number theory
greatest common divisor
symmetry

Solve the equation in integer numbers:
$x^2+3y^2=74x$

I didn't even use the hint, so I'm sure I'm wrong or at least my answer isn't full.

Wlog $y \geq 0$, we have $x \geq 0$.
Looking mod 2, $x$ and $y$ must be both even. Looking mod 4 again, $y$ must be $4 Y$

Equation to solve is the $12 Y^2 = X(37-X)$, which implies $X \leq 37$ and $Y \leq 5$. This gives only 6 cases to check manually.
Or Since $gcd (X,37-X) \neq 2,3 or 4$ we must have $X=0 \ mod\ 12$ or $X=0 \ mod\ 3$ and $37-X = 0\mod 4$, ...

Putting everything together : $0,0 ; 24,\pm 20 ; 50,\pm 20; 74,0$

[quote]Equation to solve is the $12 Y^2 = X(37-X)$, ...Or Since $gcd (X,37-X) \neq 2,3 or 4$ we must have $X=0 \ mod\ 12$ or $X=0 \ mod\ 3$ and $37-X = 0\mod 4$, ...[/quote]

Or take mod 12, and you get $0 \equiv X(1-X) \bmod 12$ which tells us $X = 0$ or $1 \bmod 12$
Also wlog  Due to symmetry one can assume $X_1<(37/2)<19$  (Every solution for $X_1$ will give two solutions, $X_1$ and $37-X_1)$

(So all you have to try are $0,1,12,13$  ($0,12$  gives other two values by subtracting from 37)


Φτιάγνω  εδώ  ένα  τόπικ  με  θέμα  την  παραστατική  και  προβολική  Γεωμετρία  που  δεν  το  κρύβω  οτι  δεν  βλέπω  να  την  περνάω.......... :(

                Τέλος  πάντων  αν  κανείς  έχει  κανα  καλό  θέμα  ας  το  βάλει   θα  ξεκινήσω  εγώ  με  μία  καλή  άσκηση  και  βλέπουμε  ...

[size=150][color=red]
    [u][b]ΠΡΟΒΛΗΜΑ 1[/b] [/u] [/color][/size]

  Δίνεται  ευθεία  α(α',α")  και  κατακόρυφη  ευθεία  Ζ(Ζ΄, Ζ").  Να  κατασκευασθεί  επίπεδο  διερχόμενο  από  την  ευθεία  α  και  το  οποίο  σχηματίζει  με  την  ευθεία  Ζ  δοσμένη    γωνία  φ.

Ρε Γιώργο δεν υπάρχει λόγος να αγχώνεσαι. Οι ασκήσεις του βιβλίου πιστεύω ότι μας καλύπτουν. Οσον αφορά την παραστατικη καλες ασκήσεις υπάρχουν στις σελίδες 31-36,112-114.Από προβολικη θα διαβάσουμε μόνο κωνικες τομές.
Μεγάλη προσοχή στα χωματουργικά και στα προσθια επίπεδα. Τα θέματα των εξετάσεων θα είναι μέσα από το βιβλίο
Ανδρέας

Ρε  παιδιά  κανείς  για  να  λύσει  την  άσκηση    :)  :)


function
number theory
totient function
prime numbers
number theory unsolved

Find all n such that the totient of n (the number of totatives that n has) is equal to 18.

How do you solve such a problem?

What is in your opinion "totative" means?

I guess you mean Euler's totient function $\varphi(n)$. There is a nice formulae $\varphi(n)=n(1-1/p_1)...(1-p_k)$, where $p_1,...,p_k$ are all distinct prime divisors of $n$. From heer you can easily solve your problem :)

Even with this formula, I was unable to solve it. Can someone show?

[b]Case 1.[/b] $k=1$, then $18=\varphi(n)=\varphi(p^a)=p^{a-1}(p-1)$. It is clear $p\neq 2$, so $p\geq 3$ and $p\mid 18$, therefore $p=3$, $a=3$ OR $p=19$, $a=1$. Thus $n=27$ or $n=19$.

[b]Case 2.[/b] $k=2$, then $18\varphi(n)=\varphi(p^aq^b)=p^{a-1}q^{b-1}(p-1)(q-1)$. If $p$ and $q$ are odd prime numbers, then $4\mid \varphi(p^aq^b)$ -- impossible. It means $q=2$ (or $p=2$ - it doesn't matter). But then $b=1$, otherwise again $4\mid \varphi(p^aq^b)$. Thus we quickly deduce $p=3$, $a=3$ OR $p=19$, $a=1$. And new answers are $n=54$ and $n=38$.

[b]Case 3.[/b] $k=3$, but then $4\mid \varphi(n)$. Contradiction.

Answers: $27,19,38,54$.

I hope I haven't missed some cases...


If $ 0.5\%$ of a number is 5, what is the number?


LaTeX
algebra
polynomial
product of roots

1.  Simplify the rational expression:
[u]2x2 – 12x + 18[/u]
       2x2 – 18

2.  Divide:
[u]x2 – 9x + 20[/u]		÷	[u]x2 – 16[/u]
   3x – 15		                              9x + 36

3.  Simplify:
√[81x12y8z10]

4.  Perform the indicated operations:
2√[48] + 7√[12] –√[27]

5.  Multiply:
(√[6] +  2√[2]) (4√[6] – 3√[2])

6.  Rationalize the denominator:
        2____
√[3] + √[2]

Where is your work to tell if you are doing the problems correctly?

[hide="#1 as an example"]$ \frac{2(x\minus{}3)^{2}}{2(x\minus{}3)(x\plus{}3)}\equal{}\frac{x\minus{}3}{x\plus{}3}$[/hide]

Also, these are much too easy for Intermediate.

1.  Simplify the rational expression:
2x2 – 12x + 18
    2x2 – 18

=2x2-12x+182x2-9 =

Use the difference of the squares formula.
=2x-6x-32x+3x-3 =

Reduce the expression.
=2x-62x+3 =

Factor the numerator.
=2x-32x+3 =

Reduce the expression.
=x-3x+3

Answer is: x-3x+3 .

2.  Divide:
x2 – 9x + 20	÷      x2 – 16
   3x – 15	        9x + 36

Replce the fraction with its reciprocal
=x2-9x+203x-159x+36x2-16 =

Use the rule for multiplying fractions.
=x2-9x+209x+363x-15x2-16 =
=x2-9x+209x+363x-5x2-16 =

Factor the numerator and the denominator.
=x-4x-59x+43x-5x-4x+4 =

Let us reduce the expression.
=93 =
=3

Answer is: 3 .

3.  Simplify:
√[81x12y8z10]

Use the property of powers.
9x6y4z52

Use the property of powers.
9x6y4z5

Use the definition of absolute value.
Break the solution into a number of particular cases.
.
9x6y4z5≥09x6y4z5
The answer is: .	The constraints on the range of variables
9x6y4z5
9x6y4z5≥0

4.  Perform the indicated operations:
2√[48] + 7√[12] –√[27]

248+712-27 =

Take the multiplier outside the radical sign.
=2•43+7•23-33 =

Arrange the terms.
=193

Answer is: 193 .

5.  Multiply:
(√[6] + 2√[2])(4√[6] – 3√[2])

Remove the brackets.
=646-632+2246-2232 =

Arrange the terms.
=624+562-2223 =

Use the property of powers.
=6•4+562-2•2•3 =

Use the property of powers.
=12+562 =

Product of roots equals the root of the product.
=12+56•2 =
=12+53•2•2 =

Property of powers.
=12+53•22 =

Take the multiplier outside the radical sign.
=12+523 =

Remove the brackets.
=12+5•23 =

Group the terms.
=12+5•23 =
=12+103

Answer is: 12+103 .

6.  Rationalize the denominator:
        2____
√[3] + √[2]

Get rid of the irrational value in the denominator.
=23-232-22 =

Use the property of powers.
=23-23-2 =

Arrange the trms.
=23-2 =

Remove the brackets.
=23-22

answer is: 23-22 .

[quote="b-flat"]Wait a sec, is there not a fraction in the first one?  Let me $ \text{\LaTeX}$ the problems to make sure I understand them correctly.  Otherwise, I have no idea what you are doing on #1...

$ \frac{2x^{2}-12x+18}{2x^{2}-18}$

$ \frac{x^{2}-9x+20}{3x-15}\divide\frac{x^{2}-15}{9x+36}$

$ \sqrt{81x\cdot 12y\cdot 8z\cdot10}$

$ 2\sqrt{48}+7\sqrt{12}-\sqrt{27}$

$ \left(\sqrt6+2\sqrt2\right)\left(4\sqrt6-3\sqrt2\right)$

$ \frac{2}{\sqrt3+\sqrt2}$[/quote]

Correct....2nd one there is a divison sign between them..

Also, what program do you use to write out your problems like that? Half my problem is that i can show the question properly..

Eh, sorry about that; I had a typo in my $ \text{\LaTeX}$.  Will be fixed.

To make learn how to make fancy math typeset, try [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url]

I'm pretty sure that what you have on #1 is wrong; see my previous post for the solution.  On #2, I can't really read your work, but the answer you have is incorrect.

Here are a few things to help you with $ \text{\LaTeX}$:

\frac{numerator}{denominator} makes a fraction

x^{2} becomes $ x^{2}$

Surround all code in dollar signs.

The better place to learn about LaTeX for purposes of the forum is http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690


Seventy-five tickets to a student performance were sold. Adult tickets sold for $ \$12.50$ and student tickets sold for $ \$8.50$. If the income from the ticket sales was $ \$865.50$, how many adult tickets were sold.

[quote="cgyao15"]Assume all children first. That would be $ \ 75*8.50\equal{}637.50$
Wow, way to small. Find the difference between cost of a ticket for a child and adult.
12.5-8.5=4
Totals subtracted give 288
288/4=72
So 72 adults went. (wow they must have like 0 children :P)[/quote]
@cgyao15, the ftw answer is not 72 but I will try it a different way.

Let $x=\text{\# of Adult Tickets Sold}$
$75-x=\text{\# of Children Tickets Sold}$

We can write the equation:

$12.50x+8.50(75-x)=865.50$

$12.50x+637.50-8.50x=865.50$

$4x+637.50=865.50$

$4x=228$

$x=57$

Therefore, the number of adult tickets sold is $\boxed{57}$.


Triangle $ ABC$ is such that $ \angle{A}\minus{}\angle{C}\equal{}60^{\circ}$. The internal angular bisector of
$B$ cuts $ AC$ in $ E$. Let $ M$ be the midpoint of $ BC$. If $ BE\equal{}2EM$, show that $ BC\equal{}2AB$.

[hide]Let $ \angle C \equal{} x$, $ \angle A \equal{} x \plus{} 60$, then we find $ \angle ABE \equal{} \angle EBC \equal{} 60 \minus{} x$. Then $ \angle BEA \equal{} 60$. Let $ K$ be the midpoint of $ BE$. Then $ KM \parallel AC$, so $ \angle EKM \equal{} \angle KEA \equal{} 60$. But $ KE \equal{} EM$, so $ \triangle KEM$ is equilateral. From here it is easy to show that $ x \equal{} 30$, so $ \triangle BME$ is a 30-60-90 triangle and $ \triangle BME \cong \triangle BEA$, so $ BC \equal{} 2BA$. [/hide]


quadratics
inequalities unsolved
inequalities

Let $a,x,y,z$ be positive real numbers.Prove that:
$\displaystyle x\frac {a+y}{a+z}+y\frac {a+z}{a+x}+z\frac {a+x}{a+y} \geq x\frac {a+z}{a+x}+y\frac {a+x}{a+y}+z\frac {a+y}{a+z}$

Just multiply with $ (a+x)(a+y)(a+z) $ then use Cauchy(or AM-GM as Lagrange said :maybe: ). :D

yep that's more or less what i did...but i looked at it as a quadratic in a which has discriminant $-3(x-y)^2(y-z)^2(z-x)^2\leq 0$ :D


trigonometry function AMC AIME geometry perimeter ratio

I'm going to pull a 4everwise and start a marathon in a subject that I, and appearently many others, are weak in: the field of complex numbers, trigonometry, and their relations.

[quote]Rules of the Marathon:
1) After somebody correctly answers a problem, they may post anywhere from 2-5 new problems. If they don't want to post, other users may post new problems at random, but make sure they are good. Nothing really easy, and nothing over USAMO #1 level. Make sure whatever problems you post are not trivially similar to any other problem in the marathon. If there is a trick or strategy that many people know that you want to share (roots of unity seive, etc), please post it.
2) Feel free to post all ideas, hints, and solutions, as long as they are hidden. We want to give everybody an equal opportunity. Try not to leave a problem unsolved.
3) You must have fun.

[/quote]

I'm hope the more experienced members (Virgil, T0r, paladin) will help out and post some really good problems and fixes or suggestions to solutions.

I'll kick it off:

[b]Interesting Technique: knowing how to derrive values of the trigonomic functions in both the 15n and 18n groups.[/b]
See: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=family+trig+18n&t=119680
[b]
Places to steal problems from:[/b]
The AoPS Problem Archive
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=family+trig+18n&t=43261
http://www.kalva.demon.co.uk
Old competition tests you may have in your possesion. The more obscure, the better!

[b]Problem 1:[/b]
Find $10 *\cot( cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21).$

Problem #2
Let $x=\frac{\sum_{n=1}^{44}\cos n^\circ}{ \sum_{n=1}^{44}\sin n^\circ}.$ What is the greatest integer that does not exceed $100x$?(AIME #11, 1997)

And PenguinIntegral, c+p didn't work right for your second link
[url]http://www.artofproblemsolving.com/Forum/resources.php[/url]

for #1, this arccot addition formula might be able to save some work
[hide]
$\cot^{-1}x+\cot^{-1}y=\cot^{-1}\frac{xy-1}{x+y}$
(could possibly be off by factor of pi, because of the definition of arccotangent)
[hide="proof"]
Start from the cotangent addition formula:
$\cot(a+b)=\frac{\cot a\cot b-1}{\cot a+\cot b}$
Taking the $\cot^{-1}$ of both sides,
$a+b=\cot^{-1}(\frac{\cot a\cot b-1}{\cot a+\cot b})$
Now substitute: $a=\cot^{-1}x, b=\cot^{-1}y$
This gives us the formula
$\cot^{-1}x+\cot^{-1}y=\cot^{-1}\frac{xy-1}{x+y}$
[/hide]
[/hide]

Junggi, is problem 2 the one where you pair terms? If so, I think that has already been posted somewhere, so why don't you add a few more problems to your post?

Yea, that's how I did it

here's a link to #2
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=393654#p393654[/url]
I guess it was revived pretty recently.... :(

edit:uh oh

holy crap really?
sorry, i didnt take it yet so I didn't know
(I got it from interm trig/complex final chall.)
you might want to edit your post then

[quote="Xaero"]All right, I'll post a problem

[b]Problem 3[/b]

Find the value of
$\cos^{2}10^{\circ}+\cos^{2}50^{\circ}-\sin40^{\circ}\sin80^{\circ}$.[/quote]

[hide]$\cos^{2}10^{\circ}+\cos^{2}50^{\circ}-\sin40^{\circ}\sin80^{\circ}=\frac{\cos 20^{\circ}+1+\cos 100^{\circ}+1}{2}+\frac{\cos 120^{\circ}-\cos 40^{\circ}}{2}=\cos 60^{\circ}\cdot \cos 40^{\circ}+1+\frac{\cos 120^{\circ}-\cos 40^{\circ}}{2}=(\cos 60^{\circ}-\frac{1}{2})\cos 40^{\circ}+1+\frac{\cos 120^{\circ}}{2}=1-\frac{1}{4}=\boxed{\frac{3}{4}}$[/hide]

Somebody else can post the next problem.

Here's an interesting problem:

Find all the values 0 ≤ θ < 360 deg for which the complex number z = cos θ + i sin θ satisfies z^28 - z^8 - 1 = 0.

Since no one's posted a problem...

The roots of $z^{n}=1$ are vertices of a regular polygon. Give the perimeter of an
$n$-sided regular polygon as a trigonometric function of $n$.

EDIT: Darn, I was beat.  :o

[quote="rnwang2"]Find all the values $0 \leq \theta < 360^{\circ}$ for which the complex number $z =\cos \theta+i \sin \theta$ satisfies $z^{2\theta}-z^\theta-1 = 0$. At least I think that's what it says...[/quote]
I think...
Find all the values $0 \leq \theta < 360^{\circ}$ for which the complex number $z =\cos \theta+i \sin \theta$ satisfies $z^{28}-z^{8}-1 = 0$.
[hide] rnwang's problem:
Each side has a length of $\sqrt{1+1-2\cos\left(\frac{2\pi}{n}\right)}$, so the answer is $\boxed{n\sqrt{2-2\cos\left(\frac{2\pi}{n}\right)}}$???? :maybe:  [/hide]

[quote="rnwang2"]That's not exactly what I was looking for.

[hide="Answer"]$2n\sin\left(\frac{180^{\circ}}{n}\right)$. Someone prove it.[/hide]

EDIT: Oh wait. Are the answers the same?[/quote]
They're probably the same. I see how you arrived at your answer: dividing the perimeter into $2n$ pieces.

[hide="New Problem"]
If $6\tan^{-1}x+4\tan^{-1}(3x)=\pi$, compute $x^2$.
Source: ARML 2009
[/hide]

[quote="pl210741"][hide="Solution"]
Let $z=re^{i\theta}$, then $z^5=r^5e^{i5\theta}$, and $\text{Im}(z)=r\sin\theta$. The expression is then equal to $\frac{\sin^5\theta}{\sin 5\theta}$, which obviously has a minimum value of 1 at $\pm\frac{\pi}{2}+2k\pi$.

[/hide][/quote]

There appear to be multiple errors in your solution.

[hide]
Along the lines of your work, we have $\frac{\Im(z^5)}{(\Im(z))^5} = \frac{\sin 5\theta}{\sin^5 \theta}$, with the condition that $\sin\theta \ne 0$.  Using the sine addition formula, we compute \[ \sin 5\theta = 5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta. \]  Dividing through by $\sin^5 \theta$ then gives \begin{align*} \frac{\Im(z^5)}{(\Im(z))^5} &= 1 - 10 \cot^2 \theta + 5 \cot^4 \theta \\ &= 5(u^2 - 2u + 1/5) \\ &= 5(u-1)^2 - 4, \end{align*} where $u = \cot^2 \theta$.  Thus, the minimum value is $-4$, attained when $u = 1$, or $\theta = \frac{(2k-1)\pi}{4}$, $k \in \mathbb{Z}$.[/hide]

[quote="thecmd999"]We divide through by 2 to obtain

3 arctan x + 2 arctan 3x = pi / 2

Note that we cannot take the tangent of both sides because the RHS will be undefined. Thus, we rearrange the equation:

2 arctan 3x = pi / 2 - 3 arctan x

When we take the tangent of both sides now, we cannot use the tangent difference because we will obtain an undefined result. Instead, we interpret this geometrically. Once we realize this key step, the problem is straightforward. Taking the tangent and interpreting the RHS geometrically, we obtain

tan 2 arctan 3x = 1 / tan 3 arctan x

Now we use the tangent double and triple angle identities:

6x / (1 - 9 x ^ 2) = (3 x - x ^ 3) / (1 - 3 x ^ 2)

The x's in the numerator cancel conveniently and we are left with a quadratic in x^2. Solving, we obtain

x^2 = (5 +- 2 sqrt 13) / 9

We note that the negative solution gives - pi instead of pi in the original equation. Thus we discard it and we are done. I probably made a silly mistake somewhere given my ugly answer; please check it.[/quote]

Hi, i got quite a different answer than this^.
using our tangent sum identities, you can prove that $\tan(3a^\circ)=(\frac{3\tan(a)-tan^{3}a}{1-3tan^{2}a}$.
Now to the problem at hand, you can divide both sides by two and obtain the equation: $3\tan^{-1}x+2\tan^{-1}(3x)=\frac{\pi}{2}$. If you take the tangent of both sides you can use your tangent sum identity and get a numerator and denominator in terms of $\tan$ and $x$, etc. Lets just look at the denominator for a second which can be simplified to $1-\tan(3tan^{-1}x)\tan(2\tan^{-1}(3x))$. The clever key to this solution is that the $\tan(\frac{\pi}{2})$ is undefined. Therefore, the denominator of our LHS will equal $0$. So we can set the denominator equal to $0$ and solve. Simplifying and using the first identity above, you will get $1=\frac{3x-x^{3}}{1-3x^2} \times \frac{2 \times 3x}{1-(3x)^{2}}$. Therefore $(1-3x^2)(1-9x^2)=(3x-x^3)(6x)$. Simplifying further yields: $33x^{4}-30x^{2}+1=0$, so using the quadratic formula, you get that $x^{2}=\frac{15 \pm 8\sqrt{3}}{33}$ and we can ignore the $"+"$ solution because it will yield a value of $3\pi$ for the LHS so the only solution is $\boxed{x^{2}=\frac{15-8\sqrt{3}}{33}}$.

I used an alternate but ultimately equivalent method of solution.  We use the identity \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \frac{a+b}{1-ab}, \] which directly follows from the tangent angle addition identity.  Then in particular, \[ 2 \tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2}, \] and \[ 3 \tan^{-1} x = \frac{3x-x^3}{1-3x^2}. \] Hence \[ \frac{\pi}{2} = \tan^{-1} \frac{6x(1-3x^2) + (3x-x^3)(1-9x^2)}{(1-3x^2)(1-9x^2) - (6x)(3x-x^3)}. \]  It follows that the denominator must be zero and the numerator positive.  Hence \begin{align*} 0 &= (1-3x^2)(1-9x^2) - 6x(3x-x^3) \\ &= 1 - 30 x^2 + 33 x^4 \\ &= \frac{(33x^2 - 15)^2 - 192}{33}. \end{align*}  This yields the solutions \[ x^2 = \frac{15 \pm 8 \sqrt{3}}{33}, \] of which we must now check which, if any, give a positive numerator.  As the numerator is $x(9 - 46x^2 + 9x^4)$, and $9x^4 = \frac{3}{11}(30x^2-1)$, we then require \[ \frac{32x}{11} ( 3 - 13x^2 ) > 0, \] or equivalently, $x^2 < 3/13$.  Since this is satisfied only by $x^2 = \frac{15 - 8\sqrt{3}}{33}$, this is the only solution.

Well since no new problem has been posted, I'll defib it back with a fairly simply problem

[hide]Compute $\sin\left(\sum\limits_{n=0}^{4} \tan^{-1}(n)\right)$.[/hide]

[hide="Hint"]You don't really want this hint... [hide="YES I DO"]Use a $\tan^{-1}$ identity.[/hide][/hide]

[hide="Possible Solution"]Since $ \tan^{-1}(0)=0$ or $\pi$, we can reduce the problem to finding $ \sin(\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)+\tan^{-1}(4))$ and analyzing $\tan^{-1}(0)$ at the end.

Applying the arctan sum identity to the first two terms and then the second two terms, we can simplify to
$ \sin\left(\tan^{-1}(-3)+\tan^{-1}(-\frac{7}{11})\right) $.

Applying the sum identity again yields
$\sin\left(\tan^{-1}\left(\frac{-3-\frac{7}{11}}{1-\frac{21}{11}}\right)\right)$
= $\sin\left(\tan^{-1}(4)\right)$

We can set $\tan^{-1}(4)=x$ and solve for $\sin(x)$ to get our desired answer.

$4=\tan(x)$
$=\frac{\sin(x)}{\cos(x)}$
We can square both sides and substitute $\cos^2(x)$ with $1-\sin^2(x)$.
$16=\frac{\sin^2(x)}{1-\sin^2(x)}$.
$16-16\sin^2(x)= \sin^2(x)$ so $\sin(x)=\pm\frac{4\sqrt{17}}{17}$.
There exists two answers since tan can lie in either the first or third quadrant, and the extra $\pi$ from $\tan^{-1}(0)$ changes the sign. [/hide]

[hide="Another solution"]The expression is
$ \sin(\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)+\tan^{-1}(4)) $

Add $ \tan^{-1}(1)+\tan^{-1}(3)$  first,using the arctan identity $ \tan^{-1}(x)+\tan^{-1}(y)$  for xy > 1 to get  $ \pi+ \tan^{-1}(-2)$  which is also $= -\tan^{-1}(2) $
Expression now is $ \sin(\pi+\tan^{-1}(4))= - \sin(\tan^{-1}(4))$
Therefore $ =- \frac{4\sqrt{17}}{17} $ [/hide]

[quote="vlchen888"]Since $ \tan^{-1}(0)=0$ or $\pi$, we can reduce the problem to finding $ \sin(\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)+\tan^{-1}(4))$ and analyzing $\tan^{-1}(0)$ at the end.
[/quote]

@vlchen888: The inverse tangent function uses the branch of the tangent function containing the origin, so $\tan^{-1}0=0$.

Anyways, another problem:

[hide="New Problem"]
Given that $\cos\theta=\frac{2}{7}$, find \[\sum_{n=0}^{\infty} \frac{\cos{n\theta}}{2^n}\]

[/hide]

[hide="Hint"]
Add $i\sin n\theta$ to the numerator and take the real part of the expression.
[/hide]

[quote="neutrinonerd3333"]Anyways, another problem:

[hide="New Problem"]
Given that $\cos\theta=\frac{2}{7}$, find \[\sum_{n=0}^{\infty} \frac{\cos{n\theta}}{2^n}\]

[/hide][/quote]

[hide="Solution"]$\sum_{n=0}^{\infty} \frac{\cos{n\theta}}{2^n}
=Re\sum_{n=0}^{\infty} \frac{e^{in\theta}}{2^n}$

${=Re\sum_{n=0}^{\infty} \left( \frac{e^{i\theta}}{2} \right )^n}$

$=Re \frac{1}{1-\frac{e^{i\theta}}{2}}
=Re \frac{2}{2-e^{i\theta}} $

$= \frac{1}{2-e^{i\theta}}  +  \frac{1}{2-e^{-i\theta}} $

$= \frac{2-e^{i\theta}  + 2-e^{-i\theta} }
{(2-e^{i\theta} )(2-e^{-i\theta} )}$

$= \frac{4 -2\cos\theta }
{4-2e^{i\theta}-2e^{-i\theta} +1}$

$= \frac{4 -2\cos\theta }
{5-4\cos\theta }$[/hide]

We are given cos theta = 2/7, so we plug in to get 8/9.

[quote="tc1729"][b]New Problem[/b]

Let $n\ge 3$ and let $P_0P_1\cdots P_{n-1}$ be a regular $n$-gon inscribed in the unit circle. Show that for all points $P$ on the unit circle, \[PP_0^2+PP_1^2+\cdots +PP_{n-1}^2\] is a constant, and find with proof this constant.[/quote]

[hide="solution"]Let: $K$ be any whole number, $\lambda=gcf(n,K)$ and $m=n/\lambda$. If $m\ge2$ then

\[P_0^K+P_1^K+\cdots +P_{n-1}^K\]
\[=\lambda(Q_0+Q_1+\cdots +Q_{m-1}) = 0\]

where $Q_0,...Q_{m-1}$ is either a regular $m$-gon inscribed in the unit circle, or 2 opposite points on the unit circle in the case that $m=2$.

In your case, $K=2$ and $n\ge3$, so $m\ge2$ and
\[PP_0^2+PP_1^2+\cdots +PP_{n-1}^2 = 0\]
regardless of $P$.[/hide]

[quote="DoctorB"][hide="Variation on tc1729's Problem"]Let $n\ge 3$, let $K$ be any whole number, and let $P_0P_1\cdots P_{n-1}$ be a regular $n$-gon inscribed in the unit circle. Show that \[|P_0^K+P_1^K+\cdots +P_{n-1}^K|\] is a independent of the choice of $P_0P_1\cdots P_{n-1}$.[/hide][/quote]

[hide="solution"]We again have:
\[P_0^K+P_1^K+\cdots +P_{n-1}^K\]\[=\lambda(Q_0+Q_1+\cdots +Q_{m-1})\]
which is zero if $m\ge2$. If $m=1$ (which is to say, $\lambda=n$, i.e. if $K$ is a multiple of $n$) then the above equation reduces to
\[P_0^K+P_1^K+\cdots +P_{n-1}^K=nP_0^K\]
Hence, $|P_0^K+P_1^K+\cdots +P_{n-1}^K|=
\begin{cases}
n &\mbox{ if }
n \mbox{ divides } K\\
0 &\mbox{otherwise}\end{cases}$
[/hide]


If $A$, $B$, and $D$ are the lengths of a side, the shortest diagonal, and the longest diagonal respectively of a regular nonagon, then which of the following is true?
\[\text{(A) }D=A+B\qquad\text{(B) }B^{2}=AD\qquad\text{(C) }D^{2}=A^{2}+AB+B^{2}\qquad\text{(D) }D=\frac{A+B}{2}\qquad \]

An isosceles trapezoid can be draw with A as one base, D as the other and B as each of the equal sides.  Join each pair of diagonals of the trapezoid and observe that they are C -- the second shortest diagpnal of the nonagon.  From an obtuse vertex of the trapezoid drop a perpendicular to the base D.  Now using Pythagoras twice the result B²=AD follows.


function
geometry
geometric transformation
reflection
analytic geometry

Prove that when one function $ f(x)$  meet her inverse $ f^{\minus{}1}(x)$  in only one point
this point there is on line y=x

If $ f$ is the set of points $ (x, y)$, then $ f^{ \minus{} 1}$ is the set of points $ (y, x)$.

So this means that $ f(x)$ and $ f^{ \minus{} 1}(x)$ are reflections over the line $ y \equal{} x$ (why?). So which points in the coordinate plane retain the same coordinates when reflected over $ y \equal{} x$?


The First Day of the 2004 Bulgarian Team Selections Tests is over. Here are the problems:

Problem 1:       http://www.mathlinks.ro/viewtopic.php?t=5903
Problem 2:       http://www.mathlinks.ro/viewtopic.php?p=19725#19725
Problem 3:       http://www.mathlinks.ro/viewtopic.php?p=19726#19726

I know that Bulgarian TST usually come together, like IMO (maybe I'm wrong?)..
So I was wondering if vess could send us the second set of problems..
i would be very grateful to you
Thanks..

I don't want to imagine what other problems were given, if problem 3 was a conjecture from 1984. Maybe problem 6 will be Catalan conjecture, who knows?  ;)  Anyway, the first three problems were very beautiful.

Sure. So, here are the problems of the First Bulgarian TST, Second Day:

Problem 1.     http://www.mathlinks.ro/viewtopic.php?p=19903#19903
Problem 2.     http://www.mathlinks.ro/viewtopic.php?p=19907#19907
Problem 3.     http://www.mathlinks.ro/viewtopic.php?p=19905#19905

when will the results be available? are these all the team selection tests? :)

Well sine there are only two weeks left before IMO, didn't the other tests take place already?
Vess, can you publish them (if you know them)?

yes, and may be inform us who is going, i know that it can be confidential but please....


function
real analysis
real analysis unsolved

Prove or disprove:

Let $ f$ be real function defined on $ [0,1]$ such that any $ x$ is a minimum. Prove that there exists an open interval
where $ f$ is constant

inspired by http://www.mathlinks.ro/Forum/viewtopic.php?t=307069. I think the result is true.

Suppose a contrary. Chose $ x_1 \equal{} 1/2$ and segment $ I_1\subset[0,1]$ where $ x_1$ is local minimum ($ x_1\in I_1$). There is $ x_2\in I_1$ s.t. $ f(x_2) > f(x_1)$. Chose segment $ I_2\subset I_1$ where $ x_2$ is local minimum. And so on. We can WLOG assume $ |I_n|\to 0$. It follows there is unique $ x_0\in I_n$ $ \forall n$. And since $ f(x_1) < f(x_2) < f(x_3) < ...$ we conclude that $ x_0$ cannot be local minimum.

Nice proof. Mine is less simple. By the way did you want to say "And since $ f(x_1) < f(x_2) < f(x_3) < ...$" ?.

Yes, right. I wanted to correct it but you were first :)


Do there exist natural number $ n $ such that $ n! $ starts with $1384  $ ? :!:

See http://www.mathlinks.ro/Forum/viewtopic.php?t=39366 or http://www.mathlinks.ro/Forum/viewtopic.php?t=46302 .

  darij


function
integration
calculus
derivative
real analysis
real analysis theorems

Let $ f$ be a continuous function then we have:

$ \int^{x}_{0}f(u)(x\minus{}u)\;du\equal{}\int^{x}_{0}\left(\int^{u}_{0}f(t)dt\right)du$

Let $ F(y)$ be the anti-derivative of $ f(y)$.

Then the lefthand side is equal to:

$ F(u)(x\minus{}u) |_{0}^{x}\plus{}\int_{0}^{x}F(u)\hspace{1 mm}du \equal{}\minus{}F(0)x\plus{}\int_{0}^{x}F(u)\hspace{1 mm}du$

And the righthand side is equal to:

$ \int_{0}^{x}F(u)\minus{}F(0)\hspace{1 mm}du \equal{}\int_{0}^{x}F(u)\hspace{1 mm}du\minus{}F(0)x$

and these two expressions are of course equal to one another.

Asking for $ f$ to be continuous is a stronger hypothesis than needed. All we really need is for $ f$ to be integrable on $ [0,x].$ (Make that Lebesgue integrable.) If we drop continuity as a hypothesis, we cannot use the antiderivative, but that's all right, because Fubini's theorem makes for a more natural proof anyway:

$ \int_{0}^{x}\int_{0}^{u}f(t)\,dt\,du\equal{}\int_{0}^{x}\int_{t}^{x}f(t)\,du\,dt \equal{}\int_{0}^{x}(x\minus{}t)f(t)\,dt.$


[quote="t0rajir0u"]Avoiding an overreliance on the $\text{cis}$ form builds algebra skills, since many complex-number problems can be done purely algebraically without any trig (edit: or without a substitution $z = a+bi$, which is also unnecessary for many problems).

[hide="This one, for example..."] $x^{3}= 1 \implies x^{14}+\frac{1}{x^{14}}= x^{2}+\frac{1}{x^{-1}}= x^{2}+x = \boxed{-1}$ [/hide][/quote]

oh hahaha opps  :blush:


function
real analysis
real analysis solved

Does there exist continuous functions f,g,h:R->R such that
f(g(x)+h(y))=max(x,y) for every x,y reals?

It's clear that g and h aren't constant (none of them). Let M1=sup[g(x)], m1=inf[g(x)], M2=sup[h(x)] and m2=inf[h(x)] when x\in R. g takes all values in (m1, M1) and h takes all values in (m2, M2). It's obviouys that we can find 2 distinct pairs (a, b) and (c,d) \in (m1, M1)x(m2, M2) s.t. a+b=c+d, and we take g(x)=a, h(y)=b, g(z)=c, h(t)=d. Then a=/=c and b=/=d, so x=/=z and y=/=t, so max(x,y)=/=max(z,t), but f(g(x)+h(y))=f(a+b)=f(c+d)=f(g(z)+h(t)), so we have a contradiction.

It looks like we don't need f to be continuous to prove the impossibility.


1) with a triangle which has 0<A,B,C<pi/2;

$\sum\frac{cosC}{cosA/2cosB/2}\geq1$

2) the same condition:

$\sum\frac{cosC}{sinC/2+1}\geq1$

I think R.H.S of the first problem have to be [tex]2[/tex]


algebra
polynomial
inequalities
induction
search
quadratics
algebra unsolved

Prove that a real polynomial P(x) is >= 0 for all x iff P(x) can be expressed as the sum of the squares of 2 other real polynomials.

The forward implication is trivial but im having trouble proving the backwards implication.

I believe this was once a putnam.
Also, wlog assume $\exists x$ such that $P(x)=0$, otherwise, take min of $P$ and scale down by that number.
Then, we have that at every zero of $P$ there is a relative min, i.e. $P'(r)=P(r)=0$. It follows that at each $r$ there is a double root, and I think that proves the assertion. It's something like that, I've never seen a solution to this before but I sure hope it's true because I use it all the time for inequalities!!!

Induction on $\deg P$.
This is clear for $P(x)$ constant $\geq 0$, so assume $\deg P\geq 1$.
Take $\lambda=a+ib\in\mathbb{C}$ some root of $P(x)$.
Then $0=\overline{P(\lambda)}=P(\overline{\lambda})$, and $\overline{\lambda}$ is also root of $P(x)$.
If $\lambda\in\mathbb{R}$, then $\lambda$ must be a double root of $P(x)$, as $P(x)\geq 0$ in the vicinity of $\lambda$.
So in both cases $(x-\lambda)(x-\overline{\lambda})|P(x)$ with $(x-\lambda)(x-\overline{\lambda})=(x-a)^{2}+b^{2}\in\mathbb{R}[x]$.
So $P(x)=(x-\lambda)(x-\overline{\lambda})Q(x)$ for some $Q(x)\in\mathbb{R}[x]$ with $\deg Q<\deg P$.
But $(x-\lambda)(x-\overline{\lambda})=(x-\lambda)(\overline{x-\lambda})=|x-\lambda|^{2}$ for all $x\in\mathbb{R}$, and then $Q(x)={P(x)\over |x-\lambda|^{2}}\geq 0$ for all $x\in\mathbb{R}\setminus\{\lambda\}$.
$Q(x)$ is continuous, and therefore $Q(x)\geq 0$ for all $x\in\mathbb{R}$
and $Q(x)=R(x)^{2}+S(x)^{2}$ for some $R(x),S(x)\in\mathbb{R}[x]$ by induction.
So
$P(x)=|x-\lambda|^{2}(R(x)^{2}+S(x)^{2})=|x-\lambda|^{2}|R(x)+iS(x)|^{2}\\\hspace*{0.5in}=|(x-\lambda)(R(x)+iS(x))|^{2}$
for all $x\in\mathbb{R}$.
$(x-\lambda)(R(x)+iS(x))=(x-a-ib)(R(x)+iS(x))\\\hspace*{1.53in}=[(x-a)R(x)+bS(x)]+i[(x-a)S(x)-bR(x)]$
and therefore
$P(x)=|(x-a)R(x)+bS(x)+i((x-a)S(x)-bR(x))|^{2}\\\hspace*{0.5in}=((x-a)R(x)+bS(x))^{2}+((x-a)S(x)-bR(x))^{2}$
for all $x\in\mathbb{R}$.
This proves the result.

Posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=15432 (there are more, search them yourself if you want).

yeah proved it without reading anyone elses soln :D.

lemma 1: If P is quadratic the result is trivialy true by completion of the square.

Assume true for all non-negative polynomials of degree 2n

a corollary of FTOA, any real polynomial of even degree can be factorised into real quadratic factors.

Let P(x) be any polynomial of deg 2n+2

=> P(x)=Q(x)R(x) where deg Q = 2n and deg R = 2.

By the inductive assumption and lemma 1:

P(x)=Q(x)R(x)

P(x)=(a^2+b^2)(c^2+d^2)

=(ac+bd)^2 + (ad-bc)^2

=> QED by induction.

oh and P(x) cannot have odd degree as that would give it a root by the intermediate value theorem.


calculus
integration
analytic geometry
calculus computations

Calculate $ \int_0^\infty x^2e^{\minus{}(x^2)/2}dx$.

By parts, since $ xe^{ - \frac {x^2}{2}} = - \frac {d}{dx}e^{ - \frac {x^2}{2}}$, we find the integral to be equal to
\[ \left.\left( - xe^{ - \frac {x^2}{2}\right)\right|_0^\infty + \int_0^\infty e^{ - \frac {x^2}{2}}dx.
}\]
As usual, calling this last integral $ I$, it is solved transforming into polar coordinates:
\[ I^2 = \int_0^\infty\int_0^\infty e^{ - \frac {x^2 + y^2}{2}}dxdy = \left(\int_0^\frac {\pi}{2}d\theta\right)\left(\int_0^\infty\rho e^{ - \frac {\rho^2}{2}}d\rho\right) = \frac {\pi}{2},
\]
and the desired result is $ \sqrt {\frac {\pi}{2}}$.


Princeton college MIT Duke Stanford calculus trigonometry

[quote="Scrambled"]are there any notable african, native american, oceanian (indonesia, new guinea, etc...), or south american mathematicians?[/quote]

Depends a lot on your definition of [b]notable[/b] mathematician. I would say I know at least two great mathematicians, one from South America, and other from Africa (living mathematicians that are well recongnized in their fields: Gambo and Caffarelli)

Also, you have Argelia Velez-Rodriguez, Cuban (and black, and woman! with a world full of prejuces, it probably was hard for here to make her way), and Alberto P Calderón, Argentinan, George Eugene Uhlenbeck from Indonesia, Claude Chevalley, South Africa, Jacob Lionel Bakst Cooper, South Africa. Probably if you look, you'll find more.

Best,

I'm sorry, but i come from Indonesia and i never hear about that man..

Anyone can tell me who is he?

This brings me to something I have been thinking about for a while. What if a mathematical genius was born in a third world country. Do you think that:

1. He/She would not be able to develop this talent because he/she does not have the resources.

or

2. He/She would still be able to develop this talent because it is just something they are born with.

[quote="djimenez"]I would say I know at least two great mathematicians, one from South America, ... Caffarelli)

... and Alberto P Calderón, Argentinan, ...[/quote]
I just noticed this post. I was personally a Ph.D. student of Alberto Calderón, so I am quite aware of his greatness. The mid-to-late 20th century saw a considerable flowering of Argentine mathematicians, quite a few with at least some connection to Calderón: Alberto Torchinsky, Felipe Zo, Nestor Riviére, Luis Caffarelli, Carlos Kenig (Kenig was a graduate student at the same time I was). I also know of a few from other parts of South America - I think both Mischa Cotlar and Cora Sadowski were from Venezuela. (And if you're wondering about that profusion of non-Spanish names, I suggest you look up some details about the demographics of Argentina and some of the other parts of South America.)

Calderón was a cultured and civilized man, and a citizen of the world at large. I think he would have found the whole premise of this topic, the idea of singling out "third world" mathematicians, as a little odd. He was a mathematician - and that needs no adjectives.

I went to a workshop this summer where Kenig spoke; he's quite an interesting speaker but most of his talk went over mine (and everyone else's?) heads.  Something about Carleman estimates...

Just a few that I know by experience at Princeton are Prof. Arlie Petters from Belize (PhD from MIT, now a professor at Duke) and Prof. Kate Okikiolu, who I believe was raised in England but has Nigerian and British parents, (BA Cambridge, PhD UCLA, now a professor at UCSD). Her Nigerian father was also a good mathematician.

I know that there are quite a few other successful mathematicians in the US originally from Nigeria and Benin. For example, Prof. Wilfred Gangbo at Georgia Tech and Prof. Idris Assani (born in Niger but from Benin) from UNC-Chapel Hill.

Also Prof. Peter Sarnak (PhD Stanford), who was the previous head of the Princeton Math Department, is originally from South Africa.

Despite being opposed to the term "third world country" -- I bet there are quite a few mathematicians from such countries in South American, Central America, and Asia also.

WHat about Ramanujan? Also, there are probably many notable chinese mathematicians. They are all from third world countries...

My professor is Argentinian (not strictly a mathematician). He says that one of his Argentinian friends use to be a professor at MIT, Caltech, and now Georgia Tech. Anybody know who I could be talking about?

[quote="blahblahblah"][quote="neelnal"]Ramanujan, Bhaskara, Megnath Shaha etc.

You can pull a lot from India, where the number system came up and differential calculus was invented.[/quote]

The latter is a stretch at best.[/quote]
Sorry to go off topic by I believe it's Not  a stretch at all!

 Madhava (Of the famous arctan ($\tan^{-1}(x) = x-x^{/}3+x^{5}/5 \dots$)  Madhava (also known as Gregory series or Madhava-Gregory series - but was published by Madhava a few centuries before)   and other kerala Mathematicians have done quite some work.

(Heck,  Yukibhasa, compiled by Jyeshtadeva (In 16th Century ?) can be (and has been  regarded by some) regarded  as one of the earliest calculus text book.

(Madhava and other's work has, for example"  has -

   - Methods of differentiation.
   - Integration.
   - Term by term integration.
    -  Numerical integration by means of infinite series.
    -  The theory that the area under a curve is its integral.
      even  integration in deriving the areas of curved surfaces and the volumes )

 Even Bhaskarcharya (much before these guys) (And I know enough Sanskrit, to have read  Suryasidhanta .. and he clearly understands and documents  things like  differentiation (well he does not call it differentiation but tells the "slope" of the curve)  of sin(x) was cos(x) etc...

Few nice books  or sites may make some  interesting reading:
( unfortunately there are few authors who are very biased who minimize or ignore non-western contributions, and also, of course,   there are  others who exaggerate  too :) - you may like to learn Sanskrit to read the originals )

Eg:
 Neither Newton nor Leibniz - The Pre-History of Calculus and Celestial Mechanics in Medieval Kerala

Link:http://www.canisius.edu/topos/rajeev.asp
Or
 An overview of Indian mathematics. Indian Maths. School of Mathematics and Statistics University of St Andrews

Science and technology in free India  (  Prof.C.G.Ramachandran Nair)

There are many other references (google shows many)... Some  among them I find  are trustworthy ...

FWIW - Just saw wikipedia article about Calculus:
Link: http://en.wikipedia.org/wiki/Calculus
[quote]i.....
In India, the mathematician-astronomer Aryabhata in 499 used infinitesimals and expressed an astronomical problem in the form of a basic differential equation.[2] Manjula, in the 10th century, elaborated on this differential equation in a commentary. This equation eventually led Bhaskara in the 12th century to develop independently a number of fundamental ideas in calculus, perhaps including an early form of the theorem now known as "Rolle's theorem".[3] He was also the first to define the notion of the derivative as a limit. In the 14th century, Madhava, along with other mathematician-astronomers of the Kerala School, studied infinite series, power series, Taylor series, differentiation, integration, and the mean value theorem.[4] Yuktibhasa, which some consider to be the first text on calculus, summarizes these results.[5][6][7][8] These developments would not be duplicated in Europe until much later.[/quote]


The area of a trapezoidal field is 1400 square yards.  Its altitude is 50 yards.  Find the two bases, if the number of yards in each base is an integer divisible by 8.  The number of solutions to this problem is:
\[ \text{(A) none} \qquad \text{(B) one} \qquad \text{(C) two} \qquad \text{(D) three} \qquad \text{(E) more than three} \]

[hide="hint"]
solve for $x+y=56$[/hide]
[hide="answer"]
D[/hide]

Wow, I can't believe no one posted a solution.
[hide]Formula for area of a trapezoid is $\frac{h(b_1+b_2)}{2}$.  So pluggin in the numbers, we have $\frac{50(b_1+b_2)}{2}=1400$.  Solving for $(b_1+b_2)$ gives us $b_1+b_2=56$, where $b_1$ and $b_2$ are integers divisible by $8$.  Listing out the multiples of $8$ less than $56$:

8, 16, 24, 32, 40, 48, 56.  We see three "pairs" of these which add up to 56, $8+48$, $16+40$, and $24+32$, so there are $3$ pairs giving us answer [b]D[/b].[/hide]

[hide]
h(b1+b2)/2=1400
50(b1+b2)=2800
b1+b2=56

8*x+8*y=56
x+y=7
(1,6)
(2,5)
(3,4)
We see 3 solutions so we point at answer D as being correct

[/hide]


calculus
integration
calculus computations

$ \sum_{n\equal{}1}^{\infty}\frac{1}{n^{3}}$

Does anyone know how to sum this? I think ive seen it done by relating it to an integral, but I can't remember.

There is no known method for suming this series. It's value is also know as Apéry's constant, because Roger Apéry (a French matematician) showed in 1979 that $ \zeta (3)$ is an irrational number.

Yes, Carcul is right
http://www.mathlinks.ro/Forum/viewtopic.php?t=144858


If $\sum^{\infty}_{n=0}\frac{\sin (nx)}{3^{n}}= \frac{a+b \sqrt c}{d}$ and $\sin x = \frac{1}{3}$ with $0 \leq x \leq \frac{\pi}{2}$, compute a+b+c+d (a,b,c,d are integers).

[hide="Hint"] As boxedexe would point out, problems of this type are trivial once you realize that $\sin nx = \frac{e^{inx}-e^{-inx}}{2i}$. [/hide]
You should also mention that of course $c$ is squarefree.

Or you can just do like

[hide]

this is the imaginary part of  and that's an arithmetic sum or something right?

[/hide]

[quote="t0rajir0u"][hide="Hint"] As boxedexe would point out, problems of this type are trivial once you realize that $\sin nx = \frac{e^{inx}-e^{-inx}}{2}$. [/hide]
You should also mention that of course $c$ is squarefree.[/quote]
You have a typo in the denominator of the identity.  It should be $2i$ instead of $2$.

[hide]
$\sum^{\infty}_{n=0}\frac{\sin nx}{3^{n}}$
$=\frac{e^{nix}-e^{-nix}}{2i}\cdot \frac{1}{3^{n}}$
$=\frac{e^{nix}-e^{-nix}}{3^{n}}\cdot \frac{1}{2i}$
$\frac{1}{2i}\sum^{\infty}_{n=0}\left(\frac{e^{ix}}{3}\right)^{n}-\frac{1}{2i}\sum^{\infty}_{n=0}\left(\frac{e^{-ix}}{3}\right)^{n}$

Which are two geometric series, which we know how to sum:

$\frac{1}{2i}\left[\left(\frac{1}{1-\frac{e^{ix}}{3}}\right)-\left(\frac{1}{1-\frac{e^{-ix}}{3}}\right) \right]$

Then just sub in the fact that $\sin x = \frac{1}{3}$ and you're almost done.

$\cos x = \frac{2\sqrt{2}}{3}$
$e^{ix}= \cos x+i \sin x = \frac{2\sqrt{2}+i}{3}$
$e^{-ix}= \frac{2\sqrt{2}-i}{3}$

Sub that in to the above eqn, rationalize denominators, and simplify.

I'm too lazy to do it now.

[/hide]


When x^9 − x is factored as completely as possible into polynomials and monomials with real integral coefficients, the number of factors is:

A. 2; B. 3; C. 4; D. 5; E. none

\begin{align*}
x^9-x&=x(x^8-1)=x(x^4-1)(x^4+1)\\
&=x(x^2-1)(x^2+1)(x^4+1)\\
&=x(x-1)(x+1)(x^2+1)(x^4+1)
\end{align*}.

Now you COULD go further, but that's 5 and that's all your question asks for.

You can't; except $ x$ they all are cyclotomic polynomials, and $ x^8 \minus{} 1 \equal{} \prod_{d \mid 8} \Phi_d(x)$. I think he was misled by your "real integral" wording (usually "integer" is enough), and with real coefficients one can go further to $ x^4 \plus{} 1 \equal{} (x^2 \plus{} \sqrt{2}x \plus{} 1)(x^2 \minus{} \sqrt{2}x \plus{} 1)$.


geometry
incenter
circumcircle
geometry unsolved

Given $\triangle ABC$ incenter $I$ and nice-point circle $(C)$ segment $IA\cap (C)={A_{1}}$ and line $IA\cap (C)={A_{2}}$ define similar $B_{1},B_{2}$ and $C_{1},C_{2}$ prove that
$cos\frac{A}{2}IA_{1}+cos\frac{B}{2}IB_{1}+cos\frac{C}{2}IC_{1}=cos\frac{A}{2}IA_{2}+cos\frac{B}{2}IB_{2}+cos\frac{C}{2}IC_{2}$
 :)

The problem can be generalized as follows

Given triangle $ABC$ and $(C)$ a circle. Suppose $P$ is a point inside the circle such that we can construct the intersections $A_{1}, B_{1}, C_{1}$ of the segment $PA, PB, PC$ with the circle. Call $A_{2}, B_{2}, C_{2}$ be the second intersections. Let $\measuredangle{BPC}= \alpha, \measuredangle{CPA}= \beta, \measuredangle{APB}= \gamma$ (here we consider angles with direction ). Then

$\sum PA_{1}sin\alpha = \sum PA_{2}sin\alpha$

[i]Proof[/i]

Let $O$ be the center of the circle. Let $A_{1}', B_{1}', C_{1}', A_{2}', B'_{2}, C'_{2}$ be respectively the symmetric of points $A_{1}, B_{1}, C_{1}, A_{2}, B_{2}, C_{2}$ wrt $O$. Then the two triangles $X_{1}Y_{1}Z_{1}$ and $X_{2}Y_{2}Z_{2}$ formed by $(A_{1}A'_{2}, B_{1}B'_{2}, C_{1}C'_{2})$ and $(A_{2}A'_{1}, B_{2}B'_{1}, C_{2}C'_{1})$  are congruent and have the same orientation. Thus their algebric areas are equal and their circumradius are equal. Call $S$ and $R$ their area and circumradius respectively.

Now, it is not hard to see that $LHS = RHS = \frac{S}{R}$. :)

yes thank treegoner for the very good solution  :D  , but I think it doesn't need angles with direction! :)


Support
geometry
power of a point
radical axis
projective geometry
geometry unsolved

Let $P$ be a point inside $\odot O$ that do not coincide with $O$. $\odot O'$ is the circle with centre $O'$ and diameter $OP$. $l$ is the radical axis of $\odot O$ and $\odot O'$. If $P$ is the intersection of the diagonals of qaudrilateral $ABCD$ that is incribed in $\odot O$, then the intersection of $AD$ and $BC$ lies on $l$.

[hide]When proving this theorem, my book used a lot of lemmas. For example, it constructed tangent lines from $A$, $B$, $C$ and $D$. Then line $AA$ and $CC$ intersect at $F$ and $BB$ and $DD$ intersect at $E$. After proving that $E$ and $F$ lie on $l$, the book then said that these two points also lie on $AC$ and $BD$ respectively. But the Geometer's Sketchpad does not support this conclusion. Moreover, the book showed another lemma: If $G \in \odot O$, let $R=EG \cap \odot O$ and $Q= FG \odot O$, then $P$, $Q$, $R$ are collinear. But the Geometer's Sketchpad still shows that this conclusion is wrong.

So I want you to help me with these problems: How do I prove the initial problem? Are the two lemmas stated above really wrong or it is just because I have made the wrong picture?

Thank you.[/hide]

I have something unclear with your problem. Can you make a picture of it? If circles have same center, then they doesn't have radical axis.

The line $l$ is the polar line of the point $P$ with respect to $\odot O$.
[hide="Here is an explanation why"]Let $A \in OA \cap l$. It's known that $OA \perp l$
Let $R$ be the radius of $\odot O$
The power of the point $A$ w.r.t. $\odot O, \odot O'$ respectively is $AO\cdot AP$ and $AO^{2}-R^{2}$
$AP\cdot AO=AO^{2}-R^{2}$
$R^{2}= AO^{2}-AP\cdot AO$
$R^{2}= AO(AO-AP)$
$R^{2}= AO\cdot OP$

This mean that the point $A$ is the inverse of point $P$.[/hide]

No we will use the theorem: [quote="Arne"][b]Theorem.[/b] Let $ABCD$ be a cyclic (and not necessarily convex) quadrilateral (or, more generally, a quadrilateral inscribed in a conic). The line passing through $AB \cap CD$ and $AD \cap BC$ is the polar of $AC \cap BD$. [/quote]([url]http://www.mathlinks.ro/Forum/viewtopic.php?t=25036[/url])

see also [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=316994#p316994]here[/url] and [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=390&forum=DCForumID6]here, Theorem 1[/url].

This theorem says that the points $AB \cap CD$ and $AD \cap BC$ are both on the polar line of $P$, which is the line $l$.

Thanks for the information about Polar. But I haven't learned anything about Inversion or Pole and Polar yet  :blush: . I shall start learning it tonight......


limit logarithms inequalities function trigonometry integration calculus

We all now that $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges.
But how many different ways can you show this?

Here is a very beautiful way.

[b]Theorem:[/b] Let $\sum_{n=1}^{\infty}s_{n}$ be convergent and $(s_{n})$ is a non-increasing sequence, then $\lim ns_{n}= 0$.

Let $H(n) = \sum_{k=1}^{n}\frac{1}{k}$.  Also, consider

$a(n) = \frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}$.

Note that for $n > 1$,

$a(n) = \frac{2n}{n^{2}-1}+\frac{1}{n}> \frac{2n}{n^{2}}+\frac{1}{n}= \frac{3}{n}$.

Therefore,

$H(3n+1) = \sum_{k=1}^{3n+1}\frac{1}{k}= 1+\sum_{j=1}^{n}a(3j) > 1+\sum_{j=1}^{n}\frac{3}{3j}= 1+H(n)$.

It follows that

$H(9n+4) = H(3(3n+1)+1)) > 2+H(n)$,
$H(27n+13) > 3+H(n)$,
$H(81n+40) > 4+H(n)$,

and in general, for any positive integers $m, n$,

$H\left(3^{m}n+\frac{3^{m}-1}{2}\right) > m+H(n)$.

Let $n = 0$ to give $H((3^{m}-1)/2) > m$, and we can immediately see that $H$ is unbounded above.  Indeed, the above proof gives us a simple (though extremely rough) asymptotic behavior to $H$.

Here you have a shorter proof.

Suppose that the series converges and let S denote its sum.
It follows that
\begin{eqnarray*}S &=& 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\ldots\\ &=& \left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}\right)+\ldots\\ &>& 1+\frac{1}{2}+\frac{1}{3}+\ldots\\ &=& S, \end{eqnarray*}
which is a contradiction.

[quote="sylow_theory"]

[b]Theorem:[/b] Let $\sum_{n=1}^{\infty}s_{n}$ be convergent and $(s_{n})$ is a non-increasing sequence, then $\lim ns_{n}= 0$.[/quote]
We can prove this by something like $ns_{2n}\le s_{n+1}+\cdots+s_{2}n$. :)

[quote="coquitao"]Here you have a shorter proof.[/quote]

Your proof lacks sufficient rigor, because by your reasoning the series $\sum_{k=0}^\infty (-1)^{k}$ converges.  One needs to treat the infinite series with care.

Nope, coquitao's proof is perfectly correct. He never said that his trick gives a [b]sufficient[/b] condition for convergence, just a necessary one. So, your objection should read "you cannot prove that $\sum_{k}(-1)^{k}$ diverges by this method", which is true, of course, but completely irrelevant to the problem. :)

My objection is not irrelevant, because his proof does not define what is meant by the expression $1+\frac{1}{2}+\frac{1}{3}+\cdots$; namely, it is the use of the mathematical ellipsis that makes the reasoning imprecise.  Manipulating it to obtain a contradiction does not prove its divergence.  Such an approach is sloppy and would not be accepted in any undergraduate analysis course.  Since when does a nonsensical statement S > S imply S is infinite?

It's very hard to see where you are coming from. The meaning of $1+\frac{1}{2}+\cdots$ is well agreed-upon, and his assumption was that the sum converged and hence took a finite value $S$. But since it follows that $S > S$ this is a contradiction, and hence the assumption must be false.

"Manipulating it to obtain a contradiction does not prove its divergence" just makes no sense... I don't know how you are reasoning.

[quote="Zhaobin"]We can prove this by something like...[/quote]

$\lim_{n\to\infty}ns_{n}= \lim_{n\to\infty}\frac{s_{n}}{\frac{1}{n}}$

and so we are comparing our convergent series $s_{n}$ with the divergent harmonic series. Suppose that this limit is a real number different from zero. Then by the comparision test $s_{n}$ would also diverge, which contradicts our hypothesis. Suppose now that the limit is $\infty$: again, by the comparision test $s_{n}$ would have to be divergent which contradicts again the assumption that $s_{n}$ converges. So, the above limit can only be zero.

Comparing the harmonic series to itself is rather pointless. If we're trying to prove divergence of the harmonic series itself, we need something like zhaobin's argument.

That idea goes back a very long way historically- look up Oresme.

[quote="atomicwedgie"]All I will say is that if you had written such a proof in my real analysis course, you would not have received credit.[/quote]
Ok, but fedja and I are saying that you are wrong. It's a completely correct mathematical argument, and any professor or TA would recognize that and give full credit for it. You seem to think that it is an incorrect proof because it includes the manipulation of series. The general idea is that you should be careful when manipulating series, particularly ones with varying sign - the idea is not that "a proof involving the manipulation of series is wrong".

There is published precedent for showing divergence of a series through rearrangement.  Take a look at the following note:

[url]http://www.rowan.edu/math/osler/Div_by_Rearrange_proof.pdf[/url]

(It's a couple of pages in.)

Additionally, this note cites another note where rearrangement of series is used explicitly to show that the harmonic series diverges.  I also seem to remember seeing this proof in a textbook once, maybe Bartle and Sherbert?

Though with that said, just because something has been published does not necessarily mean it is correct -- but the method must have survived some peer-reviewed process which is all we have sometimes.  :-)

Here is another method for proving the divergence of the harmonic series.

$\frac{1}{n}\geq \ln \frac{n+1}{n}$ for $n\geq 1$.

Hence $1+\frac{1}{2}+\cdots+\frac{1}{n}\geq \ln (n+1)$ and hence $\sum_{n\geq 1}\frac{1}{n}=\infty$.

And another one:

Let $H_{n}$ be the $n^{th}$ harmonic number, i.e. $H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$.

Assume that the series converges, i.e. $\lim_{n\rightarrow \infty}H_{n}=a$ where it is not hard to see that $a>1$.

Now for $p$ and $q$ natural numbers one has that  the following double inequality holds:

$\gamma<H_{p}+H_{q}-H_{pq}\leq 1$.

Take $p=n$ and $q=2n$ and let n converge to $\infty$ and we get that $a\leq 1$ which is definitely a contradiction.

a lot of people solved it. so i began to lose criativity.. :(

Suppose all the sums below exist and all limits exist too and are independent

${\sum_{n=1}^{\infty }\frac{1}{n}=\sum_{n=0}^{\infty }\left(\frac{1}{2 n+1}+\frac{1}{2 n+2}\right)=\sum_{n=0}^{\infty }\left(\frac{1}{3 n+1}+\frac{1}{3 n+2}+\frac{1}{3 n+3}\right)}$

${s>1}$
${\sum_{n=1}^{\infty }\frac{1}{n}=\sum_{n=0}^{\infty }\sum_{i=1}^{s}\frac{1}{i+n s}\to \lim_{s\to \infty }\, \left(\sum_{n=0}^{\infty }\sum_{i=1}^{s}\frac{1}{i+n s}\right)\to \sum_{n=0}^{\infty }\left(\lim_{s\to \infty }\, \left(\sum_{i=1}^{s}\frac{1}{i+n s}\right)\right)\to \sum_{n=0}^{\infty }\log \left(1+\frac{1}{n}\right)=\infty }$

this interaction of limit is allowed, since it is independent convergent, as the supposition?

edoo - the potential vulnerability of series to being changed by rearrangement applies to series with some positive and some negative terms. Positive term series are robust - the sum might be infinite, but if so any rearrangement will also sum to infinity. I would also accept coquitao's proof as correct.

One question for all of these: what would happen if you applied the same method, or a very similar method, to a $p$-series, that is,

$\sum_{n=1}^{\infty}\frac1{n^{p}}$ either for $p>1$ or for $p<1?$

atomicwedgie's very first post on this topic might be a good one to start with.

I posted one method above using that nice theorem. I can post a proof if you so desire.

Here is another amazing method.

Define the sequence $(s_{n})$ as follows:
$1/2$
$1/4,1/4$
$1/8,1/8,1/8,1/8$
$1/16,1/16,1/16,1/16,1/16,1/16,1/16,1/16$
....

It is easy to see that $\sum_{n=2}^{\infty}\frac{1}{n}$ dominates the series $\sum_{n=2}^{\infty}s_{n}$. The rest is trivial.

That last argument is essentially Oresme's proof, and is over 700 years old. It's also a good one to try to generalize.

Seems like we've exhausted "elementary" arguments for divergence. Let's try something "fancy" for diversity. Consider the $2\pi$ periodic function $F$ that equals $\pi-x$ on $[0,2\pi)$. It is not hard to find its Fourier series: $F(x)\asymp 2\sum_{n\ne 0}\frac{\sin nx}{n}$. If the harmonic series were convergent, $F$ would be equivalent to an everywhere continuous function. But it is clear that discontinuity at $0$ cannot be removed by changing $F$ on a set of measure $0$.

[quote="fedja"]Seems like we've exhausted "elementary" arguments for divergence. Let's try something "fancy" for diversity...[/quote]

The series of the reciprocals of the prime numbers diverges, therefore...

[quote="coquitao"]The series of the reciprocals of the prime numbers diverges, therefore...[/quote]
Do you have a proof of that that doesn't use as already known the divergence of the harmonic series?

[b]Indeed, I do have one... It is not original with me, but still...[/b]

There's a famous proof due to Erdos. Here is one particularly nice, due to Leo Moser.

Another elementary proof:

If the Harmonic series converges, then Muntz-Szasz implies that the span of $\{ x^{n}: n \in N \}$ is not dense in $C[0,1]$ which is not true -- and would be very sad for numerical analysts if it were true.  :)

[quote="sylow_theory"]I posted one method above using that nice theorem. I can post a proof if you so desire.

[/quote]

Can you please post the proof?Thanks.

I found this proof when i was the 2nd year student:
$ \frac1n\equal{}\int_0^1 x^{n\minus{}1}dx<\int_{\delta}^{1\minus{}\delta}x^{n\minus{}1}dx$, now we can sum it up and use the geometric series $ \sum x^n\equal{}\frac{1}{1\minus{}x}$, as far as $ x\in(\delta, 1\minus{}\delta)$. The conclusion is obvious.

We are approaching this topic in calculus 2, so here it goes...
For $ |x| < 1$ we have $ \log(1 \minus{} x) \equal{} \minus{} \sum_{n \equal{} 1}^{\infty}\frac {x^n}{n}$. If the harmonic series had finite sum $ S$, then by [url=http://en.wikipedia.org/wiki/Abel%27s_theorem]Abel's theorem[/url] we would have $ \lim_{x\to 1^{ \minus{} }}\log (1 \minus{} x) \equal{} \minus{} S$.


Να αποδείξετε ότι υπάρχουν άπειροι φυσικοί αριθμοί $ n$ για τους οποίους ισχύει ότι $ p| (2^{n}-n)$ για κάθε πρώτο $ p$. Μπορεί να έχει ξαναμπεί η άσκηση, οπότε ας μη διστάσει ο moderator να διαγράψει το post :)

Έτσι όπως την διαβάζω η άσκηση είναι λάθος. Μάλλον πρέπει να δείξεις ότι για κάθε πρώτο $ p$ υπάρχουν άπειροι φυσικοί αριθμοί $ n$ για τους οποίους ισχύει ότι $ p|(2^n \minus{} n)$.

[hide]
Μικρό θεώρημα του Fermat.
[/hide]

[hide]αν ο πρωτος ειναι ο 2 τοτε καθε αρτιος φυσικος $ n$ μας ικανοποιει

αν ο πρωτος δεν ειναι ο 2 τοτε ο $ n\equal{}(p\minus{}1)^{2k}$ μας ικανοποιει, k natural(με βαση το fermat little theorem)[/hide]


inequalities
analytic geometry
geometry
circumcircle
trigonometry
incenter
geometry proposed

Given are $\triangle ABC$ with any $M$ on plane prove that
$bMA^{2}+cMB^{2}+aMC^{2}\ge abc$ :)

[quote="sinajackson"]by cauchy we have: $(bMA^{2}+cMB^{2}+aMC^{2})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq (MA+MB+MC)^{2}$
and so:  $(bMA^{2}+cMB^{2}+aMC^{2}) \geq \frac{(MA+MB+MC)^{2}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\geq \frac{\frac{(a+b+c)^{2}}{4}}{\frac{ab+bc+ca}{abc}}=\frac{abc(a+b+c)^{2}}{4(ab+bc+ca)}$

we have:
$\frac{(a+b+c)^{2}}{4(ab+bc+ca)}=\frac{a^{2}+b^{2}+c^{2}+2ab+2bc+2ca}{4(ab+bc+ca)}\geq \frac{(2(ab+bc+ca)+2ab+2bc+2ca)}{4(ab+bc+ca)}=1$

As a consequence:$(bMA^{2}+cMB^{2}+aMC^{2}) \geq abc$

I hope ( and I think :D  ) my solution is correct.[/quote]

I think it is not. At least I assume you use $a^{2}+b^{2}+c^{2}\ge 2ab+2bc+2ca$ which does not hold. A bound of this type does give nothing more but $a^{2}+b^{2}+c^{2}\ge ab+bc+ca$.

Oh sorry :maybe: ,  I always make such mistakes because I always think i have solved the problem without no doubt :oops: . and that's a bad habit...

Gemath, [u]abandon du travail[/u] !

[color=darkblue]Given are a fixed triangle $ABC$ and a fixed point $P(\alpha ,\beta ,\gamma )\ ,\ \alpha+\beta+\gamma=1\ ,$ where between brackets are the normal barycentrical coordinates of the point $P$ w.r.t. the triangle $ABC\ .$  Denote the power $p_{w}(X)$ of a point $X$ w.r.t. the circumcircle $w$ of the triangle $ABC\ .$ Then for any points $X$ , $X_{1}(x_{1},y_{1},z_{1})$ , $X_{2}(x_{2},y_{2},z_{2})$ there are the following relations :

$\{\begin{array}{cc}1\blacktriangleright & \{\begin{array}{c}\alpha\cdot XA^{2}+\beta\cdot XB^{2}+\gamma\cdot XC^{2}=XP^{2}+\beta\gamma a^{2}+\gamma\alpha b^{2}+\alpha\beta c^{2}\ .\\\\ \alpha\cdot XA^{2}+\beta\cdot XB^{2}+\gamma\cdot XC^{2}\ge \beta\gamma a^{2}+\gamma\alpha b^{2}+\alpha\beta c^{2}\ .\end{array}\\\\ 2\blacktriangleright & [X_{1}X_{2}]^{2}=-(y_{1}-y_{2})(z_{1}-z_{2})\cdot a^{2}+(z_{1}-z_{2})(x_{1}-x_{2})\cdot b^{2}+(x_{1}-x_{2})(y_{1}-y_{2})\cdot c^{2}\ge 0\ .\\\\ 3\blacktriangleright &-p_{w}(P)=R^{2}-PO^{2}=\beta\gamma a^{2}+\gamma\alpha b^{2}+\alpha\beta c^{2}=\alpha\cdot PA^{2}+\beta\cdot PB^{2}+\gamma\cdot PC^{2}\le R^{2}\ .\end{array}$

Now you can choose any lower point and you can try apply the above identity (inequality) $(1)\ :$

$\{\begin{array}{cc}1\blacktriangleright & \{\begin{array}{cccc}A(1,0,0) & B(0,1,0) & C(0,0,1) & G(\frac{1}{3},\frac{1}{3},\frac{1}{3})\\\\ I(\frac{a}{2p},\frac{b}{2p},\frac{c}{2p}) & I_{a}(\frac{-a}{2(p-a)},\frac{b}{2(p-a)},\frac{c}{2(p-a)}) & I_{b}(\frac{a}{2(p-b)},\frac{-b}{2(p-b)},\frac{c}{2(p-b)}) & I_{c}(\frac{a}{2(p-c)},\frac{b}{2(p-c)},\frac{-c}{2(p-c)})\end{array}\\\\ 2\blacktriangleright\ & \{\begin{array}{cc}H(\cot B\cot C,\cot C\cot A,\cot A\cot B) & O(\frac{R^{2}\sin A}{2S},\frac{R^{2}\sin 2B}{2S},\frac{R^{2}\sin 2C}{2S})\\\\ N(\frac{p-a}{p},\frac{p-b}{p},\frac{p-c}{p}) & \Gamma(\frac{r_{a}}{4R+r},\frac{r_{b}}{4R+r},\frac{r_{c}}{4R+r})\end{array}\\\\ 3\blacktriangleright & L(\frac{a^{2}}{a^{2}+b^{2}+c^{2}},\frac{b^{2}}{a^{2}+b^{2}+c^{2}},\frac{c^{2}}{a^{2}+b^{2}+c^{2}})\end{array}\ ,$

[b][u]where : [/u][/b]$G$[b]- centroid ; [/b]$H$[b]- orthocenter ; [/b]$O$[b]- circumcenter ; [/b] $I$[b]- incenter ; [/b]                              $I_{a}$[b]-  A-exincenter a.s.o. ; [/b]$N$[b]- Nagel's point ; [/b]$\Gamma$[b]- Gergogne's point ; [/b]$L$[b]- Lemoine's point.[/b]

The powers of the all above points are presented lower down and I recommend you ascertain by themselves only by means of the above relation $(3)\ :$

$\{\begin{array}{c}p_{w}(O)=-R^{2}\ \ ;\ \ p_{w}(A)=p_{w}(B)=p_{w}(C)=0\ \ ,\ \ p_{w}(G)=-\frac{a^{2}+b^{2}+c^{2}}{9}\ .\\\\ \frac{-p_{w}(I)}{r}=\frac{p_{w}(I_{a})}{r_{a}}=\frac{p_{w}(I_{b})}{r_{b}}=\frac{p_{w}(I_{c})}{r_{c}}=2R\ \ ;\ \ p_{w}(N)=-4r(R-r)\\\\ p_{w}(\Gamma )=-r(R+r)(\frac{2p}{4R+r})^{2}\ \ ;\ \ p_{w}(L)=-3\cdot(\frac{abc}{a^{2}+b^{2}+c^{2}})^{2}\ .\end{array}$

Only with the relation $(2)$ you can ascertain the distance between any two above remarkable points
or the distance between any point $X$ and an above remarkable point. For example $\boxed{\ ON=R-2r\ }$ and the following :

$\{\begin{array}{c}PA^{2}=\gamma b^{2}+\beta c^{2}+p_{w}(P)\ge 0\ .\\\\ PI^{2}=-4Rr+(\gamma ab+\alpha bc+\beta ca)+p_{w}(P)\ge 0\ .\\\\ HI^{2}=4R^{2}+4Rr+3r^{2}-p^{2}\ge 0\ .\\\\ OH^{2}=9\cdot GO^{2}=R^{2}(1-8\cos A\cos B\cos C)=9R^{2}-(a^{2}+b^{2}+c^{2})\ge 0\ .\\\\ PG^{2}=\frac{2}{9}\cdot (a^{2}+b^{2}+c^{2})-\frac{1}{3}(\alpha a^{2}+\beta b^{2}+\gamma c^{2})+p_{w}(P)\ge 0\ .\\\\ HI^{2}=4R^{2}+4Rr+3r^{2}-p^{2}\ge 0.\\\\ I\Gamma^{2}=r^{2}\cdot[1-(\frac{p\sqrt 3}{4R=r})^{3}]\ge 0\ .\\\\ \mathrm{a.s.o.}\\\\ \mathrm{successfully\ !}\end{array}$
[/color]

Here's my solution:
Using $(b\overrightarrow{MA}+c\overrightarrow{MB}+a\overrightarrow{CM})^{2}\ge 0$
Expand; $(a+b+c)(bMA^{2}+cMB^{2}+aMC^{2}) \ge ab^{3}+bc^{3}+ca^{3}$
By Muirhead, $ab^{3}+bc^{3}+ca^{3}\ge a^{2}bc+ab^{2}c+abc^{2}=(a+b+c)abc$
Therefore, $(a+b+c)(bMA^{2}+cMB^{2}+aMC^{2}) \ge (a+b+c)abc$
Therefore, $(bMA^{2}+cMB^{2}+aMC^{2}) \ge abc$
QED :lol:

That's right msceok it is my proof too :D thank Virgil Nicula for your nice remarks

Gemath, these remarks isn't for [b]you[/b] ! I mean "you" - the reader who doesn't know it.


How many 3-digit numbers have the property that the first digit is at least twice the second digit?

Faster way than listing all those out:

If the first digit is $ n$, the second digit is $ \left \lfloor \frac{n}{2} \right \rfloor\plus{}1$.  Then we have:

$ \left \lfloor \frac{1}{2} \right \rfloor\plus{}1\plus{}\left \lfloor \frac{2}{2} \right \rfloor\plus{}1\plus{}\left \lfloor \frac{3}{2} \right \rfloor\plus{}1\plus{}\dots\plus{}\left \lfloor \frac{9}{2} \right \rfloor\plus{}1\equal{}1\plus{}2\plus{}2\plus{}3\plus{}3\plus{}4\plus{}4\plus{}5\plus{}5\equal{}29$ choices for the first two digits.

The final digit can be any digit, so our answer is $ 29 \times 10\equal{}\boxed{290}$ numbers.

Those answers are incorrect.
[hide = Solution]

The first digit doesn't matter so you have $9$ choices.
The second and third digit are grouped together
If the tens digit is $0$, then the units digit can be $0-9$, which is $\boxed{10}$ possibilities.
If the tens digit is $1$, then the units digit can be $2-9$, which is $\boxed{8}$ possibilities.
If the tens digit is $2$, then the units digit can be $4-9$, which is $\boxed{6}$ possibilities.
If the tens digit is $3$, then the units digit can be $6-9$, which is $\boxed{4}$ possibilities.
If the tens digit is $4$, then the units digit can be $8-9$, which is $\boxed{2}$ possibilities.
Every tens digit that is greater than $5$ doesn't work.
So,
$9(10+8+6+4+2) = \boxed{270}$

[quote=GameBot]How many 3-digit numbers have the property that the [b]first digit[/b] is at least twice the [b]second digit?[/b][/quote]

Please note that the first digit refers to the hundreds place, and the second digit refers to the tens digit.

[hide=Solution]
We proceed with casework.

[b]Tens Digit is 0:[/b]
When the tens digit is zero, the hundreds digit can have any value from $1$ to $9$, yielding a total of $9$ values.

[b]Tens Digit is 1:[/b]
When the tens digit is one, the hundreds digit can have any value from $2$ to $9$, yielding a total of $8$ values.

[b]Tens Digit is 2:[/b]
Similarly, we get $6$ for this case.

[b]Tens Digit is 3:[/b]
Similarly, we get $4$ for this case.

[b]Tens Digit is 4:[/b]
Similarly, we get $2$ for this case.

Now, since the hundreds digit cannot be $5\times2=10,$ and our units digit can have any value from $0-9,$ our answer is $10(9+8+6+4+2)=\boxed{290}.$

The problem now reads:
How many 3-digit numbers have the property that [b]the unit digit[/b] is at least twice [b]the tens digit[/b]?

In the answer it says if the tens digit is zero, the unit digit can also be zero. It also says if the tens digit is zero, the unit digit can be $1$ Why are these considered correct answers? (rephrasing for clarity) Why is zero considered twice as large as zero? Why is $1$ considered twice as large as zero?

0 times 2 is 0 so 0 is twice as much as 0

however 1 is not twice 0 because 0 times 2 is 0 so maybe you can report that although i do not get what you mean because i accessed this thread by looking at the page feed while on alcumus and i have not done this problem before so i dont know the real answer

The post by Gamebot should be updated, because as a previous post says, the problem now reads
[quote]How many $3$-digit numbers have the property that the units digit is at least twice the tens digit?
[/quote]
Additionally, due to this edit to the problem, it should probably be decreased from level 22 to something like level 15, because of the simplicity of the solution.

[hide=Solution]
We know that there are $9$ different possibilities for the hundreds digit. Additionally, the problem is indifferent to the hundreds digits, so we can just multiply by $9$ at the end.

Case 1: Tens digit is $0$
If this happens, then the ones digit has $10$ possibilities.
Case 2: Tens digit is $1$
If this happens, then the ones digit has $8$ possibilities.
Case 3: Tens digit is $2$
If this happens, then the ones digit has $6$ possibilities.
Case 4: Tens digit is $3$
If this happens, then the ones digit has $4$ possibilities.
Case 5: Tens digit is $4$
If this happens, then the ones digit has $2$ possibilities.
There aren't any more cases, so the solution is $9\cdot (2+4+6+8+10)=\boxed{270}$.
 [/hide]


abstract algebra
search
superior algebra
superior algebra unsolved

(i) Show that the $ \mathbb{Z}$-module $ \mathbb{Q}/\mathbb{Z}$is not finitely generated.

(ii) Show that the$ \mathbb{Z}$- module$ \mathbb{Q}/\mathbb{Z}$ is a torsion module.

i) What can you say about the denominators in a finitely generated submodule?

ii) Where are you stuck on this one?

b) all elements have finite order??
a) follows from b) by
http://www.mathlinks.ro/viewtopic.php?search_id=45024283&t=56991

If you are not finding these problems easy, I recommend trying to see i) directly, rather than by the more general approach Albanian is suggesting.

i can see how to get part b but not sure how to show that Q/Z is not finitely generated

What LydianRain is saying (if I've got this right) is that the denominators in a finitely generated submodule have a finite number of prime factors.


For all x ≠ –4, what is equivalent to the expression: (x2 + 12x + 32) / (x + 4)?

If the answer is (x + 8), what is the relevance of x ≠ –4?

The answer is $ x\plus{}8$ [b]when[/b] $ x \neq \minus{}4$, and no solution when $ x\equal{}\minus{}4$.

This is because we must revert back to the original expression to check, and if $ x\equal{}\minus{}4$, the denominator is zero, which makes the entire expression undefined.

If $ x \equal{} \minus{} 4$ then

$ \dfrac{x^2 \plus{} 12x \plus{} 32}{x \plus{} 4} \equal{} \dfrac{(x \plus{} 4)(x \plus{} 8)}{x \plus{} 4} \equal{} \dfrac{( \minus{} 4 \plus{} 4)( \minus{} 4 \plus{} 8)}{ \minus{} 4 \plus{} 4} \equal{} \dfrac{0}{0}$ which is undefined.

Edit: beaten, and on a side note "Inequation" should be "Inequality".

At point x=-4, there would be what's called a removable discontinuity. Since it's not in the domain for the not-so-simple function, it can't be in the domain for the simpler function.

Yeah in the graph, you would have a hole, where there is no defined value for that point.


Anybody intersted in posting Division E (Grade 4-6) problems?

Here are some problems in elementary level.

http://www.colstate.edu/mathcontest/problem.php?CategoryID=4&LinkID=Current

http://www.noetic-learning.com/mathkangaroo/samples.jsp

http://www.dupagechildrensmuseum.org/aunty/challenge.html

Should we have an elementary level marathon?

Here are some problems;

What is the value of x if 7x - 3 = 5x + 13?

What is the remainder when 39583285738297 is divided by 5?

Twelve people would like to join a committee, but unforunately, only 5 can join. How many possible committees are there?

What is the probability that a green marble will be drawn from a bag containing 5 green marbles, 6 red marbles, and 9 yellow marbles?

What is the area of a circle with a radius of [5/(square root of pi)]?

What is the value of [(2/5)(5/9 - 3/5 + 8/5)]/(7/3)  ?

Were those the kinds of problems you were looking for?

Last year I did MOEMS cause I was in 4th grade. It was so easy! I got 24 on E and 23 on M.
 answer to problem 1:
 7x-3 = 5x +13
7x = 5x + 16
2x = 16
x = 8
answer to problem 2:
the remainder is 2 because the last digit is 7
answer to problem 3:
 12 x 11 x 10 x 9 x 8 = 95040
answer to problem 4:
1/4
answer to problem 5:
since the radius is (5/sqrt pi), (5/sqrt pi)^2 x pi = (25/pi)(pi) = 25
answers to problem 6:
[(2/5)(5/9-3/5+7/5)]/(7/3) = [(2/5)(5/9+4/5)]/(7/3)
=(2/5)(61/45)/(7/3)
=122/525

I don't think that you were supposed to solve, because I think (s)he was going to give these to 4-6th graders.

Don't revive.  That person has one post...and hasn't been active, so I doubt there checking this trhead...


http://www.mcglaun.com/math/square.gif

If the area of the smaller square is 9 cm2, what is the area of the bigger square?
Xplain in details plz

[hide]Clearly the side length of the square is 3. Using this and the pythagorean thereom, you can find that the diagonal is $3\sqrt{2}$ cm long. Notice that this is the diameter of the circle. Then notice that the diameter of the circle is the side length of the larger square. Thus, the area of the outer square is $(3\sqrt{2})^{2}=\boxed{18\text{ cm}^{2}}$[/hide]

[hide="answer"]If the area of the small square is 9cm, then each side length is 3cm
the diagonal of this square = the diameter of the circle.

the diagonal is $\sqrt{3^{2}+3^{2}}= 3\sqrt{2}$

this diagonal also happens to be the side length of bigger square, so the area of the bigger square is

$3\sqrt{2}* 3\sqrt{2}= 18$
[/hide]

and so you know.. most of the questions youve been posing dont belong in this forum, they belong in the beginning topics forum.

[b]Attention:[/b]
Posts of this nature should go into [color=blue]Forum Index ► Middle School ► Classroom Math[/color]
Somewhat more advanced questions should go into [color=blue]Forum Index ► High School ► High School Basic[/color]  :)

[i]Anyway, this is how I used to do these things before I learned what the Pythagorean Theorem was:[/i]

Rotate the inner square along the circle 45º and you'll notice that that the square (which still has the same area) connects the midpoints of the sides of the larger square.

Now connect midpoints of opposite sides of the large square. The inside square is now made up of 4 small triangles and the big square is made out of 8 triangles - twice as many - so the big square has twice the area.

[hide="Alternatively"]
Rotate the circle $45^{\circ}$.  We can now erase the circle, leaving us with a smaller square whose vertices lie on the midpoints of the sides of the large square.  Connect the diagonals of the small square and the picture has $8$ sections, $4$ of which are part of the small square.  Thus the bigger square has twice the area of the smaller square, making the larger square's area equal $18$.
[/hide]

EDIT: Wow, the Zuton Force beat me to it, and we have the same exact proof.

[hide]The length of each side is 3 cm. The diagonal, which is he diameter of the circle, and the side length of the bigger square, is 3 root 2. Area=18 cm[sup]2[/sup]. :ninja:[/hide]

lol @ the amount of answers in such a short time


$35$ dollars...

if i didnt forget addition ;) .

[hide]Call the amount Ari, Barry, Cara, Dara, and Erin have A, B, C, D, and E, respectively. A+B+C+D+E=85. Since A+B=40 and C+D=30, we can substitute 70+E=85, and E=15. Since A+E=20, A=5 and B=[b]35[/b].[/hide]

[quote="math92"]Ari, Barry, Cara, Dara, and Erin have a total of 85 dollars. Ari and
Barry have a total of 40 dollars. Cara and Dara have a total of 30 dollars.
Erin and Ari have a total of 20 dollars. How many dollars does
Barry have? :?:  :?:[/quote]

[hide]1.A+B=40
2.C+D=30
3.E+A=20
4.A+B+C+D=85
Add sentences 1, 2, 3 and subtract 4
  2A+B+C+D+E=90
 -  A+B+C+D+E=85
A=15
Plug into sentence 1
15+B=40
B=25[/hide]

[quote]Ari, Barry, Cara, Dara, and Erin have a total of 85 dollars. Ari and
Barry have a total of 40 dollars. Cara and Dara have a total of 30 dollars.
Erin and Ari have a total of 20 dollars. How many dollars does
Barry have? [/quote]

A+B+C+D+E=$85
A+B=$40
C+D=$30
E+A=$20
That is what is given.
~~~~~~~~~~~~~~~~~
Here is what I figured out.

85-B+A=C+D+E=$45
$45-C+D=E=$15
A+$15=$20, so A=$5

So we have shown:
A=$5
B=?
C=?
D=?
E=$15

We don't need to know C or D, because we know that A+B=$40. If A is $5, then B is $35

[hide]First, we are given that Ari, Barry, Cara, Dara, and Erin have a total of 85 dollars. We get the equation $a+b+c+d+e=85$.
Next, we have that Ari and Barry together have a total of 40 dollars. So, we get $a+b=40$, and we can substitute that into our original equation; we now have $c+d+e+40=85$.
Our next piece of information is that Cara and Dara have 30 dollars. We have $c+d=30$, and again, we substitute into our original equation, produce $e+70=75$. From this, it is obvious that $e=5$.
We were also given that Ari and Erin have 20 dollars together. Since we have the value for Erin, just substitute into $a+e=20$, and we have that $a=15$.
Lastly, we go back to our fact that Ari and Barry have 40 dollars together, and that$a+b=40$. Again, we use substitution, and we determine that $b=35$. Barry has 35 dollars.[/hide]

Yeaah this is my first problem that i solved :) go math92!!!!

a+b+c+d = 70
that means erin is 15

then a+e= 20
a= 20-15
that means a is 5
b+a=40
b= 40-5
barry has 35 dollars

[quote="math92"]Ari, Barry, Cara, Dara, and Erin have a total of 85 dollars. Ari and
Barry have a total of 40 dollars. Cara and Dara have a total of 30 dollars.
Erin and Ari have a total of 20 dollars. How many dollars does
Barry have? :?:  :?:[/quote]

Lol, I'm going to have to bump this because you guys are missing something so obvious.

Ignore the first piece of data: C + D + E + A = 50 dollars, while A + B + C + D + E = 85 dollars. Isn't this method a lot easier?


calculus
integration
function
inequalities
ratio
calculus computations

$\displaystyle\int_{0}^{\infty}\frac{dx}{(1+x^{n})^{n}}= 1$
Determine all the possible values of n.

Let $P_{n}: = \frac{1}{(1+x^{n)}^{n}}$ and express this as a binomial sum by considering the reciprocal of the standard binomial formula with $a = 1$, $x = x^{n}$.

Well, from what I was suggesting it appears that the resultant improper integral converges iff n is negative. So if we require n to be a natural number, then there is no solution.

Hey didilica,
I was just wondering why you mentioned a beta function but wrote down your equality using a symbol for a gamma function? I have not covered either of the function yet in my intro to analysis course may be your notation is correct.

[quote="ashwinkrishnan"]You can easily conclude that the vlaue of n cannot be an integer.See this -
Put $n =1$ we get $I ={\infty}$
Put $n =2$ we get $I = \frac{\pi}{4}$
hence we conclude that the value of n has to lie in the interval $(1,2)$ and somewhat closer to $2$.

Just check for yourself whether $n = \frac{\sqrt{5}+1}{2}$
satisfies this .( I saw this as a result somewhere but i am not able to prove it).[/quote][hide="Index"]
by the inequalitys  we can prove that only this answer is right, but it is not so interesting. more interest is in next: How did you determine for the $n$ this value $n = \frac{\sqrt{5}+1}{2}$ ?[/hide]

[quote="didilica"]I assume that $n$ is a natural number. Use the substitution $x^{n}=y$ and you get a beta function. So the equality in question is equivalent to:

$\Gamma(\frac{1}{n})\Gamma(n-\frac{1}{n})=\Gamma(n+1)$.

so find those natural numbers for which the preceding equality holds. I think there are none.[/quote]
How do you get gamma function?
pk we use substitute $y=x^{n}$ then $\int_{0}^{\infty}{\frac{dx}{(1+x^{n})^{n}}}$=$\frac{1}{n}\int_{0}^{\infty}y^{\frac{1}{n}-1}e^{-nln(1+y)}dy$ and what next?

[quote="didilica"]Use the substitution $x^{n}=y$ and you get that

$I=\frac{1}{n}\int_{0}^{\infty}\frac{y^{\frac{1}{n}-1}}{(1+y)^{n}}dy=\frac{1}{n}\beta(\frac{1}{n},n-\frac{1}{n})$.[/quote]

Is Beta function defined by $\beta(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$

then $\beta(\frac{1}{n},n-\frac{1}{n}) = \int_{0}^{1}y^{\frac{1}{n}-1}(1-y)^{n-\frac{1}{n}-1}dy$ ??

dont see how you got $\int_{0}^{\infty}\frac{y^{\frac{1}{n}-1}}{(1+y)^{n}}dy$

[quote="DanK"][quote="Extremal"]pk we use substitute $y=x^{n}$ then $\int_{0}^{\infty}{\frac{dx}{(1+x^{n})^{n}}}$=$\frac{1}{n}\int_{0}^{\infty}y^{\frac{1}{n}-1}e^{-nln(1+y)}dy$ and what next?[/quote]

Couldn't you use integration by parts on that integral?[/quote]
 :mad:

Oh I did get $\frac{1}{n}\beta(n-\frac{1}{n},\frac{1}{n})$

subsitute $w=\frac{1}{1+x^{n}}$
then $\int_{0}^\infty (\frac{1}{1+x^{n}})^{n}dx = \frac{1}{n}\int_{0}^{1}w^{n-\frac{1}{n}-1}(1-w)^{\frac{1}{n}-1}dw$
hence $\frac{1}{n}\beta(n-\frac{1}{n},\frac{1}{n}) = 1$
$\beta(n-\frac{1}{n},\frac{1}{n}) = n$
$\Gamma(n-\frac{1}{n})\Gamma(\frac{1}{n}) = n\Gamma(n) = \Gamma(n+1)$

but then how do you get $n=\frac{1+\sqrt{5}}{2}$ ?  i dont know how yet

[quote="DanK"][quote="Extremal"]pk we use substitute $y=x^{n}$ then $\int_{0}^{\infty}{\frac{dx}{(1+x^{n})^{n}}}$=$\frac{1}{n}\int_{0}^{\infty}y^{\frac{1}{n}-1}e^{-nln(1+y)}dy$ and what next?[/quote]

Couldn't you use integration by parts on that integral?[/quote]
I know how integration by parts.

[quote="kelvin911"]but then how do you get $n=\frac{1+\sqrt{5}}{2}$ ?  i dont know how yet[/quote]

You could always just guess it, since it's the golden ratio and the topic is called the golden integral...

what if someone ask you that integral and he didnt give you any hint about what n is..  then how do u know n = golden ratio?

there must be a smart way to find n not knowing n is golden ratio


Prove that any $n$ squares can be cut up and combined into $\text{One}$ new square.

I don't quite understand the wording of the problem. Let's say we have a square with area 1 and another with area 4. 1+4=5, so we have a square of side $\sqrt{5}$.  What I have to prove is that there are infinitely many squares, that when combined, can form a square, right?  If so, isn't this problem very obvious.  Please explain the problem so I would understand.:)

Masoud Zargar

[quote="boxedexe"]What I have to prove is that there are infinitely many squares, that when combined, can form a square, right?[/quote]
Yes. But I am interested to see how you construct.

say 2 squares have sides $a$ and $b$, then construct a right triangle with legs $a$ and $b$, the side is $\sqrt{a^2+b^2}$, and a square with that side would have the area of $a^2+b^2$ so $n=2$ is covered, assume that we can do it for $n$ squares, do this, giving us a new square K, then add in another square $n+1$, we can combine $K$ and $n+1$ into another by the $n=2$, so this implies we can construct it with $n+1$ squares, proven by induction

Here is my construction. (Same idea with Altheman)


conics
geometry
parabola
circumcircle
projective geometry
complex numbers
power of a point

Here is what I've found about it:
It states that, given a (not necessarily regular, or even convex) hexagon inscribed in a conic section, the three pairs of the continuations of opposite sides meet on a straight line, called the Pascal line.

Is the converse of this theorem also true? i.e. Given a circle O and a straight line with three fixed points E, F, G on it. There exists an unique hexagon circumscribed in circle O such that E, F, G are the intersections of the three pairs of the continuations of opposite sides.

Moreover, what is the "definition" of opposite sides here? I found a strange picture here.

At last, I hope someone would tell me more about this theorem, because it is too new to me.

[quote="mecrazywong"]It [i][Pascal's Theorem - DG][/i] states that, given a (not necessarily regular, or even convex) hexagon inscribed in a conic section, the three pairs of the continuations of opposite sides meet on a straight line, called the Pascal line.[/quote]

Indeed, this is the verbose formulation of Pascal's theorem. Here is the clearest formulation: If A, B, C, D, E, F are six arbitrary points on a conic section, then the point of intersection of the lines AB and DE, the point of intersection of the lines BC and EF, and the point of intersection of the lines CD and FA lie on one line, which is called the [b]Pascal line[/b] of the hexagon ABCDEF.

Now you see what is meant by "opposite sides": The sides AB and DE are opposite to each other, so are BC and EF, and so are CD and FA.

Note that the points A, B, C, D, E, F can lie arbitrarily on the conic section. The hexagon ABCDEF need not be convex, it can be also concave or self-intersecting. Finally, Pascal's theorem even remains true if two or more of the points A, B, C, D, E, F coincide; in this case, for instance, if the points A and B coincide, then the line AB should be understood as the tangent to the conic section at the point A.

[quote="mecrazywong"]Is the converse of this theorem also true? i.e. Given a circle O and a straight line with three fixed points E, F, G on it. There exists an unique hexagon circumscribed in circle O such that E, F, G are the intersections of the three pairs of the continuations of opposite sides.[/quote]

No, au contraire: There are infinitely many such hexagons!

The following converse of Pascal's theorem is true: If A, B, C, D, E, F are six points in the plane such that the point of intersection of the lines AB and DE, the point of intersection of the lines BC and EF, and the point of intersection of the lines CD and FA lie on one line, then we can conclude that there exists a conic section passing through the six points A, B, C, D, E, F.

  Darij

Then, when the hexagon is self-intersecting, how do we define "opposite side"?
[i]If A, B, C, D, E, F are six points in the plane such that the point of intersection of the lines AB and DE, the point of intersection of the lines BC and EF, and the point of intersection of the lines CD and FA lie on one line, then we can conclude that there exists a conic section passing through the six points A, B, C, D, E, F. [/i]

Is the conic section unique? If not, can it be a circle?

[quote="mecrazywong"]Then, when the hexagon is self-intersecting, how do we define "opposite side"?[/quote]

Just as I did, abstractly: The sides AB and DE are called opposite, so are the sides BC and EF, and so are the sides CD and FA. Of course, when the hexagon is self-intersecting, these opposite sides may cross in their interior, but this doesn't matter: they are still called opposite since they are opposite in the abstract sense.

[quote="mecrazywong"][i]If A, B, C, D, E, F are six points in the plane such that the point of intersection of the lines AB and DE, the point of intersection of the lines BC and EF, and the point of intersection of the lines CD and FA lie on one line, then we can conclude that there exists a conic section passing through the six points A, B, C, D, E, F. [/i]

Is the conic section unique?[/quote]

Yes, if no two of the points A, B, C, D, E, F coincide. In fact, through any [i]five[/i] distinct points, we can draw one and only one conic (except for some degenerate cases, e. g. three of the points collinear). Pascal's theorem gives a sufficient and necessary condition for [i]six[/i] points to lie on one conic.

  Darij

darij, thank you for your replies. Here comes my last question about Pascal's Theorem:
Let E, F, G be 3 fixed points on line l, A an arbitray point on circle O. AE meets circle O at B, AG meets circle O at D, FD meets circle O at C. Show that BC passes through a fixed point on l.

I have a simple solution to this question, but someone told me I can solve it easily by Pascal Theorem. I'm just very curious about it.

[quote="mecrazywong"]Let E, F, G be 3 fixed points on line l, A an arbitray point on circle O. AE meets circle O at B, AG meets circle O at D, FD meets circle O at C. Show that BC passes through a fixed point on l.[/quote]

I see neither an elementary solution of this nor a solution using Pascal! All what I can give is a very bad projective geometry solution on $\mathbb{C}^2$:

I denote the set of all real numbers with $\mathbb{R}$ and the set of all complex numbers with $\mathbb{C}$; then I will solve the problem not in the real plane $\mathbb{R}^2$, but in the complex plane $\mathbb{C}^2$. The cool fact about the complex plane is that every circle in $\mathbb{C}^2$ intersects (or touches) every line. And we will use another property of $\mathbb{C}^2$, namely that any conic in $\mathbb{C}^2$ which is not a parabola can be mapped by an affine transformation to a circle.

Perform a projective transformation which maps the line l to infinity. The circle gets mapped to a conic. This conic is a parabola only if its preimage, the circle O, touched the line l; we consider this case as a limiting case. In the general case, the conic is not a parabola. Hence, there exists an affine transformation which maps this conic to a circle. Hereby, the infinite line remains fixed (since the transformation is affine). Then we are left with the following problem:

[i]Let E, F, G be three fixed points on the line at infinity (i. e., three fixed directions of lines), and A an arbitrary point on a circle O. The line AE meets the circle O at B, the line AG meets the circle O at D, and the line FD meets the circle O at C. Show that the line BC passes through a fixed point on the line at infinity, i. e. has a fixed direction.[/i]

Well, this is now trivial: By the chordal angle theorem, < DAB = < DCB, so that < (AG; AE) = < (FD; BC). Since the directions of the lines AG, AE and FD are fixed, the direction of the line BC must be fixed, too, and we get the required result.

But I know this is more than brutal...

  Darij

Notice that if $\ell$ intersects the circle, then this is a direct consequence of Desargue's theorem:

If the line $\ell$ cuts the circle at $X,Y$ and $BC$ at $H$, then the pairs $(X,Y),(E,F),(G,H)$ form an involution, but an involution is defined by only two pairs, and we take these to be the fixed pairs $(X,Y),(E,F)$, and since $G$ is fixed, the same can be said about $H$.

In fact, if we do what Darij did and extend the context to $\mathbb C^2$, then $\ell$ will always intersect the circle, so I guess this argument works again for the general case (although I'm not sure; I'm not really used to employing the complex plane :)).

[quote="mecrazywong"]The solution is very simple, no special theorem is needed.
Hint:
Let BC meet EG at H.
Let the circumcircle of triangle BCE meet EG at I. It's easy to prove I is a fixed point.[/quote]

I must admit that I don't understand this hint, but now I have succeeded to find a solution using the Pascal theorem:

In the following, we will denote by $g \cap h$ the point of intersection of two lines g and h.

Now we must prove that the line BC meets the line l at a fixed point. In other words, we have to show that the point $BC \cap l$ doesn't depend of the position of the point A on the circle O, i. e. if we take another point A' on the circle O, then construct the point of intersection B' of the line A'E with the circle O, then construct the point of intersection D' of the line A'G with the circle O, and then construct the point of intersection C' of the line FD' with the circle O, then the point $BC \cap l$ coincides with the point $B^{\prime}C^{\prime} \cap l$. In other words, we have to show that the lines BC, B'C' and l concur at one point, i. e. that the point $BC \cap B^{\prime}C^{\prime}$ lies on the line l.

In fact, applying the Pascal theorem to the cyclic hexagon DABD'A'B', we see that the points $DA \cap D^{\prime}A^{\prime}$, $AB \cap A^{\prime}B^{\prime}$ and $BD^{\prime} \cap B^{\prime}D$ lie on one line. But the points $DA \cap D^{\prime}A^{\prime}$ and $AB \cap A^{\prime}B^{\prime}$ are the points G and E, respectively, and lie on the line l. Hence, it follows that the point $BD^{\prime} \cap B^{\prime}D$ also lies on the line l.

Now apply the Pascal theorem to the cyclic hexagon BCDB'C'D', and you obtain that the points $BC \cap B^{\prime}C^{\prime}$, $CD \cap C^{\prime}D^{\prime}$ and $DB^{\prime} \cap D^{\prime}B$ lie on one line. But the point $CD \cap C^{\prime}D^{\prime}$ is the point F and lies on the line l; the point $DB^{\prime} \cap D^{\prime}B$ is just the point $BD^{\prime} \cap B^{\prime}D$ we have met before and also lies on the line l. Hence, the point $BC \cap B^{\prime}C^{\prime}$ must lie on the line l, too. And the problem is solved.

[i]Remark.[/i] You can rewrite the above problem in a more appealing way:

[i]Let ABCD and A'B'C'D' be two quadrilaterals inscribed into one and the same circle. Assume that the points $CD \cap C^{\prime}D^{\prime}$, $DA \cap D^{\prime}A^{\prime}$ and $AB \cap A^{\prime}B^{\prime}$ lie on one line l. Then prove that the point $BC \cap B^{\prime}C^{\prime}$ also lies on the line l.[/i]

In this form, this problem follows from a result by Chasles (in fact, Chasles formulated the same problem for arbitrary conics, not just circles). See:

[quote="JFM review, translated"][i]Neuberg, J.; Dewulf, E.[/i]
[i]Correspondance. [J] N. C. M. V. 18-22.[/i]
[i]Published: (1879)[/i]

Mr. Neuberg and Mr. Dewulf establish certain theorems in addition to some results by Mr. Folie related to the theory of involutions of n-th degree [?]. An example:
If two quadrilaterals abcd, ABCD are inscribed in one conic, and if the first three sidelines ab, bc, cd of the first one meet the first three sidelines AB, BC, CD of the second one in three points lying on one line, then [b]1)[/b] the intersection of the fourth sides da and DA also lies on this line (Chasles), and [b]2)[/b] the points ab /\ CD, AB /\ cd, bc /\ DA, BC /\ da lie on a second line.[/quote]

  Darij

[quote="mecrazywong"]The solution is very simple, no special theorem is needed.
Hint:
Let BC meet EG at H.
Let the circumcircle of triangle BCE meet EG at I. It's easy to prove I is a fixed point.[/quote]

Finally I think I have found your solution!

Our goal is to prove that the point of intersection of the lines BC and l is fixed, i. e. it doesn't depend on the actual choice of the point A on the circle O. This means that if we take another point A' on the circle O, and then perform constructions analogous to our constructions of the points B, C, D, i. e. we call B' the point where the line A'E meets the circle O, then call D' the point where the line A'G meets the circle O, and then finally call C' the point where the line FD' meets the circle O, then the point of intersection of the lines B'C' and l is the same as the point of intersection of the lines BC and l. In other words, we have to prove that the lines BC, B'C' and l concur at one point.

Let the circumcircle of triangle BCE meet the line l at a point I (apart from E). Then, using directed angles modulo 180, we have < CBE = < CIE. On the other hand, < CDA = < CBA, since the points A, B, C, D lie on one circle. Thus, < CDG = < CDA = < CBA = < CBE = < CIE = < CIG. Hence, the points C, D, G, I lie on one circle. Therefore, by the intersecting chords (or secants) theorem, we have $FD \cdot FC = FG \cdot FI$, with directed segments. Similarly, if the circumcircle of triangle B'C'E meets the line l at a point I' (apart from E), then we have $FD^{\prime} \cdot FC^{\prime} = FG \cdot FI^{\prime}$. But since the points C, D, C', D' lie on one circle, we have $FD \cdot FC = FD^{\prime} \cdot FC^{\prime}$; and thus, it follows that $FG \cdot FI = FG \cdot FI^{\prime}$, so that FI = FI', and the points I and I' must coincide. Hence, the point I lies both on the circumcircle of triangle BCE and on the circumcircle of triangle B'C'E.

Now, consider the circumcircle of triangle BCE, the circumcircle of triangle B'C'E, and the circle O.

- The circumcircle of triangle BCE meets the circle O at the two points B and C, and hence the line BC is the radical axis of these two circles.
- The circumcircle of triangle B'C'E meets the circle O at the two points B' and C', and hence the line B'C' is the radical axis of these two circles.
- The circumcircle of triangle BCE meets the circumcircle of triangle B'C'E at the two points E and I, and hence the line EI = l is the radical axis of these two circles.

But since the pairwise radical axes of three circles concur, it follows that the lines BC, B'C' and l concur. This completes the solution.

Now we have four solutions to your problem, two of them without the use of $\mathbb{C}^2$.

  Darij

i  was  trying  to  solve  the geo  problem  from the  last  iberoamerican  olympiad  and  couldn´t  see  the  application  of  pascal's  theorem.  darij  about  the  use  of  pascals  theorem ;  say  i  have  6  points  on a  circle  than  can  i  permute  the  letters (A,B,C,D,E,F)  of  each  vertex  of  the  hexagon  and  we  will  still  have  the collinearity  between  intersection  of  opposite  sides??  whaT  would  be  the  configuration  for  pascal  on  the  iberoamerican  problem??
 :blush:  :blush:

did u know that pascal theorem is the special form of the thorem papus ?

its very enjoyable... :D


algebra
polynomial
Rational Root Theorem

[quote="vishalarul"][quote="Quevvy"][quote="i_like_pie"][hide]$ a^{3}+a^{2}+4=(a+2)\left(a^{2}-a+2\right)$[/hide][/quote]
How do you figure that out?[/quote]

Just use RRT: The only possible rational roots are $ \pm 1, \pm 2, \pm 4$. Plug 'em in, and see that $ -2$ works, so $ a+2$ must be a factor. Then you can use synthetic division to divide that polynomial by $ a+2$:
$ a^{3}+a^{2}+4 = (a+2)(a^{2}-a+2)$[/quote]
What's RRT?

How do you know those are the only roots?

What is synthetic division?

[quote="Quevvy"][/quote][quote="vishalarul"][quote="Quevvy"][quote="i_like_pie"][hide]$ a^{3}+a^{2}+4=(a+2)\left(a^{2}-a+2\right)$[/hide][/quote]
How do you figure that out?[/quote]

Just use RRT: The only possible rational roots are $ \pm 1, \pm 2, \pm 4$. Plug 'em in, and see that $ -2$ works, so $ a+2$ must be a factor. Then you can use synthetic division to divide that polynomial by $ a+2$:
$ a^{3}+a^{2}+4 = (a+2)(a^{2}-a+2)$[/quote][quote="Quevvy"]
What's RRT?

How do you know those are the only roots?

What is synthetic division?[/quote]

RRT stands for the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]Rational Roots Theorem[/url].

Those aren't the only roots. Those are the only possible [b]rational[/b] roots.

[url=http://en.wikipedia.org/wiki/Synthetic_Division]Synthetic division[/url] is a quick way to divide any polynomial by a linear factor.

EDIT: More links

Rational Roots theorem
[url]http://www.artofproblemsolving.com/Wiki/index.php/Polynomial#The_Rational_Root_Theorem[/url]

[quote="vishalarul"]RRT stands for the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]Rational Roots Theorem[/url].

Those aren't the only roots. Those are the only possible [b]rational[/b] roots.

[url=http://en.wikipedia.org/wiki/Synthetic_Division]Synthetic division[/url] is a quick way to divide any polynomial by a linear factor.

EDIT: More links

Rational Roots theorem
[url]http://www.artofproblemsolving.com/Wiki/index.php/Polynomial#The_Rational_Root_Theorem[/url][/quote]
Thank you so much for the information!


A conference has 47 people attending. One woman knows 16 of the men who are attending, another knows 17, and so on up to the last woman who knows all the men who are attending. Find the number of men and women attending the conference

[hide="Solution?"]
1st woman --> 16 men
2nd woman --> 17 men
(...)
last woman = number of women --> all men = number of men

$w+m = 47 \\ m = w+15 \\ \Rightarrow w = 16 \wedge m = 31$

Answer: 16 women, 31 men.
[/hide]

I'm sorry for the bad LaTeX layout :roll:


function
algebra
domain
calculus
calculus computations

If an even function's graph has a hole on one side but not the other, would it still be even?

The notions of even and odd only make sense for functions whose domain is a set symmetric about zero. So the answer to your question is "no."


How would I go about solving this question?

What is the probability that in a random sample of size n = 25 from a $ N(20, 2^2)$ distribution the sample mean takes a value greater than 19.5?

thanks

Let $ X_i\sim N(20,2^2)$ and $ n\equal{}25\to\overline X\equal{}\sum\limits_{i\equal{}1}^{25}X_i\sim N(20,\frac{2^2}{25}).$

Then \[ \mathbb P(\overline X\geqslant 19.5)\equal{}1\minus{}\Phi(\frac{19.5\minus{}20}{2/5})\equal{}1\minus{}\Phi(\minus{}1.25)\equal{}\Phi(1.25)\equal{}0.8944\]

You need to find the probability
$ P(\frac {\xi_1 + ... + \xi_{25}} {25} > 19.5)$ where $ \xi_i ,i = 1,25$ have $ \mathfrak{N} (20,2^2)$ distribution  and they are independent, so you can easily find the distribution of the sum above and calculate the probability you need.It's clear, isn't it?


Let $ a\equal{}1.234$, $ b\equal{}2.34$, and $ c\equal{}3.4$.
Round the value of the expression $ (a\plus{}b)\div c$ to the nearest tenth.

We want to find $ \frac{1.234 \plus{} 2.34}{3.4}$

$ \frac{3.574}{3.4}$

This is a little bit more than 1. Now examine 3.4.

1/10 of it is .34, so $ 1.1 \times 3.4 \equal{} 3.74$ ; We now know our quotient will be between 1.0 and 1.1.

So which is 3.574 closer to? We see if it is greater than or less than $ 3.4 \times 1.05$.

1/20 of 3.4 is half of 1/10. 1/10 of 3.4 is .34. Half of this is .17.

So, $ 3.4 \times 1.05 \equal{} 3.4 \plus{} .17 \equal{} 3.57 < 3.574$.

Thus, our quotient is closer to $ \boxed{1.1}$ than 1.0.


How many watches purchased at 3 for 20  dollars and sold at 4 for 30 dollars are required to make a profit of 40  dollars?

[hide=" click to reveal hidden content"]

If you buy $ 12$, it will cost you $ 80$ dollars and you sell these for $ 90$ dollars, making a profit of $ 10$ dollars. Therefore to make a profit of $ 40$ dollars you need to sell $ 48$ watches.

[/hide]