vector
linear algebra
linear algebra unsolved

$ U$ and $ V$ are vector spaces over the same field $ F$ and $ W$ the subspace of $ U$. Further on $ A: W \rightarrow V$ is the linear operator. Show, that there exists such linear operator $ B: U \rightarrow V,$ that $ B(x)\equal{}A(x)$ for all $ x\in W$.

If $ U$ is finite dimensional:

Find a basis $ {w_1,w_2,\dots,w_k}$ for $ W.$ Extend this to a basis $ \{w_1,\dots,w_k,u_{k \plus{} 1},\dots,u_n\}$ for $ U.$ Define $ B(w_j) \equal{} A(w_j)$ and make some arbitrary assignment for each $ B(u_j).$

If $ U$ is infinite dimensional:

Try to use something like the same proof - only this time, you're going to need to invoke the Axiom of Choice or one of its equivalents.


who is going to be in the world series and who will win

i say its going to be yankees vs. braves and the yankees are going to win

I don't follow baseball nearly enough to make an informed prediction, but as a semi-random guess, I'll take White Sox over Cards in six.

well it wont be the braves considering the 18 inning game so i am going with Cards vs. Sox in 5 Cards win

Haha. So Houstonians are hoping for Astros, which won't happen. But it was nice anyways.

I see Cards going at it...and the Sox may be a good pick too.

Stros-Halos. Houston wins in 6.

Since the Red Sox are out of it, I will spend the next 3 weeks whining and lamenting about how playoff baseball shouldn't be shown on Tuesdays on Fox because it wipes out House MD. Grr.

And the Yankees just lost. That collision... and that call in the 5th...
It just wasn't their day, I guess. :(

angels? no way...
doesn't matter, yanks have a good shot next year, and the year after that, and the year after that....

sigh..Houston?

Here's my predictions.

Angels over White Sox, 4-3
Cardinals over Astros, 4-2
CARDINALS over Angels, 4-1

I'm pretty sure the Cardinals are going to win this year.  They just have so much talent.

[quote="yifman12"]whining and lamenting about how playoff baseball shouldn't be shown on Tuesdays on Fox[/quote]

See this story:
[url=http://www.theonion.com/content/node/41612]Fox Cancels ALCS After Just Two Episodes[/url]

[The preceding is not to be taken seriously]

i bet the white sox  :D  and the astros :D

*censored* IT. HTF DID THE ASTROS LOSE LAST NIGHT? WHAT IDIOT OF A PITCHER WOULD LET A HOMER GO IN THE FREAKING 9th INNING!!! AHHHH.

Okay *breathes*. Right so...they HAD a shot at the World Series. It's probably not happening now though.

Astros winning in St. Louis again...I mean it was a miracle the first time. I'm not sure it could happen again.

Calm down, MithsApprentice.

Albert Pujols is a great baseball player, and that was a home run that will live in legend, along with other examples like Bobby Thomson off Ralph Branca, Jack Clark off Tom Niedenfuer, Kirk Gibson off Dennis Eckersley, and so on.

Now, for some more objectivity. How likely is St. Louis to win each of the remaining games? 60%? In baseball terms, that would be a large advantage - maybe the home advantage means something, maybe the Cardinals really are the better team, maybe last night will have psychological aftereffects. Maybe 60%. If that were the case, the Cardinals would have a 36% chance of winning the series and the Astros a 64% chance. If you moved to odds on each game to 70%, which is surely too large, the odds for the whole series would only approach 50-50.

I'd like the Astros to win, but you've gotta count in the psychological hangovers with this stuff. The Yankees really didn't have a chance to win game 7 last year against the Red Sox. They had already blown it twice and gotten beaten by guy who was pretty close to a cripple.

Probably the most similar example: 1986 ALCS, Angels-Red Sox. Angels were up 3-1, a strike away from going to the world series. Dave Henderson hits a home run to win the game for the Sox. When the Angels went to Fenway for the last two games, they didn't have a chance in either.

Not saying the Astros won't change it, but it'll be tough.

i think its game 6 or bust but brad lidge had to frickin walk david eckstein just throw the heater down the middle whats the worst that could happen right?? i mean there is a chance he will miss it or not have enought power and the inning is over

Dude, Lidge walked Edmonds, not Eckstein. Eckstein hit a bleeder through the left side.

[quote="pkothari13"]yeah silly lidge.  how perfect. ONE STRIKE AWAY.   :censored:  and they're playing in busch today. super! if the astros don't go to the world series, i am going to have to resort to violent tactics...  :bomb:[/quote]

We can wreck havoc on the non-Houston people at TAMU this Saturday, PK. :D *cough*

But I doubt that'll be necessary. We're 5-1 and looking good.

EDIT:
WEE WINNNNNNNNNNNNNNNNN!!!

Dominant performance by Oswalt. Very very impressive. Should be a great World Series for pitching.

wooooooooo. looks like i wont have to be violent........

if they win the world series, that is.  :D

this should be a really good match up  :) but i think the astros can take 'em!

and i agree about the pitching comment you made whole heartedly. i can't wait to see what happens.

very happy for houstonians that their astros are finally in the WS.

go astros!


Please post your AMC12 score and your AIME score. I just want to see whether most people made it with approximately a 150/8 or more with approximately a 130/10 or so.

I got 144/10. I know someone who got 150/6 in 2002 and (I think) set the AIME floor for younger kids at 6. Many people should have been thanking him.

[quote="WarpedKlown1335"]What tests did you all take?  The AMC-10 or the AMC-12?[/quote]

Some of these folks were appear to have been in grades <=10 so their AMC scores did not really count.

Trying to remember -- I think I had a 12 on AIME, so 150/12.  I might have had a 10 or 11, though.  Actually, I could (theoretically) go look this up.  Or so could anyone else with the yellow book they sent out.

Aw..hasn't anyone ever gotten a low AMC12 score and still made it to USAMO because of a big AIME score? Like, oh...say, for example...with a 117 :wink: (searching for hope :) )

Start practicing for the AIME ...

or the AMC B :)

[quote="JBL"]
or the AMC B :)[/quote]

Our coach is fumbling with the registration stuff right now. I'm just praying that he'll get the job done and that the AMC guys will still let us register. 8-)

[quote="Ragingg"]Well this year the AMC-12 was a tid harder I think... Maybe it will be ok to get lower.[/quote]

No, I'm not going to try to fool myself.  :mrgreen:

i dont mind if i have 100.0 on amc 12. come on aime.

well it doesn't really matter
and i don't think a 150/6 will get u in, unless its 150 on da 12. otherwise, it'll be a close shave

but 2 factors are significant:
1.whether ur senior or not, underclassmen definitely get advantage
2. took the 10 or 12. 12 helps more than da 10.

And uh, AIME score, if ur underclassmen, 7 will be most likely, 8 probably definitely. Again, 6....ehhh.................close

OOO, post #1000. I had 1200 be4, but got deleted cuz they in the "other" forum haha


algebra
polynomial
inequalities
function
logarithms
number theory
prime numbers

Let $f\left(x\right)$ be a non-constant polynomial with integer coefficients, and let $u$ be an arbitrary positive integer. Prove that there is an integer $n$ such that $f\left(n\right)$ has at least $u$ distinct prime factors and $f\left(n\right) \neq 0$.

After seeing the solution you gave, I thought that it is the hardest problem of number theory ever given! In fact, we can give a very short solution, using lemma 1 (Schur theorem, I think in the literature). Indeed, consider $ p_1,...,p_k$ different large prime numbers such that there exist $ n_1,...,n_k$ for which $ p_i| f(n_i)$ and $ f(n_i)$ is non-zero (this follows from lemma 1). Choose a number $ n$ congruent to $ n_i$ mod $ p_i$. There are an infinite number of such numbers and thus in at least one of these numbers the polynomial does not vanish. Even more, the value of the polynomial in each such number is a multiple of $ p_1,...p_k$. Finish!

My solution:

I will reduce the problem to the following lemma:

[b]Lemma 1.[/b] The set of all primes p such that there is an integer x satisfying p | f(x) is infinite.

This Lemma will be proved later. Now I will solve the problem with the help of this lemma.

Since the set of all primes p such that there is an integer x satisfying p | f(x) is infinite, we can take u primes $p_1$, $p_2$, ..., $p_u$ out of this set, and there are some integers $x_1$, $x_2$, ..., $x_u$ such that $p_i \mid f\left(x_i\right)$ for every i with $1\leq i\leq u$. Now, if $y_i$ is an integer such that $y_i \equiv x_i \; \mathrm{mod} \; p_i$, then $p_i \mid y_i-x_i$, while $y_i-x_i \mid f\left(y_i\right)-f\left(x_i\right)$ (this is just because f(x) is a polynomial with integer coefficients), so that $p_i \mid f\left(y_i\right)-f\left(x_i\right)$, and since $p_i \mid f\left(x_i\right)$, this yields $p_i \mid f\left(y_i\right)$. Now, since the numbers $p_1$, $p_2$, ..., $p_u$ are prime and thus pairwisely coprime, the Chinese Remainder Theorem shows that there are infinitely many integers Y such that $Y \equiv x_i \; \mathrm{mod} \; p_i$ for every i with $1\leq i\leq u$. Among these infinitely many integers Y, there is at least one such that $f\left(Y\right) \neq 0$ (else, the polynomial f(x) would have infinitely many zeros, what is not possible for a non-constant polynomial). Denote this number Y satisfying $f\left(Y\right) \neq 0$ by n.

Hence, for these number n, we have $f\left(n\right) \neq 0$ and $n \equiv x_i \; \mathrm{mod} \; p_i$ for every i such that $1\leq i\leq u$. But the latter congruence yields $p_i \mid f\left(n\right)$ for every i with $1\leq i\leq u$. In other words, the number f(n) is divisible by all the u primes $p_1$, $p_2$, ..., $p_u$, and hence, this number f(n) has at least u prime factors. This solves the problem.

Now let's prove Lemma 1. Assume the contrary, i. e. assume that the set of all primes p such that there is an integer x satisfying p | f(x) is finite. Let this set be $\left\{p_1;\;p_2;\;...;\;p_s\right\}$, where s is a positive integer. Then, for every integer x, the value f(x) has the form $f\left(x\right)= \pm p_1^{e_1} p_2^{e_2} ... p_s^{e_s}$, where $e_1$, $e_2$, ..., $e_s$ are nonnegative integers. Hence, of course, $\left|f\left(x\right)\right| = p_1^{e_1} p_2^{e_2} ... p_s^{e_s}$.

On the other hand, let our polynomial f(x) be $f\left(x\right) = a_m x^m + a_{m-1} x^{m-1} + ... + a_1 x + a_0$, with integer coefficients $a_m$, $a_{m-1}$, ..., $a_1$, $a_0$. Hereby, of course, m is the degree of our polynomial. Since a polynomial of degree m cannot achieve one and the same value more than m times (without being a constant polynomial), our polynomial f(x) achieves every of its values at most m times. Clearly, this yields that the function $\left|f\left(x\right)\right|$ achieves every of its values at most 2m times (in fact, for each value v, we have f(x) = v for at most m different values of x and we have f(x) = -v for at most m different values of x, so that we can have $\left|f\left(x\right)\right| = v$ for at most 2m different values of x).

Let B be an arbitrary positive integer. Then, for every positive integer x with $x\leq 2mB$, the triangle inequality shows that

$\left|f\left(x\right)\right| = \left|a_m x^m + a_{m-1} x^{m-1} + ... + a_1 x + a_0\right|$
$\leq \left|a_m x^m \right| + \left|a_{m-1} x^{m-1}\right| + ... + \left|a_1 x \right| + \left|a_0\right|$
$= \left|a_m \right| x^m + \left|a_{m-1}\right| x^{m-1} + ... + \left|a_1\right| x + \left|a_0\right|$       (since x is positive)
$\leq \left|a_m \right| \cdot \left(2mB\right)^m + \left|a_{m-1}\right| \cdot \left(2mB\right)^{m-1} + ... + \left|a_1\right| \cdot 2mB + \left|a_0\right|$       (since $x\leq 2mB$).

In other words, we have $\left|f\left(x\right)\right| \leq t\left(B\right)$, where

$t\left(B\right) = \left|a_m \right| \cdot \left(2mB\right)^m + \left|a_{m-1}\right| \cdot \left(2mB\right)^{m-1} + ... + \left|a_1\right| \cdot 2mB + \left|a_0\right|$.

Of course, t(B) is a polynomial of B, and actually a polynomial which is always nonnegative for B > 0. This will become important later.

Now, consider the interval [1; 2mB]. To each of the 2mB integers in this interval, there corresponds a value of the function $\left|f\left(x\right)\right|$ at this integer. Since the function $\left|f\left(x\right)\right|$ achieves every of its values at most 2m times, at least $\frac{2mB}{2m}=B$ of these values are different. But every such value is $\leq t\left(B\right)$ on the one hand, and on the other hand, every such value can be written in the form $\left|f\left(x\right)\right| = p_1^{e_1} p_2^{e_2} ... p_s^{e_s}$. Hence, there are at least B different numbers of the form $p_1^{e_1} p_2^{e_2} ... p_s^{e_s}$ which are all $\leq t\left(B\right)$. Let r be such a number $r=p_1^{e_1} p_2^{e_2} ... p_s^{e_s}$, with $r\leq t\left(B\right)$. Then, if we would have $e_1 > \log_2 t\left(B\right)$, we would get

$r=p_1^{e_1} p_2^{e_2} ... p_s^{e_s} \geq p_1^{e_1} > p_1^{\log_2 t\left(B\right)}$
$\geq 2^{\log_2 t\left(B\right)}$      (since $p_1\geq 2$, as any prime is $\geq 2$)
$=t\left(B\right)$,

what contradicts $r\leq t\left(B\right)$. Thus, $e_1 \leq \log_2 t\left(B\right)$. Similarly, $e_2 \leq \log_2 t\left(B\right)$, ..., $e_s \leq \log_2 t\left(B\right)$. Therefore, all the integers $e_1$, $e_2$, ..., $e_s$ lie in the interval $\left[0;\;\log_2 t\left(B\right)\right]$. Consequently, the number of all possible numbers of the form $p_1^{e_1} p_2^{e_2} ... p_s^{e_s}$ which are all $\leq t\left(B\right)$ doesn't exceed $\left(\log_2 t\left(B\right) +1\right)^s$.

But remember that we have found at least B different numbers of the form $p_1^{e_1} p_2^{e_2} ... p_s^{e_s}$ which are all $\leq t\left(B\right)$. Hence, $B \leq \left(\log_2 t\left(B\right) +1\right)^s$.

And this holds for every positive integer B ! But it is not hard to show that, for every polynomial t(B) and for every integer s, there is a positive integer $B_1$ such that $B_1 > \left(\log_2 t\left(B_1\right) +1\right)^s$. In fact, this is equivalent to $\sqrt[s]{B_1} > \log_2 t\left(B_1\right) +1$, what, using the substitution $C_1=\sqrt[s]{B_1}$, takes the form $C_1 > \log_2 t\left(C_1^s\right) +1$; since $\log_2 t\left(C_1^s\right) + 1 = \log_2 \left(2t\left(C_1^s\right)\right)$, this is equivalent to $C_1 > \log_2 \left(2t\left(C_1^s\right)\right)$. And this is clear, since $2t\left(C_1^s\right)$ is a polynomial of $C_1$ (since t is a polynomial), and a linear function always overtakes the logarithm of a polynomial. Hence, we have found a positive integer $B_1$ such that $B_1 > \left(\log_2 t\left(B_1\right) +1\right)^s$, contradicting our previous result $B \leq \left(\log_2 t\left(B\right) +1\right)^s$ for every positive integer B; hence, we have a contradiction, and Lemma 1 is proven.

My apologies for the style. It's just the first time I solve and write down the solution of a slightly nontrivial number theory problem.

Indeed, I did not realize that the longest proof was for the lemma 1  :blush:  :blush: and this is because this lemma has a super short solution. I will give some hints: assuming that $ f(0)$ is non-wero, show that we can reduce to the case $ f(1)=1$ and then consider the sequence $ f(1+n!)$. But I particularly like your proof of this lemma. Although long, it has nice ideas.

Oh yes, Darij's long solution misleads about difficulty of this problem. Well known lemma 1 has short nice proof.

You probably know that well-known and known to me are different things. And I must say I couldn't see anything from Harazi's hint. Anyway, the proposed solutions have a very nice and simple proof, so I guess it should be something of the same kind...

  Darij

Here's what Harazi's hint is supposed to mean: if you have $f(1)=1$, then assume that the elements of the set $\{f(n)|n\in\mathbb Z\}$ have only finitely many prime divisors, $p_1,p_2,\ldots,p_k$. Now take $n>\max (p_i)$, and show that $p_i\not |f(1+n!)$, thus getting a contradiction. It's in fact pretty similar to the idea used to show the infinitude of primes.

Thanks, Grobber, for this. Now it remains only one question (to you, Harazi and everyone): Why can I assume f(1) = 1 ? And, please, give a detailed explanation, not just a hint.

Thanks!

  Darij

I remember Harazi's other hint was to use a polynomial transformation. I don't really know what he was talking about, so we'll just have to wait and see.

Meanwhile.. :)

Here's how I see this thing: take a number $t$ for which $f(t)\ne 0$. Let $p_1,p_2,\ldots,p_k$ be the primes dividing the elements of $f(\mathbb Z)$ (just as before). Now take $n$ so large that the exponent of $p_i$ in $n!$ is larger than the exponent of $p_i$ in $f(t)$. Now the exponent of $p_i$ in $f(n!+t)=n!\cdot N+f(t)$ is bounded by the exponent of $p_i$ in $t$, and since we can take arbitrarily large such $n$'s, we get a contradiction (the polynomial can only take a finite number of values in infinitely many points of the form $n!+t$, where $t$ is the $t$ fixed above and $n$ increases indefinitely).

This is somewhat of a change of plan, but I think it gives a better idea of what's going on: even if $f(1)\ne 1$, it can't be that important (as long as $f(1)\ne 0$).

I expect Harazi to have a different answer, since he didn't mention $f(1)\ne 0$, so maybe he has a nice way around that.

Grobber, thanks for the explanation!

[quote="grobber"]This is somewhat of a change of plan, but I think it gives a better idea of what's going on: even if $f(1)\ne 1$, it can't be that important (as long as $f(1)\ne 0$).[/quote]

Well, guarranteeing $f\left(1\right)\neq 0$ is not hard: The polynomial f can have only finitely many roots, hence it can have only finitely many integer roots, and consequently, there is a greatest integer root v of this polynomial; now, after replacing f(x) by f(x + v), it is clear that f(0) = 0 and $f\left(1\right) \neq 0$.

  Darij

[quote="Zorro"]I've posted a solution of this problem some time ago, why it isn't here??[/quote]

Probably due to the problems with database which occured last week


How many multiples of 7 are between 30 and 790?

The largest multiple of 7 less than 790 is $ 790 \div 7 \approx 112 \implies 7 \times 112 \equal{} 784$. Similarly, the smallest multiple greater than 30 is $ 35$. The difference between those numbers is $ 749$. Each multiple is separated by 7, so we divide to get $ 107$. However, that's not our final answer! We didn't count all the numbers - we still need to add one to get an answer of $ \boxed{108}$.


Vieta
analytic geometry
Pythagorean Theorem
geometry

^Title.

"Two different circles are each tangent to both the x-axis and the y-axis, as shown in the figure. One of the points of intersection of the circles is (1,5). What is the product of the lengths of their radii?"

There is a diagram--it's not much. It would be the same if you used your imagination/pencil&compass. Plus, if you're not too lazy, and the handbook is nearby, you could just grab that, too. :P

[hide="Answer"]
26
[/hide]

We looked at this problem and we kept thinking about it. I don't believe what we got was right as it was a radical and the answer was an integer value.

Thanks!

[hide]Draw two right triangles: One from the center of the big circle with center (R,R), and one from the center of the smaller circle (r,r).  Let these share a vertex at (1,5).

The side lengths of the smaller triangle are r, r-1 and 5-r.  Therefore, from the Pythagorean Theorem, $ (5\minus{}r)^2\plus{}(r\minus{}1)^2\equal{}r^2$.  After solving you get $ r^2\minus{}12r\plus{}26\equal{}0$.  Using Vieta's, you already know that the product of the solutions for $ r$ is $ \boxed{26}$.

You don't have to find anything else for the bigger circle because you can write the same set of equations for the bigger circle at R... so the length of R would just be the other solution to the equation.

[/hide]

Or... you could solve using the equations of the two circles.

$ (x\minus{}r)^2\plus{}(y\minus{}r)^2\equal{}r^2$
$ (x\minus{}R)^2\plus{}(y\minus{}R)^2\equal{}R^2$

Since one of the points on both circles is $ (1,5)$
You can plug it in.  Solve by subtracting one equation from the other.

Thanks so much for the answers!
I do have questions for both of you, though. So..I hope you both check this again. xP

@infinity4ever
[quote="infinity4ever"]
The side lengths of the smaller triangle are r, r-1 and 5-r.

Using Vieta's, you already know that the product of the solutions for is 26.
[/quote]

How do you draw the two triangles? I'm confused.
And what/how do you do Vieta's?

@mathcrazed
How do you get those equations for the circle? Is this just a general formula/equation?

Thanks in advance! :D

check the AoPS wiki for Vieta's...

As for equations of circles: If a circle has center $ (h,k)$ and radius $ r$, then the equation of the circle on the xy plane is $ (x\minus{}h)^2\plus{}(y\minus{}k)^2\equal{}r^2$.  For all $ (x,y)$ that satisfy this equation, they lie on the circumference of the circle... also check the aops wiki for a better explanation =) or go to any high school textbook

[quote="The_Scintillator"]
How do you draw the two triangles? I'm confused.
[/quote]

Start with the smaller triangle.  One vertex is at (r,r), another at (1, r) and the other is at (1, 5).  Similarly you can find the coordinates of the larger triangle.


[quote="ccy"]What is the ratio of the area of a square inscribed in a semicircle with radius 10 inches to the area of a square inscribed in a circle with radius 10 inches? Express your answer as a common fraction.[/quote]
[hide]
A line is drawn from an edge of the square of thesemicircle on the semicircle to the midpoint of the side of the square on the diameter of the semicircle. This is 10, and the length of the square is 2x. $(2x)^2+x^2=5x^2=10^2=100$. $x=2\sqrt{5}$, so the area of the square is $(4\sqrt{5})^2=80$. The circle's diameter is 20. $x^2+x^2=400$, $x=10\sqrt{2}$. The area of the square is $(10\sqrt{2})^2=200$. $\frac{80}{200}=\boxed{\frac{2}{5}}$.[/hide]

[quote="X_wing1redleader"][hide]erm, lets see, this is tricky
well, lets say x is the side of the first triangle
there is a 1:2:$\sqrt{5}$ ratio, 10 being the $\sqrt{5}$ part
so, $x=4\sqrt{5}$
the area is 80

half the diagonal of the other is 10, so the whole diagonal is 20
there is a 1:1:$\sqrt{2}$ ratio, 20 being the $\sqrt{2}$ part
therefore, the area is $10\sqrt{2}^2=200$
$\frac{80}{200}=\boxed{\frac{2}{5}}$[/hide][/quote]
wheres the triangle?


Find all natural numbers $n$ and nonnegative integers $x_{1},x_{2},...,x_{n}$ such that $\sum_{i=1}^{n}x_{i}^{2}=1+\frac{4}{4n+1}(\sum_{i=1}^{n}x_{i})^{2}$.

[quote="Rust"]$n\ge 2$.[/quote]
Why? Post your solution!

For any permutation $\sigma$, if $\{x_i\}-$ solution, then $\{x_{\sigma(i)}\}$ solution too. Therefore we can find only solutions $0\le x_1\le x_2\le ....\le x_n$.
Let $s=\sum_i x_i, y_i=nx_i-s \to \sum_i y_i=0$. Then
$\sum_i (y_i+s)^2=n^2+\frac{4n^2}{4n+1}s^2$ or
$\sum_ i y_i^2=n(n-\frac{s^2}{4n+1}).$
Therefore $(4n+1)|s^2$ and $s^2\le n(4n+1)$.
Let $P=\prod_{p|4n+1, v_p(4n+1)-odd} p$. Then $4n+1=Pq^2$ and $s=Pqk$.
Then $\sum_i y_i^2=n(n-Pk^2), Pk^2\le n, \sum y_i=0$  give solution, if $x_i=\frac{s+y_i}{n}\in Z$.
$n=1, 4n+1=5\not =pq^2$ - non solution
$n=2, 4n+1=9=1*3^2, s=3\to x_1=1,x_2=2$ - have solution.
$n=3,4,5$ - have not solution.
$n=6, 4n+1=5^2\to s=5k$ have solution $y_1=-5, y_2=...=y_6=1$ or $x_1=0,x_2=...=x_6=1$.
We can find for all $n=2+9k$, when $9|4n+1$ by add $x_i=0$ 6 times and $x_i=4$ 3 times.

Sorry,I was wrong, it won't work with that.


Find how many octogons have equal angles and lengths 1,2,3,4,5,6,7,8.


[quote="Basketball4Life"]How would you express/write the smallest number greater than 0? Would it be written like .00000bar with a 1 at the end?[/quote]

My advice to you is don't even try.  There really is no smallest number greater than 0.

Suppose that there is a smallest number greater than 0; call it $x$.  $x$ must be positive.

Now, consider the number $\textstyle{\frac{x}{2}}$.  Because $x$ is positive, $\textstyle{\frac{x}{2}}$ must be greater than 0; also, because $x$ is positive, $x>\textstyle{\frac{x}{2}}$.  This means we have found a smaller number than $x$ that is greater than 0, which contradicts the assumption we made that $x$ was the smallest number greater than 0.

Therefore, there is no smallest number greater than 0.

Magnara:

Let dy/dx = 1/2, so that dy = dx/2.  Contradiction.

[quote="ComplexZeta"]First you well-order the reals. (By the axiom of choice, any set can be well-ordered). Then it's easy.[/quote]
Nice joke ComplexZeta  :D


Find the maximum value of $ (a\minus{}b)^4\plus{}(b\minus{}c)^4\plus{}(c\minus{}a)^4$ where $ a,b,c\in [1,2]$

Obviously, we can solve this problem to find max and min of the given expression. Actually, we can deduct that the min is quite easy when $ a\equal{}b\equal{}c$
$ A\equal{} (a\minus{}b)^4\plus{}(c\minus{}b)^4\plus{}(c\minus{}a)^4 \geq 0$
Now, returning to the problem, without the loss of generality, we can assume that $ a \geq b \geq c$
That is we can conclude that $ \minus{}1 \leq a\minus{}b \leq 1$ and keep doing the same results to the other pairs (b,c) and (c,a)
So, it suffices to prove that
$ (a\minus{}b)^4\plus{}(c\minus{}b)^4\plus{}(c\minus{}a)^4 \leq (a\minus{}b)\plus{}(b\minus{}c)\plus{}(a\minus{}c)\equal{}2(a\minus{}c) \leq 2$
The equality holds if and only if $ (a,b,c)\equal{} (2,1,1)$ and its permutations.

[quote="great math"]The equality holds [b]if and only if[/b] $ (a,b,c) \equal{} (2,1,1)$ and its permutations.[/quote]
$ (2,2,1)$ :?:  :huh:

[quote="jbmorgan"]Find the maximum value of $ (a \minus{} b)^4 \plus{} (b \minus{} c)^4 \plus{} (c \minus{} a)^4$ where $ a,b,c\in [1,2]$[/quote]
see my problem http://www.mathlinks.ro/viewtopic.php?t=231147
And it is very easy,see again my hint


I am a 7th grader that won my 2009 school round. I am about to have chapter this weekend. Is there any tips on what I should do? Can someone tell me what problems to do to get ready? Please answer ASAP.   Thank You!!! :D

At least I'm not in your region...our region's top eight will most likely be out school's top eight. :P

And regionals is February 21st for us for some reason...

Get AoPS's problem solving book volume 1. Solve them all. Then you should pwn. xD

I think he wants test taking tips and stuff, because AoPS Volume 1 is really going to get to his doorstep in time for him to finish like 500 (not too trivial) problems...

Use the search function.  Recently there have been many threads asking for tips.

Just relax. It is too late to do a lot of practice and I do not recommend last minute cramming. As for tips, you probably don't need any. Just basic things sleep early, arrive early...

Considering you have chapter this weekend, there's not really much left to do. I also have it this weekend, and I'm planning on just playing FTW for a long time tomorrow :P. Go play (play, not just chat) lots of games on FTW to get into the "mood."

Well, considering that I pwn my school at countdown, I think my only hope is countdown. I have been doing FTW. I usually place 3rd 4th 5th in the mock and school.


probability search geometry Alcumus AMC AIME

[quote="batteredbutnotdefeated"]geometric probability, I say, is the hardest kind of probability... So I want to get better at it. Does anyone know some AMC tests that have geometric probability? I don't want to search all the AMC's.... Also, I'm bad at probability in general, any tips?[/quote]

Don't you have the Intro CP book?  There's an entire chapter on geometric probability.

Try Introduction to Geometry, I know it says Intro., but its still helpful, and you can skip the things you don't need. (I heard that they were coming out with Intermediate Geometry<- Is this true?)

I believe the is talking about Probability, not geometry.

Intro to C&P is quite good, but if you are looking for more, try googling. The best site I could find is Wolfram, but I'm not sure if that is what you are looking for.

http://mathworld.wolfram.com/GeometricProbability.html

Hmm...

http://lmgtfy.com/?q=geometric+probability

[quote="batteredbutnotdefeated"]I have intro to C&P, and it definitely helped, but I still don't think I would be capable of doing AMC 12 level geometric probability...[/quote]

That chapter should be enough to get you through AMC 12 geometric probability.  Otherwise, you probably should learn more in other topics.  For example, it would be pretty hard to do geometric probability if you aren't at least ok at geometry.

Actually, I really enjoy geometry and I'm pretty good at it too... That's not a problem... However, I'm horrible at probability...

[quote="cgyao15"]i mean, couldn't you just look at probablility books...  :maybe:[/quote]

or just ask a teacher.

My math teacher(school) is not the best... Anyways, I think I can ask my coach, but he coaches about 200 students, so he's really busy... Is there a probability marathon? BTW alcumus is kinda messed up on my computer...

Think about what makes you "horrible" at probability. Is it that you don't understand concepts, don't know how to approach problems, or do you just need more practice?

The fact(s) that makes me "horrible" at prob:

1. sometimes I don't know what concept(s) to apply.
2. I [b]over think[/b].
3. Sometimes I don't have a clue of how to approach the problem

On the other hand, I'm actually quite good at counting...

Er, probability [b]is[/b] counting.

In fact, the entire lecture at an AIME class I took was "Probability is the number of successful ways divided by the number of total ways".


What is the largest even integer that cannot be written as the sum of two odd composite numbers?

This problem is definitely in the Contests archive.  Search for it (since you haven't posted the year).


Let $f\in\mathbb{R}[X], \ f=aX^{n+2}+bX^{n}+2, \ n\in\mathbb{N}^{*}$ where $a,b\in\mathbb{R}$ are such that $(X-1)^{2}$ divides $f$. If $q\in\mathbb{R}[X]$ is the quotient of $f/(X-1)^{2}$, find $q(X)$.

[hide="the begining"]Denote $g=(X-1)^{2}$ so $x^{2}\equiv2X-1 (mod \ g)$. By induction we get $X^{n}\equiv nX-(n-1) (mod \ g)$. Now we plug this in $f$ and we get $f\equiv X(an+bn+2a)-a(n+1)-b(n-1)+2 (mod \ g)$. But from the hypothesis we have $f\equiv 0(mod \ g)$ hence we get $X(an+bn+2a)-a(n+1)-b(n-1)+2 \equiv 0 (mod \ g)$ Now in order to find $a,b$ we need to solve a system function of n. $\begin{cases}an+bn+2a = 0 \\ a(n+1)+b(n-1)-2 =0 \end{cases}$, and we get $a=n, \ b=-(n+2)$[/hide].

Hope it's good. But I don't know how to finish it... help  :)

you posted this problem last year http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44980  its good that you came back to it.

but did you learn calculus yet?  i am not positive, but i think apart from carrying through the long division suggested in that thread, that to finish it you'll need to differentiate and maybe use power series.

one way to simplify the work is to write $f(x)$ as $nx^{n}(x^{2}-1)-2(x^{n}-1)$, and define $g(x)=f(x)/(x-1)$, which is then $nx^{n}(x+1)-2(x^{n-1}+x^{n-2}+\ldots+1)$, then look for $g(x)/(x-1)$

I learned calculus pretty well ... I forgot that I posted this problem last year  :blush:

[EDIT]:

Looking at tetrahedr0n's solution I understand it now very well but I have the same question as you had: why did he use $ax+b$ if $f$ divides $(X-1)^{2}$? Thank you very much of reminding me of that topic  :)

[quote="cats..."]he didn't need to, it didn't help but it didn't hurt either.  of course, we know $a=b=0$.

for your current problem, i suggest trying calculus approaches.  maybe look at the answer i gave in that thread and work backwards to see some motivation.[/quote]

I understood very clearly your solution and thanks a lot for that tip :). I'll use it next time when I have to.

While I was doing another problem ... I thought we had to find here $q(1)$ which we found it by differentiation ..but what about $q(X)$?


geometry
circumcircle
Euler
parallelogram
geometric transformation
perpendicular bisector
geometry proposed

Let $ H,O$ be the orthocenter and circumcenter of the acute angled $ \Delta ABC$. The circumcircle and the Euler circle of $ \Delta HBC$ meet at $ \{M,N \}$. Let $ O^*$ be the circumcenter of $ \Delta AXY$ and $ L$ the midpoint of $ OA$. Prove that $ H,O^*,L$ are collinear

I hope you like my problem , and I hope that is not trivial   :blush: 'cause I want to see nice solutions coming from you, thanks

u can easily check that the Euler-circle of $ \Delta ABC$ and $ \Delta HBC$ are the same.
Let $ A'$ be the projection of $ A$ on $ BC$ and $ D$ the midpoint of $ AH$.
Let $ O$ be the circumcenter of $ \triangle{ABC}$.
Let $ J$ be a point on the ray $ AN$($ N$ the center of nine-points circle of $ \triangle{ABC}$) such that $ AN \equal{} NJ$.
The intersection points of the euler circle of $ \triangle{HBC}$ and its circumcenter are $ X,Y$.
u can easily prove that the circumcenter of $ \triangle{HBC}$ ,which will be called $ O'$,is in fact $ J$.
So we can conclude that $ A,O',N$ lay on the perpendicular bisector of $ XY$.
It means that $ O^{*}$ lies on $ AN$. the problem asks to prove that it lies on $ HL$.But $ AN$ and $ HL$ are the medians in
$ \triangle{AHO}$.We deduce that it remains to prove that $ O^{*}$ is the centroid of $ \triangle{AHO}$.Or in other words to prove that $ 6.O^{*}N \equal{} AO'$.
Let $ K$ be the projection of $ A$ on $ XY$.BY using (excuse me if i wrote it wrongly) Keysey's Theorem On the point $ A$ and circles $ (N),(O^{*})$ we get $ 2.AK.NO^{*} \equal{} P_{C(N)}^{A} \equal{} AD.AA'$.By using again this on the point $ A$ and circles $ (O'),(N)$ we get $ 2.AK.NO' \equal{} 2.AK.AN \equal{} |P_{C(O')}^{A} \minus{} P_{C(N)}^{A}| \equal{} |AH.2AA' \minus{} AD.AA'| \equal{} AA'.3AD$.By these two equations we have $ AO' \equal{} 6.O^{*}N$.As we wanted.It is 4:14 AM here,'MUST go to bed!

Let $ O''$ be on $ AO'$ such that $ AO'' \equal{} \frac {1}{3}AO'. (O'',O''A)\cap (O') \equal{} \{M,N\}$. We will show that $ M,N\in (E)$.
Put $ R_{BHC} \equal{} R, O''A \equal{} d$
$ O'N^2 \minus{} O''N^2 \equal{} O'K^2 \minus{} O''K^2 \equal{} R^2 \minus{} d^2 \equal{} (O'K \minus{} O''K)O''O'$
$ \equal{} (O'K \minus{} O''K)2d$
$ \Rightarrow O'K \minus{} O''K \equal{} \frac {R^2 \minus{} d^2}{2d}$
$ \Rightarrow O'K \equal{} \frac {R^2 \plus{} 3d^2}{4d}$, but $ O'E \equal{} \frac {3}{2}d$ then $ EK \equal{} \frac {R^2 \minus{} 3d^2}{4d}$
So $ O'N^2 \minus{} EN^2 \equal{} O'K^2 \minus{} EK^2 \equal{} (\frac {R^2}{4d} \plus{} \frac {3d}{4})^2 \minus{} (\frac {R^2}{4d} \minus{} \frac {3d}{4})^2 \equal{} \frac {3}{4}R^2$
$ \Rightarrow EN \equal{} \frac {1}{2}R \Rightarrow M,N$ lies on $ (E)$.
We are done!
[b]Remark:[/b]
From this nice problem, we get the result:
$ (O'',O''A)\cap (O) \equal{} \{D\}, F$ is the midpoint of $ BC$ then $ D, H, F$ are collinear.

[quote]Let $ H,O$ be the orthocenter and circumcenter of the acute angled $ \Delta ABC$. The circumcircle and the Euler circle of $ \Delta HBC$ meet at $ \{M,N \}$. Let $ O^*$ be the circumcenter of $ \Delta AXY$ and $ L$ the midpoint of $ OA$. Prove that $ H,O^*,L$ are collinear[/quote]

[hide=" Solution"]First let me introduce a lemma.

[b]LEMMA-[/b] [b][color=darkblue]Let $ ABC$ be a triangle with $ H$ is its orthocenter. Denote $ H_a,H_b,H_c$ respectively by the projections of $ H$ onto $ BC,CA,AB$. Let $ P_b,P_c$ respectively be the intersections of $ H_aH_c$ with $ HB$; $ H_aH_b$ with $ HC$. Let $ K_b,K_c$ respectively be the intersections of $ (HAB)\cap ([AC])$; $ (HAC)\cap ([AB])$. Denote $ I_a,O_a,$ $ \mathcal {E}$ respectively by the circumcenters of $ (AK_bK_c)$, $ (HBC)$ and the Euler circle wrt $ \triangle ABC$. Prove that $ \mathcal {E}$ $ ,O_a,I_a\in d_a$[/color][/b]

[b]Note-[/b] $ ([PQ])$ here means the circle with diameter $ PQ$.

[i]Proof[/i]
Consider the inversion through pole $ A$, power $ k \equal{} \overline {AH}\cdot \overline {AH_a}$ $ \equal{} \overline {AH_c}\cdot \overline {AB}$ $ \equal{} \overline {AH_b}\cdot \overline {AC}$. We have $ \mathcal {I}(A,k):$ $ H_c\mapsto B,$ $ H_b\mapsto B,$ $ H\mapsto H_a$. Therefore by $ \mathcal {I}(A,k)$, $ H_cH_a\mapsto (AHB)$, $ H_aH_b\mapsto (AHC)$, $ HB\mapsto ([AC])$ and $ HC\mapsto ([AB])$. As the result, $ P_b\mapsto K_b$ and $ P_c\mapsto K_c$ $ \Longrightarrow$ $ P_bP_c\mapsto (AK_bK_c)$, which leads to the fact that $ I_a\in d_a$ $ (*)$. In the other hand, we have $ \overline {P_bK_b}\cdot \overline{P_bA} \equal{}$ $ \overline {P_bH_c}\cdot \overline {P_bH_a} \equal{}$ $ \overline {P_bH}\cdot \overline {P_bB}$ $ \Longrightarrow$ $ \mathcal {P}_{P_b/(AK_bK_c)} \equal{}$ $ \mathcal {P}_{P_b/(H_aH_bH_c)} \equal{}$ $ \mathcal {P}_{P_b/(HBC)}$. With the same argument for $ P_c$. Then it is followed that $ P_bP_c$ is the radical axes of $ (H_aH_bH_c)$, $ (HBC)$ and $ (AK_bK_c)$ $ \Longrightarrow$ $ \mathcal {E},$ $ O_a$ and $ I_a$ are collinear and $ \overline {\mathcal {E}O_aI_a}\perp P_bP_c$ $ (**)$. From $ (*)$ and $ (**)$, we obtain result of the lemma. $ \square$

Now back to the problem. Note that the Euler circle wrt $ \triangle HBC$ is also the Euler circle wrt $ \triangle ABC$, denote this circle by $ (\mathcal {E})$ with center $ E$. Let $ K_b,K_c$ respectively be the intersections of $ (HAB)\cap ([AC])$; $ (HAC)\cap ([AB])$. Denote $ I_a,O_a$ respectively by the circumcenters of $ (AK_bK_c)$, $ (HBC)$. From the lemma above, we have learned that $ (AK_bK_c)$, $ (HBC)$ and $ (\mathcal {E})$ is an [i]eliptic bundles[/i]. Hence $ M,N\in (AK_bK_c)$. Therefore under the inversion through pole $ A$, power $ k \equal{} \overline {AH}.\overline {AH_a}$, where $ H_a$ is the projections of $ H$ onto $ BC$; we have $ M\mapsto M$ and $ N\mapsto N$, which leads to the fact that $ AN^2 \equal{} AM^2 \equal{} \overline {AB}.\overline {AH_c}$, where $ H_c$ is the projection of $ H$ onto $ AB$. Now let $ K$ be the midpoint of $ MN$ and $ M_c$ be the midpoint of $ AB$. Since $ AM,AN$, therefore $ AK\perp P_bP_c$. As the result,
$ \mathcal {P}_{A/(AMN)} \minus{} \mathcal {P}_{A/(\mathcal {E})}$ $ \equal{} 2\overline {I_aE}.\overline {KA}$ $ \Longrightarrow$ $ \mathcal {P}_{A/(\mathcal {E})}$ $ \equal{} 2\overline {I_aE}.\overline {AK}$ $ \Longrightarrow$ $ \overline {AM_c}.\overline {AH_c}$ $ \equal{} \frac {\overline {I_aE}}{\overline {AI_a}}.AN^2$. Wich is equivalent to $ \frac {1}{2}\overline {AB}.\overline {AH_c}$ $ \equal{} \frac {\overline {I_aE}}{\overline {AI_a}}.AN^2$ $ \Longrightarrow$ $ \frac {\overline {I_aE}}{\overline {I_aA}} \equal{} \minus{} \frac {1}{2}$. As the result we have $ \frac {\overline {HE}}{\overline {HO}}.\frac {\overline {LO}}{\overline {LA}}.\frac {\overline {I_aA}}{\overline {I_aE}} \equal{} 1$. Therefore by [i]Menelaus theorem[/i] for $ \triangle EOA$, we obtain $ H,I_a,L$ are collinear. Our proof is completed then. $ \square$

[b]Note-[/b] In this proof, I consider $ I_a\equiv O^*$.[/hide]

As the nice lemma mentioned by mathVNpro, Kb lies on the circle O*,
thus we can acquire that O3O*Mb are collinear
Analgously, McO*O2 are collinear  (both⊥AKb)
Notice O* is the centriod of ⊿OO3O2, denote D is the midpoint of O2O3,
Also D is the midpoint of HA because AO3HO2 is a parallelogram
=> D,O*,O are collinear =>  O* is the centroid of⊿AOH  =>  L,O*,H  are collinear

$ D,E,F$ are the feet of the altitudes on $ BC,CA,AB.$ $ M$ is the midpoint of $ BC$  and let the line $ EF$ cut $ BC$ at $ P$ and $ (O)$ at $ E',F'.$ Since $ (B,C,D,P) \equal{} \minus{} 1,$ it follows that $ PB \cdot PC \equal{} PD \cdot PM$ $ \Longrightarrow$ $ PE' \cdot PF' \equal{} PD \cdot PM$ $ \Longrightarrow$ $ M \in \odot(DE'F').$ Hence, the center of $ \odot(DE'F')$ is the intersection of the perpendicular from $ A$ to $ EF$ with the perpendicular bisector of $ DM,$ i.e. the midpoint $ L$ of $ AO.$ Inversion $ \mathcal{I}$ centered at $ H,$ that takes $ (O)$ into the 9-point circle $ \mathcal{N},$ takes the line $ EF$ into the circumcircle of $ \triangle HBC$ and $ \odot(DE'F')$ into the circle passing through $ A$ and the intersections $ M,N$ of $ \mathcal{N}$ and $ \odot(HBC),$ which are the inverse images of $ D,F',E'$ $ \Longrightarrow$ Center of the inversion $ H,$ center $ L$ of $\odot (DE'F')$ and the center $ O^*$ of its inverse $ \odot(AMN)$ are collinear.

I also found this.
[b]The Euler circle of $ \triangle HBC$ meet AX,AY the second time at the mid points of AX,AY.[/b]
Can you prove that?
Sorry, if it has been found before.


Arthur's wife picks him up at the train station and drives him home every Friday. One Friday Arthur catches the early train, arrives $ 90$ minutes early, and starts walking home. His wife, who left home at the usual time to pick him up, meets him on the way home. They arrive home $ 20$ minutes earlier than normal. How many minutes had Arthur been walking before his wife picked him up?

[hide]Let $ H$ be the home, $ S$ the station, $ X$ the meeting point.

The wife normally drives from $ H$ to $ S$ and back, but on this Friday she drove only from $ H$ to $ X$ and back, making home $ 20$ minutes earlier. Therefore from $ X$ to $ S$ is $ 10$ minutes drive.

Let's imagine the wife ignored Arthur at $ X$ and continued to the station. She would have arrived there at the normal arrival time of the train, but Arthur would have walked $ 10$ more minutes until the moment of her arrival at $ S$ (because she would have driven $ 10$ more minutes to the station), arriving at some point $ X_1$. However, Arthur started walking $ 90$ minutes before arriving at $ X_1$, because that's how much his train was early. Therefore, Arthur walked $ 80$ minutes to $ X$.[/hide]

[hide]He arrived 90 min early and started walking.

He met his wife 10 min earlier than usual.

What was he doing in the 80 min? walking![/hide]

Nice solution Farenhajt! It was exactly what I was looking for.


geometry
MATHCOUNTS
trigonometry
AMC
AIME
area of a triangle
Heron\u0027s formula

In the figure shown, circle O has a radius of 20.  AO is a radius and AB is a diameter.  $\angle ABC = 42^{\circ}, \angle DAG = 21^{\circ}, \angle AFO = 90^{\circ}$.  Find $[DGO]$.  Express your answer to the nearest tenth.  Explain your answer

Edit: changed the problem.  hope this works

[quote="236factorial"]What does the brackets around DGO mean? Area or angle?[/quote]

Area

I highly doubt that this is MathCounts level. This should probably go in the Intermediate forum for High School.

yOU can use the 180 degree rules to find many angles, but none of those information can lead to the area of a specific triangle.

If all you get is one length and a bunch of angles, and you have to find an area, you'd most likely need to use trig formulas. Unless you have like a square inscribed in a circle or something to that extent...

this should be moved to intermediate or pre-olympiad level. This is not MATHCOUNTS level.

Here is my solution. I hope you meant $\angle DGO$. So here: [img]http://4aunlimited.co.nr/AoPS/images/circles_and_geometry.gif[/img]. Te answer is 126 degrees

He already said that it means the area. You can either use some trig formulas, I think, or Heron's Formula, but for Heron's, you need to find the three sides, too.

Since AO = OD, $\angle{ADO}=21$, $\angle{AOD}=138$, $\angle{AOF}=42=\angle{DOG}$. DO = 10. How what's OG?

after infinite calculations, angle OGD is 126, DOB is 42 and ODG is 12. using some approximations and trig and pythagorean, the area of CDF should be 209. i'm stuck as of there. :(

by the way, this problem is very hard. it seems comparable to the hardest AIME geometry problems.

edit: after further calculations(i've been working for 90 minutes on this), i think the area is about 50, plus minus probably 20. :)

Knowing that, we can use the law of sines to get the other sides.
OG=16.54
DG=5.14

Heron's formula tells us the area is about 34.38.  :roll:

very nice 236! :) well, at least my approximation was correct. :D

so, what's the correct answer and full solution ccy?

I'm pm'ing LadyKn1ght to move this. We'll get discussion in high school.

[quote="h78reg"]yea, this should be moved to pre-olympiad. ;)[/quote]

Maybe intermediate is enough.

By the way, how did you get 126 and 12 for angles ODG and OGD?

well, DOB must be 42 degrees because arc DB is 42 degrees. as for ODG, triangle OAD is isosceles, meaning that ADO is 21 degrees. angle D is equal to angle B because arc AC is congruent to itself, meaning that ADO is equal to 33-21, or 12 degrees. this leaves DGO as 180-42-12, or 126 degrees.

I believe that this problem creates a contradiction, i am pretty sure of it too, where did you get this anyway?

$\angle ACB$ is right because it subtends a semicircle, given $\angle AFO$ is right
$\angle CAB$ is congruent to itself
so $\triangle AFO \sim \triangle ACB$
therefore, $\angle AOF = \angle ABC = 33^{\circ}$
$\angle AOF = 33^{\circ}$
$AO = DO$ since they are radii, making $\triangle AOD$ isosoles with vertex $O$
$\angle OAD = \angle ODA = 21^{\circ}$
by exterior angle theorem, $\angle AOF = \angle OAD + \angle ODA$
$\angle AOF = 21 + 21 = 42$
$\angle AOF = 42$
but the other given makes $\angle AOF = 33^{\circ}$ therefore one of our given angles was wrong...

Darn.  These numbers wouldn't work. :blush: I randomly selected numbers after making the diagram. :blush: Sorry.  Hope this new one won't change much.

you actually only need to give one of those angles but anyway...
$\angle DOG = 2*21 =42$ by ext. angle theorema nd vertical angles are congruent
then $\angle ABC = \angle ADC = 42$ since they intercept the same angle
$\angle ADF = \angle CDF = \frac{\angle ADC}{2} = \frac{42}{2} = 21$
then $180-21-42 = 117$
then area of a triangle is given by:
$\frac{20^{2}* \sin (42) * \sin(21)}{2 * \sin (117)} \approx 53.8 $
this formula is proven easily by the extended law of sines and $A_{\triangle} = \frac{abc}{4R}$

[hide]
Since $\angle DAB and \angle DCB$ are inscribed angles which subtend the same arc, $\angle DAB = \angle DCB = 21$.  By the sum of the angles in a triangle, $\angle CGB = 180-63 = 117$.  $\angle OGD$ is also equal to 117 because they are vertical angles.  $\angle ACB$ is right because it subtends a 180 degree arc.  Then, $\triangle AFO \cong \triangle ACB$ by AA similarity.  Since $\angle AOF = 42$, $\angle DOG = 42$, as they are vertical angles.  By the sum of the angles in a triangle, $\angle ODG = 21$.

Now, we have all the angles in the triangle ODG:
$\angle O = 42$
$\angle D = 21$
$\angle G = 117$
Additionally, OD = 20, since it is a radius.

By the (from the law of sines) area formula
 $[ABC] = \frac{a^2sinBsinC}{2sin A}$
(O becomes B, G becomes A, D becomes C)
 $[ABC] = \frac{400 \cdot sin 42 \cdot sin 21}{2 \cdot sin 117} = \frac{400 \cdot 0.66913060635885821382627333068678 \cdot 0.35836794954530027348413778941347}{2 \cdot 0.89100652418836786235970957141363} = \frac{95.917985351530991485926939356103}{1.782013048376735724719419142826} = 53.825635810525755235864580133982 \approx \boxed{53.8}$
[/hide]


You are in the downstairs lobby of a house. There are 3 swictches, all in the "off" position. Upstairs, there is a room with a lightbulb that is turned off. One and only one of the switches controls the lightbulb. You want to discover which switch controls the bulb, but your only allowed to go upstairs once.How do you do it?(No fancy strings, telescopes, etc. allowed. You cannot see the upstairs room from downstairs. The lightbulb is a standard 100-watt bulb.)

This problem really has me vexed.

[size=200]YES[/size]

can someone correct me if I'm wrong?

Turn switch no.1 on. Wait 10 years. Turn switch 1 off. Turn Switch 2 on. Wait 15 mins. Turn switch 2 off. Turn Switch 3 on. Go upstairs. If the light bulb is on, switch 3 works. If the light bulb is mildly warm, switch 2 works. If the light bulb has burned out, switch 1 works.

You don't have ten years. Turn on switch one and wait ten minutes, turn on switch two, and go up. if the light bulb is off, its switch three, if its on but cold, its two, and if its warm, its one.

Can't you just turn all three on and go upstairs? The lightbulb will be on and that's all that matters to me! :D  :D

well even if u did have 10 years, you wouldnt be able to tell if it burned out because all u would no is that it is off.  cause if 3 didnt work, then it would be either 2 or 1. if 1 didnt work, then it would be mildly warm, but if 1 did work, then it would also be warm cause u just burned it out.

hm...my solution:

flick switch 1 on and off and on and off.....for a long long time until u are sure a regular bulb would be burned out.  turn switch 2 on. go upstairs. if the light bulb exploded, switch 1 works. if its on, switch 2 works. if its not exploded or on, switch 3 works.

thank u thank u. (copyrite from sunnyray...i think....well she says it all the time)

that's the most ... unsmart (i prefer to use a different word) post i ever seen

... do u have a better one , gameworld7?

Here's mine:

You take get a blueprint for this house. Just under where the room with the lightbulb is, use a sledgehammer and break the floor of that room and the ceiling of this floor. Make the hole so that you can see the lightbulb. Try all three switches.

NoTe: This is the most expensive option... dont try it at home. (Wait... if it's your house, you already know which switch it is...oops...) I mean, dont try this at your friends house. I'd carry around with me a sledgehammer if I were at this sort of house.

Another one:
Go up there. Ask a friend to turn each switch on and off. Tell your friend whether that switch works or not. Assuming the rooms are not soundproof, this works.

A third one:

IF it is inhabited by someone, just ASK them. (DUH)

NoTe: This probably would not happen if it was not inhabited for 100 years... If you were willing , many others are not...

Oy...Jockey's on the right track but he/she has his/her solution mixed up.

Turn on switch one.  Wait five minutes.  Turn switch one off and turn on switch two.  Go into the room.  If the bulb is on, switch two is the correct switch.  If the bulb is cold and off, switch three is the correct switch.  If the bulb is warm and off, switch one is the correct switch.

Moral of the story?  If you ask a seemingly simple question...

um...yes i KNEW that...yea....but i was just being CREATIVE. but most of cardwackos ansers involve loopholes and common sense. u haf 2 ASSUME things

Fine; then try this on for size.

Turn on all three switches.  Go upstairs, smash the bulb, and electrocute yourself by licking the filament.  Now it doesn't matter.

:?  :(  :(  :shock:  :shock:  :idea: well gee thats nice. i bet it would be really painful.  but if u really had some common sense, u would just rewire the whole house and be done with it. see?

then why would u even be in it trying 2 figure out which switch turns on a lite bulb? if it wasnt urs it would either be public or private property and if it was, then u would be tresspassing.  cause if it was public property, obviously sum ppl built it for a reason so they could sell it and ur the one thats occupying it without a landlord. and if it was sumones private property, u would either be tresspassing or u could just ASK the person which one turned on the lite bulb :maybe:

They may not know, and may be paying you handsomely to solve the puzzler for them.

or they could take the easy way and go up three times, once per switch!
(what if they all work?) hahahahaha

You could try breaking the walls down and looking at the wiring...

I suggest that you break the bulb, and tie the tungsten filament in such a way that it does not burn off. Then, go downstairs, and turn switch one on. Wait a few minutes, and turn it off. Then, turn the second one on. When you hear crackling or when the house catches on fire, you know it's that one.

But still, the other guy's solution was good. Turn one on, wait. Turn second one on.

Cold, on-2
Cold, off-3
Hot, off-1

[quote="gameworld7"]HEY I KNOW THE ANSWER! THERE IS NONE! (i think)[/quote]

Well, we've been discussing the answer for a long time now. We've already come up with the correct one. The cold&on/warm&off/cold&off solution.

oh...um...i gues i don't really read all the posts...


Point $P$ is $9$ units from the center of a circle whose radius has lengh $15$. How many different chords of the circle contain $P$ and have integral lengths?

I found 2...
[hide="perhaps"]Length=30, length=24 are the min/max of length but there are 2 of each between them so we have
24
25
25
26
26
27
27
28
28
29
29
30
12 total[/hide]

Correct. The answer is 12. Nice thinking. I didn't really see that. I thought it was 7 because of 24-30. I see now how there are two each of 25-29.  :lol:


Merina's annual salary is a whole number of dollars between $ \$42,400$ and $ \$42,500$. The digits in the hundreds, tens and units places are in strictly ascending order. How many distinct possibilities exist for her annual salary?

We know that the first 3 digits of her salary is 4, 2, and 4 (again). We only need to find out the last 2 digits. If you pick any two numbers, there is one way to arrange them so that they are in ascending order (exclude the doubles like 2,2). Thus there are $ \frac{10\cdot9}{2}\equal{}\boxed{45}$ distinct possibilities for here annual salary.

Hello, the answer I found on FTW was \boxed{10}.
However, I don't know why.

Yes, LaTeX made a mistake in that the hundreds, tens, and units digit are in ascending order, not just the tens and units.  So then the tens digit would have to be at least $5$.  Hence, we have $\frac{4 \cdot 5}{2} = \boxed{10}$ ways.

Brute force to check:

$42456$

$42457$

$42458$

$42459$

$42467$

$42468$

$42469$

$42478$

$42479$

$42489$

$\boxed{10}$ ways.

[quote=Eugenis]Can someone elaborate on why the answer is $4*5/2?$[/quote]

One reason is because of the way we count numbers that work. If the tens digit is $5$, then there are $4$ possible digits for the units digit. If the tens digit is $6$, then there are $3$ possible digits for the units digit. If the tens digit is $7$, then there are $2$ possible units digits. Finally there is one possible units digit if the tens digit is $8$. Thus, the answer is $1+2+3+4$. This is the fourth triangular number, and since the formula for the $n^\text{th}$ triangular number is $\frac{n(n+1)}{2}$, we find the fourth triangular number to be $\frac{4(4+1)}{2}=\frac{4\cdot5}{2}$.

Another way to think about it is the number of ways to choose the tens and units digits. Since we know the tens digit is greater than $4$, and the tens digit $\neq$ the units digit, there are $5$ possible tens digits and $4$ possible units digits, for a total of $5\cdot4=20$ numbers. However, in exactly half of those, the units digit is greater than the tens digit, which is what we want, but in the other half, the tens digit is larger than the units digit, which doesn't work. Thus half of the $20$ work, and our answer is just $\frac{4\cdot5}{2}=10$.

[quote=Eugenis]Can someone elaborate on why the answer is $4*5/2?$[/quote]

The number has to be between 42,400 and 42,500. Therefore the hundreds digit is automatically a 4. That leaves 9-5+1=5 possible values for the digits greater than four. You then pick 2 different digits by doing 5c2=5*4/2=10. That works because you can order the two digits in whatever way you want to satisfy the conditions.


Let A-set of 2008 distinct reals. For any two elements of A- $ a,b$: $ \sqrt{2}b\plus{}a^2$- rational, prove that for any element from A-a:$ \sqrt{2}a$-rational

[hide]Let any 3 distinct elements of $ A$ be $ a$, $ b$, $ c$.
First we prove that for any two elements $ a$, $ b$, $ \dfrac{a \plus{} b}{\sqrt {2}}$ is rational. Since $ \sqrt {2}b \plus{} a^2$ is rational and $ \sqrt {2}c \plus{} a^2$ is rational, their difference $ \sqrt {2}b \minus{} \sqrt {2}c$ is rational. Since $ \sqrt {2}c \plus{} b^2$ is rational we can add this to $ \sqrt {2}b \minus{} \sqrt {2}c$ and find that $ \sqrt {2}b \plus{} b^2$ is rational. We can subtract $ \sqrt {2}b \plus{} c^2$, which is rational, to get $ b^2 \minus{} c^2$ which is rational. Dividing $ b^2 \minus{} c^2$ by $ \sqrt {2}b \minus{} \sqrt {2}c$ we get $ \dfrac{b^2 \minus{} c^2}{\sqrt {2}b \minus{} \sqrt {2}c} \equal{} \dfrac{(b \minus{} c)(b \plus{} c)}{\sqrt {2}(b \minus{} c)} \equal{} \dfrac{b \plus{} c}{\sqrt {2}}$, which is rational. Similarly, $ \dfrac{a \plus{} b}{\sqrt {2}}$ and $ \dfrac{a \plus{} c}{\sqrt {2}}$ are rational.
Adding together $ \dfrac{a \plus{} b}{\sqrt {2}}$ and $ \dfrac{a \plus{} c}{\sqrt {2}}$ we get $ \dfrac{2a \plus{} b \plus{} c}{\sqrt {2}}$, which is rational. We subtract $ \dfrac{b \plus{} c}{\sqrt {2}}$ and get $ \dfrac{2a}{\sqrt {2}} \equal{} \sqrt {2}a$, which is rational as desired.[/hide]


While I was working on some homework from my weekly math class, I came across a problem that I couldn't solve. I thought I might post this on AOPS since nobody I knew could help me. Since I came across this weeks ago, I've done my best to remember it.

Here's the problem:

The teacher whispers positive integer A to Annie, B to Brad, and C to Chris. The sum of A, B, and C is 14.  Annie says, "I know that Brad and Chris have different numbers." Then Brad says, "I already knew that all of our numbers are different." Finally Chris says, "Now I know all three of our numbers." What is the product of ABC?

[hide]For Brad and Chris to have different numbers, A must be odd. Then, B+D must be odd too and for that to happen one must be odd and the other must be even. $ \Box$
For Chris to know that all of their numbers were different, he must have gotten an $ 11,9,$ or $ 7$. Because the number is odd, he knows the other two have different numbers. Now, because they're so high, no one else can have the same number or the sum exceeds $ 14$. $ \Box$
Now let's think from Chris's point of view. If he got a $ 2$, there are $ 3$ pairs of numbers that Annie and Brad could have gotten $ [(11,1),(9,3),(7,5)]$. If Chris got a $ 4$, there are $ 2$ pairs of number that Annie and Brad could've gotten $ [(9,1),(7,3)]$, but if he got a $ 6$, there's only $ 1$! So Annie has $ 1$, Brad has $ 7$ and Chris has $ 6$. $ \Box$
\[ 1\cdot7\cdot6 \equal{} \boxed{42}\]
[/hide]


limit
real analysis
real analysis unsolved

Suppose $ \{x_{n}\}$is bounded, and $ \lim_{n\to\infty}(x_{2n}+x_{n})=A.$
Prove $ \{x_{n}\}$ converges.

Perhaps this idea would work.

By way of contradiction assume that $ x_{n}$ diverges. Since the sequence is bounded there are at least two distinct limit points.

Call them $ \alpha$ and $ \beta$, $ \alpha\neq \beta$.

Now, it follows that there are two sequences $ n_{k}$ and $ m_{p}$ such that

$ x_{n_{k}}\rightarrow \alpha$, $ k\rightarrow \infty$

and

$ x_{m_{p}}\rightarrow \beta$, as $ p\rightarrow \infty$.

On the other hand, since $ x_{2n_{k}}+x_{n_{k}}\rightarrow A$ it follows that

$ \alpha=A/2$. Simillarly $ \beta=A/2$, form which it follows that $ \alpha=\beta$, which is false.

Are you sure the problem is correct? What about the following sequence:
$ -x_{n}= \frac{A}2$, for all $ n$ that aren't powers of $ 2 \, ;$
$ -x_{2^{2k}}= c \, ;$
$ -x_{2^{2k+1}}= A-c \, ?$

Thank you for your example,perfect_radio.
what about the following problem?
Suppose $ \{x_{n}\}$is bounded, and $ \lim_{n\to\infty}(x_{2n}+2x_{n})=A.$
Prove $ {x_{n}}$ converges.

We can easily modify the same counterexample:  let $ x_{n}= A/3$ when $ n$ is not a power of 2, let $ x_{2^{2n}}= c$ and let $ x_{2^{2n+1}}= A-2c$.

No, that doesn't work. $ x_{2n}+2x_{n}$ is bad when $ n$ is $ 2^{2k+1}$.

The answer is different here; this one converges.
WLOG, $ A=0$; we can simply subtract $ \frac{A}{3}$ from the sequence.

We prove the contrapositive; if $ x_{2n}+2x_{n}$ converges to zero and the $ x_{n}$ do not converge to zero, the sequence is unbounded.
Choose $ N$ so that $ |x_{2n}+2x_{n}|<\epsilon$ for $ n>N$, and suppose that $ |x_{m}|>2\epsilon$ for some $ m>N$. Then $ |x_{2m}|>3\epsilon$, $ |x_{4m}|>5\epsilon$, $ |x_{8m}|>9\epsilon$, and so on; $ |x_{2^{k}m}|>(2^{k}+1)\epsilon$, and the sequence is unbounded.


algebra
polynomial
limit
logarithms
function
real analysis
real analysis unsolved

If $ f_n(x) \equal{} (x \minus{} 1)^n \minus{} x^{n \minus{} 2}$. Denote by $ x_n$ the largest real root of $ f_n$.
I can show easily that
$ \lim_{n\to \plus{} \infty} x_n \equal{} \plus{} \infty$;
$ \lim_{n\to \plus{} \infty} \frac {x_n}{n} \equal{} 0$
and $ x_n < \frac {n}{\log n}$, near $ \plus{} \infty$.

I can't find explicitly a simple form of $ y_n$, such that $ \lim_{n\to \plus{} \infty} \frac {x_n}{y_n} \equal{} 1$.
Any idea?

$ (x_n - 1)^n - x_n^{n - 2} = 0\Leftrightarrow x_n - 1 = x_n^{1 - \frac {2}{n}}$.  Using the series for $ x^{1 - \epsilon}$ near $ \epsilon = 0$ we find,
\[{ x_n^{1 - \frac {2}{n}} = x_n - \frac {2}{n}x_n\log x_n + \mathcal{O}(x_n n^{ - 2}})
\]
Therefore, if you have already established that $ x_n<\frac {n}{\log n}$, we have,
\[ 1 = \frac {2}{n}x_n \log x_n + \mathcal {O}(\frac {1}{n\log n}) \\
 \\
 \\
\Rightarrow x_n = \frac {n + \mathcal{O}({\frac {1}{\log n}})}{2W(\frac {n}{2} + \mathcal{O}(\frac {1}{\log n}))}\approx \frac {n}{2W(\frac {n}{2})}
\]
where $ W$ is the [url=http://mathworld.wolfram.com/LambertW-Function.html]Lambert W function[/url].  You can verify that $ \lim_{n\to \infty}\frac {W(n)}{\log n} = 1$ (though convergence is very slow), so our final result is
\[ \lim_{n\to \infty}\frac {2x_n \log n}{n} = 1
\]
In other words, in your notation, we can take $ \boxed{y_n = \frac {n}{2\log n}}$.  Numerically, the approximation $ x_n\approx \frac {n}{2W(\frac {n}{2})}$ is fairly good for large values of $ n$; for example with $ n = 2000$ the exact value is approximately $ x_{2000}\approx 190.911$, while $ \frac {1000}{W(1000)}\approx 190.491$.

Just out of curiosity, where did this problem come up?

Thanks JoeBlow
[quote="JoeBlow"]Using the series for $ x^{1 - \epsilon}$ near $ \epsilon = 0$ we find,
\[{ x_n^{1 - \frac {2}{n}} = x_n - \frac {2}{n}x_n\log x_n + \mathcal{O}(x_n n^{ - 2}})
\][/quote]
Not sure I understand that one.

[quote="JoeBlow"]
Just out of curiosity, where did this problem come up?
[/quote]
I got it from a professor.

The Taylor series for $ x^{1 \minus{} \epsilon}$ at $ \epsilon \equal{} 0$ and $ x > 0$ is: $ x^{1 \minus{} \epsilon} \equal{} xe^{ \minus{} \epsilon\log x} \equal{} x \sum_{k \equal{} 0}^\infty ( \minus{} 1)^k \frac {\left(\epsilon \log x\right)^k}{k!}$.  Substitute $ \epsilon \equal{} \frac {2}{n}, x\equal{}x_n$, collect the first few terms, and simplify.


In right $\Delta ABC$ , {right angled at B }$AB = 30 cm $ and $BC = 40cm$.Find the length of the altitude going from $B$ to $ say , D $ on $AC.$

I used similar triangles for this problem and got 24.

i think shobbers method is the easiest tho..
area=600
600/50=12
times 2=[b]24[/b]


Gind the positive integers x,y which satisfact the relation
$x^2+y^2=4385$

Hint
[hide]$4385 = 5 \cdot 877$ is the product of two primes...[/hide]

This is pretty much like [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34266]flanders 05/3[/url]...

There is also a generalized theorem in that page. :)


When we post a problem, how do I get the litle Source thing to show up by the topic title in the forum?

Depends on the forum you are located in. In some forums (most Olympiad and College ones) this option is enabled by default. For other forums (the midle school and high-school forums for example) this option is disabled to make the input easier.


For $ x,y,z>0$, Prove that

$ \frac{x^3}{x^2\minus{}xy\plus{}y^2}\plus{}\frac{y^3}{y^2\minus{}yz\plus{}z^2}\plus{}\frac{z^3}{z^2\minus{}zx\plus{}x^2}\geq x\plus{}y\plus{}z$

by AM-GM ineq,we get:
$ \sum {\frac {{{x^3}}}{{{x^2} \minus{} xy \plus{} {y^2}}}} \equal{} \sum {\left( {x \minus{} \frac {{x{y^2} \minus{} {x^2}y}}{{{x^2} \minus{} xy \plus{} {y^2}}}} \right)} \ge \sum {\left( {x \minus{} \frac {{xy(y \minus{} x)}}{{xy}}} \right) \equal{} \sum {(2x \minus{} y) \equal{} x \plus{} y \plus{} z} }$
with equal if and only if x=y=z

[quote="hollandman"]For $ x,y,z > 0$, Prove that

$ \frac {x^3}{x^2 \minus{} xy \plus{} y^2} \plus{} \frac {y^3}{y^2 \minus{} yz \plus{} z^2} \plus{} \frac {z^3}{z^2 \minus{} zx \plus{} x^2}\geq x \plus{} y \plus{} z$[/quote]
It's wrong. Try $ x\equal{}1,$ $ y\equal{}2$ and $ z\rightarrow0^\plus{}.$

[quote="arqady"][quote="hollandman"]For $ x,y,z > 0$, Prove that

$ \frac {x^3}{x^2 \minus{} xy \plus{} y^2} \plus{} \frac {y^3}{y^2 \minus{} yz \plus{} z^2} \plus{} \frac {z^3}{z^2 \minus{} zx \plus{} x^2}\geq x \plus{} y \plus{} z$[/quote]
It's wrong. Try $ x \equal{} 1,$ $ y \equal{} 2$ and $ z\rightarrow0^ \plus{} .$[/quote]
Yes,this inequality is false :wink:


linear algebra
matrix
system of equations
IMO
algebra
imo 1976

We consider the following system
with $q=2p$:

\[\begin{matrix} a_{11}x_{1}+\ldots+a_{1q}x_{q}=0,\\ a_{21}x_{1}+\ldots+a_{2q}x_{q}=0,\\ \ldots ,\\ a_{p1}x_{1}+\ldots+a_{pq}x_{q}=0,\\ \end{matrix}\]

in which every coefficient is an element from the set $\{-1,0,1\}$$.$ Prove that there exists a solution $x_{1}, \ldots,x_{q}$ for the system with the properties:

[b]a.)[/b] all $x_{j}, j=1,\ldots,q$ are integers$;$

[b]b.)[/b] there exists at least one j for which $x_{j} \neq 0;$

[b]c.)[/b] $|x_{j}| \leq q$ for any $j=1, \ldots ,q.$

[hide="Solution"]
First of all note that we have $ (q \plus{} 1)^q \minus{} 1$ possible nonzero vectors $ (x_1,\cdots,x_q)$ such that $ 0\leq x_i\leq q$ are integers.

But $ a_{j1}x_1 \plus{} \cdots \plus{} a_{jq}x_q$ can only assume $ q^2 \plus{} 1$ different values, because if it is maximized/minimized by $ (M_1,M_2,\cdots,M_q)/(m_1,m_2,\cdots,m_q)$, we have that $ \sum_{i \equal{} 1}^{q}a_{ji}(M_i \minus{} m_i)\leq q\times q \equal{} q^2$ (if $ a_{ji} \equal{} 0$, it doesn't affect the sum, if it is $ 1$, $ M_i \equal{} q,m_i \equal{} 0$, and if it is $ \minus{} 1$, $ M_i \equal{} 0,m_i \equal{} q$).

From this we conclude that there are at most $ (q^2 \plus{} 1)^p \equal{} (q^2 \plus{} 1)^{q/2}$ possible values for the vector $ (a_{11}x_{1} \plus{} \ldots \plus{} a_{1q}x_{q},\cdots,a_{p1}x_{1} \plus{} \ldots \plus{} a_{pq}x_{q})$.
But we have that:
$ (q \plus{} 1)^q \minus{} 1 \equal{} (q^2 \plus{} 1 \plus{} 2q)^{q/2} \minus{} 1 \equal{}$
$ \equal{} (q^2 \plus{} 1)^{q/2} \plus{} \left( \minus{} 1 \plus{} \sum_{j \equal{} 1}^{q/2}{{q/2}\choose j}(q^2 \plus{} 1)^{q/2 \minus{} j}(2q)^j\right) > (q^2 \plus{} 1)^{q/2}$

We conclude that by the pigeonhole principle there are two distinct vectors being mapped to the same vector. Taking their difference we have a vector with the desired properties.
[/hide]


calculus
integration
absolute value
LaTeX

How can one make the absolute value symbol bigger? I write $ |\int_C f(z) dz|$, but the absolute value symbol is too small. Thanks!

\left and \right work with more than just parentheses.
$ \left|\int_C f(z)\,dz\right|$


geometry
circumcircle
symmetry
geometry unsolved

An acute triangle $\triangle{ABC}$ is inscribed in a circle with centre $O$. The altitudes of the triangle are $AD,BE$ and $CF$. The line $EF$ cut the circumference on $P$ and $Q$.
a) Show that $OA$ is perpendicular to $PQ$.
b) If $M$ is the midpoint of $BC$, show that $AP^2=2AD\cdot{OM}$.

Part (a) is very simple. Just observe that $\angle CAO= \pi/2 -\angle ABC$ because $O$ is circumcenter, and $\angle AEF=\angle ABC$ because the quadrilateral $BCEF$ is cyclic. So $\angle CAO+\angle AEF=\pi/2$, that is, $EF$, or $PQ$, is perpendicular to $AO$.

Next, since $PQ$ is perpendicular to radius $OA$, this implies by symmetry that $AP=AQ$. Now we perform an inversion with respect to the circumference of center $A$ and radius $AP$. The circumcircle of $ABC$ inverts on line $PQ$, because it passes through the center of inversion. Also, the inverse of $B$ is the intersection of the inverses of line $AB$ and the circumcircle of $ABC$, that is, the inverse of $B$ is $F$. In the same way, we see that the inverse of $C$ is $E$. Now observe that the inverse of line $BC$ is a circumference through $A$: it must be the circumcircle of $AEF$, which passes through $H$, since quadrilateral $AEHF$ is cyclic. But, since $A$, $H$ and $D$ are collinear, this means that the inverse of $D$ is $H$. Hence we have $AD \cdot AH =AP^2$, but we know that $AH=2OM$, so $AP^2=2AD \cdot OM$, and we are done ;)

Part (a) is well known.

As Jutaro said, $AP=AQ$ by symmetry. ($PO=OQ$, $AO=AO$ and since $\Delta POQ $ is isosceles and because of part a) we have $<POA=<QOA$ which means that $\Delta AOQ$ is congruent to $\Delta AOP$ so $AP=AQ$.)

Since cuadrilateral $HECD$ is cyclic then  $AH*AD=AE*AC$.                      (1)

Since $<BFC=<BEC=90$ cuadrilateral $FECB$ is cyclic and therefore $<AEP=<FEC=180-<FBC=180-<ABC$            (2)

Clearly, cuadrilateral $APCB$ is ciclyc which gives  $<APC=180-<ABC$                         (3)

From (2) and (3),  $<APC=<AEP$ and since $<PAE=PAC$ we have that $\Delta AEP \sim \Delta APC$ and therefore:

$ AE*AC= AP^2$        (4)

From (1) and (4),  $AH*AD=AE*AC= AP^2$      (5)

It is well known that $AH=2*OM$  and substituting this in 5 we obtain   $2*OM*AD= AP^2$ as desired.

[geogebra]df180afdfe95ea51723763af24e90f59689e29e6[/geogebra]

Part A: Be K the intersection between prolongation AO and the circuncircle of ABC, <APB=90, is enoguh to show that <APQ=<PKA=b, but <PKA=<ABP=b, <PFB=<ECB=a because FECB is  cyclic, PACB is cyclic => <PAB= 180-<ACB => <PAB= 180-a but <PAB= <APQ+<QPB = <APQ + 180-a-b => <APQ + 180-a-b=180-a => <APQ=b => <APK= <PKA=b => AO is perpendicular to PQ.
Part B: as a result in Part A, AP=PQ, by the stewart teorem, AE^2 * QP = AP^2*QP - QE*EP*QP => AE^2=AP^2-QE*EP => AP^2= AE^2+QE*EP but QE*EP=AE*EC by power point, => AP^2=AE^2+AE*EC => AP^2= AE*(AE+EC) => AP^2= AE*AC, but AE*EC= AH*AD by power point, => AP^2=AH*AD, but is known that AH=2OM => AP^2= 2*AD*OM.


Let a and m are relative prime.Prove that the equation:
          a*x=b(modulo m)
has at most 1 integer root modulo m.
Sorry for my bad English.

This is sort of easy, isn't it? It's well known that in the ring Zm the invertible elements are those which are coprime with m, so a is invertible, so there is a c s.t. ac=1 (mod m), and if we multiply the equation with c to the left we get x=cb (mod m), so there is exactly one such x.


Prove that the difference of two consecutive cubes is never divisible by $ 2$.

[hide="Solution"]
We have that,
\[ (n \plus{} 1)^3 \minus{} n^3 \equal{} 3n^2 \plus{} 3n \plus{} 1\]
Also, $ 3n^2 \plus{} 3n \plus{} 1\equiv n^2 \plus{} n \plus{} 1\equiv 0 \plus{} 1\equiv 1\pmod{2}$.

This is because $ n^2 \plus{} n$ is always a multiple of $ 2$ hence it is congruent to $ 0$. Because the difference is congruent to $ 1\pmod{2}$ we can say that it is not a multiple of $ 2$, and we are done. :ninja:
[/hide]

For any two consecutive cubes, one will be odd and one will be even, so the difference will be odd. This should be in high-school basics.


In the widely-used portable mercury-cadmium battery one electrode consists of Cd metal in contact with insoluble Cd(OH)2 in a solution of NaOH. The other electrode consists of mercury metal in contact with insoluble HgO also in a solution of NaOH. If a current of 0.1 amp is drawn from this cell for one hour, what is the change in the number of moles of elemental mercury in the cell.

    (A) 5.17 x 10¯7
    (B) 1.86 x 10¯3
    (C) 3.73 x 10¯3
    (D) 360 moles

I'm not really sure, but I get:
[hide]B[/hide]

well I will do this pretty much entirely on my skill at remembering random constants from physics class...

[hide="OMG CLICK THIS"]To be perfectly honest, I have no idea why all the cadmium stuff matters. Ok, now $\text{number of moles} = \frac{\text{number of atoms}}{\text{Avogadro's number}} = \frac{1}{\text{Avogadro's number}} \cdot \frac{\text{number of electrons}}{\text{change in electrons per atom}} =$ $\frac{1}{\text{Avogadro's number} \cdot \text{change in oxidation number}}$ $\cdot \frac{\text{total charge}}{\text{charge per electron}}$ $= \frac{1}{\text{Avogadro's number} \cdot \text{change in oxidation number}}$ ${\cdot \text{charge per electron}} \cdot$ $\text{current} \cdot \text{time}$ $= \frac{.1 \text{A} \cdot 1 \text{hr} \cdot 3600 \frac{\text{s}}{\text{hr}}}{6.022 \cdot 10^{23} \frac{\text{atoms}}{\text{mole}} \cdot 2 \frac{\text{electrons}}{\text{atom}} \cdot 1.6 \cdot 10^{-19} \frac{\text{coulombs}}{\text{electron}}}$ $= 1.87 \cdot 10^{-3} \text{moles}$[/hide]

[quote="g1sk8terz"]how do you make all ur letters so neat?[/quote]Use \text{whatever you want to say} in LaTeX. :)

[quote="g1sk8terz"]am i suppose to just type in \text{whatever you want to say}??

and what's latex?

i still can't get it here's my try to make a NO3 ion

\text{NO}_3^-[/quote]
LaTeX - a way to make nice math and science expressions. See [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]this[/url].

For example, to make a nitrate ion, put NO_3^-  inbetween two dollar signs. You get $NO_3^-$. Use \text{ } for when you have a lot of words in an expression, like MysticTerminator's solution.

Hope this helps.  :)


When $ f(x)$ is divided by $ (x\minus{}a)(x\minus{}b)$, the remainder is $ Ax\plus{}B$. Apply the division algorithm to show that:

$ A\equal{}\frac{f(a)\minus{}f(b)}{a\minus{}b}$ and $ B\equal{}\frac{bf(a)\minus{}af(b)}{b\minus{}a}$

[hide]
From $ f(x) = (x - a)(x - b)g(x) + Ax + B$ we find
\begin{align} f(a) & = Aa + B,\text{ and} \\
f(b) & = Ab + B.\end{align}
Therefore subtracting $ (1)$ by $ (2)$ and dividing by $ a - b$ gives the first result. Substituting gives the second result.[/hide]


Here's a good logic puzzle:

You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.

As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.

The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister?

Nice twist on the old puzzle.
[hide="answer"]
Name them A, B, and C.  Ask sister A: "Is B truthful more often than C?"  If she says yes, choose C, if she says no choose B.
[/hide]

Wait -- I don't understand why this works. Could you please explain it to me?

[hide="explanation"]Using the answer you get on the question from the solution:

case 1: A 1 B 1/2 C 0

you pick C, the liar.

case 2: A 1 B 0 C 1/2

you pick B, the liar.

case 3: A 0 B 1 C 1/2

you pick B, the truth-teller.

case 4: A 0 B 1/2 C 1

you pick C, the truth-teller.

case 5: A 1/2

it's a win-win situation.

where 1 is truth-teller, 1/2 mischievous, 0 liar.[/hide]


trigonometry
inequalities unsolved
inequalities

give $x^2+y^2=1 $. find $min$ and $max$ of :
       $N=x\sqrt{1+y}+y\sqrt{1+x}$

Well for maximum value if by chance we are talking about$x,y >0$we have :
${x\sqrt{1+y}+y\sqrt{1+x} \le\sqrt{(x^2+y^2)(1+y+1+x)}=\sqrt{(2+x+y)}\le \sqrt{2+2\sqrt{\frac{x^2+y^2}{2}}}=\sqrt{2+\sqrt2}}$

sorry, but we don't have $x,y>0$.for $x^2+y^2=1$,we have $-1\leq x,y\leq 1$ but i think this solution is acceptable. how about $min$

Maybe we can try trig substitution. Let $x=\sin{\alpha}$, $y=\cos{\alpha}$. But I haven't found the solution yet....just a suggestion.

yeah, i follw this way,and it appear to have a new problem. solve the equation:
  $ \frac{3\cos^{2}x+2\cos{x}-1}{2\sqrt{1+\cos{x}}}- \frac{3\sin^{2}x+2\sin{x}-1}{2\sqrt{1+\sin{x}}}=0 $

yeah, i follow this way,and it appear to have a new problem. solve the equation:
  $ \frac{3\cos^{2}x+2\cos{x}-1}{2\sqrt{1+\cos{x}}}- \frac{3\sin^{2}x+2\sin{x}-1}{2\sqrt{1+\sin{x}}}=0 $ $x\in [0;2\pi] $

yes, it is true. but it's not enough. by using some way, i know it has 4 roots. so there are two roots missing.


Can anyone explain to me the difference. I tried reading the wiki article on it, but didn't quite understand. Thanks in advance for all your responses.

[b]Symmetric [/b]- sums over all permutations of the list.
Ex in 3 vars $ f(x,y,z)\plus{}f(x,z,y)\plus{}f(y,z,x)\plus{}f(y,x,z)\plus{}f(z,x,y)\plus{}f(z,y,x)$

[b]Cyclic [/b]- sums over all "cyclic" orderings, ie, moving the last entry to the front.
Ex. in 3 var $ f(x,y,z)\plus{}f(y,z,x)\plus{}f(z,x,y)$


I have a problem but i don't know the problem can solved!
[b]Problem[/b]: Find all  number reals $a,b,c,d$ satified:
$2a=(c+d)^2, 2b=(d+a)^2, 2c=(a+b)^2, 2d=(b+c)^2$

I believe, look at it mod 2/4/8, and then we have infinite descent?

but they aren't integers, so I suppose it doesn't matter.

Yes, I know! If a,b,c,d are integer then problem can be solved easier

Only solution: $(\frac12,\frac12,\frac12,\frac12)$

Obviously $a,b,c,d$ are all greater than 0.
If $a<c$, then $(c+d)^2<(a+b)^2$, $c+d<a+b$. Since $c>a$, $d<b$.
Then $(b+c)^2<(a+d)^2$, $b+c<a+d$. Since $b>d$, $c<a$, contradiction.
The arguments are similar for $c>a$, etc.

So $a=c$, $b=d$, substitute in first two equations to get $a=b$. So they are all equal.


$ \left(x_{1} x_{2} \ldots x_{1998} \right)$ shows a number with 1998 digits in decimal system. How many numbers $ \left(x_{1} x_{2} \ldots x_{1998} \right)$ are there such that $ \left(x_{1} x_{2} \ldots x_{1998} \right) \equal{} 7\cdot 10^{1996} \left(x_{1} \plus{} x_{2} \plus{} \ldots \plus{} x_{1998} \right)$ ?

$\textbf{(A)}\ 0  \qquad\textbf{(B)}\ 1  \qquad\textbf{(C)}\ 2  \qquad\textbf{(D)}\ 3  \qquad\textbf{(E)}\ 4$

[hide]Obviously, at least $ 1996$ rightmost digits of the desired number are zero, hence the problem simplifies to

$ \overline{x_1x_2} \equal{} 7(x_1 \plus{} x_2)\iff 10x_1 \plus{} x_2 \equal{} 7x_1 \plus{} 7x_2\iff x_1 \equal{} 2x_2$, so we have four solutions: $ \overline{x_1x_2}\in\{21,42,63,84\}$ and the answer is $ E$[/hide]


The sequence of numbers $.., a_{-3}, a_{-2}, a_{-1}, a_0, a_1, a_2, a_3, ..$ is defined by $a_n - (n+1)a_{2-n} = (n+3)^2$, for all integers $n$. Calculate $a_0$.

[quote="236factorial"]haha I get -17  :blush:  Is that anywhere close???[/quote]
I think that's correct.
[hide]When $n=0$, we have $a_0-a_2=9$
When $n=2$, we have $a_2-3a_0=25$

Then solve the system and done :) [/hide]

[quote="frt"][hide]When $n=0$, we have $a_0-a_2=9$
When $n=2$, we have $a_2-3a_0=25$

Then solve the system and done :) [/hide][/quote]

Exactly how I did it (except with some other unnecessary equations)

adding, we get $-2a_0=34$
$a_0=-17$


Let $k\geq 0$.
Find the maximum value of $k$ such that for every positive real $a,b,c$, satisfies
$(a^2+2)(b^2+2)(c^2+2)\geq k(a^2+b^2+c^2)+(9-k)(ab+bc+ca)$.
When $k=0$, it becomes the APMO 2004 inequality.

I can only find $k=3$ is ture.
and $k=4$ is false
 :blush:

[quote="Parkdoosung"]Find the maximum value of $ k$ such that for every positive real $ a,b,c$, satisfies

$ \left(a^2 + 2\right)\left(b^2 + 2\right)\left(c^2 + 2\right)\geq k\left(a^2 + b^2 + c^2\right) + (9 - k)(bc + ca + ab)$.[/quote][quote="zhaobin"]I can only find $ k = 3$ is ture, and $ k = 4$ is false :blush:[/quote]$ k = 4$ is the best possible :

$ \prod{\left(a^2 + 2\right) - 4\sum{a^2} - 5\sum{bc} = a^2b^2c^2 + 2\sum{b^2c^2} - 5\sum{bc} + 8\equiv xyz + 2\sum{x^2} - 5\sum{x} + 8}$

$ = \sum{\left(\frac {x - 1}{2}\right)^2\left(1 + \frac {5x + 4yz}{x + y + z}\right)} + \sum{\left(\frac {y + z}{2} - 1\right)^2\left(1 + \frac {3x}{x + y + z}\right)}\geq0.$

By the way, $ \left(a^2 + 2\right)\left(b^2 + 2\right)\left(c^2 + 2\right)\geq 4\left(a^2 + b^2 + c^2\right) + 5(bc + ca + ab)$ holds for all real numbers  $ a,b,c.$

See also : http://www.mathlinks.ro/viewtopic.php?t=21835


A truck uses 3.5 tanks of gas to drive 560 km. How many tanks of gas would this truck use to drive 640 km?

[hide="Solution"]The truck get 560/3.5=160 km per tank.
640/160=4 tanks.[/hide]

[hide="Treething says to hide this"]
this is a simple ratio problem

you state it like this

3.5 : 560 = x : 640

x = 640 * 3.5 / 560 = 4[/hide]

Excellent job both of you.
Shinwoo-hide your answers, someone else might want to solve this problem.


Given a circle of radius $1$, and $8$ points in the interior of this circle. Prove that there are two points whose distance is less than $1$.

Split the circle up into six sections, a sixth of the circle area each in 'slices' from the centre. (Each slice makes an angle of $60^{\circ}$ at the centre)
One point may lie at the centre of the circle; other than that, there must be seven points lying not in the cenre, and hence two of them in the same 'slice'.
These two points are therefore a distance less than 1 from each other.

[If each slice is $OAB$ with $A$, $B$ on the circumference, then $OA=AB=OB$ is the greatest distance between two points- but no point may lie on the circumference]

Can't we do it with $7$ points, assuming no point can lie on the circle itself. It's the same Pigeonhole argument.

Jan, when you say "two points whose distance is less than 1" what are you pertaining to? Distance from WHAT? Is it from each other? Also, are these points located randomly within the circle or is ther some "order" in how  these points are placed?

Another elegant solution is to look at the angles each point makes with the centre of the circle to its neighbour.


How many perfect square with 200 digits that the first 99 digits are 9?

This is a rather easy problem.
Restated: For how many positive integers $n$ is $10^{200}-10^{101}\le n^2<10^{200}$?
Let $n=10^{100}-m$; then $n^2=10^{200}-2m*10^{100}+m^2$. Obviously, $0 < 2m*10^{100}-m^2 < 10^{101}$ for $1 \le m \le 5$ and no other values less than $10^{100}$. The answer is therefore $5$.


greatest common divisor
number theory
number theory proposed

I'm sorry for miss typing, MathLinks everybody.I will correct as follows.

Let $p,q$ be distinct prime numbers.
Prove that the greatest common divisor is[b] 1[/b] of  the following [b]pq-1[/b] 's

\[_{pq}C_1,_{pq}C_2,\cdots\cdots,\ _{pq}C_{pq-1}\]

Best regards.

kunny

Suppose for a contradicion that $t$ is a common prime divisor of all these numbers.
Then $t$ divides ${{pq} \choose {1} }= pq$, and then $t=p$ or $t=q$.
Say $t=p$.

Now, let $q = \sum_{i=0}^k a_ip^i$ be the decomposition of $q$ in base $p$.
Since $q$ is prime and distinct from $p$ we have $a_0 \neq 0$.
Therefore $q-1 = (a_0-1) \sum_{i=1}^k a_ip^i$ is the decomposition of $q$ n base $p$.

It is well-known that for all integer $n$, the valuation of $n!$ mod $p$ (that is the greatest power of $p$ which divides $n!$) is $\nu_p(n!) = \frac {n-s_p(n)} {p-1}$ where $s_p(.)$ denotes the sum of the digits in base $p$.

From above, we deduce that $\nu_p(q!) - \nu_p((q-1)!) = 0$.

It follows that :
$\nu_p( {{pq} \choose {p}}) = \nu_p ((pq)!) - \nu_p(p!) - \nu_p((pq-p)!) = (q + \nu_p(q!) ) - 1 - ((q-1) + \nu_p( (q-1)!)) = 0$.
It follows that $ {pq} \choose {p} $ is not divisible by $p$, a contradiction.

Pierre.

Thank you very much, Pierre.

That's great. I understand the background of the problem.
The problem of [b]number theory [/b]is difficult for me,but it is seductive to me.

kunny


induction
combinatorics unsolved
combinatorics

Assume that a choclate bar consist of n squres arrenged in a rectangular pattern the bar or smaller piece of the bar can be broken along a verticle or a horizontal line seperating the squres. Assuming that only one piece can be broken at a time determine how many breaks u must succecively make to break the bar into n seperate squares. Use strongf induction to prove your answer

This really doesn't belong here; it's completely elementary.

The number of pieces increases by exactly one with each break, so exactly $n-1$ steps are required regardless of pattern.
With a one-line solution available, there's no reason to try to force another method on this problem.

I think he means you can only break along one line at a time, in which case the problem is a little more interesting. But I think I've heard it before, and it belongs in Pre-Olympiad (or Olympiad Combinatorics maybe if my memory of the difficulty of the solution is inaccurate).


Does anyone know where to get chapter and state sprint and target rounds from the 90's.

Thanks in advance!  :|

There is a topic just for this with loads of links:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189]MATHCOUNTS Resources HERE![/url]
Its a sticky so it should be at the top of the page.

I meant that the link that are there have lots of PDFs.  I think you have to buy them from mathcounts.
Take a look here:
[url]http://www.mathcounts.org/webarticles/articlefiles/store_pg7.pdf[/url]

mathcounts go to archives i think they have till 3 years back or something


analytic geometry
algebra unsolved
algebra

[quote="AbScissa"]When given a polar equation rcos (theta) - rsin (theta) = 0, is the solution
just tan (theta) = 1, or should you factor out the r and account that r = 0
in the solution?[/quote]

You must include $ r\equal{}0$, $ \theta\equal{}$any in the set of solutions.


[quote="SUPERMAN2"]The lengths of the sides of a quadrilateral are $ 2006,2007,2008$ and $ x$.If $ x$ is an integer,find the largest possible value of $ x$[/quote]

[hide="My Solution"]I believe that we want to use the triangle theorem for this one, by breaking up the quadrilateral into two triangles.

We want the diagonal to be maximized. In the triangle with sides 2006, 2007, and the diagonal being the longest, the diagonal length will be 4012.

Then, in the triangle with sides 2007, 4012, and x, maximizing x with the triangle theorem gives us the answer of [b]6019[/b][/hide]

[hide="Slight discrepancy"]

Actually, I believe that the longest diagonal length of a triangle with sides 2006 and 2007 can be anything smaller than 4013 (let's call this y). That means with the triangle with sides y, 2008, and x, x is maximized at anything smaller than 6021, namely 6020.

[/hide]

Both veezbo and mathfreak123 are really close, but WRONG.

2006 + 2007 is 4013 According to the triangle inequality the diagonal must be less than 4013. So, the maximum value of diagonal is 4012.

Now, 4012 + 2008 = 6020. According to the triangle inequality, the remainging side must be less thatn 6020. So, the maximum value of the last side is 6019!

This is the correct solution!!

$ \boxed{6020}$ is correct...think about it. The requirement is that every three sides' sum is greater than the fourth side.

[quote="JBL"]More specifically, SnowHawk assumed that the diagonals had to be integers, but this is not true.  The diagonal could have length, say, 4012.5 without there being any problem.[/quote]
Actually, it specifically said x is an integer.

You could also think of it like this:
 [geogebra]4ab722cc4b2a9111851d2d2e310a614b8b233222[/geogebra]

Similar to the triangle theorem, x has to be less than the sum of the other three sides. The sum of the other three sides is 6021, so x must be less than that or $ \boxed{6020}$.

[quote="PowerOfPi"]
Actually, it specifically said x is an integer.[/quote]

But [b]$ x$ is not the diagonal, it is the last side length[/b].


superior algebra
superior algebra unsolved

Could someone help me with this algebra problem?

Given m and n, find how many ideals of the ring Z_m={0,1,...,m-1} contain the ideal generated by n; I=(n).

Any help would be appreciated. ; )

The ideals of $Z/(mZ)$ correspond to the ideals of $Z$ containing $mZ$. They have the form $aZ$ for a unique natural number $a$, and $mZ \subseteq aZ$ means $a ~|~ m$. So, the ideals of $Z/(mZ)$ correspond to the divisors of $m$. Now it is easy to see, that the ideal of $Z/(mZ)$, which is generated by $n+mZ$, corresponds to the ideal of $Z$, which is generated of $gcd(m,n)$. So, we get $a ~|~ gcd(m,n)$. The number of ideals, we're looking for, is the number of divisors of $gcd(m,n)$.

Thanks for the great help!! :thumbup:  :)


ARML
geometry
analytic geometry
email
AMC
AIME
USA(J)MO

Greetings everyone,

This forum is open to everyone on AoPS but is intended for the use of mostly just South Carolinians.  Feel free to use this forum as a way to meet new people in your area, discuss local competitions, and coordinate various events that you might have in mind.  One major goal of this forum is to allow the members of the SC All State team to interact more than just the few times we see each other in person.

Btw, mysmarthmouth and I moderate this forum.

It would also be nice for everyone to give a brief description of theirself.  Here's mine:

My name is Joseph Laurendi.  I'm 17 years old and attend Summerville High School.  I am currently a senior.  I've gone on the Penn State/ARML trip the last two years.

If for some reason you need to contact me you can
(a) PM me
(b) email me at [edited out...]
(c) talk to me on AIM, SN is BaseballJoe2002

Arnav Tripathy at East Chapel Hill High; I went to ARML and stuffs, yeah, hi everyone.

I am Yasha B.-K. I am 15 and a sophmore at Broughton High School in Raleigh, and this year I started a rather successful Math Club at my school. I am currently taking AP Calc BC.

[clipped bragging]

I like math :-).

Joey Montoya, joml's bodyguard
Just a math enthusiast in the Charleston Area

david henshaw, another one of joml88's  body guards.

I'm John Berman from Wilmington, NC. I'm a rising sophomore at Hoggard High School, and I'll be taking Calc. 3 and Linear Algebra next year. I've been to ARML and MOSP and whatnot.

Heh...well, noone really uses them as of now.  There just aren't many people from the Carolinas on AoPS at all :( .

13 in NC already!...keep up the hard work and you will be awesome in high school :)

Hi, I'm gauss. I'm a first year teacher from South Carolina. I help to organize the South Carolina ARML Team, and generally stalk talented math students from South Carolina to throw problems at them. :)

I look forward to seeing all of you who are on the South Carolina ARML team on the bus coming soon!

great...now i have a stalker.  :D
hmmm, that makes 2 for joe.  :lol:

hey marcus, were we supposed to get more practice problems, like in the mail? not that i finished (or really started) the first set of geometry. if i finish the problems anytime soon, i'll start doing more of these online ones, i guess.

Haven't introduced myself and i've been using this forum since May.

Hi, I'm Shobhit, I go to Hammond (11th grade) and live in Columbia. I started getting into Math stuff from the beginning of the Summer. I'm on the All State Math Team and hopefully I'll be able to go to ARML this year....depending on my SAT situation :( .

Also, my goal, primarily for next year, is to win a major math contest in SC (like USC Columbia or Furman), or placing in the top three is ok too, and to make USAMO.

I'm Sean, repin' AC Flora, though I go to Crayton.  I'm in the 8th grade, but I am participating in ARML & AIME this year.  I have a long way to go, but I hope to make USAMO next year as a freshman.  I'm a moderator of this forum, along with Joe Laurendi (joml88), so if you have any questions, feel free to contact me.

[quote="Iversonfan2005"][quote="mysmartmouth"]I'm a moderator of this forum, along with Joe Laurendi (joml88), so if you have any questions, feel free to contact me.[/quote]

No, you're only a moderator of Careers in Mathematics and Other US Math Contests...not SC ;)[/quote]

No, I have just became moderator of this forum due to the overwhelming increase of post expected as ARML draws nearer.

I'm Harry Richman.  I'm 14, and a freshman at
Spring Valley High School in Columbia.  Last year I got 2nd place at the State Mathcounts competition, and 38th at Nationals.  Next year I'm taking AP Calculus BC.

Im david jensen,16,junior at James F. Byrnes (the football guys) and im in AP stats. i like to do fun things including math and sleeping

Hello everyone, I'm David Christopher, a senior at Byrnes High School. I like chocolate, sleeping, and (of course) math.

I don't have much to brag about as far as math team stuff goes since this is my first year doing it, but I did AP Stats last year and I'm in AP Calculus now - AB in school and BC on my own (at least that's the plan).

I'm on SC's State Math Team this year (just barely), and looking forward to learning what I can before ARML. I appreciate any and all help/suggestions/whatever, especially if it is accompanied by free food.  :lol:

I'm Bella Wang, sophomore from a high school you've probably never heard of in Florence. I was on the ARML team last year, and I qualified at least for AIME and hopefully for USAMO (I took the alternate AIME this morning, so I can't even get a good estimate) this year.

I haven't been doing much math recently, but I'm trying to improve my work ethic :lol:
...Yeah, we'll have to see how that one goes.

[quote="joml88"]I edited the first post since is was way out of date :D[/quote]

same here, well, not "way out of date" but still

hi i'm asif. i go to spring valley high school in columbia. i'm currently a junior.

New people from Flora:
apeman5291 - Ian
scmathman13 - Jim
SoltzElla - Ella

My name is Ankit Grover. I am a sophomore at Daniel High. I have been going to sc competitions over the last 2 years. This year I am in BC calc. I took the aime, and I am on the arml team.

hi, I'm Rono Srimani. I'm a 9th grader at Daniel High and Im on the ARML. I went to National Mathcounts in 7th and 8th grade and in 8th I got 40 something place. I took the AIME this year.

[quote="rs9999"]
Do you like my picture?[/quote]

not really ;)

but welcome to aops :)

I'm from NC, but would like to remain anonymous.

This is kind of...late but anyways...

I'm Paavan Gami from Greenville (Greer), SC. I'm part of the Riverside Middle School math team, and I'm an Eighth grader. I will be taking Pre-Calc next year.

I won our chapter's Math Counts CD and and came 2nd in written.

I'm Indian, but born in Canada...

P.S. dark_sunburst...didn't you go to RMS...Eric Kao, Brian Chou and all keep talking about you (and Tricia)

yea, bella went to riverside... cause our team beat them at state that year haha... we were sick...

Hi!   I'm Anne, a freshman at Porter-Gaud (Charleston).  PG is the most stubbornly passive-aggressive anti-math school ever.  -.-
That just about ruins my life, but hey, whatever.

I'm currently in this strange, strange course named PreCalc, which happens to cover EXACTLY the same things as last year's AlgIII/Trig.  Just that supports my theory that my school is demented.  I'm taking APCalcBC and APstat next year.

My school didn't HAVE A high school math team/club until this year, when I forced them into it.  So...basically, no one but me even got close to qualifying for AIME (I qualified on the AMC12B, not AMC10A...apparently 12B was a lot easier than 12A?).

So anyways, my school believes that the SCISA mathmeet is God, but really, we're still so pathetic that we can't place as a team.  I only got fourth...I did dominate in middle school though. ^_^ but...that's over.    >.<

Annnd we couldn't even find more than 2 people to go to CCU, so...we didn't go.  *sigh*

So basically, my school gives me a big head, and I expect to be humbled here.  ^_^
Actually, I already am... :D

And as a sidenote...I'm known for rambling incessantly.  I'm sure you've noticed.  <_<

Hello I'm Ravi and and going to 8th grade. Next year I'll probably be on Wilson High School's Math Team - and hopefully I'll do SC's ARML.


inequalities
geometry
inradius
circumcircle
triangle inequality
inequalities unsolved

$ a,b,c$ of a triangle
prove that
if $ a\plus{}b\plus{}c\equal{}1$ then $ a^2\plus{}b^2\plus{}c^2<\frac{1}{2}$

By the triangle inequality,
\[ 2\sum ab\equal{}\sum a(b\plus{}c)>\sum a^2
\]

Therefore,
\[ \sum a^2 < \frac{\sum a^2}{2}\plus{}\sum ab\equal{}\frac{(a\plus{}b\plus{}c)^2}{2}\equal{}\frac{1}{2}
\]

we know that:

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(p^2 \minus{} r^2 \minus{} 4Rr)$

where $ p,r,R$ denote the semiperimeter,inradius and circumradius of the triangle...

then by the condtion $ a \plus{} b \plus{} c \equal{} 1$ we get that $ 2p \equal{} 1$ now our inequality is equivalent to:

$ 2(p^2 \minus{} r^2 \minus{} 4Rr) < \frac {1}{2}$

$ \Rightarrow 4(p^2 \minus{} r^2 \minus{} 4Rr) < 1$

$ \Rightarrow 4p^2 \minus{} 1 < 4(r^2 \plus{} 4Rr)$

$ \Rightarrow 0 < r^2 \plus{} 4Rr$

which is obviously true since $ r,R > 0$...

[quote="mathe"]$ a,b,c$ of a triangle
prove that
if $ a \plus{} b \plus{} c \equal{} 1$ then $ a^2 \plus{} b^2 \plus{} c^2 < \frac {1}{2}$[/quote]

Stronger Beautiful Inequalitie Here:
http://www.mathlinks.ro/viewtopic.php?t=147090  :wink:


[quote="mr.k_74"]1.  A positive three-digit integer is formed by choosing three different digits form the set {0,2,4,6,8}.  The first digits cannot be 0. What is the probability that the number is divisble by 3? Express your answer as a common fraction.

16. What is the sum of all positive two-digit integers with exactly 12 factors?

21. Fifteen balls are numbered 1-15. Four balls are then selected at random without replacement. What is the probabillity that the sum of the numbers on those four balls are odd? Express your answer as a common fraction.

Thanks :)[/quote]

1. the only groups of 3 you can use are 240, 840, 246, 468 because they are divisible by 3. 240 and 840 have 4 ways each to arrange the digits so that 0 is not first. 246 and 468 have 6. Their are a total of 48 possibilities so 20/48 which is 5/12.

Sorry, can't do the latex for #16 and too lazy to do 21.

1. http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=92865699&t=232612
16. http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=448475785&t=229765
21. http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2129032618&t=258546

some of the answers might be wrong

1-) Already answered

2-) Answer : 402
Number of divisors = ( n+1)*(m+1) * (k+1)....

We are only looking for two digit integers
In order to have 12 divisors ( 12=1*12, 12=3*4, 12= 2*6 or 12=2*2*3),
then the exponents can be
1 and 11 - no 2-digit integer satisfy this ( 3^1*2^11 is too big)
2 and 3 - we can have 2^3*3^2 = 72,
2 and 6 - we can have 3*2^5 = 96
1, 1 and 2 ( 3*5*2^2 = 60, 2*5*3^2=90, 3*7*2^2=84

so the answer is 72+96+60+90+84= 402

3-) you need odd
so you can only have 3 even + 1odd, or 3odd + 1 even.
calculate the probability of getting 3 even and 1 odd
And calculate the probability of getting 3 odd one even and add them
the answer should be 32/65


On a standard clock with diameter 12 inches, the minute hand and the hour hand are the same length.  For every minute of the day there is a diameter that can be drawn on the clock that is a mirror to the minute and hour hand.  In mph, compute the speed of the tip of this diameter as it moves across the clock during the day.

[quote="miyomiyo"]...that is a mirror the the minute and hour hand. [/quote]

Ummm...what? That makes no sense.

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A day has 1440 minutes. At minute $m$, the hour hand is at $360 * \frac{2m}{1440} = \frac{m}{2}$ degrees and the minute hand is at $360 * \frac{24m}{1440} = 6m$ degrees. The "mirror hand" is at the average of these two, which is $\frac{13}{4}m$ degrees. So the tip of the "mirror hand" covers the entire circumference, $24\pi$ inches, in $\frac{1440}{13}$ minutes. So its speed is

$\frac{ 24\pi \text{ inches } }{ \frac{1440}{13}\text{ minutes }} = \frac{ 24\cdot 13 \text{ inches } }{ 24 \text{ hours }} = 13 \text{ inches per hour } = \frac{13}{5280} \text{ mph}$
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