Harvard college MIT counting distinguishability

Four students from Harvard, one of them named Jack, and five students from MIT, one of them named Jill, are going to see a Boston Celtics game. However, they found out that only $ 5$ tickets remain, so $ 4$ of them must go back. Suppose that at least one student from each school must go see the game, and at least one of Jack and Jill must go see the game, how many ways are there of choosing which $ 5$ people can see the game?

[hide="Solution"][b]Case 1:[/b] Jack goes to the game
$ \binom{5}{1}\cdot\binom{8}{3}$

[b]Case 2:[/b] Jill goes to the game
$ \binom{4}{1}\cdot\binom{8}{3}$

$ S\equal{}\binom{5}{1}\cdot\binom{8}{3}\plus{}\binom{4}{1}\cdot\binom{8}{3}\equal{}5\cdot56\plus{}4\cdot56\equal{}\boxed{504}$[/hide]

what are those things that solution is made of  ??

[quote="i_like_pie"][hide="Solution"][b]Case 1:[/b] Jack goes to the game
$ \binom{5}{1}\cdot\binom{8}{3}$

[b]Case 2:[/b] Jill goes to the game
$ \binom{4}{1}\cdot\binom{8}{3}$

$ S \equal{} \binom{5}{1}\cdot\binom{8}{3} \plus{} \binom{4}{1}\cdot\binom{8}{3} \equal{} 5\cdot56 \plus{} 4\cdot56 \equal{} \boxed{504}$[/hide][/quote]

Why am I getting a different answer?
[hide]

Case 1: Jack goes to the game, Jill does not.
$ \binom{4}{1}\binom{7}{3}$
Case 2: Jill goes to the game, Jack does not.
$ \binom{3}{1}\binom{7}{3}$
Case 3: Jack and Jill go to the game.
$ \binom{7}{4}$

280?  :(

[/hide]

[hide="And the right person is ..."]
Sorry, neither of you.  :( Don't forget about the one per school condition.

Complement counting might kill some of the casework.

Edit:
[hide="Solutions"]
If Jack goes, but Jill does not go, there are $ {7\choose 4}$ ways to determine the remaining people (note that at least one person from both MIT and Harvard). If other way around, then $ {7\choose 4}\minus{}1$, since 5 people cannot go from MIT. If both go, then there are $ {7 \choose 3} \equal{} {7\choose 4}$ ways to determine the remaining three people.

Thus, $ 3 \cdot {7\choose 4} \minus{} 1 \equal{} \boxed{104}$.

Alternatively, by complement counting. There are $ {9\choose 5} \equal{} 126$ ways to pick five people (which is an immediate sign that the two above answers are incorrect). However, it is not possible for Jill and Jack to both not go, of which there are $ {7 \choose 5} \equal{} 21$ ways, and also we cannot have 5 people from MIT going, with $ 1$ way. Thus the answer is $ 126 \minus{} 21 \minus{} 1 \equal{} 104$.
[/hide]

[/hide]

[quote="mcsquared"]Actually, I like Circle's way, which is almost right except that here:
[quote="CircleSquared"][hide]

Case 3: Jack and Jill go to the game.
$ \binom{7}{4}$

[/hide][/quote]

It's actually $ \binom{7}{3}$ , since Jack and Jill are 2 people.

FAIL[/quote]

:( it's the same thing, i swear!

and i have to agree with azjps

well here is how i did it  :) but might be wrong tho

Harvard 4 students
MIT5 students

total of 5 students are going, and at least one student from each school have to go
so
Harvard-MIT
4-1
3-2
2-3
1-4

4-1 (4 from Harvard, and 1 from MIT)
way of picking from Harvard: 4C4=1,
way of picking from MIT: Jill has to be the one from Mit, so only 1 way
and total of 1*1=1 way

3-2 (3 Harvard, 2 MIT)
Harvard: 4C3=4, but Jack must be in the group, so we have to get rid one of the group of 3 that doesnt have Jack in it, 4-1=3
MIT: 5C2=10, now how many ways we can pick 2 people with out Jill, 4C2=6 ( we are picking 2 students out of 4, assume Jill is not part of the group), so 10-6=4
and total of 3*4=12 ways

2-3 (2 Harvard, 3MIT)
Harvard: 4C2=6, assume Jack is not in the group there are 3C2=3 ways of picking 2 students without Jack, so 6-3=3 ways
MIT: 5C3=10, and there are 4C3=4 ways not picking Jill, so 10-4=6 ways
and total of 3*6=18 ways

1-4 (1 from Harvard, 4 from MIT)
Harvard: has to be Jack, 1 way
MIT: 5C4=5, and there will be one group of 4 without Jill, so 5-1=4
and total of 1*4 ways

1+12+18+4=[b]35 [/b]ways of picking students

The problem is that Jack and Jill do not need to both go, which is what you seem to be assuming. One of them suffices.  :wink:

Not quite, it's slightly closer to three times lgd's answer (consider the three cases when Jill goes, just Jack goes, and when both go; lgd's only accounts for the last one).

ehh I think I might be completely off here, but:

[hide="assuming students (other than Jack/Jill) are indistinguishable other than the fact that they go to either Harvard or MIT - nevermind"]
[b]Case 1: One of either Jack or Jill goes[/b]

the remaining four Harvard/MIT students may be:
4-0
3-1
2-2
1-3
0-4

$ 5\cdot2 \equal{} 10$

[b]Case 2: both Jack and Jill go[/b]

the remaining three Harvard/MIT students may be:
3-0
2-1
1-2
0-3

$ 4\cdot1 \equal{} 4$

so $ 10 \plus{} 4 \equal{} \boxed{14}$?
[/hide]

[hide="students are distinguishable"]
[b]Case 1: Jill (from MIT) goes[/b]

$ \binom{4}{0}\cdot\binom{4}{4} \equal{} 1$
$ \frac {\binom{4}{1}\cdot\binom{4}{3}}{4} \equal{} 4$
$ \frac {\binom{4}{2}\cdot\binom{4}{2}}{6} \equal{} 6$
$ \frac {\binom{4}{3}\cdot\binom{4}{1}}{4} \equal{} 4$
$ \binom{4}{4}\cdot\binom{4}{0} \equal{} 1$

total: $ 16$

[b]Case 2: Jack (from Harvard) goes[/b]

uh calculate as above to get $ 30$

[b]Case 3: both go[/b]

calculate as above to get $ 15$

$ 16 \plus{} 30 \plus{} 15 \equal{} \boxed{61}$

 :huh:
[/hide]

yes,this is the exact same thing that i did. I asked my dad if it was right and he said it was (he has a pHd, so i believe him)
And i think that students should be distinguishable, because otherwise they wouldn't neccaserilly give us the nemes of some of the people.

@mcsquared:

a choose b = a choose c, if b+c=a

So 7 choose 3 is the same as 7 choose 4.

[quote="undefined117"]
[hide="students are distinguishable"]
[b]Case 1: Jill (from MIT) goes[/b]

$ \binom{4}{0}\cdot\binom{4}{4} \equal{} 1$
$ \frac {\binom{4}{1}\cdot\binom{4}{3}}{4} \equal{} 4$
$ \frac {\binom{4}{2}\cdot\binom{4}{2}}{6} \equal{} 6$
$ \frac {\binom{4}{3}\cdot\binom{4}{1}}{4} \equal{} 4$
$ \binom{4}{4}\cdot\binom{4}{0} \equal{} 1$

total: $ 16$

[b]Case 2: Jack (from Harvard) goes[/b]

uh calculate as above to get $ 30$

[b]Case 3: both go[/b]

calculate as above to get $ 15$

$ 16 \plus{} 30 \plus{} 15 \equal{} \boxed{61}$

 :huh:
[/hide][/quote]

[hide="Unfortunately,"]
Your cases should be
\[ {4\choose 1} {3\choose 3} \plus{} {4\choose 2} {3\choose 2} \plus{} {4\choose 3} {3\choose 1} \equal{} 34
\]
for Jill and almost the same for the other two cases.

There are a couple of subtle problems in the casework you used: first, we aren't picking four from each school in each case, we're picking 4 from MIT and 3 from Harvard (because we don't want to pick Jack either), so there's an overcount. Second, I'm not sure where the divisions by $ 4$ and $ 6$ came from  :huh:  ... but that creates a larger undercount.
[/hide]

Edit: Not quite; think about it again. You are actually choosing from a pool of 4 MIT students and 3 Harvard students. (remember, the case you have is not "Jill is picked", it is "Jill is picked, Jack is not picked"  :wink: ).

[quote="azjps"][quote="undefined117"]
[hide="students are distinguishable"]
[b]Case 1: Jill (from MIT) goes[/b]

$ \binom{4}{0}\cdot\binom{4}{4} \equal{} 1$
$ \frac {\binom{4}{1}\cdot\binom{4}{3}}{4} \equal{} 4$
$ \frac {\binom{4}{2}\cdot\binom{4}{2}}{6} \equal{} 6$
$ \frac {\binom{4}{3}\cdot\binom{4}{1}}{4} \equal{} 4$
$ \binom{4}{4}\cdot\binom{4}{0} \equal{} 1$

total: $ 16$

[b]Case 2: Jack (from Harvard) goes[/b]

uh calculate as above to get $ 30$

[b]Case 3: both go[/b]

calculate as above to get $ 15$

$ 16 \plus{} 30 \plus{} 15 \equal{} \boxed{61}$

 :huh:
[/hide][/quote]

[hide="Unfortunately,"]
Your cases should be
\[ {4\choose 1} {3\choose 3} \plus{} {4\choose 2} {3\choose 2} \plus{} {4\choose 3} {3\choose 1} \equal{} 34
\]
for Jill and almost the same for the other two cases.

There are a couple of subtle problems in the casework you used: first, we aren't picking four from each school in each case, we're picking 4 from MIT and 3 from Harvard (because we don't want to pick Jack either), so there's an overcount. Second, I'm not sure where the divisions by $ 4$ and $ 6$ came from  :huh:  ... but that creates a larger undercount.
[/hide][/quote]
Hmm. I thought it was 4 from each school because in Case 1, Jill from MIT went (not both Jill and Jack - that's Case 3). The remaining four tickets had to be distributed to four of the [i]four MIT[/i] students and [i]four Harvard[/i] students.

I probably have other mistakes anyway (like the division by 4 and 6.. I forgot why I did that too :blush:) so I'll try it again..

EDIT: even after a few tries, I keep getting an odd number divided by an even :(

How come everyone has a different answer? Who is correct? Apparently, I got 104.

 :arrow:  :ninja:

I agree with azjps. I think the correct answer is 104, which I got by complementary counting.


geometry
trigonometry
trig identities
Law of Cosines
Law of Sines
geometry unsolved

A scalene triangle with a 35 degree angle and side lengths 33 and 17 has triangles congruent to it pasted on its sides so that touching sides are congruent. To the new triangles, more triangles are pasted on in the same fasion. After one billion triangles are pasted, the smallest circle that contains all of them is drawn. Compute the smallest possible area of this circle.

Hide solutions.

Feel free to post any partial answers (got somewhere, but not finished).

[hide="An idea - hardly anything"]There would be three steps to solving the problem:
1.  Find the 33 * 17 * x triangle with a 35 degree angle with the smallest area
2.  Find a way to tile the triangles taking up the least area
3.  Find a way to circumscribe a circle around it and find the radius[/hide]

[hide="Solution to part one"]
After quite a bit of law of cosines and absinc and law of sines, we get that the smallest triangle is the one with the 35 degree angle between the 33 and the 17, the remaining side being approximately 21.42, the angle opposite 17 being approximately 17.1 degrees and the the angle opposite 33 being 117.9 degrees.

I think part two would be done by making sort of circles with the 27.1 degree angles being at the center and the circles of triangles touching like gears, but I don't know how to count the triangles used up.

I have no idea how to do part three.
[/hide]


geometry
calculus
derivative
3D geometry
sphere
calculus computations

Area of a circle with radius r : $ \pi r^2$
Circumference of a circle with radius r : $ 2\pi r$ (Derivative of $ \pi r^2$)

Volume of a sphere with radius r : $ \frac{4\pi r^3}{3}$
Surface area of a sphere with radius r : $ 4\pi r^2$ (Derivative of $ \frac{4\pi r^3}{3}$)

Is this just a coincidence?

Every six-to-nine months, same topic appears  :)

http://www.mathlinks.ro/viewtopic.php?t=199156


Here is a problem from Erdos using nothing else but the Pigeonhole Argument. I know its solution so Im not asking for hints just to show you.

Select any $n+1$ element subset from $\{1,2,3,...,2n\}$. Show that there are two elements in the subset such that one divides the other.

As a side note, you can find a generalization [url=http://www.mathlinks.ro/viewtopic.php?p=518130#518130]here[/url].


function
number theory
totient function
number theory unsolved

Prove that if $ n\in\mathbb{N}$ satisfies $ \phi(n)\neq \phi(m)$ for all $ m\neq n$, then $ n>10^{10}$.

This is related to Carmichael's totient function conjecture ("If $ n \in \mathbb{N}$, then $ |\{x \in \mathbb{N}: \phi(x) = n\}| \ne 1$"). In fact, it has been proved that the least counterexample to this conjecture must be greater than $ 10^{10^{7}}$.

Thank you!, but do you know how to prove this problem? (not necessarily the strength of what you said but just $ 10^{10}$.)

Till now, I can just prove the least counterexample must occur for $ n > 3 \cdot 10^{6}$. Indeed, I'll show why. Before getting started, just as a notational remark, say $ \mathbb{N}= \{1, 2, \ldots\}$, $ \mathbb{N}_{0}= \mathbb{N}\cup \{0\}$ and $ \mathbb{P}= \{p \in \mathbb{N}: p \mbox{ is prime}\}$. Then suppose $ n \in \mathbb{N}$ is any actual counterexample to Carmichael's totient function conjecture.

Since $ \phi(a) = \phi(2a)$, whenever $ a \in \mathbb{N}$ is odd, it's clear we need $ n = 2^{\alpha+2 }\cdot a$, where $ a \in \mathbb{N}$ is odd too and $ \alpha \in \mathbb{N}_{0}$. Furthermore, note that $ \phi(2^{\alpha+2}a) = 2^{\alpha-1}\phi(a) = \phi(2^{\alpha-1}\cdot 3a)$, whenever $ 3 \nmid a$. Hence necessarily $ n = 2^{\alpha+2}\cdot 3^{\beta+2}\cdot b$, where $ \beta \in \mathbb{N}_{0}$ and $ b \in \mathbb{N}$ is prime with $ 6$. Now we state the following:

[b]Lemma:[/b] let $ p_{1}, p_{2}, \ldots, p_{r}\in \mathbb{P}$ be pairwise distinct and odd, where $ r \in \mathbb{N}$. If $ k, m, n \in \mathbb{N}$ are such that $ n-k \ge 1$, $ \gcd(m, p_{1}p_{2}\ldots p_{r}) = 1$ and $ q = 2^{k}\cdot \prod_{i=1}^{r}p_{i}+1$ is prime, then
\[ \forall \alpha_{1}, \alpha_{2}, \ldots, \alpha_{r}\in \mathbb{N}_{0}: \phi\!\left(2^{n}\cdot \prod_{i=1}^{r}p_{i}^{\alpha_{i}+2}\right) = \phi\!\left(2^{n-k}q \cdot \prod_{i=1}^{r}p_{i}^{\alpha_{i}+1}\right). \]
I think the proof is somehow obvious, so I won't give it. Now back to the problem note that $ 7 = 2 \cdot 3+1$ is prime. Thus according to the previous lemma and the multiplicativity of Euler's phi function: $ \phi(2^{\alpha+2}\cdot 3^{\beta+2}\cdot b) = \phi(2^{\alpha+1}\cdot 3^{\beta+1}\cdot 7b)$, whether $ 7 \nmid a$. So a fortiori $ 7^{2}\mid b$, if $ n$ contradicts Carmichaels's conjecture (as we're supposing). I.e. $ n = 2^{\alpha+2}\cdot 3^{\beta+2}\cdot 7^{\gamma+2}\cdot c$, for some $ \gamma \in \mathbb{N}_{0}$ and $ c \in \mathbb{N}$. Finally argue by the same arguments that $ 43^{2}\mid n$ as well, just considering $ 43 = 2 \cdot 3 \cdot 7+1$ is prime and applying the foregoing lemma once again. Thus $ n \ge 2^{2}\cdot 3^{2}\cdot 7^{2}\cdot 43^{2}> 3 \cdot 10^{6}$.

I made some progress till now. We have already proved that any counterexample $ n \in \mathbb{N}$ to Carmichael's totient function conjecture has necessarily the form $ n = 2^{\alpha_{1}+2}\cdot 3^{\alpha_{2}+2}\cdot 7^{\alpha_{3}+2}\cdot 43^{\alpha_{4}+2}\cdot a$, where $ \alpha_{i}\in \mathbb{N}_{\;\!0}$ (as $ i \in \overline{1;4}$) and $ a \in \mathbb{N}$ is prime with $ 2 \cdot 3 \cdot 7 \cdot 43$. Now suppose $ \alpha_{1}\ge 1$. Then clearly $ \phi(n) = \phi(5n/4)$, whenever $ 5 \nmid a$. Thus we must admit $ n = 2^{\alpha_{1}+2}\cdot 3^{\alpha_{2}+2}\cdot 7^{\alpha_{3}+2}\cdot 43^{\alpha_{4}+2}\cdot 5^{\alpha_{5}+2}\cdot b$, for some $ \alpha_{5}\in \mathbb{N}_{\;\!0}$ and $ b \in \mathbb{N}$. So, since
[list] $ 11 = 2 \cdot 5+1$,
$ 31 = 2 \cdot 3 \cdot 5+1$,
$ 71 = 2 \cdot 5 \cdot 7+1$,
$ 431 = 2 \cdot 5 \cdot 43+1$
$ 94227517231 = 2\cdot 3 \cdot 5\cdot 7 \cdot 11 \cdot 31\cdot 43 \cdot 71 \cdot 431+1$
[/list] are all primes, we get (by repetively applying the lemma in the previous post) that $ n \ge 2^{3}\cdot \prod_{p \in A}p^{2}> 10^{45}$, where $ A$ is the set
\[ \{3, 5, 7, 11, 31, 43, 71, 431, 94227517231\}. \]
In fact, note this bound can be enormously improved by searching for primes of the form $ 4t+1$, such that $ t$ is a product of pairwise distinct elements of $ A$ (e.g., $ 13 = 4 \cdot 3+1$, $ 61 = 4 \cdot 3 \cdot 5+1$ and so on) and iterating this process many and many times, while adding new elements to $ A$ at each new step, in order to obtain a larger "prime repository". Hence finally we're left with the case when $ \alpha_{1}= 0$ and $ n = 4 \cdot 3^{\alpha_{2}+2}\cdot 7^{\alpha_{3}+2}\cdot 43^{\alpha_{4}+2}\cdot a$. :wink:


function
real analysis
real analysis unsolved

Let $ a>1$ and let $ f: (1,2] \rightarrow \mathbb{R}$ be defined by $ f(x)\equal{}\frac{x}{x\minus{}1}$. Prove that $ f$ is uniformly continuous on the interval $ [a,2]$, but $ f$ is not uniformly continuous on $ (1,2]$.

I am lost on this problem. Any help is appreciated. Thanks in advance.

[quote="coquitao"][quote="eskimo343"]Let $ a > 1$ and let $ f: (1,2] \rightarrow \mathbb{R}$ be defined by $ f(x) \equal{} \frac {x}{x \minus{} 1}$. Prove that $ f$ is uniformly continuous on the interval $ [a,2].$

[/quote]

[url=http://everything2.com/e2node/uniformly%2520continuous]Any continuous function on a closed and bounded interval is uniformly continuous.[/url]  :wink:[/quote]

[i]Ahh, why didn't I see that???[/i]  Thanks.   :rotfl:


AMC AIME trigonometry LaTeX quadratics algebra polynomial

I will be posting trig problems as practice for the AIME. Most AIME's have one or two of them towards the end of the test and although they may look intimidating, they are usually solved easily with some basic tools.

[b] Problem 1 (Easy) [/b]

$\sqrt{i} \cdot \sqrt[3]{i}$ can be expressed as $\frac{i(\sqrt{a}+\sqrt{b})+\sqrt{c}-\sqrt{d}}{e}$ or $- \frac{i(\sqrt{a}+\sqrt{b})+\sqrt{c}-\sqrt{d}}{e}$ where $i=\sqrt{-1}$. Find the value of $a+b+c+d+e$.

[b] Problem 2 (Easy-Medium) [/b]

If $x+\frac{1}{x}=\sqrt{3}$ find the value of $x^{2004}+\frac{1}{x^{2004}}$

Note: All fractions are to be in simplest terms

I think figuring that out is the key to solving the problems quickest.

[quote="zanttrang"]Just out of curiosity, how are these trig problems?[/quote]

Those first two would probably be better classified as complex numbers.  But you see a lot of $\cos+i\sin$'s with complex numbers so that would be one reason why they are trig problems  :roll:

Heh, if you trying doing these problems the old fashion way (for example number 2), you will definitely not like the numbers. And yes by trig I am also including complex numbers.

[hide="Problem 1"]
$\sqrt{i}\cdot \sqrt[3]{i}=i^{5/6}=(cis 90^{\circ})^{5/6}=cis 75^{\circ}$.  Using the sine and cosine addition formulae, we find
$cis 75^{\circ}=\frac{i(\sqrt{6}+\sqrt{2})+\sqrt{6}-\sqrt{2}}{4}$
and thus, $a+b+c+d+e=20$.

NOTE: Some restrictions need to be placed on $a,b,c,d,e$.  Otherwise we could have
$\frac{i(\sqrt{6}+\sqrt{2})+\sqrt{6}-\sqrt{2}}{4}=\frac{i(\sqrt{24}+\sqrt{8})+\sqrt{24}-\sqrt{8}}{8}$
Which would give a different value for $a+b+c+d+e$.
[/hide]

[quote="DPopov"]
[b] Problem 2 (Easy-Medium) [/b]

If $x+\frac{1}{x}=\sqrt{3}$ find the value of $x^{2004}+\frac{1}{x^{2004}}$[/quote]

[hide="solution"]Simplifying and solving for $x$, we have

$x = \frac{\sqrt{3} \pm i}{2}$

$x = e^{\pm \displaystyle\frac{\pi}{6}i}$

Taking $x^{2004}$ we have

$x^{2004} = e^{\pm 338 \pi} = 1$

Therefore,

$x^{2004} + \frac{1}{x^{2004}} = 2$

$QED$[/hide]

First of all, your calculator probably gives you something incorrect for $x^{2004}$ (or rather $a+b \cdot 10^{-16}$ or something similar but manages to cancel the error after you calculate the whole expression) and these problems should not be done using a calculator! :P  As for the answer about $\displaystyle{cis}{\theta}$, yes $\displaystyle{cis}{\theta}=\cos{\theta}+i \sin{\theta}$

You found the value of $x$, like I did (and we agree on that); it was after that that I ran into trouble, but nice solution with the $e$ function. I need to remember the $e$ and $cis$ more often... that would really help me. I think these are the kinds of problems that I have the most trouble with.

Also, my calculator (a TI-89) did find the correct value for $x^{2004}$. :lol:

[quote="themonster"]Too bad, \cis doesn't work in $\LaTeX$, it would make it look less ugly.[/quote]

Just use $\text{cis} \ \theta$.  It does the same thing (it just takes a little longer to type out).

[quote="zanttrang"][quote="JesusFreak197"]Is $cis\theta=\cos\theta + i\sin\theta$?

[b]EDIT:[/b] For #2, my calculator does it just fine (after I find what $x$ is equal to on paper), but I can't figure out how to simplify the second equation.[/quote]

Yes, $cis$ is just shorthand notation for $\cos i \sin$.  As to your second question, you don't need to simplify the second equation.  Just take a look at my solution above.[/quote]

You mean $\text{cis}$ is a shorthand for $\cos + i\sin$ ;)  (in context one might figure this out, but it could also confuse some people who might think it is multiplication)

Another, less elegant way to solve #2 is with Newton sums, as follows.

[hide]First assume that $x^{2004} + \frac{1}{x^{2004}} = k$

Then

$\left(x^{2004}\right)^2 -kx^{2004} + 1 = 0$.

Assume that $x_1, x_2$ are the roots of the equation $x^2 - \sqrt{3} x + 1 =0$.  We can tell now that we want to find $x_1^{2004} + x_2^{2004}$, or, the 2004th Newton sum $s_{2004}$ for the equation.  We can now set up the recurrence equation

$a_2 \cdot s_{n+1} = - a_1 \cdot s_n - a_0 \cdot s_{n-1}$
$s_{n+1} = \sqrt{3} s_n - s_{n-1}$

with $s_1 = \sqrt{3}$ and $s_2 = 1$ (you have to know something about Newton sums to be able to set up this equation, and it's a little complicated to explain, but AOPS Volume 2 has a good section on this topic).  Now, calculating the first several $s_n$ we have

$s_2 = 1$
$s_3 = 0$
$s_4 = -1$
$s_5 = -\sqrt{3}$
$s_6 = -2$
$s_7 = -\sqrt{3}$
$s_8 = -1$
$s_9 = 0$
$s_{10} = 1$
$s_{11} = \sqrt{3}$
$s_{12} = 2$
$s_{13} = \sqrt{3}$
$s_{14} = 1$
$s_{15} = 0$

We now have a recurring pattern with period 12.  Note that $s_{12n} = 2$.  Therefore, $s_{2004} = s_{167 \cdot 12} = 2$.[/hide]

[quote="professordad"]

[b]Problem 18:[/b]
Find the positive integer $ n$ such that

$ \arctan\frac{1}{3}\plus{}\arctan\frac{1}{4}\plus{}\arctan\frac{1}{5}\plus{}\arctan\frac{1}{n}\equal{}\frac{\pi}{4}$[/quote]

[hide="Solution 18"]

arctan is relatively ugly, so let's take the tangent of both sides.

$\tan (\arctan\frac{1}{3} + \arctan\frac{1}{4} + \arctan\frac{1}{5} + \arctan\frac{1}{n}) = 1$. Let's treat $\arctan\frac{1}{3} + \arctan\frac{1}{4} = \alpha$ as on quantity and $\arctan\frac{1}{5} + \arctan\frac{1}{n} = \beta$ as another quantity, so $\tan \alpha + \tan \beta = 1 - \tan \alpha \tan \beta$.

Now,

$\tan \alpha = \tan (\arctan\frac{1}{3} + \arctan\frac{1}{4}) = \frac {\tfrac {1}{3} + \tfrac {1}{4}}{1 - (\tfrac {1}{3})(\tfrac {1}{4})} = \frac {7}{11}$. Similarly,

$\tan \beta = \tan (\arctan\frac{1}{5} + \arctan\frac{1}{n}) = \frac {\tfrac{1}{5} + \tfrac{1}{n}}{1-(\tfrac{1}{5})(\tfrac{1}{n})} = \frac {n+5}{5n-1}$.

So,

$\frac {7}{11} + \frac {n+5}{5n-1} = 1 - (\frac {7}{11})(\frac{n+5}{5n-1})$. Equivalently, $(7)(5n-1) + (n+5)(11) = (11)(5n-1) - (7)(n+5)$, so $\boxed {n=47}$.

[/hide]

[hide="Problem 19"]

If $\sin\theta+\cos\theta+\tan\theta+\cot\theta+\csc\theta+\sec\theta=7$, then find $(\cos\theta+\sin\theta)^2$

[/hide]

[quote="mhy123"]
If $\sin\theta+\cos\theta+\tan\theta+\cot\theta+\csc\theta+\sec\theta=7$, then find $(\cos\theta+\sin\theta)^2$
[/quote]

[hide="Solution to Problem 19"]
Note that $(\cos\theta+\sin\theta)^2=\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta=1+\sin2\theta$.
Writing the original equation in terms of sine and cosine, we have
$7=\sin\theta+\cos\theta+\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}+\dfrac{1}{\sin\theta}+\dfrac{1}{\cos\theta}$
Multiplying by $\sin\theta\cos\theta$ we have $7\sin\theta\cos\theta=\sin^2\theta\cos\theta+\sin\theta\cos^2\theta+\sin^2\theta+\cos^2\theta+\cos\theta+\sin\theta$
So $7\sin\theta\cos\theta=(\cos\theta+\sin\theta)(\sin\theta\cos\theta)+1+\cos\theta+\sin\theta$$=(\cos\theta+\sin\theta+1)(\sin\theta\cos\theta+1)-\sin\theta\cos\theta$
So $(\cos\theta+\sin\theta-7)(\sin\theta\cos\theta+1)=-8$. Let $x=\cos\theta+\sin\theta, y=\sin\theta\cos\theta$. So we have $(x-7)(y+1)=-8$. As noted in the first statement in this solution, $x^2=1+2y$, so substituting gives $(x^2+1)(x-7)=-16$, so $x^3-7x^2+x+9=0$. Using the Rational Root Theorem along with Descartes's Rule of Signs, we see that $-1$ is a solution to this cubic, although later we will need to check if it is extraneous for the original problem. Divide the cubic by $x+1$ to get $x^2-8x+9=0$. The solutions to this quadratic are $\dfrac{8\pm\sqrt{28}}{2}=4\pm\sqrt{7}$.
So we have our 3 possible solutions. Note that the value we are trying to find is just $x^2$. Plugging in $x=-1$ gives $1+2\sin\theta\cos\theta=1$, or $\sin\theta\cos\theta=0$. But this is impossible, because both $\sin\theta$ and $\cos\theta$ must be nonzero for $\csc\theta+\sec\theta$ to be defined in the original equation. Plugging in $x=4+\sqrt{7}$ gives $1+\sin2\theta > 23$, which is, again, absurd. The other root, $x=4-sqrt{7}$, doesn't give any such issues, so we have $x^2=\boxed{23-8\sqrt{7}}$ as the answer.
[/hide]
Wow, I almost started graph-bashing that one D: But fixed it in time :D

@Spring: The first few posts seem to say that complex numbers count as trig problems in here (I don't know why).
@professordad: You could also just derive the cotangent sum formula similar to how the tangent one is derived.

New problem:
[quote="I, not understanding why people had to hide new problems, just"]
[b]Problem 20[/b]
Find the least positive integer $n$ such that $\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.$
[/quote]

darn this marathon shouldn't die. someone who knows how to do the problem please post a solution.

my partial solution's pretty bashy, i'm trying to get some things to telescope but i can't fully get them to do so.

[hide="Solution to Problem 20"]Note that $\cot n - \cot (n+1) = \frac{\sin 1}{(\sin n)(\sin (n+1))}$. Hence,

$\cot 45^\circ - \cot 46^\circ  + \cot 47^\circ - \cot 48^\circ + \dots + \cot 133^\circ - \cot 134^\circ = \frac{\sin 1^\circ}{\sin n^\circ}$

But $\cot \theta  + \cot (180^\circ - \theta)=0$; hence,

\begin{align*}
S & = \cot 45^\circ - \cot 46^\circ + \cot 47^\circ - \cot 48^\circ + \dots + \cot 133^\circ - \cot 134^\circ
\\ & = \cot 45^\circ +(\cot 46^\circ + \cot 134^\circ) + (\cot 47^\circ + \cot 133^\circ) + \dots + (\cot 89^\circ + \cot 91^\circ) + \cot 90^\circ
\\ & = \cot 45^\circ + \cot 90^\circ = 1.
\end{align*}

Hence, $\sin n = \sin 1^\circ$, and $n=\boxed{1}$ is the least positive integer solution.[/hide]

[b]Problem 21[/b]: Let $\theta = {\pi^\pi}^\pi$. Evaluate $\frac{1}{\frac{1}{4} (\sin^4 \theta + \cos^4 \theta) - \frac{1}{6} (\sin^6 \theta+\cos^6 \theta)}$.

[hide="Solution 21"]
Let $x=\sin^2\theta$, so $\cos^2\theta=1-x$. We are looking for $\frac{1}{\frac{1}{4}(x^2+(1-x)^2)-\frac{1}{6}(x^3+(1-x)^3}$.
We see that $x^2+(1-x)^2=2x^2-2x+1$, $x^3+(1-x)^3=3x^2-3x+1$, so $\frac{1}{4}(x^2+(1-x)^2)-\frac{1}{6}(x^3+(1-x)^3=\frac{1}{2}(x^2-x-x^2+x)+\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$, so the value we want is just $\boxed{12}$, regardless of the value of $\theta$.[/hide]
[quote="The new problem, grabbing a pencil and some paper,"]
[b]Problem 22[/b]
In $\triangle ABC$, $AB = BC$, and $\overline{BD}$ is an altitude. Point $E$ is on the extension of $\overline{AC}$ such that $BE = 10$. The values of $\tan \angle CBE$, $\tan \angle DBE$, and $\tan \angle ABE$ form a geometric progression, and the values of $\cot \angle DBE,$ $\cot \angle CBE,$ $\cot \angle DBC$ form an arithmetic progression. What is six times the area of $\triangle ABC$?[/quote]

[hide="Solution 20 (nice problem)"]

First of all, note that we are given the following two crucial pieces of information:

(I) $\tan^2 DBE = \tan CBE \cdot \tan ABE$ by definition of three consecutive terms of a geometric sequence.

(II) $2\cot CBE = \cot DBE + \cot DBC$ by definition of three consecutive terms of an arithmetic sequence.

Now, note that there exists symmetry between $\angle {DBE}$ and the two angles formed by line BE. Let $\angle{DBC} = \gamma$ so we can exploit this symmetry.

By doing this, we can modify (I) in the following manner: $\tan^2 DBE = \tan (DBE - \gamma) \cdot \tan (DBE + \gamma)$. Applying tangent addition formulas and simplifying some, we see that $(\tan^4 DBE - 1)\tan^2 \gamma = 0$. But $\tan^2 \gamma \neq 0$ because this would imply that $\gamma = k\pi$ for some integer $k$, although its bound is $0<\gamma<90$. So, we come to the conclusion that $\tan^4 DBE - 1 = 0$. Of the four possible solutions, we see that only $\angle{DBE} = 45$ (*) meets the multiple conditions.

Now, let's try to modify (II). We can again apply our symmetry about $\gamma$ and (*) to get:

$2\cot (45 - \gamma) = 1 + \cot \gamma$, which after applying cotangent addition formulas and SFFT, becomes $(\cot \gamma - 3)(\cot \gamma + 1) = 0$. However, note that $\cot \gamma \neq -1$ because again we have bound restrictions that don't satisfy this criteria. So, $\cot \gamma = 3$ (**) must be true.

Now, we are asked to evaluate $6[\triangle{ABC}]$. Note that $[\triangle{ABC}] = BD \cdot DC$ because it is isosceles. We also know that because of (**), $\cot \gamma = \frac {BD}{DC} = 3 \Rightarrow DC = \tfrac {BD}{3}$. So, $[\triangle{ABC}] = \frac {BD^2}{3}$.

Now, recall from * that $\angle{DBE} = 45$. Thus, by 45-45-90 triangles, $BD = DE = 5\sqrt{2}$ because BE = 10. Thus, $6[\triangle{ABC}] = 6 \cdot \tfrac {(5\sqrt{2})^2}{3} = \boxed {100}$.

[/hide]

[quote="Problem 21"]

Rhombus $PQRS$ is inscribed in rectangle $ABCD$ so that vertices $P, Q, R$, and $S$ are interior points on sides $AB, BC, CD$, and $DA$, respectively.  It is given that $PB = 15, BQ = 20, PR = 30$, and $QS = 40$.  Let $\frac{m}{n}$, in lowest terms, denote the perimeter of $ABCD$.  Find $m + n$.

[/quote]

[hide="Trig Bash"]Let $X$ be the intersection of the diagonals of the rhombus. The diagonals of a rhombus are perpendicular bisectors, so $PX=15$ and $QX=20$. Let $\angle PQX=\alpha$. Since $\triangle QBP \cong \triangle QXP \cong \triangle QXB$ by SAS, $\angle BQP=\angle XQR=\alpha$. This means that $\angle CQR=180^{\circ}-3\alpha$. The diagram is symmetric, so $\triangle QBP \cong \triangle SDR$ and $\triangle APS \cong \triangle CRQ$. (I could prove this using some angle chasing, but I'm getting lazy.) Let $CR=x$. Then we have $\sin(180^{\circ}-3\alpha)=\sin{3\alpha}=\frac{x}{25}$. Using the angle addition identities and noting that $\sin{\alpha}=\frac{15}{25}=\frac{3}{5}$ and $\cos{\alpha}=\frac{4}{5}$, we have $x=25\sin{3\alpha}=\frac{117}{5}$. Using the Pythagorean Theorem, we find that $QC=\sqrt{25^2-\left(\frac{117}{5}\right)^2}=\frac{44}{5}$, so the perimeter of $ABCD$ is $2\left(20+15+\frac{117+44}{5}\right)=\frac{672}{5}$.[/hide]

[hide="Problem 22"]
Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$, where $x$ is measured in degrees and $100< x< 200$.

(AIME 2006)
[/hide]

First time Latexing! [hide]
On the right hand side, we use the identity \[2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta):\]
\[8\cos^34x\cos^3x=8(\cos4x\cos x)^3=8(\frac{\cos(4x+x)+\cos(4x-x)}{2})^3\]
\[=(\cos5x+\cos3x)^3=\cos^3 5x+\cos^3 3x + 3\cos5x\cos3x(\cos5x+\cos3x)\]

We can cancel out the cubed terms from both sides to arrive at:

\[3\cos5x\cos3x(\cos5x+\cos3x)=0\]

We try $5x = 90+360k, 5x=180+360k, 3x=90+360k, 3x=180+360k$ and take all the x which are between $100$ and $200$: $x=126, 162, 198, 150$  All other x are solutions of the equation $\cos5x=-\cos3x$.  We use the identity $\cos\theta=-\cos(180-\theta)$ to get \[8x=180+360k\] and once again take all x between $100$ and $200$: $x=\frac{225}{2}$ and $x=\frac{315}{2}$.
\[126+162+198+150+\frac{225}{2}+\frac{315}{2}=906\] [/hide]

New Problem: In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?  AIME 1985

[hide="Solution 23"]

Say the chord with length two and minor arc measure $\alpha$ is AB. Let CD be the one with length 3. And let EF be the one with 4. Let the midpoints of these chords respectively be X, Y, Z. Also, let the circle have center O and radius $r$.

By using our basic trig identities, we know the following things:

$\sin \tfrac {\alpha}{2} = \frac {1}{r}$.

$\sin \tfrac {\beta}{2} = \frac {3}{2r}$

$\sin  \tfrac {(\alpha + \beta)}{2} = \frac {2}{r}$.

Now, we can find $\cos \tfrac {\alpha}{2} = \sqrt{\frac {r^2 - 1}{r^2}}$ (*) and $\cos \tfrac {\beta}{2} = \sqrt {\frac {4r^2 - 9}{4r^2}}$ by using the fact that $cos^2 \theta = 1 - \sin^2 \theta$ for any $\theta$. We also eliminate the negative roots because the problem explicitly tells us $\alpha + \beta < \pi$, implying the $\tfrac {\alpha}{2}$ and $\tfrac {\beta}{2}$ must be in the first quadrant of the unit circle.

Now, we know that $\sin \tfrac {(\alpha + \beta)}{2} = \sin (\tfrac {a}{2} + \tfrac {\beta}{2}) = \sin \tfrac {\alpha}{2} \cos \tfrac {\beta}{2} + \cos \tfrac {\alpha}{2} \sin \tfrac {\beta}{2}$. Alternatively, plugging in what we found in the paragraph above,

$(\frac {1}{r})(\sqrt {\frac {4r^2 - 9}{4r^2}}) + (\sqrt{\frac {r^2 - 1}{r^2}})(\frac {3}{2r}) = \frac {2}{r}$,

which ultimately yields $\sqrt{4r^2 - 9} + 3\sqrt{r^2 - 1} = 4r$. Solving, we obtain that $r = \tfrac {8\sqrt{15}}{15}$.

Now, we want to find $\cos \alpha$. We know that:

$\cos \tfrac {\alpha}{2} = \sqrt {\frac {1 - \cos\alpha}{2}} = \frac {7}{8}$, after plugging in our value for $r$ in (*). Thus, solving, we get: $\cos \alpha = \frac {17}{32}$.

Thus, our answer is $\boxed {49}$.

[/hide]

[quote="Problem 24"]

Let $ a\equal{}\pi/2008$. Find the smallest positive integer $ n$ such that
\[ 2[\cos(a)\sin(a)\plus{}\cos(4a)\sin(2a)\plus{}\cos(9a)\sin(3a)\plus{}\cdots\plus{}\cos(n^2a)\sin(na)]\] is an integer

(2008 AIME II, #8)

[/quote]

I think there's been another numbering error after the first problem 22 and before the second problem 21 :P

So, [hide="Solution to mhy123's problem "24", which is actually problem 26"]
Notice that by product-to-sum, $2\cos(k^2a)\sin(ka)=\sin(k(k+1)a)-\sin(k(k-1)a)$, so that
$2[\cos(a)\sin(a)\plus{}\cos(4a)\sin(2a)\plus{}\cos(9a)\sin(3a)\plus{}\cdots\plus{}\cos(n^2a)\sin(na)]$
telescopes to $\sin(n(n+1)a)-\sin(0)=\sin((n^2+n)a)$. We then need to find the smallest positive integer $n$ such that $sin((n^2+n)\pi/2008)$ is an integer, so $(n^2+n)\pi/2008$ is a multiple of $\frac{\pi}{2}$, so $n^2+n=n(n+1)$ is a multiple of $1004$. The prime factorization of $1004$ is $2^2\cdot251$, so we see by checking near multiples of 251 that $\boxed{251}$ is the smallest $n$ possible, and thus our answer.
[/hide]

[quote="Problem 27, borrowing some of mhy123's LaTeX code, typed, or, more idiomatically,"]
[b]Problem 27[/b]

Find the smallest positive $\alpha$ (in degrees) for which all the numbers
\[ \cos\alpha, \cos2\alpha, \dots, \cos2^n\alpha, \dots\]are negative.
(PUMaC 2009)

[/quote]

[hide="Solution 28"]
There are only two good quadrants for $2^n\alpha$: Quadrant II and Quadrant III.  Clearly the smallest possible $\alpha$ is in Quadrant II: $90\le\alpha\le180$.  Then $180\le2\alpha\le360$.  Thus the second value, $2\alpha$ must land in Quadrant II.  (It cannot go all the way around to Quadrant I.)  [\hide] Similarly, the third value, $4\alpha$ must land in Quadrant I.  Thus each time the value doubles, the angle jumps to whichever good quadrant it is not in.

Seeing this pattern we think to try $\alpha=120$ and see that $2\alpha=240, 4\alpha=480=120$.  This repeating pattern ensures that the value will always either be 120 or 240 degrees, which both yield negative cosines.  So we just have to make sure there are not any lower $\alpha$ that are valid.  Let such an $\alpha$ be equal to $120-x$, where $0\le x\le30$.  Then $4\alpha=120-4x, 16\alpha=120-16x, 32\alpha=120-32x$ and so on.  If $30\le4^k x\le210$ for any k, the $\alpha=120-4^k x$ is invalid.  It's easy to verify that no such $\alpha$ work, so $\alpha=120$ is the smallest valid angle.
 [/hide]

[quote="Problem 28"]
What is the minimum number of arbitrary real numbers you must have in order to guarantee the following condition:  It is possible to choose some two numbers $a$ and $b$ such that $0\le\frac{a-b}{1+ab}\le\sqrt3$
[/quote]

This is coming a bit late, but I have not been able to use a computer in days =(.  Anyway,I have an alternate solution for Problem 16

Notice that $<APB = <APC = <BPC = 120$.  Using this fact we will apply it on $\triangle APB$.  Let $PC = x$.
$\Rightarrow (AB)^{2} = 136 -120\cos{120} = 196 \rightarrow AB = 14$

$\Rightarrow (AC)^{2} = 100 + x^{2} -20x\cos{120} = 100+ x^{2} +10x$
$\Rightarrow (BC)^{2} = 36 + x^{2} -12x\cos{120} = 36 + x^{2} + 6x$, subtract the two equations.
$\Rightarrow (AC)^{2} - (BC)^{2} = 64 + 4x$

Observe that $(AC)^{2} - (BC)^{2} = (AB)^{2} = 196$
Thus, $196 = 64 + 4x \rightarrow 132 = 4x \rightarrow x = 33 = PC$.

By the way, where have you all learned to do AIME level trig?  AOPS intermediate precalculus?

Nice alternate solution :)
[quote="AMC-AIME-USAMO-IMO"]By the way, where have you all learned to do AIME level trig?  AOPS intermediate precalculus?[/quote]
Well, yeah, the AoPS Precalc book helped significantly (didn't take the course), but practice doing lots of trig problems (or geometry problems that shouldn't've required trig, but I got stuck on and trig-bashed through :/ ) was probably (once I knew all the information) the biggest factor helping with increasing trig skill.
I might as well do the pending problem now that I've posted anyway :lol:
[quote="Problem 28"]
What is the minimum number of arbitrary real numbers you must have in order to guarantee the following condition:  It is possible to choose some two numbers $a$ and $b$ such that $0\le\frac{a-b}{1+ab}\le\sqrt3$
[/quote]
[hide="Solution 28"]
Since the tangent function is surjective in reals, for any $a, b$ we can write $a=\tan\alpha$, $b=\tan\beta$ for some $\alpha, \beta$. So we need $0\le\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\le\sqrt3$, or $0\le\tan(\alpha-\beta)\le\sqrt3$, or $\alpha-\beta \in [0, \frac{\pi}{3}] \pmod{\pi}$.
This is clearly satisfied unless we have a set where the positive difference between the mod $\pi$ residues of the arctangents of any two elements of the set is greater than $\frac{\pi}{3}$, which is, by the pigeonhole principle, impossible if we have 3 or more elements, as it being possible would mean more than $\pi$ radians are spanned by their arctangents $\pmod{\pi}$.
So we have shown that $3$ real numbers would satisfy the condition. To show that this is the minimum, we note that the expression can't be defined with less than $2$ real numbers, the number of real numbers must be a whole number, and we can pick a set of 2 real numbers (42, 0) to show that $2$ doesn't work. So our answer is $\boxed{3}$.[/hide]

[quote="Problem 29, not being able to think of any good puns involving writing things, simply"]
Find the maximal positive integer $n$, so that for any real number $x$ we have $\sin^n x+\cos^n x \ge \frac{1}{n}.$

(PUMaC 2009 again)[/quote]

Just a warning: My typesetting skills are wanting. :)  Could someone explain how to make it so that the math stuff in dollar signs isn't dropped a level below the regular text (at the end of my solution)?
[hide="Solution 29"]

Clearly $n=1$ and $n=2$ are valid.  We want to find the angle $\alpha$ which minimizes the quantity $\cos^n \alpha+\sin^n\alpha$.  Put the angle $\alpha$ in a right triangle with legs $x$ and $y$ and hypotenuse $r$.  The Power Mean Equality ensures that if $n\ge3$ then \[\sqrt[n]{\frac{x^n+y^n}{2}}\ge\sqrt\frac{x^2+y^2}{2} \Rightarrow\left(\frac{x^n+y^n}{2}\right)^2\ge\left(\frac{x^2+y^2}{2}\right)^n\Rightarrow\]
\[2^{n-2}(x^n+y^n)^2\ge(x^2+y^2)^n     \Rightarrow    \sqrt2^{n-2}(x^n+y^n)\ge\sqrt{x^2+y^2}^n\]
\[\frac{x^n+y^n}{\sqrt{x^2+y^2}^{n}}\ge\left(\frac{1}{\sqrt2}\right)^{n-2}\]
\[\frac{x^n+y^n}{\sqrt {x^2+y^2}^n}=\frac{x^n+y^n}{r^n}=\cos^n\alpha+\sin^n\alpha\ge\left(\dfrac{1}{\sqrt2}\right)^{n-2} \]

Note that this worst case can be obtained when $\alpha=\frac{\pi}{4}$ which implies $x=y$ and $r=x\sqrt2$:  \[\cos^n \frac{\pi}{4}+ \sin^n \frac{\pi}{4}= \frac{2x^n}{(x\sqrt2)^n}=\left(\dfrac{1}{\sqrt2}\right)^{n-2}\]

So we have found that if $n\ge3$ then the minimum value of $\cos^n\alpha+\sin^n\alpha$ occurs when $\alpha=\frac{\pi}{4}$ and this minimum value is $\left(\frac{1}{\sqrt2}\right)^{n-2}$

To find the maximum valid $n$ we set \[\left(\frac{1}{\sqrt2}\right)^{n-2}=\frac{1}{n}\]
Solving for $n$ yields $n=8$ and moreover, if $n\ge8$ then $\left(\dfrac{1}{\sqrt2}\right)^{n-2}\le\frac{1}{n}$ so the solution is $\boxed8$.
[/hide]

[quote="Problem 30"]
Find the exact value of the infinite series $\sum_{k=0}^{\infty} \mbox{Cot}^{-1} ( k^2 + k + 1 )$.

[/quote]

[quote="Problem 30"]
Find the exact value of the infinite series $\sum_{k=0}^{\infty} \mbox{Cot}^{-1} ( k^2 + k + 1 )$.[/quote]

[hide="solution#30"]$\sum_{k=0}^{\infty} \mbox{Cot}^{-1} ( k^2 + k + 1 )=\sum_{k=0}^{\infty} tan^{-1}\frac{1}{k^2+k+1}=\sum_{k=0}^{\infty} tan^{-1}\frac{(k+1)-k}{1+k(k+1)}=\sum_{k=0}^{\infty} tan^{-1}(k+1)-tan^{-1}k=\frac{\pi}{2}$[/hide]

[b]problem#31[/b] Find the least positive $x$ for which $\tan (x^{\circ}+100^{\circ})=\tan (x^{\circ}+50^{\circ})\cdot \tan x^{\circ}\cdot \tan (x^{\circ}-50^{\circ})$


Can anyone here tell me what binary [b]quote[/b] notation is? I can't find anything on google. i.e. how would you convert 29 to binary quote notation? Is there a difference between normal binary notation and binary quote notation? Thanks.

I'm a CS and econ/math major, and I've never heard of binary quote notation.

My guess is that I'd just go with the normal binary number representation... It might be a weird name for scientific notation in binary, but I doubt it. I think the last time I've seen the word "quote" written out was when I used scheme... hope that day never comes back.

Hmm, the problem is that another question I have to do asks to convert a binary quote notation number to decimal (base 10) notation. i.e. convert 1'000 to decimal notation. Thanks for the help though.

WOW. you would never believe what I discovered. I open up google, type "binary quote notation", and I get a link to what I believe must've been your assignment... So i delete the last part of the URL to get to what must be some sort of home page, with links to your assignments, and other information. I then see something that says "rational number representation and arithmetic". It just so happens that when I click on that hyperlink, I am brought to a nice file that somewhere in there describes quote notation.

And if somehow you can't find it, this is the link. http://www.cs.toronto.edu/~hehner/ratno.pdf

Thanks. Haven't checked the site lately lol.

Have you actually read the paper? I find it quite amusing - a very nice idea, really.


function
real analysis
calculus
integration
real analysis theorems

If f is a real valued function integrable on [a,b] and f(x) = g(x) almost everywhere on [a,b], then g can fail to be Riemann integrable on [a,b], even if g is defined and bounded on [a,b].

ie. give an example of such g and f

$ f(x) \equal{} 0$ and $ g(x)$ is the characteristic function of the rationals.

I was typing the same example when I looked up and saw that t0rajir0u had already posted. A dozen others would have posted the same example had they gotten there first.

I believe I learned it from one of your posts  :)

A bit harder challenge: try to find a bounded Lebesgue integrable function on $ [0,1]$ that cannot be made Riemann integrable by changing its values on a set of $ 0$ measure ;)

$ f\equal{}\pm 1$ on each component of the complement of a thick Cantor set, with $ \pm$ chosen at random.  :)

... or just the characteristic function; that's bad enough already, since the lower sums for the Riemann integral are all zero and can't be increased by modifying on a measure zero set.


If $f(x) = ax^4-bx^2+x+5$ and $f(-3) = 2$, then $f(3) =$

$\textbf{(A)}\ -5 \qquad
\textbf{(B)}\ -2 \qquad
\textbf{(C)}\ 1 \qquad
\textbf{(D)}\ 3 \qquad
\textbf{(E)}\ 8$

[hide]The first, second and fourth terms are unchange. The third term changes from -3 to 3 so a total of 6 is added, giving 2+6=8[/hide]


inequalities
algebra
polynomial
analytic geometry
symmetry

if $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}1$ an $ a\plus{}b\plus{}c\equal{}1$ prove that $ |abc|\le \frac{4}{27}$

My own proof consist in using AM-GM but I am not sure it is right :

We have $ \frac {ab \plus{} ac \plus{} bc}{3}\ge (abc)^{\frac {2}{3}}$ and since (ab+ac+bc=0) then $ (abc)^{\frac{2}{3}}\le 0$

I feel like polynomials right now so...

[hide]$ ab \plus{} bc \plus{} ca \equal{} \frac {1}{2} \left( (a \plus{} b \plus{} c)^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2) \right) \equal{} 0$.

Thus, $ a,b,c$ are the roots of the cubic polynomial $ x^3 \minus{} x^2 \minus{} abc$.

We need to find the values of $ abc$ such that this polynomial will have three real roots. At the extremes of $ abc$, the cubic will be tangent to the x-axis at some point and thus have a double-root there.

Note that $ x^3 \minus{} x^2$ has a double-root at $ x \equal{} 0$, so $ abc \equal{} 0$ is one extreme.

To find the other extreme, we note that the x-coordinate of the point of rotational symmetry of this cubic is $ \minus{} \frac {x^2 \text{ coord}}{x^3 \text{ coord}} \equal{} \frac {1}{3}$, so the other extrema occurs at $ x \equal{} \frac {2}{3}$, which gives us $ abc \equal{} \minus{} \frac {4}{27}$.

Then, $ \minus{} \frac{4}{27} \leq abc \leq 0 \implies |abc| \leq \frac{4}{27}$[/hide]

[quote="Elio (n)"]My own proof consist in using AM-GM but I am not sure it is right :

We have $ \frac {ab \plus{} ac \plus{} bc}{3}\ge (abc)^{\frac {2}{3}}$ and since (ab+ac+bc=0) then $ (abc)^{\frac {2}{3}}\le 0$[/quote]
It's wrong since you don't have a,b,c>0. :wink:

[hide="Here is my solution:"]If one of these 3 numbers equals to 0, assume that $ a\equal{}0$, then $ b\equal{}0,c\equal{}1$ or $ b\equal{}1,c\equal{}0.$
Consider $ a,b,c\neq 0$ then $ abc\neq 0.$
Without loss of generality, suppose that $ a\leq b\leq c$, so $ a\leq\frac{1}{3}.$
We have $ ab\plus{}bc\plus{}ca\equal{}0$, it follows that $ \minus{}a\equal{}\frac{bc}{b\plus{}c}\leq\frac{b\plus{}c}{4}\equal{}\frac{1\minus{}a}{4}$
Thus $ \frac{\minus{}1}{3}\leq a\leq\frac{1}{3}\implies |a|\leq\frac{1}{3}$
It is easy to see that b,c>0 since if b<0 or c<0 then c>1 or b>1, this is a contradiction.
We have $ a\plus{}b\plus{}c\geq b\plus{}c\minus{}\frac{1}{3}$, so $ b\plus{}c\leq\frac{4}{3}$ which implies $ bc\leq(\frac{b\plus{}c}{2})^2\leq \frac{4}{9}$
Therefore $ |abc|\equal{}|a|bc\leq \frac{4}{27}$
The equality holds if and only if $ a\equal{}\frac{\minus{}1}{3},b\equal{}c\equal{}\frac{2}{3}$.[/hide]


hey... When the 6th edition of the contest is going to begin???

There will be an annoucement about this very soon.

[quote="joml88"]When are the problems going to be posted?  The announcement said the eve of the 9th so maybe I just can't find them? :?[/quote]I ment eve[b]ning[/b] of 9th :P

They will be online in aprox. 6-8 hours.


linear algebra
matrix
group theory
superior algebra
superior algebra unsolved

They say that if want to solve the equation $ Ax = b$ for $ A \in \mathcal{M}^{l \times l}(\mathbb{Z}_n)$ invertible matrices, I can reduce my work to the cases $ \mathbb{Z}_{p^r}$.

How do I show prove this properly? I have some vague ideas, but can't write it down correctly:

$ \mathbb{Z}_n \cong \mathbb{Z}{}_{p{}_1^r_1} \oplus ... \oplus \mathbb{Z}{}_{p{}_j^r_j}$, and then

$ \mathcal{M}^{l \times l}(\mathbb{Z}_n) \cong \mathcal{M}^{l \times l}(\mathbb{Z}{}_{p{}_1^r_1}) \oplus ... \oplus \mathcal{M}^{l \times l}(\mathbb{Z}{}_{p{}_j^r_j})$,

so this is a group isomorphism (multiplication of matrices as operation).

Then I wanted to use group actions. Let the left group act on an element $ x \in \mathbb{Z}_n$ such that it is send it to $ b$. There must also be an element on the right hand side that acts on the same $ x$ and sends it to $ b$. And then somehow because I should conclude that finding a solution of the left thing reduces to finding solutions for all right things

Can someone do this more rigorous? :)

This feels like too much machinery for what this really is: the Chinese Remainder Theorem.

Your matrix equation is some big set of congruences $ (mod\ n)$, and finding elements of $ \mathbb{Z}_n$ that satisfy them is equivalent to finding, for any $ p^k$ dividing $ n$, elements of $ \mathbb{Z}_{p^k}$ that satisfy the same congruences $ (mod\ p^k)$.  This statement is almost exactly CRT.


From the sum ,

$1+1/2+1/3+...+1/n$

we remove all terms whose have the digit 9.

a)  Is it true that, for all n, the remaining sum is inferior to 80?
b)  Is it true that, for all n, the remaining sum is inferior to 28?


calculus
integration
calculus computations

$ \int \sqrt {\frac {4 \minus{} x}{4 \plus{} x}}\ dx$

[hide=" Hint"]

Multiply above and below by $ \sqrt{4\minus{}x}$.

[/hide]

hello, setting $ t\equal{}\sqrt{\frac{4\minus{}x}{4\plus{}x}}$ then we get the integral
$ \minus{}16\int\frac{t^2}{(1\plus{}t^2)^2}\,dt\equal{}\minus{}8\arctan(t)\plus{}\frac{8t}{1\plus{}t^2}\plus{}C$
Sonnhard.


Asymptote
geometry
perpendicular bisector

I saved the code for the cse5 package in the asymptote folder, but when I ran asymptote on an asy file, the dos window came up, displayed only the pause message, closed after I hit a key, and did not launch gsview.  Also, when I tried to run a simple program in interactive mode, I got the screenshot shown below.

Could you post the code in your asy file as there could be an error in it somewhere?

No wonder :) CR command just [b]creates[/b] a circle but doesn't [b]draw[/b] it!. You had to write [b]D(CR((1,0),1));[/b] to see any picture :lol:. As it was, Asymptote just outputted the description of the circle as a cubic spline.

That explains a lot, including why CR didn't take pen as a modifier.  I assume that L also does not actually draw the line.

[quote="roadnottaken"] I assume that L also does not actually draw the line.[/quote]
Correct. All the drawing is done by [b]D[/b] and [b]DPA[/b]. The other commands just "create" objects but do not show them on the picture. This way was chosen to allow to hide some parts of the construction and to show only the result like in
[asy]import cse5; size(150); pair A=MP("A",D((0,0))), B=MP("B",D((10,0))); path P=CP(A,B),Q=CP(B,A); MP("C",D(IP(L(A,B),D(L(IP(P,Q),OP(P,Q),0)))));[/asy]
[code]
import cse5;
size(150);
pair A=MP("A",D((0,0))), B=MP("B",D((10,0)));
path P=CP(A,B),Q=CP(B,A);
MP("C",D(IP(L(A,B),D(L(IP(P,Q),OP(P,Q),0)))));
[/code]
where the standard compass-straightedge construction of the midpoint $ C$ of an interval $ AB$ is carried out but only the perpendicular bisector and the result (the midpoint itself) are shown. Circles are used in the construction but never appear on the picture.


If we're taking a class which doesn't give a "grade", persay, can we still put that we've taken it on college applications? Mainly to show that we're interested in math. An exception might be WOOT, where we don't really get a letter grade, but our practice olympiads are graded.

You COULD, but most likely, adcoms won't really care since they probably don't have anything to measure it against.  It would, however, demonstrate a passion for math.

You certainly can mention us on your college application.  I wouldn't worry so much about putting "grades" as such, but I know the folks at MIT know who we are -- we have them host a Math Jam every year (and I think the admissions department at MIT can take a signficant amount of credit for moving MIT to (arguably) the top of the list for top math students in the US).


This isn't a bug, but recently I was sent a PM with three attachments. I've never received a PM with an attachment before, so I don't know how to download them. Anyone who knows? Help would be greatly appreciated.

It works just the way it works on the posts. You either click the link, or right click and save as ...


function
complex analysis
absolute value

Hallo All,

Consider a function $f(z)=e^{-1/|z|^2}, z\ne0, f(0)=0$ defined on the whole complex plane.

It seems that the function is differentiable at any point $z$ (including the point $z=0$ !!!), and thus holomorphic (and even entire). But it is not analytic at $z=0$, contradicting the known theorem of complex analysis that any holomorphic function is analytic. Where is the mistake? :)

[quote="Myth"]Why do you think this function is differentiable???[/quote]

I am sorry. It was of course wrong... I was thinking at first about the function $\exp(-1/z^2)$ and then took the absolute value, forgetting about the Cauchy-Riemann equations... :(


If a triangle rectangle has all its sides formed by integer numbers, then it's surface is not a perfect square
2) If a x, y are Pythagorean triple then one is multiple of 5 and the other is mnultiple of 3

[hide="1"]Let $p^{2}+q^{2}= r^{2}$. Then $p = m^{2}-n^{2}$ and $q = 2mn$. The area is $S = \frac{pq}{2}= (m^{2}-n^{2})mn = (m-n)(m+n)mn = k^{2}$. If $(m, n) = 1$ then $m-n, m+n, m, n$ are all coprimes.
If $(m, n) = d$, then $m = dx$ and $n = dy$. Hence $d^{4}(a-b)(a+b)ab = k^{2}$ with $(a, b) = 1$.
So it remains to show that $m-n, m+n, m, n$ cannot be all perfect squares.[/hide]

what do you mean by triangle rectangle?
http://en.wikipedia.org/wiki/Pythagorean_triples you can find here a lot of nice Properties of Pythagorean triples. some of them I couldn't proove, help!:
 Exactly one of a, b, (a + b), (a − b) is divisible by 7
At most one of a, b is a square.
There is no Pythagorean triple in which the hypotenuse and one leg are the legs of another Pythagorean triple.


I think we should be able to edit posts as many times as we like. I'm hosting a Tournament in the MOEMS forum, and it's annoying to constantly have to create new posts to update things like scores. It wastes space and is unnecessary. Can we change this?

Please refer to the following topic:
http://www.mathlinks.ro/viewtopic.php?t=157522


algebra
polynomial
quadratics
algebra unsolved

I don't know this problem true or not . I think it it true , because it is true for some small values .
Let $ n > p$ are two  positive  integer . Prove that the polynomial $ (x_1 \plus{} ... \plus{} x_n)^p$ can represent in the form :
\[ a_1 .\sum_{i \equal{} 1}^n x_i^p \plus{} a_2.\sum_{1\leq i_1 < i_2\leq n} (x_{i_1} \plus{} x_{i_2})^p \plus{} a_3 \sum_{1\leq i_1 < i_2 < i_3\leq n }(x_{i_1} \plus{} x_{i_2} \plus{} x_{i_3})^p \plus{} .... \plus{} a_{n \minus{} 1}\sum_{1\leq i_1 < ... < i_{n \minus{} 1}\leq n}(x_{i_1} \plus{} ... \plus{} x_{i_{n \minus{} 1}})^p
\]
where $ a_i$ are all integers .
For example $ p \equal{} 2$ and $ n > 2$ then we have :
$ (\sum_{i \equal{} 1}^nx_i)^2 \equal{} \sum_{1\leq i < j\leq n }(x_i \plus{} x_j)^2 \minus{} (n \minus{} 2)\sum_{i \equal{} 1}^n x_i^2$
...

(you meant $ x_i^p$ in the first summation?)

I am pretty sure that [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=984]this problem[/url] is a generalization, and that it gives an alternate form for your example. Did you mean for the representation to have only $ p$ terms instead?


[hide]This problem just involves getting the complicated fractions into a nicer form.

If you have , the result will be . For the first term, you would get 1/2 - 1/3 = 1/6 and 1/3 - 1/4 = 1/12.  = 12/6 or .  Thus each term in the form  can be rewritten as . When this is applied to all of the terms, all but the 100 in the numerator and 2 in the denominator cancel out. 100/2 = 50.[/hide]

Confirmed with..

1->S
For(N,2,98,2)
S*(1/N - 1/(N+1))/(1/(N+1) - 1/(N+2))->S
End
Disp S

Have to keep my programming from getting rusty. And I'm already spending too much time here. :/


For positive real numbers $a,b,c,$ prove that

1.
\[\sqrt{(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)} \geq abc+\sqrt[3]{(a^3+abc)(b^3+abc)(c^3+abc)}\]

2.
\[\sqrt{a^4+b^4+c^4}+\sqrt{a^2b^2+b^2c^2+c^2a^2} \geq \sqrt{a^3b+b^3c+c^3a}+\sqrt{ab^3+bc^3+ca^3}\]

[edit] I just realized that #1 was posted here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167

Oh well...  there's still #2 :)

I gave a solution in Old and New inequalities for 2. You just square, prove that $ a^4+b^4+c^4+a^2b^2+b^2c^2+c^2a^2\geq a^3b+b^3c+c^3a+...$ and the other one. The first inequality I wrote follows from Schur inequality, wwhile the second one will follow from Cauchy.

[quote="ThAzN1"]For positive real numbers $ x,y,z,$ prove that

$ \sqrt {\left(x^2y \plus{} y^2z \plus{} z^2x\right)\left(xy^2 \plus{} yz^2 \plus{} zx^2\right)}$

$ \geq xyz \plus{} \sqrt [3]{\left(x^3 \plus{} xyz\right)\left(y^3 \plus{} xyz\right)\left(z^3 \plus{} xyz\right)}.$[/quote]Without loss of generality, we assume $ \sum_{cyc}x^2y\leq\sum_{cyc}xy^2$. For $ k\geq 0$, using the arithmetic-mean - geometric-mean inequality,

$ \sqrt {\sum_{cyc}x^2y\sum_{cyc}xy^2}\geq\sum_{cyc}x^2y \equal{} \sum_{cyc}{\frac {kx^2y \plus{} y^2z}{k \plus{} 1}}\geq\frac {3}{k \plus{} 1}\sqrt [3]{\prod_{cyc}\left(kx^2y \plus{} y^2z\right)}$

$ \equal{} \frac {3}{k \plus{} 1}\sqrt [3]{xyz\prod_{sym}\left(kx^2 \plus{} yz\right)} \equal{} \frac {3}{k \plus{} 1}\sqrt [3]{\prod_{sym}\left(kx^3 \plus{} xyz\right)}\Longrightarrow$

$ \frac {1}{3}\sqrt {\sum_{cyc}x^2y\sum_{cyc}xy^2}\geq xyz, (k \equal{} 0)$

and

$ \frac {2}{3}\sqrt {\sum_{cyc}x^2y\sum_{cyc}xy^2}\geq \sqrt [3]{\prod_{sym}\left(x^3 \plus{} xyz\right)}, (k \equal{} 1).$

By the way, the following inequality holds :

$ x^2y \plus{} y^2z \plus{} z^2x \plus{} \left(2\sqrt [3]{4} \minus{} 3\right)xyz\geq\sqrt [3]{4\left(x^3 \plus{} xyz\right)\left(y^3 \plus{} xyz\right)\left(z^3 \plus{} xyz\right)}.$


A domino consists of two squares, each of which is marked with 0,1,2,3,4,5 or 6 dots. Is it possible to arrange them all in a circle so that the adjacent havles of neighboring dominos show the same number?

yea i did that once when i was bored. i put into a pyramid i think but on both ends were 0 dots :)

i did i t like this: first i put all the dominos with the differnce of 1 like 0-1 or 5-6 and (!!) 6-0
then the the ones with diffrence of 2 ( 0-2, 6-1, ..) and so on. the ones with 2 equal numbers can be put in between anywhere it doesnt matter.
and since 7 is prime it becomes a circle with all the dominos ...
(i hope it is understandable)

peeta


geometry
3D geometry
sphere
analytic geometry
geometry unsolved

Please can anyone tell me how to distribute n points equidistantly on the surface of a sphere of unit radius in such a way the positions of the points are defined by their Cartesian coordinates relative to the origin of the sphere.
Thanks

It might be, but considering its surface area, we bash...

Bomb


algebra
polynomial
inequalities
algebra unsolved

Find all second degree polynomial d(x)=x²+ax+b with integer coefficients, so that there exists an integer coefficient polynomial p(x) and a non-zero integer coefficient polynomial q(x) that satisfy:
                p(x)² - d(x)q(x)² = 1      for all real x

This problem was posted before, (China TST 2006) but nobody solved it. Please give me some ideas to deal with this problem.

See [url=http://en.wikipedia.org/wiki/Pell%27s_equation#Connection_to_Chebyshev_polynomials]here[/url].  I think the last idea might lead to a solution.


Euler
number theory proposed
number theory

I'm posting this question along with my solution below. I would be happy if someone could confirm my solution. I know it will take some time to read so I understand that this may not be possible. If so, I would be very interested to read other solutions to this famous problem (I already know of  Arthur Engel's solution to this problem however). The problem is actually due to Leonhard Euler. I am sorry if this solution has been seen before :blush: . I am unaware if it has and regard this solution as my own at the moment.

Show that $ 2^{n}$ = $ 7x^{2}$ $ +$ $ y^{2}$ holds for all $ n>2$ where $ n$ is an integer and $ x,y$ are odd integers.

[hide]We have two cases.

[u]Case 1[/u]
$ n$= odd integer
then we can write $ 2^{n}$ $ =$ $ 7$.$ 2^{n-3}$ $ +$ $ 2^{n-3}$ which can be proved by induction.
Since $ n$ is odd $ 2^{n-3}$ is a square, so the above statement is true for all odd $ n>1$.

[u]Case 2[/u]
$ n$= even integer
Then $ 7x^{2}+y^{2}= (2^{k})^{2}$ where $ k$ is a positive integer. We can write this as:
$ 7(u^{2}-v^{2})^{2}+(3u^{2}-3v^{2})^{2}= (4u^{2}-4v^{2})^{2}$ where $ u,v$ are positive integers.
If we can show that $ 2^{k}$ can be represented in the form $ 4u^{2}-4v^{2}$ for all $ k$ as a positive integer then the solution is complete.

We have two subcases:

[u]Subcase 1[/u]
$ k$ = even integer.
We need to show that $ 2^{k}$$ =$ $ 4u^{2}-4v^{2}$ holds for all $ k$, an even integer.
Dividing by $ 4$ we get $ 2^{k-2}= u^{2}-v^{2}$
We do as follows:
Write $ 2^{k-2}$  as $ 2^{[(k-2)/2]+1}$.$ 2^{[(k-2)/2]-1}$
Let $ S=$ $ 2^{[(k-2)/2]+1}$ $ +$ $ 2^{[(k-2)/2]-1}$
Let $ u= S/2$
Let $ v =$ $ 2^{[(k-2)/2]+1}$ $ -u$.
This holds for all $ n =$ even integer.

[u]Subcase 2[/u]
$ k=$ odd integer
Observe that for $ k=3$ we have:
$ 2^{3}=(3+1)(3-1)$
For $ 2^{5}$ we multiply $ 2^{3}$ by $ 4$ and get $ (6+2)(6-2)$
Thus if we have a solution for $ 2^{k}$ we can generate a solution for $ 2^{k}+2$ provided $ k$ is an odd integer which is true by our assumption.
Hence we have our solution for all $ n>2$ where $ n$ is an integer.[/hide]

Let $ (x_{n},y_{n})$ be the odd integers such that $ 2^{n}=7x_{n}^{2}+y_{n}^{2}$ so it is easy to see that one of $ x_{n+1}=\frac{x_{n}+y_{n}}{2}$ or $ x_{n+1}=\frac{|x_{n}-y_{n}|}{2}$ is odd.If $ x_{n+1}=\frac{x_{n}+y_{n}}{2}$ then $ y_{n+1}=7x_{n}-y_{n}$,so it is solution to:
$ 2^{n+1}=7x_{n+1}^{2}+y_{n+1}^{2}$.for $ x_{n+1}=\frac{|x_{n}-y_{n}|}{2}$ $ y_{n+1}=7x_{n}+y_{n}$.
For $ n=3$ we could take $ x_{3}=1,y_{3}=1$.
I'm sure that this problem was posted before but i can't find a link for it.

Hi Erken I spent a while trying to find a link yesterday but couldn't find it. Maybe a moderator might know where it is? Thanks for posting Engel's solution :P I'm really interested to see any other solutions to the problem and whether my approach was completely correct.

[quote="Erken"]Let $ (x_{n},y_{n})$ be the odd integers such that $ 2^{n}\equal{} 7x_{n}^{2}\plus{}y_{n}^{2}$ so it is easy to see that one of $ x_{n\plus{}1}\equal{}\frac{x_{n}\plus{}y_{n}}{2}$ or $ x_{n\plus{}1}\equal{}\frac{|x_{n}\minus{}y_{n}|}{2}$ is odd.If $ x_{n\plus{}1}\equal{}\frac{x_{n}\plus{}y_{n}}{2}$ then $ y_{n\plus{}1}\equal{} 7x_{n}\minus{}y_{n}$,so it is solution to:
$ 2^{n\plus{}1}\equal{} 7x_{n\plus{}1}^{2}\plus{}y_{n\plus{}1}^{2}$.for $ x_{n\plus{}1}\equal{}\frac{|x_{n}\minus{}y_{n}|}{2}$ $ y_{n\plus{}1}\equal{} 7x_{n}\plus{}y_{n}$.
For $ n \equal{} 3$ we could take $ x_{3}\equal{} 1,y_{3}\equal{} 1$.
I'm sure that this problem was posted before but i can't find a link for it.[/quote]If $ x_{n}$ and $ y_{n}$ are odd, then $ y_{n\plus{}1}\equal{} 7x_{n}\minus{}y_{n}$ is no longer odd? And the calculations don't work either...

[quote="Peter"][quote="Erken"]Let $ (x_{n},y_{n})$ be the odd integers such that $ 2^{n}\equal{} 7x_{n}^{2}\plus{}y_{n}^{2}$ so it is easy to see that one of $ x_{n\plus{}1}\equal{}\frac{x_{n}\plus{}y_{n}}{2}$ or $ x_{n\plus{}1}\equal{}\frac{|x_{n}\minus{}y_{n}|}{2}$ is odd.If $ x_{n\plus{}1}\equal{}\frac{x_{n}\plus{}y_{n}}{2}$ then $ y_{n\plus{}1}\equal{} 7x_{n}\minus{}y_{n}$,so it is solution to:
$ 2^{n\plus{}1}\equal{} 7x_{n\plus{}1}^{2}\plus{}y_{n\plus{}1}^{2}$.for $ x_{n\plus{}1}\equal{}\frac{|x_{n}\minus{}y_{n}|}{2}$ $ y_{n\plus{}1}\equal{} 7x_{n}\plus{}y_{n}$.
For $ n \equal{} 3$ we could take $ x_{3}\equal{} 1,y_{3}\equal{} 1$.
I'm sure that this problem was posted before but i can't find a link for it.[/quote]If $ x_{n}$ and $ y_{n}$ are odd, then $ y_{n\plus{}1}\equal{} 7x_{n}\minus{}y_{n}$ is no longer odd?[/quote]
Yes,you i right.I mean $ y_{n\plus{}1}\equal{}\frac{7x_{n}\minus{}y_{n}}{2}$,for $ x_{n\plus{}1}\equal{}\frac{x_{n}\plus{}y_{n}}{2}$ and $ y_{n\plus{}1}\equal{}\frac{7x_{n}\plus{}y_{n}}{2}$,for $ x_{n\plus{}1}\equal{}\frac{x_{n}\minus{}y_{n}}{2}$.

Sorry for reviving this old thread, but I do not want to start a new thread. I'm just curious, is there any solution without processing inductively like this?


Does anyone know where I can find solutions to the exercises in Axler's [u]Linear Algebra Done Right[/u]? The exercises are getting tough, and I'd like to be able to look up or check solutions to them.

Thanks.


Find all composite integers M such that the product 2005 x M has exactly 8 divisors.

$2005=5*401$
so this means that $2005$ has $4$ divisors.
if you want to dubble that amount, you have to multiply with any prime, but since we have to use composite integers, you should multiply with $5^2$ or $401^2$ to get to eight divisors...
$M=25$ or $160801$


probability
abstract algebra
AMC
AIME
AoPS Books

A bag contains 10 red marbles and 6 blue marbles. Three marbles are selected at random and without replacement. What is the probability that one marble is red and two are blue? Express your answer as a common fraction.

In the solution, it's worth pointing out that the three separate fraction products are really the same, so you just need to compute one and multiply by 3.  That's a lot less work (though perhaps only obvious to someone who has done a lot of these...)

the probability that we get 1 red and 2 blues is...

$ \frac{10}{16}\times\frac{6}{15}\times\frac{5}{14}$

simplify

$ \frac{5}{4}\times\frac{1}{1}\times\frac{1}{14}$

multiply

$ \frac{4}{56}$

multiply by 3 since we can pick the one red and 2 blues in any of the following ways: RBB,BRB,BBR

$ \frac{4}{56}\times3\equal{}\frac{12}{56}$

our answer is $ \boxed{\frac{12}{56}}$

[quote="vallon22"]the probability that we get 1 red and 2 blues is...

$ \frac {10}{16}\times\frac {6}{15}\times\frac {5}{14}$

simplify

$ \frac {5}{4}\times\frac {1}{1}\times\frac {1}{14}$

multiply

$ \frac {4}{56}$

multiply by 3 since we can pick the one red and 2 blues in any of the following ways: RBB,BRB,BBR

$ \frac {4}{56}\times3 \equal{} \frac {12}{56}$

our answer is $ \boxed{\frac {12}{56}}$[/quote]

~which can be simplified to $ \boxed{\frac3{14}}$

5/4 times 1/14 = 5/56 (not 4/56), so the final answer would be 3 times that, or 15/56.

[quote="lowsf"]There is also a solution to this problem using combinations.  Does anyone know how AoPS chooses what solution to display?[/quote]

There's no real reason to pick one method over another for demonstration, especially when writing up solutions.  The reason that AoPS books and modules are called "Counting and Probability" is that they really are intertwined.  Many problems can be solved using a probability lens or a combinatorics lens.

However, with more difficult problems, particularly AIME level and above, a combinatoric approach is sometimes faster, or is sometimes the only logical way to go.  It's always a good idea to have more than one way to solve a problem, because in future problems sometimes only one method will apply.  So, here's my stab at it:

[hide="Combinatoric solution"]There are $\binom{16}{3} = 560$ ways to pick three marbles from sixteen.

There are $\binom{6}{2} = 15$ ways to pick two blue marbles from six blue marbles.

There are $\binom{10}{1} = 10$ ways to pick one red marble from ten red marbles.

There are thus $15\cdot 10 = 150$ ways to pick a set of one red and two blue marbles - or three "good" marbles out of sixteen.

Thus, the overall probability is $\frac{150}{560} = \boxed{\frac{15}{56}}$ [/hide]

[hide=Quick easy combinatoric solution]
There are $\binom{10}{1}$ ways to choose one red marble and $\binom{6}{2}$ ways to choose 6 blue marbles, giving $\binom{10}{1}\cdot\binom{6}{2}=$ $10\cdot \frac{6\cdot 5}{2} =$ $10\cdot 15=$ $150$ ways to choose the three marbles satisfying the requirements.

There are $\binom{16}{3}=$ $\frac{16\cdot 15\cdot 14}{3\cdot 2}=$ $8\cdot 5\cdot 14 =$ $40\cdot 14=$ $80\cdot 7=$ $560$ ways to choose the marbles in total.

This gives $\frac{150}{560}=\boxed{\frac{15}{56}}$ as our probability.

The combinations solution seems quite painless and much easier to follow.
I found the simple version of solution one on my own, multiplying the 10/16 * 6/15 * 5/14 and then multiplying by 3. It much quicker. If I hadn't followed that I would have finished around the time that I have finished writing this!

[hide=Everyone is slow]
Ans is just $\frac{\binom{10}{1}*\binom{6}{2}}{\binom{16}{3}}=\boxed{\frac{15}{56}}$
[/hide]


[hide="Solution"]I'm pretty sure it's A.  :P [/hide]

[hide]This is (2000^1)(2000^2000). When you multiply two numbers with the same base you just add the exponents so A)2000^2001[/hide]

A
The message is too small. Please make the message longer before submitting.

By the first law of exponents, the answer is obviously A.

[quote="fireemblem13"]do we really need everyone to repeat the same thing? its just not necessary.
BTW, the answer is A. :D[/quote]
jaja lol

[quote="fireemblem13"]do we really need everyone to repeat the same thing? its just not necessary.
BTW, the answer is A. :D[/quote]

If people want to, then they can do it.  They're not forced to  :)

[hide]an additional 2000 to 2000*2000*2000*... would equal 2000^2001, meaning the answer is A[/hide]

[hide]like-base exponential multiplication--$x^m*x^n=x^{m+n}$
so it is $A$ because $2000^1*2000^{2000}=2000^{2000+1}$

[quote="ryan chan"]B. is the an :blush[/quote]
the only way to get B is by like-exponent multiplication--that is, if you had something like $2000^{2000}*2^{2000}$, etc.[/hide]


The sequence $L_1,\ L_2,\ L_3,\ \dots$ is defined by \[ L_1 = 1,\ \ L_2 = 3,\ \ L_n = L_{n - 1} + L_{n - 2}\textrm{ for }n > 2. \] Prove that $L_p - 1$ is divisible by $p$ if $p$ is prime.

By a standard method it is straightforward to obtain:

$L_n=\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\left(\dfrac{1-\sqrt{5}}{2}\right)^n$

Considering now a field $\mathbb{F}_p \cup \{\sqrt5\}$ by Fermat's Little Theorem we get:

$L_p-1 = \left(\dfrac{1+\sqrt{5}}{2}\right)^p+\left(\dfrac{1-\sqrt{5}}{2}\right)^p -1 = \left(\dfrac{1^p+\sqrt{5}^p}{2^p}\right)+\left(\dfrac{1^p-\sqrt{5}^p}{2^p}\right) -1 = \dfrac{2}{2^p}-1=\dfrac{2}{2}-1= 1 - 1 = 0$

Hence $L_p-1$ is divisible by $p$ if $p$ is a prime

Just for clarification:
You don't have $(\dfrac{1+\sqrt{5}}{2})^p = \dfrac{1+\sqrt{5}}{2}$ but $(\dfrac{1+\sqrt{5}}{2})^p = \dfrac{1^p+\sqrt{5}^p}{2^p}$ which is indeed not the same (because this would give us that $x^p=x$ has more than $p$ solutions). But the rest works as given.

[quote="ZetaX"] $(\dfrac{1+\sqrt{5}}{2})^p = \dfrac{1^p+\sqrt{5}^p}{2^p}$ which is indeed not the same ....[/quote]

But isn't $\sqrt{5}^p=\sqrt{5}$ and $\left(\dfrac{1}{2}\right)^p=\dfrac{1}{2}$ so $(\dfrac{1+\sqrt{5}}{2})^p = \dfrac{1^p+\sqrt{5}^p}{2^p}=(\dfrac{1+\sqrt{5}}{2})$ ??

 :?

Take e.g. $p=3$, then $\sqrt 5 ^{3} = 5 \sqrt 5 = - \sqrt 5 \neq \sqrt 5$ (because otherwise $-1=1$).

Thanks, I've edited my post - now it should be ok.


$ u_1,\ldots,u_m \in{}R^n$ which is linearly indepedent (so $ m\leq{}n$)
$ u \equal{} \sum^m_{i \equal{} 1}\alpha_iu_i$, $ \alpha_i\in{}R$
if $ [u,a] \leq{}max_{1\leq{}i\leq{}m}[u_i,a]$  $ \forall{}a\in{}R^n$
(where $ [u,a]$ is the inner product in $ R^n$)

proof
$ \alpha_i\geq{}0$ and $ \sum^m_{i \equal{} 1}\alpha_i \equal{} 1$

After the modification, it's easy to prove this  :)

Try to think a bit about how you state your problems. As written, this one is just false: let $ u\equal{}u_1\equal{}\dots\equal{}u_m\equal{}0$ and let each $ \alpha_j\equal{}\minus{}100$. ;) (It is not hard to guess what was meant though :) ).

Nope, that's not enough: $ u\equal{}u_1\equal{}u_2\ne 0$, $ \alpha_1\equal{}100$, $ \alpha_2\equal{}\minus{}99$ is hardly any better. ;)

:o

this is a problem i saw on a book
the constraint about $ u_i$ is not shown

now, I think  $ u_i$  should be linearly independent in $ R^n$,  (so $ m\leq{}n$)
then the claim is true

and, so,  i can prove this.

Yes, that's one way to fix the problem. The other way is to state it like the following:

If $ u,u_j\in\mathbb R^n$ and $ \langle u,a\rangle\le\max_j\langle u_j,a\rangle$ for all $ a\in\mathbb R^n$, then [color=red]there exist[/color] $ \alpha_j\ge 0$ with $ \sum_j \alpha_j=1$ such that $ u=\sum_j\alpha_j u_j$. :)

Once it is stated correctly (either way), it becomes just the standard separation lemma for a point and a convex set in $ \mathbb R^n$.


trigonometry
geometry
trig identities
Law of Cosines

1) From a barge moving with constant speed along a straigt canal, the angle of the elevation of a bridge is 10 degrees. After 10 minutes, the angle is 15 degrees. How much longer will it be before the barge reaches the bridge, to the nearest second?

2) In the triangle ABC the median through A meets BC at M. Given that AB = 8, AM = 5, and AC = 6, what is the length of BC and the measurement of angle ABC?

[hide="2a"]
I set CM=MB=x
Using the Law of Cosines, we get:
$\cos{B}=\frac{8^2+x^2-5^2}{2*8*x}$
$\cos{B}=\frac{8^2+(2x)^2-6^2}{2*8*2x}$
Set the two equal, and you get:
$\frac{64+x^2-25}{16x}=\frac{64+4x^2-36}{32x}$
$39+x^2=\frac{28+4x^2}{2}$
$39+x^2=14+2x^2$
$25=x^2$
Therefore, $x=5$ and $BC=10$[/hide]

[hide="2b"]
Using the Law of Cosines and the equation from the first part:
$\cos{B}=\frac{8^2+5^2-5^2}{2*8*5}$
$\cos{B}=\frac{64}{80}$
$\cos{B}=\frac{4}{5}$
And using a calculator:
$\angle B=36.9 \text {degrees}$
I hope that's right. Probably isn't. :) [/hide]

[quote="DanK"][hide="1"]29 minutes, 15 seconds.

[hide="method"]Let's call the height of the bridge above the canal $h$, the initial distance $d_1$, and the distance after 10 minutes $d_2$.  Then $\tan{10^\circ}=\frac{h}{d_1}$ and $\tan{15^\circ}=\frac{h}{d_2}$.  So $h=\tan{10^\circ} \times d_1=\tan{15^\circ} \times d_2$.  This means that $d_2=\frac{\tan{10^\circ}}{\tan{15^\circ}} \times d_1$, and the barge traveled $1-\frac{\tan{10^\circ}}{\tan{15^\circ}}$ of the way in 10 minutes, and will need to travel $\displaystyle\frac{1}{1-\frac{\tan{10^\circ}}{\tan{15^\circ}}} \times 10$ minutes to reach the bridge.  From there, you just need a calculator...[/hide][/hide][/quote]

I think you're forgetting to subtract the 10 minutes, because I keep getting [hide]19 minutes, 15 seconds
[/hide]

[hide="Solution to 1"]
Make a diagram with two right triangles ABC and DBC, with A as initial position, B as base of bridge, C as top of bridge, and D as position after 10 minutes. $CB=h$. $BD=x$. $DA=10$. $\angle CAB=10$ $\angle CDB=15$ Then solve like a geometry problem.
$\tan{10}=\frac{h}{x+10}\Longrightarrow h=(x+10)\tan{10}$
$\tan{15}=\frac{h}{x}\Longrightarrow h=x\tan{15}$
$(x+10)\tan{10}=x\tan{15}$
$\frac{x+10}{x}=\frac{tan{15}}{tan{10}}$
Solving for x, we get: $\frac{10\tan{10}}{\tan{15}-\tan{10}}=19.245=$
19 minutes, 15 seconds
Not very elegant, I know. :D [/hide]

[hide]

Here's my solution to it:

Make a diagram with triangle ABD and right triangle BDC. Let A denote the starting distance, and D, the distance after 10 minutes. Then, let BC represent the height, h, from the bridge. Using trigonomery, we have

DC = h \cot{15}, $AC = h cot 10$, and AD = $h(cot 10 - cot 15)$.

Then, we simply set up a proportion like this:

$(h cot 15)/(h(cot 10 - cot 15)) = t/10$

the h's cancel, and cross multiply to get

$t = 19.24500314$, which is

19 minutes and 15 seconds

[/hide]

[hide]

Here's my solution to it:

Make a diagram with triangle ABD and right triangle BDC. Let A denote the starting distance, and D, the distance after 10 minutes. Then, let BC represent the height, h, from the bridge. Using trigonomery, we have

$DC = h cot 15$, $AC = h cot 10$, and AD = $h(cot 10 - cot 15)$.

Then, we simply set up a proportion like this:

$(h cot 15)/(h(cot 10 - cot 15)) = t/10$

the h's cancel, and cross multiply to get

$t = 19.24500314$, which is

19 minutes and 15 seconds

[/hide]

[hide]You're right, my solution was the time from the start, when you subtract the 10 minutes, it's 19 minutes, 15 seconds, like everyone got...[/hide]


inequalities
trigonometry
inequalities proposed

Let $P$ is a an arbitrary point inside triangle $ABC$ with distance from $P$ to the side are $d_{a},d_{b},d_{c}$ prove that
$(\sqrt{PA}d_{a}+\sqrt{PB}d_{b}+\sqrt{PC}dc)^{2}\le\frac{9}{4}PA\cdot PB\cdot PC$

To avoid subscripts we change the notation a liitle
Let O be a point inside triangle ABC,and let OP , OQ and OR be the perpendiculars from O to sides  BC, CA , AB respectively.
We therefore need to  show $(\sqrt{OA}.OP+\sqrt{OB.}OQ+\sqrt{OC}.OR)^{2}$ $\leq$ ${\frac{9}{4}}.{(OA.OB.OC)}$.........(1)

$AROQ ,BPOR$ and $CPOQ$ are cyclic quadrilaterals.

Applying the cosine law to triangle $QOR$ we get

$QR^{2}$  =$OR^{2}+OQ^{2}+2.OR.OQ.cosA$
=$OR^{2}+OQ^{2}-2.OR.OQ.cos(B+C)$
=$(ORcosB-OQcosC)^{2}$ +$(ORsinB+OQsinC)^{2}$
$\geq$ $(ORsinB+OQsinC)^{2}$

Therefore $QR$ $\geq$  $ORsinB+OQsinC$
and similarly
$PR$ $\geq$  $OPsinC+ORsinA$
$PQ$ $\geq$  $OQ sinA+OPsinB$

Also from the sine law we have $QR =OA.sinA$, $PR=OB.sinB$ ,$PQ=OC.sinC$

It follows that for any real positive numbers r ,s and  t we have
$r.OA+s.OB+t.OC$ $\geq$ $r.({\frac{ORsinB+OQsinC}{sinA}})$+$s.({\frac{OPsinC+ORsinA}{sinB}})$+$t.({\frac{OQsinA+OPsinB}{sinC}})$

=$OP.(s.{\frac{sinC}{sinB}}+t.{\frac{sinB}{sinC}})$ +$OQ.(r{\frac{sinC}{sinA}}+t.{\frac{sinA}{sinC}})$ +$OR.(s.{\frac{sinA}{sinB}}+r.{\frac{sinB}{sinA}})$

$\geq$ $2(\sqrt{st}.OP+\sqrt{rt}.OQ+\sqrt{rs}.OR)$ Thus we have the  useful inequality

$r.OA+s.OB+t.OC$ $\geq$ $2(\sqrt{st}.OP+\sqrt{rt}.OQ+\sqrt{rs}.OR)$............(*)

Now write $r=OB.OC, s=OC.OA , t=OA.OB$ in inequality (*).Squaring the result and cancelling a factor of $OA.OB.OC$ from both sides then leads to the inequality 1.
I hope the information in this  post is useful!

You are right, thanks for your solution, it is a little trick in this my problem, thus with $\alpha,\beta\gamma>0$ we have
$\alpha MA+\beta MB+\gamma MC\ge 2(\sqrt{\alpha\beta}d_{a}+\sqrt{\beta\gamma}d_{b}+\sqrt{\gamma\alpha}d_{c}$


Prove in the easiest way the fact that if
$a+b=1$ and $a,b\neq 1$ then $\frac{a}{b^{3}-1}-\frac{b}{a^{3}-1}=\frac{2(b-a)}{a^{2}b^{2}+3}$

[quote="Beat"]Prove in the easiest way the fact that if
$a+b=1$ and $a,b\neq 1$ then $\frac{a}{b^{3}-1}+\frac{b}{a^{3}-1}=\frac{2(b-a)}{a^{2}b^{2}+3}$[/quote]

counterexample: $a=\frac{1}{3}, b=\frac{2}{3}: LHS =-\frac{288}{247}, RHS = \frac{54}{247}$

there was a typo, i have edited?
Now how it will be?


How many pairs of rational number $ \left(x,y\right)$ are there satisfying the equation $ y \equal{} \sqrt {x^{2} \plus{} \frac {1}{1999} }$?

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

$ (y\minus{}x)(y\plus{}x)\equal{}{1\over 1999}$. From $ y\minus{}x\equal{}r\land y\plus{}x\equal{}{1\over 1999r}$, where $ r$ is any positive rational number, we get one solution. Hence the answer is $ E$.


Bill can buy jags, jigs and jogs for $ 1,$2 and $ 7 each, respectively. What is the largest number of jogs he can purchase if he must buy at least one of each item and will spend exactly$50?

[hide] Let $ x$ be the number of jags, $ y$ the number of jigs, and $ z$ the number of jogs. We have

$ x\plus{}2y\plus{}7z\equal{}50$

We wish to maximize $ z$. If $ z\equal{}7$, that would mean $ y\equal{}0$, which can't happen. Hence, the maximum of $ z$ is $ \boxed{6}$.[/hide]

If 7 Jogs, condition '1 of each' goes unsatisfied
If 6 Jogs, meets criteria

Answer is 6


geometry
perimeter
complex numbers
geometry proposed

Equilateral triangles $ A'BC$ , $ AB'C$ and $ ABC'$ are drawn outside the $ ABC$ triangle and  Equilateral triangles $ A''BC$ ,$ AB''C$ and $ ABC''$ are drawn inside the triangle $ ABC$.prove that $ S_{A'B'C'}$ $ \plus{}$ $ S_{A''B''C''}$ $ \equal{}$ $ 5 S_{ABC}$

[quote="behdad.math.math"]Equilateral triangles $ A'BC$ , $ AB'C$ and $ ABC'$ are drawn outside the triangle $ ABC$ and  equilateral triangles

$ A''BC$ , $ AB''C$  and $ ABC''$ are drawn inside the triangle $ ABC$ . Prove that $ [A'B'C'] \plus{} [A''B''C''] \equal{} 5\cdot [ABC]$ .[/quote]

[b][u]Commentary[/u].[/b] I think that $ S_{XYZ} \equal{} [XYZ]$ - the area of $ \triangle XYZ$ . Is well-known that the lines $ AA'$ , $ BB'$ , $ CC'$ are concurrently and

the point $ T\in AA'\cap BB'\cap CC'$ is the [u][b]Fermat[/b]'s point[/u] or the [u][b]Torricelli[/b][/u] point and it is has many properties. See [url=http://mathworld.wolfram.com/OuterNapoleonTriangle.html][color=red][b]here[/b][/color][/url] ,  [url=http://mathworld.wolfram.com/InnerNapoleonTriangle.html][color=red][b]here[/b][/color][/url] and [url=http://en.wikipedia.org/wiki/Fermat_point][color=red][b]here[/b][/color][/url].

Thus prove easily that the following metrical relations : $ \left\|\ \begin{array}{c} TA' \equal{} TB \plus{} TC\ ,\ TB' \equal{} TC \plus{} TA\ ,\ TC' \equal{} TA \plus{} TB \\
 \\
AA' \equal{} BB' \equal{} CC' \equal{} TA \plus{} TB \plus{} TC \equal{} \sqrt {\frac 12\cdot\sum a^2 \plus{} 2S\sqrt 3} \\
 \\
TB^2 \plus{} TC^2 \plus{} TB\cdot TC \equal{} a^2\ \mathrm {a.s.o.} \\
 \\
\ TB\cdot TC \plus{} TC\cdot TA \plus{} TA\cdot TB \equal{} \frac {4S\sqrt 3}{3} \\
 \\
TA^2 \plus{} TB^2 \plus{} TC^2 \equal{} \frac 12\cdot\left(a^2 \plus{} b^2 \plus{} c^2\right) \minus{} \frac {2S\sqrt 3}{3}\end{array}\ \right\|$ .

Therefore, $ [A'B'C'] \equal{} \frac {\sqrt 3}{4}\cdot \sum TB'\cdot TC' \equal{} \frac {\sqrt 3}{4}\cdot\sum (TA \plus{} TB)(TA \plus{} TC) \equal{}$ $ \frac {\sqrt 3}{4}\cdot \left(\sum TA^2 \plus{} 3\cdot \sum TB\cdot TC\right) \equal{}$

$ \frac {\sqrt 3}{4}\cdot \left(\frac 12\cdot\left(a^2 \plus{} b^2 \plus{} c^2\right) \minus{} \frac {2S\sqrt 3}{3} \plus{} 4S\sqrt 3\right)\implies$ $ \boxed {\ [A'B'C'] \equal{} \frac {\sqrt 3}{8}\cdot\left(a^2 \plus{} b^2 \plus{} c^2\right) \plus{} \frac 52\cdot S\ }$ , where $ S \equal{} [ABC]$ .

You can find analogously $ [A''B''C'']$ . In this case is better to use the complex numbers ! I"ll come back soon in the new year.

[url=http://www.freelang.net/expressions/newyear.php][size=125][color=darkred][b]Happy New Year 2010 ![/b][/color][/size][/url] [size=125][b](Un an nou fericit / La mulţi ani !)[/b][/size] :clap:

[img]http://i263.photobucket.com/albums/ii135/stinker44/New%20Year/y01.gif[/img]


I'm not sure if this is the right section - move the topic if necessary.

Lately I'm wondering what to do (if anything) with what I could describe as "sofisticated spammers" - i.e. people who would fire from all the heavy artillery, such as complex analysis and stuff, to "prove" that $2+2=5$ (within the generally accepted axiomatics of natural numbers) and moreover, they aggressively deny any possibility that they are way off course. Naturally, they fire nothing but blanks, but anyway they DO cram up the topics and they tend to seduce less experienced posters which are prone to believe anything they're not fluent in. (And rightly so! This is MATHEMATICS forum, hence being correct for us is what following Hippocratic oath is for a medical doctor.)

The most extreme example of such behaviour is user Dr. No. Whoever's interested can search for his posts to see what he does and how that affects other posters.

The point is: What's the good measure of moderator involvement in cases like that?

I'd say PM the person if they persist in acting foolishly ban them.
If someone tells you 2+2=5 for greater values of 2 or something you tell the mod(if the do not recognise this at once.) Of course, this is only my opinion.

The user in question is not a crank, but he only partially understands what he's using.
(For an example of a crank, see the last few posts of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28607]this thread[/url])

The real problem with those threads is the nonmathematical side of the argument; moderators should try to cool down the discussion and refocus on the facts.

This needs to be in Moderator's Lounge....


Let A and B be nxn real matrices , k a positive integer , Find :
 a)       $ \lim_{x\to 0} \frac{1}{x}[(A\plus{}x.B)^{k}\minus{}A^{k}]$.
 b)       $ \frac {d}{dx}(Tr(A\plus{}xB)^{k}) (0)$ ,
([b] Fall exam ? [/b])
Thanks !!! :wink:

Suggestion: Expand $ (A \plus{} x.B)^{k}$ as a polynomial in $ x$ and look for the first order term..


function
limit
inequalities
induction
topology
triangle inequality
real analysis

Let $ (X,\rho)$ be a metric space.  Consider the space, $ C$, of all Cauchy sequences in $ X$.  Define a new function $ \rho^{*}: C\times C\to \mathbb{R}$, where  $ \rho^{*}(<x_n>,<y_n>)\equal{}\lim_{n\to\infty} \rho(x_n,y_n)$.  Consider the set of equivalence classes in $ C$ under the equivalent relation $ \rho^{*}(<x_n>,<y_n>)\equal{}0$.  Call this set $ X^{*}$ and it can be shown that $ (X^{*},\rho^{*})$ is a metric space.

Show that $ X^{*}$ is complete.

Correct me if I'm wrong, but isn't this equivalent (or else very close) to stating that the limit points of a subset of a metric space form a closed set?

Note that $ X$ is isometric to the subspace $ W \subset X^{*}$ consisting of all constant sequences corresponding to elements of $ X$ with $ T: x \mapsto (x,x,\ldots)$ being an isometry. Also note that $ W$ is dense in $ X^{*}$.

Now let $ (x_{n}^{*})$ be a Cauchy sequence in $ X^{*}$. Since $ W$ is dense in $ X^{*}$, for every $ n$ there is a $ w_{n}^{*} \in W$ such that $ \rho^{*}(x_{n}^{*},w_{n}^{*}) < 1/n$. Then by the triangle inequality, $ \rho^{*}(w_{m}^{*},w_{n}^{*})$ is less than any given $ \varepsilon>0$ for $ m,n$ sufficiently large. i.e. $ (w_{n}^{*})$ is a Cauchy sequence. Then $ (w_{n})\equal{}(T^{\minus{}1}w_{n}^{*})$ is a Cauchy sequence in $ X$ so it belongs to some equivalence class $ w^{*} \in X^{*}$. This $ w^{*}$ will be the limit of $ (x_{n}^{*})$.

[b]Step 1)[/b] Suppose $ x_n, y_n \in C$. For any $ \epsilon > 0$, we can choose $ N$ such that for all $ m, n \geq N$, we have $ \rho(x_m, x_n) < \epsilon / 2$ and $ \rho(y_m, y_n) < \epsilon / 2$. Then by triangle inequality, $ \rho(x_m, y_m) \leq \rho(x_n, y_n) + \epsilon$, or $ \rho(x_m, y_m) - \rho(x_n, y_n) \leq \epsilon$. Interchanging the role of $ m$ and $ n$, we find that $ \rho(x_n, y_n)$ is a nonnegative Cauchy sequence. Thus $ \rho^{\ast}$ is well-defined.

Define $ x_n \sim y_n$ if $ \rho^{\ast}(x_n, y_n) = 0$. It is easy to see that $ \sim$ is an equivalence relation on $ C$. Now let $ \overline{X} = C/\sim$ and denote the equivalence class of $ x_n$ by $ \overline{x_n}$. If $ x_n \sim x'_n$ and $ y_n \sim y'_n$, $ \rho(x_n, y_n) \leq \rho(x_n, x'_n) + \rho(x'_n, y'_n) + \rho(y'_n, y_n)$. Taking $ n\to\infty$ to both sides gives $ \rho^{\ast}(x_n, y_n) \leq \rho^{\ast}(x'_n, y'_n)$, so $ \rho^{\ast}(x_n, y_n) = \rho^{\ast}(x'_n, y'_n)$ by symmetry. This shows that $ \overline{\rho}(\overline{x_n}, \overline{y_n}) = \rho^{\ast}(x_n, y_n)$ is a well-defined metic on $ \overline{X}$.

[b] Step 2)[/b] We can easliy verify the following lemma:

Lemma) $ x_n$ is a Cauchy sequence in $ X$ iff for any $ \epsilon > 0$, there exists $ N$ such that for all $ n \geq N$, $ \rho(x_n, x_N) < \epsilon$.

Let $ \overline{x}_n$ be a Cauchy sequence in $ \overline{X}$ where $ \overline{x}_n = x_{n,k}$ for some $ (x_{n,k})_{k=1}^{\infty} \in C$. Choose a positive decreasing sequence $ \epsilon_n$ of real numbers such that $ \lim_{n\to\infty} \epsilon_n = 0$. Then by induction, we can choose a strictly increasing sequence $ K_n$ such that $ \rho(x_{n,k}, x_{n,K_n}) < \epsilon_n$ for all $ k \geq K_n$.

Define $ y_n = x_{n, K_n}$. We claim that $ y_n \in C$. Since $ \overline{x}_n$ is Cauchy, for $ \epsilon > 0$, there is $ N$ such that $ \epsilon_n < \epsilon/3$ and $ \overline{\rho}(\overline{x}_N, \overline{x}_n) < \epsilon / 3$ whenever $ n \geq N$. Also, $ \overline{\rho}(\overline{x}_N, \overline{x}_n) = \lim_{k\to\infty} \rho(x_{N,k}, x_{n,k})$, there is $ K$ such that $ \rho(x_{N,k}, x_{n,k}) < \epsilon/3$ for all $ k \geq K$. Then we have

\begin{eqnarray*}\rho(y_N, y_n)
&  = & \rho(x_{N, K_N}, x_{n, K_n})\\
& \leq & \rho(x_{N,K_N}, x_{N, k}) + \rho(x_{N,k}, x_{n,k}) + \rho(x_{n,k}, x_{n,K_n})\\
& < & \epsilon_N + (\epsilon/3) + \epsilon_n\\
& < & \epsilon.\end{eqnarray*}

Thus $ y_n$ is a Cauchy sequence by lemma.

Now it remains to show that $ \overline{x}_n \to \overline{y_n}$ as $ n\to\infty$. Let $ \epsilon > 0$. Choose $ N$ such that $ \epsilon_N < \epsilon/2$ and $ \rho(y_N, y_n) < \epsilon/2$ whenever $ n \geq N$. Also, for each $ n$, let $ I_n = \max\{ K_n, N\}$. Then for all $ i \geq I_n$, we have $ \rho(x_{n,i}, y_i) \leq \rho(x_{n,i}, y_n) + \rho(y_n, y_i) < \epsilon_n + (\epsilon/2) < \epsilon$, implying $ \overline{\rho}(\overline{x}_n, \overline{y_n}) < \epsilon$. Therefore, $ \overline{x}_n$ converges $ \overline{y_n}$ and $ (\overline{X}, \overline{\rho})$ is complete.


function
geometry
geometric transformation
algebra
floor function

${\alpha_{n}=3 \alpha_{n-1}-\alpha_{n-2}}$

Since this is linear, the solution is exponential

${\alpha_{n}=\gamma \theta^{n}}$

${\alpha_{n}=3 \alpha_{n-1}-\alpha_{n-2}}$
${\gamma \theta^{n}=3 \gamma \theta^{n-1}-\gamma \theta^{n-2}}$
${\gamma \theta^{n-2}\left(\theta^{2}-3 \theta+1\right)=0}$

${\theta^{n-2}\neq 0\land \gamma \neq 0}$

${\theta^{2}-3 \theta+1=0}$
${\theta =\frac{1}{2}\left(3\pm \sqrt{5}\right)}$

SO

${\alpha_{n}=\gamma_{1}\left(\frac{1}{2}\left(3-\sqrt{5}\right)\right)^{n}+\left(\frac{1}{2}\left(3+\sqrt{5}\right)\right)^{n}\gamma_{2}}$

Using

${\alpha_{1}=1}$
${\alpha_{2}=2}$

Solving and substituing

${\alpha_{n}=\frac{2^{-n}\left(\left(-5+\sqrt{5}\right) \left(3-\sqrt{5}\right)^{n}+2 \left(3+\sqrt{5}\right)^{n}\left(-5+2 \sqrt{5}\right)\right)}{5 \left(-3+\sqrt{5}\right)}}$

Just make a simple substituition and see that your last equation is wrong for $n \geq 4$

this is a little bit obviously

because the last equation is not linear

[quote="Thales418"]

Just make a simple substituition and see that your last equation is wrong for $n \geq 4$

this is a little bit obviously

because the last equation is not linear[/quote]
$[x]$ is greatest integer $\leq x$.

what you tipe is not FLoor function

the symbol is different

http://en.wikipedia.org/wiki/Floor_function

it would be good use the correct symbols

can who!?!?!?!

what expression is this!?

http://senhorcarteiro.blogspot.com/2006/12/aritmtica-modular-am.html

in this article have a equation i made

to use modular aritmetic with expressions using log

you can use this to solve

google translation

http://www.google.com/translate?u=http%3A%2F%2Fsenhorcarteiro.blogspot.com%2F2006%2F12%2Faritmtica-modular-am.html&langpair=pt%7Cen&hl=en&ie=UTF8


function
logarithms
combinatorics proposed
combinatorics

Define the function $ g(\cdot): \mathbb{Z} \to \{0,1\}$ such that $ g(n) \equal{} 0$ if $ n < 0$, and $ g(n) \equal{} 1$ otherwise. Define the function $ f(\cdot): \mathbb{Z} \to \mathbb{Z}$ such that $ f(n) \equal{} n \minus{} 1024g(n \minus{} 1024)$ for all $ n \in \mathbb{Z}$. Define also the sequence of integers $ \{a_i\}_{i \in \mathbb{N}}$ such that $ a_0 \equal{} 1$ e $ a_{n \plus{} 1} \equal{} 2f(a_n) \plus{} \ell$, where $ \ell \equal{} 0$ if $ \displaystyle \prod_{i \equal{} 0}^n{\left(2f(a_n) \plus{} 1 \minus{} a_i\right)} \equal{} 0$, and $ \ell \equal{} 1$ otherwise. How many distinct elements are in the set $ S: \equal{} \{a_0,a_1,\ldots,a_{2009}\}$?
[i](Paolo Leonetti)[/i]

A cute typo
[quote]... such that $ a_0 \equal{} 1$ [color=red]e[/color] $ a_{n \plus{} 1} \equal{} 2f(a_n) \plus{} \ell$, where ...[/quote]
It seems that everybody speaks Italian, so they figured out that the "e" means "and" (not the base of natural logarithms)  :)

LOL, sorry for the mistake..Anyway, I have not yet read solutions, has anyone completely solve this one?

Consider the obvious bijection between {1,...,2048} and the strings of 11 binary digits. The problem states that we begin with $ x_1: \equal{} 1$, and that if $ x_n \equal{} 0\ell$ or $ x_n \equal{} 1\ell$ (where $ \ell$ is a string of 10 binary digits) then $ x_{n \plus{} 1} \equal{} \ell 1$ if we can find i<n+1 such that $ x_i \equal{} \ell$, otherwise $ x_{n \plus{} 1} \equal{} \ell 0$. Define $ k: \equal{} \max\{n \in \mathbb{N}: \prod_{1 \le a < b \le n}{(x_a \minus{} x_b)} \neq 0\}$ and $ S: \equal{} \{x_1,...,x_k\}$. Suppose that $ x_k \equal{} ey$ for some e in {0,1} and a string y of 10 binary digits, then $ x_{k \plus{} 1} \equal{} y0 \in S$ by definition of k, so define $ x_i: \equal{} y0 \in S$ the other i such that 1<i<k+1. And, if $ x_k \equal{} ey$ then define j that integer such that 0<j<k+1 and $ x_j: \equal{} y1 \in S$ (it must exist since $ x_{k \plus{} 1} \equal{} y0$). If j>1 then $ \{x_{i \minus{} 1},x_{j \minus{} 1}\}$ exist and is $ \{0y,1y\}$, then $ x_k \in \{x_{i \minus{} 1},x_{j \minus{} 1}\}$, contradiction, so j=1. It means that $ 0 \equal{} y \in S$, and if $ e \equal{} 1$ then $ x_{k \minus{} 1} \in \{0,x_k\}$, contradiction, so $ x_k \equal{} 0$. Clearly $ 1 \le k \le 2048$; suppose that $ k < 2048$. A number $ a \in \{0,1\}$ and a string $ b$ of 10 binary digits exist such that $ ab \not \in S$, and suppose $ b0 \in S$ (clearly $ b \neq 0$).Define $ x_w: \equal{} b0 \in S$ for some 1<w<k+1, then there exist 0<z<k+1 such that $ x_z \equal{} b1 \in S$. As before if z>1 then $ \{x_{w \minus{} 1},x_{z \minus{} 1}\}$ exist and is $ ab \in \{0b,1b\} \subseteq S$, contradiction; and if z=1 then b=0, again contradiction. So a number $ ab$ does not belong to S only if $ b0 \not \in S$. But in this way we can define a sequence of number that do not belong to S, and that is eventually null, but $ a_k \equal{} 0$.
It shows that $ \prod_{0 \le i < j \le 2047}{(a_i \minus{} a_j)^2} > 0$.[]


I've never solved an inequality before (Edit: Not quite what I meant, but you get the idea  :P ), so I probably messed up somewhere, but anyway

[hide="Solution"]
By the reciprocal of AM-GM,
\[ \left(\sum_{i < j} \frac {1}{\sqrt {a_ia_j}}\right)^2 \ge \left(\sum_{i < j} \frac {1}{\frac {a_i \plus{} a_j}{2}}\right)^2
\]
Since there are $ {n\choose 2}$ pairs to pick for the values of $ i < j$, we can apply Cauchy
\[ \left(\sum_{i \equal{} 1}^{{n\choose 2}} 1^2 \right)\left(\sum_{i < j} \left(\frac {1}{\sqrt {a_ia_j}}\right)^2\right) \ge \left(\sum_{i < j} \frac {1}{\sqrt {a_ia_j}}\right)^2
\]
which combined give us the desired.
[/hide]

Are these the hard track 3 problems?

$ \dfrac{n(n \minus{} 1)}{2}\sum_{i < j}\dfrac{1}{a_{i}a_{j}}\ge (\sum_{i < j}\sqrt {\dfrac{1}{a_{i}a_{j}}})^2\ge (\sum_{i < j}\dfrac{2}{a_{i}\plus{} a_{j}})^2 \equal{} 4(\sum_{i < j}\dfrac{1}{a_{i} \plus{} a_{j}})^{2}$

by Cauchy and AM-HM

edit: beat

[quote="SplashD"]Are these the hard track 3 problems?[/quote]

This particular one was the easiest one (out of 3) off one of the tests.


In general one has for $ x_1,\ldots,x_n\in(0,\frac {1}{2}]$ that
\[ \frac {\sum_{i \equal{} 1}^nx_i^n}{\prod_{i \equal{} 1}^nx_i}\geq\frac {\sum_{i \equal{} 1}^n(1 \minus{} x_i)^n}{\prod_{i \equal{} 1}^n(1 \minus{} x_i)}
\]
W.l.o.g. suppose $ x_n\leq x_{n \minus{} 1}\leq\ldots\leq x_1$ and take $ \displaystyle f_n(x) \equal{} \frac {x^n \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}x_i^n}{x\cdot\prod_{i \equal{} 1}^{n \minus{} 1}x_i}$. One can verify that $ f_n'\leq0$ so $ f_n$ decreases $ \forall x\in(0,x_{n \minus{} 1}]$, in particular $ f_n(x_n)\geq f_n(1 \minus{} x_n)$ (as $ x_n\leq1 \minus{} x_n$).

The given result is obtained after $ n\minus{}1$ more "identical" steps (preserving of order & renaming might be useful).

[quote="gnosis"]In general one has for $ x_1,\ldots,x_n\in(0,\frac {1}{2}]$ that
\[ \frac {\sum_{i \equal{} 1}^nx_i^n}{\prod_{i \equal{} 1}^nx_i}\geq\frac {\sum_{i \equal{} 1}^n(1 \minus{} x_i)^n}{\prod_{i \equal{} 1}^n(1 \minus{} x_i)}
\]
W.l.o.g. suppose $ x_n\leq x_{n \minus{} 1}\leq\ldots\leq x_1$ and take $ \displaystyle f_n(x) \equal{} \frac {x^n \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}x_i^n}{x\cdot\prod_{i \equal{} 1}^{n \minus{} 1}x_i}$. One can verify that $ f_n'\leq0$ so $ f_n$ decreases $ \forall x\in(0,x_{n \minus{} 1}]$, in particular $ f_n(x_n)\geq f_n(1 \minus{} x_n)$ (as $ x_n\leq1 \minus{} x_n$).

The given result is obtained after $ n \minus{} 1$ more "identical" steps (preserving of order & renaming might be useful).[/quote]
Oh thank,my proof is quite similar to yours(I used AMGM in the first),but I dont think it is really nice (for this nice),so I post it and hope some one can kill it without Derivative.At all thank for your care


[quote="randomdragoon"][quote="junggi"]The $\Im$ symbol means the imaginary part of a number, so it doens't have much to do with logs[/quote]
What if you take logs of negative numbers? We know that $e^{\pi i}=-1$, so $\ln{(-1)}= \pi i$?[/quote]
I meant that there's nothing special about $\Im$ with logs

Thanks.  I didn't think the logs mattered, I just posted it that way because that was the context I saw it in...


Do any of you have any ideas for a science fair project suitable for a 5th grader?

My brother suggested something complicated to impress the judges.

1 computer.
2 video cards.
WHY DON'T THEY STACK?
It works.
I got an A.

How about you calculate how much air the ventilation system in your science room has to export after each kid has done thirty minutes of exercise? All you have to do is measure how much air you breath out every minute (you can buy kits for this) multiply it by the number of kids in your class, see how much space is in your classroom, calculate how long it would take for the room would take to fill up with air.

Ka-Presto.

If you play a musical instrument, try comparing sound waves from each note. Or study paramecium found in ponds( worked 4 me). Good Luck. :)

I started homeschooling about a week before the 5th grade science fair.  :lol:
I'm lucky but unlucky at the same time.

In 5th grade, I built a robot with Lego Mindstorms that could cross a small gap between tables.

Everyone said it was cool.

[quote="D3m0n Shad0w"]Does talking to plants make them grow faster? Of course we all know the answer, but my teacher said it was pretty good  :wink:[/quote]

[i]Totally[/i] my idea.

[quote="D3m0n Shad0w"]Thanks everyone, but I already got my topic which is a classic. Does talking to plants make them grow faster? Of course we all know the answer, but my teacher said it was pretty good  :wink:[/quote]

*points to the Mythbusters*

Lol, umm.... errmm
weird topic

i was going to suggest seeing which fruits were most acidic by turning them in to batteries (all you have to do is stick some copper wire and a galvanized screw in to it) and plugging them in to one of those volt measuring things, but since you already got your topic...

So what are you going to say to it? Teach your plant math!  :D

[quote="Nerd_of_the_Ages"]i was going to suggest seeing which fruits were most acidic by turning them in to batteries (all you have to do is stick some copper wire and a galvanized screw in to it) and plugging them in to one of those volt measuring things, but since you already got your topic...[/quote]

That's not a bad idea. Might take a while getting the materials, and we all know how D3m0n Shad0w procrastinates  :wink: .

[quote="Nerd_of_the_Ages"]So what are you going to say to it? Teach your plant math!  :D[/quote]
Yes, the derivitive of your photosystem 700 is P680!  :rotfl: .

maybe something with electromagnets or physics or something... i don't know

if you are entering a science fair, then you might not want to do something with human subjects, plants, or animals. it requires to much paper work and usually messes up (at least for me).

if you want to win, do something in a category where few people are participating. (if a lot of people are doing biology related projects, do chemistry.)


A computer program reduces the length of each side of an 8'' by 8'' square photograph by $ 25\%$. What is the area, in square inches, of the reduced photo?

So right now the square is an $ 8$ by $ 8$ photo. If you reduce both sides by $ 25%
$ you get a $ 6$ by $ 6$ square so the answer is $ \boxed{36}$