So the mathcounts national competition highlight video from last year has finally come out!
However, the video is kind of small....does anyone know any way to enlarge or is it possible to save the movie on my desktop?

ich mochte link(i would like a link to it :))

[quote="karatemagic7"]Is there really a video? Where can I find it, please?[/quote]

http://mathcounts.org/NETCOMMUNITY/Page.aspx?pid=535&srcid=316

do they have a video of the countdown round, i.e. the whole thing?
also, was it televised?

[quote="kl2836"][quote="abacadaea"]its not workings...its just this icon flashing...[/quote]

its probably because you didnt install flv or whatever file mathcounts used for the video[/quote]

it doesnt work on my computer at home or one at school

Watch for the people in the pink shirts...[/url]


absolute value
algebra
system of equations

Let $ a$, $ b$, $ c$ be three real numbers satisfying that \[ \left\{\begin{array}{c c c} \left(a\plus{}b\right)\left(b\plus{}c\right)\left(c\plus{}a\right)&\equal{}&abc\\ \left(a^3\plus{}b^3\right)\left(b^3\plus{}c^3\right)\left(c^3\plus{}a^3\right)&\equal{}&a^3b^3c^3\end{array}\right.\] Prove that $ abc\equal{}0$.

Lemma
We will prove that $ \left|x\plus{}\dfrac{1}{x}\minus{}1\right| \ge 1$ for all $ x \ne 0$.
If $ x>0$ then by AM-GM
$ x \plus{} \dfrac{1}{x} \ge 2$ so
$ x\plus{} \dfrac{1}{x} \minus{}1 \ge 1$, implying
$ \left|x\plus{} \dfrac{1}{x} \minus{}1\right| \ge 1$ with equality when $ x\equal{}1$.
If $ x<0$ then by AM-GM
$ |x| \plus{} \dfrac{1}{|x|} \ge 2$
so
$ x \plus{} \dfrac{1}{x}\equal{}\minus{}|x|\minus{}\dfrac{1}{|x|} \le \minus{}2$.
$ x \plus{} \dfrac{1}{x}\minus{}1 \le \minus{}3$
so
$ \left|x \plus{} \dfrac{1}{x}\minus{}1\right| \ge 3 > 1$.
--end lemma--
Assume for the sake of contradiction that $ abc \ne 0$, so $ a,b,c \ne 0$. Then we can divide the second equation by the first to get
$ (a^2\minus{}ab\plus{}b^2)(b^2\minus{}bc\plus{}c^2)(c^2\minus{}ca\plus{}a^2) \equal{} a^2b^2c^2$
and divide by $ a^2b^2c^2$ to get
$ \left(\dfrac{a}{b}\plus{}\dfrac{b}{a}\minus{}1\right)\left(\dfrac{b}{c}\plus{}\dfrac{c}{b}\minus{}1\right)\left(\dfrac{c}{a}\plus{}\dfrac{a}{c}\minus{}1\right) \equal{} 1$
Let $ x\equal{}\dfrac{a}{b}, y\equal{}\dfrac{b}{c}, z\equal{}\dfrac{c}{a}$. Then we have
$ \left(x\plus{}\dfrac{1}{x}\minus{}1\right)\left(y\plus{}\dfrac{1}{y}\minus{}1\right)\left(z\plus{}\dfrac{1}{z}\minus{}1\right) \equal{} 1$
and taking the absolute value we get
$ \left|x\plus{}\dfrac{1}{x}\minus{}1\right|\left|y\plus{}\dfrac{1}{y}\minus{}1\right|\left|z\plus{}\dfrac{1}{z}\minus{}1\right| \equal{} 1$
By the lemma $ \left|x\plus{}\dfrac{1}{x}\minus{}1\right|\left|y\plus{}\dfrac{1}{y}\minus{}1\right|\left|z\plus{}\dfrac{1}{z}\minus{}1\right| \ge 1 \cdot 1 \cdot 1 \equal{} 1$, and because we have equality
$ x\equal{}y\equal{}z\equal{}1$.
Thus $ \dfrac{a}{b}\equal{}\dfrac{b}{c}\equal{}\dfrac{c}{a}\equal{}1$ so $ a\equal{}b\equal{}c$.
Substituting into the first given equation $ \left(a\plus{}b\right)\left(b\plus{}c\right)\left(c\plus{}a\right)\equal{}abc$ we get
$ (a\plus{}a)(a\plus{}a)(a\plus{}a) \equal{} a^3$
$ 8a^3\equal{}a^3$
$ 8\equal{}1$, contradiction.

[hide="A slick solution"]We have: $ abc(a^2 \minus{} ab \plus{} b^2)(b^2 \minus{} bc \plus{} c^2)(c^2 \minus{} ac \plus{} c^2) \equal{} a^3b^3c^3$, which means either $ abc \equal{} 0$ or $ (a^2\minus{}ab\plus{}b^2)(b^2\minus{}bc\plus{}c^2)(c^2\minus{}ac\plus{}c^2)\equal{}a^2b^2c^2$
Applying AM-GM we get:
$ a^2 \plus{} b^2 \minus{} ab\geq 2|ab| \minus{} ab\geq |ab|\geq 0$
Similarly, we also have: $ b^2 \plus{} c^2 \minus{} bc\geq |bc|\geq 0$
                                $ c^2 \minus{} ac \plus{} a^2\geq |ac|\geq 0$
Therefore, $ (a^2 \plus{} b^2 \minus{} ab)(b^2 \plus{} c^2 \minus{} bc)(c^2 \plus{} a^2 \minus{} ac)\geq a^2b^2c^2$
The equality occurs when $ a \equal{} b \equal{} c$ or $ \minus{} a \equal{} \minus{} b \equal{} \minus{} c$.
Plugging this condition into the 1st equation, we easily show that $ a \equal{} b \equal{} c \equal{} 0$.It concludes that $ abc \equal{} 0( Q.E.D)$[/hide]

[hide="@Mij"]: Your last claim should be $ a^3 \equal{} 0$, not a contradiction. :wink: [/hide]

[quote="ghjk"][hide="@Mij"]: Your last claim should be $ a^3 \equal{} 0$, not a contradiction. :wink: [/hide][/quote]
He assumed for the sake of contradiction that $ a \neq 0$ ;) Then $ 8a^3 \equal{} a^3 \iff (8\minus{}1)a^3 \equal{} 0 \iff 8\minus{}1 \equal{} 0 \iff 8 \equal{} 1$ is a way to obtain a contradiction!

[hide="Solution"]

If any of $a+b, b+c, a+c$ are equal to $0$, then $abc=0$ as desired. If not, we can divide the equations to get

$(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)=a^2b^2c^2$.

Note $(a-b)^2 \geq 0 \implies a^2+b^2-ab \geq ab \implies \sum_{cyc}{a^2-ab+b^2} \geq a^2b^2c^2$. Equality occurs only when $a=b=c$, implying $8abc=abc \implies abc=0$. [/hide]

@above:
That's an incomplete solution, since you have assumed $ a,b,c $ to be positive.

Oops! . Sorry , I read it wrong .
Anyways , i think , dividing the two relation makes the problem much much simple .
As , $x^2+y^2-xy \ge xy \forall x,y \in R$ .

Hmm so while waiting in line during Black-Friday-at-eight-o-clock shopping I somehow managed to solve this one.

Suppose that $abc\neq 0$.  Dividing both sides of the first equation by $abc$ gives \[\left(1+\dfrac ba\right)\left(1+\dfrac cb\right)\left(1+\dfrac ac\right)=1,\] and a similar result can be found by dividing $a^3b^3c^3$ from the second equation.  Let $x=\tfrac ba$, $y=\tfrac cb$, $z=\tfrac ac$.  Then it suffices to prove that no $x,y,z\in\mathbb{R}$ exist such that \[\begin{cases}(1+x)(1+y)(1+z)&=1,\\(1+x^3)(1+y^3)(1+z^3)&=1,\\xyz&=1.\end{cases}\]  Once again, suppose such $x,y,z$ exist.  Then \[\prod\dfrac{1+x^3}{x(1+x)}=\prod\dfrac{1-x+x^2}x=\prod\left(x+\dfrac 1x-1\right)=1.\]  If $x,y,z$ are all positive, then $\prod\left(x+\dfrac 1x-1\right)\geq 1^3=1$ by AM-GM.  Otherwise, WLOG let $x,y<0$.  Then $x+\dfrac 1x-1\leq -2-1=-3$ and similarly for $y$, so the LHS is greater than or equal to $(-3)\cdot (-3)\cdot 1=9.$  Thus, the only equality case occurs when $x=y=z=1$, but then $(1+x)(1+y)(1+z)=8$, contradiction.  Hence such $x,y,z$ do not exist, and the only way this can happen from our original substitutions occurs when at least one of $a,b,c$ equals zero, which is what we wanted.  $\blacksquare$


[quote="bpms"][quote="4everwise"]In quadrilateral $ABCD$, $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$, $AB=8$, $BD=10$, and $BC=6$. The length $CD$ may be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.[/quote]
I think you made a typo because if ABD is congruent to BCD then CD would be 8.[/quote]

Not at all.

It is ABD the angle it does not imply that AB=CD, so yeah.

[hide="Different Solution"]Extend BC and AD to meet at E. By AA, triangle ABD is similar to triangle DCE. Thus, triangle BDE is isosceles, $\implies EB=10$. Also, since the triangles are similar, $EC=\frac{10x}{8}=\frac{5x}{4}\implies EB=10=EC-6=\frac{5x}{4}-6\implies x=\boxed{\frac{64}{5}}$.[/hide]

[quote=supercomputer]How do you get the right diagram?
Because when I drew it, I ended up with AD<BC from which the solution is impossible.[/quote]

[quote=Altheman][hide=hidden for length]This is a [i]very[/i] nice problem!!!

1) Let $AB$ intersect $CD$ at $X$

2) Let the parallel to $AD$ through $B$ intersect $CD$ at $Y$.

$DY=8$. $AY=10$. $AX=DX$

$\triangle BCX\sim\triangle AYX$
$\triangle XBY\sim\triangle XAD$

$BX=12$, $XY=12$, $CY=\frac{24}{5}$, $CD=\frac{64}{5}$.[/hide][/quote]
I understand the intuition for step 1 (since point $X$ lets us make use of isosceles triangles, which are suggested by the equal angles $BAD$ and $ADC$), but what was the intuition to make the isosceles trapezoid in step $2$? When I was working on this problem, I was able to do the first step but wasn't able to do the second one.

I have the exact same question. Maybe it's just something that is helpful when you don't really have any insights for a geometry problem?

Ugh basically a cleverly disguised but glaringly obvious isosceles triangle that took me way too long to find.

Noticing the isosceles triangle by adding $AB \cap CD = E$ makes this trivial.


vector
linear algebra
matrix
algorithm
superior algebra
superior algebra solved

Let Ax = y be an inhomogeneous system of equations. Show that if there exists a vector z in $C^n$ such that Az =y then there exists a vector w in $R^n$ such that Aw = y

Thanx for your help!

A clearer and more general version of the statement:

Suppose $F\subset E$, $A$ is an $m\times n$ matrix over $F$, $y$ is an $m\times k$ matrix over $F$, and there is an $n\times k$ matrix $v$ over $E$ such that $Av=y$. Then there is an $n\times k$ matrix $x$ over $F$ such that $Ax=y$.

Proof: By standard algebra, $E$ is a vector space over $F$. Consider the subspace spanned by 1 and the elements of $v$. Choose a basis $1,v_1,v_2,\dots,v_r$ for this subspace. Write $v=x\cdot1+x_1\cdot v_1+\cdots+x_r\cdot v_r$ with $x,x_1,\dots,x_r$ matrices over $F$. Then $Av=Ax\cdot1+Ax_1\cdot v_1+\cdots+Ax_r\cdot v_r=y\cdot1$. Since $1,v_1,\dots,v_r$ are linearly independent, $Ax=y$ and we are done.

The problem you want is a specialization of this; $k=1$, $F=\mathbb{R}$, $E=\mathbb{C}$.
One other argument you could use: the algorithm for finding solutions to linear systems and determining whether solutions exist uses only field operations.


integration
real analysis
real analysis unsolved

$ I\equal{}\int_0^{\frac{\pi}{4}} ln(1\plus{}tanx) dx$


Consider the 16 points in the first quadrant of a Cartesian coordinate plane with integer coordinates less than or equal to 4. How many squares can be formed using four of these points as vertices?

There are $ 16$ $ 1 \times 1$ squares, $ 9$ $ 2 \times 2$ squares, $ 4$ $ 3 \times 3$, and $ 1$ $ 4 \times 4$ square.  Thus, the answer is $ 16 \plus{} 9 \plus{} 4 \plus{} 1$, or $ \boxed{30}$.

I would space that more at first I thought u said 161x1 squares!

Also the sum of the first n squares is n(n+1)(2n+1)/6 so plugging in 4 gets 4x5x9/6=30

sorry, but its 20-- the maximum size is a 3 by 3, and you forgot sideways squares -- (2,1),(1,2),(2,3),(3,2) is one.


counting
combinatorics
IMO Shortlist
Recurrence
Sequences

A biologist watches a chameleon. The chameleon catches flies and rests after each catch. The biologist notices that:
[list=1][*]the first fly is caught after a resting period of one minute;
[*]the resting period before catching the $2m^\text{th}$ fly is the same as the resting period before catching the $m^\text{th}$ fly and one minute shorter than the resting period before catching the $(2m+1)^\text{th}$ fly;
[*]when the chameleon stops resting, he catches a fly instantly.[/list]

[list=a][*]How many flies were caught by the chameleon before his first resting period of $9$ minutes in a row?
[*]After how many minutes will the chameleon catch his $98^\text{th}$ fly?
[*]How many flies were caught by the chameleon after 1999 minutes have passed?[/list]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[hide="First two parts"]
Let $ f(x)$ be the function that gives the resting period (in minutes) before catching fly $ x$. We can see that this function is the sum of digits of $ x$ in binary (the "population count" of $ x$).

- How many flies were caught by the chameleon before his first resting period of 9 minutes in a row ?

The first resting period of 9 minutes in a row will therefore happen at the first number with 9 ones, or before $ 111111111_b \equal{} 511$. Therefore, he would have caught 510 flies.

- After how many minutes will the chameleon catch his 98th fly ?

To count the sum of all resting periods until the nth fly, simply count the number of ones in the binary representations. Before 64 we have $ 6 \times 2^5 \equal{} 192$ minutes. Between 64 and 96 we have $ 5 \times 2^4 \plus{} 34 \equal{} 114$. Finally, $ f(97) \plus{} f(98) \equal{} 6$, from what we have that the total time is $ 192 \plus{} 114 \plus{} 6 \equal{} 312$ minutes.
[/hide]

I believe the third one is only a matter of computation, but I don't have time to do it now...

this  one is a real deadly question . actually it is from a bulgarian maths olympiad.

well the  answers are

a)510

b)312

c)462

(Sadly, I missed the binary expansion idea, so I'm only going to post a sketch of my solution.)

Let $R_n$ denote the length of the resting period before the chameleon catches the $n$-th fly. Note that when $n$ is a power of $2$, then $R_n=1$.

    Using induction we can see that for every positive integer $k$, the largest value of $R_n$ for $2^k<=n<2^{k+1}$ is $k+1$, which happens when $n=2^{k+1}-1$. So the first $n$ such that $R_n=9$ is when $n=2^9-1=511$, and it follows that $510$ flies were caught by the chameleon before his first resting period of $9$ minutes in a row.

    For every positive integer $k$, the values of $R_{2^k}$, $R_{2^k+1}$, $\dots$, $R_{2^k+2^{k-1}-1}$ are $R_{2^{k-1}}$, $R_{2^{k-1}+1}$, $\dots$, $R_{2^k-1}$, respectively. We can see that this is true by induction. Then we can note that $$\sum_{i=2^k}^{2^{k+1}-1}R_i=2\sum_{i=2^{k-1}}^{2^k}R_i+2^{k-1},$$ so if we let $A_k$ be the value of the LHS, we get $A_k=2A_{k-1}+2^{k-1}$. By induction again, since $A_1=3$ is our base case, we can see that $$A_k=2^k+k2^{k-1}=2^{k-1}(k+2).$$ Thus we can easily find the values of $A_n$ for the first few $n$. We now have
    \begin{align*}
        \sum_{i=1}^{98} R_i&=\sum_{i=1}^{63}R_i+\sum_{i=64}^{95}R_i+(R_{96}+R_{97}+R_{98}) \\
        &=(A_0+A_1+\dots+A_5)+A_5+8,
    \end{align*}
    which is $192+112+8=312$. So the chameleon catches the $98$-th fly after $312$ minutes. The rest of the problem is just computation.

Let the number of minutes in the $x$-th resting period be $f(x)$. It is trivial by induction to see that $f(x)$ is the popcount(number of $1$'s in the binary representation) of $x$. Denote popcount by $\text{pc}$.

[b]Part a.[/b] $f(x)=9$ when the popcount of $x$ is $9$, which first happens at $x=2^9-1=511$. So the answer is $\boxed{510}$.

[b]Part b.[/b] This is just $$\left(\sum_{i=1}^{98}\text{pc}(i)\right)=\left(\sum_{i=0}^{95}\text{pc}(i)\right)+2+3+3=3\cdot\left(\sum_{i=0}^{31}\text{pc}(i)\right)+32+32+2+3+3=3\cdot(5\cdot16)+32+32+2+3+3=\boxed{312}.$$

[b]Part c.[/b] Let $g(x)=f(1)+f(2)+\dots+f(x)$. Note that $g(2^k-1)=\sum_{i=0}^{2^k-1}\text{pc}(i)=k\cdot 2^{k-1}$. So $g(511)=2304$. Also, note that $f(8a)+f(8a+1)+\dots+f(8a+7)=8f(a)+12$. So, we can make a table of popcounts right below $64$:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$x$ & 63 & 62 & 61 & 60 & 59 & 58 \\ \hline
$f(x)$ & 6 & 5 & 5 & 4 & 5 & 4 \\ \hline
\end{tabular}
So $f(58\cdot8)+f(58\cdot8+1)+\dots+f(511)=304$. So $g(58\cdot8-1)=2000$, and our answer is $58\cdot8-2=\boxed{462}$.

What's the point of this question

The waiting time before eating the $i$th fly is $s_2(i)$ where $s_p(i)$ is the sum of the digits of $i$ in base $p$. Thus, the first time he waits $9$ minutes is just before the eating the $2^9-1$th fly, or after his $\boxed{510}$th fly.

Clearly, the answer is $s_2(1)+s_2(2)+\dots + s_2(98)$. One appears on the units digit $49$ times, the $2$s digit $49$ times, the $4$s digit $48$ times, the $8$s digit $48$ times, the $16$s digit $48$ times, the $32$s digit $35$ times, the $64$s digit $35$ times. Thus, the answer is $\boxed{312}$ minutes.

Note that the number of minutes passed after $2^k-1$ flies is exactly $k\cdot 2^{k-1}$. After the $511$st fly, $2304$ minutes passed. Then, bash the rest. After the $111001111_2$th fly but before the $511$st fly, a total of $304$ minutes were waited, thus at the $1999$th minute, $462$ flies were caught.

"no it's on the aime
silly
isl stole from aime then ! ...ridiculously computational)" - jatloe cj823957y19208500239758927o30472305023874

Anyways, we note that for every 2m from m, multiplying by 2 in binary does not change the # of 1s. For 2m+1, we add a 1, and we know that the last digit was 0, so we add 1 to the # of 1s in binary representation, etc. Now for 9 minutes we have 9 ones, or 2^8+...+1=511th catch, or 510 flies were caught (almost sillied this :P)
For part b, we have 1 fly in 1 minute, 2 flies in 1 minute, etc., or the sum of the #1s in the numbers in binary representation from 1 to 98. 1 appears at the end 49 times (49 odd #s), 2nd from the back digit 49 times (is 2 or 3 mod 4), 3rd from the back digit 48 times (8,9,10,11,12,13,14,15 mod 16), etc, summing them up until mod 128, we get 312.
For part c, we know after the 511th fly 2304 minutes passed (we can bash by doing the same mod thing), so now just take reasonable guesses by subtracting. Off the top of my head, I would say we would expect each fly to take around 7 minutes so subtract (2304-1999)/7; 44 days, and we're already very reasonably close. We get the answer is 462.

I'd like to say that I didn't expect so much calculation from IMO

Let $f(n)$ be the the resting period before catching the n-th fly.
$f(1)=1$
$f(n>1)=
\begin{cases}
f(n/2) & \text{if }n\equiv0\pmod2\\
f(n-1)+1 & \text{otherwise.}
\end{cases}$

Lemma: $f(n)$ equals the number of 1s in the base 2 representation of n
Proof:
I will use induction
Base $n=1$ is true
Assume we have proven the result for all $k<n$
Now I will prove it for n
Case 1: $n\equiv0\pmod2$
$n$ and $n/2$ have the same amount of zeroes in base 2 and $f(n)=f(n/2)$. Done
Case 2:n\equiv1\pmod2
$n$ has one more one than $n-1$ in base 2 and $f(n)=f(n-1)+1$. Done

a. The answer is one less than the first number with 9 1s in base 2
Answer=$111111111_{(2)}-1_{(2)}=510$

b. The answer is $\sum_{i=1}^{98}f(i)$
Let $s(n)=\sum_{i=1}^{2^n}f(i)$
If we consider the number $k\in[1;2^{n-1}-1]$ it has one less one than $k+2^{n-1}$
and if we consider $k=2^{n-1}$ it has the same amount of ones as $k+2^{n-1}$
With this information we can say that $s(n)=2s(n-1)+2^{n-1}-1$
With induction we prove that $s(n)=n2^{n-1}+1$ for $n\geq0$
$\sum_{i=1}^{98}f(i)=s(6)+\sum_{i=65}^{98}f(i)=s(6)+34+\sum_{i=1}^{34}f(i)=s(6)+34+s(5)+\sum_{i=33}^{34}f(i)=s(6)+34+s(5)+2+\sum_{i=1}^{2}f(i)=s(6)+34+s(5)+2+s(1)=$
$=6\cdot2^{5}+1+34+5\cdot2^{4}+1+2+2=312$
$\implies$ Answer$=312$

c. The answer is the smallest n such that $\sum_{i=1}^{n}f(i)\geq1999$
Using $s(8)=1025$ and $s(9)=2305$ I estimate that $n\in[256;512]$
$\sum_{i=1}^{n}f(i)=s(8)+\sum_{i=257}^{n}f(i)=n+769+\sum_{i=1}^{n-256}f(i)$
Case 1: $n-256<128\implies n+769+\sum_{i=1}^{n-256}f(i)<384+769+s(7)=1602$ contradiction
$\implies n\in[384,512]$
$769+n+\sum_{i=1}^{n-256}f(i)=n+769+s(7)+\sum_{i=129}^{n-256}f(i)=2n+834+\sum_{i=1}^{n-384}f(i)$
Case 1: $n-384<64\implies 2n+834+\sum_{i=1}^{n-384}f(i)<2\cdot448+834+s(6)=1923$ contradition
$\implies n\in[448;512]$
Suppose $n=448$ then 1923 minutes would have passed
Manually checking for $449,450,451,...462$ by adding $f(449),f(450),f(451),...f(462)$ to 1923 we get that
$\sum_{i=1}^{461}f(i)=1994$ and $\sum_{i=1}^{462}f(i)=2000$
$\implies$ Answer$=462$

Note: If $2^{x-1}\leq n< 2^x$ then $\sum_{i=1}^{n}f(i)=s(x-1)+\sum_{i=2^{x-1}+1}^{n}f(i)=s(x-1)+n-2^{x-1}+\sum_{i=1}^{n-2^{x-1}}f(i)$


Έστω τρίγωνο ABC και το μέσο D της BC. Αν τα σημεία Μ,Ν των ΑΒ,AC αντίστοιχα (διαφορετικά από τα μέσα των πλευρών αυτών) είναι τέτοια ώστε $ AM^2 \plus{} AN^2 \equal{} BM^2 \plus{} CN^2$ και οι γωνίες $ \angle{BAC}$ και $ \angle{MDN}$  είναι ίσες, να υπολογιστεί η γωνία Α.

Μετά από μεγάλη και ωραία συζήτηση με τον φίλο μου Ηλία (Athinaios) παρουσιάζω σύντομα τη λύση:

Εστω κατ' αρχάς $ E,Z$ τα μέσα των $ AB, AC$ αντίστοιχα.
H σχέση γράφεται:
$ AM^2 \minus{} BM^2 \equal{} CN^2 \minus{} AN^2\iff 2AB*ME \equal{} 2AC*ZN\iff AB*ME \equal{} AC*ZN$ (η σχέση αυτή βγαίνει από διαφορά τετραγώνων ή από 2ο θεώρημα διαμέσων για εκφυλισμένο τρίγωνο...)
Τώρα $ \frac {EM}{ZN} \equal{} \frac {AC}{AB} \equal{} \frac {ED}{DZ}$.
Επίσης $ \angle{MDN} \equal{} A \equal{} EDZ\iff EDM \equal{} ZDN$

Αρα απ' τα τρίγωνα $ EDM,ZDN$ προκύπτει (νόμος ημιτόνων) $ \sin{ZND} \equal{} \sin{DME}$.
Αρα είτε οι γωνίες είναι παραπλρωματικές, στην οποία περίπτωση βγαίνει ότι $ M \equal{} E, N \equal{} Z$ (atopo) είτε είναι ίσες στην οποία περίπτωση $ \angle{MED} \equal{} ZND$ κι άρα το $ AEDN$ εγγράψιμο κι άρα $ A \equal{} 90$


I have no background in physics.
I haven't even taken algebra 2 yet.
I went to my school's physics club and am being taught elementary topics in calculus.
I want to learn outside the club as well since no handouts or assigned readings are being handed out during the club.

If you were me, where would you start?
What books would be good for someone with no background and little mathematical knowledge?
What resources would you recommend?

given that you have not taken algebra 2 yet, I doubt physics is any use at the moment.
mathematics is the basis of physics.
no math, no physics

so improve on your math

Well, it depends on how math-intensive you want the Physics to be...

From my experience with Physics, I would HIGHLY suggest learning techniques for algebraic manipulation and about vectors.  With math-based Physics problems, solving and setting up equations is very important -- that's where the manipulation of expressions comes in.  Since MANY quantities in Physics can be represented as vectors (vs. scalars), operations involving vectors can be very useful.  (And then, one can even apply Calculus to Physics!  But I'd suggest learning algebraic manipulation and techniques for solving/rearranging equations before even learning Calculus...)

However, it *is* possible to learn about Physics concepts without using much math -- it's just going to involve a different way of approaching things, probably a much more intuitive sense of things.

So, I would first start out making sure that I had a solid base in understanding elementary functions, how to manipulate them, how to solve functions involving them, and what relationships they imply between variables.  Basically, I would suggest learning the techniques typically introduced in Algebra 2 -- which can be self-taught, given enough effort and the right materials.  Start out looking through an Algebra 2 book and then look at a Physics book.


linear algebra
matrix
vector
algebra
polynomial
linear algebra unsolved

Prove that a complex matrix is similar to a real matrix if and only if $A$ is similar to $A^{*}$.

So, in particular, every real matrix is normal? This is obviously not true.. :?

Is $A^*$ the conjugate transpose? Because if it is, there are still counterexamples. Take $A=\begin{pmatrix}0&0\\1&0\end{pmatrix}$. It's similar to its transpose (which is also its conjugate transpose, since it's real), but it isn't normal.

Perhaps you meant unitarily similar instead of simply similar? :?

Grobber, are you sure about your example? I don't have the time now to do the computations, but the book is a very prestigious one... :?

What computations? :) The matrix is similar to its transpose by multiplication to the left and right with $\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Surely, you can see immediately that $AA^t\ne A^tA$.

harazi,excuse for this silly question,but is a normal matrix? I want only the definition. ;)

The matrix, which commutes with its adjoint, is called normal. As I remember, the matrix over $\mathbb C$ is normal iff it has an orthonormal eigenbasis.

Well, If A is normal then $A$ and $A^t$ comute.If $A$ is similar with $A^*$ then $rank(A)=rank(A^*)$ If $A$ is singular,well the statement is clear,because $rank(A^*)=0 or 1$ but $0$ it can't be.If $det(A)\neq0$..... ;)

Cezar, it is not the romanian adjoint. It is defined as $ A^{*}_{ij}=conj(a_{ji}$ where $conj(x)$ is the conjugate of x.

I thought I was mistaken,cause I did not use that $A$ and $A^t$ comute.Why it is called adjoint? :huh:

Grobber, excuse me, I must have been drunk when writing this yesterday evening. I am terribly sorry. I will edit the statement. What a shame... :blush:

I do not know why, but it is called so. Adjoint matrix (or adjoint operator in a space with scalar product) is defined by equality $(Ax,y)=(x,A^{*}y)$ for any vectors $x,\,y$. It is equivalent to $a_{ij}=\overline{a_{ji}}$.

:rotfl: So it had absolutely nothing to do with normality..

Anyway, now it's easy. Since the Jordan forms of $A$ and $A^*$ are the same, it means that the Jordan cells of $A$ can be arranged in pairs of cojugates, and it suffices to prove that a matrix with two conjugate Jordan cells (i.e. a matrix with minimal polynomial $m(x)=(x-\lambda)^k(x-\bar\lambda)^k$ with non-real $\lambda$) is similar to a real matrix. We can take this real matrix to be the companion of the real polynomial $m(x)$.

Always Jordan matrix... Anyway, I think we had to show that I'm not normal to write such stupidities. This could be a connection with the normal matrix. Except that, I don't see any connection either. :rotfl:


limit
function
real analysis
real analysis unsolved

i) let $C$ be a subset of $\mathbb{R}, |C| \neq 0$, and $\{x_n\}_n$ a sequence of real numbers with the following properties:

 a) $\displaystyle \lim_{n\to \infty} |x_n-y| $ exists and it's finite for every $y\in C$;

 b) the limit for every convergent subsequence of $x_n$ is in $C$

Prove that $x_n$ is convergent.

ii) Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous and increasing function with the property that $f(x)=0$ has at most one solution.
Let $\{x_n\}_n$ be a sequence s.t. $x_{n+1}+c_nf(x_{n+1})=x_n$, for every $n\geq 0$, where $x_0 \in \mathbb{R}$  and $c_n>0$ for every $n \geq 0$.
Prove that $\{x_n\}$ is convergent.

For (i):

Assume that xn doesn't tend to a certain y\in C. Then for any t>0, from a starting nt onwards, the terms of (xn)n are all in one of the intervals (y-c-t,y-c+t) and (y+c-t,y+c+t) (infinitely many in each of the intervals), and if z\in C and z<>y it's easy to choose t so small that |xn-z| has no limit, so y is the only element in C, so, from the second condition, we get that xn->y, contradiction, so our first assumption was false, so xn->y for all y\in C, so C has only one element and this element is the limit of xn.


Let $\lambda$ be an eigenvalue of the real matrix $A$.

(a) Prove that if $\lambda$ is simple then $y^T x \neq 0$ for every right/left eigenvectors $y^T,x$.

(b) Prove that if $A$ if diagonalizable, then there are right/left eigenvectors $y^T,x$ such that $y^T x \neq 0$.


Let $ a,b,c$ be three sides of a triangle, show that
  \[ \frac{a}{b\plus{}c}(\frac{a\minus{}b\minus{}c}{a\plus{}b\plus{}c})^2\plus{}\frac{b}{c\plus{}a}(\frac{b\minus{}c\minus{}a}{a\plus{}b\plus{}c})^2\plus{}\frac{c}{a\plus{}b}(\frac{c\minus{}a\minus{}b}{a\plus{}b\plus{}c})^2\leq\frac{1}{6}.\]

[quote="miwalin"]Let $ a,b,c$ be three sides of a triangle, show that
\[ \frac {a}{b + c}(\frac {a - b - c}{a + b + c})^2 + \frac {b}{c + a}(\frac {b - c - a}{a + b + c})^2 + \frac {c}{a + b}(\frac {c - a - b}{a + b + c})^2\leq\frac {1}{6}.
\]
[/quote]

Let's $ a=y+z$, $ b=x+z$ , $ c=x+y$ then we will have:
$ \frac{y+z}{2x+y+z}\cdot \frac{x^2}{(x+y+z)^2}+\frac{x+z}{x+2y+z} \cdot \frac{y^2}{(x+y+z)^2}+\frac{x+y}{x+y+2z} \cdot \frac{z^2}{(x+y+z)^2} \le \frac{1}{6}$
or it equivolent to
$ \sum_{sym}{x^5}+11\sum_{sym}{x^4y}+23\sum_{sym}{x^3y^2}+24\sum_{sym}{x^3yz}+37\sum_{sym}{x^2y^2z} \ge 6\sum_{sym}{x^4y}+30\sum_{sym}{x^3y^2}+18\sum_{sym}{x^3yz}+42\sum_{sym}{x^2y^2z}$
$ \leftrightarrow$
$ \sum_{sym}{x^5}+5\sum_{sym}{x^4y}+6\sum_{sym}{x^3yz} \ge 7\sum_{sym}{x^3y^2}+5\sum_{sym}{x^2y^2z}$
which is right becouse:
$ \sum_{sym}{x^5}+\sum_{sym}{x^3yz} \ge 2\sum_{sym}{x^{\frac{10}{3}}y^{\frac{5}{3}} \ge 2\sum_{sym}{x^3y^2}}$
and
$ 5\sum_{sym}{x^4y}+5\sum_{sym}{x^3yz} \ge 5\sum_{sym}{x^3y^2}+5\sum_{sym}{x^2y^2z}$

[quote="vuductuong"]I think Sayat use muirhead to prove  last two inequality[/quote]
More exactly?

[quote="miwalin"][quote="SaYaT"][quote="miwalin"]Let $ a,b,c$ be three sides of a triangle, show that
\[ \frac {a}{b + c}(\frac {a - b - c}{a + b + c})^2 + \frac {b}{c + a}(\frac {b - c - a}{a + b + c})^2 + \frac {c}{a + b}(\frac {c - a - b}{a + b + c})^2\leq\frac {1}{6}.
\]
[/quote]

Let's $ a = y + z$, $ b = x + z$ , $ c = x + y$ then we will have:
$ \frac {y + z}{2x + y + z}\cdot \frac {x^2}{(x + y + z)^2} + \frac {x + z}{x + 2y + z} \cdot \frac {y^2}{(x + y + z)^2} + \frac {x + y}{x + y + 2z} \cdot \frac {z^2}{(x + y + z)^2} \le \frac {1}{6}$
or it equivolent to
$ \sum_{sym}{x^5} + 11\sum_{sym}{x^4y} + 23\sum_{sym}{x^3y^2} + 24\sum_{sym}{x^3yz} + 37\sum_{sym}{x^2y^2z} \ge 6\sum_{sym}{x^4y} + 30\sum_{sym}{x^3y^2} + 18\sum_{sym}{x^3yz} + 42\sum_{sym}{x^2y^2z}$
$ \leftrightarrow$
$ \sum_{sym}{x^5} + 5\sum_{sym}{x^4y} + 6\sum_{sym}{x^3yz} \ge 7\sum_{sym}{x^3y^2} + 5\sum_{sym}{x^2y^2z}$
which is right becouse:
$ \sum_{sym}{x^5} + \sum_{sym}{x^3yz} \ge 2\sum_{sym}{x^{\frac {10}{3}}y^{\frac {5}{3}} \ge 2\sum_{sym}{x^3y^2}}$
and
$ 5\sum_{sym}{x^4y} + 5\sum_{sym}{x^3yz} \ge 5\sum_{sym}{x^3y^2} + 5\sum_{sym}{x^2y^2z}$[/quote]
Can you explain me last two inequalities?[/quote]

First one:
at first use  AM-GM inequality:
$ \sum_{sym}{x^5} + \sum_{sym}{x^3yz} \ge 2(x^5+y^5+z^5+3(xyz)^{\frac{5}{3}})$
then Schur inequality:
$ 2(x^5+y^5+z^5+3(xyz)^{\frac{5}{3}}) \ge 2\sum_{sym}{x^{\frac {10}{3}}y^{\frac {5}{3}}}$
and finally use Muirhead inequality:
$ 2\sum_{sym}{x^{\frac {10}{3}}y^{\frac {5}{3}} \ge 2\sum_{sym}{x^3y^2}}$(becouse $ (\frac{10}{3},\frac{5}{3},0) \succ (3,2,0)$)

Second one is directly from Muirhead inequality:
$ 5\sum_{sym}{x^4y} \ge 5\sum_{sym}{x^3y^2}$(becouse $ (4,1,0) \succ (3,2,0)$)
$ 5\sum_{sym}{x^3yz} \ge 5\sum_{sym}{x^2y^2z}$(becouse $ (3,1,1) \succ (2,2,1)$)


Let ABC be a triangle with $ cos B\plus{} cos C \le cos(C\minus{}B)$. Prove that;
$ R\plus{}r \le max(h_a,h_b,h_c)$

[quote="NguyenDungTN"]Let ABC be a triangle with $ cos B + cos C \le cos(C - B)$. Prove that;
$ R + r \le max(h_a,h_b,h_c)$[/quote]
[hide="Hint"] We have:
$ r=4Rsin{\frac{A}{2}}sin{\frac{B}{2}}sin{\frac{C}{2}} \Rightarrow R+r=R(1+4sin{\frac{A}{2}}sin{\frac{B}{2}}sin{\frac{C}{2}})=R(cos A+cos B+ cosC)$
We may assume that $ A\le B \le C \Rightarrow h_a \geh_b\ge h_c$ or $ max(h_a,h_b,h_c)=h_a$
Thus
$ R+r \le max(h_a,h_b,h_c) \Leftrightarrow R(cos A+cos B+cos C) \le h_a$
$ \Leftrightarrow R(cos A+cos B+cos C) \le csin B$
$ \Leftrightarrow R(cos A+cos B+cos C) \le 2Rsin C sin B$
$ \Leftrightarrow cos A+cos B+cos C \le cos (C-B)-cos (C+B)$
$ \Leftrightarrow cos B+cos C \le cos (C-B)$ (because $ cos(B+C)=-cos A$)
Which is clearly true!
[/hide]


Prove that
 [img]http://img225.imageshack.us/img225/9847/40d638fbc888f2981e06e82yl1.gif[/img]
When
[img]http://img233.imageshack.us/img233/33/59e1fb9969e7e1d2950d915yl1.gif[/img]

[img]http://img181.imageshack.us/img181/617/1c0e7372360204de65849c0rx5.gif[/img]

Nobody can help me with this test ? Don't you know someone that can resolve it ? I've thinked for 2 months , and not even my teacher couldn't resolve it   :(

[quote="KaSmIR"]Prove that
 $ \frac {a_1}{a_2} \plus{} \frac {a_2}{a_3} \plus{} ... \plus{} \frac {a_n}{a_1}\leq2n\left(m_a \minus{} m_h \plus{} \frac {1}{2}\right)$
When
\[ m_a \equal{} \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n}
\]
$ m_h \equal{} \frac {n}{\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ... \plus{} \frac {1}{a_n}}$
[/quote]
I think, it's wrong. Try $ n \equal{} 2,$ $ a_1 \equal{} \frac {1}{2}$ and $ a_2 \equal{} \frac {1}{3}.$ :wink:


Let ABC be an acute-angled triangle with $ \angle B \equal{} \angle C$. Let the circumcentre be O and the orthocentre be H. Prove that the centre of the circle BOH lies on the line AB. The circumcentre of a triangle is the centre of its circumcircle. The orthocentre of a triangle is the point where its three altitudes meet.

Let the circumcenter of $ \triangle{BOH}$ be $ O'$. Then $ \angle{BO'H}\equal{}2\angle{BOH}\equal{}2\angle{A}$. Hence $ \angle{O'BH}\equal{}90^o\minus{}\angle{A}$.

Since $ \angle{CBH}\equal{}\frac{1}{2}\angle{A}$ and $ \angle{ABC}\equal{}90^o \minus{} \frac{1}{2} \angle{A}$, so $ \angle{ABH}\equal{}90^o\minus{}\angle{A}$.

Therefore we have $ \angle{ABH}\equal{}\angle{O'BH}$, which gives that $ O'$ lies on $ AB$.

Let M the second intersection of the circumcircle of the triangle BHO with AB. We note that <MBO=MHO=A/2, where A is the measure of the <BAC. Let D the intersection of  AH and BC. We note that <BDH=BCA=90- A/2. Hence <MHB=180- A/2- (90-A/2)=90º, So MB is diameter of the circumcircle of BHO, and the circumcenter lies on AB.


This one's actually been bugging me for quite some time and I would like some closure to it...

"How many ways are there to add up six even, positive numbers (NOT including zero) to get a sum of 26?"

Be sure to show your work and enjoy!  :huh:

[quote="randomdragoon"][hide]$2a+2b+2c+2d+2e+2f = 26$
$a+b+c+d+e+f = 13$
$a' = a-1$, etc.
$a'+b'+c'+d'+e'+f' = 7$

Balls and urns argument ==>

$\binom{12}{5}=792$.

Not in the mood for coherent English sentences right now, maybe someone else will explains?

---===---

Basically, divide both sides by two, create a new variable for each variable that is one less than each variable, and you are left with the problem of how many nonnegative solutions there are to a+b+c+d+e+f=7, which has a well-known solution

EDIT: If order doesn't matter, then this solution is screwed[/hide][/quote]

yeah, your answer is WAAAYYYYY to large


function
vector
calculus
calculus computations

If $ F: \mathbb{R}^3 \to \mathbb{R}^3$ is a function such that $ \bigtriangledown \cdot F(x,y,z) \equal{} 0$, does there exist a $ G: \mathbb{R}^3 \to \mathbb{R}^3$ such that $ F \equal{} \bigtriangledown \times G(x,y,z)$?

Such $ \vec G$ is called a [url=http://en.wikipedia.org/wiki/Vector_potential]vector potential[/url] of $ \vec F$. The Wikipedia article states one existence result for vector potentials. The more general result is Poincare's lemma for [url=http://en.wikipedia.org/wiki/Closed_and_exact_differential_forms]closed differential form[/url], but I don't remember what degree of smoothness of $ F$ it requires.


function
calculus
integration
trigonometry
real analysis
real analysis unsolved

Hello, can anyone help me on this question please :(

Solve the initial boundary-value problem:

du/dt=d^2u/dx^2, for 0<x<pi ,t>0

u(0,t)=0, (left boundary condition)

du/dx(pi,t)=0, (right boundary condition)

u(x,0)=f(x), initial condition

This models the temperature variation in a rod of length pi, initially at a temperature given by f(x), then positioned with one end in ice and the other end insulated.

a) Determine the steady-state solution, u bar (x), to the problem.

b) Using the method of separation of variables, find the general solution u(x,t) to the PDE plus boundary conditions as an infinite series involving unknown coefficients A(subscript n).

c) Find an expression for A(subscript n) in terms of the initial condition function f(x). You may assume the result

Intergral with upper intergrand=pi and lower intergrand=0 sin((2n+1)/2*x)sin((2m+1)/2*x)dx={0, n is not equal to m,  pi/2, n=m) } for n, m intergers.

d) Find u(x,t) for the case where f(x)=100.

Thanks in advance

The heat equation (Fourier's favorite).

The boundary conditions are homogeneous and the steady-state solution (what is asked for in (a)) is simply zero.

Separation of variables will lead you to the eigenvalue problem:

$\frac{d^{2}y}{dx^{2}}+\lambda y=0;\ y(0)=0,\,y'(\pi)=0.$

Our solutions are of the form $y(x)=a\cos \left(\sqrt{\lambda}x\right)+b\sin \left(\sqrt{\lambda}x\right).$

The condition $y(0)=0$ knocks out the cosine term. Now we have

$b\sqrt{\lambda}\cos \left(\sqrt{\lambda}\pi\right)=0$ from which $\sqrt{\lambda}=k-\frac12$ for some integer $k.$

That's just one piece of the whole solution - I was getting you started. This looks like it would be a standard textbook exercise.

Thanks for your reply Kent Merryfield, you see, the textbook that I have don't give you examples of how to do the problems they teach you and even the exercises they give you to do at the end of the chapter don't have answers at the back of the book  :(  So I am really having trouble doing it.

With regards to part b) I can understand some of the first part of the solution that you gave but that's it, I can't go any further because I am stuck. If it's possible, can you or anyone show me how the whole thing is done please ?

Since this is a standard computation, you can find it worked out on the Internet, [url=http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/PDE/node21.html]even with examples[/url]


This is a test page for students at A.C. Flora.

What is this? You guys do realize there is a latex testing forum on this site ;) , someone lock this!

here is where you should go http://www.artofproblemsolving.com/Forum/index.php?f=224

i don't mind the test page, but why is it a.c. flora specific? i know it also says "any other new people" but that seems to be a afterthought. this looks like discrimination against non-a.c. flora newbies who want to test out $\LaTeX$ or other features  :P

This thread was locked, and now it's unlcoked?  I'm not gonna interfere; make up your mind Joe!

It is A.C. Flora specific because I am trying to get a group of kids to learn latex so that they can post.  I will have five or six new people on the site by the end of the week.  Lighten up Shobhit.

Yes Shobhit, let's be courteous and understanding toward our new members. We were once new to AoPS as well, you know (and I'm still fairly new, anyway).


Prove, that in non-isosceles triangle points $ K$,$ U$,$ U_1$ lies on one line, where:
$ K$ – Lemoine point;
$ U$,$ U_1$ - centres isogones.

This is quite known. See here for a more general result: [url]http://www.mathlinks.ro/viewtopic.php?t=169837[/url].


It is now permitted to discuss the AMC 10B/12B contests.

Remember: we will be having the AMC 10B/12B Math Jam on Friday, February 29, at 7 PM Eastern / 4 PM Pacific.  We will discuss the solutions to the last five problems on each contest, and selected other problems by request.

I also want to repost this reminder.

We frequently gets asked:
[quote="cognos599"]when will we know:

1) our official scores
2) the cutoff to qualify for AIME.[/quote]
Basically, you'll know when the AMC does.  The AMC works very hard to accurately process all the answer sheets, prepare the score reports, and determine the AIME qualification scores.  Please do not rush them.  When they have this information, they will make it public and we will announce it here too.  Until then, please do not continually ask when this information will be available.

I will quote from page 6 the AMC 10A/12A Teacher's Manual:
[quote="AMC Teacher's Manual"]V. School Results
The AMC office will send results by email (if available) and first
class mail as soon as the answer forms are scored. If you have
not received your results from our office within 30 days after the
AMC 10/AMC 12 please contact us to verify that your answer
forms were in fact received...
Results are not official until you receive the postal mailed paper
copy of your report, and that should be no longer than 30 days
after AMC 10/AMC 12.[/quote]The full Teacher's Manual is at
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf
If you have questions about the administrative aspects of the AMC, I suggest that you read it.

(Note: I would going to post a link to the 10B/12B manual, but I couldn't find it on the web site.  Steve Dunbar, if you're reading this, could you post the link?  Thanks.)

There is also an FAQ list on the AMC web site:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml
[quote="AMC FAQ"]Q, AMC10/12 #4. When will the results come back by email, roughly?

    A. It takes about 3 weeks to score and report the AMC10 and AMC12 contests.  The email report is sent to the school contest manager.  The official hard copy report is mailed at the same time.[/quote]

(I believe that the 10B/12B reporting is a little bit faster, due to the need to determine AIME qualifiers in time for the AIME on 3/18.  But I couldn't find anything definitive on the AMC web site about this.)

AIME qualifications will not be possible until the majority of AMC 10B/12B papers are scored.  This won't happen until the first week of March at the earliest.

If you missed the 10B/12B Math Jam, the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=219]transcript[/url] is now available.


Let $ f: \mathbb{Z\rightarrow\mathbb{Z}}$ a function such that $ f(0)\equal{}0$, $ f(1)\equal{}1$, $ f(2)\equal{}2$ and $ f(x\plus{}12)\equal{}f(x\plus{}21)\equal{}f(x)$ to all $ x\epsilon\mathbb{Z}$ . So $ f(2009)$ is:

A) 0
B) 1
C) 2
D) 3
E) 2009

How to do ?
Thanks.

[hide="Solution"]$ 2009 \equal{} 2 \plus{} 3 \times 669$.

$ f(x\plus{}24) \equal{} f(x\plus{}12) \equal{} f(x) \equal{} f(x\plus{}21)$, so $ f(x) \equal{} f(x\plus{}3)$

Therefore, $ f(2009) \equal{} f(2) \equal{} \boxed{2}$.

Answer: C[/hide]

[quote="luiseduardo"]Let $ f: \mathbb{Z\rightarrow\mathbb{Z}}$ a function such that $ f(0) \equal{} 0$, $ f(1) \equal{} 1$, $ f(2) \equal{} 2$ and $ f(x \plus{} 12) \equal{} f(x \plus{} 21) \equal{} f(x)$ to all $ x\epsilon\mathbb{Z}$ . So $ f(2009)$ is:

A) 0
B) 1
C) 2
D) 3
E) 2009

How to do ?
Thanks.[/quote]
[hide]$ 2009\equiv 5\bmod{12}$ so $ f(2009)\equal{}f(5)\equal{}f(26)\equal{}f(2)\equal{}2$, C[/hide]

[quote="darkdieuguerre"][hide="Solution"]$ 2009 \equal{} 2 \plus{} 3 \times 669$.

$ f(x \plus{} 24) \equal{} f(x \plus{} 12) \equal{} f(x) \equal{} f(x \plus{} 21)$, so $ f(x) \equal{} f(x \plus{} 3)$

Therefore, $ f(2009) \equal{} f(2) \equal{} \boxed{2}$.

Answer: C[/hide][/quote]

Thanks, but I don't understand. How did you find $ f(x) \equal{} f(x \plus{} 3)$ and $ f(x \plus{} 24) \equal{} f(x \plus{} 12) \equal{} f(x) \equal{} f(x \plus{} 21)$ ?
Why did you think that $ f(2009) \equal{} f(2) \equal{} \boxed{2}$ ?
:(

We know that $ f(x\plus{}12) \equal{} f(x)$. Therefore, let $ y \equal{} x\plus{}12$. We know that $ f(y\plus{}12) \equal{} f(y)$, or $ f(x\plus{}24) \equal{} f(x\plus{}12)$. Therefore, by transitivity, $ f(x\plus{}24) \equal{} f(x)$.

Furthermore, we know that $ f(x\plus{}21) \equal{} f(x)$, so again by transitivity, $ f(x\plus{}24) \equal{} f(x\plus{}21)$. Let $ z \equal{} x\plus{}21$, so $ f(z\plus{}3) \equal{} f(z)$. Therefore, $ f(x\plus{}3) \equal{} f(x)$.

Therefore, $ f(2009) \equal{} f(2006) \equal{} f(2003) \equal{} ... \equal{} f(5) \equal{} f(2) \equal{} 2$.

Thanks, darkdieuguerre. I understand now your method !  :)
Very good.

[quote="luiseduardo"][quote="ckck"][quote="luiseduardo"]Let $ f: \mathbb{Z\rightarrow\mathbb{Z}}$ a function such that $ f(0) \equal{} 0$, $ f(1) \equal{} 1$, $ f(2) \equal{} 2$ and $ f(x \plus{} 12) \equal{} f(x \plus{} 21) \equal{} f(x)$ to all $ x\epsilon\mathbb{Z}$ . So $ f(2009)$ is:

A) 0
B) 1
C) 2
D) 3
E) 2009

How to do ?
Thanks.[/quote]
[hide]$ 2009\equiv 5\bmod{12}$ so $ f(2009) \equal{} f(5) \equal{} f(26) \equal{} f(2) \equal{} 2$, C[/hide][/quote]

Hello, but how did you know that $ f(5) \equal{} f(26) \equal{} f(2) \equal{} 2$ ?[/quote]
$ f(5)\equal{}f(26)$ because $ f(x)\equal{}f(x\plus{}21)$, so in that case $ x\equal{}5$. $ f(26)\equal{}f(2)$ because $ f(x\plus{}12)\equal{}f(x)$, so using it twice we get it to $ f(2)$.


[quote="k00lperson"]hi guys. can i join this forum?[/quote]

Where do you live? (What state?)

Also, if you've posted... um... you sort of joined already

Technically, this forum is only for Indiana residents


$f: \mathbb{K} \rightarrow\mathbb{K}: x\rightarrow \sum^n_{k=0}a_n x^n$

it's more like mimetex like this...

Did it get changed recently? Now even simple equations look weird. It looked better when it was sorted bolded.

$x = 1+1$

Yes, it looks really ugly... $a_k\in\mathbb{K}$...

There is a bigger discussion here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=88819[/url]

 :)


linear algebra
matrix
linear algebra unsolved

Let $ v_{ij}\in C^n$, where $ i,j\in\{1,2,\cdots, n\}$, and $ v_{ij}\circ v_{ji}\equal{}v_{ii}\circ v_{jj}$ for any $ i,j\in\{1,2,\cdots, n\}$, where $ \circ$ denotes the usual entrywise product. Show that

\[ A\equal{}\left(\begin{array}{cccc}(v_{11},v_{11})&(v_{12},v_{21})&\cdots & (v_{1n},v_{n1})\\(v_{21},v_{12})&(v_{22},v_{22})&\cdots & (v_{2n},v_{n2})\\\vdots&\ \cdots&\cdots& \vdots \\(v_{n1},v_{1n})&(v_{n2},v_{2n})&\cdots & (v_{nn},v_{nn})
\end{array}\right)\]

is positive semidefinite. :rotfl:

[quote="mlok"]Why don't you just replace $ v_{ij}\circ v_{ji}$ with $ v_{ii}\circ v_{jj}$ in the matrix?[/quote]

$ (x,y)\equal{}y^*x$ rather than $ x\circ y$.


If you take any given positive integer and add an integer whose digits are the reverse of the first integer (i.e. if your integer was 5638, then add 8365 to it), then add an integer whose digits are the reverse of the sum to the sum (i.e. the sum was 14003, so add 30041), and by continuing the adding process, you will get a palindrome eventually no matter what integer you start with and how big that integer is.

This is a property of palindromes that I remember reading about.  Is there a way to prove this for all positive integers, regardless of size?

Start with 89. You may be surprised by what happens.


[quote="indybar"]Prove that the number of integral solutions $(x,y)$ to $x^2+y^2=n$ is equal to $4(d_1-d_3)$, where $d_1$ is the number of divisors of $n$ of the form $4k+1$ and $d_3$ is the number of divisors of $n$ of the form $4k+3$.[/quote]

Well: Assuming above, prove:

$\frac {\pi} 4 =  1 -\frac 1 3 + \frac 1 5 - \frac 1 7 + \frac 1 9 \dots $

(Hint: Its not  very hard), if you assume the above well-known  number theory theorem)

[quote="Pascual2005"]It is well known but i dont considere it pre olympiad![/quote]

And how to prove it?


Let a and b be digits of a three–digit number “6ab” that is greater than 600. if “6ab” is divisible by 4 , how many possibilities are there for the two- digit number “ab” ?

[hide="Solution"]There are $ 25$ numbers $ ab$ divisible by $ 4$. However, $ ab \equal{} 00$ is invalid. Therefore, there are $ \boxed{24}$ such numbers.

[/hide]

[hide]Explaining darks solution - because 100/4=25, there are 25 possibilities for ab.  However, we must disregard 00.  Therefore, there are 24.[/hide]


algebra
function
domain
inequalities
search
analytic geometry

Let $ D$ be the domain expressed by the system inequality:
\[ y \minus{} 2x \plus{} 4\geq 0,\ 4x \minus{} y^2\geq 0\]
.

When the point $ P(x,\ y)$ move in the domain $ D$, find the maximum and minimum value of $ 2x \plus{} y$.

hello, we get
$ \minus{} 2\le2x \plus{} y\le12$ ,the minimum will obtained in the point $ P(1;\minus{}2)$ and the maximum in $ P(4;4)$.
Sonnhard.

hello kunny, at first we compute the intersection points of both functions, we get
$ P_1(1; \minus{} 2)$ and $ P_2(4,4)$, since the objective function has no extrema we must search for the extrema on the curves of $ y \equal{} 2x \minus{} 4$ and $ y^2 \equal{} 4x$.
Inserting $ y \equal{} 2x \minus{} 4$ in our function we get $ h(x) \equal{} 4x \minus{} 4$ and since $ h'(x) \equal{} 4 > 0$ our function is strictly monotonically increasung on $ y \equal{} 2x \minus{} 4$ and the same result we get by inserting $ y \equal{} 2\sqrt {x}$ so we get the maximum in $ P(4;4)$ and we get $ f(4;4) \equal{} 12$.
Inserting $ y \equal{} \minus{} 2\sqrt {x}$ in our objectiv function we get $ g(x) \equal{} 2x \minus{} 2\sqrt {x}$,therefore we have the minimum at the point $ P(\frac {1}{4}, \minus{} \frac {1}{2})$, with $ f(\frac {1}{4},\minus{}\frac {1}{2}) \equal{} 0$, but $ f(1; \minus{} 2) \equal{} 2 \minus{} 4 \equal{} \minus{} 2 < 0$ is our
minimum of the objectiv function.
Sonnhard.

$ 2y \plus{} 8 \ge 4x \ge y^2 \Longrightarrow 0 \ge y^2 \minus{} 2y \minus{} 8 \Longrightarrow \minus{} 2 \le y \le 4$

Therefore $ 2x \plus{} y \le 2y\plus{}4\plus{}y \le 12$, equality at $ (4,4)$

And $ 2x\plus{}y \ge \frac{y^2}{2}\plus{}y \ge \frac{\minus{}1}{2}$ equality at $ \left(\frac{1}{4},\minus{}1\right)$

One approach utilizes the following fact:
[i]Linear functions of the coordinates in a contiguous region are optimized on the boundary of the region.[/i]

If it's not immediately obvious to you why this should be the case, you should try to prove it to yourself.

[quote="Dr Sonnhard Graubner"]hello kunny, at first we compute the intersection points of both functions, we get
$ P_1(1; \minus{} 2)$ and $ P_2(4,4)$, since the objective function has no extrema we must search for the extrema on the curves of $ y \equal{} 2x \minus{} 4$ and $ y^2 \equal{} 4x$.
Inserting $ y \equal{} 2x \minus{} 4$ in our function we get $ h(x) \equal{} 4x \minus{} 4$ and since $ h'(x) \equal{} 4 > 0$ our function is strictly monotonically increasung on $ y \equal{} 2x \minus{} 4$ and the same result we get by inserting $ y \equal{} 2\sqrt {x}$ so we get the maximum in $ P(4;4)$ and we get $ f(4;4) \equal{} 12$.
Inserting $ y \equal{} \minus{} 2\sqrt {x}$ in our objectiv function we get $ g(x) \equal{} 2x \minus{} 2\sqrt {x}$,therefore we have the minimum at the point $ P(\frac {1}{4}, \minus{} \frac {1}{2})$, with $ f(\frac {1}{4}, \minus{} \frac {1}{2}) \equal{} 0$, but $ f(1; \minus{} 2) \equal{} 2 \minus{} 4 \equal{} \minus{} 2 < 0$ is our
minimum of the objectiv function.
Sonnhard.[/quote]

That's correct answer.

Let me summarize the answer,

Maximum value is $ 12$ when $ (x,\ y)\equal{}(4,\ 4)$.

minimum value is $ \minus{}\frac 12$ when $ (x,\ y)\equal{}\left(\frac 14,\minus{}1\right)$.


Let f is a derative real functon from R to R such that satesfies

[f(x)=f(x/2)+x/2f´(x)]

show that f is linear[/b]

Hi!

This problem has already been posted in http://www.mathlinks.ro/viewtopic.php?t=100225


A 5 by 5 square is composed of 25 unit squares. How many unit squares are entirely contained within the inscribed circle of diameter 5?

how would you do this problem?

[quote="sonny"][quote="kstan013"][hide="Unsure of this"]Take it 1 row at a time. The top row and the bottom row will each have 0. The middle row will have all except for the ends, 3. The 2nd and 4th rows will each have 3 because the arc comes and cuts through the corner (1,1) (I think!) making 3 each there so our answer is 3+3+3=9 (I think)[/hide][/quote]

What about 2 by 2s? [u][b][i]I think[/i] [/b][/u]with  2 by 2 squares and 3 by 3 squares the answer is 14.[/quote]

It says unit squares in the problem, so the answer is 9.

Here's a picture. Sorry it's in a word document. So it is 9 squares.

[size=67]Picture made in GeoGebra[/size]

[quote="nutz_for2.718281828"][hide="solution"]When a 5*5 square is inscribed in a circle, the diameter is 5rt2. When a 4*4 is inscribed, the diameter is 4rt2. 4rt2 is jujst a hair more than 5, so 3rt2 is the greatest diameter that can be inscribed in a square with integer sides.

9[/hide][/quote]
That is a very interesting way to think about it because most people just visualize the picture in their head, but you make cases.


A carpenter draws a $ \frac{1}{10}$-size scale drawing of the top view of a coffee table's rectangular top. The scale drawing of the table's top has an area of 3 square inches. What is the actual area of the table's top going to be?

Since the scale size is $ \frac {1}{10}$, the scaling for area is $ (\frac {1}{10})^2 \equal{} \frac {1}{100}$. Thus, the area of the tabletop is $ 3(100) \equal{} \boxed{300}$.

[quote="pi man"]Because you are multiplying eachside by 10, the area increases to 100 times the original size. The rectangle with area 3 now becomes 300 square units.

Therefore, 300 is the answer.

_-_-_-_-_-_-_-_-_-_-_-_-_-_-_

~pi man!! :rotfl:[/quote]

nice username dude

Hello Trueshot. If you had to revive this please don't revive to congratulate someone on their username. Please pm... maybe I should be... but I have a solution!

Whenever you have length it is $\times x$
area $\rightarrow$ $\times x^2$
volume $\rightarrow$ $\times x^3$

So in this case we have area so it must be $\times x^2$
So you need to multiply by $10^2 = 100$
$3 \cdot 100 = \fbox{300}$

:ninja: How do you know every single post I make???


The Round 1 scores and feedback are now available on the USAMTS website.

A couple of notes:

1. If you wish to contest a score, follow the instructions on the website.  Don't post on the forum and don't send email.  (You may send email for something unusual, for example, a problem that you submitted is completely missing.)

2. The leaderboard and stats page only includes data from students who have submitted their Entry Form.  If you have not submitted one, make sure that you do so with your Round 2 solutions.

i got an 18 also woot i was so surprised when i got a 5 for Q 1 and a 2 for Q 5 cause i thought it would be vice versa
i just didnt mention one little obvious (i thought it would be infered) thing in Q 5 so i missed the three pionts  :dry:  and i completely misunderstodd Q3 but i still got one point  :lol:

[quote="gauss1181"]If we want to clarify a solution to a problem without using up a protest, is that possible?[/quote]I don't know what you mean...what do you mean by "clarify a solution"?

[quote="xpmath"]I believe that gauss means to say that she wishes to explain what her solution meant and maybe clear up a misunderstanding that one may have had with it.[/quote]If she tells me that that's what she meant, then I'll answer her.  Until then, I'd rather not speculate on what another person's question might have meant.

So should I just email solution clarification at usamts@usamts.org?

oh my gawd. 2 extra points after protests.... :o


inequalities trigonometry function symmetry

can anyone write the basic trigonometric inequalities?

thanks!!!

There are inequalities? I know the basic equalities...

If you mean the basic trig functions, then they are

sin=o/h
cos=a/h
tan=o/a
csc=h/o
sec=h/a
cot=a/o

Some other basic equalities are

$ \sin^2\theta\plus{}\cos^2\theta\equal{}1$

$ \tan^2\theta\plus{}1\equal{}sec^2\theta$

$ 1\plus{}\cot^2\theta\equal{}\csc^2\theta$

There are others involving symmetry (the even and odd functions), cofunctions, etc.

If I didn't cover what you meant, please tell me.  :)

[quote="DIMITRIS T"]I mean the basic inequalities of trigonometry such as sinx<=1 or |sinx|<=|x|

I can't find a full catalogue of these inequalities  :([/quote]

We did not know such a catalogue existed, but I bet that if someone post such a list, many in this forum can prove them. (not me)


Find the last three digits of $ {987^{654^{321}}}\minus{}{123^{456^{789}}}$

Please give me elegant solution  :lol:

[quote="jonysatie"]I think you have to look at mod 100[/quote]

The question is find the last 3 digits ...

But are you sure that we need to find 3 ?

anyway, I already have this: $ 13^{654^{321}}\equal{} 561^{29}mod(1000)$ and $ 123^{456^{789}}\equal{} 161^{19}mod(1000)$

With Combinatoric, I found the answer is 400.

Hint :

$ 3^{2}\equal{}9\equal{}10\minus{}1$

$ 7^{2}\equal{}49\equal{}5.10\minus{}1$

:D

[quote="abacadaea"][quote="jonysatie"]I think you have to look at mod 100[/quote]

are you sure its mod 100 not mod 1000?[/quote]

Yeah, You have to look at mod 1000.  :blush:


geometry
incenter
circumcircle
geometry unsolved

Let $ I$ be the incenter of $ \Delta ABC$ and let $ A_1,B_1,C_1$ be arbitrary points on the segments $ AI,BI,CI$ respectively. The perpendicular bisectors of $ AA_1,BB_1$ and $ CC_1$ intersect at $ A_2,B_2$ and $ C_2$.
Prove that the circumcenter of $ A_2B_2C_2$ coincides with the circumcenter of $ ABC$ if and only if $ I$ is the orthocenter of $ A_1B_1C_1$

call $ O_2$ the circumcenter of $ A_2B_2C_3$ and O of ABC. Let $ A_2B_2$ meet BC ot K and $ A_2O_2$ meet BC ot H. When $ A_1$ is moving on AI the locus of $ O_2$ is a line perpendicolar to BC cause $ \angle O_2A_2B_2 \plus{} \angle JKA_2 \equal{}$ $ 90 \minus{} \angle A_2C_2B_2 \plus{}  \angle A_2B_2C_2 \minus{} 90 \plus{} \frac{A}{2} \plus{} B \equal{} \frac{C}{2} \minus{}90 \plus{} 90 \minus{} \frac{B}{2} \plus{} \frac{A}{2} \plus{} B \equal{}  90$. But then if $ O_2 \equiv O$ we have that $ A_2$ stay on the perpendicolar axis of BC and so that $ BCC_1B_1$ is cyclic and so by angle chasing that $ B_1C_1 \perp AI$. Equivalently we obtain $ C_1A_1 \perp BI$ and  $ A_1B_1 \perp CI$ and so if $ O_2 \equiv O$ I is the orthocenter of $ A_1B_1C_1$. Also if is the orthocenter of $ A_1B_1C_1$ we have that $ BCC_1B_1$ is cyclic and then that $ A_2$ sta on the perpendicolar axes of BC and equal for $ B_2$ and $ C_2$. So O is the circumcenter of $ A_2B_2C_2$.


suppose we have polynom with two variable $ x,y$and we know that
$ P(x,y)>0$ for all $ x,y$
then can we say that there exist $ \epsilon >0$ such as $ P(x,y)>\epsilon$ for all $ x,y$ ??
I think thta we can't, but I don't know why .. :huh:

That is a very classical one. Take $ P(x,y)\equal{}x^2\plus{}(1\minus{}xy)^2$. Of course, plenty of other choices are possible.


The latex is really weird... Check this out
http://www.artofproblemsolving.com/Wiki/index.php/1986_AIME_Problems/Problem_1

[quote="i_like_pie"]If there is no square root, square, fraction, or other object that cannot be typed without LaTeX, then it stays in normal next (although the size becomes smaller), even if you put <math></math> around it.

If you look at the code in the example you gave, somebody made it bold and LaTeX, so it came out to be small, bold text.[/quote]
The latex was all messed up ten minutes ago, but now it's better. Someone can delete this, somehow the latex is perfect now...
EDIT: Wait now it's messed up again. It is messed up when I am logged in and looks like
[code]Let $ y = \sqrt[4]{x} $. Then we have $ \displaystyle y(7 - y) = 12 $, or, by simplifying, $ \displaystyle y^2 - 7y + 12 = (y - 3)(y - 4) = 0 $. This means that $ \sqrt[4]{x} = y = 3 $ or $ \displaystyle 4 $. Thus the sum of the possible solutions for $ \displaystyle x $ is $ \displaystyle 4^4 + 3^4 = 337 $, the answer.[/code]

Check your personal Wiki settings, you might have chosen to display text instead of images when $\LaTeX$ appears :)


Determine the explicit formula and the lim(n-->00) of the recursive sequence a_n which is defined as follows:

(1) a_1 = 4 and (2) a_(n+1) = 2^(n+1.5)*(1-(1-(a_n/2^(n+1))^2)^0.5)^0.5

Use induction to prove that a_n = 2 n+1 *sin( \pi /2 n ).
From the well-knwon limit sin(x)/x -> 1 as x tends to 0, we deduce that (a_n) is convergent to 2 \pi .


HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:

Happy new year to everybody!

This year wil beee..... oh, I don't know how it will be! But it's certainly easier to factor, since $2007 = 2^{3}\cdot 223$!  :P

[quote="edriv"]

But it's certainly easier to factor, since $2007 = 2^{3}\cdot 223$!  :P[/quote]

$2007=3^{2}*223$ :P

Wow, $2007$ is a cool number!

It has two threes and three twos, when factored!
And it also contradicts the Fundamental Theorem of Arithmetic, as you can see from my factorization.  :D


geometry
parallelogram
circumcircle
geometry unsolved

Given a parallelogram $ABCD$. Let $E$ be a point on the side $BC$ and $F$ be a point on the side $CD$ such that the triangles $ABE$ and $BCF$ have the same area. The diaogonal $BD$ intersects $AE$ at $M$ and intersects $AF$ at $N$. Prove that:

[b]I. [/b] There exists a triangle, three sides of which are equal to $BM, MN, ND$.
[b]II.[/b] When $E, F$ vary such that the length of $MN$ decreases, the radius of the circumcircle of the above mentioned triangle also decreases.

excuse me I have changed the name of the points.so the promlem is:

[color=blue]Given a parallelogram$ABCD$ . Let $E$ be a point on the side$AB$  and $F$ be a point on the side$BC$  such that the triangles$ADE$  and$ABF$  have the same area. The diaogonal $AC$ intersects $DE$ at$M$  and intersects  $DF$at $N$. Prove that:

[b]I[/b]. There exists a triangle, three sides of which are equal to$AM,MN,NC$ .
[b]II[/b]. When $E,F$ vary such that the length of $MN$ decreases, the radius of the circumcircle of the above mentioned triangle also decreases.[/color]
[i][color=red]Solution[/color]:[/i]
let $AB=a,AD=b ,AC=d$       $AE.AD=BF.BA \longrightarrow \frac{AE}{BF}=\frac{AB}{AD} $ let$ \frac{AE}{AB}=k$ therefor $AE=ka,BF=kb$

$\triangle AEM \sim \triangle MDC$ so$ \frac{AM}{MC}=k \rightarrow \frac{AM}{AC}=\frac{k}{k+1} \longrightarrow AM=\frac{d.k}{k+1}=\frac{d(k^2-2k)}{k^2-k-2}$

$\triangle AND \sim \triangle NCF $so $\frac{NC}{AN}=1-k \rightarrow \frac{NC}{AC}=\frac{k-1}{k-2} \longrightarrow NC=\frac{d(k-1)}{k-2}=\frac{d(k^2-1)}{k^2-k-2}$

so:$MN=\frac{d.(k-k^2-1)}{-k^2+k-2}$

notice that$ 0<k<1$ therefore$ k^2-k-2<0$

$AM+NC=\frac{d(2k^2-2k-1}{k^2-k-2}$

$AM+MN=\frac{d(-k-1)}{k^2-k-2}$

$MN+NC=\frac{d(k-2)}{k^2-k-2}$

so we have :$AM+NC>MN$ , $AM+MN>NC$ , $MN+NC>AM$

Have you solved II? It looks really difficult.  ;)

I am not able to remember the way I solved the first question, therefore I shall only continue the previous solution:
See that $ MN^2 \equal{} AM^2 \plus{} CN^2 \minus{} AM\cdot CN$, therefore the angle opposite to $ MN$ length is $ 60^\circ$, hence its circumradius is $ MN\frac{\sqrt 3}{3}$.

Best regards,
sunken rock


I have no idea whether it's obvious or not (tough I guess it's not), but is it true that any complex (real, integer,...) matrix is product of idempotent matrices?

i guess not

it is a theorem that idempotent matrices are just those that represent projections
the only projections that are invertible is the identity

so if you want an invertible matrix to be a product of idempotent matrices, they all have to be invertible, thus your matrix has to be the identity matrix!
so  i think what you are saying is not true over any field

if you want a proof of my first statement
if T is idempotent and invertible,it is surjective
so every v can be written as v=T(a)

then T(v)=T(T(a))=T(a)=v
so T is identity

greetz

fred

of course harazi meant "any non invertible complex matrix is ... " (see my previous post about nilpotent matrices  :)  )

ah of course
so uhm is it true
my hunch is no, but how to prove?


function search algebra polynomial quadratics ratio geometry

How can I generate an explicit formula out of any recursive formula, such as if $ a_{n}\equal{}4a_{n\minus{}1}\minus{}5$?

Using generating functions may be a good approach.  Search for the free e-book "generatingfunctionology" by Wilf.

[quote="DiscreetFourierTransform"]If it's a linear recursion, you can generally find the characteristic polynomial and solve a system of equations to get the coefficients on the terms of the explicit formula. *Famous example: Fibonacci and Lucas sequences*[/quote]

Hmm... could you elaborate on that?

Let $ a_{n}\equal{}b_{n}\plus{}k$. Then

$ b_{n}\plus{}k\equal{}4*b_{n\minus{}1}\plus{}4k\plus{}5$
$ b_{n}\minus{}4b_{n\minus{}1}\equal{}3k\plus{}5$.

It would be convenient to choose $ 3k\plus{}5\equal{}0$, so lets do that. Then $ b_{n}\equal{}4^{n}b_{0}$ [it is a geometric sequence....]

Then $ a_{n}\equal{}4^{n}*\left( a_{0}\plus{}\frac{5}{3}\right)\minus{}\frac{5}{3}$.

I still dont get it - can you explain a bit further?   :blush:

Part of your confusion may be the fact that there have been three proposed methods, of varying difficulties and suitabilities.  In rough order of difficulty:

[hide="Method 1 (Altheman)"] This is specifically suited to recursions of the form $ a_{n}\equal{} pa_{n\minus{}1}\plus{}q$.  In the case where $ q \equal{} 0$, this is very easy, because the solution to the recursion $ a_{n}\equal{} pa_{n\minus{}1}$ is just the geometric series $ a_{n}\equal{} p^{n}a_{0}$.

When $ q\neq 0$, the recursion "looks like" (but is not quite) a geometric series.  Altheman's method posits that there exists a constant $ k$ such that $ a_{n}\minus{}k$ is a geometric series.  Letting $ b_{n}\equal{} a_{n}\minus{}k$ and plugging in this hypothesis,

$ b_{n}\plus{}k \equal{} pb_{n\minus{}1}\plus{}pk\plus{}q\implies$
$ b_{n}\equal{} pb_{n\minus{}1}\plus{}(p\minus{}1)k\plus{}q$

If $ k \equal{}\minus{}\frac{q}{p\minus{}1}$, then

$ b_{n}\equal{} pb_{n\minus{}1}$

Which is a geometric series as we claimed.  So that gives us Altheman's result,

$ a_{n}\equal{} p^{n}(a_{0}\minus{}k)\plus{}k$ [/hide]

[hide="Method 2 (DiscreetFourierTransform)"] The simplest nontrivial example of this method is the Fibonacci sequence, which is given by a recursion of the form

$ f_{n}\equal{} f_{n\minus{}1}\plus{}f_{n\minus{}2}$

As before, such sequences "look like" geometric sequences.  Suppose $ f_{n}\equal{} x^{n}$ for some $ x$.  Plugging in,

$ x^{n}\equal{} x^{n\minus{}1}\plus{}x^{n\minus{}2}\Leftrightarrow$
$ x^{2}\equal{} x\plus{}1$

So $ x \equal{}\frac{1\pm\sqrt{5}}{2}$.  We will refer to the two roots of this quadratic, called the [i]characteristic polynomial[/i], with the symbols $ \phi,\varphi$ (the golden ratio and... the other golden ratio).

So we know that $ f_{n}\equal{}\phi^{n}, f_{n}\equal{}\varphi^{n}$ are sequences that satisfy this recursion.  But the Fibonacci sequence has the initial conditions $ f_{0}\equal{} 0, f_{1}\equal{} 1$, and clearly neither of the above sequences satisfies these conditions.  Now it is time for some lemmas.

[b]Lemma:[/b]  If a sequence $ a_{n}$ satisfies a [i]homogeneous[/i] linear recursion (in which there are no constants and no term in the sequence is raised to any power), then so does $ ca_{n}$ for arbitrary $ c$.

[b]Lemma:[/b]  If two sequences $ a_{n}, b_{n}$ satisfy the same homogeneous linear recursion, then so does $ a_{n}\plus{}b_{n}$.

Now we know that any sequence of the form $ f_{n}\equal{} A\phi^{n}\plus{}B\varphi^{n}$ also satisfies the Fibonacci recursion.  (In fact, it turns out that every solution to the recursion takes this form.)  Given our starting conditions, we have the system of equations

$ f_{0}\equal{} A\plus{}B \equal{} 0$
$ f_{1}\equal{} A\phi\plus{}B\varphi \equal{} 1$

It turns out that

$ f_{n}\equal{}\frac{\phi^{n}\minus{}\varphi^{n}}{\phi\minus{}\varphi}$

Keep in mind that this method [b]does not work[/b] if there are constants in the recursion, such as $ f_{n}\equal{} f_{n\minus{}1}\plus{}f_{n\minus{}2}\plus{}1$.  However, a modification of Altheman's idea will allow you to solve such recursions in a similar way. [/hide]

[hide="Method 3 (illcrowflu)"] As I understand what illcrowflu is trying to say, this method is the most complicated and you really shouldn't bother with it.  Roughly, a standard technique in generating functions is to study a sequence $ a_{n}$ by considering the function

$ f(x) \equal{}\sum_{k \equal{} 0}^{\infty}\frac{a_{n}x^{n}}{n!}$

The recursion $ a_{n}\equal{} 4a_{n\minus{}1}\minus{}5$ corresponds to the [i]differential equation[/i]

$ f(x) \equal{} 4 f'(x)\minus{}5 e^{x}$

Which is a subject fairly far beyond the consideration of HSB.  There are several standard methods for dealing with such a differential equation which are actually analogous to the easier methods of solving the original recurrence. [/hide]

[quote="Altheman"]Let $ a_{n}\equal{} b_{n}\plus{}k$. Then

$ b_{n}\plus{}k \equal{} 4*b_{n\minus{}1}\plus{}4k\plus{}5$
$ b_{n}\minus{}4b_{n\minus{}1}\equal{} 3k\plus{}5$.

It would be convenient to choose $ 3k\plus{}5 \equal{} 0$, so lets do that. Then $ b_{n}\equal{} 4^{n}b_{0}$ [it is a geometric sequence....]

Then $ a_{n}\equal{} 4^{n}*\left( a_{0}\plus{}\frac{5}{3}\right)\minus{}\frac{5}{3}$.[/quote]

I still don't understand the explanations so well, by everybody, but Altheman, do you mean $ a_{n}\equal{} 4^{n}*\left( a_{0}\minus{}\frac{5}{3}\right)\plus{}\frac{5}{3}$?

Also, t0rajir0u, what do you mean by "looks" like a geometric sequence?

[quote="123456789"]t0rajir0u, what do you mean by "looks" like a geometric sequence?[/quote]

Well, one sense of what I mean is the fact that the ratio $ \frac{a_{n\plus{}1}}{a_{n}}$ approaches a limit.  For your given sequence it is $ 4$ and for the Fibonacci sequence it is $ \phi$.

Another way I mean it is precisely what the equations say - that such sequences are geometric series to which some other operation has been performed (adding a constant, or adding another geometric series).

Yet a third way I mean it is that expanding them out produces geometric-like progressions.  Take the sequence $ a_{n}\equal{} 2a_{n\minus{}1}\plus{}1$ with the initial condition $ a_{0}\equal{} 0$; we have

$ a_{1}\equal{} 1$
$ a_{2}\equal{} 2\plus{}1 \equal{} 3$
$ a_{3}\equal{} 2(2\plus{}1)\plus{}1 \equal{} 4\plus{}2\plus{}1 \equal{} 7$
$ a_{4}\equal{} 2(2(2\plus{}1)\plus{}1)\plus{}1 \equal{} 8\plus{}4\plus{}2\plus{}1 \equal{} 15$

In fact, these are sums of geometric series!  It's not hard to see that the general term is given by $ a_{n}\equal{} 2^{n}\minus{}1$, which is the same answer you'd get from Altheman's method.

In order to get $ k \equal{}\minus{}\frac{q}{p\minus{}1}$, did you just choose a value of k for which the equation would resemble a geometric sequence?

By the way, where do you guys learn all these things about math? What's a good book?

Also, could you explain the lemmas in DiscreetFourierTransform's solution? I don't understand his solution much at all. Thanks.

Also, the ways to find the explicit formulas all used the knowledge that the sequence resembled a geometric sequence. What if the sequence didn't? What if the recursive formula included square roots, or squares, or cubes of previous terms? Would you need to find a sequence that it resembled?

[quote="t0rajir0u"][quote="123456789"]In order to get $ k \equal{}\minus{}\frac{q}{p\minus{}1}$, did you just choose a value of k for which the equation would resemble a geometric sequence?[/quote]

Yep.  It's the value that makes things simplest.

[quote="123456789"]By the way, where do you guys learn all these things about math? What's a good book?[/quote]

Actually, I picked up most of what I know about recursion from reading posts on the forums.  Adam Hesterberg taught me generating functions :P

[quote="123456789"]Also, could you explain the lemmas in DiscreetFourierTransform's solution? I don't understand his solution much at all. Thanks.[/quote]

[hide="Let me be clearer"] A homogeneous linear recursion is a recursion of the form

$ a_{n}\equal{} A a_{n\minus{}1}\plus{}B a_{n\minus{}2}\plus{}C a_{n\minus{}3}\plus{}...$

Going back however many terms.  Now, suppose there is a sequence $ a_{n}$ that satisfied such a recursion.  Then

$ ca_{n}\equal{} A ca_{n\minus{}1}\plus{}B ca_{n\minus{}2}\plus{}C ca_{n\minus{}3}\plus{}...$

So $ ca_{n}$ also satisfies the recursion.  This is the first lemma.  If there are two sequences $ a_{n}, b_{n}$ that satisfy the recursion, then

$ a_{n}\plus{}b_{n}\equal{} A (a_{n\minus{}1}\plus{}b_{n\minus{}1})\plus{}B (a_{n\minus{}2}\plus{}b_{n\minus{}2})\plus{}C( a_{n\minus{}3}\plus{}b_{n\minus{}3})\plus{}...$

(Easily obtained by adding the two recursions for $ a_{n}, b_{n}$ together.)  Hence $ a_{n}\plus{}b_{n}$ also satisfies the recursion.  This is the second lemma.  [/hide]

[quote="123456789"]What if the recursive formula included square roots, or squares, or cubes of previous terms?[/quote]

There isn't a general method here.  There are several specific methods for several specific types, including recognizing trigonometric identities or manipulating the recursion to get it into a manageable form.  For example, the simple nonlinear recursion $ a_{n}\equal{} a_{n\minus{}1}^{2}$ has the solution

$ a_{n}\equal{} a_{0}^{ 2^{n}}$

Which is not hard to see.  One way to think about this problem is to consider the sequence $ b_{n}\equal{}\ln a_{n}$, which has the recursion $ b_{n}\equal{} 2b_{n\minus{}1}$, and then the answer is even more obvious.  This is one general method; rather than study a given sequence $ a_{n}$, study related sequences $ b_{n}$ defined however is most convenient.

[quote="Anaxerzia"]Also in the last steps of Altheman's solution, how did you get the final step?[/quote]

[hide="Discussion"] Since $ b_{n}\equal{} a_{n}\minus{}k$ has the recursion

$ b_{n}\equal{} pb_{n\minus{}1}$

We know that it has the closed form

$ b_{n}\equal{} p^{n}b_{0}$

Where $ b_{0}\equal{} a_{0}\minus{}k$.  Hence

$ a_{n}\equal{} b_{n}\plus{}k \equal{} (p^{n}b_{0})\plus{}k \equal{} p^{n}(a_{0}\minus{}k)\plus{}k$ [/hide][/quote]

In $ a_{n}\equal{} A a_{n\minus{}1}\plus{}B a_{n\minus{}2}\plus{}C a_{n\minus{}3}\plus{}...$, are A, B, C just different constants that have nothing to do with the previous terms?

Also, I mean in general, where did you learn everything? Just from reading posts here?

[quote="rrusczyk"]Our Intermediate Counting & Probability book, which will have a thorough introduction to recursion and generating functions (and lots of other stuff), will be available in about a month.  The corresponding class starts at the end of October.[/quote]

I'm not sure if you can tell what level I'm at, but if I got a 112.5 on the AMC 10 in 9th grade, do you know what book I should start off with?

[quote="123456789"]In $ a_{n}\equal{} A a_{n\minus{}1}\plus{}B a_{n\minus{}2}\plus{}C a_{n\minus{}3}\plus{}...$, are A, B, C just different constants that have nothing to do with the previous terms?[/quote]

Yep.  The Fibonacci and Lucas sequences both take the form $ a_{n}\equal{} a_{n\minus{}1}\plus{}a_{n\minus{}2}$, which corresponds to the constants $ A \equal{} B \equal{} 1, C \equal{} D \equal{} ... \equal{} 0$.

[quote="123456789"]Also, I mean in general, where did you learn everything? Just from reading posts here?[/quote]

I learned a surprising amount just reading posts here, yes.

T0r, this would be a place to apply the ordinary generating function $ A(x) \equal{}\sum_{n\geq 0}a_{n}x^{n}$ rather than the exponential generation function $ \sum_{n\geq 0}\frac{a_{n}}{n!}x^{n}$.  This replaces the differential equation you named with a far simpler relationship that can be easily solved.

Oh, I see.  You mean that we now have

$ xA(x) \equal{} 4A(x)\plus{}\frac{5}{1\minus{}x}\implies$
$ A(x) \equal{}\frac{5}{(1\minus{}x)(x\minus{}4)}\equal{}\frac{5}{3}\left(\frac{1}{4\minus{}x}\minus{}\frac{1}{1\minus{}x}\right)$

That is much simpler, isn't it...  :oops:

Yes, or something very much like that (didn't check the details).  In general, figuring out in advance whether an ordinary or exponential generating function best suits your needs seems to me like something of an arcane art (of course, one at which the solver improves with practice).

You can use the formula repeatedly, and look for a special form... then you prove it by induction.

In this case you get something like this:

           $ a_{n}\equal{}4^{n}a_{0}\plus{}5\sum^{n}_{j\equal{}1}{4^{j}}\equal{}4^{n}a_{0}\plus{}5\frac{4^{n\plus{}1}\minus{}1}{3}$

Here it worked. But if the recursion were not linear it would probably get ugly and it may be easier to use one of the above methods.

 :)

I saw from somewhere that Mathematica used to use Rsolve for generating functions? What's the new function? And how is it used?


For all n prove thart

             (exp(2.5sqrtn))/13n<P(n) where p(n) is partition funtion.

Bomb

bomb,what is the partition funtion? :?:  :?:  :?:

Partiation function is the number of ways a positive integer can be partitioned.
Like 3=2+1,3,1+1+1 etc...

Bomb

Usually when people discuss Rademacher and the partition function, the reference is to his  "exact formula" for $p(n)$, a rapidly converging series whose first term gives the asymptotic formula of Ramanujan and Hardy.

i think the most interessant expression of the partition function is:
$ p\left(n\right) \equal{} \frac {1}{n}\sum_{k\equal{}0}^{n\minus{}1} p\left(k\right)s\left(n \minus{} k\right)$.

Hereby, $ s$ is the divisor function defined by $ s(n)\equal{}\sum_{d\mid n} d$.


a,b are numbers such that $ \left\{ \begin{array}{l}
 a^3  \minus{} 3b^2 a \equal{} 19 \\
 b^3  \minus{} 3a^2 b \equal{} 98 \\
 \end{array} \right.$

Find the value of $ S \equal{} a^2  \plus{} b^2$.

You could've picked nicer numbers  :wink: 26 and 18, perhaps?

[hide="Solution"]$ \blacktriangleright (b \minus{} ai)^3 \equal{} (b^3 \minus{} 3a^2b) \plus{} (a^3 \minus{} 3ab^2) i \equal{} 98 \plus{} 19i$

$ \blacktriangleright (b \plus{} ai)^3 \equal{} (b^3 \minus{} 3a^2b) \minus{} (a^3 \minus{} 3ab^2) i \equal{} 98 \minus{} 19i$

$ \therefore [(b \plus{} ai)(b \minus{} ai)]^3 \equal{} (b^2 \plus{} a^2)^3 \equal{} (98 \plus{} 19i)(98 \minus{} 19i) \equal{} 9965$

$ \therefore \boxed{a^2 \plus{} b^2 \equal{} e^{\frac {2k \pi i}{3}} \cdot \sqrt [3]{9965}} \qquad k \in \{ 0,1,2 \}$

[color=darkblue]where $ k$ depends on which solution we use.[/color][/hide]

Please not to use CMPLX number, I amnot used that!!!  :(

The complex solution is basically motivated by the expansion of $ (x \plus{} y)^3$, so I think it is easy to see. But if you want to avoid that, then you can take the following, less motivated approach:
[hide]
1. Square the equations and add: $ (a^3 \minus{} 3 b^2 a)^2 \plus{} (b^3 \minus{} 3 a^2 b)^2 \equal{} 19^2 \plus{} 98^2$

2. Expand both sides: $ a^6 \plus{} 3 a^4 b^2 \plus{} 3 a^2 b^4 \plus{} b^6 \equal{} 9965$

3. Factor left side: $ (a^2 \plus{} b^2)^3 \equal{} 9965$

And now, you take one of the three cuberoots.[/hide]


Can somebody prove the $Bertrand$ $postulate$: For every positive integer $n$, there exist a prime $p$ such that $n\leq p\leq 2n$


modular arithmetic
number theory unsolved
number theory

Find all $ x,y,z \in Z_\plus{}$ such that : $ x^{z \plus{} 1} \minus{} y^{z \plus{} 1} \equal{} 2^{100}$
[color=red]Edited by moderator.[/color]


Hi - You can use the same method shown [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16905]here ( your other problem)[/url] just use the integer part  (instead of fractional par) and yes,  subtract N  and add 1 for luck.

That is:

Number of digits = [0.565705518 )*N+.0.4+(N+.5)log N -N +1

And this time just keep the integer part.. eg for N=100

You get   56.57+0.4+201-100+1 = 158 .something  so, 158 digits...

Waiting for  Ravi to spoil my fun and say it is nothing but:

(N+1/2) log (N) +  (1/2) (log (2*pi)) - N log(e)
 :)   (And I bet he is  going to use latex to make it look pretty too  :)

BTW:

some what less popular formula gives a little better approx for n!

 ( :sqrt:{(2n+1/3)*pi))  *   (n/e)^n

(Yea .. I know it is  just a little trick to add  that additional 1/12n..)

P.S -    My son  tells me  that easiest method is to just type 100!
in  [url=http://www.google.com/search?hl=en&lr=&ie=UTF-8&q=100%21&btnG=Search]google[/url]  and that gives:
100 ! = 9.33262154  10^157    :lol: :)


For a positive integer $n$ let $d(n)$ be the sum of its digits.when written in demical form.Prove that for each positive integer $k$ there exists a positive integer $m$ such that the equation    $x+d(x)=m$ has exactly $k$ positive integer solutions.


inequalities
trigonometry
inequalities proposed

$ 0<x<y<z<\frac{\pi}{2}$
Prove that:$ \sqrt{2}\plus{}2\sin x\cos y\plus{}2\sin y\cos z>\sin 2x\plus{}\sin 2y\plus{}\sin 2z$

[quote="hxy09"]$ 0 < x < y < z < \frac {\pi}{2}$
Prove that:$ \sqrt {2} \plus{} 2\sin x\cos y \plus{} 2\sin y\cos z > \sin 2x \plus{} \sin 2y \plus{} \sin 2z$[/quote]

long time ago I have seen this problem in this link:
http://www.52aosai.com/bbs/dispbbs.asp?BoardID=8&replyID=8858&id=2055&skin=0

[quote="can_hang2007"][quote="hxy09"]I think it indeed a nice and hard problem :)[/quote]
Your proof is very nice, hxy09. And here is an ugly one (maybe you and some others dont like this proof):

Setting $ a \equal{} \sin x,$ $ b \equal{} \sin y$ and $ c \equal{} \sin z.$ Since $ 0 < x < y < z < \frac {\pi }{2},$ we have $ 0 < a < b < c < 1.$ The inequality becomes
\[ a\sqrt {1 \minus{} a^{2}} \plus{} (b \minus{} a)\sqrt {1 \minus{} b^{2}} \plus{} (c \minus{} b)\sqrt {1 \minus{} c^{2}} < \frac {\sqrt {2}}{2}.\]
By the AM-GM Inequality, we have
\[ \sqrt {1 \minus{} a^{2}}\leq \frac {2(1 \minus{} a^{2}) \plus{} 1}{2\sqrt {2}},\quad\sqrt {1 \minus{} b^{2}}\leq \frac {2(1 \minus{} b^{2}) \plus{} 1}{2\sqrt {2}}, \\
\sqrt {1 \minus{} c^{2}}\leq \frac {2(1 \minus{} c^{2}) \plus{} 1}{2\sqrt {2}}.\]
Therefore, it suffices to prove that
\[ a(3 \minus{} 2a^{2}) \plus{} (b \minus{} a)(3 \minus{} 2b^{2}) \plus{} (c \minus{} b)(3 \minus{} 2c^{2}) < 2,\]
or
\[ 2a(b^{2} \minus{} a^{2}) \plus{} b(3 \minus{} 2b^{2}) \plus{} (c \minus{} b)(3 \minus{} 2c^{2}) < 2.\]
Fix $ b,$ we will find the upper bounds of the two expressions
\[ f(a) \equal{} a(b^{2} \minus{} a^{2})\]
and
\[ g(c) \equal{} (c \minus{} b)(3 \minus{} 2c^{2})\]
for $ 0 < a < b < 1$ and $ c > b.$

From the AM-GM Inequality, we have
\[ a(b^{2} \minus{} a^{2}) \equal{} \sqrt {\frac {1}{2}\cdot 2a^{2}\cdot (b^{2} \minus{} a^{2})^{2}} \leq \sqrt {\frac {1}{2}\cdot \left[ \frac {2a^{2} \plus{} 2(b^{2} \minus{} a^{2})}{3}\right]^{3}} \equal{} \frac {2}{3\sqrt {3}}b^{3}.\]
Hence
\[ f(a)\leq \frac {2}{3\sqrt {3}}b^{3} < \frac {7}{18}b^{3}. \eqno (1)\]
Now, we will find the upper bound of $ g(c).$ From
\[ g^{\prime }(c) \equal{} \minus{} 6c^{2} \plus{} 4bc \plus{} 3,\]
we find that $ g^{\prime }(c) \equal{} 0$ for $ c \equal{} c_{0} \equal{} \frac {2b \plus{} \sqrt {4b^{2} \plus{} 18}}{6}$, $ g^{\prime }(c) > 0$ for $ b < c < c_{0}$ and $ g^{\prime }(c) < 0$ for $ c > c_{0}.$ This means that $ g(c)$ is increasing on $ (b,c_{0}]$ and decreasing on $ [c_{0}, \plus{} \infty ).$ According to this result, we get $ g(c)\leq g(c_{0})$ for $ c > b,$ that is
\[ g(c)\leq \frac {4}{27}b^{3} \minus{} 2b \plus{} \frac {1}{27}(2b^{2} \plus{} 9)\sqrt {4b^{2} \plus{} 18}. \eqno(2)\]
From (1) and (2), we see that it suffices to prove that
\[ \frac {7}{9}b^{3} \plus{} b(3 \minus{} 2b^{2}) \plus{} \frac {4}{27}b^{3} \minus{} 2b \plus{} \frac {1}{27}(2b^{2} \plus{} 9)\sqrt {4b^{2} \plus{} 18} < 2,\]
or
\[ b(27 \minus{} 29b^{2}) \plus{} (2b^{2} \plus{} 9)\sqrt {4b^{2} \plus{} 18} < 54.\]
In the nontrivial case $ 27 > 29b^{2},$ using the AM-GM Inequality, we get
\[ b\leq \frac {16b^{2} \plus{} 9}{24}\quad\text{and}\quad\sqrt {4b^{2} \plus{} 18}\leq \frac {(2b^{2} \plus{} 9) \plus{} 8}{4},\]
and hence, it suffices to show that
\[ \frac {16b^{2} \plus{} 9}{24}(27 \minus{} 29b^{2}) \plus{} (2b^{2} \plus{} 9)\frac {2b^{2} \plus{} 17}{4} < 54.\]
This inequality reduces to
\[ 440b^{4} \minus{} 483b^{2} \plus{} 135 > 0,\]
which is true according to the AM-GM Inequality.[/quote]

Hi friend,I like your incredible proof so much :!:
Your proof leave me a strong impression of power of [b]Am-Gm[/b]
By the way,I didn't solve this problem.The skillful approach was offered by my Olympiads book :D


On the unit circle the point ((-(sqrt 21)/5), 2/5) is the endpoint of arc 'z'. Find the coordinates of the endpoint of the arc pi+z

Consider the point as a complex number :-sqrt21/5+2/5i and multiply it by i^2


Hi, I've been reading around and some sources seem to suggest that the use of Lagrange multipliers and certain other techniques are actually penalised at the IMO. Is this true, even when a solution is perfectly valid and proof correct, or even when Lagrange multipliers are the most elegant way to solve a problem?

This would seem a little random and illogical.

If the solution is [b]perfectly valid[/b] then you will always get a 7/7.

However, if you use some ugly calculus method then you need a lot of 'analytic good manners'. Coordinators could point to something that isn't really an omission in your solution but is slightly unclear, and insist on taking marks away- though if you use a more traditional method they may not make such a fuss over such slight errors.

It is best to avoid using any calculus in the IMO, (likewise cartesian coordinates) unless you have no choice or are sure of of the clarity in your write up.

Perhaps someone with more experience could clarify this.

I think most of the IMO problems can not be solved by the use of Lagrange multipliers.The people who draw out theme will avoid it.


Let $a_1,a_2...a_n$ and $b_1,...,b_n$ be integer numbers. And for any i, $0\le i \le n$ the following equality holds:
\[ \sum_{i_1+i_2+...+i_n \equiv i (mod n)}a_{i_1}a_{i_2}..a_{i_n}= \sum_{i_1+i_2+...+i_n \equiv i (mod n)}b_{i_1}b_{i_2}..b_{i_n}  \]
where $i_j \in (1,2,...,n) \forall j \in (1,2..,n)$ may not be distinct.
Prove that $a_i=b_i, \forall i \in (1,2,...,n)$.

Still no $k$...

Did you mean

$\sum_{i_1+i_2+\cdots+i_n \equiv k(mod n)} a_{i_1}a_{i_2} \cdots a_{i_n} = \sum_{i_1+i_2+\cdots+i_n\equiv k(mod n)} b_{i_1}b_{i_2} \cdots b_{i_n}$

$\forall k \in \{0,1, \cdots ,n-1\}$

I found a solution, but i think that n must be prime. So I will use p instead of n from now on. Denote by $\xi_p$ the p-th root of the unity. Consider polynomials $f=\sum_{i=0}^{p-1} a_iX^i,g=\sum_{i=0}^{p-1} b_iX^i$. Condition of the problem can be written as:
\[ f^p(\xi_p^k)=g^p(\xi_p^k). \].Extracting p-th rootfrom this equality we have:
\[ f(\xi_p^k)=g(\xi_p^k)\xi_p^{j_k}. (1) \]
Assume that there exist $i,m$ such that $j_i%Error. "neqj" is a bad command.
_m$. If there not exist we have that $f-\xi_p^kg=0$ for p values:$\xi_p^k,k\in\{1,..p\}$ and have degree ${p-1}$. So $f=\xi_p^kg$. So f=g, because f,g have integer coeff. and $a_0\neq b_i\xi_p^k$.
Now  $j_i \neq j_m$. Using the fact that numbers of the form $\xi_p^k, \forall k\in \{1,..p\}$ are linear independent over $\mathbb{R}$ we can indentify the coeff in (1) and obtain $a_t=b_{t+j_m}=b_{t+j_i}\forall t \in \{1,..p\}$. Now it is obvious that all $b_i$s are equal, and then, from (1) all $a_i$s are equal.
Now from (1) we have that $a_i=b_j$ for some i and j. So, we have $a_i=b_i \forall i\in \{1,..p\}$
Sorry for mistakes :blush: .


Will any of the AoPS staff be coming this year? I'd really like to meet them. If so, when?

The instructor list is not yet finalized.

[quote="DrKevin"][quote="Dojo"]Surely, it must be finalized now?[/quote]
You can visit the Math Zoom website ([url]http://www.mathzoom.org/[/url]) for faculty information.  Besides myself, we have Dr. Reiter, Dr. Dreyer, Professor Davis (AMC problem writer), Sean Markan, Kelly Ren, and a few more.  Invited speakers include Arthur Benjamin from Harvey Mudd College (The Mathemagician) and Richard Rusczyk.[/quote]

I think you should distinguish more clearly in the website between teachers from previous years and teachers for this year.

[quote="DrKevin"]You can visit the Math Zoom website ([url]http://www.mathzoom.org/[/url]) for faculty information.  Besides myself, we have Dr. Reiter, Dr. Dreyer, Professor Davis (AMC problem writer), Sean Markan, Kelly Ren, and a few more.  Invited speakers include Arthur Benjamin from Harvey Mudd College (The Mathemagician) and Richard Rusczyk.[/quote]

Yes.

[quote="Dojo"]Are they coming for all weeks?[/quote]
Most of the instructors will be there for all 3 weeks.  Dr. Reiter will be there for two weeks.

[quote="helagha"]I agree with Differ.
Is Mr. Rusczyk coming this year?[/quote]
Richard will give a talk on Sunday 8/9 afternoon.

[quote="Differ"][quote="DrKevin"]Richard will give a talk on Sunday 8/9 afternoon.[/quote]

My guess is that it will have something to do with trading and stocks.

Hmm...I think I've heard this speech before somewhere... :roll:[/quote]
You will learn something new each time :)


function
trigonometry
vector
inequalities proposed
inequalities

Find the minimum and maximum of $A{\cos}^{2}(x)+B{\sin}(x){\cos}(x)+C{\sin}^{2}(x)$. For what $x$ is equality achieved?

[quote="kunny"]The answer is $(A-C,\ B)\parallel (\cos 2x,\ \sin 2x).$[/quote]
What do you mean? :?:

This means that two vectors are parallel.


From the first equation

$b_n=\frac{a_n-a_{n+1}}{4}\qquad (*)$

implying

$b_{n-1}=\frac{a_{n-1}-a_n}{4}\qquad(**)$

From the second equation $b_n=2a_{n-1}+7b_{n-1}$. Substituting that into the first, we get $a_{n+1}=a_n-8a_{n-1}-28b_{n-1}$. Plugging $(**)$ in, we obtain

$a_{n+1}-8a_n+15a_{n-1}=0$

The characteristic equation is $t^2-8t+15=0$, and the solutions are $t_1=5$ and $t_2=3$. Hence

$a_n=A\cdot 5^n+B\cdot 3^n$ where $A,B$ are some constants.

Now from $(*)$ we get

$b_n=-A\cdot 5^n-{B\over 2}\cdot 3^n$

(using, of course, $5^{n+1}=5\cdot 5^n$ and $3^{n+1}=3\cdot 3^n$)

From the initial conditions we get

\begin{eqnarray*}5A+3B &=& 1\\-5A-{3\over 2}B &=& 2\end{eqnarray*}

The solution is $A=-1, B=2$

Therefore

\begin{eqnarray*}a_n &=& 2\cdot 3^n-5^n\\b_n &=& 5^n-3^n\end{eqnarray*}

It's easy to check that this solution satisfies the given equations.

That's right. :) Here is another approach.

We can determine the coefficients $p,q$ such that $a_{n+1}+pb_{n+1}=q(a_n+pb_n).$


We have a triangle ABC. We draw the incircle, with center O. We mark where the incircle hits AB, BC, and CA as X, Y, and Z, respectively. Prove that $ [ABC] \equal{} AX * BY * CZ / OY$

[hide]
Remember the formula of area to inradius...

[/hide]

I had a lot of trouble creating the problem but obviously there's a faster way than the one I used.

[hide] Let the distance from each of the vertices tangent to the circle be x,y, and z. Then, the lengths of the sides are x+y, y+z, and x+z. By Heron's we have A=$ \sqrt(sxyz)$, where s is the semiperimeter and A is the area, so $ A^2 \equal{} sxyz$. However, we know A=rs, where r is the inradius. So we have $ A(A/s) \equal{} xyz$, $ Ar \equal{} xyz$, and $ A \equal{} xyz/r$, which is what we wanted to prove. [/hide]


geometry
circumcircle
power of a point
radical axis
geometry proposed

In triangle $ ABC$ let $ AM,BN,CK$ be the altitudes. Let $ P$ be a point not on any altitudes. Show that the circumcircles of triangles $ AMP,BNP,CKP$ have a common point other than $ P$.

It is true (at least if you read "the circumcircles of triangles $ AMP,BNP,CKP$ have a common point other than $ P$" as "... or touch each other at $ P$").

  darij

The orthocenter of $ ABC$ has the same power with respect to all three circles.  So the three circles have a common radical axis and you're done.


:?: two bikes are 100 miles apart, facing directly at each other.  bike A travels at 6 mph and bike b travels at 4 mph.  a bug is flying from the front of bike A to the front of bike B and back, over and over again until the two bikes meet.  If the bug flys at 2 mph, and the bikes start toards each other at the same time, how far will the bug have traveled when the two bikes meet?  the bug starts on bike A. :?:

[hide]The bikes cover 10 mph together, so 100 miles takes 10 hours.  The bug flies at 2 mph for 10 hours, so the distance is 20 miles.[/hide]

my SCIENCE teacher asked me this!!!

lol, it is hard if you aren't having a good day

jorian

[quote="zserf"]If the bug is flying at a speed slower then the speed of either bike how can it fly back and forth?[/quote]

It obviously doesn't fly back and forth before bike b travels to it...

Since it is a physical problem, it would eventually become impossible; in acutality the problem it is an infinite sum.  This sum will have a limit of 20, as said above.

there is nothing wrong with the question.  it merely asks how much will the bug have traveled.  the first solution posted is legitimate.  there is no statement that says the bug must be in between the bikes or crushed.

[quote]Funny... I could have sworn I saw a problem like this on a nat practice somewhere... think I got it wrong too...[/quote]
yeah after i posted this i found a similar problem in the art of problem solving book "volume 1: the basics" in chapter 3 problem number 51

I summed the infinite series! :blush: that took a bit. :wallbash:

Later I tried using time and got the same answer.....

What was the infinite series that you summed, philB?

Distances the bug flew... it took me forever to find the right numbers...

I meant the actual infinite set, not its description.


superior algebra
superior algebra unsolved

Hi,

I was just browsing through Langs "Algebra" to learn more about the free product of groups. In order to understand the whole section I was wondering how to rigorously construct the disjoint union / coproduct in the category of sets. For two arbitrary sets $ S$ and $ S'$ he constitutes the existence of a set $ T$ having the same cardinality as $ S'$ but being disjoint from $ S$. Why is that so? How do you technically construct this set $ T$?

laxton

Usually the disjoint union of sets is described as $ X\coprod Y\equal{}X \times \{0\} \cup Y\times \{1\} \subset \left(X\cup Y\right)\times \{0,1\}$


Find the general solution to the first order differential equation

xy^2(dy)-(x^3+y^3)(dx)=0

Don't be so quick to dismiss kunny's hint; it does help. This is a so-called "homogenous" equation (not the linear meaning of that word) and kunny's hint is simply the standard solution technique for such equations.

Write the hint as $y=xz$ and use the product rule in computing $dy.$ The equation should simplify down to a separable equation in $x$ and $z.$


Find all functions $f: R\to R$ such that:
\[ f(x^2)-f(y^2)=(x+y)(f(x)-f(y)), x,y \in R \]

If $f$ is solution so is $f - cst$ so we may assume for now $f(0) = 0$

Then $f(x^2) = xf(x)$ and equation is equivalent to $f(y)/y = f(x)/x = a$ for $x,y \neq 0$

Solutions are $f(x) = a*x + b$

why can we assume that $ f(0) \equal{} 0$ ? I don't understand you...

By "cst" he meant "constant".

In other words, a particular function $ F(x)$ is a solution iff. every function of the form $ F(x) \plus{} k \; \; \forall k \in \mathbb{R}$ is a solution.

Then, for every such family of functions, it suffices to proceed using the member of the family (or the value of $ k$) that makes $ f(0) \equal{} 0$.


limit
logarithms
integration
calculus
real analysis
real analysis unsolved

A very nice problem proposed by O. Skaskiv, Ukrainian Mathematical Olympiad for university students 2007 is the following one:

$ \lim_{x\rightarrow \infty}\frac{x}{\ln x}\sum_{n=0}^{\infty}\frac{1}{2007^{n}+x}=\frac{1}{\ln 2007}$.

Here is a generalization of his result:

Let $ a>1$. Prove that

$ \frac{1}{\ln a}=\lim_{x\rightarrow \infty}\frac{x}{\ln x}\sum_{n=0}^{\infty}\frac{1}{a^{n}+x}$.

One has just to note that $ \int_{1}^{\infty}\frac{dy}{x+a^{y-1}}\leq \sum_{0}^{\infty}\frac{1}{x+a^{n}}\leq \int_{0}^{\infty}\frac{dy}{x+a^{y}}$. The integral can be easily computed after substitution $ t=x+a^{y}$.

That's not really a generalization; I can't imagine a proof for 2007 that wouldn't also work for some other large number. The year here isn't doing anything special; it's just an arbitrary large parameter.

Yustas,

You are right. That is the trick.

Let $ k>0$ and $ a>1$. Then

1) $ \lim_{x\rightarrow \infty}\frac{x^{k}}{\ln x}\sum_{n=0}^{\infty}\frac{1}{a^{n}+x^{k}}=\frac{k}{\ln a}$.

and

2) $ \lim_{x\rightarrow \infty}\ln \ln x\left(\frac{x^{k}}{\ln x}\sum_{n=0}^{\infty}\frac{1}{a^{n}+x^{k}}-\frac{k}{\ln a}\right)=0$.

The solution is based on the estimate

$ \frac{\ln(1+x^{k})}{\ln a\ln x}\leq \frac{x^{k}}{\ln x}\sum_{n=0}^{\infty}\frac{1}{a^{n}+x^{k}}\leq \frac{\ln(1+x^{k})}{\ln a\ln x}+\frac{x^{k}}{1+x^{k}}\frac{1}{\ln x}$,

which follows, as you observed, from the following one:

$ \int_{0}^{\infty}\frac{dt}{a^{t}+x^{k}}\leq \sum_{0}^{\infty}\frac{1}{a^{n}+x^{k}}\leq \int_{0}^{\infty}\frac{dt}{a^{t}+x^{k}}+\frac{1}{1+x^{k}}$.

[quote="didilica"]Let $ k>0$ and $ a>1$. Then

$ \lim_{x\rightarrow \infty}\frac{x^{k}}{\ln x}\sum_{n=0}^{\infty}\frac{1}{a^{n}+x^{k}}=\frac{k}{\ln a}$.
[/quote]

that's a nice problem.   :thumbup:

i will post an [i]alternate [/i]solution later. ( atm, i am debating
whether to ask this question on an exam or not... the thing is i don't
think anyone in the class will be able to prove it... it's quite tricky imo...  :| )


induction
logarithms
probability
expected value
probability and stats

I must find expected value $ E(M)$.

I know that $ E(M)\equal{}2P(M\equal{}1)$

I need a $ P(M\equal{}1)$

I know that $ \frac{P(M\equal{}k)}{P(M\equal{}k\minus{}1)}\equal{}\frac{1}{2} (1\minus{}\frac{1}{k})$, for k=2,3,4,...

Is there a smart way to count $ P(M\equal{}1)$?

Thank you for help :)

I am assuming that the state space of the random variable $ M$ is $ \mathbb{N}$.

Show by induction that ${ P(M = k) = \frac {P(M = 1)}{2^{k - 1}k}}$ and hence use the normalization condition $ 1=\sum_{k = 1}^\infty P(M = k)$ to show that $ P(M = 1) = \frac {1}{\sum _{k = 1}^\infty \frac {1}{2^{k - 1}k}} = \frac {1}{2\log 2}$