Find $ x;y \in Z$ such that
$ 8(54x^3\minus{}y^3)\plus{}28\equal{}9y(6\minus{}4y)$

$ 8 \left(54 x^3\minus{}y^3\right)\plus{}28\equal{}9 (6\minus{}4 y) y$

$ y\equal{}\frac{b\plus{}3}{2}\land (b\equiv 1) \bmod 2$

Expanding and simplifying...

$ b^3\equal{}432 x^3\plus{}1$ $ \Longrightarrow$ $ b^3\equal{}2^4 3^3 x^3\plus{}1\equal{}2 (6 x)^3\plus{}1\equal{}2 z^3\plus{}1$

$ b^3\equal{}2 z^3\plus{}1$

you can continue from here, eh?

Maybe you could try this:

take a congruence modulo y and you'll obtain:

54·8x3 is congruent to -28 ==> 108x3 is congruent to -7 ==> 108x3=ay-7 for some integer a ==> x3=(ay-7)/108 then you can substitute into the original equation and you get a cubic --which you can solve-- that will get you to a solution for y involving a, then you have to determine the values of a for which y is an integer.

After you get these values, you have to determine the values of a for which y is an integer and satisfies that (ay - 7)/108 is a perfect cube. Getting these values leads to solution. (im lazy tonight).

[quote="pi3.14"]Find $ x;y \in Z$ such that
$ 8(54x^3 \minus{} y^3) \plus{} 28 \equal{} 9y(6 \minus{} 4y)$[/quote]

Sorry but it is a problem from VM site contest. Please discuss it in next week. Thank all  :)


Let $E$ be a normed space of finite dimension over the field of real numbers and let $N$ be a norm. Suppose that $F$ is a subspace of $E$ , fixed. Prove that there is at most one vectorial subspace of $E$, called $G$ such that $N(x+y)=N(x-y)$ for all $x\in F,y\in G$ and $E=F+G$.

First we reduce it to the case $\dim E=2$:

Suppose we can find two different such spaces $G,\ \widetilde G$, and take $\tilde g\in \widetilde G\setminus G$. Then $\tilde g=f+g$ for some $f\in F,\ g\in G$. If we can prove the assertion for $\dim E=2$, then we can apply it to the plane spanned by $f,g$, which also contains $\tilde g$, and we get a contradiction.

Now we solve the case $\dim E=2$. Of course, the only interesting case is $\dim F=1$:

Let $f$ be a basis for $F$, and suppose we can find two different complements $G,\widetilde G$ of $F$ with the required proeprty, spanned by $g$ and $tf+g,\ t\in\mathbb R\setminus\{0\}$ respectively. We know that $\|\alpha f+\beta g\|=\|\alpha f-\beta g\|,\ \forall \alpha,\beta\in\mathbb R\ (*)$, and also $\|\gamma f+\delta(tf+g)\|=\|\gamma f-\delta(tf+g)\|,\ \forall \gamma,\delta\in\mathbb R$. This second relation is equivalent to $\|\alpha f+\beta g\|=\|(\alpha-2\beta t)f-\beta g\|,\ \forall \alpha,\beta\in\mathbb R\ (**)$.

Applying $(*)$ to $(**)$ repeatedly, with $\alpha=\beta=1$ for instance, gives $\|f+g\|=\|(1-2nt)f+g\|,\ \forall n\in\mathbb N$. Clearly, this is a contradiction, since the LHS is fixed, while the RHS goes to $\infty$ with $n$.


Assume that $(x-2)$ is a factor of the polynomial $f(x)=x^3 + ax^2 + bx +2$ and that $f(x)$ gives a remainder of -3 when it is divided by x+1. Then what is a and b on this polynomial?


modular arithmetic
number theory unsolved
number theory

Find the first digit to the left and the first digit to the right of the decimal point in the expansion of $ (\sqrt{2}\plus{}\sqrt{5})^{2000}.$

well $ (\sqrt{5}\plus{}\sqrt{2})^{2000}\plus{}(\sqrt{5}\minus{}\sqrt{2})^{2000}$ is an integer and the last digit is $ 2(2^{1000}\plus{}5^{1000})\equiv 2 \pmod{10}$
but $ (\sqrt{5}\minus{}\sqrt{2})^{2000}$ is a very very small number :D so the answer to our problem is $ ...1.9...$


Well, for all you math junkies out there USAMTS round 1 problems are up!!! Check 'em out here -

http://www.nsa.gov/programs/mepp/usamts/problems/problems15R1.pdf

And here is a site where you can find more about USAMTS if you don't know about it! Please spread the USAMTS gospel!

http://www.nsa.gov/programs/mepp/usamts.html

Enjoy!
IW.OAV

[quote]"No, I'm a USAMTS virgin"[/quote]

Well, I've never thought of it that way.

BTW, I voted yes, just in case you were wondering.

You  better damn right think of it that way...

 :twisted: Just playin'... The problems don't seem that bad this time, although there is some number theory from what I saw with my 30s glance at the problems. And I hate the greatest integer function... it annoys me for some reason...

Peace...
IW.OAV

Speaking of USAMTS, I haven't seen any feedback from round three of last year...

Anyway, thanks for the notification.

Charles

Maybe we ought to write to USAMTS, AMC, Mathcounts, Mu Alpha Theta, etc. webmasters and ask them to link to this site.  That would increase our community size.

The problems are not that diffictult, but I think after each USAMTS round of probs come out, someone can make an informative post, but we don't discuss them until after the due date, just to make sure we don't cheat. How's that?[/hide]

Yes, Please do not discuss the USAMTS problems with anyone... Remember no one is making you do them.. They are just problems for your own benefit so do not cheat... Anyways it destroys that mini-rush you get after solving one of those problems.

[quote]Yes, Please do not discuss the USAMTS problems with anyone[/quote]

Naturally.  And HS (Happy Solving)


In how many ways can a president and a 2-person committee be chosen from a group of 8 people (where the order in which we choose the 2 people doesn't matter)?  (The president cannot be on the committee.)

[hide="Solution"]
The number of ways to choose a president is $ \binom81\equal{}8$ while the number of ways to choose the group of 2 people is [u]now[/u] $ binom72\equal{}21.$ Multiply these together to get $ 8\times21\equal{}\boxed{168}.$
[hide="@lightning"]
Your final answer is correct, but the way you got to the answer is...[/hide][/hide]

@lightning:  Since I did it RoFl's way, I'll use his equation, and change it into your equation.  We'll use $ n$ people, so make sure this works in ALL cases.  We do $ \binom{n}{1}\cdot\binom{n\minus{}1}{2}\equal{}\frac{n\cdot(n\minus{}1)!}{(n\minus{}1\minus{}2)!(2)!}\equal{}\frac{n!}{2(n\minus{}3)!}\equal{}3\cdot\binom{n}{3}\equal{}\binom{3}{1}\cdot\binom{n}{3}$.  We see that in the end, both solutions are equal.  lightning, this is a valid solution, not just luck  :D .

[quote="lightning"]i'm just wondering, is my solution valid? or can it run into mistakes, and this was just a lucky run?

i did

[hide](8C3) to get the amount of total people (2 members and 1 pres.)

then i multiplied that by (3C1) to get that amount of ways you can pick a pres. from that group of 3.[/hide][/quote]

Sorry to reply 4 years later...

This was my reasoning too, basically:

[hide]$ \binom83=56 $ ways to choose three people from 8; $ \binom31=3 $ ways to partition those three into a president and two committee members; $ 3\times56=\boxed{168}. $[/hide]

Thanks jxl28 for formalizing! That added a lot of clarity.

a You have 8 choices for president, and have 7*6/2 ways for committee (by counting pairs), you get 8*21 which is equal to 168

Lightning I believe your solution is valid.

[hide=Solution]
We have $8$ choices for president, and $\binom{7}{2}=21$ choices for the committee (because there are only $7$ people after they've chosen the president). That's $8\times21=\boxed{168}$ ways to choose the president and the committee.
[/hide]

[hide=My solution][color=#9a00ff][b]8[/b] ways to chose a president [b]• 7[/b] ways left to chose the first person on the committee[b] • 6[/b] ways left to chose the second person on the committee [b]/ 2 [/b]because it does not matter what order they are chosen in[b] =[/color][/b][hide=...][color=#9a00ff]168[/color][/hide][/hide]


For abc=8
prove that 1<1/(1+a)^1/2+1/(1+b)^1/2+1/(1+c)^1/2<2
I've already known how to prove the left part but the right one seems to be very difficult.... :maybe:

[quote="kuing"]this is the 2008 college entrance examination questions in China...
so many solution...
you can see here
http://forum.cnool.net/topic_show.jsp?id=4760099&oldpage=1&thesisid=494&flag=topic1[/quote]

I've seen this solution but I don't like it....

Then I find another link on that page...thanks :maybe:


function
calculus
derivative
algebra
polynomial
inequalities
inequalities unsolved

Let $a,b,c,d$ be nonegative real numbers such that $a+b+c+d=4$, find the maximum of
\[f=10abc+bcd+cda+dab.\]

[hide]
Clearly the given expression is not greater than
\[10\left( a+\frac{d}{3}\right) \left( b+\frac{d}{3}\right) \left( c+\frac{d}{3}\right)\]
since when expanding out, we get $10abc+\frac{10}{3}(bcd+cda+dab)$ plus terms with at least 2 factors of $d$. But the above expression is less than or equal to
\[10\left(\frac{a+\frac{d}{3}+b+\frac{d}{3}+c+\frac{d}{3}}{3}\right)^{3}=\frac{640}{27}\]
by AM-GM. Therefore $\boxed{\frac{640}{27}}$ is the maximum, equality holding for $a=b=c=\frac{4}{3},d=0$.
[/hide]

Hmm... Clever solution :)

Fix $d$, so that $a+b+c = s \in [0,4]$ is also fixed. We then have that $f(a,b,c,d) = 10abc+d(ab+bc+ca)$. But, as is well-known, both $abc$ and $ab+bc+ca$ are maximized when $a=b=c$.

So it suffices to find the maximum of $10 x^{3}+3(4-x)x^{2}$ on $[0,4]$, where $x=a=b=c$, which is kind of easy, since that function is increasing.

OK, they are nice. Last night I came up with one similar to perfect-radio's  :blush: What if
\[f=9abc+10bcd+cda+dab. \]

[quote="perfect_radio"]So it suffices to find the maximum of $10 x^{3}+3(4-x)x^{2}$ on $[0,4]$, where $x=a=b=c$, which is kind of easy, since that function is increasing.[/quote]
[color=red]Here, I obviously meant that [/color][color=blue][i]it suffices to find the maximum of $10 x^{3}+3 \left( 4-3x \right) x^{2}$ on $\left[ 0, \frac43 \right]$[/i][/color][color=blue].[/color]

Now, back to the front:
[quote="pvthuan"]What if
\[f=9abc+10bcd+cda+dab ? \]
[/quote]
Write $f = bc(9a+10d)+ad(b+c)$; from here it is obvious that
\[f(a,b,c,d) \leq f \left( a, \frac{b+c}2, \frac{b+c}2, d \right) , \]
with equality iff $b=c$.
Hence, it suffices to find the maximum of $g(x,y,z) = 9 x y^{2}+10 y^{2}z+2xyz$, where $x,y,z$ are non-negative numbers such that $x+2y+z = 4$ (I took $x : = a, \, y : = b = c, \, z : = d$).

[hide="Case 1."]
$x=0$: $g(0,y,z) = 10 y^{2}z$, where $2y+z = 4$. Make the substitution $z = 4-2y$, so that $g(0,y,z) = 20 \left( 2 y^{2}-y^{3}\right)$. The maximum in this case is $\frac{640}{27}$ and is reached for $\left( 0, \frac43, \frac43 \right)$.
[/hide]
[hide="Case 2."]
$y=0$: $g(x,0,z) = 0$.
[/hide]
[hide="Case 3."]
$z=0$: $g(x,y,0) = 9 x y^{2}$, where $x+2y = 4$. Make the substitution $x = 4-2y$, so that $g(x,y,0) = 18 \left( 2 y^{2}-y^{3}\right)$. The maximum in this case is $\frac{576}{27}$ and is reached for $\left( \frac43, \frac43, 0 \right)$.
[/hide]
[hide="Case 4."]
$xyz \neq 0$: We shall use the [color=green]Lagrange Multiplier[/color] $\lambda$.
\[\begin{cases}9 y^{2}+2yz = \lambda \\ 18 xy+20 yz+2xz = 2 \lambda \\ 10 y^{2}+2xy = \lambda \end{cases}\]
From the first and third equation one gets that $y^{2}+2xy = 2yz$. Since $y \neq 0$, it follows that $y+2x = 2z$, i.e. $y = 2z-2x$.
Substituting that into $x+2y+z = 4$, we get that $z = \frac{4+3x}5$, so that $y = \frac{8-4x}5$.
Hence, $g = 9 x \left( \frac{8-4x}5 \right)^{2}+10 \left( \frac{8-4x}5 \right)^{2}\cdot \frac{4+3x}5+2 x \cdot \frac{8-4x}5 \cdot \frac{4+3x}5$, where $x \in [0,2]$.
Mathematica tells me that $g = \frac8{25}\cdot \left( 27 x^{3}-102 x^{2}+64x+64 \right)$. The derivative of that is $\frac8{25}\cdot \left( 81 x^{2}-204 x+64 \right)$; its roots are $\frac{2 \left( 17 \pm \sqrt{145}\right)}{27}$. The bigger one is clearly $>2$, whilst the smaller one is in the right interval, i.e. $[0,2]$. Hence, the third degree polynomial reaches its maximum on $[0,2]$ at $x = \frac{2 \left( 17-\sqrt{145}\right)}{27}$. That implies $y = \frac8{135}\left( 10+\sqrt{145}\right), \, z = \frac2{45}\left( 35-\sqrt{145}\right)$. Not amazingly, this verifies those three equations, so it's an extremum point.
Note that $g \left( \frac{2 \left( 17-\sqrt{145}\right)}{27}, \frac{8 \left( 10+\sqrt{145}\right)}{135}, \frac{2 \left( 35-\sqrt{145}\right)}{45}\right) = \frac{128 \left( 335+29 \sqrt{145}\right)}{3645}$.
[/hide]
[hide="The last step."]
All that's left to do is to prove the following inequality
\[\frac{128 \left( 335+29 \sqrt{145}\right)}{3645}> \frac{640}{27}, \]
which is by using Mathematica's function [i]Simplify[/i] reduces to
\[29 \sqrt{145}> 340 , \]
i.e.
\[121,945 > 115,600 , \]
which is clear.
[/hide]
Therefore, $\max f = \max g = \frac{128 \left( 335+29 \sqrt{145}\right)}{3645}\approx 24.026995$.

[color=magenta]This number is beautifulnot![/color]


A circle is inscribed in a square. A second circle, inside the square, is externally tangent to the first circle, and is also tangent to two sides of the square. Compute the ratio of the radius of the smaller circle to the radius of the first circle

The edges of a cube have length 1, and A is a vertex of the cube. Compute the length of the segement joiniing A to the center of a face that does not contain A.

[quote="fanaticsm"]The edges of a cube have length 1, and A is a vertex of the cube. Compute the length of the segement joiniing A to the center of a face that does not contain A.[/quote]

Form a triangle with the center of the face, A, and the vertex next to A on the mentioned face. That triangle is a right triangle. It's legs are 1 and sqrt(2)/2. So the length is sqrt(3)/sqrt(2)

[hide]let x be the radius of the smaller circle
let y be the radius of the larger circle
draw lines from the radius of the smaller circle to meet the radius of the larger circle such that they form a right triangle with legs (y-x) and hypotenuse (y+x)
:sqrt: 2 (y-x) = (y+x)
y :sqrt: 2 - x :sqrt: 2 = y + x
y ( :sqrt: 2 - 1) = x ( :sqrt: 2 + 1)
( :sqrt: 2 - 1) : ( :sqrt: 2 + 1) = x : y
3 - 2 :sqrt: 2 : 1 = x : y[/hide]


trigonometry
limit
calculus
calculus computations

Hello,
let's consider the following series $ \sum_{n \equal{} 1}^{ \plus{} \infty} n \sin n$.
And the problem is that despite not having a limit in infinity ( $ \lim_{n \to \plus{} \infty} n \sin n$ doesn't exist) the sum is equal to 0.
It intrigues me a bit because it's against n-th term test ( http://en.wikipedia.org/wiki/N-th_term_test ).
What can you say about this situation  :?:

[quote="@petko"]the sum is equal to 0[/quote]  This is wrong: as you note, the sum fails the $ n$th-term test and so doesn't converge.  Why do you think it is equal to 0?

Well, I must say that Mathematica gives $ 0$ to be the answer. But, as we all know, Mathematica isn't always correct.

What about:
$ \sum_{n \equal{} 1}^{\plus{}\infty} n \sin n \equal{} \Im \left\{ \sum_{n \equal{} 1}^{\plus{}\infty} n \textbf e^{\imath n} \right\} \equal{} \Im \left\{ \frac{\textbf e^\imath}{(\textbf e^\imath \minus{} 1)^2} \right\} \equal{} 0$
 :huh:

$ \sum_{n \equal{} 1}^{ \plus{} \infty} n \textbf e^{\imath n} \equal{} \frac {\textbf e^\imath}{(\textbf e^\imath \minus{} 1)^2}$ in exactly the same sense as $ \sum_{n \geq 1} n \cdot 2^n \equal{} \frac {2}{(2 \minus{} 1)^2}$.  You're plugging in values outside the set of convergence of the power series in question.

Is it because $ | \textbf e ^{\imath n} | \equal{} 1$ ?

... exactly, and hence $ |ne^{in}|$ diverges, and the sum $ \sum_{n \equal{} 1}^\infty ne^{in}$ does not converge...  Actually, it would suffice that $ |e^{in}|$ would tend to 0 slower than $ \frac {1}{n}$, which is obviously the case since $ |e^{in}|$ is constant.


There is no such thing as a negative angle.

really?? cause i saw them online and i didnt get it.

um type it in on google and something should pop up

Actually, no, why don't YOU google it?  I googled it, and the first page had a very nice explanation.

just did and do you mean the one with the trig? like the sine -40 one?

Polar coordinates. negative angles mean go clockwise.


quadratics
Vieta
algebra
quadratic formula

[quote]
and one more problem
x+y=5
xy=6 [/quote]

Solve for x or y in both of the equations.

Next, use the real number property that if any number a = b and b = c, then a = c.

If a = b and b = c,

then a + (-b) = 0 and b + (-c) = 0

=> -a = -b and -c = -b (from the property a + (-a) = 0, for any numer a).

so a + (-b) = a + (-c) = 0

=> a = c.

And also a second real number property:

If for any two numbers, a*b = 0, then a = 0 or b = 0.

This is taken from the property that for any number a, a*0 = 0.
-----------------------------------

x = 6 / y

and

x = 5 - y

=> (5 - y) = 6 / y

=> -y^2 + 5y = 6

=> -y^2 + 5y - 6 = 0 = y^2 - 5y + 6

This is a factorable quadratic equation, so

y^2 - 5y + 6 = (y - 2)(y - 3) = 0

=> (y - 2) = 0 or (y - 3) = 0

=> y = 2 or y = 3

Good luck with your studies!
 :lol:

[quote="icx"]i don't remember how to solve this (just an example)

$x^{2}+x=12$[/quote]

or you can simply...

[hide]$x(x+1)=12$
and clearly...
$x=3$ or $x=-4$[/hide]

however, this may not work with more complicated numbers.

[quote="icx"]i don't remember how to solve this (just an example)

$x^{2}+x=12$[/quote]
To solve this quickly, you can do
$x^{2}+x=x(x+1)=12$,
and obviously $x=3$ works, where you have
$3\times 4=12$.
You can also notice that if you have $-3$ and $-4$ as the two numbers, this also works, which occurs when $x=-4$.

For a general quadratic, you can use the quadratic formula:
If $ax^{2}+bx+c=0$, then $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$.

You can also complete the square, as Chocolate milk did.

[quote="icx"]and one more problem

$x+y=5$
$xy=6$

please help[/quote]

Square the first equation to get
$x^{2}+2xy+y^{2}=25$.
Subtract $4xy$ to get
$(x-y)^{2}=x^{2}-2xy+y^{2}=25-4(6)=1$.
Take the square root to get
$x-y=1$.
Add this to $x+y=5$ to get
$2x=6$.
From here we can easily get $x=3$, $y=2$.
(However, notice that $x=2$, $y=3$ also works, so don't forget about switching around the variable values if the equations are symmetric [e.g. $x+y$, $xy$, $x^{2}+y^{2}$, etc.])

[quote="buzzer11"][quote="icx"]i don't remember how to solve this (just an example)

$x^{2}+x=12$[/quote]

or you can simply...

[hide]$x(x+1)=12$
and clearly...
$x=3$[/hide]

however, this may not work with more complicated numbers.[/quote]
And that only gave you one solution ;).

There exists a formula for two numbers given their sum and product.
[hide="Formula"]
Let $a+b=S$ and $ab=P$.  Consider $(x-a)(x-b)=0$.  It has roots $a$ and $b$.  Factoring gets us $x^{2}-(a+b)x+ab=0$, or $x^{2}-Sx+P=0$.  So the formula is $\boxed{\frac{S\pm\sqrt{S^{2}-4P}}{2}}$.[/hide]

[quote="icx"]i don't remember how to solve this (just an example)

$x^{2}+x=12$[/quote]

[hide]THis is just factoring
$x=\{-4, 3\}$[/hide]

[quote="buzzer11"]haha. I forgot a solution.
 :blush:
its edited.[/quote]
That method isn't very good because it is very easy to miss a solution...

[quote="kyyuanmathcount"][quote="anirudh"][quote="icx"]and one more problem

$x+y=5$
$xy=6$

please help[/quote]

[hide]
Substitute:
$x=5-y$
$(5-y)y=6$
$5y-y^{2}=6$
$y^{2}-5y+6=0$
$(y-2)(y-3)=0$
Therefore, $y=\{2, 3\}$[/hide][/quote]

Actually, I'm a lot more fond of the following method for solving these systems.

[hide]Substitute $x=a+b$ and $y=a-b$.

This means that $2a=5$ and $a^{2}-b^{2}=6$.

From here, we can just plug in $a=\frac{5}{2}$ and chug the rest of the way.  [/hide][/quote]Did you come up with that method???

[quote="bpms"][/quote][quote="kyyuanmathcount"][quote="anirudh"][quote="icx"]and one more problem

$x+y=5$
$xy=6$

please help[/quote]

[hide]
Substitute:
$x=5-y$
$(5-y)y=6$
$5y-y^{2}=6$
$y^{2}-5y+6=0$
$(y-2)(y-3)=0$
Therefore, $y=\{2, 3\}$[/hide][/quote]

Actually, I'm a lot more fond of the following method for solving these systems.

[hide]Substitute $x=a+b$ and $y=a-b$.

This means that $2a=5$ and $a^{2}-b^{2}=6$.

From here, we can just plug in $a=\frac{5}{2}$ and chug the rest of the way.  [/hide][/quote][quote="bpms"]Did you come up with that method???[/quote]

Not really, I'm not that beastly.  I saw rcv post it a while before.

heh that method is nice... too bad when x+y=m and xy=n the problem's usually trivial even without a calculator, you could probably use the same plugging in method on a more complicated question though.

Just thought about it. Couldn't Newton's Sums give you instantaneous results?

[quote="bpms"][quote="kyyuanmathcount"]Newton sums, like, writing $x,y$ as the roots of a quadratic?  (I haven't learned those).

It would make the same quadratic as just substituting, though that is a possible shortcut, I think.[/quote] You don't really need them for something as simple as this. But the sum of the roots of a quadratic (say x and y) is well known to be $\frac{-b}{a}$ and the product is $\frac{c}{a}$. So this can be rewritten as $x^{2}-5x+6=0$ whose roots are clearly 2 and 3.
EDIT:
[quote="bpms"]Here's the reason. I think it is somewhat related to the binomial theorem's explanation.
The polynomial can be factored to the product of the following with one number chosen from each row to find each part of the equation when multiplied out.
$(x+r_{1})$
$(x+r_{2})$
$(x+r_{3})$
.
.
.
$(x+r_{n})$
So with this when we want to find the coefficient of $x^{n-1}$ we know that every possible combination is a one where we chose all of the x's and one of the r's. So in the polynomial without combining like terms we get that every coefficient of $x^{n-1}$ is $r_{1},r_{2},...r_{n}$. So the sum of them is the coeffecient of $x^{n-1}$ when you combine like terms. This is where newtons sums come from. You can use similar reasoning to find the coeffecient of any terms  in a polynomial if you know the roots.[/quote]
To add to this if we are looking for the coefficient of $x^{n-2}$ it isn't hard to see that we need to choose x for all but two of the factors. So every possible way to pick two of the factors is the coefficient. I think this is a good uncomplicated explanation of how newton's sums work.[/quote]
I think you are referring to vieta's formulas?
newton's sums are these stuff
[url]http://mathworld.wolfram.com/Newton-GirardFormulas.html[/url]

[hide][quote="icx"]i don't remember how to solve this (just an example)

$x^{2}+x=12$[/quote]

use quadratic formula? or factor?
x^2+x-12=0
(x+4)(x-3)
x=-4,3[/hide]


Here is the question:

I alternately toss a fair coin and throw a fair die until
I either toss a head or throw a 2. If I toss the coin first,
what is the probability that I throw a 2 before I toss a
head ??

The probability of getting a 2 before a head the first round is $\frac{1}{12}$ throwing a tail and a two.  The probabilty of the game contnuing is $\frac{5}{12}$ throwing a tail and not a two.  Therefore the probability of getting a 2 before a head in the second round is $\frac{1}{12}\frac{5}{12}=\frac{5}{144}$ And so on and so forth. Therefore, the overall probability is $\frac{5^0}{12^1}+\frac{5^1}{12^2}+\frac{5^2}{12^3}...$  This is known as an infinite geometric sequence with common ratio $\frac{5}{12}$ The formula for the sum of an infinite geometric sequence is $\frac{a_1}{1-r}$ Plugging in the above values the probability is $\frac{\frac{1}{12}}{\frac{7}{12}}=\frac{1}{7}$

I think this problem is above the difficulty level of a middle school forum.  I am moving this to the high school forums area.

This problem was already discussed here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30044.
It might help for different solutions.


I apologize if this is in the wrong forum but I have been asked to find the flaw in this proof for my "intro to proofs" class but I really can't find anything wrong. Anyways, here it is.

Result: If n is even and m is odd, then nm+1 is odd.

Proof: Suppose n is even and m is odd. Then for some integer k, we know n=2k and m=2k+1. Thus nm+1 =2k(2k+1)+1 = 4k^2+2k+1 = 2(2k^2+k)+1. Since 2k^2+k is an integer, nm+1 is odd.

I assume so. It doesn't explicitly say. Is that the error in the proof?

Yes, that's it. By overloading $ k$, the alleged proof makes an assumption that probably isn't true. $ 2k\cdot(2k+1)$ isn't just the product of an even number and an odd number- it's the product of two consecutive integers.


LaTeX
geometry
geometric transformation
algebra
binomial theorem

[quote="sludgethrower"]I am sorry about this...but I don't know how to write maths properly on the computer :) . Anyway. Here is the problem.

Problem:

Express the binomial expansion of (x - 2/x)^4 in the form ax^4 + bx^2 + c + d/(x^2) + e/(x^4) for integers a, b, c, d and e.[/code][/quote]
Latex translation: $(x-\frac{2}{x})^{4}$.

[hide="My solution"]Just square it, and square it again.
$(x-\frac{2}{x})^{2}=x^{2}+\frac{4}{x^{2}}-4$
$(x^{2}+\frac{4}{x^{2}}-4)^{2}=x^{4}-8x^{2}-\frac{32}{x^{2}}+\frac{16}{x^{4}}+24$
I might have done this wrong because I did it all in my head. Can someone check?[/hide]

It is far easier to do it using the binomial theorem. But you gat the same answer as me anyway :)

[quote="sludgethrower"]It is far easier to do it using the binomial theorem. It isn't the same answer as I got.[/quote]
Yea I know, I just didn't have any paper or pencil at the moment, so I squared it in my head.  :D

[quote="sludgethrower"]It is correct...I just read it wrongly because you put the constant at the end...[/quote]
Oh, heh, sorry, I thought I saw the constant at the end when I was reading the problem.


inequalities
trigonometry
Pythagorean Theorem
geometry
geometry proposed

$ ABC$ is a right triangle in $ C$,and $ a$ is the measure of the angle between the median that pass trough $ A$ and the hypotenuse.

prove that        $ sin (a)  \le \frac{1}{3}$

Let $ M$ be the midpoint of $ BC$. We have
$ \sin(a) \equal{} \sin(\angle AMC \minus{} \angle CBA) \equal{}$

$ \sin(\angle AMC)\cos(\angle CBA) \minus{} \sin(\angle CBA)\cos(\angle AMC)$.

Let $ AC \equal{} b, AB \equal{} c$. Then
\[ \sin(a) \equal{} \frac {2b}{\sqrt {4b^2 \plus{} a^2}} \cdot \frac {a}{\sqrt {a^2 \plus{} b^2}} \minus{} \frac {b}{\sqrt {a^2 \plus{} b^2}} \cdot \frac {a}{\sqrt {4b^2 \plus{} a^2}} \equal{} \frac {ab}{(\sqrt {4b^2 \plus{} a^2})(\sqrt {a^2 \plus{} b^2})}
\]
Hence, it suffices to show that
\[ \frac {ab}{(\sqrt {4b^2 \plus{} a^2})(\sqrt {a^2 \plus{} b^2})} \le \frac {1}{3} \Leftrightarrow 9a^2b^2 \le 4a^4 \plus{} b^4 \plus{} 5a^2b^2 \Leftrightarrow 4a^4 \plus{} b^4 \ge 4a^2b^2
\]
which is true by AM-GM.

[quote="rachid"] [color=darkred]$ ABC$ is a $ C$ - right triangle , $ M$ is the midpoint of $ [BC]$ and $ \phi \equal{} m(\widehat {BAM})$ . Prove that $ \sin\phi \le \frac 13$ .[/color] [/quote]

[color=darkblue][b][u]Method I.[/u][/b] Observe that $ \left\{\begin{array}{c} m_a^2 \equal{} a^2 \plus{} 4b^2 \\
 \\
ab \equal{} 2cm_a\sin \phi\end{array}\right\|$ $ \implies$ $ \sin \phi \equal{} \frac {ab}{\sqrt {(a^2 \plus{} b^2)(a^2 \plus{} 4b^2)}}$ . Therefore,

$ \sin\phi\le\frac 13\Longleftrightarrow 9a^2b^2\le a^4 \plus{} 5a^2b^2 \plus{} 4b^4$ $ \Longleftrightarrow$ $ a^4 \minus{} 4a^2b^2 \plus{} 4b^4\ge 0$ $ \Longleftrightarrow$ $ (a^2 \minus{} 2b^2)^2\ge 0$ ,

what is truly. We have equality if and only if $ a \equal{} b\sqrt 2$ .

[b][u]Method II.[/u][/b] $ b\sin (A \minus{} \phi ) \equal{} c\sin\phi$ $ \implies$  $ \tan\phi \equal{} \frac {b\sin A}{b\cos A \plus{} c} \equal{} \frac {ab}{a^2 \plus{} 2b^2}$ . Thus, $ \sin \phi \equal{} f(t) \equal{} \frac {t}{\sqrt {t^2 \plus{} (t^2 \plus{} 2)^2}}$ ,

where $ \boxed {\ t \equal{} \frac ab > 0\ }$ . Since $ f'(t)\ .s.s.\ (\sqrt 2 \minus{} t)$ obtain $ f(t)\le f(\sqrt 2) \equal{} \frac 13\ \implies\ \sin\phi\le\frac 13$ .

I defined $ X\ .s.s.\ Y\ \Longleftrightarrow\ XY > 0\ \vee\ X \equal{} Y \equal{} 0$ , i.e. $ X$ , $ Y$ have [b]S[/b]ame [b]S[/b]ign.[/color]

[quote="rachid"] [color=darkred]$ ABC$ is a $ C$ - right triangle , $ M$ is the midpoint of $ [BC]$ and $ \phi \equal{} m(\widehat {BAM})$ . Prove that $ \sin\phi \le \frac 13$ .[/color] [/quote]
Let $ AB \equal{} 2$, W.L.O.G. and Let $ H$ is the foot of the perpendicular from $ M$, we have $ BM \equal{} MC \equal{} \cos \phi$ and $ AC \equal{} 2\sin \phi$. Then by Pythagorean theorem we have

$ AM \equal{} \sqrt {MC^2 \plus{} AC^2} \equal{} \sqrt {\cos ^ 2 \phi \plus{} 4\sin ^ 2 \phi} \equal{} \sqrt {3\sin ^ 2 \phi \plus{} 1}$. Since $ MH \equal{} BM\sin \phi \equal{} \sin \phi \cos \phi\ \therefore \sin \alpha \equal{} \frac {MH}{AM} \equal{} \frac {\sin \phi}{\sqrt {3\sin ^ 2 \phi \plus{} 1}}$.

By A.M.-G.M. we have

$ 9\sin ^ 2 \alpha \equal{} \frac {9\sin ^ 2 \phi \cos ^ 2 \phi}{3\sin ^ 2 \phi \plus{} 1} \equal{} \frac {9\sin ^ 2 \phi (1 \minus{} \sin ^2 \phi)}{3\sin ^ 2 \phi \plus{} 1}$

$ \equal{} 5 \minus{} \left(3\sin ^ 2 \phi \plus{} 1 \plus{} \frac {4}{3\sin ^ 2 \phi \plus{} 1}\right)\leq 5 \minus{} 2\sqrt {(3\sin ^ 2 \phi \plus{} 1)\cdot \frac {4}{3\sin ^ 2 \phi \plus{} 1}} \equal{} 1$,

The equality holds when $ 3\sin ^ 2 \phi \plus{} 1 \equal{} \frac {4}{3\sin ^ 2 \phi \plus{} 1} \ \left(0 < \phi < \frac {\pi}{2}\right)\Longleftrightarrow \sin \phi \equal{} \frac {1}{\sqrt {3}}$.

from $ 0 < \alpha < \frac {\pi}{2}$, $ \sin \alpha > 0$, yielding $ \sin \alpha \leq \frac {1}{3}$. $ Q.E.D.$

[quote="rachid"] [color=darkred]$ ABC$ is a $ C$ - right triangle , $ M$ is the midpoint of $ [BC]$ and $ \phi \equal{} m(\widehat {BAM})$ . Prove that $ \sin\phi \le \frac 13$ .[/color] [/quote]
Here is my solution:

with the law of sinus in triangle $ ABM$ we have:

$ \frac{2sin(a)}{a}\equal{}\frac{sin(B)}{AM}\equal{}\frac{2b}{c \sqrt{4b^2\plus{}a^2}},$

it suffices to prove that:

$ \frac{ab}{c \sqrt{4b^2\plus{}a^2}} \le \frac{1}{3},$

which is equivalent to:

$ (a^2\plus{}b^2)(4b^2\plus{}a^2) \ge 9a^2b^2,$

$ \Longleftrightarrow\ a^4\plus{}4b^4 \ge 4a^2b^2$ which is true by AM-GM  :)


X, Y normed spaces, $T \in L(X,Y), T^{*}$ is adjoint of $T$, prove:

1. T* is injective iff range of T is dense in Y.
2. If range of T* is dense in X*, then T is injective. Converse is true if X is reflexive.

Can someone help? thanks.

1. Let $T^{*}$ is injective. Suppose, that $R(T)$ is not dense in $Y$. Then $R=\overline{R(T)}\neq Y$ and we can find $f_{1},f_{2}\in Y^{*}: \quad f_{1}|_{R}=f_{2}|_{R}$ but $f_{1}\neq f_{2}$. the last statement is follows from Ghan-Banach theorem. Then $g_{1}= T^{*}f_{1}$ , $g_{2}=T^{*}f_{2}$  and $g_{1}\neq g_{2}$. But $\forall x\in X: g_{1}(x)=f_{1}(Ax)=f_{2}(Ax)=g_{2}(x)$. Contradiction. So $R=Y$.

Let $R=Y$. Suppose that $T^{*}f_{1}=T^{*}f_{2}=g$ then $g(x)=f_{1}(Tx)=f_{2}(Tx) \forall x \in X$. So $f_{1}(y)=f_{2}(y), y\in R(T)$. But $f_{1},f_{2}\in Y^{*}$ so $f_{1}(y)=f_{2}(y)$ when $y\in \overline{R(T)}=Y$. Contradiction. Then $T$- is injective

2. Suppose, that $T$ is not injective. Then $\exists x_{1}\neq x_{2}$, such that $Tx_{1}=Tx_{2}$. Thus $\exists x=x_{1}-x_{2}\neq 0$, and $Tx=0$.
Use Ghan-Banach theorem and we obtain $f\in X^{*}$ such that $f(x)=\parallel x\parallel \neq 0$. Now $\overline{R(T^{*})}=X^{*}$, so $\exists f_{0}\in R(T^{*}),f_{0}(x)\neq 0$. (Suppose that $\forall h\in R(T^{*}): h(x)=0$. One can find $h_{n}\to f$ in $\parallel \cdot\parallel _{T^{*}}$ where $h_{n}\in R(T^{*})$, so $0=h_{n}(x)\to f(x)\neq 0$. )

Now: $f_{0}=T^{*}g_{0}$, and \[0=g(0)=g_{0}(Tx)=T^{*}(g_{0}(x))=f_{0}(x)\neq 0\]
contradiction

Now let $T^{*}$ - is injective.
Suppose $R(T^{*})$ is not dence in $X^{*}$. We can find $F\in X^{**}$, such that $F|_{\overline{R(T^{*})}}=0$ and $F(g)\neq0$ where $g\in X^{*}\setminus \overline{R(T^{*})}$. $X$ - is reflexive. $\exists x\in X: f(x)=F(f)$, or $f(x)=0,f\in R$, and $g(x)\neq 0$ for some $g\in X^{*}$.
We have $\forall h\in Y^{*}: 0=T^{*}(h)(x)=h(Tx)$. So $Tx=0$ and $x=0$, because $T$-is injective. But $g(x)\neq 0$.contradiction

Is there a counterexample in non-reflexive spaces when the converse of 2. is not true?

Found one myself:

Let $J : \ell^\infty \to c_0,\ J(x_n)=(\frac{1}{n}x_n)$.

Then $J$ is injective. We have $J^* : (c_0)^* \to (\ell^\infty)^*$. Now we have the following facts:
[list]$(c_0)^*\simeq \ell^1$;
$\ell^1$ is separable;
$\overline{R(J^*)}$ is a separable subspace of $\ell^\infty$;
$\ell^\infty$ is not separable.
[/list]

Therefore $\overline{R(J^*)}\neq\ell^\infty$;


the answers of below are finite:
$n!=u^a-u^b$
(u does not need to be fixed)
is this  correct?

This has been posted before at least twice. The first thread was started by sam-n (it's in the "Number Theory Unsolved Problems" section), and harazi posted a non-elementary solution. I can't remember seeing an elementary solution on the forum.

You should have enough good keywords to find it, I think.

Grobber, do you know where i can find Harazi's solution? :?


geometry
incenter
geometric transformation
rotation
geometry proposed

Given triangle $ABC$ $A',B',C'$ are midpoint of $BC,CA,AB$ resp. $E$ is nine-point center (of $ABC$) $A_{1},B_{1},C_{1}$ are symmetric point of $A',B',C'$ through $E$, $H$ is orthocenter of $ABC$, $A_{1},B_{2},C_{2}$ is midpoint of $HA,HB,HC$ resp.Called $d_{a}$ is line connecting midpoint of $B_{1}C_{2}$ and $B_{2}C_{1}$, similar $d_{b},d_{c}$ prove that $d_{a},d_{b},d_{c}$ are concurrent! :)

There is a typo in the notations, because of $A_{1}\equiv A_{2},$ $B_{1}\equiv B_{2},$ $C_{1}\equiv C_{2},$ as well known.

Please clarify your problem.

Kostas Vittas.

I don't think this is correct either.  I get that the midpoints of $B_{1}C_{2}$ and $B_{2}C_{1}$ always coincide, etc.

[quote="gemath"]oh you are right, it must be
Given triangle $ABC$ $A',B',C'$ are midpoint of $BC,CA,AB$ resp. $I$ is incenter $A_{1},B_{1},C_{1}$ are symmetric point of $A',B',C'$ through $I$, $H$ is orthocenter of $ABC$, $A_{1},B_{2},C_{2}$ is midpoint of $HA,HB,HC$ resp.Called $d_{a}$ is line connecting midpoint of $B_{1}C_{2}$ and $B_{2}C_{1}$, similar $d_{b},d_{c}$ prove that $d_{a},d_{b},d_{c}$ are concurrent! :)
Thanks for your remark![/quote]

Ok nsato sorry for all mistake, In my note book the wrtting is so sloppy :P
I think here is correct
Given triangle $ABC$ $A',B',C'$ are midpoint of $BC,CA,AB$ resp. $E$is nine-point center (of $ABC$), $A_{1},B_{1},C_{1}$ are image of $A',B',C'$ through an arbitrary rotation center $E$, $P$ is arbitrary point, $A_{2},B_{2},C_{2}$ is midpoint of $PA,PB,PC$ resp. Called $d_{a}$ is line connecting midpoint of $B_{1}C_{2}$ and $B_{2}C_{1}$, similar $d_{b},d_{c}$ prove that $d_{a},d_{b},d_{c}$ are concurrent! :)

Define Q such that $4\vec{GQ}=\vec{GP}$
than we can show that $d_{a}, d_{b}, d_{c}$ are pass through Q.
now, it is obvious!

That's right a nice solution but actually you can see we change $E$ by an arbitrary point this problem is still true  :D


a radio station is going to construct a 9-foot tower for a new antenna. the tower will be supported by three cables, each attached to the top of the tower and to points on the roof of the building that are 12 feet from the base of the tower. find the total length of the three cables

When you draw this picture out, you will get 3 right triangles, each with legs of 9 and 12.  Use the Pythagorean theorem to get the hypotenuse of 15.   Because you have $ 3$ cables, the answer is $ 15 \times 3 \equal{} \boxed{45}$

Pythagreon theorm states that a^2+b^2= c^2. So when you illustrate this, the legs of the triangle is 9 and 12. If you calculate it, it comes out to 15. Therefore the length is 15 times 3 which equals 45.


I am applying for math graduate school, and I have a question about the personal statement for anyone who has experience in this.

Who exactly reads these things? Is it still the dreaded amorphous blob of an admissions committee, forcing me to throw in pretentious anecdotes and as many A+ words as I can cull up? Or do actual math professors get to see any of these statements?

Since I am applying to a technical department, I would like to be quite technical in my statement about my work. It would of course go over the head of a typical admissions officer, so it would be a waste of time and space.

Does anyone have concrete information on this?

There are two different committees which look at a graduate school application.  One is within the department and one is the graduate school (which is administratively a different unit).  Often they have separate parts to the application.  If the personal essay is required by the department, then actual professors will definitely read it.  If it's required by the graduate school, it may or may not be forwarded to the department... that probably depends on the institution.  I would guess that more often than not, it's forwarded.

Here's one example: I am certain that at Maryland the department sees the personal essay (or at least they did 15 years ago), because when I got a response from the department it was personalized and mentioned something I'd said in my essay.

By the way, you have to be accepted by both the department and the graduate school in order to be admitted.  Sometimes you'll get two separate letters and other times not, depending on the administrative structure of the school.

When I was a professor at NYU, I was on the graduate admissions committee of our department.  We were a committee of 3 professors who made the admissions decisions, and we definitely read each personal statement.

Thanks for the replies.

So as someone who is experienced with this process, would you say it is ok to throw in some technical aspects of the work I have done? I don't want to talk about grades, courses and any of that stuff because everyone has all the same crap. Instead I would like to discuss some of the details of the various research projects I have worked on and independent study I have undertaken.

What is your opinion on this?

Thanks again for your replies.

Makes sense to me.  If you could tie your previous work into the area(s) of research that you would like to investigate further, and if that area is represented by at least one faculty member, then all the better.

[quote="pmlbowt"]...would you say it is ok to throw in some technical aspects of the work I have done?... Instead I would like to discuss some of the details of the various research projects I have worked on and independent study I have undertaken.[/quote]

I think it's a good idea to discuss your independent study and research projects.  But make sure that your desire to do mathematics comes across through that discussion; that is, use those topics as a means to document your progress toward your academic goals, rather than as ends in themselves.  Also, don't make the essay [i]too[/i] long or folks won't want to read it.  (I've been on hiring committees...)

Thank you very much smbelcas. You have been incredibly helpful, and this thread might be of help to others applying to graduate school (I notice that most discussions here involve undergrad). If you have any other nuggets of wisdom I am all ears... for the most part, the graduate application process has seem a little mystified compared to undergrad.


That's right. :)

Can anyone solve without using Landau sign?


floor function
geometry
geometric transformation
rotation
combinatorics proposed
combinatorics

A regular n-agon with $p$ vertex white and $n-p$ vertex white is given, $n-p\geq p\geq 3$. Prove there exist a $\left\lfloor p/2\right\rfloor+1$ white poligon congruent to a black one.

[quote="Pascual2005"]A regular n-agon with $p$ vertex white and $n-p$ vertex white [/quote]
Ok, which are white and which are black? :)

Indeed, we are to show that there is a rotation of the white set s.t. we get at least $[p/2]+1$ intersections with black set. Suppose the contrary, then each rotation gives us at most $[p/2]$ intersection, so $(n-1)[p/2]$ totally. But finally each white vertex will meet each black vertex, so there are $p(n-p)$ intersections. Thus $(n-1)[p/2]\geq p(n-p)$ -- contradiction.


Hey, I am having some trouble with a problem here. I dont know if I have got it, assistance needed..

It goes like this:

Find the $\beta$ matrix for $T$, when $\beta$ is the basis $\{1,1+x,1+x+x^{2},1+x+x^{2}+x^{3}\}$

$T(a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}) = (a_{0}+a_{1})+(a_{1}+a_{2})x+(a_{2}+a_{3})x^{2}+(a_{3}+a_{0})x^{3}$

I got:(sorry dont know how to make a matrix)

Mathlab way: $[T]_\beta =$ [1 0 0 1;1 1 0 0;0 1 1 0;0 0 1 1]

Is this right?

Sorry for double post, was unable to edit my previous post.

What I ment was something like this $[T]_\beta$ =
1  1  0  0
                                                                           0  1  1  0
                                                                           0  0  1  1
                                                                           1  0  0  1

Edit: I found it out! There was more info on the problem which I did not see. Sooo just forget me asking in the first place.


calculus
integration
real analysis
real analysis unsolved

Give an equivalent for an integral $\int_0^{+\infty}(\frac{te}{x})^xdx$ when $t \to +\infty$
I've got something like $et \sqrt{2 \pi t}$-can it be near to true equivalent

It seems strange as if you consider only the integral $\int_2^3 (\frac{te}{x})^xdx)$ it is superior to ct^2 where C is a constant.


let $ s(n)$ denote the sum of the digits of n. For how many integers n is
$ n \plus{} s(n) \plus{} s(s(n)) \equal{} 2007$?
[b]
A) 1    B)2    C)3    D)4    E)5[/b]


The only proof I know for the following problem is very long and difficult. Does anyone have an easier one?
    Prove that there exists a constant C such that any natural number greater than 1 is the sum of at most C prime numbers. Thanks.

[url=http://mathworld.wolfram.com/MannsTheorem.html]This[/url] might be of interest. And, related to it, is [url=http://mathworld.wolfram.com/SchnirelmannsTheorem.html]this[/url].

Indeed, this is the way I found to prove the statement, but my question was not refering to this proof, but to an eventually easier one.


linear algebra
matrix
function
linear algebra unsolved

a) Prove that if $\begin {small}X=\left[ \begin{array}{cccccccc} X_1&|& X_2&|& \dots &|& X_n \end{array}\right]\end{small}$ is a  $n\times n$ matrix and $A\in \mathcal M_n(\mathbb F)$. Prove that :
\[ \begin {small}\det \left[ \begin{array}{cccccccc} AX_1&|& \dots &|&X_n \end{array}\right]+\ \dots\ +\det \left[\begin{array}{cccccccc} X_1&|& \dots &|&AX_n \end{array}\right]= \mbox {Tr}(A)\det (X)\end{small}  \]

b) Suppose that $A,X$ are $n\times n$ matrices which $X$ is invertible and $\mbox{Tr}(A)\neq 0$. Prove one of the matrices $\begin {small}\left[ \begin{array}{ccccccc} AX_1&|& \dots &|&X_n \end{array}\right]\end{small}$, ... , $\begin {small}\left[ \begin{array}{ccccc} X_1&|& \dots &|&AX_n \end{array}\right]\end{small}$ are invertible.

c) Can we make the last statement stronger?

Well the second statement is obvious with the first one. And for the first one you must do some clever computations.

But why b) follows from a)? Probably you meant that $X$ is invertible... And for a), one doesn't need computations: just observe that the LHS is a n-linear alternated application and thus we know it is proportional with the determinant application. Then one reduces to the case when $X_i$ are the canonical basis, case which is trivial (this is exactly the argument used to find the wronskian in linear differential equations).

Great Proof Harazai !!! :) I didn't even thought that this function is n-linear alternated. For the last one I think we could make a weaker assumption and prove the same thing.


u r able 2 find the middle value

by doing 283+119-380=22

4 the highest value of x, assume that in a set of {a,b,c,d,x}, a=1, b=2, c=22, d=23, sum(a through d)= 58
subtract that from 380, thus, highest value of x = 322

4 the lowest value of x, assume that in a set of {a,b,c,d,x}, a=20,b=21,c=22, sum(a through c)= 63
subtract this value from 380, thus, d+x=317

2 find the lowest value of x, divide 317 by 2 (158.5) this is the mean of d and x.

the lowest value of x must be 159.

322-159=163

idk if this is right.

if u mem this :bomb:, dont assume this is the right answer, u will pay...

:harhar:

Sorry, that is incorrect.

First we need to organize our information.

Let the numbers be, in this order, $ a, b, c, d, x$.

1) $ a \plus{} b \plus{} c \plus{} d \plus{} x \equal{} 320$

2) $ c \plus{} d \plus{} x \equal{} 283$

3) $ a \plus{} x \equal{} 119$

We see that the middle number, $ c$, is $ 283 \plus{} 119 \minus{} 320 \equal{} 82$. Now, for the greatest value of $ x$, we let $ a \equal{} 1$, $ b \equal{} 2$, $ c \equal{} 82$, and $ d \equal{} 83$. This means that $ x \equal{} 152$.

With some brute force, we find that the smallest possible value of $ x$ is 135, so our answer is $ 152 \minus{} 135 \equal{} \boxed{17}$.

I have a few questions...

First, why is the middle number 283+119-320. Shouldn't that just be b-x?
Second, if x is 152, then and a is 1 then the equation a+x=119 is false right?

I got that the answer was 22 because:

If a=13, b=24, c=93, d=94, and x=96 it looks to me that all the conditions are met and x is the lowest it can be.
And if a=1, b=37, c=82, d=83, and x=118 the conditions seem to be fulfilled as well and x is the highest i can be.

[quote="GameBot"]The sum of five different positive integers is 320. The sum of
the greatest three integers in this set is 283. The sum of the
greatest and least integers is 119. If $ x$ is the greatest integer
in the set, what is the positive difference between the greatest
possible value and least possible value for $ x$?[/quote]

Use inequalities.

$ x\equal{}320\minus{}(a\plus{}b\plus{}c\plus{}d)$
$ a<b<c<d<320\minus{}(a\plus{}b\plus{}c\plus{}d)$
$ 320\minus{}(a\plus{}b)\equal{}283\implies a\plus{}b\equal{}37$ (1)
$ 320\minus{}(b\plus{}c\plus{}d)\equal{}119\implies b\plus{}c\plus{}d\equal{}201$ (2)
Deal with (1) first. Substituting this into the equation for $ x$, we have $ x\equal{}283\minus{}(c\plus{}d)$.
Since $ a<b$, we also have that $ 19\leq b\leq36$, which in turn means that $ 165\leq c\plus{}d\leq182$, and $ 101\leq x\leq118$. Notice that for each possible value of $ c\plus{}d$, we can have $ a,b,c,d\leq\frac{182}2\equal{}91\leq101$, so our answer is $ 118\minus{}101\equal{}\boxed{17}$.

wait if 101 <= x <= 118 then shouldn't there be 18 values?

[quote="GameBot"]...what is the positive difference between the greatest
possible value and least possible value for $ x$?[/quote]

Read carefully.  :P


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nam định:đỗ đức hòa(12)
thái bình:vũ hữu tiệp(12)
thanh hóa:vũ văn tân(12)
đhsp hà nội:nguyễn hữu kiên(12)

4)giải khuyến khích:

bắc giang:nguyễn thành hải(11)
bắc ninh:nguyễn ngọc khôi(12)
đắc lắc:trần hoàng giang(12);lê đức quang(12)
đhkhtn-đhqg hà nội:đỗ ngọc tuấn(12);hà minh tuấn (12)
ptnk hcm:đỗ tuấn hoàng anh(12)
gia lai:nguyễn bá tiến(11)
hà nam:nguyễn viết xuân(11)
hà nội:đào đức chính(12)
hà tây:phạm văn trường(12);lã vĩnh hòa(12);
hà tĩnh:nguyễn nhật linh(12)
hải dương:vũ đình quyền(12);trương ngọc sơn(12);phạm thành thái (12)
nam định :vũ ngọc đào(11)
ninh bình:đinh xuân thực(11)
quảng bình:đặng ngọc thanh(12)
thái nguyên:nguyễn trung kiên(11)
quôc học huế:lê hữu điền khuê(12);huỳnh đình tuân(12)
đhsp hà nôi:trần thanh hải(12)
vĩnh phúc:hoàng thị thúy hồng(11)

Đây là đề Ngày 1: http://diendantoanhoc.net/forum/index.php?showtopic=30345&hl=


calculus
derivative
calculus computations


MATHCOUNTS
geometry
calculus
integration
perimeter
area of a triangle

I know that this maybe a dumb question to post, but I couldn't find a nice solution during the test.  Could someone explain how they did it?

Four friends took a multiple-choice test.  Their papers are shown below.  Each of them has exactly two correct answers.  Using their answers, what are the correct answers to questions 1,2,3 and 4, in that order?

Jerry
1. A 2. B 3. A 4. C

Jim
1. B 2. A 3. D 4. B

John
1. B 2. A 3. A 4. C

Jeremy
1. A 2. C 3. D 4. A

Thanks
p.s.
I found a solution afterwards but it took me more than 6 minutes so it doesn't count.

Haha. I remember that question. Lets just say I was lucky.  :D

Yeah, well brute forcing only took 1 and a half to 2 minutes, if you were doing it correctly. And #4 wasn't that difficult so it wasn't too bad.

However, I am interested to see an elegant solution for this kind of problem.

[hide="Is the answer"]
A
A
D
C

Jerry got 1 and 4
Jim got 2 and 3
John got 2 and 4
Jerry got 1 and 3

That works

[hide="solution"]
just by looking at this problem and knowing that is mathcounts difficulty, i figured that in this case not only did each person get two correct, but there was a good chance that there were two correct answers to each problem (maybe that has to be true...i don't know)
so
the four given answers to #2 were B, A, A, C. So A is probably the answer
the four given answers to #4 were C, B, C, A. So C is probablty correct
if this is true,
then the answer to #3 must be D, if it were A then John would have 3 right
and the answer to #1 must be A, if it were B, then Jim would have 3 right
it all checks
so the answer must be A, A, D, C

I did it this way in 1:16
but of course i didnt make nats this year so it really doesnt matter
because of two STUPID mistakes on state target
problem 4: calculator mistake
problem 7: I put 15
oh well[/hide][/hide]

Yes that is correct.  I do believe that it has to be 2 the same from each person, but I haven't proved it yet.

Thanks.

Proving that sort of thing has shown up on IMOs, except it is a much bigger proof, much larger numbers.

[quote="nebula42"][quote="Ignite168"]Since #4 was easy, I think it was change? I just brute forced it, there were only 6 possibilities.[/quote]

I'm pretty sure that was #6, and there were actually more like 4 possibilites you had to write out brute force equations for because if the first number was a.bc and the second one started with a, they'd both have to be 5.00, which doesn't work because they can't be the same.[/quote]

I mean I brute forced this problem because there were 6 possibilities and yeah, #6 was change. #4 was something like what is the largest area of a triangle with integral side lengths and perimeter 20?

[quote="nebula42"][quote="Ignite168"]Since #4 was easy, I think it was change? I just brute forced it, there were only 6 possibilities.[/quote]

I'm pretty sure that was #6, and there were actually more like 4 possibilites you had to write out brute force equations for because if the first number was a.bc and the second one started with a, they'd both have to be 5.00, which doesn't work because they can't be the same.[/quote]

There were 8 possibilities for the change question weren't there.   Either way, for number 3, the test questions, is it true what sapphyre571 said?  About exactly 2 people have to get each question correct.

I hope that is true.  I haven't found any thing that contradicts Sapphrye's statement yet.

This post has been edited by an administrator for offensive content. -- MC

[hide="Hint"]There are 8 total correct answers (2 from each person), but no three people had the same answer to any question.[/hide]

OK, that makes sense for the number of questions correct is 8 and no three have the same answer.

thanks.

If three people got 1 question correct then one person would have to be the only person to get a question right. Try to make a test and four potential answer sets and see if you can create a scenario in which more or less than two people get one question correct, but no one gets more then two correct.

So, basically you are saying that it does not necessarily mean exactly 2 people for each question, right?

If you can make such a set then that would be disproving the conjecture.

Okay

A A A A
B B B B
C C C C
D D D D

Correct answers A, A, C, C.  That does disprove it doesn't it.

Ok that is a little bit harder.  CAn't figure it out now.

[quote="bpms"]But the trick is to find one where three people get a question right.[/quote]
$A$ is a correct answer. $B$ is an incorrect answer.
Person 1: A, B, B, A
Person 2: A, B, A, B
Person 3: A, A, B, B
Person 4: B, A, A, B

Each person has 2 correct. 3 people got Question 1 and 1 person got question 4. :D


How many integers $ \minus{}11 \leq n \leq 11$ satisfy $ (n\minus{}2)(n\plus{}4)(n \plus{} 8)<0$?

This means that an odd number of the following expressions are negative: n-2, n+4, n+8.

If there is 1 negative expression, then n is between -4 and 2  (not inclusive). 5.
If there are 3 negative expressions, then n is less than -8 (not inclusive). 2.

So, (5+2)/23=7/23

Tsk tsk tsk james4l. You have to learn how to read  :P . It asks for the number of integers that satisfy, not the probability. The answer is $ \boxed{7}$.

[quote="james4l"]If there are 3 negative expressions, then n is less than -8 (not inclusive). 2.[/quote]
Actually, there are $3$ negative expressions, then $n<-8$.  There are not $2$ possible values for $n$, but $3$.  Therefore, the answer  $5+3=\boxed{8}$.

Tsk tsk tsk RoFlLoLcOpT, you need to learn how to check your answer  :P .

[quote="JSGandora"][quote="NeurologyX"]Please do not post any demotivating phrases to others.[/quote]
Obviously we're only playing as hinted by the " :P ".[/quote]
Still, it would be better if you are not "playing."


induction
trigonometry
algebra solved
algebra

The sequence of numbers $u_n$ is defined by:
    \[ u_1=\sqrt2\\u_n=\sqrt{2+u_{n-1}}\]
and the sequence of numbers $s_n$ is defined by:
     \[ s_n=\frac {u_1u_2...u_n}{2^n}\]
Find the limit of $s_n$

You use induction to prove that: $u_{n}=2\cos( \frac{\pi}{2^{n+1}})$ and then replacing in the formula for $s_{n}$ and using the fact that $ \prod_{k=1}^{n} \cos \frac{\pi}{2^{k}}=\frac{\sin(\pi)}{ 2^{n}\cdot \sin(\frac{\pi}{2^{n}})}$ you get the limit you want!
cheers!


Given $ x,y,z>0$ find the best const $ k$ such that
$ (1\minus{}k)(x^{2}\plus{}y^{2}\plus{}z^{2})\plus{}(k\plus{}1)(xy\plus{}yz\plus{}zx)\ge 2\sqrt{3xyz(x\plus{}y\plus{}z)}$


AMC
AIME
AMC 12
AMC 12 A
trigonometry
geometry
AIME I

What does everyone do the night before taking the AIME?

: |

another option: do pre-math competition rituals
(making my packet of doom;spinning my arms around;assembling my costume for the next day)

AKA putting pen(cil), protractor, paper and compass into a bag; yes, i do spin my arms around; lay my clothes out

1.  Eat
2.  Sleep
3.  Poop

I'm not even kidding about the third one.  I had a horrible score on the AMC12A because my bowels wouldn't shut up for the second half of the test.  It's affected my performance on a lot of math competitions, and it was probably part of the reason why I did the math wrong on #14 on the AIME.  And the first two are pretty self-explanatory.  They're a lot more important than last-minute practice.

My plan was to go out and get my favorite pie to relax, then come home and get to bed early.

But nobody wanted to go get pie with me and I couldn't fall asleep.   :o

I don't remember quite what I used to do the night before AIME (I never gave it too much thought), but were I taking it again, I would definitely place a high priority on sleeping.  This is mainly because I'm very used to taking competitions early in the morning, so I've gotten good at it over the years.  Lack of sleep, however, could affect this negatively, especially with a relatively long competition like the AIME.

Bascially I just did what I normally do - homework, did a few AIME problems, and I slept at around midnight.

I slacked the whole day and slept at 1AM. seriously

I also tried to relieve my bowels, because that has been a major problem in performance and when you feel tension, it screws up concentration quite horribly.

Find a few extremely easy AIME problems or ones that you already know how to do and solve them the night before or the morning of the test.  DONT try to learn everything there is to know about complex numbers or trig the night before the test.  DONT forget to make sure your pencils are #2 and not #2 5/10.  Yes, this has happened to me before.  :rotfl:

Yeah seriously, just do what you normally do the day before.  I just had track practice and went out for a bit (gotta love IMSA dorms).  Then, it was just a few AIME problems, mental preparation, and a lotta sleep -- and everything turned out just fine.  :)

this year,
i had a biology test,
then i came home,
watched everybody loves Raymond,
eat
sleep

half an hour before the test:
go for a run (it really helps)

I was actually having a date the evening of March 6. I thought a relaxing environment would ease up the tension a bit. I turned out to be right I guess.

I wish I would have gone on a date.  But instead I just got to hang out with her, which is just as good.

Then I memorized the prime factorization of 2005, 2006, and 2007.

[quote="CheesyP18"]Then I memorized the prime factorization of 2005, 2006, and 2007.[/quote]

Same here :D . It always comes up on the tests.

Well specifically for the AIME.. cramming will not help if you're trying to make USAMO.  So therefore I suggest you relax and review your like geometry notes or something in the morning because your short term memory is good in the morning.  Relaxation is good in general, but I wouldn't spend more than two days before the test doing that.  The second to last day before the competition should be the last day you do any serious preparation.

I did HW for the 2 classes I had that day. (I missed French and English by taking the AIME.) Then I sharpened 10 pencils, put 360 deg protractor, compass, and ruler in bag, Then listened to music and went to bed at 1 AM. Going to bed at 1 AM along with the constant announcements for club pics probably made me make 3 careless errors.

Who commits suicide before the AIME?

Most people that make the AIME don't have dates or very good social lifes.
Why would anyone get a tatoo before the AIME?

[quote="Philip_Leszczynski"]If you go on a date before the AIME, then you're probably not enough of a loser to make it to the USAMO.  :P[/quote]

Me. Check and mate.

Woo, I'm on date number 3 lol.

I was trying to sleep that night. And I indeed slept very early. But.......... I couldn't fall asleep until 12 or 1. :(  :mad:
So, of course, I was not in good condition when taking AIME. However, not that bad, I think I still get the chance to get in.

Since we got to miss 4 classes for the AIME, I barely had any homework  :)

That opened up alot of time (though pre-season crew practice and orchestra burned up a lot of time anyway)

I went to sleep at 10:30. I've found through experience that if I go to bed before then, the extra sleep backfires on me.

I perform my top-secret math ritual. I think it helps. It's become a tradition now :)

You can tatoo yourselves some trig identities or times tables?

[quote="pieterminate"]Sleep is definitely paramount....
I didn't sleep enough and I virtually fell asleep by the time i got to 8 .........4 stupid mistakes :mad:  :([/quote]

Definitely....curse english papers!  :mad:


quadratics
AMC
AIME
algebra
polynomial
sum of roots

[hide]Multiplying both sides of the equation by $7-\sqrt[4]x$ and simplifying, $\sqrt x-7\sqrt[4]x+12=(\sqrt[4]x-4)(\sqrt[4]x-3)=0$. Thus, $\sqrt[4]x=4$ or $\sqrt[4]x$ and the sum of the solutions is $4^4+3^4=\boxed{337}$.[/hide]

[hide] This gives a quadratic $0=(\sqrt[4]x)^2-7(\sqrt[4]x)+12$, with solutions $x=81$ and $x=256$. So the answer is $81+256=\boxed{337}$.[/hide]

[hide]How do all of you keep getting $12$ on the same side as $\sqrt[4]x(7-\sqrt[4]x)$?  Shouldn't it be $-12$ when you bring it over, so that you get $(\sqrt[4]x)^2-7(\sqrt[4]x)-12$?[/hide]

Um, no...they multiplied it by $-1$ after. :)

[quote="DanK"]:blush: Wow, that's pretty sad that I didn't catch that, even if I have been up for over 18 hours...[/quote]
heh...i know the feeling

[hide][hide][hide] :D [/hide][/hide][/hide]

[quote="shyong"]What would you think would be the sum of roots if the question is rewrite as

[$\sqrt[4]{x}=\frac{13}{7-\sqrt[4]x}$ for any complex $x$ .

 :)[/quote]

[hide]For that equation,
$\sqrt[4]x=\frac{7\pm\sqrt{-3}}{2}$
$x=\frac{(7+ i\sqrt{3})^4}{16}$, $x=\frac{(7- i\sqrt{3})^4}{16}$.

Seeing as the terms containing $i$ cancel out when the two expressions are expanded and added, we are left with $\frac{2(7^4)+12(-147)+2(9)}{16}=\boxed{191}$[/hide]

(edit: oops! forgot to hide)

For some reason this problem seems too easy to be an AIME problem.(Even for a number 1)

[quote="Altheman"][quote="shyong"]What would you think would be the sum of roots if the question is rewrite as

$\sqrt[4]{x}=\frac{13}{7-\sqrt[4]x}$ for any complex $x$ .

 :)[/quote]

this would not work since when we have the fourth root of a complex number, there is not one "primary root"

[quote="cincodemayo5590"][quote="chess64"]Look at AIME I 2006 #1 and say that again, please.[/quote]
:rotfl:  that was pretty bad...[/quote]

i got that one wrong...  :mad:[/quote]

i dont think it matters that there isnt a primary root b/c we not finding it, it cancels out when you add the roots. I guess as long as we use the same number as the root it doesnt matter.

[quote="Altheman"][quote="cincodemayo5590"]

i dont think it matters that there isnt a primary root b/c we not finding it, it cancels out when you add the roots. I guess as long as we use the same number as the root it doesnt matter.[/quote]

it is not that you cannot do it, but it makes the problem ambigious because when you plug that root back in, 3 of the complex roots will not work, and 1 will, and there is no real rule for choosing the one that works, so you are assuming something that you cannot necessarily do, whereas with real numbers, it is the convension to choose the real root
ex, $\sqrt[2]{2i}=1+i$ but we could just as well say that it is equal to $-1-i$[/quote]
merr.... ok.

[quote="chess64"]Look at AIME I 2006 #1 and say that again, please.[/quote]

look at 2003 AIME 1 #1, it doesn't get worse than that ;)

[quote=4everwise]What is the sum of the solutions to the equation $\sqrt[4]x =\displaystyle \frac{12}{7-\sqrt[4]x}$?[/quote]

you can substitute the the fourth root of x by y to make stuff
so y=3 or y=4
so 4^4+3^4 so 337

[hide=Solution]We make a helpful substitution, $a=\sqrt[4]{x}$. Therefore, $a=\frac{12}{7-a}$, or $7a-a^2=12$. This simplifies to $a^2-7a+12=0$, or $(a-3)(a-4)=0$. If $a=3$, then we have $x=81$, and if $a=4$, then we have $x=256$. The sum of these two is $\boxed{337}$.[/hide]

[quote=cincodemayo5590][quote="4everwise"][quote="shyong"]What would you think would be the sum of roots if the question is rewrite as

$\sqrt[4]{x}=\frac{13}{7-\sqrt[4]x}$ for any complex $x$ .

 :)[/quote]Or you can use newton's sums.  :)[/quote]

What are newton's sums?[/quote]
Its an ezy way to find the nth power sum of the root of a polynomial

[hide=sol]Let $y = \sqrt[4]{x}$. Simplifying the equation, we get $(y-3)(y-4) = 0$. Solving for $x$ when $y$ equals $3$ and $4$, respectively, we get $x =81$ and $256$. Summing these yields $\boxed{337}$ as the answer.[/hide]

[hide=Solution]Let $\alpha=\sqrt[4]{x}$. Thus, we have
$$\alpha=\frac{12}{7-\alpha} \implies -\alpha^2+7\alpha=12 \implies \alpha^2-7\alpha+12=0 \implies (\alpha-3)(\alpha-4)=0$$
Therefore, $$\sqrt[4]{x}=3,4$$
Thus, $$x=3^4+4^4=81+256=\boxed{337}$$

[hide=Solution]Let $y=\sqrt[4]{x}$. Then, $$y=\frac{12}{7-y} \implies y=3,~4.$$ Re-substituting, we get $\sqrt[4]{x}=3 \implies x=81$ or $\sqrt[4]{x}=4 \implies x=256$. The desired answer extraction is $81+256=\boxed{337}$.[/hide]

[hide="My Sol"]\begin{align*}
\sqrt[4]x &=\displaystyle \frac{12}{7-\sqrt[4]x}\\
7\sqrt[4]x&=12+\sqrt[2]x\\
49\sqrt x&=144+ x + 24\sqrt x\\
625x&=144^2+ 288x+x^2 \\
0&=x^2-337x+144^2
\end{align*}
Thus, by Vieta's the sum of the solutions will be $\boxed{337}$[/hide]

[hide=Sol]Let $y = \sqrt[4]{x}$

$$y = \frac{12}{7-y}$$
$$y^2 - 7y + 12 = 0$$
$$(y-4)(y-3) = 0$$

Therefore, $y = 4$ and $y=3$ are our solutions. Plugging these values back into our original expression yields the solutions $256$ and $81$, respectively, the sum of which is $\boxed{337}$, as desired.[/hide]

[quote=MC413551]Wow this thread has been going for a long time![/quote]

When you find one of your old posts ;-;

[hide]LaTeX code below[/hide]
$First, /say/  \sqrt[4]{x}=a$
[hide]LaTeX code below[/hide]
$ a=\frac{12}{7-a}$
then we'll get;

[hide]LaTeX code below[/hide]
$a^2-7a+12=0$
if we solve this, we will get two roots.
a1=3 and a2=4
Then,
[hide]LaTeX code below[/hide]
$ \sqrt[4]{x}=3  /and x=81$
[hide]LaTeX code below[/hide]
$ second/ root /is/ a=4/  4^4=256$

256+81=[b]337[/b]


A 5x5x5 cube is formed by assembling 125 unit cubes. Nine unit squares are painted on each of the six faces of the cube according to the pattern shown. How many of the 125 unit cubes have no paint on them?
[asy]fill((2,0)--(3,0)--(3,5)--(2,5)--cycle,gray(.7));
fill((0,2)--(0,3)--(5,3)--(5,2)--cycle,gray(.7));
path p=(0,0)--(5,0);

draw(p,linewidth(1));
draw(shift(0,1)*p,linewidth(1));
draw(shift(0,2)*p,linewidth(1));
draw(shift(0,3)*p,linewidth(1));
draw(shift(0,4)*p,linewidth(1));
draw(shift(0,5)*p,linewidth(1));

draw((0,0)--(0,5),linewidth(1));
draw((1,0)--(1,5),linewidth(1));
draw((2,0)--(2,5),linewidth(1));
draw((3,0)--(3,5),linewidth(1));
draw((4,0)--(4,5),linewidth(1));
draw((5,0)--(5,5),linewidth(1));[/asy]

Since there are 9 squares painted on each face, we know there are $ 25 \minus{} 9 \equal{} 16$ that aren't painted. There are 6 faces on a cube, so we simply have $ 16 \times 6 \equal{} 96$.

Isabella, I think you overcounted.  If you see the left square on one side, that cube is also painted on another square: the right side on an adjacent face.  This is true on the middle square of all the edges.  You have 9 squares painted on each face*6 faces=54, but you must subtract the overlaps on each edge (12), so 42 cubes are really painted.  Therefore, 125-42=[b]83[/b] unit cubes have no paint on them.


logarithms
search
number theory unsolved
number theory

[b]numbers like(1/n) that n is a natural number are named egyption numbers.
for any k  what are the k  egyption numbers  with the minimum sum less than  1?
it looks like a greedy way. :D [/b]

I don't understand exactly the statement, but I think he asks for those $x_1,...,x_k$ which maximize $\frac{1}{x_1}+...+\frac{1}{x_k}$ and such that $\frac{1}{x_1}+...+\frac{1}{x_k}<1$. This is a difficult problem. As far as I remember, they are the consecutive terms of the sequence $a_1=2, a_{n+1}=a_n^2-a_n+1$.

[quote="Rodolphe2005"]$\frac{1}{k}+...+\frac{1}{2k}$ ??[/quote]
But it is about $\ln 2\approx 0.69$, i.e. too small to be the answer.

We need to find for which $k$ we can have $k$ egyptian numbers with sum equals to 1.
For others $k$ we just have to take the first $k$ inverse of the elements of the sequence defined by :
$x_0=2$
$x_{n+1}=\Bigg[\frac{1}{1-\displaystyle \sum_{k=0}^n \frac{1}{x_k}}\Bigg]+1$

2 hours that I search, it looks so hard to prove that I think it's not correct

Don't worry, there was a period when not even Erdos could prove it!

harazi was correct.
what are the k egyption numbers with the max sum less than 1.
 remark;
it looks like a greedy way

if k=3;

(1/2)+(1/3)+(1/7)

dashmiz do u want we solve this problem?

well,it was ever a problem in Chinese Team Selection Test.
Answer is:
put a(1)=2,a(n+1)=(a(n))^2-a(n)+1


Find the number of natural solution of the equation $ [\frac{x^3}{103x \minus{}1}]\equal{}[\frac{x^3}{103x\plus{}1}]\plus{}1$

___________________________
Azerbaijan Land of the Fire   :ninja:  :roll:

[quote="Pirkuliyev Rovsen"]Find the number of natural solution of the equation $ [\frac {x^3}{103x - 1}] = [\frac {x^3}{103x + 1}] + 1$ [/quote]

So $ n - 1\le \frac {x^3}{103x + 1} < n\le \frac {x^3}{103x - 1} < n + 1$. As a consequence $ \frac {x^3}{103x - 1} - \frac {x^3}{103x + 1} < 2$ and so $ x < 103^2$

So it is possible to find the solution with a computer. I did it and found exactly $ 5304$ solutions :

The $ 20$ first : $ 103,104,205,206,207,$ $ 244,271,308,309,310,$ $ 319,347,374,402,411,412,413,414,422,450$

The $ 20$ last : $ 10159,10187,10195,10196,10197,10198,$ $ 10207,10235,10262,10290,10299,10300,$ $ 10301,10338,10365,10402,10403,$ $ 10404,10505,10506$

And I would very interested in a simple method giving this number ($ 5304$) (I think such a method may exist since ${ 5304=\frac{103^2-1}2}$ :) )

Please, Pirkuliyev Rovsen, could you kindly tell us from where is this problem coming and give us your solution (or at least a hint towards a clever solution) ?

This example has been given in the kolmoqorov competition.Solution  been in me am approximate such !!


The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r$, where $a$ and $r$ are positive integers.  Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r)$.

[hide="Solution"]By properties of logs, it simplifies to
$log_8(a_1a_2a_3...a_{12}) = 2006$
which becomes
$a_1a_2a_3...a_{12} = 8^{2006}$

Since both a and r are integers, we know that the sequence is increasing. The first term is $a$, the next is $ar$, the next is $ar^2$, $ar^3$, etc. When we multiply the first twelve terms we get $a^{12}r^{66}$.

Our new equation is
$a^{12}r^{66} = 2^{6018}$

Obviously both a and r have to be powers of 2. Let a = 2^x and r = 2^y. Then
$12x + 66y = 6018$
$2x + 11y = 1003$

All that needs to be done from here is finding how many integers (x,y) satisfy that equation, where both are > 0.
There are 46 pairs.

(I made the mistake of having it equal 1002 instead of 1003, and got 45 pairs)
[/hide]

I did the same thing as you guys, except I found the solution set to be, for integer $k$, $(x,y)=(1-11k,91+2k)$, where $a=2^x$ and $r=2^y$. Then I said $-45\le k\le -1$, and forgot the possibility $k=0$... :wallbash_red:

But this was a really really easy problem for a #9.

Even worse. I was rushing, for some reason, I don't know (I still had 2 hours left), and instead of writing $8^{2006}$, I dunno how, but I wrote $2006^8$ instead...  :wallbash:

I did essentially the same thing and once I noticed that the first set was $(x=1, y=91)$ and the second set was $(x=12, y=89)$, I established that all solutions had the form $(x=1+11n, y=91-2n)$ for all values of n from 0 to 45 inclusive.  Indeed an easy problem.

[quote="pieterminate"][quote="me@home"]When I added up the powers of $r$, I got $78$ instead of $66$ and proceeded to solve the problem with the wrong numbers... :oops_sign:  :wallbash_red:  :wallbash:[/quote]

Me too! $1+2+3+.....+11+12=......$  :wallbash_red:
Darn... :([/quote]
What [i]does[/i] $1+2+3+.....+11+12$ add up to, anyway? I thought it is $\frac{12*13}{2}=78$

Yeah, and I should have did $0+1+2+.....11=66$  :(

[hide="Solution"]
First, note that $\log_8 a_1 + \log_8 a_2 + \dots + \log_8 a_{12} = 2006$ implies that $\log_8 a_1a_2\dots a_{12} = 2006$ since $\log_b X + \log_b Y = \log_b XY$. Then we have \[a_1a_2\dots a_{12} = 8^{2006} = (2^3)^{2006} = 2^{6018}.\] Since $a_1 = a$ and the common ratio is $r$, the product of the terms is \[a * ar * ar^2 * \dots * ar^{11} = a^{12}r^{\dfrac{11 \times 12}{2}} = a^{12}r^{66}.\] Then $a^{12}r^{66} = 2^{6018} \implies a^2r^{11} = 2^{1003}$ by taking the sixth root of both sides. Now, note that $a$ and $r$ must both be powers of two since the LHS only has factors of $2$. Thus, we can write $a = 2^m, r = 2^n$. Then \[2^{2m}2^{11n} = 2^{1003} \implies 2^{2m+11n} = 2^{1003} \iff 2m +11n = 1003.\] Note that we only want solutions where $m, n \in \mathbb{Z}^+$. This only occurs when $n \equiv 1 \pmod{2}, n \le 91$. We can see then that $n \in \mathcal{S} = \{1, 3, \dots, 89, 91\}$. Then, our answer is $|\mathcal{S}| = \dfrac{91-1}{2} + 1 = \boxed{046}$.
[/hide]

This thread is like a mini time capsule. Thanks for the bump, happiface!

Wow, i'm surprised you actually found this, considering you were in the world of math competitions eight years ago. I guess even after our contest careers end, we all come back to take a glance at our old communities.

^
That contest career might not be over! All one does is move to the Putnam forum :)

Well,
\begin{align*}
\log_8 a_1 + \log_8 a_2 + \cdots + \log_8 a_{12} =& \log_8 a + \log_8 ar + \cdots + \log_8 ar^{11}\\
=& \log_8 (a \cdot ar \cdots ar^{11})\\
=& \log_8 (a^{12}r^{66}).
\end{align*}
This implies,
\[a^{12}r^{66} = 8^{2006}.\]
I haven't gone much farther than this. May I get a hint?

[hide=Hint]Rewrite as $2^{6018}.$ Thus, $a$ is a power of $2$, and so is $r$. Can you find a way to make a useful substitution from here?[/hide]

@below nice

[quote=A1001][hide=Hint]Rewrite as $2^{6018}.$ Thus, $a$ is a power of $2$, and so is $r$. Can you find a way to make a useful substitution from here?[/hide][/quote]

Thanks for the hint! So, we can substitute $a = 2^x$ and $r = 2^y,$ giving
\begin{align*}
12x + 66y = 6018\\
\implies 2x + 11y = 1003.\\
\end{align*}
$46$ valid pairs for $(x, y),$ and that's our answer.

[hide] This wasn't too bad..

So just substitute $a_1$ for $a$ and $a_2$ as $ar$ and so on until you get $a_{12}$ as $a{r^{11}}$. Then you can simply the expression by multiplying $(a_1)(a_2)(a_3)(a_4)...(a_{11})(a_{12})$ and in terms of $a$ and $r$ it's equal to $a^{12}r^{66}$.

So our expression is $\log_8 a^{12}r^{66} = 2006$ or $8^{2006} = a^{12}r^{66}$.

Taking the $2$nd root of each side and simplifying the $8$ to $2^3$ gives... $2^{3009} = a^{6}r^{33}$. We can make a Diophantine equation...
$6x + 33y = 3009$ which is equal to $2x + 11y = 1003.$

An (even number) $+$ (odd number) is equal to an odd number, and it follows that $y$ is a odd number.

So we have the sequence $1, 3, 5, \cdots 91$ which is $46$ numbers therefore we have $\boxed{046}$ pairs.
[/hide]

i remember this problem being my favorite on this test (mocked it an year ago)

Solved with [b]sarun[/b], [b]js104[/b], [b]crazyeyemoody[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLclesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b]

We have $a_{6.5}^{12}=8^{2006}$ so $a_{6.5}=2^{501.5}$, which implies $a^2r^{11}=2^{1003}$. Therefore, $a=2^x$ and $r=2^y$ where $2x+11y=1003$, so there are $\boxed{046}$ solutions.


Convert the repeated decimal  1.26 (with a bar over ONLY the six)

to a simplified fraction.

1. how do you do this? ( i think there is a way where you make the number $ 1 \frac{26}{99}$ but im not sure)

2. how do you make the repeating line over number in latex?

[hide="Solution"][code]\overline{x}[/code] comes out $ \overline{x}$.

$ 1.2\overline{6}\equal{}1.2\plus{}0.0\overline{6}$.

Let $ x\equal{}0.0\overline{6}$.
$ 10x\equal{}0.6\overline{6}$.

Thus, $ 9x\equal{}0.6 \implies x\equal{}\frac{1}{15}$.

$ \therefore 1.2\overline{6}\equal{}1.2\plus{}\frac{1}{15}\equal{}\frac{19}{15}$.

Answer: $ \frac{19}{15}$[/hide]

let x = 1.26 (bar over 6)
then 10x = 12.66 (bar over 6)

subtract x from 10x and we get

9x = 11.4
x = 19/15

EDIT: it appears fantasylover answered it before me.  :(


Hi All,

I have a puzzle let say i get a random 10 numbers, how can i find which 4 numbers will add up or multiply to the answer i have?
and in condition i can only use each numbers once.

Example:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Q1: which 4 numbers a+b+c+d=19
Q2: which 4 numbers a+(bXc)-d=28

I only know how to try n error to get the answer which is taking too long   :blush:  any help or advice is appreciated.

There is no algebraic way.  As a general rule, you need an equal amount of equations as you have variables.

[quote="percy314"]Yeah, they're right. For example; the answer to Q1 could be (1,2,6,10) or (2,3,4,10). There are multiple solutions.[/quote]

percy314,

your examples do show that algebra couldn't do it
because equations/inequalities can give more than
one solution.

Those equations have conditions, so I state it is algebra.

The first question is a type of partitioning of $ 19$, but for
allowing repeats of numbers because it stated that the numbers
are randomly picked (but not distinct).

In either case, you can have inequalities associated with the
conditions on the variables in the equations.  You would be
using algebra to streamline the process of getting the different
(multiple) solutions.  Your *algebraic* method would keep it from
being just trial-and-error/guess-and-check.

Unless I'm mistaken, the only solutions are (2,3,10,4) and (2,10,3,4).

hahahahaha correction done  :blush:

Well yes my try n error method only gives me this (2,3,10,4) & (2,10,3,4).

So if algebra cannot help me solve these questions "smartly",
may i know what other method can use other then my try n error  :P

[quote="grinchlygreen"]Depending on the number they add up to or mutliply to,
there could be 10 choose 4 answers, or 0, or anything in between. There's no
algebra to this, so there could be many ways to do this. But to your first
question, no. Btw, say "an" if the next word starts with a vowel, and "a" if
it doesn't.[/quote]

No.  You are *choosing* to do it and/or think about doing it a
non-algebraic way.  At mot is you can state that you personally
don't know of an algebraic way.
Think outside of your quote box.

[quote="RoFlLoLcOpT"]you have 4 variables and only 1 equation, so there is no algebraic way to solve.[/quote]
This question is written unclear.
Let me rewrite it

For postive integers $ a,b,c,d$ less than $ 11$, Find all that satisfies
$ a\plus{}b\plus{}c\plus{}d\equal{}19$
and
$ a\plus{}bc\minus{}d\equal{}28$

My mistake a few posts up- other cases such as (9,4,5,1) and (6,3,8,2) also work.

[hide="hint"]18<bc<28
b and c cannot both be even.[/hide]


IChO problem is most about identification of compound(hide-seek game).What's the strategies to  solve them efficiently. :maybe:

For example ,this problem ,please show the processes of thinking lead to the answer

  [b][i] An equimolar mixture X of three colorless organic liquids A, B, C was treated by water with a
drop of hydrochloric acid at heating to give, after separation from water, a 1:2 (molar ratio)
mixture of acetic acid and ethanol without any other components. To the mixture after hydrolysis
a catalytic amount (one-two drops) of concentrated sulfuric acid was added, and after long reflux
(boiling with reflux condenser) a compound D, a volatile liquid with pleasant smell, was formed
in 85% yield. Compound D is not identical to any ofA, B, C.
1.Draw the structure of compound D?
2. Which class of organic compounds does D belong to? Choose the proper variant from
those given in the Answer Sheet.
3. Even if the reflux is continued twice as long, the yield of D would not exceed 85%.
Calculate the expected yield of D if 1:1 (molar ratio) mixture of ethanol and acetic acid is taken.
Assume that: a) volumes do not change during the reactions; b) all concomitant factors, such as
solvent effects, non-additivity of volumes, variation of temperature, etc. are negligible. If you
cannot make a quantitative estimate, please indicate whether the yield will be: a) the same
(85%); b) higher than 85%; c) lower than 85%.

1H NMR spectra of compounds A, B,C look very similar and each shows singlet, triplet and
quartet with the ratio of integral intensities equal to 1:3:2.
The same mixture X was subjected to alkaline hydrolysis. A remained unchanged, and was
separated. The remaining solution after acidification and short boiling gave 2:3 (molar ratio)
mixture of acetic acid and ethanol with evolution of gas.
The mixture X (3.92 g) was dissolved in diethyl ether and underwent hydrogenation in the
presence of Pd on charcoal catalyst. 0.448 L (standard conditions) of hydrogen were absorbed,
but after the reactionA and C were isolated unchanged (3.22 g of mixture were recovered) while
neither B, nor any other organic compounds except diethyl ether could be identified after
hydrogenation.
4.Determine and draw the structures of A, B, andC.[/i][/b]

       (Reference: Theoretical problem 39th IChO,Moscow,Russian Federation,official version)

[hide]
1. Obviously the ester ethyl ethanoate
2. ester
[/hide]

Thank you, In 1),2),3) is O.K. .But in 4) ,to solve them,It's difficult to see whether is  corresponding to the conditions.

From the given information we can conclude that each of A, B, C has an isolated ethyl group (or more than one), together with a carbon (also isolated) with at least one hydrogen.

[u]3) Is expected yield 80.075%?[/u]

No,the expect yield is 67%

Yes, i rechecked my answer with a calculator and it is indeed 67%.

It is an acis catalysed esterification reaction, hence it is a simple equation you get when you equate the equilibrium constants. :)

Haha i never thought i'd see that problem again (Silver medal at IChO 07).  The last part was terrible, especially under the time pressure of that exam.  Practice past problems under time pressure if you want to improve.

Lawl.

My experience at IChO: spend 5 hours trying to figure out what problems you are able to do. Then, life gets slightly easier. I don't think practicing individual problems would help necessarily, although it gives you experience in solving problems. muahahahA

But I don't good at in practical part ,especially the ability to manage the time to do each of the part in Practical task.Therefore I'll do the Theoretical task best.Please give me for suggestions.


function
calculus
integration
floor function
limit
calculus computations

Can you guys post problems involving floor and ceiling function? Like integrals, series, etc.

Thanks.  :lol:

$ 1.\int^1_{\minus{}2} {\left\lfloor {x\left\lfloor{1\plus{}cos(\frac{x \pi}{2}})\right\rfloor  \plus{} 1}\right\rfloor }dx$
 [hide="Answer"]
[b]$ 2$[/b]
[/hide]
$ 2.$Evaluate::
 $ \lim_{n\to \infty} { \left\lfloor {\frac{1}{n^2} sec^2(\frac{\pi}{3n^2})\plus{}\frac{2}{n^2} sec^2(\frac{4\pi}{3n^2})\plus{}..........\plus{}\frac{1}{n}sec^2(\frac{\pi}{3})} \right\rfloor }$
[hide="Answer"]
[b]$ \frac{3\sqrt3}{2\pi}$[/b]
[/hide]
I could remember these two only.I will try to post some good question when i will reach home if u need more.
thank u


inequalities
limit
inequalities proposed

[quote="mohammad-mahmoodi"]Let $ a_1, a_2, ........, a_n$be positive numbers such that $ a_1 \plus{} a_2 \plus{} .......... \plus{} a_n\leq\frac {1}{2}$.
 Prove that
$ (1 \plus{} a_1)(1 \plus{} a_2)...........(1 \plus{} a_n) < 2.$[/quote]
Yes, $ \prod_{i\equal{}1}^n(1\plus{}a_i)<\sqrt{e}.$

[quote="Flakky"][quote="arqady"][quote="mohammad-mahmoodi"]Let $ a_1, a_2, ........, a_n$be positive numbers such that $ a_1 \plus{} a_2 \plus{} .......... \plus{} a_n\leq\frac {1}{2}$.
 Prove that
$ (1 \plus{} a_1)(1 \plus{} a_2)...........(1 \plus{} a_n) < 2.$[/quote]
Yes, $ \prod_{i \equal{} 1}^n(1 \plus{} a_i) < \sqrt {e}.$[/quote]
This follows from [b]Maclaurin's Inequality[/b] right?[/quote]
Yes, but AM-GM is easier.

[quote="mohammad-mahmoodi"]Let $ a_1, a_2, ........, a_n$be positive numbers such that $ a_1 \plus{} a_2 \plus{} .......... \plus{} a_n\leq\frac {1}{2}$.
 Prove that
$ (1 \plus{} a_1)(1 \plus{} a_2)...........(1 \plus{} a_n) < 2.$[/quote]
By AM-GM: $ (1 \plus{} a_1) ... (1 \plus{} a_n) \le \left (1 \plus{} \frac {1}{2n} \right)^n$. Thus it is left to prove that $ \left (1 \plus{} \frac {1}{2n} \right )^n < 2$.. I heard something about $ e^a \equal{} \lim_{n \to \infty} \left(1 \plus{} \frac {a}{n} \right) ^n$. If this is true, it follows directly from this, but could someone provide a proof?

[quote="Mathias_DK"][quote="mohammad-mahmoodi"]Let $ a_1, a_2, ........, a_n$be positive numbers such that $ a_1 \plus{} a_2 \plus{} .......... \plus{} a_n\leq\frac {1}{2}$.
 Prove that
$ (1 \plus{} a_1)(1 \plus{} a_2)...........(1 \plus{} a_n) < 2.$[/quote][/quote]
By AM-GM: $ (1 \plus{} a_1) ... (1 \plus{} a_n) \le \left (1 \plus{} \frac {1}{2n} \right)^n$.
Is it should be $ \equal{} < (1 \plus{} \frac {1}{2n})^n$??
Follow this, we need to prove: $ (2n)^n.2 > \equal{} (2n \plus{} 1)^n$, which is obviously true since $ 1/2n < 1$ with any $ n > \equal{} 1$.

yes, it should.
$ (1 \plus{} a_1) ... (1 \plus{} a_n) \le \left( \frac{n\plus{}\sum a_i}{n}\right)^n \le \left(1 \plus{} \frac {1}{2n} \right)^n$

[quote="BG Yoda"][quote="Mathias_DK"]I heard something about $ e^a \equal{} \lim_{n \to \infty} \left(1 \plus{} \frac {a}{n} \right) ^n$. If this is true, it follows directly from this, but could someone provide a proof?[/quote]

$ \Bigg(\lim_{n \to \infty} \left(1 \plus{} \frac {1}{\frac {n}{a}} \right) ^{\frac {n}{a}}\Bigg)^a \equal{} e^a$[/quote]
Thank you :)
I'm still confused, since there is so many definitions of $ e$: Apparently $ e \equal{} \lim_{n \to \infty}(1\plus{}\frac{1}{n})^n$ and $ e$ is also defined as the number for which $ f(x) \equal{} e^x \Rightarrow f'(x) \equal{} e^x$, and i know some more. It's a bit confusing :) Does anyone know how we can link them together?

$ f(x) \equal{} e^x \Rightarrow f'(x) \equal{} e^x$ is a consequence of the definition for $ e$.

$ \lim_{\Delta x\rightarrow 0}\frac {e^{x \plus{} \Delta x} \minus{} e^x}{\Delta x} \equal{} e^x\lim_{\Delta x\rightarrow 0}\frac {e^{\Delta x} \minus{} 1}{\Delta x}$ and since $ e \equal{} \lim_{x\rightarrow 0}(x \plus{} 1)^{1/x}\Rightarrow \lim_{x\rightarrow 0}\frac {e^x \minus{} 1}{x} \equal{} 1$, then $ (e^x)' \equal{} e^x$. :wink:

Another way:

1) Expand out everything
2) Group all monoms of same degree together
3) Use AM-GM once on every group
Q.E.D.


Using the Frenet-Serret formula, Prove:

$\tau = \frac{(r'{\times}r''){\bullet}r'''}{|r'{\times}r''|^{2}}$


We find IMO nothing compared to Hopmon!! This really challenging game is installed on every computer here in Slovenia. We've already managed to complete all the 45 levels.

What about you, guys??

 :rotfl:  :rotfl:  :rotfl:

Can this game be found online elsewhere in the world?

[quote="Dave Vather"]Do some google  :wink:[/quote]

Yes, google loves you....... http://www.google.com.ar/search?hl=es&q=Hopmon&btnG=B%C3%BAsqueda+en+Google&meta=

[quote="José"]Yes, google loves you....... http://www.google.com.ar/search?hl=es&q=Hopmon&btnG=B%C3%BAsqueda+en+Google&meta=[/quote]

Yeah, that's it!   :)

Download and enjoy this beautiful game!!  :D


geometry trapezoid search ratio quadratics LaTeX

1. A circle with radius 1 unit lies in the first quadrant and is tangent to both the x- and y-axes. A second larger circle lies in the first quadrant, is tangent to both axes and is externally tangent to the first circle. What is the radius of the second circle? Express your answer in simples radical form.

2. Equilateral triangle ABC has a side length of 6 units. Point D lies on segment BC such that DC=2(BD). What is the length of the altitude of triangle ADC from point C? Express your answer as a common fraction in simplest radical form.

3. In trapezoid ABCD, bases AB and CD are 13 and 39 units, respectively. Legs BC and DA are 24 and 10 units, respectively, and sides BC and DA lie on lines that are perpendicular to teach other. What is the area of ABCD?

I only have time for 1, so==>
[hide="#2"]==>We draw AD, and see that CD=(2/3)(6)=4.  Also, we see that [ADC]=(2/3)([ABC])=2/3(36sqrt3/4)=2/3(9sqrt3)=6sqrt3.  Now, to get the altitude from C we know that first, we have to get the length of AD.  So, we draw the altitude from A to CB to E, and quickly find that its length is given by:
4h/2=6sqrt3==>h=3sqrt3.  We also search for the base of this formed right triangle, DE, which we find by first determining 4-DE.  Letting this length be b, we get:
b^2+(3sqrt3)^2=36==>b^2+27=36==>b^2=9==>b=3.  Hence, DE=4-3=1, and AD=sqrt((3sqrt3)^2+1^2)=sqrt(28)=2sqrt7.  Hence, we get==>
(h(C))(2sqrt7)/2=6sqrt3==>h(C)=(6sqrt3)/sqrt7==>h(C)=[u](6sqrt21)/7[/u][/hide]

Hmm...your solution looks correct.

I thought #1 was the easiest of the 3.

[hide="#3"]==>Since BC and AD lie on lines that are perpendicular to each other, they must intersect, so AB and CD must be the parallel lines (this is trapezoid, keep in mind).  Extend CB and AD to where they meet, at E, a right angle.  We see that triangle AEB is similar to DEC by AA, with ratio 39/13=3.  Letting AE be s, we have that (10+s)/s=3; this yields s=5.  We know that BE has to be 12, from simple Pythagorean Triples, and a quick check reveals that it is.  Hence, [ABCD]=[CED]-[BEA]=(36)(15)/2-(12)(5)/2=(18)(15)-(6)(5)=270-30=[u]240[/u][/hide]

circle with radius 1 unit lies in the first quadrant and is tangent to both the x- and y-axes. A second larger circle lies in the first quadrant, is tangent to both axes and is externally tangent to the first circle. What is the radius of the second circle? Express your answer in simples radical form.
[hide="#1"]==>We're looking for a length, so we look to make a right triangle.  Let the center of the second circle be B, and the center of the first circle be A.  Drawing the perpendicular segments from A to both the x and y axes, we find that the distance from A to the origin (0,0) is sqrt2.  Now, we have enough information to make a right triangle.  The distance from B to A is equal to r (the radius of the big circle)+1, so the distance from B to the origin is r+1+sqrt2.  Since the distance from B to the x axis is r, as is the distance to the y axis, we have:
r^2+r^2=(r+1+sqrt2)^2.  Now, we could solve this very messy quadratic, but instead, we notice that we can determine the distance from B to the orgin in 2 ways, r(sqrt2), or r+1+sqrt2.  Setting these equal yields r((sqrt2)-1)=1+sqrt2.  Dviding and simplifying yields r=[u]3+2sqrt2[/u][/hide]

Both #1 and #3 are right.

I was doing #1 quickly (I do Warm-Ups in about 13 minutes) so I never noticed that "quick way."

Thanks for bringing that up.

Yeah, took me like 8 minutes with a graphing calculator.

 :ninja:

Whenever I see a long quadratic that I know will yield the answer, I very often look for another way, because long quadratics yield long and complex solutions.  Numerous times I've solved a problem very algebraically, and usually my answer looks like it might equal the real answer, but there's no way to tell how.  Complex methods usually lead to complex answers, usually seemingly, but not exactly, correct.

[hide] For #1, if the center of the small circle is A and the center of the large circle is B, then A to origin is sqrt2, since all circles are the same, and only size differs, I made a ratio: (sqrt2-1)/1=(sqrt2+1)/x, which simplifies to x=(sqrt2+1)/(sqrt2-1), which simplifies to x=3+2(sqrt2)[/hide]

[quote="psxter"][hide] For #1, if the center of the small circle is A and the center of the large circle is B, then A to origin is sqrt2, since all circles are the same, and only size differs, I made a ratio: (sqrt2-1)/1=(sqrt2+1)/x, which simplifies to x=(sqrt2+1)/(sqrt2-1), which simplifies to 3+2(sqrt2)[/hide][/quote]

Um, the exact same solution was already posted, but yeah, its right.

Use "\sqrt{blah}" in latex to create radicals.

Well I used what I thought to be a simpler method, and also z-coli must have simplified wrong.


geometry
3D geometry
algebra
polynomial
ratio
factorization
sum of cubes

Factor $ x^5\plus{}y^5$ so that it includes $ (x\plus{}y)$ as one of the two factors. I need the exact steps, since I already tried using the sum of cubes with $ x^3x^2\plus{}y^3y^2$, but that didn't work.

Here are two good factorizations to remember:
\[ x^n\minus{}y^n\equal{}(x\minus{}y)(x^{n\minus{}1}\plus{}x^{n\minus{}2}y\plus{}x^{n\minus{}3}y^2\plus{}\cdots\plus{}x^2y^{n\minus{}3}\plus{}xy^{n\minus{}2}\plus{}y^{n\minus{}1})\] \[ x^{2n\plus{}1}\plus{}y^{2n\plus{}1}\equal{}(x\plus{}y)(x^{2n}\minus{}x^{2n\minus{}1}y\plus{}x^{2n\minus{}2}y^2\plus{}\cdots\plus{}x^2y^{2n\minus{}2}\minus{}xy^{2n\minus{}1}\plus{}y^{2n}),\] where the second term in the second product alternates in signs. See if you can convince yourself why these two identities are true. You might want to multiply a couple of examples and look at the way the extraneous terms cancel.

More specifically, \[ x^5\plus{}y^5\equal{}(x\plus{}y)(x^4\minus{}x^3y\plus{}x^2y^2\minus{}xy^3\plus{}y^4).\]

Sum of cubes doesn't work because this isn't a sum of cubes.  What does work, however, is polynomial division (treating $ y$ as a constant), which gets you the answer immediately.  Note that the second factorization towersfreak2006 gives is a special case of the first (where $ \minus{}y$ is used instead of $ y$) and that the first is just the sum of a geometric series with first term $ x^{n\minus{}1}$ and common ratio $ \frac{y}{x}$.


geometry
geometric transformation
rotation
geometry unsolved

In acute triangle $ ABC$, let $ X,X'$ on side $ BC$, $ Y,Y'$ on side $ CA$, $ Z,Z'$ on side $ AB$ be such that triangles $ XYZ$ and $ X'Y'Z'$ are congruent (with $ X,X'$ being the corresponding vertices, etc.) Prove that $ X\equal{}X',Y\equal{}Y',Z\equal{}Z'$

I think there is something wrong with your problem.

I will give an anti-example:

Let $ \triangle XYZ$ be an arbitrary triangle and $ O$ be a point differs from $ X,Y,Z$. Now consider the rotation with center $ O$ and arbitrary angle $ \alpha \in \left(0;\frac{\pi}{2}\right)$ turn: $ X\mapsto X',Y\mapsto Y',Z\mapsto Z'$. Let $ \{A\}\equal{}YY'\cap ZZ', \{B\}\equal{}ZZ'\cap XX', \{C\}\equal{}ZZ'\cap XX'$ then the two triangle $ \triangle XYZ$ and $ \triangle X'Y'Z'$ inscribed in $ \triangle ABC$ and are congruent to but differs from each other.


limit
induction
calculus
integration
logarithms
real analysis
real analysis unsolved

The sequence $(a_{n})_{n\geq 1}$ satisfies $a_{n} >0$ and $a_{n-1}a_{n}=n$ for all $n\geq 2.$
 Prove that $\lim_{n\to\infty} a_{n}=\infty.$ ;)

We can considerate $f: N\to R$ and $g: N\to R$ such that $0<f(1)\leq a_1$, $g(1)\geq a_1$, $g(n)=\frac{n}{f(n-1)}$ and $f(n)=\frac{n}{g(n-1)}$, for all natural n such that $n\geq 2$. Now we can prove by induction that $f(n)\leq a_n\leq g(n)$. And now it´s just to prove that $\lim_{n\to +\infty}f(n)=+\infty$ and to see this it´s just note that $f(n+1)=\frac{n+1}{n}f(n-1)$. And thus we have $f(2n+1)=e^{\sum_{k=1}^{n}ln(\frac{1+2k}{2k})}f(1)$. But, by integral test, we can see that $\lim_{n\to +\infty}\sum_{k=1}^{n}ln(\frac{1+2k}{2k}) = +\infty$, and so it´s proved.

[quote="cezar lupu"]The sequence $(a_{n})_{n\geq 1}$ satisfies $a_{n} >0$ and $a_{n-1}a_{n}=n$ for all $n\geq 2.$
 Prove that $\lim_{n\to\infty} a_{n}=\infty.$[/quote]

Let $(d_n)_{n \ge 1}$ be a sequence such that $d_n - d_{n-1} \ge n^{-1} > 0$ for all $n \ge N$. Suppose the sequence $c_n$ is defined by $c_0 = 0$, $c_n +c_{n-1} = d_n$ for $n \ge 1$. Then $c_n \to \infty$.

The assertion follows by taking $d_n = \log n, \, c_n = \log a_n$.

To show this, first note that $c_n = \sum_{i=1}^n (-1)^{n-i} d_i = d_n - d_{n-1} + d_{n-2} + - ...$, by induction. Let $n = 2k$ be even, then this can be written as $c_n = \sum_{j=1}^k (d_{2j} - d_{2j-1})$. If $n = 2k+1$ is odd, then $c_{2k+1} = d_1 + \sum_{j=1}^k (d_{2j+1} - d_{2j})$. Clearly, if $n > N$, then $c_{n+1} \ge c_{n-1} + n^{-1}$, and the statement follows.


Are there any significant differences/changes between an earlier edition of a textbook and a later one? Are there different problems, or are there only minor edits with the majority of the problems staying the same?

I'm asking this because I am interested in purchasing a 4th edition chemistry textbook (specifically), which was published 1988. Online searches have told me that no available copy exists, but there are a few later editions that could be bought. I am just wondering if it would be more or less the same as the 4th edition (same problems, same format, etc.) or would it be completely different.

Thanks!

Normally, they just add content and correct errors. There are a couple of exceptional cases where content was lost, but that's rare. If it's relevant for you, maybe tho most dangerous part would be that the enumerations may be different (so any pointer to something like "page 37" or "chapter 2.1" may be wrong).

[quote="TZF"]Changes can range from fixing typos to including new content to switching around/splitting/merging sections, and from adding a few extra problems to changing several of the problems. You should see if you can get a hold of the two editions at a library and compare them, or ask about the differences between the two books online -- I'm sure there will be plenty of people who have seen several editions of that book and can comment.[/quote]
Normally, a test book will tell the readers what is new in this edition at the preface. They basically list all the new stuff they added or changes they made. They have to do that so the readers get to know how different and good the new book is compared to the old editions. Otherwise no one will be interested in the new book.

Whether or not it matters depends on the purpose behind purchasing the textbook.  Different editions of a standard textbook *can* be significantly different, but typically the differences are microscopic, due to the following theorem:

New edition of book = More money for publisher.