Hello AoPS users. This post is the post for storing records on AoPS for example there are some records below. Anybody that posts after me can either add a record, or change one that has been broken. It starts now. Here are the first two records;

Highest number of posts on AoPS that weren't in Games and Fun Factory;

t0rajir0u - 8873
JBL - 8813
Kent Merryfield - 8602
grobber - 7847
kunny - 7274

Highest All Time Highest Ratings on FTW;

mewto55555 - 1907
AIME15 - 1878
superllamel - 1860
kl2836 - 1776
isabella2296 - 1775

And here are some ideas of more records that you guys can start;

Longest Post
Highest Point Scores on the Point Game
Highest Scores on the 'Guess who will post next' game.
Oldest AoPS users
Newest AoPS users

Only do the top 5 records, unless there is a tie for fifth.

fastest mem(no glitch) starting September 21,2008
1. program4 1.038
2.
3.
4.
5.

lol
most number of 6's in username
1.6466466464
2.68656c6c6f00
3.1361661460
4.3236666
5.fddf6666

hmm... surprisingly they have a total of...
[hide]
0
[/hide]
posts :D

fastest mem(no glitch)
1. ernie 0.734 :P
2. program4 1.038
3.
4.
5.

fastest mem(no glitch)
1. AIME15 0.249 :D
2. ernie 0.734 :P
3. program4 1.038
4.
5.

Lol i was randomly guessing "e" and i got one right :D

some1 post the records 4 maximum posts per day--

1.madness
.
.
.rt_08

(3 mnths back)

1.madness
2.rituraj007
3.shruthibaskaran

wait you could only post four posts per day?

Okay here are the records so far, please keep them all on each post. You can change a record too, if it has changed;

Highest number of posts on AoPS that weren't in Games and Fun Factory;

t0rajir0u - 8873
JBL - 8813
Kent Merryfield - 8602
grobber - 7847
kunny - 7274

Highest All Time Highest Ratings on FTW;

mewto55555 - 1907
AIME15 - 1878
superllamel - 1860
kl2836 - 1776
isabella2296 - 1775

Most number of 6's in username;
1.6466466464
2.68656c6c6f00
3.1361661460
4.3236666
5.fddf6666

Most visited AoPSWiki pages:

1. Main Page (141371)
2. Asymptote (Vector Graphics Language) (27235)
3. LaTeX (19092)
4. Mathematics (13174)
5. AoPSWiki:Words of the Week (9291)

fastest mem(no glitch)
1. AIME15 0.249
2. ernie 0.734
3. program4 1.038
4.
5.

Largest Number of Topics Made by a User
1.
2.
3.
4.
5.

Most on FTW in one day. being afk counts. I guess

1.jjx1: 11 hours about

[quote="program4"]AIME15's 0.2 sec  is not a mem, but a guess[/quote]

If you count only mems...

Okay here are the records so far, please keep them all on each post. You can change a record too, if it has changed;

Highest number of posts on AoPS that weren't in Games and Fun Factory;

t0rajir0u - 8873
JBL - 8813
Kent Merryfield - 8602
grobber - 7847
kunny - 7274

Highest All Time Highest Ratings on FTW;

mewto55555 - 1907
AIME15 - 1878
superllamel - 1860
kl2836 - 1776
isabella2296 - 1775

Most number of 6's in username;
1.6466466464
2.68656c6c6f00
3.1361661460
4.3236666
5.fddf6666

Most visited AoPSWiki pages:

1. Main Page (141371)
2. Asymptote (Vector Graphics Language) (27235)
3. LaTeX (19092)
4. Mathematics (13174)
5. AoPSWiki:Words of the Week (9291)

fastest mem(no glitch)
1. AIME15 0.628
2. ernie 0.734
3. program4 1.038
4.
5.

Highest number of posts on AoPS that weren't in Games and Fun Factory;

t0rajir0u - 9023
JBL - 8880
Kent Merryfield - 8691
grobber - 7852
kunny - 7315

Highest All Time Highest Ratings on FTW;

mewto55555 - 1907
AIME15 - 1878
superllamel - 1860
kl2836 - 1776
isabella2296 - 1775

Most number of 6's in username;
1.6466466464
2.68656c6c6f00
3.1361661460
4.3236666
5.fddf6666

Most visited AoPSWiki pages:

1. Main Page (141371)
2. Asymptote (Vector Graphics Language) (27235)
3. LaTeX (19092)
4. Mathematics (13174)
5. AoPSWiki:Words of the Week (9291)

fastest mem(no glitch)
1. AIME15 0.628
2. ernie 0.734
3. program4 1.038
4.
5.

the number of posts changed

Highest average posts per day:
1. myyellowducky82--11.88 (as of now)

Highest number of posts on AoPS that weren't in Games and Fun Factory;

t0rajir0u - 9023
JBL - 8880
Kent Merryfield - 8691
grobber - 7852
kunny - 7315

Highest All Time Highest Ratings on FTW;

mewto55555 - 1907
AIME15 - 1878
superllamel - 1860
kl2836 - 1776
isabella2296 - 1775

Most number of 6's in username;
1.6466466464
2.68656c6c6f00
3.1361661460
4.3236666
5.fddf6666

Most visited AoPSWiki pages:

1. Main Page (141371)
2. Asymptote (Vector Graphics Language) (27235)
3. LaTeX (19092)
4. Mathematics (13174)
5. AoPSWiki:Words of the Week (9291)

fastest mem(no glitch)
1. AIME15 0.628
2. ernie 0.734
3. program4 1.038
4.
5.

most number of posts per day
1. isabella2296 - 13.21
2. myyellowducky82 - 11.79
3. Wickedestjr - 3.27
4.
5.

[hide="Highest number of posts on AoPS that werent in Games and Fun Factory"]t0rajir0u - 9023
JBL - 8880
Kent Merryfield - 8691
grobber - 7852
kunny - 7315[/hide][hide="Highest All Time Highest Ratings on FTW"]mewto55555 - 1907
AIME15 - 1878
superllamel - 1860
kl2836 - 1776
isabella2296 - 1775[/hide][hide="Most number of 6s in username"]1.6466466464
2.68656c6c6f00
3.1361661460
4.3236666
5.fddf6666 [/hide][hide="Most visited AoPSWiki pages"]1. Main Page (141371)
2. Asymptote (Vector Graphics Language) (27235)
3. LaTeX (19092)
4. Mathematics (13174)
5. AoPSWiki:Words of the Week (9291) [/hide][hide="fastest mem(no glitch)"]1. AIME15 0.628
2. ernie 0.734
3. program4 1.038
4.
5. [/hide][hide="most number of posts per day"]1. isabella2296 - 13.21
2. myyellowducky82 - 11.79
3. 1=2 - 4.14
4. Wickedestjr - 3.27
5.[/hide]

[hide="Highest number of posts on AoPS that werent in Games and Fun Factory"]t0rajir0u - 9032
JBL - 8883
Kent Merryfield - 8694
grobber - 7852
kunny - 7315[/hide][hide="Highest All Time Highest Ratings on FTW"]mewto55555 - 1907
AIME15 - 1878
superllamel - 1860
kl2836 - 1776
isabella2296 - 1775[/hide][hide="Most number of 6s in username"]1.6466466464
2.68656c6c6f00
3.1361661460
4.3236666
5.fddf6666 [/hide][hide="Most visited AoPSWiki pages"]1. Main Page (141371)
2. Asymptote (Vector Graphics Language) (27235)
3. LaTeX (19092)
4. Mathematics (13174)
5. AoPSWiki:Words of the Week (9291) [/hide][hide="fastest mem(no glitch)"]1. AIME15 0.628
2. ernie 0.734
3. program4 1.038
4.
5. [/hide][hide="most number of posts per day"]1. isabella2296 - 13.21
2. myyellowducky82 - 11.79
3. t0rajir0u - 8.55
4. AIME15 - 5.28
5. JBL - 4.61
[/hide]

fastest mem
1. program4 0.614
2. AIME15 0.628
3. ernie 0.734
4. program4 1.038
5. program4 1.156

ahem

fastest mem should be killed bcuz of the 1.4 sec rule

thus, ur mem is fake

What does it even mean when it says fastest mem?


vector
algebra
polynomial
integration
function
abstract algebra
group theory

If $(A,+,\cdot)$ is a ring and for $a,b\in A$ such that $ab=1$ is it true that $ba=1$ ? [if the answer is no, please give an example]

$\mathcal{M}_{2}(\mathbb{R})$

The ring of $2 \times 2$ matrices with real entries.

That's not a counterexample.
The easiest counterexamples are infinite dimensional- look at shift operators in rings of linear transformations on infinite-dimensional vector spaces.
For example, let $A$ be the space of all real polynomials, and let $R$ be the ring of linear transformations from $A$ to $A$. Set $a(p)(x)=p'(x)$ and $b(p)(x)=\int_{0}^{x}p(t)\,dt$.

Right. I must have been drunk when I read the question. I totally misinterpreted it... :)

Hopefully this one will work:

Consider $A^{A}$: all functions from an abelian group $A$ into $A$, with pointwise addition and function composition. This is a ring with $1 = Id_{A}$. If $A$ is infinite, then there exists some function $f: A \to A$ that is surjective but not injective (a basic fact about sets). But $f$ is injective iff it has a left-inverse function $g$, meaning $gf = Id_{A}$, while $f$ is surjective iff it has a right-inverse function $h$ meaning $fh = Id_{A}$. So for this particular $f$ of ours (surjective but not injective), there exists an $h$ with $fh = 1 \neq hf$, because $f$ has no left inverse at all (so $h$ isn't one of them).


function
LaTeX
quadratics
algebra
polynomial
geometry
similar triangles

I hope you saw the warning before looking my question. I have four questions but I'm not quite sure whether they're for intermediate or advanced (definitely not getting started).  Anyway, big help here! :lol:

1. Prove that a function of the following form is odd.
y = a_(2n+1) x^2n+1 + a_(2n-1) x^(2n-1) + ..... a_3 x^3 + a_1 x
2. Prove that a function of the following form is even.
y = a_2n x^2n + a_(2n-2) x^(2n-2) + .... a_2 x^2 + a_0

Much nicer if I can do it with Latex but too lazy :P.

What's this mean?? (see the next question)
3. Use a graphing utility to graph the function f (x) = x^4 - 4x^2 + k for different values of k.  Find values of k such that the zeros of f satisfy the specified characteristics. (Some parts do not have the unique answers.)
a) Four real zeros
b) Two real zeros each of multiplicity 2
c) Two real zeros and two complex roots
d) Four complex zeros

Um.. Next one might be stupid question but I don't seem to get an approach toward the answer.

4. To obtain the height of a flagpole you measure the flagpole's shadow and find that it is 10 feet long.  You also measure the shadow of a yardstick and find that it is 15 inches long.  How tall is the flagpole?

Thanks!!! :D

1.  You can use induction.
Base case: n=1
$f(x)=a_1x$
$f(-x)=-a_1s$   Check

Assume it is true for n=k
$a_{2k+1}x^{2k+1}+a_{2k-1}x^{2k-1}+\ldots+a_3x^3+a_1x$

Prove for n=k+1
\[f(x)=a_{2k+3}x^{2k+3}+a_{2k+1}x^{2k+1}+\ldots+a_3x^3+a_1x\]
If you plug in -x for the equation you get
\[a_{2k+3}(-x)^{2k+3}-({a_{2k+1}x^{2k+1}+\ldots+a_3x^3+a_1x)}\]

We used what we assumed in the second step to make that substitution.

\[a_{2k+3}(-x)^{2k+3}=-a_{2k+3}x^{2k+3}\]
We know this because 2k+3 is an odd power.

Pluggin back in and factoring a -1 we get:
\[-(a_{2k+3}x^{2k+3}+a_{2k+1}x^{2k+1}+\ldots+a_3x^3+a_1x)\]
which means that the function is odd.

4. Similar triangles yields: x/10=3/1.25
x=24 ft.

None of this is advanced.  If you had defined even and odd functions, I think most of it would have been beginner.

#3
$f(x)=x^{4}-4x^{2}+k$ is a quadratic in $x^{2}$, so $x^{2}=\displaystyle\frac{4\pm\sqrt{16-4k}}{2}=2\pm\sqrt{4-k}$.
For a)
If $x$ is real, $x^{2}$ is real and non-negative. In order for that to be true, $\sqrt{4-k}$ must be real and $2-\sqrt{4-k}$ must be non-negative, so:
$\sqrt{4-k}\in\Re\Rightarrow 4-k\geq0\Rightarrow k\leq 4$
and
$2-\sqrt{4-k}\geq0\Rightarrow4\geq4-k\Rightarrow k\geq 0$
So, for a), $0\leq k\leq4$.
For b)
The roots are $\pm\sqrt{2\pm\sqrt{4-k}}$, so their will only be two distinct real roots if $2\pm\sqrt{4-k}$ is a distinct real number. $2\pm\sqrt{4-k}$ is one distinct number $iff$ $\sqrt{4-k}=0\Rightarrow k = 4$.
For c)
The two roots $\pm\sqrt{2+\sqrt{4-k}}$ are real if $2+\sqrt{4-k}$ is non-negative and real and the other two roots $\pm\sqrt{2-\sqrt{4-k}}$ are complex if $2-\sqrt{4-k}$ is negative and real.
So:
$4-k\geq0\Rightarrow k\leq4$
$+\sqrt{4-k}$ is always positive $\Rightarrow 2+\sqrt{4-k}$ is always positib
and
$2-\sqrt{4-k}<0\Rightarrow 4<4-k$ and $k<0$
So, $k<0$
For d)
All roots are always complex: $\sqrt{2}+0i$ is complex. But I think he meant complex but not real, so, the roots $\pm\sqrt{2\pm\sqrt{4-k}}$ are complex if $2\pm\sqrt{4-k}$ is complex or negative.
They are complex if $4-k<0\Rightarrow k>4$. Since $+\sqrt{4-k}$ is always positive or complex, $2+\sqrt{4-k}$ can never be negative and thus, all the roots can be complex only when $2+\sqrt{4-k}$ is complex, so: $k>4$.

[quote="Silverfalcon"]1. Prove that a function of the following form is odd.
y = a_(2n+1) x^2n+1 + a_(2n-1) x^(2n-1) + ..... a_3 x^3 + a_1 x
2. Prove that a function of the following form is even.
y = a_2n x^2n + a_(2n-2) x^(2n-2) + .... a_2 x^2 + a_0

...

Um.. Next one might be stupid question but I don't seem to get an approach toward the answer.

4. To obtain the height of a flagpole you measure the flagpole's shadow and find that it is 10 feet long.  You also measure the shadow of a yardstick and find that it is 15 inches long.  How tall is the flagpole?

Thanks!!! :D[/quote]

1.  First remember that (-x)^(2n+1) = -(x^(2n+1)).  Plugging in -x into the polynomial, we can factor out a -1 since all the exponents are odd.  So f(-x) = -f(x) and the function is odd.
2.  (-x)^(2n) = x^(2n)  All the exponents are even, so f(-x) = f(x).

4.
(length of yardstick shadow)/(length of yardstick) = (length of flagpole shadow)/(length of flagpole)

5/12=10/x
120 = 5x
x = 24


abstract algebra
limit
integration
calculus
real analysis
real analysis unsolved

let $\delta>0$ then this holds for the kernel (one of the three basic assumptions)
\[\lim_{n\to \infty}\int_{\delta<|s|<a}K_{n}(s)ds=0 \]
Where $K_{n}$ is an summation kernel defined on an interval (-a,a).

[color=red]Given that show that[/color] the kernel $K_{n}= n, \, \mbox{if}\,|s|<1/2n\quad \mbox{or}\,0, \, \mbox{if}\,|s|>1/2n$
defined on R to R, satisfy the condition above.

My try was ty split the interval up so that we have
\[\int_{\delta<|s|<a}\ldots =\int_{\delta<|s|<1/2n}+\int_{|s|>1/2n}\]
Here i assumed a is infinite since its all of R. but I could not show that the first term on the right is zero, while the other is obvious. [color=green]Please help[/color]

[quote="mlok"]For $n>2/\delta$ the kernel vanishes on the region of integration.[/quote]

Why? How? Please show.

[quote="mlok"]Actually, $n>\frac{1}{2\delta}$ suffices. Indeed, $K_{n}(x)=0$ when $|x|>1/(2n)$. The integral only involves $x$ with $|x|>\delta$. If $1/(2n)<\delta$, $K_{n}$ is zero everywhere on the region of integration.[/quote]

So does your diskussion apply to my integral-split above.
You men that we choose delta so that the first itegral on the  right is goes to zero?


Let x,y,z positive integer such that

gcd(x,y)= 1 and x^2 +y^2 = z^4

Prove that 7 divide x.y

I think it ca be proved by using Pythagoras: Z^2=k, then x^2+y^2=k^2, and because (x,y)=1 we must hhave x=m^2-n^2, y=2mn, z^2=k=m^2+n^2. We apply this again for the equation z^2=m^2+n^2, and we get m=t(a^2-b^2), n=t*2ab, z=t*(a^2+b^2), we replace m and n in the expressions of x and y and we have to get what we want. I haven't actually checked it, but I think it works.

Notice that t=1, because (x,y)=1, so m=a^2-b^2, n=2ab (it could of course be the other way around, m=2ab and n=a^2-b^2, but it doesn't matter).

hihi moubi it remained me one of the problem i posted before its great for me you use it !


Can somebony to show idea of solving?
$ x^4\plus{}4x\equal{}1$ :roll:

Sketching $ y\equal{}x^{4} \plus{} 4x$ and $ y\equal{}1$, we see that there are two distinct real solutions at about $ \minus{}1.6$ and $ 0.25$ and two complex conjugate roots.

$ x^4 \plus{} 4x \equal{} 1$

factoring we get:

$ (x^2 \plus{} 1)^2 \minus{} 2x^2 \minus{} 1 \plus{} 4x \equal{} 1$
$ (x^2 \plus{} 1)^2 \minus{} 2(x^2 \plus{}1 \minus{} 2x) \equal{} 0$

$ (x^2 \plus{} 1)^2 \minus{} 2(x \minus{} 1)^2 \equal{} (x^2 \plus{} 1 \minus{} \sqrt{2}(x\minus{}1))(x^2 \plus{} 1 \plus{} \sqrt{2}(x\minus{}1)) \equal{} 0$

The four roots can now be found by the quadratic formula.

hello, i have done the following
$ x^4\equal{}1\minus{}4x$
$ x^4\plus{}2x^2\equal{}1\plus{}2x^2\minus{}4x$
$ x^4\plus{}2x^2\plus{}1\equal{}2x^2\minus{}4x\plus{}2$
$ (x^2\plus{}1)^2\equal{}2(x^2\minus{}2x\plus{}1)$
$ (x^2\plus{}1)^2\equal{}2(x\minus{}1)^2$
$ x^2\plus{}1\equal{}\sqrt{2}|x\minus{}1|$
This equation can be solved.
Sonnhard.


inequalities
geometry
inequalities proposed

In any triangle with sides a, b, c and S area of the triangle prove that

$ S \le  \frac{3}{4} \left (\frac{a^k\plus{}b^k\plus{}c^k}{3}  \right  )^{\frac{2}{k}}$, $ k>0$.

In any triangle with sides a, b, c and S area of the triangle prove that

$ S \le  \frac{\sqrt3}{4} \left (\frac{a^k\plus{}b^k\plus{}c^k}{3}  \right  )^{\frac{2}{k}}$, $ k>0$.

Remark:
1)If k=4 obtain $ 4S \le  \sqrt3 \left (a^4\plus{}b^4\plus{}c^4  \right  )^{\frac{1}{2}}$ inequality F.Goldner.
2)If k=1 obtain $ S \le \frac{\sqrt3}{4} \frac {\left ({a\plus{}b\plus{}c} \right )^2}{9} \equal{}\frac{p^2}{3 \cdot \sqrt 3}$


$a^{b}+a^{c}+1=p^{k}$  $(b=2n+1, c=n+1)$ $a,n,p,k\in Z^{+}$ and $k>1$
  find all solutions to the equation...

[quote="sinankaral53"](a^b)+(a^c)+1=p^k  (b=2n+1, c=n+1) a,n,p,k are positive integers...and k>1
  find all solutions to the equation...[/quote]

is p a prime?

in fact the question is 2^(2n+1) +2^n+1=p^k where p doesn't have to be a prime.it's not a difficult problem(the answer is n=4) but i am really curious about the general solution.

I'd be glad if you post a solution.I solved it for m - odd but when m is even the only solution which i was able to find is n=0 but this also is not positive! How you managed to find the solution n=4?  :blush:

I think there's something missing in the problem.
But if P is prime, There exist no solutions.
Can someone guarantee the authenticity of the problem?
Thanks!


o it said square of the square of the square. that's $2^{3}$

[quote="7h3.D3m0n.117"]no that would mean

$(3^{6})$....

keep trying  :lol:[/quote]
no, the square of square of square is
$((3^{2})^{2})^{2}$, which is $3^{2^{3}}$, or $3^{8}$
so ilikepie's answer shuld be the most reasonable one
keep trying  :lol:

[quote="buzzer11"][quote="i_like_pie"][quote="7h3.D3m0n.117"]The square of the square root of the square of a positive integer is equal to the square of the square of the square of 3. What is the integer?[/quote]
[hide]$\left(\sqrt{x^{2}}\right)^{2}=3^{8}$

$x^{2}=81^{2}$

$x=\boxed{81}$[/hide][/quote]

The square of the square of 3 would be $3^{4}$.  :wink:[/quote]
[quote="7h3.D3m0n.117"]The square of the square root of the square of a positive integer is equal to [b]the square of the square of the square[/b] of 3. What is the integer?[/quote]

[hide][quote="7h3.D3m0n.117"][b]The square of the square root[/b] of the square of a positive integer is equal to [b]the square of the square of the square[/b] of 3. What is the integer?[/quote]

The square of the square root cancels out. So:
[b]The square[/b] of a positive integer is equal to [b]the square of the square of the square[/b] of 3.

Canceling out the "squares",
The positive integer is equal to [b]the square of the square[/b] of 3.

So the positive integer equals ${3^{2}}^{2}=81$.

This is a good strategy on word problems.[/hide]

[quote="davidlizeng"][quote="jhredsox"][hide]
first is basically $x^{2}$ and the second is basically $3^{4}$

so.... $x=9$[/hide]
jorian[/quote]

no... the first is not $x^{2}$, its $x$!!!
the sqrt and sqr cancel to leave with $x^{2}$, but you need to take the sqrt of $3^{2^{2^{2}}}$ to get $3^{4}$ and the sqrt of $x^{2}$ to get $x$!!!!!![/quote]

a little aggressive there david???

jorian

guys, $3^{2^{2^{2}}}$ is not the same as $((3^{2})^{2})^{2}$
the first one is $3^{2^{4}}$, which is $3^{16}$ (innermost exponents have precedence)
the second one is $3^{2*2*2}$, which is $3^{8}$
the square of the square of the square of 3 is $((3^{2})^{2})^{2}$, not $3^{2^{2^{2}}}$.
the second case is three to the two to the two squared th th power
[quote="mathgeniuse^ln(x)"]How di dyou know his name was David?[/quote]
he simply called him by the first 5 letters. he didn't know that it was david. just like if someone's name is protestant id call him by the first 3 letters (pro) for short, or someone else's name is catdog id call him by the first 3 letters (cat) for short.

no, he just told me in an email (sry davidlizeng) - if u look closely enough u can find out his whole name :P
jorian

[quote="7h3.D3m0n.117"]The square of the square root of the square of a positive integer is equal to the square of the square of the square of 3. What is the integer?[/quote]

[hide="Solution"]$(\sqrt{x^{2}})^{2}=3^{2^{2^{2}}}$
So, by expansion
$x^{2}=81^{2}$
$x=9^{2}$[/hide]

[quote="myspacesucks"]Listen, square and square root of x cancel out so it is x squared.  The square of the square of the square of 3 is 81 squared.  Therefore x squared is equal to 81 squared and thus you get the answer.  The answer is 81 not 9!!!!!!!!!!!![/quote]
You know you don't have to be mean about it. He probably didn't know how to do it. Thats why you come here. You learn stuff by posting problems that you don't know.


A non-negative integer $f(n)$ is assigned to each positive integer $n$ in such a way that the following conditions are satisfied:
(a) $f(mn) = f(m)+f(n)$ for all positive integers $m,n$
(b) $f(n) = 0$ whenever $n$ ends in a 3 (in base 10)
(c) $f(10) = 0$
Prove that $f(n) = 0$ for all positive integers $n$.

It is easy to see that if $f(mn)$ is 0, then so are $f(m)$ and $f(n)$, because ie. $f(m) \geq 0$. (*)

Now $f(3) = 0$ by (b), and $f(2) = f(5) = 0$ by (*) and (c).

It is enough to prove $f(p) = 0$ for all primes $p > 5$.  We show that there exists some numbers $y$ [where y is 3 mod 10] for which $p | y$.  In which case, $0 = f(y) = f(pk) = f(p) + f(k)$ implying $f(p) = 0$.

It is enough to show that $10k + 3 = 0$ (mod p) for some k, and any prime $p > 5$.  Clearly, $10$ does not divide $p$, in which case $10k$ cycles through the residues $1,2,\dots,p$.  Thus it meets the residue $-3$.  QED.


Would anyone be interested in creating solution manuals for old USC Math Contest problems?  If we could get enough interest, we could create them and send them to the USC people...

Well, I recently did 1988, and I know how to do all the problems on that contest (except #28 :mad: ) so I'll put up all those solutions AFTER friday (internship). I'll do 1986 by Saturday too so I'll put those solutions up as well. Also, I don't think there's any point posting all those USC questions here as mysmartmouth has been doing. It makes this forum look really messy, Just make threads of EACH of the years and post solutions there. It'd be good if we split the number of tests up between the three of us, and we'll put the solutions to all the problems for that test.

So like, I can do all of them 1986-1989, and then you two split the other 6 tests among yourselves. If there are questions you don't know how to do, just don't post a solution and someone else can or just post which solutions are needed.

Oh, I was posting those problems because I needed help with them.  :blush: But the threads sound good.

Does anyone actually need me to do 1-15 on any of those USC contests? I mean, those problems are pretty easy and I don't want to waste my time posting solutions if no one needs them. :wink:

I will try to post 1-15 for those contests.  I can usually do #s 1-15, but a lot of them don't have nice solutions.  For example, casework, synthetic division, etc.


AMC
AIME
number theory
least common multiple
greatest common divisor
prime factorization

Three planets revolve about a star in coplanar circular orbits with the star at the center. All planets revolve in the same direction, each at a constant speed, and the periods of their orbits are 60, 84, 140 years each. The positions of the star and all three planets are currently collinear. They will next be collinear after n years. Find n.

[quote="life=tennis+math"]Well the AMC topic is unlocked now...
[hide]
Isn't this just an LCM problem, or am I missing something?
Well we can rewrite all 3 numbers in prime factorization
$60=(2^{2})(3)(5) 84=(2^{2})(3)(7) 140=(2^{2})(5)(7)$
The LCM would therefore be $(2^{2})(3)(5)(7)=(105)(4)=420$
After 420 years they would all be in the same place where they started with. However, since the GCF is $4$ you should take the LCM and divide by $4$. Thus after $\frac{420}{4}=105$ years, while they will not be in the same place, they will still be collinear[/hide][/quote]

I don't think that's right. For example, if their orbital periods were 3, 3, and 9, you would get $n=3$, which would not work.

[hide]I essentially proved that it must be a multiple of 105. It turns out that 105 does work, but they're on opposite sides (I missed this and put 210).[/hide]

Opposite sides? They're evil. Evil. Evil.

LCM does work. LCM(3,3,9)=LCM(3,3,3^2)=3^(max(1,1,2)=3^2=9.

Bill The Maniac, what calc rulz was saying is that life=tennis+math's method of dividing LCM by GCF doesn't work in general.  Neither does simply taking LCM in general.  Instead of guessing a formula, try to solve the problem from scratch.

Dunno about this notation but the concept is the same.

The three planets must be congruent mod one-half their orbital position.

$\frac{t}{60}= \frac{t}{84}= \frac{t}{140}$  mod $\frac{1}{2}$

Multiply through the whole thing by 2.

$\frac{t}{30}= \frac{t}{42}= \frac{t}{70}$ mod 1

Break these up into two-sided equations and simplify to get

$\frac{2t}{210}= 0$ mod 1, $\frac{2t}{210}= 0$ mod 1, and $\frac{4t}{210}= 0$ mod 1.

0 mod 1 means it's an integer, so 105 is the smallest positive number that works.

A very similar problem was on the latest Rocket City Math League.

Yes, that's how I solved it.  When you break it up into three two-sided equations, you need only two of those two-sided equations: The third follows from the other two by transitivity.

[quote="panda888"]Yeah, I put down 210 because I didn't realize the planets could be on opposite sides..... :([/quote]

Same here.

i put down 420 just from the beginning
then i worked on it for a while and realized that 210 worked
then i happened to check 105 for the heck of it and realized that it worked even though it was on an opposite side and thats why i hadn't gotten it before
but i got it so im happy :D

Can someone explain to me it again why its 105 not 210 or 420? thanks.

Study Cicatriz's solution above.  If you don't fully understand it, let us know which step you're having trouble with.


geometry
geometric transformation
vector
trigonometry
dilation
ratio
real analysis

Assume that $C$ is a convex subset of $\mathbb R^{d}$. Suppose that $C_{1},C_{2},\dots,C_{n}$ are translations of $C$ that $C_{i}\cap C\neq\emptyset$ but $C_{i}\cap C_{j}=\emptyset$. Prove that \[n\leq 3^{d}-1\] Prove that $3^{d}-1$ is the best bound.
P.S. In the exam problem was given for $n=3$.

$C$ should be measurable and its measure should not be zero.

But every convex set is measurable.  I don't think measure zero is necessary, since then it will just be a lower dimensional convex set.

This is just a try of partial-pseudo-proof.

Call the space in which we are moving, whose elements are vectors, $S$.

Let $C_{i}=C+v_{i}=\{x\in S| \exists y\in C,\; y+v_{i}=x \}$. We will change all the $C_{i}$ and $C$ by $C^{*}= \{ \frac{x-y}{2}|x,y\in C\}|$, $C^{*}_{i}=C^{*}+v_{i}$; it's easy to see that $C^{*}$ is simmetrical with respect to the origin (in fact I have seen this tecnique in "Proofs from the Book" and is called Minkowsky simmetrization, I think).
$C^{*}$ is also convex, since for any $0<p,q<1$ such that $p+q=1$, for any $x,y,w,z\in C$, $p\frac{x-y}{2}+q\frac{w-z}{2}=\frac{(px+qw)-(py+qz)}{2}\in C*$, which comes from the fact that $(px+qw),(py+qz)\in C$. Furthermore $C_{i}\cap C_{j}=\emptyset \Leftrightarrow C^{*}_{i}\cap C^{*}_{j}=\emptyset$ (consider $C_{0}=C$ with $v_{0}=0$). Indeed for $x,y,z,v\in C$,  $\frac{x-y}{2}+v_{i}=\frac{w-z}{2}+v_{j}\Leftrightarrow \frac{x+z}{2}+v_{i}=\frac{w+y}{2}+v_{j}$, and the convexity of $C$ implies $\frac{x+z}{2}\in C, \;\frac{w+y}{2}\in C$. So the new sets continue to satisfy the hipothesis conditions.
Now for $1\leq i\leq n$, for any $\epsilon$ with $1>\epsilon>0$ we can change in the previous formula for the $C^{*}_{i}$,  $v_{i}$ for a vector $v'_{i}=k_{i}v_{i}$ with $k_{i}\geq 1$ such that the resulting $C'_{i}$ has still nonempty intersection with $C^{*}$, but (V: d-volume)  $V(C^{*}\cup \bigcup_{i=1}^{n}C'_{i})\geq (n+1-\epsilon)V(C^{*})$ . Of course, since the distance between the centers of any two $C'_{i},\;C_'{j}$ is greater than that between the centers of $C^{*}_{i},\;C^{*}_{j}$ as their respective distances to the origin are greater (due to cosine theorem), we will still have $C'_{i}\cap C'_{j}=\emptyset$. Now, since the sets are simmetrical, for each $v'_{i}$, ${\frac{v'_{i}}{2}}\in C^{*}$, and so for each $x\in C^{*}$, $\frac{v'_{i}+x}{3}=\frac{2}{3}\frac{v'_{i}}{2}+\frac{1}{3}x\in C^{*}$, since it belongs to one segment joining two points of $C^{*}$. This means that if we make a dilation of ratio $\frac{1}{3}$ from the origin, the union of all the resulting sets will be contained in $C^{*}$, and so:
$V(C^{*})\geq (\frac{1}{3})^{d}V(C^{*}\cup (\bigcup_{i=1}^{n}C'_{i}) )=(\frac{1}{3})^{d}(n+1-\epsilon)V(C^{*})$
And by just taking $\epsilon$<1 (actully it is somehow superfluous now that I think of it) this means:
$n\leq 3^{d}-1$

[Hope I wont edit it any more. ]


trigonometry
geometry
rectangle
analytic geometry
vector
graphing lines
slope

ABCD,CDEF and EFGH are squares.Now,BH and AF are drawn.BH and AF intersects each other at O.Prove that angle AOB =45 degrees ???

[hide="Solution"] [geogebra]4c4a9807e9c3df2c4d4a2409a3585309c01e1c9f[/geogebra]
First of all, extend $ HG$ and $ AF$ and let its intersection be $ P$.

Denote $ \angle OHP \equal{} \theta$ and $ \angle OPH \equal{} \theta '$.

Obviously $ \tan \theta \equal{} 3$ and $ \tan \theta ' \equal{} 2$.

$ \tan \angle AOB \equal{} \tan \angle HOP \equal{} \tan (180 \minus{} \theta \minus{} \theta')$.

Note that $ \tan x \equal{} \tan (180 \minus{} x)$ and $ \tan x \equal{} \minus{} \tan ( \minus{} x)$.

Hence, $ \tan \angle HOP \equal{} \minus{} \tan (\theta \plus{} \theta ')$.

Using tangent angle sum formula, we have $ \tan \angle HOP \equal{} \minus{} \frac {5}{1 \minus{} 3\cdot2} \equal{} 1$.

Thusly, $ \angle HOP \equal{} 45^{\circ}$.

Q.E.D.[/hide]

I'm sure there is more elegant solution, but this is indeed a solution :)

[hide="Nice Pure Geo"]
Draw rectangle AHIJ, with HI=GH and I is not G.Then $ \triangle AFI$ is right isoceles, so $ 45\equal{}\angle FAI\equal{}\angle FAH\plus{}\angle HAI\equal{}\angle OAH\plus{}\angle OHA\equal{}\angle AOB$.[/hide]

this is indeed good.but,can you send me some more elegant proofs of this problem using plane geometry only???
do post similar problems...

These ARE plane geometry.

[hide="analytic geometry"]Let $ B$ be the origin and let $ A\equal{}(0,a)$, where $ a$ is a positive real number. Therefore $ C\equal{}(a,0)$, $ D\equal{}(a,a)$, $ E\equal{}(2a,a)$, $ F\equal{}(2a,0)$, $ G\equal{}(3a,0)$, and $ H\equal{}(3a,a)$. The line that contains $ AF$ is $ y_1\equal{}\minus{}\frac{1}{2}x\plus{}a$ and the line that contains $ BH$ is $ y_2\equal{}\frac{1}{3}x$. The intersection is $ O\equal{}(6a/5,2a/5)$. Now that we know the coordinates of $ A$, $ O$, and $ B$, we can now find the lengths of each of $ AO$, $ AB$, and $ OB$, then use the Law of Cosines to find $ \angle AOB$.[/hide]

This is possible with vectors and imaginary numbers, I believe.

can we do it with pure euclidean geometry,apart from the above shown ways???

[quote="dgreenb801"][hide="Nice Pure Geo"]
Draw rectangle AHIJ, with HI=GH and I is not G.Then $ \triangle AFI$ is right isoceles, so $ 45 \equal{} \angle FAI \equal{} \angle FAH \plus{} \angle HAI \equal{} \angle OAH \plus{} \angle OHA \equal{} \angle AOB$.[/hide][/quote]

I am sure that angle chasing is included in pure euclidean geometry...

[hide="Solution"]
Another approach of analytic geometry is to notice that if you were to slide the rectangle into the corner of the Cartesian plane then line AF is of the equation -2x+3=y and line BH is of the equation 3x=y. If we want to show that angle AOB is 45 degrees then we could construct a triangle with one of its angles being AOB then we can make the other angle be a right angle and if the triangle is isosceles then obviously all its non right angles must be 45 degrees. So we can see that AF and BH intersect at -2x+3=3x $ \rightarrow$ x=.6 and 3(.6)=y $ \rightarrow$ y=1.8. Now to construct our right angle we can simply take a line perpendicular to the line AF and its slope is 1/2 and we solve the equation for that line intersecting line AF at point (1,1) (just because its an easy point to work with) (1/2)x+.5=y and this intersects line BH at (1/2)x+.5=3x $ \rightarrow$ x=.2 and 3(.2)=.6 so y=.6. and if we use the point (1,1) as our right angle then we can see that both of the legs have length of $ 4\sqrt{5}$ (distance formula) and so the triangle must be isosceles and right meaning that two of its angles are 45 and we can see that it would mean angle HOF is 45 which in turn mean angle AOB is 45.

[/hide]

maybe we could show CO is perpendicular to AF. also, CO produced to DF meets at its mp.

[quote="dgreenb801"][hide="Nice Pure Geo"]
Draw rectangle AHIJ, with HI=GH and I is not G.Then $ \triangle AFI$ is right isoceles, so $ 45 \equal{} \angle FAI \equal{} \angle FAH \plus{} \angle HAI \equal{} \angle OAH \plus{} \angle OHA \equal{} \angle AOB$.[/hide][/quote]
Or you can simply note $ AI\parallel{}BH$ and the rest follows. Nice construction!

Let $ C_1$ be the circumcircle of $ ABCD$ and $ C_2$ be the circle centered at $ E$ that passes through $ D,F,H$. Let $ O'$ be the second intersection of these two circles. Then \[ \angle AO'F\equal{}\angle AO'D\plus{}\angle DO'H\plus{}\angle HO'F\equal{}\angle ADB\plus{}\angle DFH\plus{}\angle HDF\equal{}45\plus{}90\plus{}45\equal{}180\] Consequently, $ O'$ lies on $ AF$. Also \[ \angle HO'B\equal{}\angle HO'D\plus{}\angle DO'B\equal{}\angle DFH\plus{}\angle DAB\equal{}90\plus{}90\equal{}180\] So $ O'$ lies on $ BH$. But then $ O\equiv O'$. So obviously, $ \angle HOF\equal{}\angle HO'F\equal{}\angle HDF\equal{}45$, as desired.

[quote="Altheman"]\[ \angle AO'F \equal{} \angle AO'D \plus{} \angle DO'H \plus{} \angle HO'F \]
[/quote]
If I'm interpreting the diagram you have in mind correctly, I'm not sure how you deduced this (without using the collinearity of $ A,O',F$ which follows from this equality - so if we use that, it's circular reasoning)... would you mind explaining please?

[quote="mathwizarddude"][quote="Altheman"]
\[ \angle AO'F \equal{} \angle AO'D \plus{} \angle DO'H \plus{} \angle HO'F
\]
[/quote]
If I'm interpreting the diagram you have in mind correctly, I'm not sure how you deduced this (without using the collinearity of $ A,O',F$ which follows from this equality - so if we use that, it's circular reasoning)... would you mind explaining please?[/quote]

$ \angle XOY\plus{}\angle YOZ\equal{}\angle XOZ$ always if we are speaking in directed angles. I'm not sure what the problem is.

My bad... I interpreted $ \angle AO'F$ in your solution as $ 360^{\circ}\minus{}\angle AO'F$.
Is there a way to avoid such confusion though (I guess the usual interpretation would be the angle that is less than 180 degrees, but what if the two angles are not known yet)?


Εστω $ y_i>a_i>0$  $ \forall i\equal{}1,2,...,n$
ΝΔΟ:
$ \prod_{i\equal{}1}^n \left(\frac{y_i}{y_i\minus{}a_i}\right) \geq 1\plus{}\sum_{i\equal{}1}^n\left(\frac{a_i}{y_i\minus{}a_i}\right)$

$ 0<x_i \equal{} \frac{a_i}{y_i} <1$ για να μην χολοσκάμε.

Μετά επαγωγή, $ x_n \plus{} x_{n\plus{}1} \minus{} x_n x_{n\plus{}1} < 1$ και... :lol:

Αναλυση με λογαρίθμους κανείς?

Cheerio,

Durandal 1707

Μια άλλη λύση είναι η παρακάτω:
Θέτουμε $ x_i\equal{}\frac{1}{y_i\minus{}a_i}$  (1)
τότε από την ανισότητα weisrtas έχουμε $ \prod_{i\equal{}1}^n(a_ix_i\plus{}1)\geq 1\plus{}\sum_{i\equal{}1}^n(a_ix_i)$
και αντικαθιστούμε το $ x_i$ με το ίσον του συμφωνα με την (1) και προκύπτει το ζητούμενο.

H πιο φυσιλογική λυση vangelismak  :thumbup:


How many distinct sums can be made from three different numbers from the set
{1, 2, 3, 11, 12, 13, 21, 22, 23}?
I feel like making a list....

Ugh... I don't like this set. I can't figure out a rule for how the sums overlap, so I guess I'll just write them out. Maybe I'll see something as I go along...

[hide="Answer"]What we do first is split the set into three subsets: $(1,2,3)$, $(11,12,13)$, and $(21,22,23)$. Using $(1,2)$, we have the numbers $(6,14,15,16,24,25,26)$. Using $(1,3)$, we have the new numbers $(17, 27)$. Using $(2,3)$, we have the new numbers $(18,28)$. Now, we move up to the next set. Using $(11,12)$, we have the new numbers $(36,44,45,46)$. Using $(11,13)$, we have $47$, and using $(12,13)$, we have $48$. Move up to the final set. Using $(21,22)$, we have $(54,55,56,66)$, $(21,23)$ gives $57$, and $(22,23)$ gives $58$.

Next, we need to combine numbers between sets. $(1,11)$ gives the new numbers $(33,34,35)$. $(1,12)$ gives no new numbers, but $(1,13)$ gives $37$. $(1,21)$, $(1,22)$, and $(1,23)$ give no new numbers. $(2,11)$ and $(2,12)$ give no new numbers, but $(2,13)$ gives $38$. $(3,11)$ and $(3,12)$ give no new numbers, but $(3,13)$ gives $39$. No other combinations give new numbers, so we are done, and we must simply see how many numbers we have, which turns out to be $\boxed{29}$.[/hide]

[b]EDIT:[/b] I realized I made a mistake, but I still get 27. I assume it doesn't tell you what the sums are?

[b]EDIT:[/b] Odd... I somehow didn't find $(57,58)$. Okay, that makes $29$. Fixed.

Okay, I get the pattern now. You have the minimum and maximum sums of $6$ and $66$, but you can't get any others near those. Obviously, the minimum sum of any three terms can end in anything from a $3$ to a $9$, but not a $1$ or $2$. We have $14$ through $18$, because $13$ would require $11$ and two $1$'s, which is not possible, and $19$ would require $13$ and two $3$'s, which is also not possible. We have the same situation with $24$ through $28$, $44$ through $48$, and $54$ through $58$, but we can get $33$ and $39$ because they use the smallest and largest numbers from each set of three. Therefore, we have a total of $5\cdot 4+7+1+1=\boxed{29}$ distinct sums. :)

Are you sure this will work for every problem? It's sorta a coincidence....
Maybe lists are better still?  :D

Dunno... I only found this after I made a list. :P

You basically have to look at the possible units digits, then the possible tens digits, and add up the number of sums.

I mean for this one: {2, 4, 6, 8, 10, 12, 13}, it's pretty easy, but for one's that have random numbers, it's much harder. I think the hardest thing about the sets is that they have a lot of repeats.


whats the equation for Cartesian to Cylindrical Coordinates (3-D) ??

This should provide you what you're looking for :
[url=http://en.wikipedia.org/wiki/Cylindrical_coordinate_system#Conversion_from_Cartesian_to_cylindrical_coordinates]Wiki[/url]
http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/cylindrical/body.htm
http://mathforum.org/dr.math/faq/formulas/faq.cylindrical.html
http://mathworld.wolfram.com/CylindricalCoordinates.html

:)


function
algebra
domain
trigonometry
calculus
calculus computations

What does 'limit' really mean? Does it mean when x approaches to some point, then this function is undefined or it no longer works? g(x)=(x^2-1)/(x-1)  The function g has limit 2 as x-->1 even though g(1) is not equal to 2. This is what my book says. But Why g(1) is not equal to 2?! I think IT IS 2 since we know the limit 2 by canceling out the factor (x-1), so what was left was (x+1). On the graph, it was like a linear graph. There was [b]a hollow circle/point (1,2) [/b]on the line, which i guess it represented the limit-- as x approaches to 1, g has limit 2. Then there was a [b]solid circle or a point (1,1) outside the line [/b]-- its x is also 1 but the g is 1. The book seems to say that that's the real answer when x=1. BUT this point is NOT on the graph? What does this circle mean? I am totally confused.
In general, can a hollow circle and solid circle both represent limit? but then what's the difference? For example, imagine a hollow circle/point (3,3) as an endpoint of a radial and it is toward left. Then, there is another hollow point (3,-2) as an endpoint of another radial but its direction is to the right. Then in between these two hollow points, ther is a solid point (3,1).  Then what does this solid point represent?
I HAVE CHANGED MY QUESTION AT BIT SO THAT YOU CAN UNDERSTAND ME BETTER. Hope you can tell me, thanks.

$ g(x)\equal{}\frac{x^{2}\minus{}1}{x\minus{}1}.$

This formula is not defined for $ x\equal{}1,$ so if this formula is the only thing we know, then we have to say that $ 1$ lies outside the domain of the function.

Now consider a new function, with a definition given pointwise:

$ h(x)\equal{}\begin{cases}\frac{x^{2}\minus{}1}{x\minus{}1},&x\ne1\\ 2,&x\equal{}1\end{cases}.$

The function $ h(x)$ is not identical to the function $ g(x).$ It is not identical because it has a different domain. $ h$ is defined for all real numbers; $ g$ fails to be defined at $ 1.$

We can perform algebraic manipulations, such as canceling like factors, but [b]only if the algebraic expression exists[/b]. Hence, for $ x\ne1,$ we can show that $ g(x)\equal{}x\plus{}1.$ But we cannot show that for $ x\equal{}1.$ We can't show anything about the function $ g$ for $ x\equal{}1$ because it's not defined there.

In fact, we can show that for all $ x,$ $ h(x)\equal{}x\plus{}1.$ But that's $ h,$ not $ g.$ $ h$ is a continuous function everywhere, including at $ 1.$ (Note, of course, that I had no freedom about assigning $ h(1).$ It had to be $ 2$ if I wanted the function to be continuous.) $ g$ is not continuous at $ 1$ because it's not even defined there.

----

Down the road, there's a somewhat more sophisticated attitude which I'll call "removable singularities deserve to be removed." In that mode of thinking, I will casually refer to $ \frac{\sin x}{x}$ as having the value $ 1$ for $ x\equal{}0.$ But the function I'm thinking doesn't really have that formula everywhere; if forced to express it formally, I'll admit that I'm really talking about the function

$ \begin{cases}\frac{\sin x}{x},&x\ne0\\1,&x\equal{}0\end{cases},$

and that that function is continuous at zero.

The empty circle is on the graph of the function to indicate that the function is not defined at that point. You see, the functions $ g(x)\equal{}\frac{x^{2}\minus{}1}{x\minus{}1}$ and $ f(x)\equal{}x\plus{}1$ are not the same, but they are equal at every value of $ x$ except $ x\equal{}1$, in particular we have that $ g(1)$ does not exist but $ f(1)\equal{}2$. This is because when you cancel the factor of $ x\minus{}1$ on top and bottom you must first assume that $ x\neq 1$.

Thank you guys. I sort of understand now. But what happened to the other solid circle/point that is outside the graph if you can understand what I mean by that.

[quote="Kent Merryfield"]$ g(x) \equal{}\frac{x^{2}\minus{}1}{x\minus{}1}.$

In fact, we can show that for all $ x,$ $ h(x) \equal{} x\plus{}1.$ But that's $ h,$ not $ g.$ $ h$ is a continuous function everywhere, including at $ 1.$ (Note, of course, that I had no freedom about assigning $ h(1).$ It had to be $ 2$ if I wanted the function to be continuous.) $ g$ is not continuous at $ 1$ because it's not even defined there.

[/quote]

thanks. Is h(x)=(x^2-1)/(x-1) or h(x)=2 which is like a constant? Which is the function for h(x)? Does h(x) have two functions? Do you mean like it depends on the value of x? If x is not equal to 1, then h(x) could be (x^2-1)/(x-1); however, if x=1, then we create another function of h(x)? If we create another one,should we change the name? Like if h(x)=(x^2-1)/(x-1), so if we want something =2, then we should name it j(x)=2? hope you can understand me, i know what i am trying to say sounds confusing.

It's not two functions! There's no "which"! It's [b]ONE[/b] function, which has a branched, piecewise definition. The statement that $ h(1)\equal{}2$ has nothing to do with any constants - it's just a single value of the function.

Let's try this again.

Let $ f(x)\equal{}\begin{cases}x\plus{}1,&x\ge 3\\ 4\minus{}x,&x<3\end{cases}$

This is ONE function, whose domain is all real numbers. Give me any real number $ x,$ and I can tell you what $ f(x)$ is - that's what we need for a function.

What is $ f(4)?$ Since $ 4\ge 3,$ we have $ f(4)\equal{}4\plus{}1\equal{}5.$
What is $ f(2)?$ Since $ 2<3,$ we have $ f(2)\equal{}4\minus{}2\equal{}2.$
What is $ f(3)?$ Since $ 3\ge 3,$ we have $ f(3)\equal{}3\plus{}1\equal{}4.$
What is $ f(2.999)?$ Since $ 2.999<3,$ we have $ f(2.999)\equal{}4\minus{}2.999\equal{}1.001.$

(Oh, by the way, this function has no limit at $ 3$ and is discontinuous at $ 3.$ That's not really my point.)

[quote="Kent Merryfield"]It's not two functions! There's no "which"! It's [b]ONE[/b] function, which has a branched, piecewise definition. The statement that $ h(1) \equal{} 2$ has nothing to do with any constants - it's just a single value of the function.

Let's try this again.

Let $ f(x) \equal{}\begin{cases}x\plus{}1, & x\ge 3\\ 4\minus{}x, & x < 3\end{cases}$

(Oh, by the way, this function has no limit at $ 3$ and is discontinuous at $ 3.$ That's not really my point.)[/quote]
Maybe I have never learned about this kind of 'branched' function before, so I don't know. Since in my mind, I thought the graph of a function is supposed to be a single curve or a single line, but now maybe there are some other graphs of branched function. Some graphs I saw were weird for me. BTW, how do you know there is no limit at 3? What does it really mean by 'limit'? What does it mean when something has limit? It means the value makes this function undefined?

[quote]BTW, how do you know there is no limit at 3? What does it really mean by 'limit'? What does it mean when something has limit? It means the value makes this function undefined?[/quote]
I know it doesn't have a limit at $ 3$ because its right hand limit and left hand limit don't match. And it is, very much, defined at $ 3$ - being defined is only one part of the definition of a limit. See the other topic that's part of this same conversation.

[quote]Hope you can explain it to me,[/quote]
I think not, as long as you don't have the slightest idea what a function is. For now, I give up.

[quote="Kent Merryfield"]
as long as you don't have the slightest idea what a function is. [/quote]
Well, I know what a function is. I just want to know what does limit actually mean? If a function has a limit, what does it mean? If a function doesn't have a limit, what does it mean? I have heard a question in the class today that between a number and the other number (range), every point on a curve is a limit. If I thought limit was something that made a function undefined, then could it be like all the points on a curve make a function undefined? That's impossible. I must have wrong conception for 'limit'!


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

Show that for all i, the subgroup Zi(G) which appears in the upper central series
of G is a characteristic subgroups of G.

I'm thinking of do this by induction.
But I'm not sure how.

To use induction, relate $ Z_{n\plus{}1}(G)$ to $ Z_n(G)$ by using the fact that $ Z_{n\plus{}1}(G)$ consists of precisely those elements $ a$ such that $ aZ_n(G)$ is in the center of $ G/Z_n(G)$. Equivalently, $ aga^{\minus{}1}g^{\minus{}1}\in Z_n(G)$ for all $ g\in G$.

[hide]
Induct on $ n$. For simplicity, let $ Z_0(G)$ be the trivial subgroup (then $ Z_1(G)$ will still be $ Z(G)$) so the base case is trivial.

For the inductive step, let $ \alpha$ be an automorphism of $ G$. For any $ a\in G$, for all $ g\in G$ we have
\[ a\in Z_{n\plus{}1}(G)\iff aga^{\minus{}1}g^{\minus{}1}\in Z_n(G)\]
\[ \iff \alpha (a)\alpha (g)\alpha (a)^{\minus{}1}\alpha (g)^{\minus{}1} \in \alpha (Z_n(G))\]
Now by the inductive hypothesis, $ \alpha (Z_n(G))\equal{}Z_n(G)$. As $ \alpha$ is a bijection, $ \alpha (a)\alpha (g)\alpha (a)^{\minus{}1}\alpha (g)^{\minus{}1}\in Z_n(G)$ for all $ g\in G$ if and only if $ \alpha (a)h\alpha (a)^{\minus{}1}h^{\minus{}1}\in Z_n(G)$ for all $ h\in G$.

Therefore, $ a\in Z_{n\plus{}1}(G)$ if and only if $ \alpha (a)\in Z_{n\plus{}1}(G)$. So $ \alpha (Z_{n\plus{}1}(G))\equal{}Z_{n\plus{}1}(G)$ for all automorphisms $ \alpha$, making $ Z_{n\plus{}1}(G)$ characteristic.
[/hide]


[hide]
The side of length 12 will divided into segments with length $5x$ and $13x$ by the Angle-Bisector Theorem, so $x=\frac{12}{18}=\frac{2}{3}\iff 5x=\frac{10}{3}$, which is the smaller segment at the right angle. Then, $5^{2}+\left(\frac{10}{3}\right)^{2}=h^{2}\Longrightarrow h=\frac{5 \sqrt{13}}{3}$.[/hide]

isn't the formula for length to side c like

2/(a+b)   x    rt [abs(s-c)]

[quote="Hoborobo"][hide]Since it is a right triangle, the bisector bisects side 12 at its midpoint. So it is $\sqrt{5^{2}+6^{2}}=\sqrt{61}$[/hide][/quote]
True for the median, not the angle bisector.
[hide]5-12-13 is right.
Let the length from the right angle to the intersection of the angle bisector to side 12 be x. From Angle Bisector Theorem, $\frac{5}{x}=\frac{13}{12-x}\implies x=\frac{10}{3}$(?).
Then, by Pythagorean, the angle bisector $=\sqrt{5^{2}+(\frac{10}{3})^{3}}=\sqrt{\frac{325}{9}}=\frac{5\sqrt{13}}{3}$?.[/hide]

the median is

1/2 rt( 2a^+2b^-c^)

the angle bisector is the one i listed earlier.

[quote="Ignite168"]Find the length of the angle bisector drawn to the side of length 12 in a triangle with side lengths 5, 12, and 13.[/quote]

[hide]By angle bisector the answer is

$\frac{5^{2}+(\frac{15}{2})^{2}}=\frac{5\sqrt{13}}{2}$[/hide]


If x=2^12*3^6 and y=2^8*3^8, what integer z satisfies x^x*y^y=z^z? (Note: Your answer should not be expanded. Instead, write your answer as a product of powers of primes.)

Has already been posted at least twice.

[hide="Solution"] Since the LHS is only divisible by $2$ and $3$ we must have $z = 2^{a}3^{b}$.  Then equating powers on the LHS and RHS we have

$\text{Number of powers of two}= 12 \cdot 2^{12}3^{6}+8 \cdot 2^{8}3^{8}= a \cdot 2^{a}3^{b}\Leftrightarrow$
$2^{14}3^{7}+2^{11}3^{8}= a \cdot 2^{a}3^{b}\Leftrightarrow$
$(2^{3}+3)2^{11}3^{7}= a \cdot 2^{a}3^{b}$

Which suggests $a = 11, b = 7$.  To check,

$\text{Number of powers of three}= 6 \cdot 2^{12}3^{6}+8 \cdot 2^{8}3^{8}= b \cdot 2^{a}3^{b}\Leftrightarrow$
$2^{13}3^{7}+2^{11}3^{8}= b \cdot 2^{a}3^{b}\Leftrightarrow$
$(2^{2}+3) 2^{11}3^{7}= b \cdot 2^{a}3^{b}$

So $z = \boxed{ 2^{11}3^{7}}$. [/hide]


[size=125]Let $x, y$ be real numbers such that ${x^2}$+$y^2$= 1  and  ${x \neq}$y[/size]

[size=125]Find the mimimum value of  [/size]  [size=142]$\log_{2} \displaystyle\frac{x^{11} - y^{11}}{x-y}$ [/size]

I solved this problem today. Check my solution.


logarithms
number theory proposed
number theory

We examine the following two sequences: The Fibonacci sequence: $F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 }$ for $n \geq 2$; The Lucas sequence: $L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2}$ for $n \geq 2$. It is known that for all $n \geq 0$ \[F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n},\] where $\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}$. These formulae can be used without proof.

We call a nonnegative integer $r$-Fibonacci number if it is a sum of $r$ (not necessarily distinct) Fibonacci numbers. Show that there inﬁnitely many positive integers that are not $r$-Fibonacci numbers for any $r, 1 \leq r\leq 5.$

There are $n=[\frac{ln(x)}{ln{\alpha }}+\frac{1}{2}]$ Fibonacci numbers not great, than x.
Therefore $O(ln(x)^{r})$  r - Fibonacci numbers, less than x. For big x $O(lnx^{r})<x$.

Rust,
Why you allway post as that? I don't understand this solution at all!  :oops:

Calculate number of r - Fibonacci numbers, less than x  $N_{r}(x)=\#\{a<x| a-\ is \ r \ Fibonacci \ number\}$, obviosly $N_{r}(x)\le N_{1}(x)^{r}$. Because $N_{1}(x)=O(\ln x)$ we get, that densyty of r Fibonacci numbers is 0.


geometry
geometric transformation
ratio
geometry proposed

Given a triangle $\triangle ABC$, construct a point $X$ such that the three lines drawn through $X$, each parallel to a side of the triangle and limited by the other two sides, are equal.

Let $d_a$ be the length of the relevant segment of the line through $X$, parallel to $a$. I believe I am correct in saying that the locus of all points such that $d_b=d_c$ is a line, since if we consider a translation parallel to $b$, it will change $d_c$ by a fixed amount while leaving $d_b$ unchanged; similarly for one parallel to $c$, so we can find a composite translation which changes both by the same amount. To find this line, we must get two points. We can find one as the intersection of the line through $B$ parallel to $b$ and the line through $C$ parallel to $c$, and another by dividing $a$ in the ratio $b:c$.

[quote="pestich"]Just for completeness it would be nice to have the length
        of this segment as f(a,b,c).

        Cpt.   Pestich[/quote]

The length of equal parallel intercept is

\[ \frac{{2abc}}{{ab + bc + ca}}\]

Leon


Imagine that your friend was put in jail.
your friend is given 1 chance at freedom:
there are 3 doors, 1 that leads to freedom, 2 that lead to DEATH *muahahaha*first, your friend picks a door and the jailor will choose one of the remaining door(s) that lead(s) to death.
your friend picks the first door and the jailor reveals that the 3rd door leads to death.
Of the first and second doors, Which is better, and what are the probabilities of each dfoor leading to freedom?

[hide="answer/solution:"]
[hide="heh u rly think id give u the answer?"]
[hide="Are you sure you want to see?"]

2/3...second door

the picking of the first door and third door is hypothetical- the first door could be anything and the door leading to death could be any of the rest.
the chance of the formally chosen door leading to freedom is 1/3
and the chance of the formerly chosen door leading to death is 2/3
since this case means that one of the other doors leads to freedom, and the door leading to death will be revealed, the door that was not chosen or revealed will lead to freedom in this case.

[/hide]
[/hide]
[/hide]

[hide]Since if you switch, you win if you chose a door of death in the first place, which occurs 2/3 of the time.[/hide]

Wow someone really took a specific type of problem and ran with it!
all the entries and articles might take me a good hour to read!
thx 2 andersonw for the link :D

This is a really cool problem!

It appears to not make sense to begin with, but mathematically, it's correct.

Mathematics solves everything...

 :lol:


[quote="me@home"]Hang on I'm working on this one... anyway is the $\frac{\pi}{4}$ inside or outside the sin?[/quote]

Inside, of course - since $(2\sin\pm 1)\cdot{\pi\over 4}\neq \pm\sqrt{2}$, whilst $2\sin(\pm 1\cdot{\pi\over 4})=\pm\sqrt{2}$

Sorry I didn't realize what a stupid question that was...

This has been left unsolved. Any takers? Or it's time for me to post the proof?

oh i haven't been working on this :oops: hang on one more day...is that fine?

OK then, we mustn't leave this unsolved.

First, note that if $a=\pm 1$, then $\sin ax=a\sin x$. For the main statement we'll use induction.

$1^\circ\qquad 2\sin{a_{1}{\pi\over 4}}=2a_{1}\sin{\pi\over 4}=a_{1}\sqrt{2}$

For shortness, put $f(c_{1},\dots,c_{n})=c_{1}+\frac{c_{1}c_{2}}{2}+\dots+\frac{c_{1}c_{2}\dots c_{n}}{2^{n-1}}$ and $g(c_{1},\dots,c_{n})=c_{1}\sqrt{2+c_{2}\sqrt{2+\dots+c_{n}\sqrt{2}}}$. Then we have to prove $2\sin f(a_{1},\dots,a_{n}){\pi\over 4}=g(a_{1},\dots,a_{n})$. Assume it holds for some natural $n$. Then

\begin{eqnarray*}2\sin f(a_{1},\dots,a_{n+1}){\pi\over 4}&=& 2\sin \left({a_{1}}+{a_{1}\over 2}f(a_{2},\dots,a_{n+1})\right){\pi\over 4}\\ &=& 2a_{1}\sin\frac{(2+f(a_{2},\dots,a_{n+1})){\pi\over 4}}{2}\\ &=& 2a_{1}\sqrt{\frac{1-\cos({\pi\over 2}+f(a_{2},\dots,a_{n+1}){\pi\over 4})}{2}}\\ &=& 2a_{1}\sqrt{\frac{1+\sin f(a_{2},\dots,a_{n+1}){\pi\over 4}}{2}}\\ &=& a_{1}\sqrt{2+2\sin f(a_{2},\dots,a_{n+1}){\pi\over 4}}\\ &=& a_{1}\sqrt{2+g(a_{2},\dots,a_{n+1})}\\ &=& g(a_{1},\dots,a_{n+1})\end{eqnarray*}

and the statement is thus proven.


algebra
polynomial
trigonometry
integration
function
geometry
analytic geometry

Hi, could someone give me some light on these problems please? Please give explanations. Thanks in advanced.

1: At how many points in the xy-plane do the graphs of $y = x^{12}$ and $y = 2^x$ intersect?
a - None
b - One
c - Two
d - Three
e - Four

2: Let p(x) be the polynomial $x^3 + ax^2 + bx + c$ where a, b and c are real constants. If p(-3) = p(2) = 0 and p'(-3) < 0, which of the following is a possible value of c?
a: -27
b: -18
c: -6
d: -3
e: $-1/2$

3: In the xy-plane, if C is the circle $x^2 + y^2 = 9$, oriented counterclockwise, then $\oint_{C} -2y dx + x^2 dy$ =
a: 0
b: 6pi
c: 9pi
d: 12pi
e: 18pi

Let $x=3\cos t,y=3\sin t$,$t\in [0,2\pi]$. Then $dx=-3\sin t dt,dy=3\cos t dt$ and we obtain

$\int_{C} -2ydx+x^2dy=\int_{0}^{2\pi}(-6\sin t (-3\sin t) + 9\cos^2 t 3\cos t ) dt = \int_{0}^{2\pi} 18\sin^2t+27\cos^3t dt = |\int_{0}^{2\pi}\cos^{2n+1} = 0| = \int_{0}^{2\pi} 18 \sin^2t=$
$=4\cdot 18 \cdot \int_{0}^{\pi/2}\sin^2tdt=18\pi.$
Maybe , we can use Green formula, but i can't remember it :blush:

if $p(-3)=p(2)=0$ then $p(x)=(x+3)(x-2)(x-\alpha).$ then $c=6\alpha$ and $p'(-3)=15+5\alpha <0$. We have $\alpha < -3$ then $c=6\alpha <-18$ . Thus it can be inly $-27$

if $x<0$ then functions are monotonic and it is easy to see that intersect in one point. If $x>0$ then in $x=0$ we have $x^2<2^x$ and in $x=2$ we have $x^2>2^x$. But $2^x-x^2 \to \infty,x\to \infty$  so they have at least two points of intersect. And they are convex. So answer is $3$

Thanks Andrew  :)

Still, I am confused with this part.

How did you get $\oint^{2\pi}_0 cos^{2n+1}(t)dt = 0$?

And when I tried to use Green's theorem, somehow I got a problem. This one is in the form $\oint_C Mdx + Ndy = \int\int_R (N_x - M_y)dxdy$
So M = $-2y$ and N = $x^2$ => $M_y = -2$ and $N_x = 2x$. We have $\int\int_R (2x + 2) dxdy = 2\int\int_R (x+1)dxdy$. Now the area of R is $9\pi$, but what about (x+1)?

Thanks again,

[quote="futurus"]
 This one is in the form $\oint_C Mdx + Ndy = \int\int_R (N_x - M_y)dxdy$
So M = $-2y$ and N = $x^2$ => $M_y = -2$ and $N_x = 2x$. We have $\int\int_R (2x + 2) dxdy = 2\int\int_R (x+1)dxdy$.
[/quote]

Now, we need to use polar coordinates and evaluate this integral.
 I thought that if we use Green formula, then the integral will be simple, (But, as i writed, i could not remember it.. ) but i  was wrong. In any way, you can evaluate this integral as you like  :)


superior algebra
superior algebra solved

how can i find a ring with two ideals which union is not an ideal itself?

Almost anything would work ... try to look in $\mathbb {Z}$.

the only ideals in Z are nZ so i don t think it works

It does. Remember that an ideal must be a group in particular ...

yes u r right i sorted it out
i was just checking closure under multiplication
sorry


[quote="pda"]What is the maximum number of acute interior angles that can be on a 2001-gon?[/quote]

So this is any type right?  It can be concave?

Answer (Partial)

All angles are in degrees.

Let $ n$ = number of acute angles

Let $ m$ = number of non-acute angles

Let $ {\theta}_{n}$ = average of acute angles

Let $ {\theta}_{m}$ = average of non-acute angles

The sum of all the internal angles in a 2001-gon is given by:

$ \Sigma \theta = 180\cdot 2001-360 =$

$ \Rightarrow n{\theta}_{n}+m{\theta}_{m}= 180\cdot 2001-360$

But $ m+2 = 2001$

$ \Rightarrow n{\theta}_{n}+(2001-n){\theta}_{m}= 180\cdot 2001-360$

$ \Rightarrow n({\theta}_{n}-{\theta}_{m}) = 180\cdot 2001-360-2001{\theta}_{m}$

$ \Rightarrow n = \frac{180\cdot 2001-360-2001{\theta}_{m}}{{\theta}_{n}-{\theta}_{m}}$

$ \Rightarrow n = \frac{2001{\theta}_{m}-180\cdot 2001+360 }{{\theta}_{m}-{\theta}_{n}}$

Now we are going to diverge from geometry for a while and find the maximum possible $ n$ algebraicly (since if it's not possible algebraicly, then it certaintly isn't going to be possible geometrically).

Let $ {\theta}_{n}\to 90$ as this will minimise the denominator, so will help maximise $ n$. Now we need to consider what value of $ {\theta}_{m}$ maximises n. To do this I am going to turn to calculus using the qoutient rule.

$ \frac{dn}{d{\theta}_{m}}= \frac{2001({\theta}_{m}-{\theta}_{n})-(2001{\theta}_{m}-180\cdot 2001+360)\cdot 1}{({\theta}_{m}-{\theta}_{n})^{2}}$

$ \frac{dn}{d{\theta}_{m}}= \frac{180\cdot 2001-360-2001{\theta_{n}}}{({\theta}_{m}-{\theta}_{n})^{2}}$

Since $ {\theta}_{n}< 90$ then:

$ 180\cdot 2001-360-2001{\theta}_{m}> 180\cdot 2001-360-2001\cdot 90 = 90\cdot 2001-360 > 0$

$ \Rightarrow \frac{dn}{d{\theta}_{m}}> 0\,\, \forall{\theta}_{m}$

Hence $ n$ strictly increases as $ {\theta}_{m}$ increases.

$ \Rightarrow{\theta}_{m}\to 360$ maximises $ n$.

$ \Rightarrow n < \frac{2001\cdot 360-180\cdot 2001+360 }{360-90}$

$ \Rightarrow n < \frac{2003\cdot 180 }{270}$

$ \Rightarrow n < \frac{2}{3}2003$

$ \Rightarrow n \leq 1335$

Now we have to work down from $ n = 1335$ either showing that a 2001-gon exists with this many acute angles, or prove that no 2001-gon exists for this many acute angles. This is where I got stuck. I realised that we can draw a 2001-gon with 1334 sides. Since 1334 is two thirds of 2001, we can draw a "windmill" shaped polgon with 667 spokes.

I expect this is the largest polygon, but I failed in showing that a 2001-gon can or cannot have 1335 acute angles. (However, I did manage to prove that if such a polgon did exist, it would have no interior angles which are between 90 and 270 degrees Perhaps this would pave a way for a proof by contradiction?)


Let $2^{n}=6(mod \ n)$. Prove, that $193\not |n$.

i solved this!

my solution is very ugly. you have an nice solution Rust?

Original problem (see russian forum) was if $n|2^{n}-3$, then $1489\not | n$. I don't shure for my calculations about 193. Therefore I give solution original problem (same idea). One calculate, that $1489|2^{n}-3$ then $n=637(mod \ 744)$. Therefore n is odd and $n=5(mod \ 8), \ n=1(mod \ 12)$, Jakobi simbols $(\frac{2}{n})=-1, \ (\frac{3}{n})=1$. Last can be if only n is even, but n is odd.

I remenber to there's a problem :In sequence $12.k+1$ has infinite prime number such that $p$ is not divises $2^{n}-3$ for all $n$


Like yeah, gl and stuff. of course, this was made like really early, since it's not for like 2.0 weeks.

oh and WHO IS THIS PIANOFORTE IMPOSTERATOR?

Well, I took the I
historically (for me at least)
I fail at II's and i own I's.
I got a total of 3 something teen's on the I's including one unofficial official one, but i probably only got scores that end in en but not teen on the II's.


function
logarithms
algebra unsolved
algebra

$ t_n \equal{} \frac {\prod _{i \equal{} 1}^n \alpha _i}{\sum _{i \equal{} 1}^n \frac {\alpha _i}{n}}$

Deriving this expression 2 times in function of $ \alpha _i$

[hide]

$ f\equal{}\prod _{i\equal{}1}^n \alpha _i\land g\equal{}\sum _{i\equal{}1}^n \frac{\alpha _i}{n}$

$ \frac{\partial f}{\partial \alpha _i}\equal{}\frac{f}{\alpha _i}\land \frac{\partial ^2f}{\partial \alpha
   _i^2}\equal{}0$

$ \frac{\partial g}{\partial \alpha _i}\equal{}\frac{1}{n}\land \frac{\partial ^2g}{\partial \alpha _i^2}\equal{}0$

$ \frac{\partial ^2\frac{f}{g}}{\partial \alpha _i^2}\equal{}\frac{2 f \frac{\partial g}{\partial \alpha
   _i}^2}{g^3}\minus{}\frac{2 \frac{\partial f}{\partial \alpha _i} \frac{\partial g}{\partial \alpha
   _i}}{g^2}\plus{}\frac{\frac{\partial ^2f}{\partial \alpha _i^2}}{g}\minus{}\frac{f \frac{\partial ^2g}{\partial
   \alpha _i^2}}{g^2}$

$ \frac{\partial ^2\frac{f}{g}}{\partial \alpha _i^2}\equal{}\frac{2 f}{g^3 n^2}\minus{}\frac{2 f}{g^2 n \alpha _i}$

[/hide]

$ \frac {\partial ^2t_n}{\partial \alpha _i^2} \equal{} \frac {2}{n^2} \frac {\prod _{i \equal{} 1}^n \alpha _i}{\left(\sum _{i \equal{} 1}^n \frac {\alpha _i}{n}\right)^3} \minus{} \frac {1}{\alpha _i} \frac {2}{n} \frac {\prod _{i \equal{} 1}^n \alpha _i}{\left(\sum _{i \equal{} 1}^n \frac {\alpha _i}{n}\right)^2}$

It's easy to proove that the last expression is less than zero.

$ \frac {2}{n^2} \frac {\prod _{i \equal{} 1}^n \alpha _i}{\left(\sum _{i \equal{} 1}^n \frac {\alpha _i}{n}\right)^3}\leq \frac {1}{\alpha _i} \frac {2}{n} \frac {\prod _{i \equal{} 1}^n \alpha _i}{\left(\sum _{i \equal{} 1}^n \frac {\alpha _i}{n}\right)^2}\Longleftrightarrow \sum _{i \equal{} 1}^n \frac {\alpha _i}{n}\geq \frac {\alpha _i}{n}$

So $ t_n$ is a crescent function (for the variables $ \alpha _i$) So it have a maximum at $ \alpha _i \equal{} 1$. But since it is not definide for this numer, lets make:

$ \alpha _i \equal{} \epsilon$

$ \sum _{i \equal{} 1}^n \frac {\log \left(t_n\right)}{\log \left(\alpha _i\right)}\geq \sum _{i \equal{} 1}^n \frac {\log \left(\epsilon ^{n \minus{} 1}\right)}{\log (\epsilon )} \equal{} \sum _{i \equal{} 1}^n (n \minus{} 1) \equal{} (n \minus{} 1) n$


Let A,B,C,D be distinct points on a line , in that order .The circles with diameters AC and BD intersect at X  and Y . O is an arbitrary point on the line XY but not on AD. CO intersects  the circle with diameter AC again at M , and BO intersects the other circle at N.Prove that the lines AM,DN ,XY are concurrent.  :)

Let $ H,H'$ be the orthocenters of $ \triangle AOC$ and $ \triangle DOB,$ respectively. $ H \equiv {} AM \cap XY$ and $ H' \equiv{} DN \cap XY.$ If $ T \equiv {} XY \cap AD,$ it follows that $ OT \cdot HT \equal{} TC \cdot TA$ and $ OT \cdot H'T \equal{} TB \cdot TD .$ Therefore, we conclude that $H \equiv H',$ due to $TA \cdot TC \equal{} TB \cdot TD$ $\Longrightarrow$ Lines $ XY,AM,DN$ concur.


Let $a,b,c \geq 0 , \sqrt{a}+\sqrt{b}+\sqrt{c}=1$.Prove that:
$\frac{\sqrt{ab+c}}{a+b}+\frac{\sqrt{bc+a}}{b+c}+\frac{\sqrt{ca+b}}{c+a}\geq \frac{9}{2}$

Has anyone solved it?

Thank you very much. :)


search
combinatorics unsolved
combinatorics

A radio receiver receives radio signals consisting of a binary sequence of length 10. The receiver is broken and sometimes confuses two signals if they resemble each other at five or more places. What is the largest number of signals that can be sent to the receiver without any risk of two signals getting mixed up?

Have a look here for a more general statement
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=52217384&t=4535

Pierre.


analytic geometry
geometry
MATHCOUNTS
graphing lines
slope
rectangle

can it be done without drawing it or using distance formulas, etc.?

jorian

[quote="ckck"][quote="davidli"]The vertices of a tirangle have coordinates (-3,1), (-2,-3), (x,0). Find x, given x is positive and the area of the triangle is 7.5 square units.[/quote]
[hide]Using shoelace theorem on it you have the following equation $\frac{4x+11}{2}=7.5$. Solving for x we get that x=1.[/hide][/quote]
What is Shoelace theorem?

A person from my school taught it to me at the MathCounts Competition but I can't remember it... :P

[quote="Quevvy"][quote="bpms"]Instead of shoelace, consider this hint:
[hide="hint"]The slope of a perpendicular line=the opposite reciprocal of the slope of the line it's perpendicular to.[/hide][/quote]
It never says that it is a right triangle...[/quote]
I know...
Here is what I had in mind.
[hide]To avoid confusing my tiny mind, I'm going to use a instead of the initial x.
The slope of the line from a,0 to -2,-3 is $\frac{3}{a-2}$
The slope of a line perpendicular to this is $\frac{2-a}{3}$.
Plugging the fact that this hits -3,1 in, it's equation is
$y=\frac{2-a}{3}x+3-a$
The equation of the line between
a,0 and -2,-3 is
$y=\frac{3}{a-2}x-3+\frac{6}{a-2}$
Equating our two formulas, we find that the intersection point is
$9x+9=-a^{2}x+4ax-4x+9-3a$
$a^{2}x+13x+3a-4ax=0$[/hide]
Okay, apparantley my idea was messier than I thought :D
Instead of that, sketch the problem then draw a rectangle around our triangle, then subtract out the area of the four smaller triangles.
That should work much much better.


What is your blood type? :)

(I know the theoretical ratio, but I just want to see if there is any interesting trend in the AoPS/Mathlinks.)

i don't know what my blood type is.  How can i find out?

math92, your blood type should be on your birth certificate. :)

frt, what is the theoretical ratio? Just curious.

Nice questio, but what about the rhesus system, you can have plus or minus.
A minus can donate to everyone, a plus can only donate to plus.  Can the poll be changed somehow?  It would be interesting?

I live in Europe where 16% is minus, but in Asia it appears to be less than 1 procent.  Rhesus minus travellers should watch out in those countries as the minus kind of blood is low in supply sometimes.

I have O minus, only 7 percent of the population has it, meaning I can donate to everyone  :lol:  but I can only receive from other O minus people  :|

Here are some numbers, I believe they are for Europeans (and as I pointed out, it is not the same all over the world, although O dominates pretty much everywhere)

O+ 38%
A+ 34%
B+ 9%
O- 7%
A- 6%
AB+ 3%
B- 2%
AB- 1%

I had a biology teacher in high school who told us he once let everyone check up on their blood type and their parents

he claims, dunno if it is true, hope he was joking, that he once actually had a student for whom it genetically was impossible :o
not a nice test :|

0-  ....I'll call fredbel if i'm in trouble with blood...ok? :lol:

They took a little statistical thing with bloodtypes in China.

When the nationalists captured communists, they tortured them.  Here are the results.

O-  The people would rather die than give up communism.

B- People just gave up and were released.

A- People said that they would rather die than give up communism, but changed their mind at the last minute and were released near death.

AB-  Mixed.

Seriously , is that correct?
I had never heard of a study like that? :huh:

Depends if you want to believe chinese news sites  :P

You  mean the Kuomintang did that, to the communists?

Think so....

Dont think you can read chinese, but you can find it in http://www.sohu.com search, i think...

hummmm......

*state of confusion*

Random statistics

I thought -'s could receive once from +'s and that's it.

I don't know what my blood type is and won't bother to find out.

[quote="monoversketcher"][quote="Kalimdor"]Oh, very nice-- I'm a B-, so that I'm going to give up at the very first second to the Kuomintang? I dunno, but my grandpa had a friend who faced that with blood type B-, who was killed since he refused to give up communism.
How do you explain that, then
(P.S., to those FBI and CIA guys, better check everyone's(including theirselves and those who are at their hands) blood types first. haha

Frank[/quote]

dude just because of that one study doesn't make every single people with B-type blood will automatically give up in those situations...

it's just an estimate, so who knows what will the results be if every single people on earth was tested?  :?  just one of those nature thingies that nobody can know entirely of...[/quote]
Actually, I post those information is just because I wanna  give out the idea that: no statistic is perfect! :lol:
Frank


These are the scores I know of so far

SUN 128 + 1A,1B to go (<= 21 more points)
Vert 76
Simon 17
Yi 15
Keone 12
Cinjon ?
Kristen 8
Matt ?

ACT 40 + 3A,1V to go (<= 142 more points)
Vert ?
Arnav 20
Max 10
Andrew 10

STL 45
Vert 18
Joseph 4
?

...and this could get pretty tedious, since I only really know some individual scores (from the top scorer e-mail).  Can anyone contribute their teams' scores?  (Thanks joml88.)

I think the scores for Acton are:

ACT 141
Vert 65
Arnav 20
Joel 12
Feiqi 12
Fred 12
Max 10
Andrew 10

Interesting.  What was the score on #2, out of curiosity?

mathfanatic, can you update sunnyvale's scores?

I would if I knew them exactly, but I'll see what I can do.

...what on earth was SL thinking, not challenging #2?  I don't want to be mean, I'm just saying, you guys did [i]not[/i] deserve a 5 for what you did...

Haha, well, what did they get on theirs?  A tree?  This is definitely not a tree problem.

The Sugarland team felt that they did not deserve their score on #2 and hence didn't wish to challenge #2.

Yes, but I project that SL and Acton will come out with approximately equal scores on #1 and #3, judging by their proofs.  (SL may come out a few points ahead, but it won't be significant.)  Since Acton has a higher individual, they will then win.  The only way SL has a decent chance is if they challenge #2 and pray that their score doesn't go down very much, since Acton's would undoubtedly go way, way down on that one.


well i dont remember the problem number on spoj..
well consider a finite sequence a_1,a_2...  constructed according to the given rules

1) a_1=0
2) a_k-a_(k+1)=1 or a_k-a_(k+1)=-1
every term is either greater or less than the previous term by 1.

so  the input is
S=sum of the terms in the sequence.
K=length of the sequence.

find a sequence which satisfies the above properties ...or say it's impossible...

this is what i've come up with now..
if
K=4n then it represents all even numbers <=2n(4n-1)
K=4n+1 then it represents all even numbers <= 2n(4n+1)
K=4n+2 then it represents all odd numbers <=(2n+1)(4n+1)
K=4n+3 then it represents all odd numbers <= (2n+1)(4n+3)
otherwise its impossible

now i am difficulting in constructing solutionss:)
please help
best regards

What size of numbers are we talking about here?

If you start with the sequence 0,1,..,(K-1), then you can continually subtract 2 from the highest number to give another valid sequence. (That is actually a proof of attainability or not; the parity is invariant). So just do that enough times until you get the sum you want. That may be too slow though, depending on the size of the numbers..

[edit] OK, I found it on sphere, and the sum is probably a bit too high to keep finding the maximum. However, you should be able to find a very close approximation to the sum, and then you can subtract 2 from max / add 2 to min to get the exact answer. (And don't forget negative sums, though you handle them in exactly the same way).

It's problem 176 ( Sum of one-sequence) on SPOJ.

I just got AC on SPOJ with just 0.03 s runtime. Algorithm is very simple, If D = (n*(n-1)/2) - S, I first check if it's possible to build sequence. Then starting from most significant number ( with index 1) I try to substract (n-i) from D, where i is index of current number. That works in O(n) time and seems to be the fastest you can do.

sorry dude but i dont get u
i know the conditions when a solution exists...how are u contructing it...cud u be more clear...with an example preferably please
thanks again
vasu

ok for the moment looks like i have a construction myself
will get back if it doesnt work :D

Let's consider sample input/output, n = 8, S = 4.

You start with a sequence: 0 1 2 3 4 5 6 7
and D = 8*7/2 - 4 = 24

Start processing from the second position( number 1 ). You know that decreasing it by 2 would change sum of entire sequence by 14 ( 2*7, number of elements after it +1 ). Since 14 < 24, you decrease all numbers from 1-7 by 2. Now D-14 = 10. Your sequence:
0 -1 0 1 2 3 4 5.
You go for the next position, it decreases sum by 12, uff, too much. The next position decreases sum by 10. We use it, D = 0 so we're done. Sequence:
0 -1 0 -1 0 1 2 3.
That gives us sum of 6.

I hope it was informative :D

sure twas informative:)
and i managed to come up with another algorithm to generate the answers..

thanks a lot
kind regards
vasu

ok got my AC as well ..in .23 sec...i dont see why it's so slow:(

You're probably doing something really stupid or you just using strange algorithm. I got 0.03 without work at all. I just implemented my idea. PM me your code so I can look for a problem.

well stupid is not the correct word dude..it's too harsh(as i am still a learner) :):)
anyways got AC in .04 secs
thx a lot

Well what would you like? Wise mistake :D

In general, when your code works and you get WA, it's pretty obvious that you mistyped or forgot about something.

Anyway, Congratulations and Good luck  :)


The numbers 1, 2, 3, 4, ..., 30 are arranged on a circle in arbitrary way. Somebody has done several operations such that on each operation he traded places of two adjacent numbers (that is $...axyb... \to ...ayxb...$). It turned out that after several operations each number moved to diametrically opposite place. Prove that there was a moment when the numbers with sum 31 were traded.

Suppose the assertion is false. Assign a value $a_i$ to the number $i$ to keep track of how it has traveled: every time it moves clockwise, raise $a_i$ by $1$, if it moves counterclockwise, decrease $a_i$ by $1$. If a number $j$ moves to the diametrically opposite position, $a_j = 15n$, were $n$ is an odd integer. If the numbers $k$ and $31-k$ never pass, then $a_k=a_{31-k}$ at the end. $a_1 + a_2 + \dots + a_{30}=2( a_1+\dots+a_{15})$. But $a_1+\dots+a_{15}$ is odd, so $a_1 + a_2 + \dots + a_{30}$ is nonzero. But this is impossible, because this quantity is zero and invariant.


linear algebra
matrix
abstract algebra
vector

I'm a beginner to linear algebra (I'm a senior in high school, going to college next year), and was wondering if this was a valid (yet unneedingly long) proof of the fact that the rowspace of a matrix is always orthogonal to the nullspace of the matrix. The purpose of this is just to gain comfort with linear algebra manipulation. So anyways, here it goes:

[b]Problem: Prove that for any matrix $ A,$ the rowspace and nullspace of $ A$ are orthogonal.[/b]

[b]Proof[/b]

The nullspace (or kernel) of said matrix $ A$ may be described as all vectors $ \vec{x}$ which satisfy the property that
\[ A\vec{x} \equal{} \vec{0}
\]
Also, the rowspace of $ A$ may be described as all $ \vec{b}$ which satisfy the property that
\[ A^T \vec{y} \equal{} \vec{b}
\]
for some valid $ \vec{y}.$ The proposed statement is that $ x^T b \equal{} b x^T \equal{} 0.$ Now, note that (we will drop the vector arrows for now...)
\[ Ax \equal{} 0 \Leftrightarrow x^T A^T \equal{} 0
\]
so we have that
\[ (A^T y)(x^T A^T) \equal{} (b) (0) \equal{} 0 \Leftrightarrow (A^T y x^T) A^T \equal{} 0
\]
If we consider $ A^T y x^T$ to be a linear transformation, we find that it must be equivalent to the zero matrix, as there exists one and only one matrix which takes $ A^T$ to zero, assuming that $ A$ is not the zero matrix itself (in which case, the problem would be trivial... zero vectors are orthogonal to literally life itself  :P ). Thus,
\[ A^T y x^T \equal{} 0 \Leftrightarrow y x^T \equal{} 0 \Leftrightarrow x y^T \equal{} 0,
\]
by the same argument and we have what we desired. $ \blacksquare$

I understand that the proof is extremely sloppy (because of the $ 0$ on the right hand side... for example going from $ y x^T \equal{} 0$ to $ x y^T \equal{} 0$... whenever I said $ 0,$ i meant that all entries in the appropriate matrix on the right hand side was $ 0.$).

A much cleaner approach: $ x^Tb\equal{}x^T(Ay)\equal{}(x^TA)y\equal{}(A^Tx)^Ty\equal{}0$.

I would actually prefer writing the row space as $ \{x^TA\}$, which would shorten the above by another step.

In linear algebra you'll find it useful to consider the bilinear form $ y^T Ax$ for column vectors $ x, y$ (or the equivalent for row vectors).  $ y^T A$ belongs to the row space and $ Ax \equal{} 0$, which is equivalent to jmerry's solution.

When $ A$ is positive-definite and symmetric, the above defines a (real) inner product.  So the above result can be interpreted as meaning that $ y^T$ (an arbitrary vector) and $ x$ (a vector in the nullspace) are orthogonal with respect to this product.

[quote="jmerry"]A much cleaner approach: $ x^Tb = x^T(Ay) = (x^TA)y = (A^Tx)^Ty = 0$.

I would actually prefer writing the row space as $ \{x^TA\}$, which would shorten the above by another step.[/quote]

Do you mean

\[ x^T b = x^T (A^T y) = (Ax)^T y = 0?
\]

Because $ A^T y = b,$ not $ Ay = b.$

Also, to t0r, could you explain what bilinear form is? Also, I don't understand the second paragraph of your last post.  :maybe:

Sorry, I typed that post on a computer with IE so I couldn't open up tabs to give the relevant links.  See [url=http://en.wikipedia.org/wiki/Bilinear_form]bilinear form[/url], [url=http://en.wikipedia.org/wiki/Positive-definite_matrix]positive-definite[/url], and [url=http://en.wikipedia.org/wiki/Inner_product_space]inner product[/url].

I had another question... suppose we have $ Ax \equal{} Bx$ where $ A,B$ are matrices and $ x$ is a vector. Does this imply that $ A \equal{} B$? I think the answer is no, and here is why (correct me if I'm wrong):
\[ Ax \equal{} Bx \Leftrightarrow x \equal{} A^{ \minus{} 1} B x \equal{} x
\]
Let $ A^{ \minus{} 1} B \equal{} C.$ So, we have $ Cx \equal{} x.$ So assuming that $ C$ has an eigenvalue of $ 1,$ then $ C$ doesn't have to be the identity? By the way, this is an obvious way of proving (I think) that the identity matrix must have an eigenvalue of $ 1$ (it seems pretty obvious, and it is, but it's one of the few theorems I've proven in my 4 or so days of learning linear algebra  :P ).

[quote="t0rajir0u"]All it implies is that $ x$ is in the nullspace of $ A \minus{} B$.  Note that you didn't specify whether $ A, B$ were invertible or even square.  You're playing a little fast and loose with dimensions and invertibility - be more careful.

As for the identity matrix, $ 1$ is its only eigenvalue, which is somewhat obvious from its definition.[/quote]

Thanks  :) I'll keep that in mind studying more linear algebra. However, should $ A$ and $ B$ be invertible and square, this would work, right?

[quote="Karth"]If we consider $ A^T y x^T$ to be a linear transformation, we find that it must be equivalent to the zero matrix, as there exists one and only one matrix which takes $ A^T$ to zero, assuming that $ A$ is not the zero matrix itself (in which case, the problem would be trivial... zero vectors are orthogonal to literally life itself  :P ). [/quote]
This is plainly not true, and you need to be much more careful here.  First of all, the ring of matrices has many zero divisors; in other words, many pairs of nonzero matrices $ A, B$ exist such that $ AB \equal{} 0$.  (Examples of such matrices include powers of [url=http://en.wikipedia.org/wiki/Nilpotent_matrix]nilpotent matrices[/url]).  Second of all, linear transformations usually aren't applied to matrices; we think of them as matrices and apply them to vectors.  (The statement you cite is, of course, not true for vectors either.)  Matrix multiplication is a composition of linear transformations (from this perspective), not the application of a linear transformation to a vector.  We can consider matrix multiplication as the application of a linear transformation to a set of vectors, but the point is we need to be more careful not to use theorems from one situation in another without making sure they apply.

[quote="Karth"]Thanks  :) I'll keep that in mind studying more linear algebra. However, should $ A$ and $ B$ be invertible and square, this would work, right?[/quote]
Yes; then $ C$ is any matrix with eigenvalue $ 1$ and $ x$ is a corresponding eigenvector.  Keep in mind that for most of the general theorems you'll prove about matrices, this is a small subcase.

It looks like you need to do some more computations with matrices to get a feel for the ways in which matrix arithmetic differs from ordinary arithmetic.

[b]Exercises:[/b]

- $ AB \equal{} I$.  Is it true that $ BA \equal{} I$?
- Is it true that $ (A \plus{} B)(A \minus{} B) \equal{} A^2 \minus{} B^2$?
- Find a $ 2 \times 2$ matrix $ A$, not equal to zero, such that $ A^2 \equal{} 0$.

Your correction of me was correct- I transposed A, and didn't check it properly.


Is anybodi here going to MathPath?

I've already posted this is the Getting Started forum but I'm wondering if I might get more responses here.

I'm almost done the quiz, but I'm not sure if i'll do it anyway: it's so expensive!

I made it into MathPath.
I am really glad that I did! :lol:

Not if you make it into the top 50 :D ...

[quote="aznness"]what's mathpath?[/quote]

http://mathpath.org

See also this year's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22358]AoPS thread on MathPath[/url]. Several AoPSers are going, it appears.

I would go, but I'm not paying $3500 to go.

How is everyone paying for it? It says fundraising, but who can you fundraise from and what can you do, etc? Thanks in advance.

You can also do well at Nats or on the Qualifying Quiz to get a scholarship...

I think they took away the scholarship for high scorers on the quiz this year :(

Really? I didn't see anything on their site about that...

I think so... and you won't have to compete with a lot of returning students because they might not want to put in the effort since they don't have to do it again.

How many problems do you have to do to get accepted? Is there a guaranteed number? I think it's five, but I'm not sure...

It says on the site that the number of seats  is unlimited, so if you qualify, you're accepted. So if I do five, and I know they're all right, I'm automatically in?

It's recommended that you do all the qualifying quiz problems that you can - if you only do five and happen to mistakenly think that they're all right while in reality you missed part of a question then you likely won't be admitted, best not to cut it so close.

The scholarship plan is different this year from last year.

For those who think MathPath is just another summer camp, I would like to mention that six of the twelve students in this year's National Countdown Round attended MathPath last summer.  Applications are due in about a week.

As it turns out, none of the students in the National Masters Round attended MathPath.  One has to wonder what a MathPathian might have been able to do with this year's Masters topic.

Yeah, at first I thought Patricia Li was in it and that she would win despite being fourth written. BTW, both the girl semifinalists went to MathPath, so if any girls want to make it to Countdown next year, you know where to go ;)

Note that a lot of MathPath alumni who made Countdown didn't even qualify for Nationals last year, so there's hope for everyone.

A question and an answer concerning how to write a response have been deleted.  You folks should take notice that MathCamp either disqualified or was considering disqualification of a number of applicants for violating their rules.  I am not aware that MathPath has taken such drastic action.

The question that was asked and answered seems innocent, but it goes to the heart of academic integrity.  When MathPath says no help writing your answers, they mean NO HELP!  It is not up to you, the student applicant, to decide that this little bit of help would be OK.  If it didn't matter, you needn't have asked.  If it did matter, you shouldn't have asked.

Any further discussion on the matter will be forwarded to my good friend, Dr. George Thomas, Founder of MathPath.