers

YES!!!!! that's how u do it

In fact, this trick is very useful in trying to find how many times more complex numbers divide into factorials also :)

[quote="Meghrabi"][quote="amcavoy"]What is the highest power of $2$ that divides $2005!$?

I just said that the answer would be $\sum_{n}\left\lfloor\frac{2005}{2^{n}}\right\rfloor=1997$, so it would be $2^{1997}$.  At the moment I don't have a way of checking this.  Is that correct?[/quote]

How do you get that 1997 actually, i can't see it. :mad:[/quote]

$[\frac{2005}{2}]=1002$
$[\frac{1002}{2}]=501$
$[\frac{501}{2}]=250$
$[\frac{250}{2}]=125$
$[\frac{125}{2}]=62$
$[\frac{62}{2}]=31$
$[\frac{31}{2}]=15$
$[\frac{15}{2}]=7$
$[\frac{7}{2}]=3$
$[\frac{3}{2}]=1$

Add up all those values and get 1997

[quote="amcavoy"]What is the highest power of $2$ that divides $2005!$?

I just said that the answer would be $\sum_{n}\left\lfloor\frac{2005}{2^{n}}\right\rfloor=1997$, so it would be $2^{1997}$.  At the moment I don't have a way of checking this.  Is that correct?[/quote]

That's a really cool trick! So let me just get this clear...
The highest power of $a$ that divides $b!$ is $\sum_{n}\left\lfloor\frac{b}{a^{n}}\right\rfloor$ ?
Sweet.  :D

That's the way I've been doing it; it seems to work.  Alternatively you could ask what is the smallest number that is divisible by $a^{n}$ (I think this is a bit more guess-and-check to begin with, but the same basic method).


Let $p(x)$ be a polynomial.

Show that $p(x)-x | p^{n}(x)-x$ where $n = 1,2,3,...$
(here index indicates multiple compositions)
Or in other words, that $p(x)-x$ is a factor of $p^{n}(x)-x$.

[hide="The relevant lemma"] $a-b | P(a)-P(b)$ [/hide]
[hide="Hence"] $p(x)-x | p^{n}(x)-p^{n-1}(x)$, and the result is obvious from induction. [/hide]

[quote="t0rajir0u"][quote="iluvmath1"]$p(ag+x) = ag+(x+g) = ag+p(x)$[/quote]

I am having trouble seeing how this step is justified.  In particular, if $p(x)$ is of degree $n$ then the LHS is of degree $n(k+1)$ but the RHS is of degree $nk$.

(By the way, $a$ is not an integer; it is a polynomial.)[/quote]

Correct. I got too wishful thinking a was an integer.

\[f(x)-g(x)|P(f(x))-P(g(x))\]
$\sum_{k=0}^{n}p_{k}x^{k}= P(x)$ with degree $n$.
Then, \[P(f(x))-G(f(x) = \sum_{k=0}^{n}p_{k}\left( f(x)^{k}-g(x)^{k}\right) \]
$S_{k}=f(x)^{k}-g(x)^{k}$.
If $S_{1}|S_{K}$ for $K=k,k-1$
then $S_{1}|[f(x)+g(x)]S_{k}-f(x)g(x)S_{k-1}=S_{k+1}$ by the Euclidean Algorithm.
Since $S_{1}|S_{1},S_{2}$ is trivial the result follows by induction that $S_{1}|S_{n}$ for integer $n$ of course.
Thus, \[P(f(x))-G(f(x) = \sum_{k=0}^{n}A_{k}\left( f(x)-g(x) \right) =\left(\sum A_{k}\right)S_{1}\] for some coefficients $A_{k}$ which establishes the lemma.


factorial
modular arithmetic
number theory unsolved
number theory

Suppose that x=a 2 +b 2 =c 2 +d 2 where a,b,c,d are distinct natural numers. Prove that x is not a prime.

Actually, this follows quickly from the fact that each prime $p$ which is of the form $4k+1$ can be decomposed in $\mathbb Z[i]$ as a product of primes $(a+bi)(a-bi)$ in a unique way (because $\mathbb Z[i]$ is Euclidean, and the unique prime decomposition theorem holds).

Wow, that was fast! And me, who wanted to show that I have finally understood the thing with $ Z[i]$... ;) :(  :(  Why you...  ;)  ;) (just kiddin, I'm in holiday...).

A second WOW, it took me  almost an hour to find a synthetic prove but you solved it in 3 minutes. Where can I find information about properties of Z[i]?

I can't give you a specific source, but I'm sure you can find loads of info if you google things like "Euclidea ring" or "factorial ring" or "Gaussian ring" (which is just another name for a factorial ring).

another WOW - that's magnificent solution, but I do have one question - why $x$ is of $4k+1$ form ?

That's because no number of the form $4k+3$ can be the sum of two perfect squares (just try all the cases to see what $a^2+b^2\pmod 4$ can be).

Notice that even though all primes of the form $4k+1$ can be written as the sum of two squares (this has been discussed several times here on the forum), but this is not important to us. We just need the fact that if $p$ is a prime which can be written as the sum of two squares, then this decomposition is unique.

Ok, now replace the squares by $k$-th powers (where $k$ is fixed), and prove the same conclusion.

Pierre.

Unfortunately my synthetic solution does't work here anymore :( , perhaps Grobber can help you.

I'd certainly have to put a lot more thought into that :).

Probably I do not see obvious things, but if p is a prime and $p=a^k+b^k$ , then k must be a power of 2 and we go back to the first problem. No, it can't be that...

It is a really good joke, Pierre!  :D  :D  :D
I like your humor!

Actually, if $k$ is even, then we're done, and if $k$ is odd, then $a+b,c+d$ are divisors of $p$. I suppose you did pretty much the same thing.

Hey, grobber, you know what? We both were catched in Pierre's trap! Evil Pierre!  :evil:  ;)  ;)  ;)  At least I did not think more than 1 minute...

I do not understand. Is it not a 'wow'-generalisation? :D

Pierre.

No, this is just a $w$ generalization, since I solved it in 1 minute, while the initial problem took grobber 3 minutes!  :D  :D. At least you could have tried to create a $ wo$ generalization... Silly me, I was so naive  :maybe:  :maybe:

I seem not to be able to follow the discussion with all these $w$'s and $wo$'s and $wow$'s flying around :D.

The elementary solution still is not posted. It might be as follows: if $p=a^2+b^2=c^2+d^2$,
wlog $a<c<d<b$ and
$(bc-ad)(ad+bc)=b^2c^2-a^2d^2=b^2c^2-(p-b^2)(p-c^2)$ is divisible by $p$, hence either $bc-ad$, or $ad+bc$ is divisible by $p$. But $0<bc-ad<b^2<p$
and $ad+bc<(a^2+d^2)/2+(b^2+c^2)/2=p$ --- a contradiction.


Suppose that $ a$ and $ b$ are distinct real numbers such that
$ a\minus{}b,a^2\minus{}b^2,.....a^k\minus{}b^k,......$
are all integers.Show that $ a,b$ are integers

$ \frac {a^2 \minus{} b^2}{a \minus{} b}$ is rational, hence $ a, b$ are rational.  Write $ a \equal{} \frac {p \plus{} qd}{d}, b \equal{} \frac {p}{d}, (p, d) \equal{} 1$, where $ q \equal{} s_1$ is an integer.  Define $ s_k \equal{} (ad)^k \minus{} (bd)^k \equal{} (p \plus{} qd)^k \minus{} p^k$.  The problem statement claims $ d^k | s_k \forall k$.  Note that

$ s_{k \plus{} 1} \equal{} (2p \plus{} qd) s_{k} \minus{} p(p \plus{} qd) s_{k \minus{} 1} \implies$
$ d^{k \plus{} 1} | (2p \plus{} qd) s_{k} \minus{} p(p \plus{} qd) s_{k \minus{} 1} \implies$
$ d^k | p^2 s_{k \minus{} 1} \implies$
$ d^k | s_{k \minus{} 1} \implies$
$ d^2 | s_1 \equal{} q$.  But then

$ d^{k \plus{} 2} | 2p s_k \minus{} p^2 s_{k \minus{} 1} \implies$
$ d^{k \plus{} 1} | s_{k \minus{} 1}$,

etc.  Contradiction unless $ d \equal{} 1$.  (This could be phrased more nicely.)

[quote="t0rajir0u"]$ \frac {a^2 \minus{} b^2}{a \minus{} b}$ is rational, hence $ a, b$ are rational. [/quote]
Sorry t0rajir0u are you wrong the Contradiction is $ a,b \in R$ for example $ a\equal{}\sqrt2 \minus{}1$ and $ b\equal{}1\minus{}\sqrt 2$
Perhaps,i didnt understand your ideal can you detail it,please

$ (\sqrt{2} \minus{} 1) \minus{} (1 \minus{} \sqrt{2}) \equal{} 2 \sqrt{2} \minus{} 2$ is clearly not rational.  We are given that $ a \minus{} b, a^2 \minus{} b^2$ are integers, so $ \frac{a^2 \minus{} b^2}{a \minus{} b} \equal{} a \plus{} b$ is rational, which implies $ a \equal{} \frac{(a \minus{} b) \plus{} (a \plus{} b)}{2}$ and $ b \equal{} \frac{(a \plus{} b) \minus{} (a \minus{} b)}{2}$ are also rational.

i am wrong ,I am really sorry,your proof is very nice ,thanks a lot    :blush:  :blush:  :blush:

Posted often enough, locked! And stop using this nonsense titles!


VCR
   +  VCCT
___________
      SERVE

sorry if its not very lined up    :(

cool thanks!!! I could have looked at the answer in the book but i wanna see if someone here can do it.  :)

Well, I can tell you that $ S \equal{} 1$ and $ V \equal{} 9$.

I'll get you more when I find it.

$ S \equal{} 1$ because if you add two numbers with 4 or less digits, and the result is a 5-digit number, the ten thousands digit must be one.

$ V \equal{} 9$ because the only possible way for $ VAB$ and $ VCDE$ to add up to a 5-digit number is if $ V \equal{} 9$.

I think $ T \plus{} R \equal{} 10$, $ E \equal{} 0$, $ C \equal{} 4$.

So, based on my presumtions, we have:

944T
+94R
-----
10R90

$ T \plus{} R \equal{} 10$.

$ R \equal{} 3, \text{therefore}, T \equal{} 7$

So I think that:

$ S \equal{} 1$
$ R \equal{} 3$
$ T \equal{} 7$
$ V \equal{} 9$

Correct me if I am wrong.

[quote="myyellowducky82"]i though T was 5??[/quote]

I substituted the letters with my numbers.

$ 943 \plus{} 9447$ does equal $ 10,390$.  Thus, my solution is correct.

And I'm not sure, but $ T$ can never equal $ 5$, because then $ R$ would equal $ 5$, and I think that is illegal in these problems.


Cauchy-Schwarz.Master.Class.(Steele.2004)

Ενα πολυ καλο βιβλιο (αν οχι το μοναδικο) για τις μαθηματικες ανισοτητες.
Απο την συλληψη μεχρι την αποδειξη και γενικευση.

Download: http://www.mediafire.com/?6jsandddnwn

Κώστα ,
σε καλωσορίζουμε σε αυτόν τον χώρο και είμαστε σίγουροι ότι οι κοινές μαθηματικές ολυμπιακές ανησυχίες και συγκινήσεις θα μας κάνουν συντροφιά για πολύ καιρό.
 Ελπίζουμε και στη δική σου συνεισφορά και βοήθεια.
 Σε ευχαριστούμε και για το εξαιρετικό βιβλίο. Μακάρι να ξαναφτιάξουμε την ολυμπιακή ιστοσελίδα με τα ολυμπιακά βιβλία , όπου θα το έχουμε και αυτό!

 [u]Μπάμπης[/u]


I would really like to take the AMC 12 B, but my high school (Arcadia High School in California) doesn't offer it.  Does anyone know of a place where I can take it in my area?  I've contacted PCC and Polytechnic and neither of them offer it.

http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml

If you're going to try Chapman University, you just email mbvajiac@chapman.edu and tell her you want to sign up.

If you're going for another university, you'll have to contact them and figure out how to register.


Today I received Saturday school.

I am taking university courses, but they have not started yet, so I have a free block. I thought it would be fine to work in one of my school's computer labs for 3rd block, but later in the school day, I got a call from the main office and was handed Saturday school for not reporting to a room that I was supposed to be in. It was labeled "skipping".

I have tried getting out of the Saturday School, but it looks like no luck. My main concern is if this will come up on any type of transcript or anything, so that the schools I'm sending a mid- year report to will see "skipping." Would they do anything?

The schools are MIT, Caltech, Nebraska.

Last time I checked for my school district at least, a single "cut" does not go on a transcript -- only suspensions and expulsions. But you may want to be sure by going in to the office and nicely asking what happened -- if anyone could verify you were on campus for 3rd block I hope everything would be ok.

And even if it went on your transcript I doubt Caltech and MIT would care -- if it got that far, just send them a letter explaining.

Good luck with this!

They knew I was on campus though. They don't seem to be budging. I also found out that the saturday school is assigned on a day that our math team was supposed to be going to a big math tournament. This sucks.  :(

Usually for SAT, math tournament, or athletic competition, my school moves around Saturday school... not that I've gotten one, but my friend got his mom to give the school a signed note because he was taking SAT Chinese that day (only offered once a year) and the school delayed his saturday school for a week.

[quote="juicybooty911"]Take it up with the principal if you know him/her well.[/quote]

definetely.

Like aznphatso said, it's doubtful that MIT or Caltech will do anything if they see you got in trouble once.  On the other hand, if you started failing all your classes, got caught skipping on a regular basis, and got busted for dealing drugs...well :D

If you don't get off the hook and they refuse to reschedule or whatnot, make a ruckus.  You shouldn't be forced to miss an academic event on account of some bogus skipping charge.

[quote="Pocket Sand"][quote="drunner2007"]
I am taking university courses, but they have not started yet, so I have a free block. I thought it would be fine to work in one of my school's computer labs for 3rd block, but later in the school day, I got a call from the main office and was handed Saturday school for not reporting to [b]a room that I was supposed to be in[/b]. It was labeled "skipping".
[/quote]

I don't think I understand exactly what you're saying... were you supposed to be in a room or not?[/quote]

I was supposed to be in a certain room, but I was in another one.
They're letting me change the date of the Saturday School so that I don't miss the competition. At least that makes it more bearable.


Determine the number of odd binomial coefficients in the expansion of $(x+y)^{1000}$.

1000=111101000 (2)
by Lucas's Theorem, The answer is 2^6

Already posted at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=102805[/url].


Given 2 circles C1 and C2 external to each other and externally tangent to a circle C at pts A and B respectively, and given a common tangent t of C1 and C2 touching them at D and C respectively (such that t and C have no common point), show that ABCD is cyclic.

Do you mean C1 and C2 are also tangent to each other, or don't they touch at all ?

It doesn't matter in fact. The solution seems to be quite straightforward.
You can suppose C1 and C2 don't touch/intersect each other.
I will firstly reformulate the problem.
You know, you've got a circle C and a point C ...

Given two circles C1 and C2 which do not intersect or touch.
(Probably the problem is also correct if C1 and C2 do intersect.)
A third circle C touches C1 and C2 externally in A and B.
A common tangent t to C1 and C2 (which doesn't intersect C) touches C1 and C2 in Q en P.
Show that ABPQ is a cyclic quadrilateral.

Let O1, O2, O be the centers of C1, C2, C. We know that O, A, O1 and O, B, O2 are collinear.
Let N be the intersection of OO1 with t. Let M be the intersection of OO2 with t.
We know that triangles AO1Q and BO2P are isosceles.
So suppose angle O1AQ = angle O1QA = x and angle O2BP = angle O2PB = y.
Then, since angle O1QP = angle O2QP = 90, we have angle AQP = 90 - x, angle BPQ = 90 - y.
Also angle O1NQ = angle ONM = 90 - 2x, and angle OMN = 90 - 2y.
So angle MON = 2x + 2y. Now triangle AOB is isosceles.
Hence angle OBA = angle OAB = 90 - x - y.
Since O, A, O1 are collinear we have angles OAB + BAQ + O1AQ = 180,
so angle BAQ = 90 + y.
Hence QAB + BPQ = (90 + y) + (90 - y) = 180. So ABPQ is cyclic.

Thanks! I have a soln using.. guess what? That's right, inversion :D I'm pretty obsessed with inversion. Try to prove the following fact: Given 2 circles which do not intersect (I think it also works if they do) and for which their 2 common tangents meet at I and given a circle tangent to the 2 circles at A and B respectively, prove that A, B, I are collinear. It helped me in my soln.

OK, but which two common tangents do you mean ?
I thought two non - intersecting circles had 4 common tangents.

Can you prove this with inversion ?
If so, how ? Because I never manage to use inversion the right way ...

I'm looking for a non - inversion proof.

I found an elementary non - inversion proof. It uses homotheties and some angle - chasing. I hope I didn't make a mistake because you must always be careful in proving collinearity. I can post my proof if you want to ...

Btw, what's your inversion proof ?

Sorry I forgot to mention which tangents :D I meant the external tangents. I'll post my soln a bit later.

An inversion with center $A$, and an arbitrary radius will solve this easy problem.name the tangency of

$(C1),(C2)$, $O$.and consider that points $B,C,D,O$, turn into points$B',C',D',O',$ respectively.

Circles $(C),(C1)$, turn into two parallel lines.

Circle$(C2)$, turnes into a circle which is tangent to the two parallel lines at $B',O'$.

Line $CD$, turnes into a circle which tangent to $(C1),(C2)$, at $D,C$ respectively, and also goes through

Point$A$(the center of inversion)

Now we should prove that points,$B',C',D'$ are collinear.

It is clear that $\angle D'C'O'=90$

And incircle $(C2)$, $B'O'$ is diameter.then we have: $\angle B'C'O'=90$

And according to the last two line, we find that:

$\angle D'C'B'=180$, and $D',C',B'$ are collinear.

And it shows us that the quadrilateral $ABCD$, is cyclic.


Find the minimum possible value of:
\[ |z|^2  \plus{} |z \minus{} 3|^2  \plus{} |z \minus{} 6i|^2
\]
where $ z$ is a complex number and $ i\equal{}\sqrt{\minus{}1}$

[hide]Putting $ z\equal{}x\plus{}yi$, we get $ E\equal{}3(x\minus{}1)^2\plus{}3(y\minus{}2)^2\plus{}21$, hence $ E_{\min}\equal{}21$ for $ z\equal{}1\plus{}2i$.[/hide]


Show that for any $ \{b,c,d,x,y,z\}\subset\mathrm R$ exists the implication

$ \left\|\begin{array}{c} (z \plus{} d)(b \plus{} c) \equal{} 2(bc \plus{} zd) \\
\ (b \plus{} d)(x \plus{} c) \equal{} 2(bd \plus{} xc) \\
\ (c \plus{} d)(y \plus{} b) \equal{} 2(cd \plus{} yb)\end{array}\right\|\ \implies (z \plus{} d)(x \plus{} y) \equal{} 2(zd \plus{} xy)$ .

Ha I'm pretty sure I found this as well. The motivation is geometric!  I wonder if you arrived at the problem in the same way.

It is easy to brute-force because one can solve for $ x,y,z$ in terms of $ b,c,d$ and plug them in and you get an identity.


Solve

$ \frac{x}{y} \plus{} \frac{y}{x} \equal{} \frac{13}{6}$

$ x\plus{}y\equal{}5$.

hello, our first equation is for $ x\ne0$ and $ y\ne0$ equivalnet with
$ x^2\plus{}y^2\equal{}\frac{13}{6}xy$
plugging $ y\equal{}5\minus{}x$ in the equation above we get
$ x^2\plus{}(5\minus{}x)^2\equal{}\frac{13}{6}x(5\minus{}x)$
simplifying and factoring this equation we get $ (x\minus{}2)(x\minus{}3)\equal{}0$ and from here we get the possible solutions of our system.
Sonnhard.


induction
linear algebra
matrix
LaTeX
function

There once was a man named Herbert. Now Herbert isn't necessarily a [b]good[/b] mathmatician, but he's moderately... well, he kinda sucks actually. So once, he came up with a couple of proofs. But his proofs are usually unstable, incomplete, or just plain nonsense. Now here is the challenge: exploit his errors, pinpoint them, explain how he might have gotten there, and give an alternative bridge to the conclusion. Now he has OCD, which causes him to understand ONLY proof by induction, but we want to help Herbert, so all suggestions [b]MUST[/b] be inductive. WARNING: NOT ALL PROOFS ARE BAD!!!!!!!!!! Each one, if all criteria are met, is worth one point. Most points wins a special suprise!

Proof #1: $\sum_{x=1}^{n}x=\frac{n}{2}(n+1)$
Well, in case $n=1$, we have $1=\frac{1}{2}(1+1)=\frac{2}{2}=1$. $\sqrt{}$
Now let's assume that: $\sum_{x=1}^{k}x=\frac{k}{2}(k+1)$.
So IF $k+1$ worked, then it would be $\frac{k+1}{2}(k+2)$, so that's what we want.
So we add $k+1$ to both sides, so it would look like $\frac{k}{2}(k+1)+(n+1)$.
ITWORKS =O. $Q.E.D.$ (I know, it's a lame one to catch.)

Next couple of faulty proofs coming soon! (cuz I'm too tired to show his to you now >D.)[/b]

[quote="GlenthemanN"]I believe the real answer to the first would be something like:
[hide]
$\sum_{x=1}^{n}x=\frac{n}{2}(n+1)= 1+2+3+\ldots+n$

Case1: $n=1$
$\frac{n}{2}(n+1) = \frac{1}{2}(2) = 1$
which is true,
now assume for some k
$p(k) = 1+2+3+\ldots+k = \frac{k}{2}(k+1)$
and for k + 1
$p(k+1) = 1+2+4+\ldots+(k+1) = \frac{k+1}{2}(k+2)$

now assuming for $k$ is true, it must be shown that $k+1$ is also true

adding $k+1$ to each side of $p(k)$
$1+2+3+\ldots+k+(k+1) = \frac{k}{2}(k+1)+(k+1)$
taking the right side we have
$\frac{k}{2}(k+1)+(k+1) = (k+1)(\frac{k}{2}+1) = (k+1)(\frac{k+2}{2}) = \frac{(k+1)(k+2)}{2}$

This is of the form $\frac{n}{2}(n+1)$ when $n = k+1$. $p(k)$ implies $p(k+1)$. Since it is true for $n = 1$, and $n = k$ and well as $n = k+1$ then by induction the function is true.
[/hide][/quote]

I think you should have defined $p(n)$ first such that $p(n)=\sum_{x=1}^{n}x=\frac{n}{2}(n+1)= 1+2+3+\ldots+n$

And also:
[quote]and for k + 1
$p(k+1) = 1+2+4+\ldots+(k+1) = \frac{k+1}{2}(k+2)$
[/quote]
No, you can't assume that. You should just delete that part since it's kinda pointless.

dont you have to assume that it is true as well, and then prove it is true based on induction from p(k)?

[quote="kohjhsd"]I learned that you derive it from the case $n=k$... it depends on your math teacher :lol:[/quote]
My math teacher would usually say to state it as what you want to get, not an assumption. But oh well, we can all figure out how to fix that according to our teacher...
Proof#2:$1^{2}+2^{2}+\cdots+n^{2}=\frac{n(n+1)(2n+1)}{6}$.
Assume: $\sum_{i=1}^{k}i^{2}=\frac{k(k+1)(2k+1)}{6}$.
Then $1^{2}+2^{2}+\cdots+k^{2}+(k+1)=\frac{k(k+1)(2k+1)}{6}+k^{2}+1$.
Now, we want to get $\frac{(k+1)(k+2)(2k+3)}{6}$,
so $\frac{k(k+1)(2k+1)}{6}+k^{2}+1\\=\frac{(k+1)(2k^{2}+5k+6)}{6}\\=\frac{(k+1)(k+2)(2k+3)}{6}$. $\sqrt{}$
Therefore, $1^{2}+2^{2}+\cdots+n^{2}=\frac{n(n+1)(2n+1)}{6}$. $Q.E.D.$

[quote="bigboy82892"]Proof#3:$n<2^{n}$ whenever $n$ is a positive integer.
$1<2^{1}$. $\sqrt{}$
Assume:$k<2^{k}$.
Then, $k+1<2^{k}+1\le2(2^{k})=2^{k+1}$. $\sqrt{}$
Therefore, $n<2^{n}$. $Q.E.D.$

Score:
GlenthemanN:$\frac{27}{29}$
kohjhsd:$\frac{2}{29}$(for that extra peice. :P )[/quote]

This one isn't too bad, though it definitely could have been written up more nicely.

Base case:  $1<2^{1}$, which is trivially so.

Assume that for $n=k$,

$k<2^{k}$

Then, we have $k+1<2^{k}$ and because $2^{k}>0$, we have $2^{k}<2^{k+1}$, so combining this, we have:

$k+1<2^{k+1}$

Therefore, by mathematical induction, $n<2^{n}$, whenever $n$ is a positive integer.

Proof #5: If $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$, then $A^{n}=\begin{bmatrix}1&n\\0&1\end{bmatrix}$.
$A^{1}=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. $\sqrt{}$
Assume:$A^{k}=\begin{bmatrix}1&k\\0&1\end{bmatrix}$.
We want:$A^{k+1}=\begin{bmatrix}1&k+1\\0&1\end{bmatrix}$.
So,$A^{k}\times A=A^{k+1}=\begin{bmatrix}1&k\\0&1\end{bmatrix}\times\begin{bmatrix}1&1\\0&1\end{bmatrix}$,
$A^{k+1}=\begin{bmatrix}1(1)+0(k)&1(1)+1(k)\\1(0)+0(1)&0(1)+1(1)\end{bmatrix}$,
$A^{k+1}=\begin{bmatrix}1&k+1\\0&1\end{bmatrix}$. $\sqrt{}$
$\therefore A^{n}=\begin{bmatrix}1&n\\0&1\end{bmatrix}$. $Q.E.D.$

Interestingly, the powers of $B = \begin{bmatrix}1&1\\1&0\\ \end{bmatrix}$ follow a very different pattern.  Can you conjecture and prove it, and then prove an interesting identity?

Edit:  [quote="b-flat"]Assume that for $n=k$,

$k<2^{k}$

Then, we have $k+1<2^{k}$[/quote]

This is actually less valid than the original proof.

[b]bigboy82892:[/b]  Is Proof #5 supposed to be invalid?  It looks fine to me, aside from the minor error in $\LaTeX$ in line 5.

[quote="t0rajir0u"]Interestingly, the powers of $B = \begin{bmatrix}1&1\\1&0\\ \end{bmatrix}$ follow a very different pattern.  Can you conjecture and prove it, and then prove an interesting identity?[/quote]

I conjecture
[hide]For $X=\begin{bmatrix}1&1\\1&0\\ \end{bmatrix}$, $X^{k}= \begin{bmatrix}F_{k}&F_{k-1}\\1&0\\ \end{bmatrix}$
when $F_{0}=F_{1}=1$.[/hide]

Not quite.  Try more numerical examples.

Then I think it is[hide]
$X^{k}= \begin{bmatrix}F_{k}&F_{k-1}\\ F_{k-1}&F_{k-2}\\ \end{bmatrix}$ if $F_{-1}=0$...
which I could prove by induction most likely[/hide]
I forgot all about matrices and I can't even multiply right :maybe:

what is wrong with this proof: [for real, a teacher at my school did this]

lets prove that 1+2+...+n=n(n+1)/2

ok so base case...n=1...1=(1)(1+1)/2 check

ok write out the k+1 case

1+2+...+k+1=(k)(k+1)/2
multiply by 2

2+4+...+(2k+2)=(k)(k+1)

divide by 2

1+2+...+(k+1)=(k)(k+1)/2

therefore the statement for n=k implies the statement for n=k+1, and we are done

[i am totally serious about this too...she really did write this on the board...and half the class was like ok...and i was just like WTF you don't know induction]

[quote="t0rajir0u"]The standard is to let $F_{0}= 0, F_{1}= 1$.  Now, what about the second part?  What interesting identity immediately results?

[hide="Hint"] Think determinants. [/hide][/quote]

[hide]Okay, then I will change the expression to
$X^{k}= \begin{bmatrix}F_{k+1}&F_{k}\\ F_{k}&F_{k-1}\\ \end{bmatrix}$
and its determinant is $F_{k+1}F_{k-1}-F_{k}^{2}$. The determinant of $X$ is $-1$, so I guess $F_{k+1}F_{k-1}-F_{k}^{2}= (-1)^{k}$.[/hide]

[quote="Altheman"]ok write out the k+1 case
....
therefore the statement for n=k implies the statement for n=k+1, and we are done
[/quote] Well there was no showing of $n=k$... $\times 2 \div 2$ does not prove anything... I don't think any teacher would really do that.

1) Does anyone have any ideas about any errors with #'s 2&4?
2)[quote="t0rarjir0u"]
Edit:  [quote="b-flat"]Assume that for $n=k$,

$k<2^{k}$

Then, we have $k+1<2^{k}$[/quote]

This is actually less valid than the original proof.[/quote]
He's right, cuz if you tried that to $k=1$, then you would have $1+1<2^{1}$, which is obviously false. So until someone can find the problems with that, no one gets points for that.
3)Good job catching the one error in the proof; man, gotta make these errors more invisible...
4)GlenthemanN: $\frac{27}{29}$
kohjhsd:$\frac{2}{27}$
t0rarjir0u:$1!!!!!!!!!!$;)

[quote="Anaxerzia"][quote="bigboy82892"]
Then, $(k+1)!=k!(k+1)=k!+k^{2}(k-1)$,
[/quote]

How does $k!(k+1) = k!+k^{2}(k-1)$ ?

That seems to be the error...[/quote]
Always remember your basic algebraic mathematical properties: Refliexive, Associative, Commutative, ___________, etc.:P

EDIT: Oh, I see, heh, I guess I'm worse than Herbert, :P. Assume it says $k!+k^{2}(k-1)!$.;)

Distributive?

Even through that I got $k!+k^{2}(k-1)!$ ...

Was it a latex error?

edit: oh, lol..

Does anyone have any ideas for 2,3,4? Free points up for grabs!=gXrXaXbXs. :lol:

[quote="bigboy82892"]GlentheMAn:$\frac{27}{29}$
kohjokds:$\frac{2}{29}$
b-flat:1

Proof #4:$n^{2}<n!$ for $n\ge4$
Well, we can't start at 1, so we start at 4, so...
$(4)^{2}=16<(4)!=24$. $\sqrt{}$
Assume: $k^{2}<k!$.
We want $(k+1)^{2}<(k+1)!$.
So... Since $k^{2}+2k+1=(k+1)^{2}$,
We have $(k+1)^{2}<k!+2k+1$.
Then, $(k+1)!=k!(k+1)=k!+k^{2}(k-1)$,
So $k!+2k+1?<?k!+k^{2}(k-1)$.
$2k+1?<?k^{2}(k-1)\\2k+1<k^{2}\\k^{2}<k^{2}(k+1)\\(k+1)^{2}<(k+1)!$.
$Q.E.D.$[/quote]

I still don't quite understand this. What do the question marks indicate? Was it another LaTeX error?

If they mean nothing than, it is mainly here I believe-

$k!+2k+1<k!+k^{2}(k-1)!$.

There is some chaos nonsense at the bottom that I can't seem to understand either-

$2k+1?<?k^{2}(k-1)\\2k+1<k^{2}\\k^{2}<k^{2}(k+1)\\(k+1)^{2}<(k+1)!$.
$Q.E.D.$

You're close... think about the order of steps... then fix it.

[quote="kohjhsd"]

[quote="Altheman"]ok write out the k+1 case
....
therefore the statement for n=k implies the statement for n=k+1, and we are done
[/quote] Well there was no showing of $n=k$... $\times 2 \div 2$ does not prove anything... I don't think any teacher would really do that.[/quote]

yes she did...she did not understand induction...sadly she has tenure...there have been many complaints to the department head, but nothing is really happening...

[quote="t0rajir0u"]The question mark is a tactic where you are manipulating two sides of an equation to determine which side is greater.  It's not rigorous, and should be done in the other direction.[/quote]
So basically, you're saying that herbert is correct on #4 except he made a LaTeX error and should reverse the ?? steps?

[quote="t0rajir0u"]It's not a $\LaTeX$ error.  That particular tactic is an example of working backwards from a conclusion to a premise, when of course the order of the steps should be the other way around.  It's not logically sound.[/quote]
I mean on the k^{2}(k+1)[!] thingy.

[quote="bigboy82892"]
Then $1^{2}+2^{2}+\cdots+k^{2}+(k+1)=\frac{k(k+1)(2k+1)}{6}+k^{2}+1$.
[/quote]

To find the $k+1 = n$ case, you would say on the LHS $(k+1)^{2}$ and on the RHS $k^{2}+2x+1$, or just $(k+1)^{2}$ (as I would write it...)

I guess that's the error for #3, what else do you want me to do with it?


Assume $A=\{a_{1},a_{2},...,a_{12}\}$ is a set of positive integers such that for each positive integer $n \leq 2500$  there is a subset $S$ of $A$ whose sum of elements is $n$. If $a_{1}<a_{2}<...<a_{12}$ , what is the smallest possible value of $a_{1}$?

[quote="N.T.TUAN"][quote="aviateurpilot"]we have $1\le 2500$, then there is subset $S$ of $A$ whose sum of elements is $1$
so, $card(S)=1$   =>   $S=\{a_{1}\}$    => $a_{1}=1$[/quote]
Are you sure ? :D[/quote]
we have $a_{1}<a_{2}<..<a_{12}$, then  $min(\{sum\ of\ elements \ of\ S/ \ S\in P(A)\})=a_{1}$
then there isn't subset $S\in P(A)$ whose sum of elements is n, if $n< a_{1}$
then, $a_{1}=0$ :huh:

[quote="aviateurpilot"]
we have $a_{1}<a_{2}<..<a_{12}$, then  $min(\{sum\ of\ elements \ of\ S/ \ S\in P(A)\})=a_{1}$
then there isn't subset $S\in P(A)$ whose sum of elements is n, if $n< a_{1}$
then, $a_{1}=0$ :huh:[/quote]

Are you sure? again :D

I am null in English, wanted you to explain me the question ?  :blush:

read, please!  :wink:
[quote="N.T.TUAN"]each positive integer $n \leq 2500$  there is a subset $S$ of $A$ whose sum of elements is $n$. [/quote]

I agreee with aviateurpilot that $a_1=1$. Didn't you want to ask to $a_{12}$? Or didn't you want to say $a_1>a_2>\dots >a_{12}$?


symmetry
geometry
parallelogram
rectangle
rhombus
geometric transformation
rotation

[quote="qwertythecucumber"]Draw a figure that has rotational symmetry but not line symmetry.

This question appeared in my little brother's 5th grade math book. I don't think it's possible, and it may be a trick question, I want to be sure.

EDIT: Never mind. It could be a  :yinyang: .  :blush:[/quote]
 :blush: funny!!

I think it might be a trick question as u said

can  anyone explain me  what is  line symetry and what is rational symetry?

Just stopped by to recommend this article:
[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Questions.php[/url]

If you're missing something you feel is simple, there's no shame in learning.

[quote="Akashnil"]How about a parallelogram? (which is not a rectangle or a rhombus)

[hide="To b555:"]A figure is said to have [u]rotational symmetry[/u] if there exists a point in the same plane around which there exists a rotation which maps the figure to itself.

(rotations with angles which are integral multiples of $ 360^0$ are not applicable)

A figure is said to have [u]line symmetry[/u] if it coincides with itself upon reflection on a straight line on the same plane.[/hide][/quote]

Non-square rectangles also work, as well as the letter 'S'.

[quote="dragon96"]Non-square rectangles also work, as well as the letter 'S'.[/quote]
I thought all rectangles have line symmetry?


inequalities
IMO Shortlist
inequalities unsolved

[quote="Kirill"]Prove that if $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ac \equal{} 1$, then $ ^3\sqrt {\frac {1}{a} \plus{} 6b} \plus{} ^3\sqrt {\frac {1}{b} \plus{} 6c} \plus{} ^3\sqrt {\frac {1}{c} \plus{} 6a}\leq \frac {1}{abc}$.[/quote]

 $ LHS^3 \equal{} (\sqrt [3]{\frac {1}{a}(1 \plus{} 6ab)} \plus{} \sqrt [3]{\frac {1}{b}(1 \plus{} 6bc)} \plus{} \sqrt [3]{\frac {1}{c}(1 \plus{} 6ca)})^3$

$ \rightarrow LHS^3 \leq 3(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})[3 \plus{} 6(ab \plus{} bc \plus{} ca)]$

$ \equal{} 3[\frac {ab \plus{} bc \plus{} ca}{abc}][3 \plus{} 6(ab \plus{} bc \plus{} ca)] \equal{} \frac {27}{abc}$

$ \rightarrow LHS \ge\ \sqrt [3]{\frac {27}{abc} } \le\ \frac {1}{abc}$

We have done!

[quote="Kirill"]Prove that if $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ac \equal{} 1$, then $ ^3\sqrt {\frac {1}{a} \plus{} 6b} \plus{} ^3\sqrt {\frac {1}{b} \plus{} 6c} \plus{} ^3\sqrt {\frac {1}{c} \plus{} 6a}\leq \frac {1}{abc}$.[/quote]
It is not new Kirill,This is a problem of Imo shortlist 2004

Yes! It's here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30865

[quote="Allnames"]
It is not new Kirill,This is a problem of Imo shortlist 2004[/quote]
Sorry, "New equality" - I was sure that in this forum this inequality is new, but I was wrong. And I didn't knew that this problem from IMO 2004 shortlist.


calculus
integration
trigonometry
logarithms
calculus computations

$ \boxed1$ $ \int {\frac{{1 \plus{} x}}
{{x(1 \plus{} xe^x )^2 }}} dx$

$ \boxed2$ $ \int {\frac{{x \minus{} 1}}
{{(x \plus{} 1)^3 }}e^x } dx$

$ \boxed3$ $ \int {\sqrt {\frac{{1 \minus{} \sqrt x }}
{{1 \plus{} \sqrt x }}} } dx$

$ \boxed4$ $ \int {\frac{1}
{x}\sqrt {\frac{{1 \minus{} \sqrt x }}
{{1 \plus{} \sqrt x }}} } dx$

$ \boxed5$ $ \int {\left( {\sqrt {1 \plus{} \sin \frac{x}
{2}} } \right)} dx$

[hide="3"]
$ \int{\sqrt{\frac{{1\minus{}\sqrt x }}{{1\plus{}\sqrt x }}}}dx$
$ \equal{}\int \frac{1\minus{}\sqrt{x}}{\sqrt{1\minus{}x}} dx$
put $ x\equal{}cos^{2}(\theta)$
$ \equal{}\int (\minus{}cos(\theta) \plus{}1\plus{}cos(2\theta)) d(\theta)$
[/hide]

for 4 : put $ x \equal{}cos^2(\theta)$

[hide="5"]
$ \int{\left({\sqrt{1\plus{}\sin\frac{x}{2}}}\right)}dx$
u just need to realize $ 1 \plus{} sin(2*\frac{x}{4}) \equal{} (sin(\frac{x}{4})\plus{}cos(\frac{x}{4})^2$

[/hide]

(4)

Let $ \sqrt{\frac{1\minus{}\sqrt{x}}{1\plus{}\sqrt{x}}}\equal{}u$

$ \int \frac{1}{x}\sqrt{\frac{1\minus{}\sqrt{x}}{1\plus{}\sqrt{x}}}dx$

$ \equal{}\allowbreak \int \,\left( \frac{2}{u\plus{}1}\minus{}\frac{2}{u\minus{}1}\minus{}\frac{4}{1\plus{}u^{2}}\minus{}8u^{2}\right) du\allowbreak$

$ \equal{}\allowbreak \minus{}\frac{8}{3}\left( \sqrt{\frac{1\minus{}\sqrt{x}}{1\plus{}\sqrt{x}}}\right) ^{3}\minus{}2\ln \left( \sqrt{\frac{1\minus{}\sqrt{x}}{1\plus{}\sqrt{x}}}\minus{}1\right) \plus{}2\ln \left( \sqrt{\frac{1\minus{}\sqrt{x}}{1\plus{}\sqrt{x}}}\plus{}1\right) \minus{}4\tan ^{\minus{}1}\sqrt{\frac{1\minus{}\sqrt{x}}{1\plus{}\sqrt{x}}}\plus{}C$


Hello,

I've checked out most of the Mock AIME's, and I was wondering how you prepare for them..

Some of the problems I don't even understand, not mentioning the problems, some of the SOLUTIONS are even hard to follow...

So how do you guys do it? lol..

Thanks in advance

Well for the mock AIMEs:

1) Sometimes they're harder than normal AIMES.
2) Sometimes they're not written clearly.
3) Sometimes the solutions people write are unclear.

so if you want a better idea about the real AIME (that's what you're asking about, right?), you should check out the [url=http://www.artofproblemsolving.com/Forum/resources.php]resources[/url] section and look at some past AIME problems. Or, better yet, you could order some old tests (which are not too expensive, but you'll have to weigh your options here) with the exact wording and fairly thorough solutions.

Now if you still find those problems intimidating, you should 1) not be afraid to read some solutions, 2) learn the basic material that is covered if you haven't already, and 3) try some easier problems first and work your way up from there.

Oh, BTW, why'd you start preparing already!? AIME is tomorrow! You don't need to start preparing yet... \end{sarcasm} It's a little too late to do serious preparation for the AIME this year, but you can start preparing for next year. ;)

yeaaa haha im a bit late on that im going to try out for it next year given that i'll make AMC again.. i think i scored 125 this year im not sure..

thanks guys for your thoughts... it just seems like a really big jump from easy AMC questions and fermat level questions to AIME... is there certain ways to approach a problem? Some of the solutions it seems weird to even start it that way (probably because I just don't know what it's really trying to ask)

Eat a good breakfast before taking the AIM Exam. Eg. 2 boiled eggs 3 fried eggs. 1 fried rice. Some chicken beef. Preserved Veggies. IN short doen't eat too much what I put down here will be fine


http://www.ebaumsworld.com/marioguitar.html

I want to hear your response.

http://www.ebaumsworld.com is a great site!

I loved that old Mario game, lol.  BTW, have you watched End of the World on that site?  Funniest thing ever.

ebaumsworld is awesome... yeah i discovered End of the World like a year ago watched it too many times to count... "..and australia is just like WTF m8.."


probability
pifinity is ze best
g1zq is even better

Nine people sit down at random seats around a round table. Four of them are math majors, three are physics majors, and two are biology majors. What is the probability that all four math majors sit in consecutive seats?

Let's say those 4 people are just 1 person. That would give us 8 people in total. There are $ 7!$ ways or 5040 to arrange those people. There are 8! total ways, so the answer is $ \frac{1}{8}$.

If the four people are just 1 person, there would be 6 people.

there are $ 8!$ divided by $ 2! * 3! * 4!$ ways to arrange them. So the denominator is $ 140$. Then we make the math majors one person and we have $ 6$ people (1 big math major  :D 3 physics 2 bio) so the way we can arrange them is $ \frac{5!}{2!*3!} \equal{} 10 \implies \boxed{\frac{10}{140}}$ which is just $ \frac {1}{14}.$
QED

Maybe I forgot my basics or something, but I'm not sure how there are $\binom{9}{4}$ ways total, not $9!$...that's what I did

[quote=Element15-Phosphorus][hide="Alcumus Solution"]
[quote=Alcumus Solution]There are $\binom{9}{4}=126$ ways to choose the seats for the four math majors. Of these ways, there are only 9 where the four math majors sit in consecutive seats. Therefore, the probability that the math majors sit is consecutive seats in $\frac{9}{126}=\boxed{\frac{1}{14}}$.[/quote]
[/hide]
But I'm still confused as in why there are 9 ways for the consecutive seats though.[/quote]

I think that's because there are 9 choices of places to sit the math majors can use.

Does anybody have a solution for this problem?

what don't you understand do you not know the choose function

No, I understand that, but I am not sure exactly [i][b]why[/b][/i] there are 9 choices for the consecutive seats.

I understand why the solution is correct but I don't understand why mine isn't
[hide=inccorectttt soloootion]
pick one mathematician
call his chair the special chair and look at everything relative to him
there are 8! ways to arrange the people around him by going to the right of him

now how many ways are there to get math men to sit next to him
3 could be to the right of him
2 could be to the right of him
1 could be to the right of him
0 could be to the right of him
within these possibilities there are 3! ways to arrange the rest of the mathmeticans
so the answer is $\frac{4*3!}{8!}=\frac{4!}{8!}=\frac{1}{8*7*6*5}=\frac{1}{1680}$
[/hide]

I think what's wrong with your solution is that the people with the same majors are not distinguishable. So, there would not be $8!$ different ways to place the rest of the people (though fixing one at the top is correct). Instead, since there are $8$ seats available and $3$ math majors, there are $\binom{8}{3}$ ways to place the other math majors. Then, there are $5$ seats left and $\binom{5}{2}$ ways to place the biology majors and the physics majors just go in the remaining seats. Note that you could place the physics majors first too since $\binom{5}{2} = \binom{5}{3}.$ So, the total number of ways would be $\binom{8}{3} \cdot \binom{5}{2}.$

[hide]Treat the 4 mathematicians as a single person, we therefore have 9 different ways of seating them consecutively. Amongst the mathematicians, there are 4! ways of arranging themselves, and once they are arranged, we have 5! ways of arranging the other majors.

In total, they can be arranged 9! ways,
So the desired probability is (9*4!*5!)/9! = (4!*5!)/8! = 1/14. :)[/hide]


inequalities
function
inequalities unsolved

Let $a\ge b\ge c\ge 0$ and $0\le x\le y\le z$. Prove that \[\frac{a}{1+x}+\frac{b}{1+y}+\frac{c}{1+z}\ge \frac{a+b+c}{1+\sqrt[3]{xyz}}\]
[hide="a question"]is the function $f(x)=\frac{x}{x+1}$ convex? sorry but i don't know much calc so can't check :blush: . but i tried drawing a graph and it seemed to be concave.[/hide]

It's wrong, consider $a=b=c$ and $x=0$.

Your conditions imply
\[\frac{a}{1+x}+\frac{b}{1+y}+\frac{c}{1+z}\geq \frac{1}{3}(a+b+c) \left( \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\]
by Chebyshev. This gets a factor of $a+b+c$ and attains equality when $a=b=c$ no matter what $x,y,z$ are, so we only really need to worry about $\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$. However, Jensening this to prove
\[\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\geq \frac{3}{1+\sqrt[3]{xyz}}\]
depends not on $\frac{1}{1+t}$ but $\frac{1}{1+e^{t}}$. This is concave up for $t\geq 0$ and concave down for $t\leq 0$, so the above will be true if you change the condition to $1\leq x\leq y\leq z$.

[quote="nayel"]thank you all very much for helping me out. :lol: but my latter question was unanswered. i'll be really greatful if someone answers that.[/quote]

The function $f(x) = \frac x{x+1}$ is concave, which follows from  $f'' \le 0$.

I don`t know yet derivates so I will give a solution at my level.
chose a>b then f((a+b)/2) > (f(a)+f(b))/2 which after some computations gives (a-b)^2>0 which is obviously true. so f is concave


logarithms
search
integration
calculus
calculus computations

Could anyone help me with these ones:

Show that the series

$\sum_{=2}^{\infty}{\frac{1}{n\log{n}}}$ is divergent.

And

$\sum_{n=2}^{\infty}{\frac{1}{n(\log{n})^r}}$ is convergent (where $r>1$).

$\bf Lema$ If $a_n>0$ then $\sum_{k=1}^{\infty} a_k <\infty \Leftrightarrow \sum_{k=1}^{\infty} 2^k a_{2^k}<\infty$

Just a small (but necessary) correction to anik_andrew's post: it should be "if $a_n$ is a [b]decreasing[/b] sequence of positive numbers"... :)

Thanks fedja, i forget for this condition...

Or, with a closely related hypothesis: suppose that for some $m>0,\,f(x)$ is positive and decreasing on $[m,\infty).$ Suppose that $a_k=f(k).$ Then

$\sum_{k=m}^{\infty}a_k$ converges iff $\int_m^{\infty}f(x)\,dx$ converges.

The condensation test (what anik_andrew presented) and the integral test (what I just presented) tend to work efficiently in the same cases, and indeed, both work efficiently for these examples.


equation
algebra
parametric equation
IMO
IMO 1963

Find all real roots of the equation \[ \sqrt{x^2-p}+2\sqrt{x^2-1}=x  \] where $p$ is a real parameter.

[quote="Jaswinder"]Let ${x^2=p+t^2}$

so after transforming the given condition and some manipulations we arrive at

${|x|=|p-4|/\sqrt(8(2-p))}$

with condition that p<2[/quote]

Little correction : $x=\frac{|p-4|}{\sqrt{8(2-p)}}$ and not $|x|=\frac{|p-4|}{\sqrt{8(2-p)}}$ : from the original equation, we get $x>0$ :)

In fact the solution is valid iff $0\le p\le \frac{4}{3}.$
This is problem 1 from the fifth IMO :)
(We have $x^2\ge p$ and $2\sqrt{x^2-1}\le x$ which leads to $x^2\le \frac43$.
It is no difficult to check that for these values of $p$ the solution is valid.)

You can see the 5 Solutions here^^

Post # 8, 9, 11 in
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=500396[/url]

[hide=Solution]Squaring both sides, we have $x^2-p+4(x^2-1)+4\sqrt{x^2-p}\sqrt{x^2-1}=x^2$, or $4x^2-4-p=-4\sqrt{x^2-p}\sqrt{x^2-1}$. Square both sides and plug in a variable for $x^2$ when necessary. After this the result is $\boxed{x = \pm \frac{a-4}{2\sqrt2\sqrt{2-a}}}$.[/hide]

Whew!

Because of the radicals, we will naturally have several restrictions on $x$ and $p$. The first few are $x^2\geq p$ $(1)$ since we must have $\sqrt{x^2-p}\in\mathbb{R}$, $x^2\geq1$ $(2)$ since we must have $\sqrt{x^2-1}\in\mathbb{R}$, and $x\geq0$ $(3)$ since we must have $\sqrt{x^2-p}+2\sqrt{x^2-1}\geq0$. Now, squaring the given equation,
$$(x^2-p)+4(x^2-1)+4\sqrt{(x^2-p)(x^2-1)}=5x^2-p-4+4\sqrt{x^4-(p+1)x^2+p}=x^2$$
or
$$4\sqrt{x^4-(p+1)x^2+p}=p+4-4x^2\iff\sqrt{x^4-(p+1)x^2+p}=\frac{p}{4}+1-x^2.$$
Then, we must have $\tfrac{p}{4}+1-x^2\geq0$ $(4)$, and assuming this, we can square to obtain
$$x^4-(p+1)x^2+p=x^4+\left(\frac{p}{4}+1\right)^2-\left(\frac{p}{2}+2\right)x^2$$
or
$$\left(1-\frac{p}{2}\right)x^2=\left(\frac{p}{4}-1\right)^2\iff x^2=\frac{\left(\frac{p}{4}-1\right)^2}{1-\frac{p}{2}},$$
where $p\neq2$; taking the square root, $x=|\tfrac{p}{4}-1|/\sqrt{1-\tfrac{p}{2}}=\tfrac{|p-4|}{2\sqrt{4-2p}}$, where we take the positive root because of $(3)$. Since $p<2$ in order to have $\sqrt{4-2p}\in\mathbb{R}/\{0\}$, we have $\underline{x=\frac{4-p}{2\sqrt{4-2p}}}$. Finally, we check criteria $(1)$, $(2)$, and $(4)$. For $(4)$,
$$\frac{p}{4}+1-\left(\frac{4-p}{2\sqrt{4-2p}}\right)^2=\frac{p}{4}+1-\frac{(p-4)^2}{16-8p}=\frac{4p-2p^2}{16-8p}+\frac{16-8p}{16-8p}-\frac{p^2-8p+16}{16-8p}=\frac{-3p^2+4p}{16-8p}\geq0$$
or $\tfrac{p(3p-4)}{2-p}\leq0\iff\underline{p\in[0,\tfrac{4}{3}]}$; for $(2)$,
$$\left(\frac{4-p}{2\sqrt{4-2p}}\right)^2=\frac{p^2-8p+16}{16-8p}\geq1\iff p^2-8p+16\geq16-8p\geq p^2\geq0,$$
always true because $p\in\mathbb{R}$; and for $(1)$,
$$\left(\frac{4-p}{2\sqrt{4-2p}}\right)^2=\frac{p^2-8p+16}{16-8p}\geq p\iff p^2-8p+16\geq16p-8p^2\iff9p^2-24p+16=(3p-4)^2\geq0,$$
again always true. Henceforth, taking the restriction from $(4)$, we have a requested answer of
$$\boxed{x=\frac{4-p}{2\sqrt{4-2p}}\text{ if }p\in[0,\tfrac{4}{3}],\text{ no solutions otherwise}}.$$
$\blacksquare$


A 420-page book contains an average of 600 words per page, and Roslyn read the book at the rate of 360 words per minute. How many hours did it take her to read the book? Express your answer as a mixed number.


Slove equation:
$ 4\cos x\cos\frac {x}{2}\cos\frac {5x}{2} \equal{} \sin\frac {3x}{2}$

[color=green]LaTeX fixed by mod.[/color]

[hide="A Start?"]
Repeatedly using the fact that $ \cos\alpha\cos\beta\equal{}\frac{\cos(\alpha\plus{}\beta)\plus{}\cos(\alpha\minus{}\beta)}2$, we get

$ \cos x\plus{}\cos2x\plus{}\cos3x\plus{}\cos4x\equal{}\sin\frac{3x}2$.

I suck at trig, though... :(
[/hide]

Multiply both sides with $ 2\sin\frac{x}{2}$ ?


In bonds and overlap, do all nodal planes have to be mutually perpendicular?

Don't you mean $ \sigma$ and $ \pi$ bonds?

I'd think not.  Think about it: since space is 3D, if all planes had to be perpendicular you could have at most 3 nodal planes.

And $ \phi$ and $ \gamma$ bonds are actually valid terms for higher level bonds (where you start messing with $ f$ and $ g$ orbitals).  I don't think they actually occur in nature though.


trigonometry
algebra
polynomial
Vieta
algebra unsolved

At least it can be proved this way :

$\tan (\pi - x) = - \tan x$ so if we can compute $\prod_{i=1}^{2n}\tan i\theta$ we are done.

$(\cos \theta + i \sin \theta)^{2n+1} = 1$
Expanding with binomial, dividing by $(cos \theta)^{2n+1}$, we obtain an equation whose every roots are the $\tan \theta$.
Using Vièta relations gives $\prod_{i=1}^{2n}\tan i\theta = \binom{2n+1}{1} = 2n+1$

can you further explain how binomial expansion and vieta formula work here ? I cant seem to get you  :blush:

[quote="shyong"]If $\theta=\frac{\pi}{2n+1}$ , $n\in\mathbb{N}$ , prove or disprove that

$\prod_{i=1}^n\tan i\theta=\sqrt{2n+1}$[/quote]

I would like to give you the source, where you can find the information required to solve this problem.

You can refer to the book
"Selected Problems and Theorems in Elementary Mathematics" by
D. O. Shklyarsky
N. N. Chentsov
I. M. Yaglom

Mir Publishers, Moscow.

In fact, This problem is solved in the book.


vertices of equilateral triangle is located distance 1, 2, 3 from origin. Find its side length. (any clever solutions?)

Let an equilateral triangle ABC in the plane with origin O such that OA=1,OB=2,OC=3. Find the length of AB.
Solution: For any point M in that plane, $ MA\plus{}MB\ge MC$. The quality holds iff M lies on the minor arc AB of the circumcircle (ABC).
This implies that O lies on the minor AB of circle (ABC). So, $ AB^2\equal{}MA^2\plus{}MB^2\plus{}MA.MB\equal{}7\Rightarrow AB\equal{}\sqrt{7}$


geometry
circumcircle
absolute value
geometry solved

Given a triangle $ABC$ with circumradius $R$, and circumcenter $O$. Let $Q$ be a point inside the triangle. Drop perpendiculars from $Q$ to the three sides $BC$, $CA$, $AB$, and call their feet $D$, $E$, $F$, respectively. Prove that
\[\text{Area}\left(DEF\right)=\frac{R^{2}-\left|OQ\right|^{2}}{4R^{2}}\text{Area}\left(ABC\right) \]

Since R is the radius and O is the center of the circumcircle of triangle ABC, the expression $R^2-\left|OQ\right|^2$ is the absolute value of the power of the point Q with respect to the circumcircle of triangle ABC, and hence, your formula is equivalent to the formula $T=\frac{Sp}{4R^2}$ from http://www.mathlinks.ro/Forum/viewtopic.php?t=5658 post #2, to which I gave a proof in post #14.

  darij


floor function
combinatorics proposed
combinatorics

I have solved the problem and I am interested in second opinion(s).

[u]problem[/u] Any two elements of the finite set $ S\equal{}\{1,\dots,n\}$ are either related or unrelated, we write $ a\sim b$ if a relates to b, $ a,b\in S$. $ R$ has the following properties:
(1) $ a\sim b \Leftrightarrow b\sim a$ (Reflexivity)
(2) $ a\sim b \quad and \quad b\sim c \Rightarrow a\sim c$ (Transitivity)

A relation $ R$ on the set $ S$ is written as $ (S,R)$,the set of the unordered (due to refelxivity) pairs.
example: $ \{(1;2),(2;3),(5;6)\}$
The above example is extend able to $ \{(1;2),(2;3),(5;6),(1;3)\}$ by transitivity. Find the maximum cardinality of a non extend able $ (S,R)$.

[hide]$ \lfloor \frac{n}{2} \rfloor \plus{} \lfloor \frac{n}{6} \rfloor$ in a few cases.

$ \lfloor \frac{n}{2} \rfloor \plus{} \lfloor \frac{n}{6} \rfloor \plus{} 1$ in rest of the cases.

[/hide]

What do you mean by a non-extendable set?
Isn't the complete set, of all such pairs $ (i,j)$ not extendable, since you can't add anything to it?

If you mean that the set does not have a triple of sets: $ (i,j), (j,k), (i,j)$, then i get a different answer: Draw a graph with vertices labelled $ (1,2 .. n)$ and connect the two elements $ i,j$ if they are related. Each vertex has degree at most $ 1$ (if some vertex has degree $ 2$, i.e. then $ (i,j)$ and $ (i,k)$ are related, and then $ (j,k)$ are related giving a bad triple of subsets) - so this means the number of edges, which is the number of sets, is at most $ \frac {n}{2}$.

What am I missing?

Yes ,I meant we cannot have triples of sort $ (i;j),(j;k),(i;k)$ and non extendible.

Consider three points $ i,j,k$. We can have atmost one pair connected(related), so we get $ \lfloor \frac{n}{3} \rfloor$. Say we have $ i$ and $ j$ connected and $ k$ is not connected to $ i$ and $ j$, now connect all such $ k$'s pairwise from each possible triples.That gives $ \lfloor \frac {\lfloor\frac{n}{3} \rfloor}{2} \rfloor\equal{}\lfloor \frac{n}{6} \rfloor$. plus 1 gets added in certain cases depending upon divisibity by 3 and 6.


Alice and Bob are playing a game. Alice starts first. On Alice's turn, she flips a coin. If she gets a heads, she wins. If not, it becomes Bob's turn. On Bob's turn, he flips a coin. If he gets a tails, he wins. If not, it becomes Alice's turn. What is the probability that Alice wins the game?

This would be equal to:

$ \frac12 \plus{} \left(\frac12 \times \frac12 \times \frac12 \right) \plus{} \left(\frac12 \times \frac12 \times \frac12 \times \frac12 \times \frac12 \right) \plus{} \left(\frac{1}{2^5} \times \frac12 \times \frac12 \right)  \plus{} ... \implies$

$ \frac12 \plus{} \frac1{2^3} \plus{} \frac1{2^5} \plus{} \frac1{2^7} \plus{} ...$

The common ratio is $ \frac14$.

Let's say that the sum of this series is $ s$.

$ s \equal{} \frac12 \plus{} \frac1{2^3} \plus{} \frac1{2^5} \plus{} \frac1{2^7} \plus{} ... \implies$

$ \frac{s}{4} \equal{} \frac1{2^3} \plus{} \frac1{2^5} \plus{} \frac1{2^7} \plus{} \frac1{2^9} \plus{} ... \implies$

$ s \minus{} \frac{s}{4} \equal{} \frac12 \implies \frac{3s}{4} \equal{} \frac12 \implies 6s \equal{} 4 \implies s \equal{} \boxed{\frac23}$


Question - Find the probability that the sum of n dice (rolled) is a multiple of 7.

What do you mean by "sum of n dice is a multipleof 7"

He means that if you role n dice, the number of dots on the top of them will be a multiple of 7.


Prove that: for each integer $ a$ and two differrent primes $ p$ and $ q$ we have:
$ pq|a^{pq}\minus{}a^{p}\minus{}a^{q}\plus{}a$

It is obviosly $ p|x^{p}\minus{}x$, therefore $ p|(a^{q})^{p}\minus{}a^{q}$ and $ p|a^{pq}\minus{}a^{q}\minus{}(a^{p}\minus{}a)$. Same way $ q|f^{pq}\minus{}a^{q}\minus{}a^{p}\plus{}a$.

Here is more detailed solution to this verz simple problem.

I wiil prove that both $ p$ and $ q$ divide the $ a^{pq}-a^{p}-a^{q}+a$

If a has a multiple $ p$ or $ q$ then obviously $ a^{pq}-a^{p}-a^{q}+a$ is divisible by $ p$ (or $ q$) so it is enough to consider case when $ (a,p)=(a,q)=1$

We have: $ a^{pq}-a^{p}-a^{q}+a=a^{q}(a^{(p-1)q}-1)-a(a^{p-1}-1)=a^{q}(a^{p-1})^{q}-1)-a(a^{p-1}-1)$  =

= ${ a^{q}(1^{q}}-1)-a(1-1)=0$ ($ mod$ $ p)$

Last results follows from Fermat's Little theorem becouse it gives:
$ a^{p-1}=1$ ($ mod$ $ p)$.

In same way we prove that $ q$ divides $ a^{pq}-a^{p}-a^{q}+a$.


Show that $ 8x^{3}+y^{3}+z^{3}-6xyz\geq0$, with $ x$,$ y$, and $ z$ being nonnegative.

[hide="Hint"]
$ a^{3}+b^{3}+c^{3}-3abc=\frac12[a+b+c][(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$]
[/hide]

[hide="Suggestion"]
(Try proving it in more than one way. Like you could use AM-GM, Muirhead, Adam Hesterberg's factorization, etc.)
[/hide]

[quote="Lazarus"]Show that $ 8x^{3}+y^{3}+z^{3}-6xyz\geq0$, with $ x$,$ y$, and $ z$ being nonnegative.

[/quote]

By Am-Gm, \[ \frac{(2x)^{3}+y^{3}+z^{3}}{3}\geq (2x)(y)(z) \] :P

LIES I TELL YOU. LIES!

What is this bs about the "Adam Hesterburg" factorization? Among other places, it is in Math Olympiad Treasures [which was published before any of you MOP kids decided to give it to somebody.] Obviously this effort is a copy of SFFT...

*gets rated low post ratings from those who like spreading nonsense"

Geez, it's just a name...
Simon didn't invent his factoring trick...

Adam Hesterb[b]e[/b]rg credits it to Dorin Andreescu, who might credit it to somebody else. Sheesh, it's just a name.

And it's undeniably a useful factorization.

Eh.  Adam would probably just be amused that anyone bothered to name such a common factorization after him.


group theory
abstract algebra
induction
superior algebra
superior algebra theorems

Hi,

   I have a problem that i just can't figure out, and would appreciate any help or guidance someone can give. The problem is:

Let [i]G[/i] be a finite group with more than one element. Show that [i]G[/i] has an element of prime order.

I have just started studying groups, and so far in my book (Contemporary Abstract Algebra by Joseph Gallian) he has presented basic group properties, subgroups, centers of groups, order of groups, and subgroup tests. The problem above is from the book. I don't yet have Cauchy's theorem/etc...Thanks,

Patrick

Just consider an element $g\neq e$. Now we consider the subgroup generated by this element: $e,g,...,g^{k-1}$. Let $p$ be a prime which divides $k$. Then $g^{k/p}$ has obviously order $p$.

Misha

thanks, i guess I was making it too hard!

it is also easily follows from Cauchy theorem: for any prime $p$ dividing $|G|$ there exists an element of order $p$

Of course it is clear with Cauchy. But he wanted a solution without bigger theorems. I don't know how you prove Cauchy's theorem but our professor introduced it as a corollary to the Sylow theorems (after 6 weeks of group theory :) ).

Misha

Of course, Cauchy follows from Sylow's Theoirems, but here's a simple proof which I never forgot :).

Let $p$ be a prime dividing the order $n$ of the group $G$. Let $\mathcal S$ be the set of all products of $p$ elements which are equal to $1$ (the identity of $G$). Clearly, this set has $n^{p-1}$ elements, because we may freely select the first $p-1$ elements in a product from the $n$ elements of $G$, and then the $p$'th element is uniquely determined by the condition that all $p$ of them have product $1$. In particular, we get $p|n^{p-1}||\mathcal S|\ (*)$.

Now, let $\mathcal T$ be the set of those $x$ s.t. $x^p=1$. We have $1\in \mathcal T$, so $|T|\ge 1$. If $x_1x_2\ldots x_p\in\mathcal S\setminus\mathcal T$, then $x_{\sigma(1)}x_{\sigma(2)}\ldots x_{\sigma(p)}\in\mathcal S\setminus \mathcal T,\ \forall$ cyclic permutations $\sigma$. It's easy to see from here that we can partition $\mathcal S\setminus\mathcal T$ in sets of $p$ elements each, meaning that $p||\mathcal S\setminus\mathcal T|\ (**)$.

$(*)$ and $(**)$ show that $p||T|$, and, since $|T|\ge 1$, we find that there is at least a non-trivial $x\in G$ s.t. $x^p=1$.

I think it can be very easily prooved using a proper action of the group on a suitably chosen set. If someone's very interested in it i'll try to find/proove it myself.

Really easier than grobber did¿
Then I want to see it!

sorry, actually i think it's essentially the same argument and i don't consider it easier at all  :?

kgbto can you plz post the proof of Cauchy theorem using  group action.
I also know a proof of Cauchy theorem using class eqn.The proof is in Herstein.

well, it's actually the same argument as grobber's proof above or maybe slightly different: consider p being a prime divisor of |G| and then take the group Zp acting on the set X={(g1,g2,...,gp) in G^p | g1.g2.....gp = 1} by shifting:
k.(g1,g2,...,gp) = (g(k+1), g(k+2),...,gp,g1,g2,...,gk)  - all coefficients are taken modulo p of course (thus p fixes all elements of X). Obviously |X| = |G|^(p-1) is divisible by p, and X can be represented as the union of all different and non-intersecting orbits of this action, each one of them having number of elements, which is an index of Zp by the corresponding stabilizer, thus clearly 1 or p. The orbit containing (1,1,...,1) (which is an element of X ) is a singleton, so it contributes with a 1 in the equality : |X| = Sum |Ox|  taken over all orbits Ox. All the members of the sum in the right are equal to 1 or p and the whole sum is divisible by p. Clearly, there must be another one orbit which contributes with 1 (actually even p-1 more !), so there is an element (g1,g2,...,gp) in which however shifted stays the same. That means g1=g2=...=gp=g and g^p = 1. Eventually, obviously g is not the 1 of G, so it has an order of p.
the game, i haven't read Herstein so far but i know there is a proof using induction on the order of G. Actually, a well-known technique for proving some certain facts about groups. Is the proof you are talking above of this kind?

Yes you are absolutely right.Its on the induction on the order of group & after that using the class eqn.The proof is quite elegant


function
floor function
algebra proposed
algebra

A generalization of an easy question I saw today-

$ \displaystyle\sum_{i\equal{}1}^{n^{2}} \lfloor \frac{i}{t} \rfloor \equal{} \frac{(n^2)(n^2\minus{}(t\minus{}2))}{2t}$.

I think it can be generalized even further... Is it right for every natural t? And is there any elegant not-algebraic proof?

I think we'd prefer to write $ \sum_{i\equal{}1}^{n} \left\lfloor \frac{i}{t} \right\rfloor \equal{} \frac{n(n \minus{} (t\minus{}2))}{2t}$.  What do you mean by a "non-algebraic proof"?  As far as I can see, we can get a straightforward proof just by counting the number of terms with value $ k$ on the LHS.

This generalization is wrong - take t=3, n=5. But my identity *can* be strengthen.

4 days and no one pointed out that this is just plain wrong... what was I thinking? (and why didn't you all say something?)
when we plug n=1 we get: $ \lfloor \frac{1}{t} \rfloor \equal{} \frac{3\minus{}t}{2t}, 2t|3\minus{}t,$  $ t|3$ so t=$ \pm 1,\pm3$. t=-3,-1 don't work, and it's easy to show that the identity holds for t=3 (2 cases, $ 3|n^{2}$ and $ 3|n^{2}\minus{}1$).
if t=1, $ LHS\equal{}\sum_{i\equal{}1}^{n^{2}} i \equal{} RHS$. so, the identity is true iff t=1 or t=3. how sad.


my computer keeps locking when i am about to post and i have to copy the post stuff onto a file, click off the internet, then i have to load everything all over again. Can anyone help?  (My computer only has been locking during the past week, and about once every 5 posts) :(

Above is my original post before locking, my comp locked again and wouldn't let me post this very message :mad:

Could you be more explicit about what "locking" means?!  :? What computer do you use? (OS, Browser version, etc.)

I think he means either crash or a page called "page not displayed". It happens on mine too lately.

i mean that it freezes and will not go anywhere on AoPS unless i use the back button on my toolbar.  It still will go to other sites, but will not go back to AoPS unless i restart internet explorer :(

Well good news then ... if you act quickly you can get your free registration to Opera 8! :) http://my.opera.com/community/party/reg.dml


How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entires in each column is 0?

[hide="Solution"]Hmm......interesting problem.

Well, there are $\binom{4}{2}=6$ ways to place the 1's and the -1's in the top row. Now some casework for the second row:
[b]Case 1:[/b]
The numbers in the each column of the second row are the opposite of the numbers of the column in the first row. This can occur in only one way, and then the next row can be placed independently of the first, or in $\binom{4}{2}=6$ ways.
So this is a total of $6*1*6*1=36$ possibilities.
[b]Case 2:[/b]
The numbers in the each column of the second row are the same as the numbers of the column in the first row. This can occur in only one way, and the next two rows are uniquely determined by the arrangement in the first two rows.
So this is a total of $6*1*1*1=6$ possibilities.
[b]Case 3:[/b]
Under one of the 1's in the first row, there is a 1, under one of the -1's in the first row, there is a -1. Under the other 1, there is a -1. Under the other -1, there is a 1.
So this can occur in 4 ways (2 ways for 1, 2 ways for -1), and in the next row, two of the spots are uniquely determined, but the other two aren't (since the sum of the column above them so far is 0 ). So there are 2 ways to fill these two spots, and the last row is uniquely determined.
So this is a total of $6*4*2*1=48$ possibilities.

Total= $36+6+48=\boxed{90}$ arrangements[/hide]

hmmm....now doesn't this look familiar...too bad i did not look at the solution after i took this as a practice last night....

Another way:

WLOG fill in first four boxes $ 1,1,\minus{}1,\minus{}1$, and there are $ 6$ possible arrangements.
WLOG fill in the rest of the first column, and there are $ 3$ possible arrangements.
Now, consider the bottom right 2x2 square. If we choose any one of these squares to be $ \minus{}1$, it uniquely determines an arrangement and none of the other squares can be $ \minus{}1$, so there are $ 4$ possibilities.
If none of these squares are $ \minus{}1$, it is clear there is only one arrangement for this.

So total is $ 3*6*(4\plus{}1) \equal{} 90$.

Here's another solution.  I don't mean to revive the thread, just posting so that people working on problems in the Contests archive have access to another solution.

[hide]Call two rows "complementary" if they contain different values in each column.  For example, a row containing 1, 1, -1, -1 would be complementary to a row containing -1, -1, 1, 1.

We claim that every row has a complementary row.  Consider some row, WLOG containing 1, 1, -1, -1.  Suppose for the sake of contradiction that this row does not have a complementary row.  Then each other row contributes at least as much to the sum of the entries in the first two columns as it does to the sum of the entries in the last two columns, so the sum of the entries in the first two columns is strictly greater than the sum of the entries in the last two columns.  But both of these sums are zero, a contradiction.  Therefore, every row has a complementary row, so the rows must occur in complementary pairs.

Also, if we have an 4x4 array of 1's and -1's with all row sums 0, and the rows occur in complementary pairs, then each complementary pair contributes 0 to each column sum, so the column sums are also all 0.

So we've proved that a 4x4 array of 1's and -1's is a valid array if and only if:
(1) all row sums are 0
(2) the rows occur in complementary pairs

Now we count.  There are $\binom{4}{2}/2 = 3$ ways to choose which pairs of rows are complimentary.  For each pair of complementary rows, there are $\binom{4}{2} = 6$ ways to fill in the higher row, and the lower row is uniquely determined by the higher row.  This gives us $3\cdot 6\cdot 6 = 108$.

But we've overcounted the cases where a row has more than one complementary row.  These cases have two pairs of identical rows, and each row in one pair is complementary to each row in the other pair.

How many such arrays are there?  There are $\binom{4}{2}/2 = 3$ ways to choose which pairs of rows are identical.  Then there are $\binom{4}{2} = 6$ ways to fill in the first row of the array, and the rest of the rows are uniquely determined by the first row.  Therefore, there are $3\cdot 6 = 18$ such arrays.

How many times did we count each of these?  Given such an array, there are two different complementary pairings, and we counted the array once for each such pairing.  Therefore, we counted each of these arrays twice, so we need to subtract one for each array.

Therefore, the answer is $108 - 18 = 90$.[/hide]

[hide=think this is different]
Observe that the problem is equivalent to placing $8$ ones into the array such that each row and column has $2$ ones.

WLOG let the first column have ones in the top and second-to-top squares; we can multiply by $6$ to account for this later.

Case 1: The next column has ones in the same rows.

Then everything is determined, giving us $1$ possibility.

Case 2: The next column has ones in the bottom two rows.

Case 2a: The third column is identical to the 2nd or 1st. Then we have two possibilities and everything else is fixed, giving us $2$.

Case 2b: The third column is not identical to 2nd or first. Then we have $4$ possibilities.

Case 3: The next column has a one in a row from the first column and a one without. This gives us $4$ possibilities. WLOG we have $\begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$. This gives us $2$ more options, so we have $4 \cdot 2 = 8$ possibilities.

This gives us $\boxed{090}$.

The only way to get a sum of 0 is 1 1 -1 -1 in some order. Now we do cases.

Case 1: Bottom left + top left = 2, bottom right + top right = 2

Then the middle two in both of those columns are -1, -1 and so the middle four have to be all ones. Also, the bottom left, top left, bottom right, and top right all have to be 1, so there is only 1 way here.

Case 2: Bottom left + top left = 2, bottom right + top right = 0

Then the middle two in the left column is -1 -1 and the middle two in the right column is -1 1 in some order. WLOG 1 is above -1, so there is then $2$ ways to fill the middle four. There are $2$ ways to satisfy the original condition, for a total of $2 \cdot 2 = 4$.

Case 3: Bottom left + top left = 0, bottom right + top right = 2

Same as the above case, so $4$ again.

Case 4: Bottom left + top left = 0, bottom right + top right = 0

Subcase 1: -1 1
                      1 -1
OR

1 -1
-1 1

These each give $4$ ways to fill out the middle four, so $4 \cdot 2 = 8$ ways.

Subcase 2: Anything else

There are two other cases, each with $1$ way to fill out the middle four, so $1 \cdot 2 = 2$ ways.

Summing gives $10$.

The answer is $1+4+4+10 = 19$.


calculus
integration
derivative
calculus computations

Calculate $\int_{0}^{1}\!\!\frac{d^{x}(sin(t))}{(dt)^{x}}dx$ with respect to $x$.

Let's start by defining $\frac{d^{x}(sin(t))}{(dt)^{x}}$.

:) I agree. People usually write this operator as $\mathbb{D}_{t}^{x}$, it figuratively means $x^{th}$ derivative.
There are many ways to define this operator. Important and useful result is that $\mathbb{D}_{t}^{x}e^{\alpha t}= \alpha^{x}e^{\alpha t}$.[hide="step2"]$\mathbb{D}_{t}^{x}\sin(t) = sin(t+\frac{x\pi}{2})$[/hide] [hide="answer"]$=\frac{2}{\pi}(\sin(t)+\cos(t))$[/hide]
I never saw integral over degree of derivative, I wonder if this has any physical application/significance?


[hide="Solution"][list=1][*] Qg8+, Rxg8
[*] Nf7++ (mate)[/list][/hide]
Are there going to be more chess problems?

[quote="Patterns_34"]Are there going to be more chess problems?[/quote]

You can start a Marathon, I guess. Be sure to post an image, and how many moves it takes to mate.

Smother mate is awesome. Unfortunately I've never got it to use in a game yet

[quote="serialk11r"]Classical smothered mate...no offense but this was a little too easy to be worth posting in my opinion...[/quote]
Yes, it's quite simple; I just wanted to see if anybody here was interested in this sort of stuff.

You're welcome to post harder problems in the marathon. :)

Here's a trickier problem, one of many favorites of mine.
[hide="White to move and win in..."]5 moves (at most) (I find exposing the amount of moves needed to win to be too helpful in several cases).[/hide]

Patterns_34:  See the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193263]chess marathon thread[/url] to solve and post new chess problems.

:oops_sign: I'm surprised I didn't see the other thread.

Just to answer that problem,
Start by throwing a check Qe7+. There's too many possible combinations for me to list, but eventually black must move his queen away, and White checkmates with Qg3++ (or Qxg3++ if he moves his queen there).

1. Qe7+

Pawn blocks:
1. ... g5
2. Qe1+ Qg3
3. Qxg3#

Queen blocks:
1. ... Qg5
2. Qe4+ Qg4
3. Qe3 g5
4. Qf2+ Qg3
5. Qxg3#

I seriously remember this problem... it's a zugzwang, isn't it?

New one:

CHECKMATE IN HALF A MOVEMENT!!!!!

Black king A8
White knight C6
White bishop: D5
White rook: E4
White queen: F3
White king: G2

What is half a movement? You can't have half a move

I don't understand, too :P

The solution is stupid :P

Here is the smallest puzzle of a certain idea I could come up with:
White:Kb6,Pb7,Nf6,Ph3
Black:Kb8,Nf8

White to move and mate in 2.

Please post other problems [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193263&start=40]here[/url]. :)


Find all integer solutions (x,y) of the equation:
$ x^3\plus{}y^4\equal{}7$


be vozooh adade [ 2 be tavane d ] -1 do adade [2 be tavane n]-1  va [2 be tavane m] -1 ra aad mikond pas kafie sabet konim age adde t ee bashad ke do adade mazkoor ra aad konad dar in soorat hatman adade [ 2 be tavane d ] -1 ra niz aad khahad kard .ke baraye in kar kafie ghaziyeye bezo ra vase do adade n,m neveshte va az oon estefade konim. omidvaram tozih kamel bashe.

emkanesh has iekam bishtar tozih bedin?
man nemitoonam oonaii ke latex kardin dorost bebinam.lotfan :oops: