calculus
integration
trigonometry
logarithms
function
calculus computations

Calculate the Integral :
   $ I \equal{} \int_{0}^{1}{\frac {ln(x^2 \minus{} 2xcos(a) \plus{} 1)}{x}dx} \equal{} ? a \in \mathbb{R}$

$ \displaystyle\int \frac {\ln(x^2 \minus{} 2x\cos a \plus{} 1)}{x}\,dx \equal{} \displaystyle\int \frac {\ln((x \minus{} \cos a)^2 \plus{} 1 \minus{} cos^2 a)}{x}\,dx \equal{} \displaystyle\int \frac {\ln((x \minus{} \cos a)^2 \plus{} sin^2 a)}{x}\,dx \equal{} \displaystyle\int \frac {\ln(x \minus{} \cos a)^2\times\ln(sin^2 a)}{x}\,dx \equal{} 2\ln(sin a)\displaystyle\int \frac {2\ln(x \minus{} cos a)}{x}\,dx \equal{} 4\frac {\ln(sin a)}{\ln(cos a)}\displaystyle\int \frac {\ln x}{x}\,dx$
now substitute $ \ln x$  with $ t$ . so $ \frac {dx}{x} \equal{} dt \rightarrow dx \equal{} x.dt$ and $ e^{t} \equal{} x$ .. so result is  $ 2\frac {\ln(sin a)}{\ln(cos a)} \times \ln^2 x$  but from $ 0$ to $ 1$ .... I dont know :-?? :?:

Let $ \delta$ be a sufficiently small positive number and let $ a \in [\delta, \pi]$.

Then $ I$, considered as a function of $ a$ behaves nice in the sense that $ \frac {dI}{da}$ exists and is equal to $ \int_{0}^{1} \frac {d}{da} \frac {\log (x^2 \minus{} 2x \cos a \plus{} 1)}{x} \, dx$ by Leibniz's integral rule.

Then $ \frac {dI}{da} \equal{} 2 \sin a \int_{0}^{1} \frac {dx}{x^2 \minus{} 2x \cos a \plus{} 1}$, thus we have

$ \frac {dI}{da} \equal{} 2 \left[ \arctan \left( \frac {x \minus{} \cos a}{\sin a} \right) \right]_{0}^{1} \equal{} \pi \minus{} a$

This implies $ I \equal{} \pi a \minus{} \frac {a^2}{2} \plus{} C$ for some $ C$, and since

$ I(\tfrac{\pi}{2}) \equal{} \int_{0}^{1} \sum_{n \equal{} 1}^{\infty} \frac {( \minus{} 1)^{n \minus{} 1}}{n} x^{2n \minus{} 1} \, dx \equal{} \frac {1}{4} \zeta(2) \equal{} \frac {\pi^2}{24}$.

This determines $ C$, giving $ I \equal{} \minus{} \frac {\pi^2}{3} \plus{} \pi a \plus{} \frac {a^2}{2}$ for $ a \in [\delta, \pi]$.

Since $ \delta$ is arbitrary, this identity actually holds for $ a \in (0, \pi)$.

The value of $ I$ for other values of $ a$ is completely determined by the peroidicity of cosine function.


If $a,b,c,x,y,z$ are real numbers and $0 < a,b,c < 1$, and $x,y,z$ are real positive numbers such that $a^{x}=bc$, $b^{y}=ac$, $c^{z}=ab$. Prove that:

$\frac{1}{2+x}+\frac{1}{2+y}+\frac{1}{2+z}\leq \frac{3}{4}$

F.D.  :wink:

By substitution$p = ln(a); q = ln(b); r = ln(c);$, we get

$\frac{p}{2p+q+r}+\frac{q}{2q+p+r}+\frac{r}{2p+p+q}\leq \frac{3}{4}$

what is true for p,q,r all positiv or negative.


[quote="bearkiller"]Find all reals x and y with:
$ (1 \plus{} x^3)(1 \plus{} y^3) \equal{} 1$
$ \sqrt [3]{\left|(1 \plus{} x)(1 \plus{} y)\right|}$ is a rational number
$ \sqrt [3]{\left|xy(x \plus{} y)\right|}$ is a rational number[/quote]

Let $ (1\plus{}x)(1\plus{}y)\equal{}a^3$ with $ a\in\mathbb Q$
Let $ xy(x\plus{}y)\equal{}b^3$ with $ b\in\mathbb Q$
Let $ p\equal{}x\plus{}y$
Let $ s\equal{}xy$

$ (1\plus{}x)(1\plus{}y)\equal{}a^3$ $ \iff$ $ s\plus{}p\equal{}a^3\minus{}1$
$ xy(x\plus{}y)\equal{}b^3$ $ \iff$ $ ps\equal{}b^3$
$ (1\plus{}x^3)(1\plus{}y^3)\equal{}1$ $ \iff$ $ x^3\plus{}y^3\plus{}x^3y^3\equal{}0$ $ \iff$ $ s^3\plus{}p^3\equal{}3ps$

So we have the system :

$ p^3\plus{}s^3\equal{}3b^3$
$ p\plus{}s\equal{}a^3\minus{}1$
$ ps\equal{}b^3$

Since $ p^3\plus{}s^3\equal{}(p\plus{}s)^3\minus{}3ps(p\plus{}s)$, the first equation implies $ (a^3\minus{}1)^3\minus{}3b^3(a^3\minus{}1)\equal{}3b^3$ and so $ (a^3\minus{}1)^3\equal{}3a^3b^3$

$ a\equal{}0$ would imply $ (\minus{}1)^3\equal{}0$, impossible
$ b\equal{}0$ implies $ (a^3\minus{}1)^3\equal{}0$ and so $ a\equal{}1$. So $ ps\equal{}0$ and $ p\plus{}s\equal{}0$ and $ p\equal{}s\equal{}0$ and so $ x\equal{}y\equal{}0$

$ ab\neq 0$ implies $ \left(\frac{a^3\minus{}1}{ab}\right)^3\equal{}3$, which is impossible, since $ \frac{a^3\minus{}1}{ab}\in\mathbb Q$ while $ \sqrt[3]3\notin\mathbb Q$

Hence the unique solution $ x\equal{}y\equal{}0$


calculus
integration
function
geometry
trigonometry
calculus computations

Could someone explain to me how to solve these problems?

16. A city is built around a circular lake that has a radius of 1 mile. The population density of the city is f(r) people per square mile, where r is the distance from the center of the lake, in miles. Which of the following expressions gives the number of people who live within 1 mile of the lake?

Answer: 2pi * ( definite integral of rf(r) from 1 to 2)

18. Let f be a function such that f"(x)<0 for all x in the closed interval [1,2]. Selected values of f are shown in the table below. Which of the following must be true about f'(1.2)?
x     1.1     1.2      1.3       1.4
f(x)  4.18   4.38    4.56      4.73

(a) f'(1.2)<0
(b) 0<f'(1.2)<1.6
(c) 1.6<f'(1.2)<1.8
(d) 1.8<f'(1.2)<2.0
(e) f'(1.2)>2.0
Answer: d

24. If f'(x) = sin((pi*e^x)/2) and f(0)=1, then f(2)=

Answer: 1.157

For the first question, you're basically summing up infinitesimal populations-that is the number of people who live within each radius.  For example, if the radius is $ dr$, you'd expect to find $ 2\pi\cdot dr\cdot f(r)$ people in the region (the $ 2\pi$ is of course the area factor).  Just match units to check.

For 24, you'd probably just want to plug into your calculator.

By the fundamental theorem, $ \int_0^2 f'(x)\ dx\equal{}f(2)\minus{}f(0)\equal{}f(2)\minus{}1$, so $ f(2)\equal{}1\plus{}\int_0^2 \sin\left(\frac{\pi e^x}{2}\right)\approx 1.157$.


Square $ ABCD$ has area $ 36,$ and $ \overline{AB}$ is parallel to the x-axis. Vertices $ A,$ $ B,$ and $ C$ are on the graphs of $ y \equal{} \log_{a}x,$ $ y \equal{} 2\log_{a}x,$ and $ y \equal{} 3\log_{a}x,$ respectively. What is $ a?$

$ \textbf{(A)}\ \sqrt [6]{3}\qquad \textbf{(B)}\ \sqrt {3}\qquad \textbf{(C)}\ \sqrt [3]{6}\qquad \textbf{(D)}\ \sqrt {6}\qquad \textbf{(E)}\ 6$

[hide]
Say that $A=(x,y)$.  Then $B=(x+6,y)$ or $(x-6,y)$.  $\log_{a}x < 2\log_{a}x+6$, clearly, so $B=(x-6,y)$.  By the same reasoning, $C=(x-6,y+6)$. We thus have

$y = \log_{a}x$

$y = 2\log_{a}x-6$

$y+6 = 3\log_{a}x-6$

Subtracting the second equation from the third, we get

$6 = \log_{a}x-6\Rightarrow 2\log_{a}x-6 = y = 12$

With the first equation, this gives $x = a^{12}$.

The first two equations give $(x-6)^{2}=x$, for which the solutions are 9 and 4.  4 runs into problems with $\log_{a}x-6$, so $x=9$.

Thus $a^{12}=9=3^{2}$ and $a=\sqrt[6]{3}$.  A.

[/hide]

[hide]
...
$y = 3\log_{a}x$ corresponds to $C$
$y = 2\log_{a}x$ to $B$
and $y = \log_{a}x$ to $A$

So $2\log_{a}x = \log_{a}{x+6}$, since the $y$ value is the same for both.

This is a quadratic: $x^{2}-x-6 = 0$, so $x=3$.

Then $3\log_{a}x-2\log_{a}x = 6$, since the area is $36$, so $\log_{a}x = 6$.

Substituting gives: $a^{6}= 3 \Rightarrow a = \sqrt[6]3$.[/hide]

to white_horse_king88
i think the line you solve for x is not correct...

Sorry... that's supposed to say $x^{2}-x-6$. I just messed up a sign.

i mean the log one..

2*log_{a}x=log_{a}x^2
               =log_{a}x+log_{a}a^6
               =log_{a}x*a^6

so x^2=x*a^6
therefore, x should equal to a^6

....i think that is the real mistake...

and to E^(pi*i)=-1

when you subtract the second line from the third one,
i think you should get : log_{a}x=6,
so you would get x=a^6 as well

Alright, if you followed my solution at all, you'll realize that there's just a missing set of parentheses...  :wink:
The method is valid.

[quote="joseph14i"]and to E^(pi*i)=-1

when you subtract the second line from the third one,
i think you should get : log_{a}x=6,
so you would get x=a^6 as well[/quote]

Okay, I'll put parentheses around $x-6$ to make the equation clearer.  But white_horse_king88 and I both got the right answer, so there is no mistake.

EDIT: Or at least I would have, if I could still edit my initial post.

sorryy... :oops:
that was my foolish mistake....
please forgive my stupidity....

[quote="Differ"]Why can't $ a$ be $ \frac1{\sqrt [6]3}$?[/quote]

Bump for an answer.

[quote="E^(pi*i)=-1"]  But white_horse_king88 and I both got the right answer, so there is no mistake.

[/quote]
No, that is a false statement.  Two (or more people) arriving at the same answer does not
imply that there is no mistake.  Instead it could point to a wrong method, wrong assumptions,
and/or errors canceling each other out, for instance.

@[b]Differ[/b],
$a=\frac{1}{\sqrt[6]{3}}$ is not found as a solution, once the quadratic and log equations are solved.

[b]Solution:[/b] Since we know that $B$ and $C$ have the same $x$-coordinate and are 6 units apart, we have that $3\log_ax = 2\log_ax + 6$, so that said $x$-coordinate is $a^6$. Now, we know that the $x$-coordinate of $A$ is $x = a^6 + 6$. Since $A$ and $B$ have the same $y$-coordinate, $\log_a(a^6 + 6) = 12$, which gives us our answer of $a = \sqrt[6]{3}$.


[hide]6^3=206
5^3=125
So we need side lenghs containing 6 and 5.
The other must be 4.
2(30+24+20)
2(74)
148
[/hide]

[quote="rcv"]A rectangular prism, with integer edge-lengths, has a volume of 120 cubic centimeters.  What is the smallest possible surface area of this solid?[/quote]

Is it four-dimensional? If so, the surface area can be 6 (or theoretically, 0), but I think I won't go there =P .

[hide="sane solution"]
The closest you can get to a cube is 4 x 5 x 6.
2(20 + 24 + 30)
2(74)
148
[/hide]

[hide="my answer"]the side lenghts musdt be close to eachother
4X5X6
2(4*5+4*6+5*6)
2(20+24+30)
2(74)
148 [/hide]

[quote="AznXtreme418"]this is state level? doesn't seem bad at all, but some of the last 10 problems are really hard since there's so little time left[/quote]
Actually sometimes the earlier problems can be the hardest.. b/c you aren't warmed up and so you make careless errors... but typically the last page is always the hardest in terms of difficulty (here in CT the chapters start pretty early in terms of time, so you're half asleep when you're starting the sprint :lol:) as you said..


[hide="Solution"]Since $ a \plus{} b \equal{} 7$, $ (a \plus{} b)^3 \equal{} a^3 \plus{} b^3 \plus{} 3ab (a \plus{} b) \equal{} 343$. Substituting in $ a^3 \plus{} b^3 \equal{} 42$ and $ a \plus{} b \equal{} 7$, we find that $ ab \equal{} \frac{43}{3}$. This implies that $ \frac1a \plus{} \frac1b \equal{} \frac{a \plus{} b}{ab} \equal{} \frac{21}{43}$.[/hide]

[quote="12markkram34"]Can someone please tell me how to solve this:

25. If a + b = 7 and a^3 + b^3 = 42, what is the value of the sum 1/a + 1/b? Express your answer as a commmon fraction.

Thanks.
-12markkram34[/quote]

Cube the first equation. Then, we get

$ a^3+b^3+3ab(a+b)=343 \implies 42+3ab(7)=343 \implies
ab=\frac{43}{3}$.

So, $ \frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{7}{\frac{43}{3}}=\frac{21}{43}$.

[hide="Click here to win a Free Nintendo Wii!!!!  :) "](jk) Squaring the a+b=7 equation gives $ (a \plus{} b)^3 \equal{} (a \plus{} b)^2(a \plus{} b) \equal{} (a^2 \plus{} 2ab \plus{} b^2)(a \plus{} b) \equal{} a^3 \plus{} 2a^2b \plus{} b^2a \plus{} a^2b \plus{} 2ab^2 \plus{} b^3 \equal{} a^3 \plus{} 3a^2b \plus{} 3b^2a \plus{} b^3 \equal{} a^3 \plus{} 3(a^2b \plus{} b^2a) \plus{} b^3 \equal{} a^3 \plus{} 3ab(a \plus{} b) \plus{} b^3$ All of this stuff equals 7^3, or 343.

Okay, 1/a + 1/b could also be $ \frac {a \plus{} b}{ab}$, after finding the common denominator.

a+b=7, so we have $ \frac {7}{ab}$

We are given that a^3+b^3=42, so we can substitute that into the $ a^3 \plus{} 3ab(a \plus{} b) \plus{} b^3 \equal{} 343$ equation to get 42+3ab(a+b)=343. a+b=7, so we have 42+3ab(7)=343. subtracting both sides by 42 and then dividing by 21 gives ab=43/3. So now we have $ \frac {7}{\frac {43}{3}} \equal{} \frac {21}{43}$  :)

The Intro to Alg. book talks a lot about this.  :lol: [/hide]

Okay so A + B = 7, and A^3 + B^3 = 42.

Okay well you want to know what 1/A + 1/B is equivalent to. Well in order to get from A + B to 1/A + 1/B, you need to divide by AB. So you need to find the value of AB. In order to do this you need to square (A + B) to get A^2 + B^2 + 2AB = 49. So, now you need to know what A^2 + B^2 is equivalent to. To get this find (A + B)^3 because we know what A^3 + B^3 equals.

(A + B)^3 equals A^3 + B^3 + 3A(B^2) + 3B(A^2) = 343. And A^3 + B^3 equals 42 so subtract 42 to get

3A(B^2) + 3B(A^2) = 301. So if you factor out an AB.

AB(3B + 3A) = 301. 3B + 3A equals 21. So 21AB equals 301. So AB equals 43/3. So now in order to solve the problem you need to divide 7 by 43/3 to get 21/43 as your answer.

On a scale of 1 to 10, what did you think of this solution.

Be honest!

You shouldn't really have posted here...

But anyway,

LaTeX would really help people understand your solution.

Your solution is not phrased very well. A better way to phrase the first

sentence is:

We first find an equivalent expression for $ \frac{1}{a}\plus{}\frac{1}{b}$.

We know that $ \frac{1}{a}\equal{}\frac{b}{ab}$, and $ \frac{1}{b}\equal{}\frac{a}{ab}$,

so the sum is $ \frac{a\plus{}b}{ab}$.

The full solution should look like:

We first find an equivalent expression for $ \frac{1}{a}\plus{}\frac{1}{b}$.

We know that $ \frac{1}{a}\equal{}\frac{b}{ab}$, and $ \frac{1}{b}\equal{}\frac{a}{ab}$,

so the sum is $ \frac{a\plus{}b}{ab}$.

Since $ a\plus{}b\equal{}7$, cubing gives us $ a^3\plus{}b^3\plus{}3ab(a\plus{}b)\equal{}343$.

We are given that $ a\plus{}b\equal{}7$, so $ a^3\plus{}b^3\plus{}21ab\equal{}343$ (1).

We are given that $ a^3\plus{}b^3\equal{}42$ (2),

so subtracting (2) from (1) gives $ 21ab\equal{}301$.

Dividing by $ 21$ gives $ ab\equal{}\frac{301}{21}\equal{}\frac{43}{3}$.

So, plugging in to our original expression, $ \frac{a\plus{}b}{ab}\equal{}\frac{7}{\frac{43}{3}}\equal{}\frac{21}{43}$, which is our answer.

This is just a restatement of your solution, in easier to understand terms.

on a scale of 1-10, I give you a 7.25

Hope that helps!

Well, I guess my solution included a little bit of irrevelant information. I think that that would be a good rating. Thanks!

$ a^3\plus{}b^3\equal{}(a\plus{}b)(a^2\minus{}ab\plus{}b^2)\equal{}7(a^2\minus{}ab\plus{}b^2)\Rightarrow a^2\plus{}b^2\minus{}ab\equal{}6$
$ a^2\plus{}b^2\plus{}2ab\equal{}49$ as well.
So $ 3ab\equal{}43$

Factoring is easier.

That was a really good solution, it is always a good idea to factor. Btw - there is so many different ways to solve these types of Algebra problems!


search
number theory proposed
number theory

The positive integers $ a$ and $ b$ are such that $ 15a\plus{}16b$ and $ 16a\minus{}15b$ are squares of
integers.What is the least possible value that can be taken on by the smaller of these $ 2$ squares?

This problem and solutions can be found here:
http://www.mathlinks.ro/viewtopic.php?search_id=1570042646&t=60430


Can someone please help me?  I thought it was really confusing...

[hide="Hopefully I read this correctly..."]
WLOG, assume that the radius of the gas pipe is $ 1$. So I think it's the area of the region above the circle in the trapezoid over the area of the total region, which is $ \frac{9\minus{}\frac\pi2}{9\plus{}\frac\pi2}$ which is $ \frac{9\minus{}\frac{11}7}{9\plus{}\frac{11}7}\equal{}\boxed{\frac{26}{37}}$ in this problem.
[/hide]


logarithms
factorial
number theory proposed
number theory

Hi,

I have discoverd something very strange.
I do not know how to explain it for now but it deserve to be shared with the mathematician community.
Here is my discovery

If we compute

(m+1!)^2 mod (2m+1) with 2m+1 prime

we obtain almost P(n)
There is a little difference but it is stonishing!!!
I do not have the tools and the skills to compute the sequence for the values of more than 24! but the sequence seems very strange.

(2!)^2mod(3)=1
(3!)^2mod(5)=1
(4!)^2mod(7)=2
(6!)^2mod(11)=3
(7!)^2mod(13)=3
(9!)^2mod(17)=4
(10!)^2mod(19)=5
(12!)^2mod(23)=6
(15!)^2mod(29)=7
(16!)^2mod(31)=8
(19!)^2mod(37)=9
(21!)^2mod(41)=10
(22!)^2mod(43)=11
(24!)^2mod(47)=12

[b]Please learn some number theory.[/b]  The same technique I showed you in the previous post gives

$ (m \plus{} 1)^2 (m!)^2 \equiv (m \plus{} 1)^2 ( \minus{} 1)^{m \plus{} 1} \equiv \frac {( \minus{} 1)^{m \plus{} 1}}{4} \bmod 2m \plus{} 1$

which is, as you have seen, going to be about $ \frac {2m \plus{} 1}{4}$.  The relationship you describe [b]cannot continue to hold[/b].  Rather than $ P(n) \sim 4n$, the correct asymptotic is $ P(n) \sim n \ln n$.

I have tried to teach you a few things in your threads.  Perhaps I should just spell them out:

1.  Until you discover an efficient way to compute factorials, they will [b]not[/b] lead anywhere interesting.  There is a reason you can't compute anything you try to compute to larger terms - it's because the things you're trying to compute [b]are hard to compute[/b] and mathematicians know of much better ways to test for primality.

2.  Patterns in the small primes can be [b]very deceptive[/b] (like the above), although it can take a [b]very long time[/b] to figure this out.  That's why you need techniques that still work on very large primes, and nothing you've posted satisfies this criterion.

[quote="t0rajir0u"][b]1.  Until you discover an efficient way to compute factorials, they will [b]not[/b] lead anywhere interesting.  There is a reason you can't compute anything you try to compute to larger terms - it's because the things you're trying to compute [b]are hard to compute[/b] and mathematicians know of much better ways to test for primality.

2.  Patterns in the small primes can be [b]very deceptive[/b] (like the above), although it can take a [b]very long time[/b] to figure this out.  That's why you need techniques that still work on very large primes, and nothing you've posted satisfies this criterion.[/quote]

Thank you for your comments.
I'm still trying to find a way to make factorials more affordable for computing.
I have an idea a good one but I lack the tools to develop it. Only ideas can get rid of some problems the tools are helpful however.

Someone told me that there is a same sequence in Sloane Encyclopedia but [u]without a way to produce it.[/u]
http://www.mymathforum.com/viewtopic.php?f=40&t=7307
the sequence A024698

I know that [i]patterns in the small primes can be very deceptive [/i] because it is not the first time that it happens to me.
My philosophy is try always try no matter what it could be said.
Try try until finding the solution.
Every day I learn something new.
There is no hard problem for people who try. It takes only to go in the right direction and to find the right gate (le bon chemin comme on dit chez nous).
Thank you again for your useful comments.


linear algebra
matrix
algebra
polynomial

HI,
       how can i complete a vandermond matriz like this kind:

1      1      1

a^2 b^2 c^2

a^3 b^3 c^3

i want tho solve this with vandermond, but we dont have sequences of expoents to use vandermond, so we need do a change... after sucessufull in this case, i want generalizated to a matriz nxn

its very unusual i think, because all of my teachers and olympic friens(IMO) dont know how to do it.  thanks  :)

(Edit:  My apologies for misreading the question  :oops: )

It would be a Vandermonde matrix if those were first and second powers rather than third powers.

The traditional trick works to a point: $ (a\minus{}b)(b\minus{}c)(a\minus{}c)$ must be a factor of the determinant. After that, we need a degree-2 homogeneous polynomial as the second factor, and it must be symmetric since the overall determinant has alternating symmetry. That makes it a linear combination of $ a^2\plus{}b^2\plus{}c^2$ and $ ab\plus{}bc\plus{}ac$. Since there are no terms in the determinant like $ a^4b$, we can't have the $ a^2\plus{}b^2\plus{}c^2$ factor, and the determinant is $ (a\minus{}b)(b\minus{}c)(c\minus{}a)(ab\plus{}bc\plus{}ca)$, with sign and scale verified by the $ b^2c^3$ term.

sorry, eu falei em portugues matriz,!!! u.u

NOOO, i dont want the transpost or inverse, i want the determinant ,BUT i know that is a easy determinant if you solve anyaway
i want that you solv this using vandermond, i dont know if you understand bu for a example:

   1     1      1               the determinat of this is by vandermond:
  a      b      c
a^2  b^2  c^2

=  (b-c).(c-a).(b-a)

you can use Xio or laplace to solve yes , but i want  a method to complete a incomplet vandermond without sequences of  expoentes.

the easily:     1   1   1
                    a^2 b^2 c^2
                    a^3  b^3  c^3

in this case you need complete like this   a^0 , a^1, a^2 .. in order to use vandermond. how can  complete this

  :lol:

*anyway

and do a generalization to a nxn

dont  u  use addcitions and multiplications in the matrix?
linear combinations?  i see a polynomial think /o/

This is a problem in polynomial algebra, whether you like it or not.

All right, here's a generalization:
For an $ n$-tuple $ x\equal{}(x_1,x_2,\dots,x_n)$, let $ s_k(x)$ be the sum of all products of $ k$ different $ x_i$; $ s_0(x)\equal{}1,s_1(x)\equal{}x_1\plus{}x_2\plus{}\cdots\plus{}x_n,\dots,s_n(x)\equal{}x_1x_2\cdots x_n$.
The determinant of the $ n\times n$ matrix $ V\equal{}\begin{bmatrix}1&1&\cdots&1\\x_1&x_2&\cdots&x_n\\ \vdots&\vdots&\ddots&\vdots\\ x_1^{j\minus{}1}&x_2^{j\minus{}1}&\cdots&x_n^{j\minus{}1}\\ x_1^{j\plus{}1}&x_2^{j\plus{}1}&\cdots&x_n^{j\plus{}1}\\ \vdots&\vdots&\ddots&\vdots\\ x_1^n&x_2^n&\cdots&x_n^n\end{bmatrix}$ is $ s_{n\minus{}j}(x)\cdot\prod_{i<j}(x_j\minus{}x_i)$.

The proof is very similar to the traditional proof of the Vandermonde identity. For the $ s_{n\minus{}j}$ factor, consider the polynomial $ (t\minus{}x_1)(t\minus{}x_2)\cdots (t\minus{}x_n)$, and how a linear combination of the rows of $ V$ might be zero.

As a polynomial identity, this is of course true over any field.

My apologies.  Thinking of the Vandermonde matrix as a question of polynomial interpolation, this question can be answered the same way:  the determinant of the matrix given in the OP is a factor that appears in the solution to the interpolation problem

$ p(a) \equal{} t$
$ p(b) \equal{} u$
$ p(c) \equal{} v$

where $ p(x) \equal{} a_3 x^3 \plus{} a_1 x^2 \plus{} a_0$.  This interpolation can be solved, like the original Vandermonde problem, by considering the solutions $ p_1, p_2, p_3$ to

$ p_1(a) \equal{} 1, p_1(b) \equal{} 0, p_2(c) \equal{} 0$
$ p_2(a) \equal{} 0, p_2(b) \equal{} 1, p_2(c) \equal{} 0$
$ p_3(a) \equal{} 0, p_3(b) \equal{} 0, p_3(c) \equal{} 1$.

As an example calculation, $ p_1(x) \equal{} (rx \plus{} s)(x \minus{} b)(x \minus{} c)$ where the linear coefficient is equal to $ rbc \minus{} cs \minus{} bs \equal{} 0$ and $ p_1(a) \equal{} (ra \plus{} s)(a \minus{} b)(a \minus{} c) \equal{} 1$.  The first condition gives $ r \equal{} s \left( \frac{b \plus{} c}{bc} \right)$ and the second condition can be used to conclude.

The matrix you are looking at is a minor in the larger matrix
\[ \left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\
a & b & c & d \\
a^2 & b^2 & c^2 & d^2 \\
a^3 & b^3 & c^3 & d^3 \end{array} \right]
\]
and so its determinant can be extracted as the coefficient of $ d$ in the determinant of the above Vandermonde matrix. This approach works equally well on the generalization posted.

That gets the fully factored form neatly, too. I like it about as well as my own method.


calculus
function
conics
hyperbola
calculus computations

this problem has got me stumped:

list the point(s) on the curve [b]x^2 - y^2 = 4[/b] closest to the point [b](6,0)[/b]

Do you want the Calculus I version (find a way to make it depend on one variable) or the Calculus III version (Lagrange multipliers)?

Calculus I please, that's currently what i'm taking :D

sh0ebox

See the figure for a sketch of the function.

$x^2-y^2=4 \Longleftrightarrow y=\pm\sqrt{x^2-4}$

Suppose the point on the curve is $(x,\sqrt{x^2-4})$. The closest point indicates the least distance between $(6,0)$  and $(x,\sqrt{x^2-4})$

$D(x) = \sqrt{ (x-6)^2 + (\sqrt{x^2-4}-0)^2} = \sqrt{2x^2-12x+32}$

We need to minimize this function, by deriving we get: $D'(x)=\frac{2(x-3)}{\sqrt{2x^2-12x+32}}$

So the function decreases in $[2,3)$ and increases in $(3,\infty) \implies \boxed{x=3}$ is a local minimum, the closest points are $(3,\sqrt{5})$ and $(3,-\sqrt{5})$

You're trying to minimize $D(x,y)=\sqrt{(x-6)^2+y^2}.$ You might as well square that; you'd get the same thing by minimizing $(x-6)^2+y^2.$

We have the constraint $x^2-y^2=4.$ (The point lies on a hyperbola.) Let's solve it for one variable in terms of the other. I have a hunch that things will be a little nicer if we work in $y.$

$x^2=y^2+4,$ or $x=\pm\sqrt{y^2+4}$.

If we sketch the graph, we see that the hyperbola has two branches, and we only want the right-hand branch, so $x=\sqrt{y^2+4}.$

Subsitute that. We're now trying to minimize

$f(y)=\left(\sqrt{y^2+4}-6\right)^2+y^2$

There's still a ways to go, but that's a start, albeit a clumsy one.

It would have been better to use a different parameterization. Try this:

$x=2\cosh t,\,y=2\sinh t.$

Then we're trying to minimize $g(t)=(2\cosh t-6)^2+4\sinh^2t.$

That would be another approach.


algebra
polynomial
number theory
relatively prime
number theory unsolved

For two relatively prime $a,m$ integers, a polynomial $f$ with integer coefficients is calles euclidean with respect to $a,b$ if the prime factors of $ f(n)$ , with only finitely many exceptions, are congruent to 1 or a mod m and there are infinitely many primes of the second kind. Prove that there is an euclidean polynomial with respect to $a,m$ if and only if $ a^2=1 (mod m)$. This problem is related to Dirichlet theorem> Do you see how?

I found this link [url=http://www.math.uconn.edu/~kconrad/blurbs/dirichleteuclid.pdf]www.math.uconn.edu/~kconrad/blurbs/dirichleteuclid.pdf[/url]... do you believe an elementary proof is possible ?


A circle with chord $ BC$ is given. $ A$ is an arbitrary point on the circle.

Prove that

$ i)$ The apollonius circles of each angle of $ \triangle ABC$ coincide at two points. (It is a known property)
$ ii)$ When $ A$ varies, the loci of intersection points of these three circles are a pair of circles.
$ iii)$ Let $ R$ be the radius of the given circle, $ R_1$ and $ R_2$ be the radii of the locus circles. $ | \frac {1}{R_1} \pm \frac {1}{R_2} | \equal{} \frac {1}{R}$

i) Let $ P$ and $ Q$ be the intersection of any two apollonius circle, e.g. apollonius circles of $ \angle BAC$ and $ \angle ABC$ intersect at $ P$ and $ Q$.
Because $ A$ and $ P$ are on same apollonius circle of $ \angle BAC$. $ \frac{BP}{PC}\equal{}\frac{AB}{AC}$. Similarly for $ \angle ABC$, we have $ \frac{PC}{AP}\equal{}\frac{BC}{AB}$. While multiplying two equations, we get $ \frac{BP}{AP}\equal{}\frac{BC}{AC}$. This means $ P$ is on the apollonius circle of $ \angle BCA$. Same situation holds for $ Q$. Then three apollonius circles of $ \triangle ABC$ intersects each other at only two points $ P$ and $ Q$.

So for the $ ii$ case, we are asked to find the locus of $ P$ and $ Q$.


function
topology
real analysis
real analysis unsolved

Let $ X$ uncountable set, $ \mathcal{U}$ topology on $ X$ that consists of set $ \emptyset$, and complements of countable subsets of $ X$; $ \mathcal{V}$ discrete topology on $ X$. Prove that indentity function ( $ i: X\rightarrow X,i(x) \equal{} x$ ), from $ (X,\mathcal{U})$ to $ (X,\mathcal{V})$ isn't a continuous function even though for each sequence $ (x_{n})$ in $ X$, convergence $ (x_{n})\rightarrow x_{0}$ with respect to $ \mathcal{U}$ implies convergence with respect to $ \mathcal{V}$.

The first part of the statement isn't hard, since all you need is an open set in $ (X,\mathcal{V})$ such that the preimage isn't open. Take any point $ x\in X$, then the singleton $ \{x\}\in\mathcal{V}$, but the preimage $ i^{\leftarrow}(\{x\})=\{x\}$ and the complement of that is clearly uncountable, thus it is not an open set. Therefore the function is not continuous. I'm definitely going to look at the next part, since I haven't worked this counterexample before.

Are you sure about that? Every point is an isolated point in the discrete topology, so no sequence converges. So if $ (x_{n})\to x_{0}$ in $ \mathcal{U}$, then since $ x_{0}$ is an isolated point in $ \mathcal{V}$ that sequence doesn't converge to $ x_{0}$ in $ \mathcal{V}$. Or maybe I'm missing something?  :maybe:

The sequence can be stationary, what is the generally the case if we have a sequence on a discrete space  :wink: . Refer to [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=161107[/url], maybe that will help. Unfortunately no one solved the problem on provided link.

Ah, that makes sense. Some people adopt different conventions concerning that. In that case it sounds as if we are in agreement that an equivalent way to prove this to prove that if $ (x_{n})\to x_{0}$ in $ \mathcal{U}$, then $ \exists N\in\mathbb{N}$ such that $ x_{n}\equal{}x_{0}$ $ \forall n>N$.

Suppose not. Then take $ S$ to be the set of every point in that sequence. $ S$ is clearly a countable set. If you examine $ L\equal{}\left(X\setminus S\right)\cup\{x_{0}\}$ the complement of which is $ S$, then $ L$ an open neighborhood about $ x_{0}$ that doesn't contain a single element from the sequence. Therefore $ (x_{n})$ does not converge to $ x_{0}$, a contradiction.

Thus, every convergent series in $ \mathcal{U}$ is of a form that converges in $ \mathcal{V}$.


I've been reading a lot of the recent posts on registration issues. I apologize, however, if some of the answers to my concerns have already been presented.

So for me, getting a group of friends and a teacher is not an option. However, there is the same "private" school that MIMath mentioned earlier. They are willing to proctor me. However, they have asked me to register with AMC directly.

So I attempted online registration, but it requires a CEEB number which ICAE (the school in mention) does not have, since it isn't really a high school. Similarly, the paper form also requires a CEEB number. What should I do about this? How do I specifically indicate that I am a visitor that will be taking the exam with another school?

Many thanks.

I ran into the same problem with a homeschool group registration. Phone the AMC 800 number and then a helpful staff member will assign your "school" a CEEB number. You will need to assure them that the school is really an independent educational institution and not your own home. AMC is tightening up its rules about proctoring the high-stakes AMC tests a little, and I think that is great. It has taken me some getting used to this year, the first year my homeschool group has ever arranged 10/12 testing, but it is worth it to have scores that are above suspicion and that count for advancement to AIME and USAMO.

I've got a couple of questions if its ok, (sorry to take over your thread :D but i didnt really want to start a new one).
1) If I took A can I take B also?
2) If so, my school doesn't offer it, are there any other options?  Is it possible to not take it through a school at all?

Thanks.

Just do what we do at my school: Order the tests ourselves, and ask a school official (testing coordinator for us) to proctor it.  Of course that'll cost $74 ($52 late registration, $15 for a 10-test bundle, and $7 for shipping).


The Word Problem: A sporting goods manufacturer makes a 5.00 profit on soccer balls and a 4.00 profit on volleyballs. Cutting requires 2 hours to make 75 soccer balls and 3 hours to make 60 volleyballs. Sewing needs 3 hours to make 75 soccer balls and 2 hours to make 60 volley balls. Cutting has 500 hours available, and Sewing has 450 hours available. How many soccer balls and volley balls should be made to maximize the profit?

Please explain this in detail as to how you get the answer. Thanks.

[hide="Solution"]For easier calculation, we'll express time in minutes:

Cutting:

For a soccer ball ${2\over 75}\cdot 60\mathrm{\ min}=1.6\mathrm{\ min}$
For a volleyball ${3\over 60}\cdot 60\mathrm{\ min}=3\mathrm{\ min}$

Sowing:

For a soccer ball ${3\over 75}\cdot 60\mathrm{\ min}=2.4\mathrm{\ min}$
For a volleyball ${2\over 60}\cdot 60\mathrm{\ min}=2\mathrm{\ min}$

Therefore, if we have $s$ soccer balls and $v$ volleyballs, then

$1.6s+3v\leqslant 30000$ ($500$ hours converted into minutes)
$2.4s+2v\leqslant 27000$ ($450$ hours converted into minutes)

For the profit, we know that $p=5s+4v$

From the first inequality we get $v\leqslant{30000-1.6s\over 3}=10000-{8\over 15}s$, hence $p\leqslant 5s+4\left(10000-{8\over 15}s\right)={43\over 15}s+40000$

From the second inequality we get $v\leqslant{27000-2.4s\over 2}=13500-{6\over 5}s$, hence $p\leqslant 5s+4\left(13500-{6\over 5}s\right)={1\over 5}s+54000$

Now, those conditions for $p$ must be satisfied simultaneously, therefore it would be the best if those expressions have the same value (if possible), because otherwise we'd be limited by the smaller of the two (graphically, it means that we're looking for the intersection of the two lines):

${43\over 15}s+40000={1\over 5}s+54000\iff{8\over 3}s=14000\iff s=5250$.

Now from $v\leqslant 10000-{8\over 15}s$ we get $v\leqslant 7200$, and from $v\leqslant 13500-{6\over 5}s$ we get $v\leqslant 7200$. Therefore, we can produce $v=7200$ volleyballs.

For those values, maximal profit will be $p_{max}=\$ 5\cdot 5250+\$ 4\cdot 7200=\$55050$[/hide]


Given $ a, b, c \geq\ 0$ .
Prove that: $ \frac{a(b\plus{}c)}{\sqrt[]{8b^2\plus{}ca}} \plus{} \frac{b(c\plus{}a)}{\sqrt[]{8c^2\plus{}ab}} \plus{} \frac{c(a\plus{}b)}{\sqrt[]{8a^2\plus{}bc}} \leq\ \frac{a\plus{}b\plus{}c}{\sqrt[]{2} }$
PS: Bài này tặng anh Cẩn đúng lúc diễn đàn tạm đóng cửa. Không biết anh đã nhận đc chưa. Em cũng đã hứa là từ nay không ghi tên mình vào sau bài toán nữa  :roll:
(Sr everybody because I write VietNamese)

Try $ a\equal{}0.2,b\equal{}c\equal{}1$ ;)

What a pity!!! I thought it's very interersting.
You should try an another problem: (easy)
$ (a^2\minus{}ab\plus{}b^2)(b^2\minus{}bc\plus{}c^2)(c^2\minus{}ca\plus{}a^2) \geq\ \frac{1}{3}abc(a^3\plus{}b^3\plus{}c^3)$


function
conics
ellipse
geometry
geometric transformation
reflection
AMC

How do I graph the function below?

f(x) = -(sqrt{4 - 4x^2})

well you could use your graphing calculator, but agebraically, that might take a little more time!

well, it looks like the equation for the bottom half of an ellipse with minor axis 1 and major axis 2.  I might be wrong, though.

[hide="solution"]We just need to graph $ \sqrt{4\minus{}4x^2}$ and reflect over the x axis.

$ \sqrt{4\minus{}4x^2}\equal{}2\sqrt{1\minus{}x^2}$, so the range is [-1,1], with maximum at 0. Now just plug in some other numbers(1/2, -1/2, 1/4, -1/4, you get the idea, right?), equate, and then graph. Since the graph is symmetrical at the y-axis, we can just "turn the paper over" and graph $ \sqrt{4\minus{}4x^2}$ as if the axis are right-side up, then turn upside-down. Voila![/hide]

Sharkman ALWAYS posts homework problems. Anyway, that looks like a homework problem anyway.

[quote="JBL"]Can we please get a moratorium on answering what are obviously people's homework problems?  For example, every single post sharkman has ever made in this forum is homework.  Please just ignore him.  (In general, force the people whose work you're doing to show how much of the problem they can do, then give a hint or a nudge in the right direction, etc.)[/quote]

Gotcha.

[quote="zolojetto"]yes i dont like doing other peoples homework either, but how do we know if its homework or maybe a challenging problem that they cant figure out from like AIME or something? There is definetely a difference between helping with an AIME problem than doing some ones homework![/quote]

This is definitely not AIME level. AIME problems are actually pretty difficult (especially 10-15)

[quote="zolojetto"]yes i dont like doing other peoples homework either, but how do we know if its homework or maybe a challenging problem that they cant figure out from like AIME or something? There is definetely a difference between helping with an AIME problem than doing some ones homework![/quote]
By the problem's difficulty and method of solving.

If you did similar problems in a classroom or you could solve the problem in one step with a school-learned formula, it's probably homework.

If you can't discern whether this is the case or not for topics in a forum, you might  be reading the wrong forum.

[quote="zolojetto"]yes i dont like doing other peoples homework either, but how do we know if its homework or maybe a challenging problem that they cant figure out from like AIME or something? There is definetely a difference between helping with an AIME problem than doing some ones homework![/quote]

People rarely get contest math problems for homework; contest math tends to emphasize creative thinking while homework math usually drills a certain concept into the brain.

[quote="perfect628"]My teacher does use contest math, but this is clearly straight from a standard textbook.[/quote]

You most likely have a very intelligent and open-minded teacher. Or you're in a math class that caters specifically to really smart students =p .


Instead of music videos, this thread concerns everybody's favorite [i]songs[/i].  You just have to write the title and artist.  Okay, I'll start:

[i]Tomorrow Never Knows[/i], The Beatles

this doesnt work b/c its not guaranteed that the person will even know the song
to back up the claim, i dont know what that song is

Okay, just post songs you like, not necessarily better than the previous.

My top 10

10. Night of the Alebrijes - Alquimia
9. Balance Dance - Steve Gordon
8. Tarnima - Nasser Kilada
7. Akwaba - TYA
6. Fireside Conversations - Hennie Bekker
5. What is Right - Marcator
4. Why - Tya

Special ones below...

3. Samah - Nasser Kilada
2. Northern Lights - Stella Maris

BEST of mine...

1. Earth Meets Sky Pt. 2 (Buddha Edit) - Steve Gordon

OK. Now you can laugh.

[quote="1=2"]OK. Now you can laugh.[/quote]Hahahahaha.  Okay, now let me read your post.


If a,b,c is  positive integers  such that $ 0 \leq a^{2}+b^{2}+c^{2}-abc \leq c$
Prove that $ a^{2}+b^{2}+c^{2}-abc$ is a square
I think it is not hard and I'm trying to prove it  :oops:

[quote="dtdong91"]If a,b,c is  positive integers  such that $ 0 \leq a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} abc \leq c$
Prove that $ a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} abc$ is a square
I think it is not hard and I'm trying to prove it  :oops:[/quote]

Wrong for $ a \equal{} 2$, $ b \equal{} 5$, $ c \equal{} 6$. Maybe you want to assume $ 0 \leq a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} abc \leq a,b,c$, or maybe you just meant http://www.mathlinks.ro/Forum/viewtopic.php?t=150372 ?

[b]EDIT:[/b] With the condition $ 0 \leq a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} abc \leq a,b,c$ the problem would be wrong as well - e. g. take $ a\equal{}40$, $ b\equal{}7$, $ c\equal{}6$.

  darij


Let $ ABC$ be an equilateral triangle with it`s side $ 1$ and $ M\in \text{Int(ABC)}$ . Prove that $ MA\plus{}MB\plus{}MC<2$.

Three lines through M and parallel to sides BC,CA,AB cut sides, they will create three small equilateral triangles whose sides equal x,y,z and x+y+z=1. So, MA+MB+MC<2(x+y+z)=2.
We have that $ \sqrt{3}\le MA\plus{}MB\plus{}MC<2$


function
logarithms
algebra unsolved
algebra

Let $A$ be a real number and $(a_{n})$ be a sequence of real numbers such that $a_{1}=1$ and \[1<\frac{a_{n+1}}{a_{n}}\leq A \mbox{ for all }n\in\mathbb{N}.\]

$(a)$ Show that there is a unique non-decreasing surjective function $f: \mathbb{N}\rightarrow \mathbb{N}$ such that $1<A^{k(n)}/a_{n}\leq A$ for all $n\in \mathbb{N}$.

$(b)$ If $k$ takes every value at most $m$ times, show that there is a real number $C>1$ such that $Aa_{n}\geq C^{n}$ for all $n\in \mathbb{N}$.

$k(n)=[\frac{\ln a_{n}}{\ln A }]+1$. It prove all.

First, there are a few typos to be fixed.
$1)$ The sequence must be unbounded from above.
$2)$ Part $(b)$ preamble should be $A^{f(n)}$, rather than $A^{k(n)}$.

We will proceed as follows.
$a)$ We will first explicitly recover $k(n)$, and prove that it is surjective and non-decreasing. From the preamble,
$$
\log a_n < k(n)\log A \leq \log a_n + \log A \implies \frac{\log a_n}{\log A}<f(n)\leq \frac{\log a_n}{\log A}+1.
$$
Hence,
$$
\boxed{f(n)=\left\lfloor\frac{\log a_n}{\log A}+1\right\rfloor},
$$
since $f(n)$ is integer-valued. Clearly, $f(n)$ is uniquely defined, and non-decreasing. For surjectivity, note that $f(n)=k$ if and only if, $A^{k-1}\leq a_n < A^k$. Now, we will show that for every given $k$, there exists an $n$ such that this will hold (and this is where unboundedness comes into the picture). Fix a $k$ and let $n_k$ be the smallest integer such that $a_{n_k}\geq A^{k-1}$. Since $a_{n_k-1}<A^{k-1}$, we have $a_{n_k}\leq Aa_{n_k-1}<A^k$, and therefore, $f(a_{n_k})=k$.

$b)$ For this part, since $f(\cdot)$ can take every value at most $m$ times, we have $f(m+1)\geq 2$, $f(2m+1)\geq 3$, etc. Hence, $f(nm+1)\geq n+1$, for every $n\in\mathbb{N}$.

Now,
$$
f(nm+1)\geq n+1 \implies \frac{\log a_{nm+1}}{\log A}\geq n \implies a_{nm+1}\geq A^n=\left(A^{\frac{n}{nm+1}}\right)^{nm+1}.
$$
Hence, it suffices to take $C=A^{t}$, where, $t$ is defined as,
$$
t=\inf\left\{\frac{n}{nm+k}:n,k\in\mathbb{N},k\leq m\right\},
$$
which is (easy to check) equal to $1/2m$.


Bill, Frank and Grete are play a game. Bill is three times as likely to win as Frank, and Frank is two times as likely to win as Grete. Assuming no ties, what is the probability that Grete will win? Express your answer as a common fraction.

Let $ p$ be the probability that Grete wins.  Then, Frank has probability $ 2p$ of winning and Bill has probability $ 3(2p)\equal{}6p$ of winning.  These probabilities add up to $ 1$, so \[ p\plus{}2p\plus{}6p\equal{}1\] \[ 9p\equal{}1\] \[ p\equal{}\boxed{\frac{1}{9}}\]


inequalities
linear algebra
matrix
vector
invariant
quaternions

[b]Let[/b] $ A,B,C \in M_n(\mathbb{R})$ Such that $ AB\equal{}BA,AC\equal{}CA,BC\equal{}CB$ prove that :$ \det(A^2\plus{}B^2\plus{}C^2) \geq 0$

You can actually use a fourth matrix $ D$ (if it commutes with the first 3 as well) and have $ \det(A^2 \plus{}B^2\plus{}C^2\plus{}D^2)\geq 0$.
To prove it consider $ \det(A\plus{}iB\plus{}jC\plus{}kD)\det (A\minus{}iB\minus{}jC\minus{}kD)$ where $ 1,i,j,k$ is the usual basis of $ \mathbb{H}$ for $ \mathbb{R}^4$.

[quote="JC_math"]You can actually use a fourth matrix $ D$ (if it commutes with the first 3 as well) and have $ \det(A^2 \plus{} B^2 \plus{} C^2 \plus{} D^2)\geq 0$.
To prove it consider $ \det(A \plus{} iB \plus{} jC \plus{} kD)\det (A \minus{} iB \minus{} jC \minus{} kD)$ where $ 1,i,j,k$ is the usual basis of $ \mathbb{H}$ for $ \mathbb{R}^4$.[/quote]
  I will read many time and try to understand ..... :read:
 Now , you can thinks about If we use simultaneously triangularizable  matrices  $ A,B,C$ ?
Begin example implies $ A,B \in M_n(\mathbb{R})$ such that $ AB\equal{}BA$ then $ \det(A^2\plus{}B^2) \geq 0$
 we use simultaneously triangularizable  matrices  $ A,B$ Interested in to  arrange eigenvalues of $ A,B$ ....
I hope you help me .Thanks !

[quote="QuyBac"]..Begin example implies $ A,B \in M_n(\mathbb{R})$ such that $ AB \equal{} BA$ then $ \det(A^2 \plus{} B^2) \geq 0$
 we use simultaneously triangularizable  matrices  $ A,B$ Interested in to  arrange eigenvalues of $ A,B$ ....
[/quote]
No need for that, you have $ \det(A^2\plus{}B^2)\equal{}\det((A\plus{}iB)(A\minus{}iB))\equal{}\det(A\plus{}iB) \overline{\det(A\plus{}iB)}$

[quote="JC_math"]
No need for that, you have $ \det(A^2 \plus{} B^2) \equal{} \det((A \plus{} iB)(A \minus{} iB)) \equal{} \det(A \plus{} iB) \overline{\det(A \plus{} iB)}$[/quote]
 I known , here I want understand to arrange eigenvalues of $ A,B$ in simultaneously triangularizable.

1) Hi quybak,
 $ A,B,C,D$ are simultaneously triangularizable over $ \mathbb{C}$ that is we may assume that $ A,B,C,D$ are upper triangular and there exist ordering of the spectra of $ A,B,C,D$ s.t. their diagonals are $ diag(\lambda_i),diag(\mu_i),diag(\nu_i),diag(\delta_i)$. But I do not know how to show that $ \Pi_i({\lambda_i}^2+{\mu_i}^2+{\nu_i}^2+{\delta_i}^2)\geq{0}$

2) Hi JC_math,
$ \mathbb{H}\approx\mathbb{R}^4$, the set of quaternions (see Wikipedia), is a skew field and an algebra of dimension $ 4$ over $ %Error. "mathh" is a bad command.
{R}$ ; there exists a basis $ \{1,i,j,k\}$ s.t. $ i^2=j^2=k^2=ijk=-1$ ; thus any of its element can be writtten $ q=a+bi+cj+dk$. Its conjugate $ \bar{q}=a-bi-cj-dk$ satisfies ${ q\bar{q}}=\bar{q}q=a^2+b^2+c^2+d^2$ and $ \parallel{}q\parallel{}=\sqrt{q\bar{q}}$ is a norm over $ \mathbb{H}$ s.t. $ \parallel{}pq\parallel{}=\parallel{}p\parallel{}\parallel{}q\parallel{}$.
In $ \mathcal{M}_n(\mathbb{H})$, let $ M=A+iB+jC+kD$. Then $ \bar{M}=A-iB-jC-kD$ and $ M\bar{M}=\bar{M}M=A^2+B^2+C^2+D^2$ because the matrices $ A,B,C,D$ pairwise commute AND THEY ARE REAL.
  Yet if $ U\in\mathcal{M}_n(\mathbb{H})$ then I do not know how to define $ det(U)$ ; even  in the case where $ U$ is diagonal.

[quote="loup blanc"]  Yet if $ U\in\mathcal{M}_n(\mathbb{H})$ then I do not know how to define $ det(U)$ ; even  in the case where $ U$ is diagonal.[/quote]
Hi loup blanc,
You could define an injective map $ \Phi_n : \mathcal{M}_n(\mathbb{H}) \to \mathcal{M}_{2n}(\mathbb{C})$ and then for $ M \in \mathcal{M}_n(\mathbb{H})$ define $ \det M: \equal{} \det \circ \Phi_n(M)$.

Hi JC_math,
we can write a quaternion $ q\equal{}a\plus{}ib\plus{}jc\plus{}kd$ in the form $ q\equal{}(a\plus{}ib)\plus{}(c\plus{}id)r$ where $ r^2\equal{}\minus{}1,ri\equal{}\minus{}ir$ (that is $ \mathbb{H}\approx\mathbb{C}^2$ as a vector space). $ det$ is a holomorphic function that commutes with COMPLEX conjugation. Let $ U\in\mathcal{M}_n(\mathbb{C})$ ; then $ det(U)\equal{}x(U)\plus{}iy(U)$ where $ x,y$ are real functions. If $ V\in\mathcal{M}_n(\mathbb{H})$ then let $ det(V)\equal{}x(V)\plus{}y(V)r$. I think that it is a correct definition of the $ det$ of a "quaternionic" matrix but I don't think that $ det(\bar{V})\equal{}\bar{det(V)}$  ??

[quote="loup blanc"]I don't think that $ det(\bar{V}) \equal{} \bar{det(V)}$  ??[/quote]
You are right loup blanc, it is the same with my definition, we don't have $ \det (\overline{M})\equal{}\overline{\det(M)}$.

That why I talk that You are Interested in to arrange eigenvalues of $ A,B$ by if you prove that
  $ A,B,C \in M_n{\mathbb{(R)}}$ such that $ AB \equal{} BA,BC \equal{} CB,CA \equal{} AC$ then $ \det(A\otimes{{A}} \plus{} B\otimes{{B}} \plus{} C\otimes{{C}}) \geq 0$  It isn't implies ..... If You haven't have  $ \det(A\otimes{\overline{A}} \plus{} B\otimes{\overline{B}} \plus{} C\otimes{\overline{C}}) \geq 0$ ......
 I hope that you are interested this matrix theory .....  :play_ball:

Ok we are wedged .
QuyBac are you sure that the result is valid with $ 3$ matrices ?

Assume the contrary. Consider the smallest $ n$ for which the statement is false; consider the corresponding counterexample. Observe that $ A$, $ B$ and $ C$ don't have a common non-trivial invariant (real) subspace. Indeed, assume that $ U$ is an invariant subspace. Denote the restrictions of $ A$, $ B$, $ C$ to $ U$ by $ A_U$, $ B_U$, $ C_U$. Note that since $ U$ is an invariant subspace for $ A$, $ B$, $ C$, they act on the quotient space $ V/U$ (here $ V$ is the whole space). We have
\[ \det(A^2 \plus{} B^2 \plus{} C^2) \equal{} \det(A^2_U \plus{} B^2_U \plus{} C^2_U)\det(A^2_{V/U} \plus{} B^2_{V/U} \plus{} C^2_{V/U}).\]
But since $ \dim U < n$ and $ \dim V/U <n$, both determinants in the righ-hand side are non-negative.

Now consider the ring of fractions $ \cal A$ generated by $ I$, $ A$, $ B$ and $ C$. First of all, $ \cal A$ is a commutative associative algebra of linear operators. Consider a non-zero element $ X$ of $ \cal A$; $ \ker X$ is an invariant subspace for $ A$, $ B$ and $ C$, thus $ \ker X$ is trivial. That is, $ X$ is non-singular. We conclude that $ \cal A$ is a commutative division algebra. Thus $ \cal A\simeq \mathbb R$ or $ \cal A\simeq \mathbb C$ ( [url]http://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)[/url] ). In the former case,  $ n\equal{}1$ and the statement trivially holds. In the latter case, $ n\equal{}2$, operators $ A$, $ B$, $ C$ are of the form
\[ \begin{pmatrix} a& \minus{}b\\b&a\end{pmatrix}.\]
The sum of their squares also has this form. The determinant is $ a^2\plus{}b^2 \geq 0$.

Well done Yuri.
Three remarks:
1) Your $ A_{V/U}$ is in fact $ \phi.A_{V/U}$ where $ \phi: V\rightarrow{V/U}$ is the canonical surjection.
2) The result is still valid for $ k$ matrices $ A_1,\cdots,A_k$ that pairwise commute.
2) The Frobenius theorem is useless:
if $ det(A_1^2\plus{}\cdots\plus{}A_k^2)<0$ then there exists $ t\in[0,1)$ s.t. $ det((tA_1)^2\plus{}\cdots\plus{}(tA_{k\minus{}1})^2\plus{}A_k^2)\equal{}0$. But $ (tA_1)^2\plus{}\cdots\plus{}(tA_{k\minus{}1})^2\plus{}A_k^2$ is in the algebra spanned by $ I,A_1,\cdots,A_k$ and is not invertible ; then it is $ 0$. $ A_1^2\plus{}\cdots\plus{}A_k^2\equal{}(1\minus{}t^2)(A_1^2\plus{}\cdots\plus{}A_{k\minus{}1}^2)$ and $ det(A_1^2\plus{}\cdots\plus{}A_{k\minus{}1}^2)<0$; we have reduced the problem to the case of $ k\minus{}1$ matrices.

[quote="loup blanc"]
1) Your $ A_{V/U}$ is in fact $ \phi.A_{V/U}$ where $ \phi: V\rightarrow{V/U}$ is the canonical surjection.
[/quote]
I am not sure what you mean :maybe: $ A_{V/U}$ is defined by the following commutative diagram:
\[ \begin{array}{ccc}
V&\stackrel{A}{\longrightarrow} & V \\
\downarrow\lefteqn{\phi}&&\downarrow\lefteqn{\phi}\\
V/U&\stackrel{A_{V/U}}{\longrightarrow}& V/U
\end{array}\]
[quote="loup blanc"]
2) The result is still valid for $ k$ matrices $ A_1,\cdots,A_k$ that pairwise commute.
[/quote]
Of course. :D

[quote="loup blanc"]
2) The Frobenius theorem is useless:
if $ det(A_1^2 \plus{} \cdots \plus{} A_k^2) < 0$ then there exists $ t\in[0,1)$ s.t. $ det((tA_1)^2 \plus{} \cdots \plus{} (tA_{k \minus{} 1})^2 \plus{} A_k^2) \equal{} 0$. But $ (tA_1)^2 \plus{} \cdots \plus{} (tA_{k \minus{} 1})^2 \plus{} A_k^2$ is in the algebra spanned by $ I,A_1,\cdots,A_k$ and is not invertible ; then it is $ 0$. $ A_1^2 \plus{} \cdots \plus{} A_k^2 \equal{} (1 \minus{} t^2)(A_1^2 \plus{} \cdots \plus{} A_{k \minus{} 1}^2)$ and $ det(A_1^2 \plus{} \cdots \plus{} A_{k \minus{} 1}^2) < 0$; we have reduced the problem to the case of $ k \minus{} 1$ matrices.
[/quote]
This is a nice alternative solution! However, the version of the Frobenius theorem I use is really simple: every finite dimensional commutative unital associative division algebra over $ \mathbb R$ is either $ \mathbb R$ or $ \mathbb C$. Indeed, first, any such algebra is a field. This field is either an algebraic or transcendental extension of $ \mathbb R$. In the former case, the algebra is clearly either $ \mathbb R$ or $ \mathbb C$. In the latter case, the algebra is infinite dimensional, which contradicts to our assumption.

[quote="loup blanc"]Well done Yuri.
Three remarks:
1) Your $ A_{V/U}$ is in fact $ \phi.A_{V/U}$ where $ \phi: V\rightarrow{V/U}$ is the canonical surjection.
2) The result is still valid for $ k$ matrices $ A_1,\cdots,A_k$ that pairwise commute.
2) The Frobenius theorem is useless:
if $ det(A_1^2 \plus{} \cdots \plus{} A_k^2) < 0$ then there exists $ t\in[0,1)$ s.t. $ det((tA_1)^2 \plus{} \cdots \plus{} (tA_{k \minus{} 1})^2 \plus{} A_k^2) \equal{} 0$. But $ (tA_1)^2 \plus{} \cdots \plus{} (tA_{k \minus{} 1})^2 \plus{} A_k^2$ is in the algebra spanned by $ I,A_1,\cdots,A_k$ and is not invertible ; then it is $ 0$. $ A_1^2 \plus{} \cdots \plus{} A_k^2 \equal{} (1 \minus{} t^2)(A_1^2 \plus{} \cdots \plus{} A_{k \minus{} 1}^2)$ and $ det(A_1^2 \plus{} \cdots \plus{} A_{k \minus{} 1}^2) < 0$; we have reduced the problem to the case of $ k \minus{} 1$ matrices.[/quote]
[quote="loup blanc"]2) The result is still valid for $ k$ matrices $ A_1,\cdots,A_k$ that pairwise commute.[/quote]
 Happy  :roll:  !!! I read many time proof of[b] loup blanc[/b] and  [b]Yury[/b]
A some inequality about It for $ A_k, 1 \leq k \leq m$ can be function a some examples to remember :
 $ \det(e^{A_1} \plus{} e^{A_2} \plus{} ... \plus{} e^{A_m}) \geq  0,\det(ch(A_1) \plus{} ch(A_2) \plus{} ... \plus{} ch(A_m)) \geq 0$ ,$ \det(sin(A_1) \plus{} sin(A_2) \plus{} ... \plus{} sin(A_m) \plus{} mI_n) \geq 0,\det(ch(A_1) \plus{} ch(A_2) \plus{} ... \plus{} ch(A_m)) \geq 0$ ,$ \det(cos(A_1) \plus{} cos(A_2) \plus{} ... \plus{} cos(A_m) \plus{} mI_n) \geq 0$, .....
 I want understand this theory can be from http://www.mathlinks.ro/viewtopic.php?t=288605 here isn't commute.
Add : Let $ A,B \in M_n{\mathbb{(R)}}$ such that : $ A^2B^2 \equal{} B^2A^2$ (or $ A^2B \equal{} BA^2$) then $ \det(A^2 \plus{} B^2) \geq 0$ It isn't hard
But with $ A,B \in M_n{\mathbb{(C)}}$ , $ A\overline{A} \equal{} B\overline{B}$ then $ \det(A\overline{A} \plus{} B\overline{B}) \in \mathbb{R}, \leq 0$ ,...... :rotfl:

[quote=QuyBac]
$\det(ch(A_1) \plus{} ch(A_2) \plus{} ... \plus{} ch(A_m)) \geq 0$
[/quote]
what is ch? i think it is hyperbolic cosine.


Javier is going to Disneyland during spring break. He plans on visiting four particular attractions all before lunch. In how many orders could he visit all four attractions once?

Javier can't visit two attraction at a time so the answer would be
$ 4*3*2*1\equal{}24$.


[url=http://gowers.wordpress.com/]The Polymath1 Project[/url] has apparently been a success!  Read about the project [url=http://michaelnielsen.org/polymath1/index.php?title=Main_Page]on the wiki[/url].

I think this will turn out to be a historic moment in mathematics.  I only wish that I'd had the background to contribute :)

This is very interesting! I hope it works out


Segment CD is tangent to the circle. Segmant AB has length 6, BC has length 10, HD has length 20, and AI has length 4. What is the length of segment CD?

[hide="hint"]
Use POP twice.[/hide]

[hide="solution"]Use POP on A to get:

$ 6\cdot16 \equal{} 4AH$
$ AH \equal{} 24$

Then use POP on D to get:

$ (CD)^2 \equal{} 20\cdot40$
$ CD \equal{} 20\sqrt {2}$[/hide]


I was wondering if someone could help me with the following two problems:

1.) A number N, divided by 6 gives a non-zero quotient and a remainder of 2; the quotient divided by 4 gives a non-zero quotient and a remainder of 3; this second quotient divided by 7 gives a non-zero quotient and a remainder of 5. Compute the smallest possible interger N fulfilling these conditions.

2.) A right triangle has sides of length 12, 16 and 20. The bisector of the larger acute triangle intersects the opposite side at point P. Compute the distance from point P to the hypotenuse of the triangle

[hide]2. let the vertices be A, B, and C such A is opposite the leg of length 12, B is opposite the hypotenuse, and C is opposite the leg of length 16.
AP/BP = 20/12 = 5/3
AP = 10 and BP = 6
CP =  :sqrt: (6 2 + 12 2) = 6 :sqrt: 5

area of APC =  :sqrt: (-3 :sqrt: 5 + 15)(-3 :sqrt: 5 + 5)(-3 :sqrt: 5 - 5)(3 :sqrt 5: + 15)
= :sqrt: (225-45)(45-25) = :sqrt: 3600 = 60
let the altitude from P meet AC at D
20(PD)/2 = 60
PD = 6[/hide]

1)

[hide]
We will work work backwards. The smallest non-zero quotient is 1, so we set the quotient after dividing by 7 equal to 1. Add the remainder (5) to it, and we have the quotient after dividing by 4. Multiply by 4, add the remainder (3) to get the quotient after dividing by 5. Multiply by 5, add the remainder (2), and we are done.

$((1\cdot7+5)\cdot4+3)\cdot5+2=257$[/hide]


geometry
geometric transformation
reflection
analytic geometry
linear algebra
matrix
vector

My math class is doing reflections. We learned how to reflect points and shapes over x-axis, y-axis, line y=x, and line y=-x. I wonder how to reflect those things over other lines, for example y=-x+6. My teacher said she does not know how to do that, only the lines I said in second sentence. How do you reflect over other lines? Thank you.

I think you plug in the x and y coordinates in the equation. For example,
reflecting point $ (1,3)$ over line $ y\equal{}2x\minus{}1$. Plug in 1, $ y\equal{}2(1)\minus{}1 \equal{} 1$ so the y coordinate for the new point will be 1. Plug in 3, $ 3\equal{}2x\minus{}1$, so x = 2. The new point will be $ (2,1)$.

Consider a line $ y \equal{} mx \plus{} b$ and a point $ (x_0, y_0)$. The line $ y \minus{} y_0 \equal{} \minus{} \frac {1}{m}(x \minus{} x_0)$ goes through $ (x_0, y_0)$ and is perpendicular to $ y \equal{} mx \plus{} b$. The intersection of those two lines, which (if I didn't mess up) is

$ P \equal{} \left(\frac {\frac{x_0}{m} \plus{} y_0 \minus{} b}{m \plus{} \frac {1}{m}},\frac {x_0 \plus{} m y_0 \plus{} \frac {b}{m}}{m \plus{} \frac {1}{m}} \right)$

So, to find the reflection, we should go to $ P$ and along the line $ y \minus{} y_0 \equal{} \minus{} \frac {1}{m}(x \minus{} x_0)$ in the direction away from $ (x_0, y_0)$ that for a distance equal to the distance from $ P$ to $ (x_0, y_0)$.

So, that would make the reflection $ 2P \minus{} (x_0, y_0)$.

No, that doesn't work. The line that contains a point and its reflection needs to be perpendicular to the line that you reflect over. The line that contains $ (1,3)$ and $ (2,1)$ is $ y \equal{} \minus{}2x\plus{} 5$ which is not perpendicular to $ y \equal{} 2x \minus{} 1$.

This might help. It shows four different ways of finding the distance from a point to a line.

[url]http://www.worsleyschool.net/science/files/linepoint/distance.html[/url]

Can't you do the pythag theorem and then double the length that you obtain?

[hide="To reflect a point over a line in general..."]
Let $ \vec v \equal{} \begin{pmatrix}x & y\end{pmatrix}$ and the line be $ Ax \plus{} By \plus{} C \equal{} 0$ so we want the reflection $ \vec v' \equal{} \begin{pmatrix}x' & y'\end{pmatrix}$. The distance from the point to the line is $ \frac {|Ax \plus{} By \plus{} C|}{\sqrt {A^2 \plus{} B^2}}$, and therefore $ \|\vec v' \minus{} \vec v\| \equal{} \frac {2|Ax \plus{} By \plus{} C|}{\sqrt {A^2 \plus{} B^2}}$.

We need a direction for $ \vec v' \minus{} \vec v$: the vector that is perpendicular to the line $ Ax \plus{} By \plus{} C \equal{} 0$ is $ \begin{pmatrix}A & B\end{pmatrix}$, so the unit normal in this direction is $ \frac {\begin{pmatrix}A & B\end{pmatrix}}{\|\begin{pmatrix}A & B\end{pmatrix}\|} \equal{} \frac {\begin{pmatrix}A & B\end{pmatrix}}{\sqrt {A^2 \plus{} B^2}}$**.

Finally, we have $ \vec v' \minus{} \vec v \equal{} \begin{pmatrix}x' \minus{} x & y' \minus{} y\end{pmatrix} \equal{} \frac {2|Ax \plus{} By \plus{} C|}{\sqrt {A^2 \plus{} B^2}}\cdot\frac {\begin{pmatrix}A & B\end{pmatrix}}{\sqrt {A^2 \plus{} B^2}}$

$ \equal{} \frac {2|Ax \plus{} By \plus{} C|\begin{pmatrix}A & B\end{pmatrix}}{A^2 \plus{} B^2}$

$ \implies(x',y') \equal{} \boxed{\left(x \plus{} \frac {2A|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2},y \plus{} \frac {2B|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2}\right)}$.

**The unit normal is only $ \frac {\begin{pmatrix}A & B\end{pmatrix}}{\sqrt {A^2 \plus{} B^2}}$ if the point that we are reflecting is above the line of reflection. If it's below it, then the unit normal is instead $ \frac {\begin{pmatrix} \minus{} A & \minus{} B\end{pmatrix}}{\sqrt {A^2 \plus{} B^2}}$, yielding $ \implies(x',y') \equal{} \boxed{\left(x \minus{} \frac {2A|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2},y \minus{} \frac {2B|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2}\right)}$.
[/hide]

Since $ (1,3)$ is above $ y \equal{} 2x \minus{} 1\implies2x \minus{} y \minus{} 1 \equal{} 0$, we get the reflection as $ (2.6,2.2) \equal{} \left(\frac {13}5,\frac {11}5\right)$, I believe.

[b]jonathanchou711:[/b] Yes you can, but there are nicer ways with which you can derive the distance from a point to a line, plane, etc.

[hide="reflections non-wikipedia style"]You have a line $ k$ and a point $ A$. Then you need to draw a line $ l$ perpendicular to line $ k$ through the point $ A$. Label the intersection of lines $ k$ and $ l$ $ B$. Let the distance between $ A$ and $ B$ (which is the length of segment $ AB$, usually denoted $ \overline{AB}$) be $ a$. Now if we draw a circle with center $ B$ and radius $ a$, it will go through point $ A$. It will also go through another point on line $ l$. Let's call it $ C$. We now have $ \overline{AB}\equal{}\overline{BC}\equal{}a$. We define $ C$ to be the [b]image[/b] when $ A$ is reflected over line $ k$. When graphed on the cartesian coordinate plane, the slope of $ l$ is the additive inverse of the reciprocal of $ k$ (for example, if the slope of line $ k$ is $ \minus{}2$, then the slope of line $ l$ is $ \minus{}\frac{1}{\minus{}2}\equal{}\frac{1}{2}$).

Let's take your example: reflecting over the line $ y\equal{}\minus{}x\plus{}6$. Let's say we want to reflect the point $ (4,\minus{}5)$ over the line $ y\equal{}\minus{}x\plus{}6$. Let the perpendicular line through $ (4,\minus{}5)$ to line $ y\equal{}\minus{}x\plus{}6$ be $ y\equal{}mx\plus{}b$. The slope is $ \minus{}\frac{1}{\minus{}1}\equal{}1$, so we have $ y\equal{}x\plus{}b$. We also know that the point $ (4,\minus{}5)$ is on the line, so we know that $ \minus{}5\equal{}4\plus{}b$. Solving for $ b$ yields $ b\equal{}\minus{}9$, and the perpendicular line we are looking for is $ y\equal{}x\minus{}9$.

The intersection of these lines (which I'm sure that you know how to figure out) is $ (15/2, \minus{}3/2)$. Now we draw a circle with this center that goes through $ (4,\minus{}5)$. The radius of this circle is the distance between these two points, which, from the [[Pythagorean Theorem]], is $ \sqrt{(\frac{15}{2}\minus{}4)^2\plus{}(\minus{}\frac{3}{2}\minus{}(\minus{}5))^2}\equal{}\sqrt{\frac{49}{2}}.$ We now have a line with equation $ y\equal{}x\minus{}9$ and a circle with center $ (15/2,\minus{}3/2)$ and radius $ \sqrt{\frac{49}{2}}$. If you know about the equations of circles, you could set up an equation for the given circle, then substitute $ y\equal{}x\minus{}9$ into that equation and solve (you would get (4,-5) as a solution, you would want the other solution), or you could do it the non-rigorous way.[/hide]

[hide="non-rigorous way"]After you find the intersection of the lines, you would want to see how much to the right and how much up you have to go from the original point to the intersection. Use negative values if you aren't going up and to the right. For example, if the original point is $ (x,y)$ and the intersection of the lines is $ (a,b)$, then you go $ a\minus{}x$ to the right and $ b\minus{}y$ up. Once you get those values, add them to $ (a,b)$. This would essentially move $ a\minus{}x$ to the right and $ b\minus{}y$ up from $ (a,b)$. This is the image when $ (x,y)$ is reflected over the original line.[/hide]

Sorry if this isn't that clear to you, but I wanted to put up something that didn't involve vectors. -.-


If $ a,b,c$ are nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$, then

$ ab^3\plus{}bc^3\plus{}ca^3 \plus{}(ab\plus{}bc\plus{}ca)^2 \le 12$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ ab^3 \plus{} bc^3 \plus{} ca^3 \plus{} (ab \plus{} bc \plus{} ca)^2 \le 12$.[/quote]
It's equivalent to$ ab^2\plus{}bc^2\plus{}ca^2\plus{}abc \le 4$ :)

[quote="can_hang2007"] It's equivalent to $ ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc \le 4$ :)[/quote]
Of course, Yeah!  :)


inequalities trigonometry quadratics function geometry trapezoid

I've heard that there are 14 proofs for this inequaltiy:
\[(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2\]
So can you please post your proof for this one? I wanna see if there are more proofs.

I'll post the most basic one:
[color=red]Proof 1[/color]
\[(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2\]
\[\Longleftrightarrow a^2c^2+a^2d^2+b^2c^2+b^2d^2 \geq a^2c^2+2abcd+b^2d^2\]
\[\Longleftrightarrow a^2d^2-2abcd+b^2c^2 \geq 0\]
\[\Longleftrightarrow (ad-bc)^2 \geq 0\]

I will use dot product.
[color=red]Proof 2[/color]

Let $\overrightarrow {OA}=(a,b), \overrightarrow {OB}=(c,d)$ and $\angle AOB=\theta\ (0\leq \theta \leq \pi)$.

We have $\overrightarrow {OA}\cdot \overrightarrow {OB}=|\overrightarrow {OA}||\overrightarrow {OB}|\cos \theta$, thus
${(\overrightarrow {OA}\cdot \overrightarrow {OB})^2}=|\overrightarrow {OA}|^2|\overrightarrow {OB}|^2 \cos^ 2 \theta \leq |\overrightarrow {OA}|^2|\overrightarrow {OB}|^2$
$\Longleftrightarrow (ac+bd)^2\leq (a^2+b^2)(c^2+d^2)$

The equality holds when $\cos ^ 2 \theta=1 \Longleftrightarrow \theta =0,\pi\Longleftrightarrow \overrightarrow {OA} \parallel \overrightarrow {OB}$

[color=red]Proof 3[/color]

For real numbers $a,b,c,d$ and $t$, we have $(at+c)^2+(bt+d)^2\geq 0$
$\Longleftrightarrow (a^2+b^2)t^2+2(ac+bd)t+c^2+d^2\geq 0\ \cdots [\#], \forall t\in\mathbb{R}$
$\Longleftrightarrow$ Case 1:$a^2+b^2=0 \Longleftrightarrow a=b=0$, we have $[\#]\Longleftrightarrow c^2+d^2\geq 0$ which holds for all real numbers $t$ automatically.or
Case 2:$a^2+b^2\neq 0$, consider the graph of quadratic function, under the condition $a^2+b^2>0$,we have  $Discriminant/4=(ac+bd)^2-(a^2+b^2)(c^2+d^2)\leq 0$
$\Longleftrightarrow (a^2+b^2)(c^2+d^2)\geq (ac+bd)^2$

[color=red]Proof 4[/color]
Let $a^2+b^2=x^2$ and $c^2+d^2=y^2$. Set $\frac{a}{x}=\sin{\alpha}$, $\frac{b}{x}=\cos{\alpha}$, $\frac{c}{y}=\sin{\beta}$, $\frac{d}{y}=\cos{\beta}$.

Hence: $\frac{ac}{xy}+\frac{bd}{xy}=\sin{\alpha}\sin{\beta}+\cos{\alpha}\cos{\beta}=\cos{(\alpha-\beta)} \leq1$.

So: $(ac+bd) \leq xy$.

Well , it is quite similar to the first one which we need to expand out

[color=red]Proof 5[/color]

$RHS=a^2c^2+a^2d^2+b^2c^2+b^2d^2=(ac+bd)^2+(ad-bc)^2\geq (ac+bd)^2$

 :)

I found a proof using geometry. I don't know if it is original...

[color=red]Proof 6[/color]
Construct two right triangles with side length $a, b, x$ and $c, d, y$. Hence $a^2+b^2=x^2$, $c^2+d^2=y^2$.

The area of the large trapezoid is $\frac12 (a+c)(b+c)$. So:
\[\frac12 (a+c)(b+d) = \frac12 ab + \frac12 cd + \frac12 xy \sin{\angle{DBA}}\]
\[\Longleftrightarrow ac+bd =xy \sin{\angle{DBA}}  \leq xy\]
\[\Longleftrightarrow (ac+bd)^2 \leq x^2y^2 = (a^2+b^2)(c^2+d^2)\]

Another proof using complex number

[color=red]Proof 7[/color]

Define complex number $w,z$ as $w=a-bi$ ; $z=c+di$

Since $|z||w|=|zw|=|(ac+bd)+(ad-bc)i|=\sqrt{(ac+bd)^2+(ad-bc)^2}\geq \sqrt{(ac+bd)^2}$

But $|w||z|=\sqrt{(a^2+b^2)(c^2+d^2)}$ . Hence we are done

 :)

[i]Congratulations, Shobber ![/i] The your message is a nice provocation.
And I am one from the "victims" of this provocation !. \[\overline {\underline {\left| The\ 8^{th}\ solution.\right| }}\] [b]I suppose w.l.o.g. that [/b]$ac+bd\ge 0 .$

$M_1 (a,b)\in C_1\equiv C(O,r_1),\ M_2(c,d)\in C_2\equiv$ $C(O,r_2);\ a^2+b^2=r_1^2,\ c^2+d^2=r_2^2.$

$|M_1M_2|\ge |r_1-r_2|\Longrightarrow (a-c)^2+(b-d)^2\ge r_1^2+r_2^2-2r_1r_2\Longrightarrow ac+bd\le r_1r_2\Longrightarrow$

$ac+bd\le \sqrt {(a^2+b^2)(c^2+d^2)}\Longleftrightarrow (ac+bd)^2\le (a^2+b^2)(c^2+d^2).$

[b]We have the equality iff [/b] $O\in M_1M_2$ [b], i.e. exists[/b] $k\in R$ [b]such that [/b] $a=k\cdot c,\ b=k\cdot d.$

[b]Remark.[/b] I propose the realization a [u][b]TOP-10[/b][/u] with these solutions. Therefore, we wait at least two solutions. May be anyone who will offer a geometrical interpretation of the C.B.S.'s inequality for any $n\in N^*$.


find all function from (0,inf)---->(0,inf) increasing on (1,inf)  such that
f(ab)f(bc)f(ca)=f(a2)+f(b2)+f(c2)
f(a2b2c2)
from wot i find f(x)=f(1/x) and sum more weird relation but i'm still not able to finish off the problem that f(x)= x+1/x
plz help

A SMALL hint: it follows allmost immediately from an already solved problem: a challenging equation by Orl. It is exactly the same idea and moreover this form can be very quickly changed into that one. Try it!

First prove f(x) is not less than 2 for any x.
Then Suppose f(2)=y+1/y prove for any rational r f(2^r)=y^r+1/y^r
From this because f is increasing on [1,infinity) so f(2^r)=y^r+1/y^r for any real r.
So f(x)=x^k+1/x^k taht r is real

[quote="pco"][quote="yunustuncbilek"]$ \text{\LaTeX} $[/quote]
I think that in june $2004$, $\text{\LaTeX}$ was just just available on forum and most of posts around this beginning did not use it.[/quote]
@2 made his mission :D


geometry
incenter
geometric transformation
homothety
circumcircle
geometry proposed

$ 3$ congruent circles have a common point $ P$ and are in the interior of triangle $ ABC$.Each circle is tangent to $ 2$ sides of the triangle.Prove that the incenter , the circuncenter of $ ABC$ and $ P$ are collinear. :wink:

If $ O_1,O_2,O_3$ denote the centers of the said circles, then it's clear that $ \triangle O_1O_2O_3$ is homothetic to $ \triangle ABC$ through a homothety with center  $I.$ But $ P$ is the circumcenter of $ \triangle O_1O_2O_3$ since $P$ is equidistant from $O_1,O_2,O_3.$ Hence, circumcenters of $\triangle ABC$ and $\triangle O_1O_2O_3$ are collinear with the homothetic center $I.$ Further, $P$ is $X_{55}$ of $\triangle ABC.$


geometry
geometric transformation
reflection
power of a point
radical axis

A generalization for : http://www.mathlinks.ro/Forum/viewtopic.php?t=106916

[color=blue]Problem.[/color]Let $\mathcal{C}_{1},\mathcal{C}_{2}$ and $\l$ be two circles and a line in a plane .
Find a point like $\mathcal{X}$ on $\l$ such that \[\mathcal{XC}_{1}+\mathcal{XC}_{2}\] Takes its minimum possible value.

 $(\mathcal{XC}_{i}$,means the lenght of tangent from $\mathcal{X}$ to $\mathcal{C}_{i})$

Just draw all the possible common tangent segments. If a tangent segemnt intersects the line at a point, then that is the required point. Otherwise reflect one of the circles and do the same thing. :)

For Lomos Lupin, from the his neighbour, Virgil Nicula - Yakima, WA.

The your extension is a very nice geometrical interpretation of the remarkable Minkowschi's inequality.
Suppose w.l.o.g. that the line $l$ separates the circles $C_{1}=C(O_{1},R_{1})$ and $C_{2}=C(O_{2},R_{2})$ (the interesting case).
Denote the projections $A_{1}$, $A_{2}$ and the distancies $d_{1}$, $d_{2}$ of the points $O_{1}$, $O_{2}$ respectively to the line $l\ .$
Then the required point is $M\in [A_{1}A_{2}]$ for which $\frac{MA_{1}}{MA_{2}}=\frac{\sqrt{d^{2}_{1}-R^{2}_{1}}}{\sqrt{d^{2}_{2}-R^{2}_{2}}}\ .$

Precise Solution Mr.Nicula (Master of Latexed Geometry )
Here Goes :
[hide="My Solutoin"]We wil reduce the problem to the old one by replacing the circles by points(Assume that circles dont meet the line )
Let $\mathcal{C}_{1}=(\mathcal{O}_{1},R_{1})$
We will show that there exist a point like $\mathcal{P}_{1}$ such that for any point like $\mathcal{X}$ on $\l$ we have \[\mathcal{X}\mathcal{P}_{1}=\mathcal{X}\mathcal{C}_{1}\] or $\l$ is the radical axis of $\mathcal{P}_{1}$and$\mathcal{C}_{1}$

Let $\mathcal{N}$ be a line passing through $\mathcal{O}_{1}$ and perpendicular to $\l$ .
Because of continuity such a point like $\mathcal{P}_{1}$ exists on $\mathcal{N}$(While $\mathcal{C}_{1}$ doesnt meet $\l$,this holds)
Thus we can replace the $\mathcal{C}_{1}$ with $\mathcal{P}_{1}$ ,In a similar way we can replace the second circle by a point .Therefore problem is reduced to the old one .[/hide]


[quote="M4RI0"]In a triangle $ABC$, let $G$ be the baricenter, $AP$, $BN$ and $CM$ are the medians. $\measuredangle MGA=\alpha$, $\measuredangle CGN =\theta$, $\measuredangle PGB= \beta.$ Determine the equivalent expression for $cotA+cotB+cotC$.

$a)tan\alpha tan\beta tan\theta$
$b)cot\alpha+cot\beta+cot\theta$
$c)tan\alpha+cot\beta+tan\theta$
$d)cot\alpha+tan\beta+cot\theta$
$e)cot\alpha cot\beta cot\theta$[/quote]

what is the baricenter :?:  :oops:

It seems to be the centre of gravity. Which in a triangle is th centroid.

You can use a slash in front of your trig functions to make them look better. i.e.

$\tan x$ is cooler than $tan x$.


geometry
3D geometry
algorithm
ARML
videos

I want to learn to solve Rubik's cube faster. My average time for solving is about 1:10. I think I can easily practice FL(final layer), since it's all about memorizing algorithms. But, I have no idea how to practice for F2L(first two layers).

Someone in the MathLinks/AoPS is bound to be better than me, so give me some advice. :)

BTW, is this an appropriate forum for this topic?

im pretty sure it is the appropriate forum becuz it fits under "game" sort of

You should PM litoxe2718281828 , he's the world record holder for rubix cube.

yea i know its crazy u might want to check one of the last pages on introduce yourself up on the stickies

[quote="ambierona"]Wait, you can memorize the algorithms easily? That's gonna be the hard part for me.. but for doing F2L.. You just need to practice a lot, so you don't hafta stop and look at the cube.. And do finger tricks and stuff too, to go faster.. And there's some algorithms for F2L if you wanna memorize those. Blah, memorizing.. :dry:[/quote]
I did not say memorizing is easy. I meant that the memorizing part can be practiced by myself. But I think I need some help with F2L.

Ah okay. Well, you can also practice F2L.. there are some algorithms for it. Have you PMed Macky? (litoxe2718281828) He has this site with stuff on it, and there are also lotsa other sites with speedcubing methods and algorithms and stuffs. Yeah, i'm not an expert at speedcubing.. haven't learned a faster method yet, and I'm not good with F2L either, but I'm sure if you practice enough, you'll get good at it. And eventually you'll be able to tell what you need to do quickly.

wow...now I want to know how many AoPSers cube.  :lol:

i can solve a rubix cube, but not too fast. i take somewhere around 2-3 minutes. it's been awhile since i've messed w/ the cube. never tried speedcubing.

Oh, I take around 1:10-1:20. 12:11 is really amazing.

I'm basically the same as furious--I can solve the cube in about 2 minutes, but I haven't learned speed cubing.  I want to learn, though...

[quote="LynnelleYe"]I can quickly solve the first two layers (never timed myself) but I've never learned how to solve the third layer and I can't figure it out. Grr.[/quote]

Same here. I can probably memorize the combinations for the third layer without much difficulty, but I am too lazy to.  :D

Wow that was a lot of replies. I can solve the Rubik's cube, but I can't speedcube yet.. my fastest is 47.50 seconds (layer by layer), but I don't do official timing rules or anything, I just time myself.

Yay for Rubik's Cubes. BTW, Macky (litoxe) you want to wear something to identify yourself at ARML? Because I really want to meet you :D Or just bring around a ton of cubes lol.

well cant u identify him from the video??

Sometimes, it takes me more than 12.11 seconds to make a cross! :rotfl:

That is so amazing I wish I could do that.  My times are usually somewhere between 1:05 and (on bad days) 1:25 (mins not hours lol).

[quote="peter"]At San Jose State ARML we got to see him juggle two cubes in one hand and solve one in the other hand.[/quote]

Yeah. That was brilliant. :winner_first:

[quote="peter"]I suck at cubing.[/quote]Yeah me too :( I can only square :icecream:

well, not only can I double a cube, I can also square a circle.

[quote="Derek"]I've attempted to solve one 2 years ago... and my time is well.... 2 years and counting.[/quote]

I've gotta second this, heh. Never was into it heh.

I've been cubing for a while.  My best is probably around 50 sec.  if you want to practice the first layers, just learn algorithms so you can solve edge and corner pairs together after getting the cross.  They aren't too hard and you can figure out alot by yourself, and knowing a few common ones can definitely improve your time.

Well I've been into cubing since January 2004 and I can say I like it. I use the Petrus method with some kind of variation of my own, that is with algs of my own. You can read more about it on my website (a welcome advertisement - I get about 2 visitors every day)

[url=http://www.geocities.com/portoseb/]www.geocities.com/portoseb/[/url]

I'm especially interested in solving the cube using fewest possible moves. I'm proud to hold the unofficial world record for that: 21 moves. My speedcubing records are far from any kind of records though: 21.56 seconds for single solve and 28.31 seconds for an average of 12 solves (best and worst solves removed).

I think it would be really cool if we would make a Rubik's cube meeting at IMO 2005. Three members of the Finnish team (including me) can solve the cube in under one minute, and one of them is even faster than me.

This student assistant coach for MI's ARML team could solve a cube an average of 14 seconds or something. He could do it blindfolded too. It was amazing. He just memorized the cube. He's ranked like 3rd in the nation and has a few thousand dollars worth of cubes.

Anyone here know anything about the 4d Rubik's Cube? Apparently it's a computer program analog thingy and much, much harder than the usual 3d, for obvious reasons.

[quote="ambierona"]I saw that... it's crazy..  [url=http://www.plunk.org/~hatch/MagicCube4dApplet/]here's a java thing with it.[/url][/quote]

Hey! I solved the 2x2x2x2! Now I need to find enough time (around a day) when I can quit doing math and try the big one. I rarely play with it, so I always have to find new algorithms.


Find all positive integers $\ a$ such that $\ 5a^2 + 4$ is a perfect square.  Are there an infinite number?

[quote="shyong"]yes there are .

write it as

$5a^2+4=b^2$

replace $a=2x$ , $b=2y$ then we have

$5x^2+1=y^2$ or $y^2-5x^2=1$ which guarantee infinity solution by Pell's equation since $n$ is square-free .[/quote]

i suppose that seems correct, but it seems that you are assuming something about the nature of the solutions to the pell equation, i.e. that there are an infinite number of pairs that are (even,even), i know this seems pretty obvious that is true, but i don't understand what situations that substitution is allowed,

ex. take the situation:

$3^{2x}=y*2^{2x}+1$

let $a=3^x$, and $b=2^x$

then the equation becomes

$a^2-y*b^2=1$

so there are infinite solutions by pell...but this is wrong...as from this topic:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31947[/url]

there are finite solutions

generally, what substitutions can we make and still use pell?

Linear substitutions, also with rational coefficients, are ok (but you need one solution of that type then). But above, substitution wasn't even necessary as it is everytime when they make sense ;)
And only those are allowed, especially not exponentials or polynomials of degree higher than one.

But: Dirichlets theorem on Units give a lot of more cases where the idea works, e.g. for $a^3+2b^3+4c^3=6abc+1$ (see http://www.mathlinks.ro/Forum/viewtopic.php?t=66669 ).
Indeed there are such polynomials of any degree with that property.


Find the integer-pair solutions to $ x^2\plus{}y^2\equal{}20080104$.

No solutions!
THe right hand side is 0 mod3.
So lhs is also 0 mod3
Only possibility is 0+0 mod3
so lhs is divisible by 9
but rhs not divisible by 9; not possible.

[quote="123456789"]Why can't the two squares on the left side by 1 mod 3 and 2 mod 3?[/quote]

Because every square is 0 or 1 mod3

Oh yes...I made a mistake and checked $ 2^{3}$ instead of $ 2^{2}$ mod 3.

[quote="cf249"]0+0 mod 3 doesn't mean a number has to be divisible by 9.
Example: 3+3=6 (not divisible by 9)[/quote]
hell_ever never claimed this, but that every square is $ 0 \mod 3$. And if a square is divisible by $ 3$, then it is divisble by $ 9$.


How do i do this? (it on the first search result)
http://search.txstate.edu/search?q=PMWC+problems&service=site&x=0&y=0&site=txstate_no_users&output=xml_no_dtd&client=txstate&proxystylesheet=txstate&sitesearch=www.txstate.edu%2Fmathworks
i am talking about the second one :)

That is one of my least favorite problems. There is not really any method to those, you just have to use logic. It is often hard to tell when to start brute forcing it...

To get you started, the units digit of the product is 5. Do you see why?

blah, thats the most obvious thing.

$ 2*3\equal{}6, 2*5\equal{}0, 2*7\equal{}4, 3*5\equal{}5, 3*7\equal{}1, 5*7\equal{}5$

only $ 3*5$ and $ 5*7$'s units digit is one of 2, 3, 5, 7.

well, considering that that was a world level test, that seemed pretty easy...
what is the age maximum on that?

[quote="FantasyLover"]blah, thats the most obvious thing.

$ 2*3 \equal{} 6, 2*5 \equal{} 0, 2*7 \equal{} 4, 3*5 \equal{} 5, 3*7 \equal{} 1, 5*7 \equal{} 5$

only $ 3*5$ and $ 5*7$'s units digit is one of 2, 3, 5, 7.[/quote]

And 5x5 :wink: ...blah you knew but just saying...there isn't any good way to do this problem.

I DID NOT THINK THAT WAS EASY!!!
But, i suck. I can do all the other problems though, now those problems are very easy and leonixln you can't be in more than 8th grade


number theory
relatively prime
number theory unsolved

1. Let $ a$ and $ d$ be relatively prime nos. and consider the sequence: $ a\plus{}d$, $ a\plus{}4d$, $ a\plus{}6d$,......, $ a\plus{}n^2 d$.
Given a positive integer $ b$  prove that one can always find an integer in the sequence relatively prime to $ b$.

2.  A positive integer is called monotic if its digits in base 10, read from left to right  are in non decreasing order. Prove for a natural no. $ n$, there exists an $ n$-didgit montic no. which is a perfect square.

2. For odd n we have $ 16...67^{2}\equal{} 277...88..89$, where the number of 8 = (number of 7) + 1 = (number of 6) + 2.
For even n we have $ 33...35^{2}\equal{}11..22..25$ where the number of 2 = (number of 1) + 1 = (number of 3) + 1.

What is n in the other problem?


If by learning how to read you can increase your scores on the SATs by 10% and if by learning how to pay attention you can increase your scores on the SATs by a further 10%, by what percentage do your scores increase on the SATs if you learn to read and learn to pay attention? Round to the nearest tenth.

I'm betting that there's like some trick to this or I'm missing something... but...

Is it
[hide]21.0 %[/hide]
?

What about this?

[hide]I got a 0 on the SAT this year. If I learned to read, it would become 0. If I learned to pay attention, it would be, 0. Divide by 0 to get $\boxed{Undefined}$.[/hide]

[quote="Treething"][quote="Ignite168"][hide]I got a 2400 on the SAT this year. If I learned to read, it would become 2640. If I learned to pay attention, it would be, 2904. Divide by 2400 to get $\boxed{21\%}$.[/hide][/quote]

Well, no, you only got a 2640.[/quote]

Hahaha. Read the problems more carefully. Plz. Ggkthxbai.

[hide]
The answer is 21.0% because the 10%'s stack and you have to round to the nearest tenth.
[/hide]

[quote="Ignite168"][quote="Treething"][quote="Ignite168"][hide]I got a 2400 on the SAT this year. If I learned to read, it would become 2640. If I learned to pay attention, it would be, 2904. Divide by 2400 to get $\boxed{21\%}$.[/hide][/quote]

Well, no, you only got a 2640.[/quote]

That used to say Ignite168 could use a 10% on his SAT scores.[/quote]

Yes, well, this is more fitting =p .

this is the answer[hide="i think"] 21%[/hide]

More systematically

[hide]$(1.1x)*1.1=1.21x$.  Thus, it is 21.[/hide]

[quote="Ignite168"][quote="Treething"]Don't say I didn't warn you when all of except for Klebian you end up with 727.3s (BIG HINT) on the math portion of the SATs...[/quote]

On the MathCounts rule things for what forms of answers will be expected, it even says after rounding, if it is 0, to omit it. D'OH point for Daesun, now you have 5.[/quote]

The standard rule for rounding is to use zeroes...

Where is t0rajir0u when you need him?

[quote="Knuth's_check"]You just do [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/6758a07da54008e5debd0f5c4de00df0.gif[/img]and after that[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/6d2947227e983b09914f1ae6d1bc9aa4.gif[/img], so it is 21.0%[/quote]

$.1 \cdot.1 \neq=1.21$

$1.1 \cdot 1.1=1.21$

[quote="Knuth's_check"]You just do [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/6758a07da54008e5debd0f5c4de00df0.gif[/img]and after that[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/6d2947227e983b09914f1ae6d1bc9aa4.gif[/img], so it is 21.0%[/quote]

Why are you guys  all using 2400?  100 is easier- At first you get a 100, by learning to read you get a 100+10=110, and by paying attention you get a 110+11=121, so 121-100=21

[quote="Remorseless Thinking"]Hmn...are you sure? Maybe that's why they give you the automatic 200 points...so they have something to deduct from when you start getting every single problem wrong.[/quote]
And since every grade is curved, that 200 RAW SCORE becomes a 260 or so. That's why a 1600 on the SAT in like, 1992 would be equivalent to around 1700 in 2000.

Okaaaay...so is it actually possible to get a 0 on the SAT, if you intentionally got every single one wrong, or would you get a scaled score of like...10?

Just curious.

The lowest score possible on each section is a 200. That's if you get every single question wrong. If you every one blank you are looking at about a 300 :D.

Actually if you leave every question blank you'll get about a 210.

If you leave every question blank, it will not be scored. Leaving every question blank is a sign of cancellation.

[hide]I get 21.0%
Nice way to trick people Treething[/hide]

[quote="Ignite168"][quote="Treething"]Don't say I didn't warn you when all of except for Klebian you end up with 727.3s (BIG HINT) on the math portion of the SATs...[/quote]

On the MathCounts rule things for what forms of answers will be expected, it even says after rounding, if it is 0, to omit it. D'OH point for Daesun, now you have 5.[/quote]
Actually, I was looking at the instructions in the one year comp set of 01 and it said "21.0 will [b] not [/b] be accepted for 21 [b]and vice versa[/b]. D'OH point for ignite.


geometry
parallelogram
geometric transformation
rotation
ratio
homothety
geometry proposed

Let a convex quadrilateral ABCD. Let M,N inside and P,Q outside the quadrilateral ABCD such that MAD,NCB,PAB,QCD are similar to others. Prove that M,N,P,Q form a parallelogram.

MQ/AC=MD/AD     PN/AC=BN/BC      but  we have MD/AD  =BN/BC
so  MQ/AC=PN/AC    MQ=PN

analogous, we acquire  MP=QN

So MPNQ is a parallelogram

Using the plane geometry's drawing:
A rotation around D of angle <ADM followed by a homotethy of ratio AD/DM will map the triangles DMQ and DAC.
A rotation around B of angle <ABP followed by a homothety of ratio AB/BP will map the triangles BPN and BAC.
Since the rotation angles and direction, and the ratios are equal, MQ and PN are equal and parallel.

Best regards,
sunken rock


Hi,

I am typing out some of my notes (in LaTeX) that I have made this term (using pdflatex)

At one point I have a strange tab which I would like to get rid of

\begin{eqnarray*}
	ax^2+bx+c \\
	&= a[x^2 + \frac{bx}{a} + \frac{c}{a}]\\
	&= a[(x+\frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2]\\
	&= a(x+\frac{b}{2a})^2 + c - a(\frac{b}{2a})^2
\end{eqnarray*}

[code]\begin{eqnarray*}
	ax^2+bx+c \\
	&= a[x^2 + \frac{bx}{a} + \frac{c}{a}]\\
	&= a[(x+\frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2]\\
	&= a(x+\frac{b}{2a})^2 + c - a(\frac{b}{2a})^2
\end{eqnarray*}[/code]

Thanks

You are using eqnarray incorrectly which needs 2 &'s for tabs. But please don't use eqnarray (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url] number 5). Use align instead (with the amsmath package loaded). That only requires one & for each tab.
Also make sure () and [] are resized properly by using \left( \right) and \left[ \right]

You can either use small fractions using \tfrac like this:
[code]\begin{align*}
   ax^2+bx+c \\
   &= a\left[x^2 + \tfrac{bx}{a} + \tfrac{c}{a}\right]\\
   &= a\left[(x+\tfrac{b}{2a})^2 + \tfrac{c}{a} - (\tfrac{b}{2a})^2\right]\\
   &= a\left(x+\tfrac{b}{2a}\right)^2 + c - a\left(\tfrac{b}{2a}\right)^2
\end{align*}[/code]
\begin{align*}
   ax^2+bx+c \\
   &= a\left[x^2 + \tfrac{bx}{a} + \tfrac{c}{a}\right]\\
   &= a\left[(x+\tfrac{b}{2a})^2 + \tfrac{c}{a} - (\tfrac{b}{2a})^2\right]\\
   &= a\left(x+\tfrac{b}{2a}\right)^2 + c - a\left(\tfrac{b}{2a}\right)^2
\end{align*}

or large fractions with resizable (),[] like this
[code]\begin{align*}
   ax^2+bx+c \\
   &= a\left[x^2 + \frac{bx}{a} + \frac{c}{a}\right]\\
   &= a\left[\left(x+\frac{b}{2a}\right)^2 + \frac{c}{a} - \left(\frac{b}{2a}\right)^2\right]\\
   &= a\left(x+\frac{b}{2a}\right)^2 + c - a\left(\frac{b}{2a}\right)^2
\end{align*}[/code]
\begin{align*}
   ax^2+bx+c \\
   &= a\left[x^2 + \frac{bx}{a} + \frac{c}{a}\right]\\
   &= a\left[\left(x+\frac{b}{2a}\right)^2 + \frac{c}{a} - \left(\frac{b}{2a}\right)^2\right]\\
   &= a\left(x+\frac{b}{2a}\right)^2 + c - a\left(\frac{b}{2a}\right)^2
\end{align*}

EDIT:

Thanks, worked a treat

EDIT EDIT:

Is is possible to align two equations, one above the other, so terms are "aligned", if you know what I mean. To add them together

[quote="stevem"]Yes you can use more &s in a line. Any symbol after & will be lined up but you can't use \left & \right across the &, so you will have to use things like \big(.
The best way to find out what can be done is to look at [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math Mode[/url]. The default align environment is on page 52 but there's lots more round it, well illustrated.[/quote]

Thanks for the quick responses

I'm now having a problem with the whole many &s on a line.

Demo

\begin{align*} S_n & = a & + (a + d) & + \ldots & + (a + (n - 2)d) & + (a + (n - 1)d) \\
S_n & = (a + (n - 1)d) & + (a + (n - 2)d) & + \ldots & + (a + d) & + a \\
\end{align*}

Its almost right, but some of them are aligned the wrong way

[quote="stevem"]That example makes it clear what you want.
Unfortunately align is designed to alternate the alignment so it goes rl rl rl ... where r=right aligned and l= left aligned. But you can miss out the wrong alignments by using two && together (so the wrong alignment is essentially used for nothing).
\begin{align*} S_n & = a & & + (a + d) & + \ldots & + (a + (n - 2)d) & & + (a + (n - 1)d) \\
S_n & = (a + (n - 1)d) & & + (a + (n - 2)d) & + \ldots & + (a + d) & & + a \\
\end{align*}
Align spreads the formula across the page. You can improve things a bit by using alignat (explained in Math Mode linked in a post above). I can't show it here but this is the code:
[code]\begin{alignat*}{4} S_n & = a && + (a + d) & + \ldots & + (a + (n - 2)d) && + (a + (n - 1)d) \\
S_n & = (a + (n - 1)d) && + (a + (n - 2)d) & + \ldots & + (a + d) && + a \\
\end{alignat*}[/code][/quote]

Worked brilliantly, thanks for your time (on Christmas day as well).

Merry Christmas


[quote="Lepuslapis"]Let $x,y$ be positive reals. Find the minimal value of $\frac{xy+1}{x}+\frac{xy+1}{y}$.[/quote]
$\frac{xy+1}{x}+\frac{xy+1}{y}$
=$x+\frac{1}{x}+y+\frac{1}{y}$
$\geq 2+2$ (AM-GM)
$=4$
(x=y=1)


I have hard geometric problem.
  Let line OS intersect incircle at F and line HF intersect circumcircle at F'. Then prove that |HF|=|FF'|.
 ( H is the orthocenter, O is the center of circumcircle, S is the center of incircle)
  See the figure that I attached. And note that it is true for only one point of intersection

This is not true, unfortunately. It would be true, however, if instead of the line $OS$ wou would consider the line $NS$, where $N$ is the nine-point ccenter of $ABC$ (and you have to consider one of the intersection points between $NS$ and the incircle, since it doesn't hold for both of them; this point is the Feuerbach point of $ABC$).

grobber: I think you are right. I made some mistakes while posting this post


There are $10$ teams in a football league. Each team plays every other team exactly once.

What is the maximum number of undefeated teams at the end of the season?
What is the maximum number of 1-loss teams at the end of the season?
Let $x$ and $y$ be the number of undefeated and 1-loss teams, respectively. What is the maximum value of $x+y$?

[hide]There can only be $1$ undefeated team at the end of the season. It has defeated each other team, so it is the only undefeated team.

There can be $2$ one-loss teams at the end of the season. Each has won 8 times, so those 8 other teams have lost at least twice.

The maximum value of $x+y$ is $2,$ because you can't have and undefeated team and two one-loss teams.[/hide]

[quote="ragnarok23"]Couldn't there be ten undefeated teams? If they all tie with each other....Technically, that's not a loss.
Or are we not counting ties at all?[/quote]
If ties are allowed, then the answer is 10.

[quote="LordoftheMorons"][hide]$1$ (trivial)
$3$ (A loses only to B, B loses only to C, C loses only to A)
$3$ (trivial given the second part)[/hide][/quote]
Oops. :blush:

I think LordoftheMorons is correct. I was thinking that graph theory applies really well.

For example, let's apply it to the second question. Represent teams by nodes, and let directed edges (arrows) from one node to another represent games, with one team beating another team (an arrow from A to B signifies that A beats B). Since there are $10$ teams and each team plays every other team once, we have a complete graph.
Thus focusing our attention to any subset of the 10 nodes, we have another complete graph (every one of the teams we look at will play every other team we look at exactly once). Suppose we have $n$ 1-loss teams, where $n>3$. Then there are $\binom{n}{2}$ games among the $n$ 1-loss teams, indicating that there are a total of $\binom{n}{2}$ losses in these games (1 loss per game). If these teams are indeed 1-loss teams, there should be at most $n$ losses among the $n$ teams we are looking at. However, since $\binom{n}{2}>n$ for $n>3$, we have reached a contradiction. So from there you can check that $n=3$ works, and thus it is the largest possible number of 1-loss teams in the league.

$3$ is maximum for an arbitrary number of teams right?


Find the largest positive $p$ ($p>1$) such, that $\forall a,b,c\in[\frac1p,p]$ the following inequality takes place
\[9(ab+bc+ca)(a^2+b^2+c^2)\geq(a+b+c)^4\]

I hope my proof is correct.
First, write the ineq in an equivalent form
\[
\begin{gathered}
  \frac{{3\left( {\sum {a^2 } } \right)}}
{{\left( {\sum a } \right)^2 }} \geqslant \frac{{\left( {\sum a } \right)^2 }}
{{3\left( {\sum {ab} } \right)}} \hfill \\
   \Leftrightarrow \frac{{3\left( {\sum {a^2 } } \right)}}
{{\left( {\sum a } \right)^2 }} - 1 \geqslant \frac{{\left( {\sum a } \right)^2 }}
{{3\left( {\sum {ab} } \right)}} - 1 \hfill \\
   \Leftrightarrow \frac{{2\left( {\sum {a^2 }  - \sum {ab} } \right)}}
{{\left( {\sum a } \right)^2 }} \geqslant \frac{{\left( {\sum {a^2 }  - \sum {ab} } \right)}}
{{3\left( {\sum {ab} } \right)}} \hfill \\
   \Leftrightarrow 4\left( {\sum {ab} } \right) \geqslant \sum {a^2 }  \hfill \\
\end{gathered}
\]
Now, put $c = p,a = b = \frac{1}{p}$, then \[
\begin{gathered}
  4\left( {\frac{1}
{{p^2 }} + 2} \right) \geqslant \frac{2}
{{p^2 }} + p^2  \hfill \\
   \Leftrightarrow p^4  - 8p - 2 \leqslant 0 \hfill \\
   \Leftrightarrow p \leqslant \sqrt {4 + 3\sqrt 2 }  \hfill \\
\end{gathered}
\]
We will show that $p \le \sqrt {4 + 3\sqrt 2 } $
Let \[
f(a,b,c) = 4(ab + bc + ca) - a^2  - b^2  - c^2
\]
and let \[
t = \sqrt {(a + b)(a + c)}  - a
\]
WLOG assume c to be the largest, it's easy to prove that $f(a,b,c) \ge f(t,t,c)$.
Then consider
\[
\begin{gathered}
  f(t,t,c) - f(tk,tk,\frac{c}
{k}) \hfill \\
   = 2t^2  - 2t^2 k^2  - c^2  + \frac{{c^2 }}
{{k^2 }} \hfill \\
   = 2t^2 (1 - k^2 ) + \frac{{c^2 }}
{{k^2 }}(1 - k^2 ) \hfill \\
   = \left( {2t^2  + \frac{{c^2 }}
{{k^2 }}} \right)(1 - k^2 ) \hfill \\
\end{gathered}
\]
Put $k \le 1$, we have the above inequality $\ge 0$.
That means, if c is larger and a,b are smaller, the minimum is attained.
Put $c=p, a=b=\frac{1}{p}$, and we obtain the answer: $p=\sqrt{4+3\sqrt{2}}$

[quote="Myth"]Maybe $6\sum ab\geq \sum a^2$ ?[/quote]

 $6\sum ab\geq (\sum a)^2$

Let $E(a,b,c)=a^2+b^2+c^2-4(ab+bc+ca)$. Since $E(x,b,c)$ is convex, it is maximal for $x\in \{\frac{1}{p},p\}$.
Moreover, we can easily prove that $E(a,b,c) \le max \{E(\frac{1}{p},b,c), E(p,b,c)\}$.
Thus, $E(a,b,c)$ is maximal for $a,b,c\in \{\frac{1}{p},p\}$, and the problem is practical solved.


I know it is kind of a long time ago, but I wanted to bring up the topic nevertheless.

Did the AMC 12 and AIME seem much easier in 2006 than in previous years? It seemed like that.

There have been many topics on this in the AMC forum where this belongs.

Yes, please make sure you put a topic in the right board before starting one. Locked. And no, I will not move it because there are some many threads on this issue already.


[quote="mihail911"][b]Prove that no polynomial in $ x$ can ever produce primes for all $ x$.[/b][/quote]

Consider, $ f(x)$ to be a polynomial. If it is a constant, then it may be prime for all $ x$. Consider $ f(0)$ for non constant functions

Case $ f(0)\equal{}0,\pm1$ Then $ f(x)$ is not prime for all $ x$.

Case $ f(0)\equal{}m$, if m is non prime then it is not prime for all $ x$

Case $ f(0)\equal{}p$, then it is of the form $ a_n x^n \plus{} \ldots \plus{} a_1 x \plus{} p$. Then $ p|f(p)$ Hence it is not prime for all $ x$

You're close though.

[hide="Consider..."] $ f(kp)$ for sufficiently large $ k$.  :) [/hide]

[quote="JBL"]That is not sufficient: there are plenty of polynomials for which $ f(0) \equal{} p \equal{} f(p)$.[/quote]

Case $ f(0)\equal{}f(p)\equal{}p$

Then $ f(kp) \equiv 0 (\mbox{mod }p)$

Therefore, it is a constant function

The problem should be stated to disallow constant functions.  And you're still not quite there.  What you should say is that for sufficiently large $ k$ we have $ f(kp) > p$, hence $ f(kp)$ is composite, a contradiction.

If you add in some missing steps his method (or what I think it is) works. It looks like he was using the idea that since $ f(x) \minus{} p$ has infinite roots $ f(x)$ is constant, which is the way I first saw it.

[hide="Is this not quite straightforward?"]Suppose that $ f$ is such a polynomial and $ f(k)\equal{}p$ for some integer $ k$ and prime $ p$. Then $ p\mid f(k\plus{}np)\minus{}f(k)\implies p\mid f(k\plus{}np)$ for all $ n\in\mathbb Z$. This implies $ f(k\plus{}np)\equal{}p$ for all $ n\in\mathbb Z$ and therefore $ f$ is constant.[/hide]


My problem is:
For $a,b,c>0$ prove that:
$(a^2+b^2+c^2)(ab+bc+ca) \geq 3(a^2b^2+b^2c^2+c^2a^2)$
can you show me is it wrong? :)  :)

Indeed, it is wrong for a = b = 3 and c = 1, for instance...

  Darij

Ok thanks Darij It is truly wrong, it is still wrong when a,b,c is three side of triangle,  :( therefore I have to check my supposition .