How do you do this problem? I dont see an easy way to do it

Define functions $f$ and $g$ by $f(x)=x+\sqrt{x}$ and $g(x)=x+\frac{1}{4}$. Compute $g(f(g(f(g(f(g(f(3)))))))).$ (Your answer must be in the form $a+b\sqrt{c}$, where $a,b,$ and $c$ are rational).

Well, brute-forcing it isn't very hard, as far as I can tell

Calculate $g(f(x))$, as that seems to be the core function. Call $g(f(x)) = x+\sqrt{x}+1/4$ $h(x)$.

So, the ugly expression becomes $h^{4}(x)$.

To calculate this, first find $h^{2}(x)$.

Uh, yeah, except [hide]$(g \circ f)(x) = x+\sqrt{x}+\frac{1}{4}= (\sqrt{x}+\frac{1}{2})^{2}$[/hide]
and then you put square root signs around everything and the answer is $(\sqrt{3}+2)^{2}$ or something...

[quote="not_trig"]Yeah I know... my dad had to tell me, and then he's like "don't post it..."[/quote]

er, what?

[quote="not_trig"]My dad told me the answer, and he wouldn't let me post it because he's anti-AoPS[/quote]

er, what?

Is his name Bradley by chance?

nvm, you had to have been a part of the biggest (mostly one-sided) flame war on AoPS.


[quote="diophantient"]I heard from someone that Bertrand's Postulate can be greatly strengthened. How so?[/quote]

For every integer $ n>3$ there exists a prime $ p$ so that $ n<p<2n\minus{}2$.

The stonger version is like that:
For each k>1 (k is real) theri is such $ N_{k}$ that for $ n>N_{k}$ there is allways a prime p such that n<p<kn.
Bertrand postulate is special case for k=2.

For $ k\ge2$ $ N_{k}\equal{}\frac{2}{k}$, because then for $ N_{k}<n<2$ 2 is such prime that: $ n<p<kn$, and for $ n\ge2$, there is also such p, because it's between $ n$ and $ 2n$.
For $ 1<k<2$ I can only estimate. All of such n that: $ n!\plus{}1<n<n!\plus{}n\plus{}1$ aren't primes. So if $ k<\frac{n!\plus{}n\plus{}1}{n!\plus{}1}\equal{}1\plus{}\frac{n}{n!\plus{}1}$ then $ N_{k}>n$


Let $ ABCDEF$ be a convex cyclic hexagon. Prove that $ AD, BE, CF$ are concurrent if and only if $ AB\cdot CD\cdot EF \equal{} BC\cdot DE\cdot FA$.


Dood, 10A was like a month ago.  Do you mean 10B?

well whatever you mean.
10A: 121.5
10B: 126

AMC 10A- 118.5ish

AMC 10B- was in florida, couldn't take it :)

I guess people who didn't pass and weren't satisfied didn't bother voting. I wouldn't if I go less than 110. And I don't know anyone who got less than 50. No offense to those who did, especially if because of bubbling errors or mistakes on the form. My score was 126, didn't take B.

AMC 10A-132
Didn't take a B test

I doubt there are many low scorers on AoPS vs. # high scorers. Take a look at the MathCounts polls...

um 138

was hoping for perfect

misread 2 problems

I think I've already said this once

oh well?

so I blew my last chance for perfect on the 10

argh

Yes, I do mean the A. I got a 150 as a seventh grader.

I got 112.5 only because I made four silly mistakes on four of the problems  :mad:  :blush:

AMC 10A: 145.5

Haha, I screwed up, didn't want to take get #18 wrong, so I decided against guessing.
Hopefully, I'll do better on AIME.

[quote="Zmastr"]Do you think I should've gotten my official results back by now?  It's been a month since I took the 10A, and the AIME is in a week...[/quote]

yes i think you should ask your teacher/proctor who adminstered the test.

Infact, B results should be coming this week...


algorithm
modular arithmetic
number theory
Euclidean algorithm
Diophantine equation
greatest common divisor

How to use Euclidean Algorithm to find 74x + 58y = 14?

I don't quite understand what you're asking...are you asking how to solve diophantine equations?

yes, find the integer solutions of the equation

Euclidean Algorithm:
$ 74\equal{}58 (1) \plus{} 16$
$ 58\equal{}16 (3) \plus{} 10$
$ 16\equal{}10 (1) \plus{} 6$
$ 10\equal{}6 (1) \plus{} 4$
$ 6\equal{}4 (1) \plus{}2$
$ 4\equal{} 2 (2) \plus{} 0$

The last non-zero remainder is the greatest common factor $ 2$.

After this, use backward substitution to find the initial solution for solving the diophantine equation, and then you can find the general form for all the solutions.

erm cyra? can u show the full workings pls?i wanna know more abt solving tis sorta eqns. Thx alot. sorry for my noobness.

I didn't understand his solution...

$ 74x \plus{} 58y \equal{} 14\implies37x \plus{} 29y \equal{} 7$
Taking this $ \bmod{29}$, we have $ 37x\equiv8x\equiv7\equiv152\pmod{29}$.
This means that $ x\equiv19\pmod{29}$, so $ x \equal{} 29m \plus{} 19$.
Substituting this into our original equation, we find that $ y \equal{}\minus{}37m \minus{} 24$.

Thus, our solutions are of the form $ (x,y) \equal{} (29m \plus{} 19,\minus{}37m \minus{} 24)$ for all integers $ m$. Not sure about the Euclidean Algorithm though... :(

[quote="Cyrazeno"]Euclidean Algorithm:
$ 74 \equal{} 58 (1) \plus{} 16$ Eqn 1
$ 58 \equal{} 16 (3) \plus{} 10$ Eqn 2
$ 16 \equal{} 10 (1) \plus{} 6$ Eqn 3
$ 10 \equal{} 6 (1) \plus{} 4$ Eqn 4
$ 6 \equal{} 4 (1) \plus{} 2$ Eqn 5
$ 4 \equal{} 2 (2) \plus{} 0$
[/quote]

Now you just substitute backwards. (actually, I prefer a putting what Cyrazeno had in parentheses as coefficients and Cyrazeno's coefficients in parentheses, but I'll be consistent with Cyrazeno's notation)

$ 2 \equal{} 6 \minus{} 4(1)$ From Eqn 5
$ 2 \equal{} 6 \minus{} [10 \minus{} 6(1)](1)$ From Eqn 4 (subbed the $ 4$)
$ 2 \equal{} 6(2) \minus{} 10$ careful to keep it in terms of $ 6$ and $ 10$
$ 2 \equal{} [16 \minus{} 10(1)](2) \minus{} 10$ From Eqn 3 (subbed the $ 3$)
$ 2 \equal{} 16(2) \minus{} 10(3)$ careful to keep it in terms of $ 10$ and $ 16$
$ 2 \equal{} 16(2) \minus{} [58 \minus{} 16(3)](3)$ From Eqn 2 (subbed the $ 10$)
$ 2 \equal{} 16(11) \minus{} 58(3)$ careful to keep it in terms of $ 16$ and $ 58$
$ 2 \equal{} [74 \minus{} 58(1)](11) \minus{} 58(3)$ From Eqn 1 (subbed the $ 16$)
$ 2 \equal{} 74(11) \minus{} 58(14)$ Eqn 6 (careful to keep it in terms of $ 74$ and $ 58$)

EDIT: MESSED UP HERE, FIXED Thanks math154.  :)

We don't need $ 2 \equal{} 74x \plus{} 58y$, we need $ 14 \equal{} 74x \plus{} 58y$. So multiply everything in Eqn 6 by $ 7$ to get $ 14 \equal{} 74(77) \minus{} 58(98)$. So one solution is $ (x,y) \equal{} (77,\minus{}98)$. $ (77 \plus{} \frac {58}{\gcd(74,58)}t,\minus{}98 \minus{} \frac {74}{\gcd(74,58)}t) \equal{} (77 \plus{} 29t, \minus{}98 \minus{} 37t)$ describes all integer solutions.

-------------------

Just another method to solving linear Diophantines in two variables.

Eh, I think you meant $ (77\plus{}29t,\minus{}98\minus{}37t)$.

Okay, first I'll do the Euclidean algorithm (after dividing $ 74x + 58y = 14$ by $ 2$)

This is basically used to determine the [b]greatest common factor[/b] of the two numbers.

$ \dfrac{74x+58y}{2}=\dfrac{14}{2}$

$ 37x + 29y = 7$     [This is in the form $ ax + by = c$]

[b]Euclidean Algorithm: [/b]

All the numbers at the end of the equations that are being added are the remainders.

$ 37 = 29(1) + 8$  [$ 37$ equals $ 29*1$ plus the remainder of $ 8$]

$ 29 = 8(3) + 5$    [We bring the number being multiplied (in the previous equation it was $ 29$) to the number in parentheses to the other side of the equal sign in the next equation. $ 29$ equals $ 8*3$ plus the remainder of $ 5$. The remainder is the number that is kept outside the parentheses in the following line.]

$ 8 = 5(1) + 3$      [$ 8$ equals $ 5*1$ plus the remainder of $ 3$]
$ 5 = 3(1) + 2$
$ 3 = 2(1) + 1$
$ 2 = 2(1) + 0$

Alright, now that's done.

I'm going to move the equations around a bit so everything is in terms of the remainders (from algorithm). $ (8, 5, 3, 2, 1, 0)$
This helps with the backward substitution:

$ 8 = 37 - 29(1)$
$ 5 = 27 - 8(3)$
$ 3 = 8 - 5(1)$
$ 2 = 5 - 3(1)$
$ 1 = 3 - 2(1)$
$ 0 = 2 - 1(2)$

Okay, now, let's start the backward substitution (this can get pretty messy sometimes). I am going to use backward substitution to find the initial solutions which will help us to find the general form of all solutions.

This is basically just plugging everything in so it's in terms of $ 37$ and $ 29$.

[b]Backward substitution: [/b]

This is backward; so, we are obviously starting from bottom equation and going up. (We're not counting the $ 0 = 2 - 1(2)$)

Start with the other two bottom equations:

$ 1 = 3 - 2(1)$ and
$ 2 = 5 - 3(1)$

$ 1 = 3 - (5 - 3(1))$       [Over here I just substituted what $ 2$ equals into the first equation: $ 1 = 3 - 2(1)$]
$ 1 = 5( - 1) + 3(2)$      [Now, you don't want to multiply all of this out because then you won't be able to find your initial solutions. So, just leave it so it shows how many 5's you have: $ - 1$ and how many 3's: $ 2$.]

The next equation up is: $ 3 = 8 - 5(1)$

$ 1 = 5( - 1) + (8 - 5(1))(2)$
$ 1 = 5( - 3) + 8(2)$

Next equation up is: $ 5 = 29 - 8(3)$

$ 1 = (29 - 8(3))( - 3) + 8(2)$
$ 1 = 29( - 3) + 8(11)$

$ 1 = 29( - 3) + (37 - 29)(11)$

Finally,

$ 1 = 37(11) + 29( - 14)$

[b]Using initial solutions to find general form of all solutions:[/b]

So, you see that your initial solutions are what $ a$ and $ b$ are being multiplied by: $ 11$ and $ - 14$, respectively.

Now, $ x_0$ and $ y_0$ are going to be initial solutions ($ 11$ and $ - 14$)

$ x = x_0 + (b/d)t$

and

$ y = y_0 - (a/d)t$

$ x_0$ is $ 11$, b is $ 29$, d is the gcd(a,b), and t is an arbitrary integer.

The $ gcd(37, 29) = 1$ so, $ d = 1$

So, that turns out to be: $ x = 11 + 29t$

and

$ y_0$ is $ - 14$, a is $ 37$, and t, once again, is an arbitrary integer.

So, $ y = - 14 - 37t$

[b]Solution: [/b]$ (11 + 29t, - 14 - 37t)$

My answer differs from the above poster(s) so I may have made a minor mistake somewhere... Even so, this should hopefully give you a better idea on how to approach the problem.  :)

Sorry for the bad explanation in my previous post.

To do subscripts, use the code
[code]
x_0
[/code]
to produce $ x_0$.

Also, I'm sorry about assuming that you were a boy. It's just kind of hard to tell sometimes. :)

i don't understand the euclidean algorithm section.

Alright, let's look at the Euclidean Algorithm again:

Our equation is $ 37x\plus{}29y\equal{}7$, correct?

Now, I am going to take the $ 37$ and the $ 29$ in the equation and do the Euclidean alg. with the two numbers.
$ 37$ is larger than $ 29$ so it's going to be placed on the other side of the equal sign.

$ 37\equal{}29(1)\plus{}8$

Multiply $ 29$ by the largest integer that when multiplied with $ 29$ is still less than $ 37$ (or equal to). Of course, we are going to have a remainder since $ 37$ is not a multiple of $ 29$; and in this case, the remainder is $ 8$.

For the next step, I'm going to take $ 29$ to the other side of the equal sign and use the remainder ($ 8$) to  be the number being multiplied by another integer.

So, it's:

$ 29\equal{} 8(3)\plus{}5$

I multiplied $ 8$ by $ 3$ giving me $ 24$. If I had multiplied it by any integer greater than $ 3$ it would have been greater than $ 29$ which is, of course, not what we want. $ 8(3)\equal{}24$ plus the remainder of $ 5$ which gives us the $ 29$.

Next one:

I'm going to take the 8 to the other side this time and 5 will be multiplied with another number since it was the remainder in the last equation.

$ 8\equal{}5(1)\plus{}3$

I'm repeating these steps over and over again until there is no remainder, or $ 0$. Or you could just stop at the point where you have a remainder of $ 1$ since the equation with no remainder doesn't really matter in the problem.

$ 5\equal{}3(1)\plus{}2$

$ 3\equal{}2(1)\plus{}1$

$ 2\equal{}1(2)\plus{}0$

Let's try another one.

Let's take the numbers $ 57$ and $ 16$.

$ 57\equal{}16(3)\plus{}94$
$ 16\equal{}9(1)\plus{}7$
$ 9\equal{}7(1)\plus{}2$
$ 7\equal{}2(3)\plus{}1$
$ 2\equal{}1(2)\plus{}0$

I hope this helps.  :)


logarithms
real analysis
real analysis unsolved

$ \sum_{n\equal{}1}^{\infty}\left(a^{\frac{1}{n}}\minus{}\frac{b^{\frac{1}{n}}\plus{}c^{\frac{1}{n}}}{2}\right)$

where $ (a,b,c > 0)$

Convergence? :idea:

Write the terms as $ e^{\frac1n\ln a}\minus{}\frac12e^{\frac1n\ln b}\minus{}\frac12e^{\frac1n\ln c}.$ Then expand out the power series of the exponential. We get that the terms are

$ \frac1n\left(\ln a\minus{}\frac{\ln b\plus{}\ln c}2\right)\plus{}O(n^{\minus{}2}).$

So there are two possibilities:

(1) If $ a\equal{}\sqrt{bc}$ (that is, $ a$ is the geometric mean of $ b$ and $ c,$) then the series converges by limit comparison to $ \sum\frac1{n^2}.$

(2) In all other cases, the series diverges by limit comparison to the harmonic series.


geometry
circumcircle
geometric transformation
reflection
trigonometry
rhombus
parallelogram

Hello guys,

Determine with proof the answer to this question. Need not write up a rigorous proof with all steps but atleast give an outline of the proof.

Consider a triangle ABC with circumcenter O and orthocenter H. Given that OH = 1 , BO = 2 , BC = 3, BH = 2.75 and CH = 1.25 Find the measure of angle A in this triangle.

You need the answer to the first question to do the answer for the second question.

Let A/4 = m. Here A is measure of angle A.

Second question :

Let the sides of triangle DEF be m-2,m-1 and m. Let the circumcenter of triangle DEF be G. Reflect G about the three sides DEF and call the reflections XYZ. Find the circumradius of triangle XYZ.

It was fun making up this problem. Hope you guys like it. When i have more time, i will post some more interesting details on the motivation and extension of the second problem.

Thanks,
Befuddlers

[hide="Hint"]
CO = 2 because the circumcenter of the triangle is equidistant from all the vertices of the triangle. And already given that BO = 2. Now you know all the side lenghts for the quadrilateral BCOH, try proving cyclicness by using Ptolemy's. And then for some fun angle chasing. :lol:  :D  :starwars: . Sorry for being random in my choice of emoticons.[/hide]

[hide="Problem with Problem#1"]
I don't think that those measurements are possible.

Using the law of cosines and the given measurements,

$ \cos(\angle OBH) \equal{} \frac {2^2 \plus{} 2.75^2 \minus{} 1^2}{2 \cdot 2 \cdot 2.75} \equal{} \frac {169}{176}$,

and $ \cos(\angle HBC) \equal{} \frac {2.75^2 \plus{} 3^2 \minus{} 1.25^2}{2 \cdot 2.75 \cdot 3} \equal{} \frac {10}{11}$.

So, $ \sin(\angle OBH) \equal{} \frac {\sqrt {2415}}{176}$, and $ \sin(\angle HBC) \equal{} \frac {\sqrt {21}}{11}$.

Either $ \angle OBC \equal{} \angle OBH \plus{} \angle HBC$, or, $ \angle OBC \equal{} \angle OBH \minus{} \angle HBC$

If $ \angle OBC \equal{} \angle OBH \plus{} \angle HBC$, then:

$ \cos(\angle OBC) \equal{} \cos(\angle OBH \plus{} \angle HBC) \equal{} \frac {169}{176}\cdot\frac {10}{11} \minus{} \frac {\sqrt {2415}}{176}\cdot\frac {\sqrt {21}}{11}$ $ \equal{} \frac {1690 \minus{} 21\sqrt {115}}{1936} \approx 0.7566116161$

If $ \angle OBC \equal{} \angle OBH \minus{} \angle HBC$, then:

$ \cos(\angle OBC) \equal{} \cos(\angle OBH \minus{} \angle HBC) \equal{} \frac {169}{176}\cdot\frac {10}{11} \plus{} \frac {\sqrt {2415}}{176}\cdot\frac {\sqrt {21}}{11}$ $ \equal{} \frac {1690 \plus{} 21\sqrt {115}}{1936} \approx 0.9892561525$

However, since O is the circumcenter, $ CO \equal{} BO \equal{} 2$, and:
$ \cos(\angle OBC) \equal{} \frac {2^2 \plus{} 3^2 \minus{} 2^2}{2\cdot2\cdot3} \equal{} \frac {3}{4} \equal{} 0.75$

This is a contradiction.
[/hide]

[hide="Problem#2"]
Is $ A$ in degrees or radians? I'll assume degrees. Also, I'm going to guess that $ A \equal{} 60^\circ$ for the purpose of solving the problem.

Thus, $ m \equal{} 15$, and $ \Delta DEF$ is a $ 13 \minus{} 14 \minus{} 15$ triangle.

It is well known that the area is $ A \equal{} \frac {1}{2}\cdot12\cdot14 \equal{} 84$.

Thus, the circumradius of $ \Delta DEF$ is: $ R \equal{} \frac {abc}{4A} \equal{} \frac {(13)(14)(15)}{4(84)} \equal{} \frac {65}{8}$

Since $ \Delta DEF$ is non-obtuse, $ \Delta XYZ \cong \Delta DEF$

Proof: Since $ G$ is the circumcenter, $ G$ lies on the perpendicular bisectors of $ DE$, $ EF$, and $ FD$. So, $ G$ is reflected about the midpoints of these sides. Connect the midpoints of $ \Delta DEF$. Clearly, the triangle formed by the midpoints is similar to $ \Delta DEF$ by a factor of $ \frac {1}{2}$. So, reflecting $ G$ about each midpoint will yield a point that is twice the distance from $ G$ as the midpoint was. Since $ \Delta XYZ$ and the triangle formed by the midpoints are homothetic about $ G$, $ \Delta XYZ \cong \Delta DEF$

Since $ \Delta XYZ \cong \Delta DEF$, the circumradius of $ \Delta XYZ$ is also $ \frac {65}{8}$.
[/hide]

[hide="#1?"]Connect $ OC$. We have $ OC \equal{} OB \equal{} 2$ and $ BC \equal{} 3$. Note $ \angle BOC \equal{} \angle BAC \plus{} \angle ABO \plus{} \angle ACO$. Since $ OB \equal{} OA \equal{} OC$, we have $ \angle ABO \equal{} \angle AOB$ and $ \angle ACO \equal{} \angle CAO$, so $ \angle BOC \equal{} 2\angle BAC$. Then we can calculate $ \angle BAC \equal{} \sin ^{ \minus{} 1} (3/4)$.

This doesn't use the orthocenter condition at all...
[/hide]

[hide="#2"]Let the reflection over side $ a, b, c$ be $ A', B', C'$ respectively. Note that $ AOBC'$ is a rhombus since $ AC'\equal{}AO\equal{}OB\equal{}BC'$, so $ BC'//AO$ and $ BC'\equal{}AO$. Similarly $ CB'//AO$ and $ CB'\equal{}AO$. Then $ CB'//BC'$ and $ CB'\equal{}BC$, so $ BC'B'C$ is a parallelogram, so $ B'C'\equal{}BC$. Similarly, $ A'B'\equal{}AB$, $ C'A'\equal{}CA$. Therefore $ ABC$ is congruent to $ A'B'C'$, so they have the same circumradius. We can calculate this by $ K\equal{}abc/(4R)$.[/hide]

Sorry  :blush: for hastily posting numbers without checking the trigonometry part. The main motivation for the first problem was the cyclicness of quadrilateral BCOH.

Once u know BCOH is cyclic quad then angle A is indeed 60 degrees as said by Xantos.
This follows at once from the fact that angle $ BOC$ is $ 2A$ and $ BHC$ is $ 180\minus{}A$. For cyclicness $ 2A \equal{} 180\minus{}A$. Implying $ A$ is 60 degrees.

Sorry Xantos and timwu for the wrong measurements.


The empty set is closed and otherwise every sequence has a constant sequence as a subsequence.

What definition of compactness are you using?

That's not a correct definition for general metric spaces; it's true for subspaces of $ \mathbb{R}^n$ though.

See http://en.wikipedia.org/wiki/Heine-borel_theorem

[quote="pardesi"][quote="hebby"]Let (X, d) be a metric space.

Show that any finite subset is compact.[/quote]
Let $ U_{i}$ be a cover for $ P \subset X$ where $ P$ is finite.Then $ p_{i}$ belongs to some $ U_{i}$ for some $ i$ If there are two of them lets choose by some pre-defined agreement. Consider the collection $ U_{i}$ which is now a finite cover for the set.[/quote]

Being a metric space isn't needed. In any topological space the same is true with the same proof.


Ring Theory
superior algebra
superior algebra unsolved

Let $ R$ be a commutative ring with 1 and $ S$ a subset closed under multiplication, not containing 0.

Prove that there exists a prime ideal $ I$ s.t. $ I \cap S \equal{} \phi$

Since S does not contain 0, the zero ideal does not contain any element from S. Then your other exercise implies that there exists a prime ideal that misses S. We are done.

The symbol you're looking for is $ \emptyset$.


Determine all functions $ f: \mathbb{R} \mapsto \mathbb{R}$ with $ x,y \in \mathbb{R}$ such that

\[ f(x \minus{} f(y)) \equal{} f(x\plus{}y) \plus{} f(y)\]

Are we allowed, in these sorts of competitions, to assume that the function is continuous?

No... It is written "find all functions". Did you find weird functions? I found...

We take an additive function $ f$.

$ f(x \minus{} f(y)) \equal{} f(x \plus{} y) \plus{} f(y) \Longleftrightarrow f(f(y) \plus{} 2y) \equal{} 0$

Define $ f$ for a basis $ B$ of the reals over the rationals suffices. But we can take randomly $ f(b) \equal{} \minus{} 2b$ or $ f(b) \equal{} 0$, for $ b \in B$... The condition will be respected in both cases (if $ f(b) \equal{} 0$ then $ f(f(b) \plus{} 2b) \equal{} f(2b) \equal{} f(b) \plus{} f(b) \equal{} 0$ and if $ f(b) \equal{} \minus{} 2b$ we get $ f(0) \equal{} 0$).

Is there a mistake?

Anyway, I proved $ f$ is additive.

[hide="Proof"]If $ x \minus{} f(y) \equal{} y$ then $ f(x \plus{} y) \equal{} 0$. In particular, there is $ t$ such that $ f(t) \equal{} 0$. Setting $ x : \equal{} 0$ and $ y : \equal{} t$ in the main equation we get

$ f( \minus{} f(t)) \equal{} 2f(t) \Rightarrow f(0) \equal{} 0$

For $ x \equal{} \minus{} y$ the (main) equation is

$ f( \minus{} y \minus{} f(y)) \equal{} f(0) \plus{} f(y) \equal{} f(y)$

Then it is a good idea to put $ y : \equal{} \minus{} y \minus{} f(y)$ (in the main equation).

$ f(x \minus{} f(y)) \equal{} f(x \minus{} y \minus{} f(y)) \plus{} f(y)$

We know that $ f(x \minus{} f(y)) \equal{} f(x \plus{} y) \plus{} f(y)$. Thus, $ f(x \minus{} y \minus{} f(y)) \equal{} f(x \plus{} y)$. In other words, $ 2y \plus{} f(y)$ is a period of $ f$ for any $ y$, which implies $ f(x \minus{} f(y)) \equal{} f(x \plus{} 2y)$. Using this the main equation becomes

$ f(x \plus{} 2y) \equal{} f(x \plus{} y) \plus{} f(y)$[/hide]

Let x=f(y), y=0 then f(f(0))=0
Let y=f(0), x=0 then f(0)=f(f(0))=0
Let x=f(x), y=x then 0=f(f(x)+x)+f(x), or x=f(f(x)+x)+f(x)+x
Let g(x)=f(x)+x, then x=g(g(x))
So g is a function whose inverse is itself, and f(x)=g(x)-x. But f(0)=g(0)=0. Can anyone else finish this?

[quote="feliz"]

Define $ f$ for a basis $ B$ of the reals over the rationals suffices. But we can take randomly $ f(b) \equal{} \minus{} 2b$ or $ f(b) \equal{} 0$, for $ b \in B$... The condition will be respected in both cases (if $ f(b) \equal{} 0$ then $ f(f(b) \plus{} 2b) \equal{} f(2b) \equal{} f(b) \plus{} f(b) \equal{} 0$ and if $ f(b) \equal{} \minus{} 2b$ we get $ f(0) \equal{} 0$).

[/quote]

Not completely randomly. Suppose $ f(b) \equal{} 0$, for $ b \ne 0$. Then set $ y \equal{} b$:

$ f(x \minus{} f(b)) \equal{} f(x \plus{} b) \plus{} f(b)$
$ f(x) \equal{} f(x \plus{} b)$

Then suppose $ f(a) \equal{} \minus{} 2a$ for some $ a \ne 0$. We have that $ f(a) \equal{} f(a \plus{} b) \equal{} f(a \plus{} 2b) \equal{} \minus{} 2a$. Now set $ y \equal{} b$ and $ x \equal{} a \plus{} b$:

$ f(a \plus{} b \minus{} f(b)) \equal{} f(a \plus{} 2b) \plus{} f(b)$
$ f(a \plus{} b) \equal{} f(a \plus{} 2b) \plus{} f(b)$
$ \minus{} 2a \equal{} \minus{} 4a$, contradiction because $ a \ne 0$.

So the only functions should be $ f(x) \equal{} 0$ and $ f(x) \equal{} \minus{} 2x$ for all $ x$ (but we are still disregarding non-continuous solutions to Cauchy's equation in the proof).

@xxp: I noticed it after I posted, but I didn't have time to correct it. I shouldn't post when I'm sleepy, really... But if it is possible to prove that $ f$ is additive, you don't actually need the $ g$ function.

ffao,
$ f(a\plus{}2b)\plus{}f(b)\equal{}\minus{}2a\plus{}0\equal{}f(a\plus{}b)$. You will not get contradiction!

This problem is classic. Define $ g(x)\equal{}x\plus{}f(x)$. We will end up with
$ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ and $ g(g(x))\equal{}x$.

[quote="xxp2000"]ffao,
$ f(a \plus{} 2b) \plus{} f(b) \equal{} \minus{} 2a \plus{} 0 \equal{} f(a \plus{} b)$. You will not get contradiction!

This problem is classic. Define $ g(x) \equal{} x \plus{} f(x)$. We will end up with
$ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ and $ g(g(x)) \equal{} x$.[/quote]
Can you explain how to solve the equation $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ and $ g(g(x))\equal{}x$.I cannot solve this :maybe: .Thanks.

[quote="SUPERMAN2"]Can you explain how to solve the equation $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ and $ g(g(x)) \equal{} x$.I cannot solve this :maybe: .Thanks.[/quote]here: http://www.mathlinks.ro/viewtopic.php?t=313709[quote="pco"]so all solutions are involutary solutions of Cauchy's equation. [/quote]

Let $h(x)=x+f(x)$ and $z=x+y$ then $h(z-h(y))=h(z)-y.$ Clearly $h$ is bijective and so $h(0)=0.$ Setting $z=h(y)$ we get $h(h(y))=y$ and now setting $y=h(y)$ we get $h(x-y)=h(x)-h(y),$ so $h$ is additive. If $h$ is continuous then, $f(x)=cx-x,$ which works.


Prove that there are infinitely many solutions for the equation

$ w^{2} \plus{} x^{3} \plus{} y^{4} \plus{} z^{5} \equal{} h^{6}$

if

w,x,y,z,h $ \in \mathbb{N}$

First find one solution like (2,3,1,2,2).

Now it's easy:

$ (w,x,y,z,h) \equal{} (2k^{30},3k^{20},k^{15},2k^{12},2k^{10})$ for $ k\in\mathbb{N}$ gives infinitely many solutions.


Determine whether the given vectors span R3:

v1=(3, 1, 4); v2=(2, -3, 5); v3=(5, -2, 9); v4=(1, 4, -1)

Now, I have no trouble showing whether or not three vectors span R3, or two vectors spanning R2, etc...but the four unknowns in three equations is giving me trouble, since the coefficient matrix is not square and thus I can't check it by verifying the determinant...can someone walk me through this? Thanks a lot.

[quote="Xevarion"]Obviously, if those four vectors span $R^3$, one of them is redundant. You can do the four 3x3 determinants that arise from ignoring each of the vectors in turn. This probably isn't the most efficient way, but it'll work.[/quote]

That's what I was thinking, but I didn't want to take the time to do four different 3x3 determinants only to find out that you can't even do it that way. Thanks, I'll try it later. In the meanwhile, if anyone knows of a quicker way, please post.

[quote="Ramanujan"]Use Gauss-Jordan Elimination.

 :)[/quote]

So would the matrix look like this:

3  2  5  1  b1
1 -3 -2  4  b2
4  5  9  -1 b3

?

And what am I looking for? Sorry if I'm a little slow, but I missed a few classes near the end.

I deleted the previous post because i wanted to explain you what i mean but you was faster.
Here it is:

I write $a =(3,1,4) \, , b=(2,-3,5) \, , c=(5,-2,9) \, ,d =(1,4,-1)$ .

$\\ \left(\begin{array}{ccc|c} 3&1&4&a\\\ 2&-3&5&b\\\ 5&-2&9&c\\\ 1&4&-1&d\end{array}\right) \sim \left(\begin{array}{ccc|c} 0&-11&7&a-3d\\\ 0&-11&7&b-2d\\\ 0&-22&14&c-5d\\\ 1&4&-1&d\end{array}\right) \sim \left(\begin{array}{ccc|c} 0&-11&7&a-3d\\\ 0&0&0&b+d-a\\\ 0&0&0&c+d-2a\\\ 1&4&-1&d\end{array}\right) \\ \\ \\ \sim \left(\begin{array}{ccc|c}1&4&-1&d\\\ 0&-11&7&a-3d\\\ 0&0&0&b+d-a\\\ 0&0&0&c+d-2a \end{array}\right)$

The vectors $d, \, a-3d$ form a basis for $S (a,b,c,d)$ and $dimS =2$ therefore the original vectors does not span $\mathbb{R}^3$ .

I think it is ok now  :)

[b]Edit[/b] : a different way to see that: $b=a-d$ and $c=2a-d$ therefore $b,c \in S (a,b,c,d)$ therefore $S (a,b,c,d)=S(a,d)$
Also $a,d$ are linearly independent. Check the computations yourself. Maybe there are errors (i corrected one i found ) .

[quote="jpps1"]And what am I looking for? [/quote]
That can be stated fairly simply: you're looking for a pivot in every row.

Consider your set of vectors to be column vectors, make a matrix out of them, and do Gaussian elimination.

If you get a pivot in every column, the set of vectors is independent. (That was clearly impossible here: a $3\times 4$ matrix can have at most 3 pivots, so it can't have one in every column - the vectors must be dependent.)

If you get a pivot in every row, then the matrices span the space they're contained in. Knowing you had four vectors in a three dimsensional space, this was certainly possible, even likely - but there's no way to know for sure without actually doing the work. ramanujan performed the Gaussian elimination (and I'm suggesting that he didn't have to do the divided-off last column), and found only two pivots. (In the last matrix he wrote, the $1$ in the (1,1) position and the $-11$ in the (2,2) position are pivots.) So these vectors span only a two-dimensional subspace of $\mathbb{R}^3,$ not the whole thing.

The exact details of the work in ramanujan's post and the row-echelon form he ended with are not unique; another person's steps would likely look different. But anyone doing it correctly should find two pivots.

[quote="Xevarion"]This [using determinants] probably isn't the most efficient way ... [/quote]
Yup. Not efficient at all. Let's not even mention determinants.


Hey guys,

I already own Mathematical Olympiad Challenges, but then, a friend told me that maybe Math Olympiad Treasures was better.  Can anybody here compare the two?  Which do you think I should study from first?  Should I do all the problems in Math Olympiad Challenges and then move to Treasures, or should I switch off?  Please help.

Math Olympiad Treasures is from first then Mathematical Olympiad Challenges, i think so.

treasure is much easier than challenge, both of them are good for national olympiads


Warum sieht man in dem Spiegel links und rechts verkehrt, aba nicht oben und unten??

weil was wir im Spiegel sehen, unsere Spiegelbild ist.....hast du lust zu spammen oder was? :huh:

Ja, unser spiegelbild ist links und rechts verkehrt.

Das war eine absolut ernstgemeinte Frage. Es hat an sich weniger mit Mathematik zu tun, aber warum koennen/sollen Mathematiker nicht trotzdem auch sich manchen Fragen stellen, die realitaetsnaeher sind.

Vielleicht koennt ihr mir sagen. Ihr seid die Jugend in Deutschland, die spaeter an der Spitze der mathematischen Wissenschaft stehen werdet. Was ist Mathematik ueberhaupt? Die Wissenschaftler der anderen Branchen sagen, Mathematik sei nur Werkzeug. Wenn das Rechnen nur Werkzeug sei, dann gibt es keine Mathematik, oder? Vielleicht koennt ihr mir sagen, was die Mathematik ausmacht, was es ist.

Tut mir leid, wenn es etwas protestierend klingt. Es ist nicht so gemeint. Mag mich nur etwas informieren.

Danke.

nein du hast recht....und ich moechte mich entshuldigung wenn du gedacht hast dass ich dir nicht antowrten wollte...ich dachte nur dass es noch spam war.....
ich denke auch dass wir sollten mehrere Fragen stellen..

willkommen in mathlinks  :lol: und
entschuldigung für mein deutsch :oops:

Ich denke dass der Spiegel nichts verdreht sondern es uns nur so vorkommt.
Man hebt die rechte Hand und das Spiegelbild hebt die (für uns) linke Hand (vllt meinst du das). Außerdem woher sollte denn der Spiegel wissen wo oben und unten ist?  :wink:

Physisch vertauscht der Spiegel vor und hinten, also nicht links-rechts oder oben-unten.
Psychisch vertauscht der Spiegel links und rechts.
Links-rechts nimmt tatsächlich eine besondere Lage ein. Links-rechts ist abgeleitet von den übrigen Richtungen.
An den anderen Richtungen können wir Objecte verbinden (Oben->Kopf; Unten->Füße; Vor->Nase; Hinten->Rücken, Links->???; Rechts->???).
Wenn eine Koordinatachse die Richtung wechselt, ändert sich ein rechtshändiges Koordinatensystem in ein linkshändiges.
Wenn eine der drei Achsen die Richtung wechselt, dann ist es nachher nicht mehr möglich herauszubekommen welche Achse sich die Richtung geändert hat.
In einem Spiegel wird die Achse vor-hinten geändert in hinten-vor. Man merkt zunächst nichts denn die Richtungen vor-hinten und oben-unten sind verbunden mit Objecten die im Spiegel genau so aussehen wie im Original (Nase, Rücken,Kopf,Füße).
Die abstrakte unsichtbare (das heisst, nicht mit Objecten assoziierte) Richtung links-rechts bekommt dann die Schult. Ein psychologischer Effekt.


limit
real analysis
real analysis unsolved

Hi!

compute
\[ \lim_{n\to +\infty}\sum_{k=1}^n\left({k\over n}\right)^n \]

it seems it was posted long time ago, mathlinkers solvers found e/(e-1)

Yes $\frac e {e-1}$ is the answer.

Using the fact that $\left (1+ \frac 1 x\right)^{x+1}\ge e,\forall x>1$ we get that the limit is less than
$\sum \frac 1 {e^k}=\frac e {e-1}$
Then we have : $\lim \sum_{k=1}^n \left ( 1-\frac k n\right)^n\ge\lim \sum_{k=1}^N \left ( 1-\frac k n\right)^n=\frac{1}{e^1}+\frac{1}{e^2}+\frac{1}{e^3}+...+\frac{1}{e^N},\forall N\in \mathbb{N}.$
Hence $\lim \sum_{k=1}^n \left ( 1-\frac k n\right)^n\ge \sup_{N \in \mathbb{N}} \frac{1}{e^1}+\frac{1}{e^2}+\frac{1}{e^3}+...+\frac{1}{e^N}=\frac e {e-1}$

Nice problem :lol:

Try this one too:

\[ \lim_{n \to \infty } \frac{1} n \sum_{k=1}^n\left(\frac k n\right)^{\frac n k} \]
it is simple and nice.


If $ a,b,c>0$ and distinct, then

$ \frac{a^2\plus{}b^2}{(b\minus{}c)^2}\plus{}\frac{b^2\plus{}c^2}{(c\minus{}a)^2}\plus{}\frac{c^2\plus{}a^2}{(a\minus{}b)^2}\geq5\minus{}\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}$


calculus
integration
trigonometry
function
derivative
logarithms
calculus computations

Did you make these up yourself? From a cursory glance it looks like a bunch of substitution problems.

I skipped from Regular Calculus to Calc BC, so some content I covered on my own. This is a worksheet my teacher gave me for practice. I'm pretty sure I can do them all, but I think I need some assisstance. This was Regular Calc content - I didn't practice it for quite a while

#1, 4 and 5: $ u \equal{} ln(x)$
#6 and 7: $ u \equal{} e^{x}$
#8: $ u \equal{} \cos(x)$
#9: re-write the problem as $ \int \sec{x}\csc{x} dx$ and use integration by parts?

Try and get used to seeing certain functions accompanied by their derivatives (eg/ $ \ln(x)$ and $ 1/x$) so you can quickly figure out the applicable u-substitution.

[hide="Number 9"]$ \int \frac{\tan x}{\sin^2x}\,dx \equal{} \int \frac{1}{\sin x \cos x}\,dx \equal{} 2 \int\frac{1}{\sin(2x)}\,dx \equal{} 2 \int \csc(2x)\,dx$

$ \equal{} \ln|\tan x| \plus{} C$.

Another way to see this:

$ \int \frac{\tan x}{\sin^2x}\,dx \equal{} \int \frac{1}{\sin x \cos x}\,dx \equal{} \int \frac{\frac{1}{\cos^2x}}{\frac{\sin x}{\cos x}}\,dx \equal{} \int \frac{\sec^2x}{\tan x}\,dx \equal{} \ln|\tan x| \plus{} C$.[/hide]

[hide="Number 7"]Because $ (1\plus{}e^x)' \equal{} e^x$ we have

$ \int_{0}^{\ln 3} \frac {e^{x}}{1 \plus{} e^{x}}\,dx \equal{} [\ln(1\plus{}e^x)]_0^{\ln 3} \equal{} \ln 4 \minus{} \ln 2 \equal{} \ln 2.$[/hide]

[hide="Number 6"]Here we have a similar case: $ (e^x\plus{}2)' \equal{} e^x$. Therefore:

$ \int \frac{e^x}{(e^x\plus{}2)^2}\,dx \equal{} \int e^x(e^x\plus{}2)^{\minus{}2}\,dx \equal{} \frac{(e^x\plus{}2)^{\minus{}1}}{\minus{}1} \plus{} C \equal{} \minus{} \frac{1}{e^x\plus{}2} \plus{} C$.[/hide]

[hide="Number 5"]Observe that $ [(\ln x)^2]' \equal{} 2 \ln x \cdot \frac{1}{x} \equal{} 2 \frac{\ln x}{x}$. With this in mind,

$ \int \frac {ln x}{x (1 \plus{} (ln x)^{2})}\,dx \equal{} \frac12 \int \frac{2 \frac{\ln x}{x}}{1 \plus{} (\ln x)^2}\,dx \equal{} \frac12 \ln[1 \plus{} (\ln x)^2] \plus{} C$.[/hide]

[hide="Number 4"]$ \int \frac{1}{x \sqrt{\ln x}}\,dx \equal{} \int \frac1x (\ln x)^{\minus{}1/2}\,dx \equal{} \frac{(\ln x)^{1/2}}{1/2} \plus{} C \equal{} 2 \sqrt{\ln x} \plus{} C$.[/hide]

Thank you, I've got everything right now (it all checks out).

Thanks for your help.

onlyrevolutions:

I understand that this topic is "finished" in the sense that it's unlikely to generate any more posts. But our practices around here are that we do not delete topics that have drawn answers - and please don't edit away the original questions. Just leave it as you wrote it.

We also understand that this was a classroom topic and problems similar to these might well be on this week's Calculus II exam. That's OK - that's permitted here.


In $ \triangle{ABC}$, $ \angle{A}\equal{}30^o$, $ \angle{B}\equal{}50^o$. $ P$ is inside the triangle and $ \angle{PAB}\equal{}20^o$, $ \angle{PCA}\equal{}40^o$.

Find $ \angle{PBA}$.

[geogebra]a5bab4a3c0e27999d1183a00cc08b329e0b965ff[/geogebra]

By Ceva's theorem (with help of my calculator), I got $ \boxed{30^{\circ}}$.


Let $ a_1,a_2,\ldots, a_n$ $ (n \ge 4)$ be nonnegative real numbers, not all are zero. Prove that
$ (n\minus{}3) \sum^n_{i\equal{}1} a^3_i \plus{}\frac{2n^{n\minus{}2}a_1a_2\cdots a_n}{\left( \sum^n_{i\equal{}1} a_i\right)^{n\minus{}3}} \ge \frac{6}{n\minus{}2} \sum_{1\le i <j<k \le n} a_ia_ja_k.$
Equality holds for $ a_1\equal{}a_2\equal{}\cdots \equal{}a_n$ and again for $ a_1\equal{}a_2\equal{}\cdots \equal{}a_{n\minus{}1},a_n\equal{}0$ up to the permutations. :)

[quote="Vasc"][quote="can_hang2007"]Let $ a_1,a_2,\ldots, a_n$ $ (n \ge 4)$ be nonnegative real numbers, not all are zero. Prove that
$ (n \minus{} 3) \sum^n_{i \equal{} 1} a^3_i \plus{} \frac {2n^{n \minus{} 2}a_1a_2\cdots a_n}{\left( \sum^n_{i \equal{} 1} a_i\right)^{n \minus{} 3}} \ge \frac {6}{n \minus{} 2} \sum_{1\le i < j < k \le n} a_ia_ja_k.$
Equality holds for $ a_1 \equal{} a_2 \equal{} \cdots \equal{} a_n$ and again for $ a_1 \equal{} a_2 \equal{} \cdots \equal{} a_{n \minus{} 1},a_n \equal{} 0$ up to the permutations. :)[/quote]
Since
$ \sum_{1\le i < j < k \le n} a_ia_ja_k \equal{} (\sum a_1)^3 \minus{} 3(\sum a_1)(\sum a_1^2) \plus{} 2\sum a_1^3$,
we can apply EV-Theorem:
If $ \sum a_1 \equal{} constant$ and $ a_1a_2...a_n \equal{} constant$, then the sums $ \sum a_1^2$ and $ \sum a_1^3$ are minimal for $ a_1\le a_2 \equal{} ... \equal{} a_n$.
Therefore, it suffices to prove the inequality for $ a_2 \equal{} ... \equal{} a_n$.[/quote]
Very nice, Vasc. :) I also used EV-theorem to prove the inequality for $ n$ variables. But for $ n\equal{}4$, there is a nice solution without using EV. ;) :)


calculus
function
ARML
trigonometry
parameterization
parametric equation

I don't know if this belongs here, but it's PREcalc instead of calculus, so...

Given a parametric equation, how would you find where the curve intersects itself? That is, given x=f(t) and y=g(t), when you graph the function in x and y, how would you find the ordered pair (x, y) that satisfies two values of t?

This was on a certain ARML and I'd like to know how to do it before I take that test. My friend says it's so obvious it's trivial, but I seem to be missing something.

well, if the function is periodic like the sine or cosine function, then $f(t) = f(t + 360)$
or if you had some absolute values
basically you would need to find some a and b such that $f(a) = f(b)$, and $g(a) = g(b)$

My friend says even that isn't necessary. Is he bluffing?

On a side note... I noticed that there was a ME in team long ago, but not that there's no U. Nice. ;)


limit
Gauss
real analysis
real analysis unsolved

[quote="flip2004"][b]Find [/b]             $\lim\limits_{n\to \infty}n^2\left(\displaystyle \frac{{16}^n}{\pi \displaystyle {2n \choose n}^2} - \frac{32n^2+8n+1}{32n}\right)$ .[/quote]

Could you please check the problem, Flip2004, because it is strange i've got $\infty$.
Using Stirling's formula we can get ${\displaystyle {2n \choose n}^2} \sim \frac{1}{\pi n} 16^n$, so
$\lim\limits_{n\to \infty}n^2\left(\displaystyle \frac{{16}^n}{\pi \displaystyle {2n \choose n}^2} - \frac{32n^2+8n+1}{32n}\right)= =lim_{n \to \infty} n^2( n-\frac{32n^2+8n+1}{32n} )=\infty$.

[quote="eugene"] ... Could you please check the problem, Flip2004, because it is strange i've got $\infty$.
[/quote]    Hi eugene, thanks for interest. After me, the limit is finite.

[quote="Moubinool"]I also found an infinite limit[/quote]
[b] Solution by proposer.[/b]
Let us denote
\[ (1)\; \; \; \; \left\{\begin{array}{ccl} (z)_k&: =&z(z+1)\cdots(z+k-1)\; \; , (z)_0: =1\; \; ,\; \; z\in {\mathbb C}\; ,\; k\in {\mathbb N}^* \\ &&\\ {}_2F_1\left(a,b;c;z\right)&: =&\displaystyle \sum\limits_{k=0}^{\infty}\frac{(a)_k(b)_k}{(c)_k}\frac{z^k}{k!} \end{array}\right.  \]
According to Gauss
\[ (2)\; \; \; \; {}_2F_1(a,b;c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\; \; ,\; c-a-b>0 \; .  \]
If in (2) we select $a=b=-\; \frac{1}{2} \; , \; c=n ,$ one finds  (see also (1))
\[ \frac{{16}^n}{\pi {2n \choose n}^2}=n\sum\limits_{k=0}^{\infty}\frac{(-1/2)_k(-1/2)_k}{(n)_k}\cdot\frac{1}{k!}=  \]
\[ =n+\frac{1}{4}+\frac{1}{32(n+1)} +\frac{3}{128(n+1)(n+2)}+\frac{75}{2048(n+1)(n+2)(n+3)}+\ldots  \]
Therefore
\[ \begin{array}{c} \lim\limits_{n\to \infty}n^2\left(\displaystyle \frac{{16}^n}{\pi \displaystyle {2n \choose n}^2} - \frac{32n^2+8n+1}{32n}\right)=\\ = \lim\limits_{n\to \infty}n^2\left( n+\frac{1}{4}+\frac{1}{32(n+1)} +\frac{3}{128(n+1)(n+2)}+\frac{75}{2048(n+1)(n+2)(n+3)}+\ldots - n - \frac{1}{4}-\frac{1}{32n} \right) =\\ =\lim\limits_{n\to \infty}\left[n^2\left(-\frac{1}{32n(n+1)}+ \frac{3}{128(n+1)(n+2)} \right)+{\mathcal O}\left(\frac{1}{n}\right)\right]=-\; \frac{1}{128} \end{array}  \]
(sorry for possible errors !)


Let $f\in \mathbb Z[X]$ with $\deg f$ odd. Is it possible to find and interval $I\subset\mathbb R$ s.t. $f(I)\subseteq I$?

Yes, if you consider that a single point is also an interval.  :D

:) thx harazi :) but I was thinking to a larger interval ... for example one which contains an integer point ... but... now I tihnk that it does not exist such an $I$ only for some polynomials :(


Discuss what you play. or at least answer the description question.

May I make a suggestion? Don't post 3 of the same topic. One is enough.
On topic: I used to play WoW and before that Star Wars Galaxies. Then, high school came.

Yes, I love playing with My Massively Overpowered Rocket-Propelled Grenades!  :P

[/pointless post]

And I think he probably just clicked "submit" three times by accident because it wasn't loading or something.

Ugh. I don't even want to start discussing my RS addiction.

I'm more of a first-person shooter guy; but I've played some RPGs.

Most notably, I've spent quite a portion of my middle school life on RuneScape.

I've also played Fable II and Oblivion on the Xbox 360.

hmmm...

I was addicted to Runescape in 5th Grade... (played 3-5 hours each weekend)

[quote="zephyredx"]Maple Story. It was a terribly addicting time. But likewise, high school came  :(  (I actually quit before high school because there were too many hackers, and too much lag)[/quote]
There are too many hackers, and i suck.

[quote="andersonw"][quote="1=2"][quote="nick42"]KOL[/quote]

so fun.[/quote]

Nethack>KOL[/quote]

[url=http://www.animecubed.com/billy/?Gary]BvS[/url]>Nethack>KoL

[quote="Garyzx"][/quote][quote="andersonw"][quote="1=2"][quote="nick42"]KOL[/quote]

so fun.[/quote]

Nethack>KOL[/quote][quote="Garyzx"]

[url=http://www.animecubed.com/billy/?Gary]BvS[/url]>Nethack>KoL[/quote]

AoPS Mafia !>BvS>Nethack>KoL

Uh oh it is starting to look like a quote stack

Wait...does !> mean not > or does it mean >?

>.

! for emphasis. More like:

AoPS Mafia >> BvS>Nethack>KoL  :)

[quote="james4l"][/quote][quote="Garyzx"][/quote][quote="andersonw"][quote="1=2"][quote="nick42"]KOL[/quote]

so fun.[/quote]

Nethack>KOL[/quote][quote="Garyzx"]

[url=http://www.animecubed.com/billy/?Gary]BvS[/url]>Nethack>KoL[/quote][quote="james4l"]

AoPS Mafia !>BvS>Nethack>KoL

Uh oh it is starting to look like a quote stack[/quote]

KOL>AoPS Mafia !>BvS>Nethack>KoL

Yeah it is.

In KoL, you can't morph into a dragon, lay eggs, morph back, and use those eggs as grenades later.

Nethack is free.
BvS... don't know about that
AoPS MAFIA is free (long wait time)...

Nethack>KOL>AoPS Mafia !>BvS>Nethack>KoL

Nethack is cool.

Nethack>Nethack???

james4l, your logic is flawed.

Nethack>KOL<AoPS Mafia !>BvS>Nethack>KoL

That works, right?

Is this the new PWNAGE game¿ Otherwise,please stop just posting orderings of MMORPGs and [b]discuss[/b] them (more seriously than "I suck").

:rotfl:

KoL is FREE.
Membership is FREE.
Donation is OPTIONAL.
Humor is EVERYWHERE.
Clans are NUMEROUS.
Opportunities are VAST.
Quests are INTERESTING.
This game is POSSIBLE.
PvP is ALLOWED.
This game is for TEENAGERS AT LEAST.
This game is AWESOME.

[quote="1=2"]:rotfl:

KoL is FREE.
Membership is FREE.
Donation is OPTIONAL.
Humor is EVERYWHERE
Clans are NUMEROUS.
Opportunities are VAST.
Quests are INTERESTING.
This game is POSSIBLE.
PvP is ALLOWED.
This game is for TEENAGERS AT LEAST.
This game is AWESOME.[/quote]
Nethack is FREE. In fact, NetHack is OPEN SOURCE (although it is not in public domain).
You don't NEED membership.
Donations will NOT help you in the game (unlike KoL)
Humor is EVERYWHERE. (in fact, a bunch of KoL humor was taken from NetHack, although I have to admit KoL is funnier)
No clans.
Opportunities are VASTER.  :P
Quests are INTERESTING.
This game is POSSIBLE WITH SPOILERS (1=2: Have you ascended without using spoilers? If so, I congratulate you)
No PvP, not an online game. Oh wait, that means it's not an MMORPG. Hmmm....
This game is AWESOME(r).

I should also add that the concepts of ascensions, restrictions, and turns come from NetHack.

And, I have played KoL before and thought that it was okay, and I got one ascension (with spoilers, of course), before I decided it was getting too repetitive, and my account was deleted for inactivity.

About Runescape, I think this game is pretty pointless (who names a game "run escape" anyways?). There isn't really a goal in this game, either. The main point is to kill monsters, fulfill quests (which I should also say, are pretty hard without spoilers), and beat everyone else. At least there is an actual goal in both KoL and Nethack (ascend). I used to play runescape too, but got pretty bored of it (yay, I cut down a billion trees, so now I can cut more different kinds of trees!).
Oh, and I know the actual name is "rune scape", I like to make fun of it by calling it "run escape".

I'm in Cir Senam's Provocable Nice Guys.

I just started KOL, 1=2, and I had goofballs.  :blush: Now I have Goofball Withdrawal, which sucks.  :cursing:

What are quest#'s??? Oh yeah, I HATE GOOFBALLS. I just got out of Mt.Noob. I=Seal Clubber (does this suck?)

I ate mine and I'ma not get any more.
Quests:
The Goblin Who Wouldn't Be King
Cryptic Emanations
Me and My Nemesis
Angry oneequalstwo, this is Azazel in Hell.
A Bounty Hunter Is You!

oneequalstwo
Level 7
Macaroni Magician

[quote="eggylv999"]What are quest#'s?[/quote]
It's the level at which the council will start to give you the quest.

[quote="andersonw"][quote="1=2"]I'm gonna try to ascend without spoilers.[/quote]

Good luck. Which quest are you on right now?

I also have no ideas what quests 12 and 13 are like, since I ascended before they changed it from 11 quests to 13 quests.[/quote]

Quest 13 is the old quest 11.  Quests 11 and 12 are both really, REALLY long.  The level 11 quest involves 5 mini-quests tied together by a MacGuffin.  The level 12 one has a lot of grinding.

1=2, how many ascensions have you done?

[quote="Garyzx"]Quest 13 is the old quest 11.  Quests 11 and 12 are both really, REALLY long.  The level 11 quest involves 5 mini-quests tied together by a MacGuffin.  The level 12 one has a lot of grinding.
[/quote]

That doesn't sound very fun.

And apparently, my friend in school just finished an oxycore ascension :O

so wait...who plays RS again?  I'm just curious cuz i play it sometimes if im really bored.  its so addicting.  yeah it sounds like i have no life playing RS but i cant help it.

Dead, no need to revive it.  You thread necromancer :P

[quote="coolts"]actually i only posted that for 100 posts. Still, the topic continues.[/quote]
Do that again and your postcount my suddenly decreasy drastically  :wink:

I like the Artix Entertainment games.........

Dragonfable (RPG), Battleon (RPG), AQ Worlds (MMORPG), Mechquest (RPG)......

They are original, with good graphics....

AND, they add new stuff every Friday!


[quote="The Navigator"]How do I quickly compute
\[ 99^3 \plus{} 3\left(99^2\right) \plus{} 3\left(99\right) \plus{} 1
\]
[/quote]

We note that this is in the form

$ x^3\plus{}3x^2y\plus{}3xy^2\plus{}y^3\equal{}(x\plus{}y)^3$.

In this case, $ x\equal{}99$ and $ y\equal{}1$. So, we get $ 100^3\equal{}1000000$

yes, for most mathcounts computational problems there is a "trick"
this one is from 2006(or 2007?) states cd

what is the value of
(109^2-100^2)/9

well the numerator is equal to (109+100)(109-100)/9
=209(9)/9=209. So the moral of the story ( :P ) is to look for shortcuts.


group theory
abstract algebra
linear algebra
matrix
linear algebra unsolved

Let $ G$ be a subgroup of $ GL_n(K)$ such that every $ g \in G$ has a fixed vector. Does there exist a basis such that $ G$ is a subgroup of $ \begin{pmatrix}1 & *\\
0 & *\end{pmatrix}$ or $ \begin{pmatrix}1 & 0\\
* & *\end{pmatrix}$?

This cannot be true in general: Take $ G\equal{}SO(3)$, the group of real orthogonal $ 3\times 3$-matrices with determinant 1.

$ G$ is a subgroup of $ GL_2(K)$ s.t. if $ g\in{G}$ then $ g$ has an eigenvalue which is $ 1$.
The answer is yes.
Assume $ G\not\equal{}\{I_2\}$.
case 1: $ G$ admits a diagonalizable matrix $ A$ wich is not $ I_2$; then we may assume $ A\equal{}\begin{pmatrix}1&0\\0&\alpha\end{pmatrix}$ where $ \alpha\not\equal{}{1}$. If all matrices of $ G$ are diagonal then we are done; else let $ B\equal{}\begin{pmatrix}a&b\\c&d\end{pmatrix}\in{G}$ be a non diagonal matrix; necessarily $ ad\minus{}bc\equal{}a\plus{}d\minus{}1$. $ AB\in{G}$ gives $ (a\minus{}1)(\alpha\minus{}1)\equal{}0$; thus $ a\equal{}1,bc\equal{}0,d\not\equal{}0$.
For example $ c\equal{}0,b\not\equal{}0$ and $ B\equal{}\begin{pmatrix}1&b\\0&d\end{pmatrix}$. It remains to prove that if $ C\equal{}\begin{pmatrix}1&f\\g&h\end{pmatrix}\in{G}$ is not a diagonal matrix then $ C$ is in the preceding form. Necessarily $ fg\equal{}0,h\not\equal{}0$; assume $ f\equal{}0$; we show that $ g\equal{}0$. $ BC\equal{}\begin{pmatrix}1\plus{}bg&bh\\dg&dh\end{pmatrix}\in{G}$; thus $ bg\equal{}0$ and $ g\equal{}0$.  QED.
Case 2: Except $ I_2$, the matrices of $ G$ are not diagonalizable. Then we may assume $ A\equal{}\begin{pmatrix}1&1\\0&1\end{pmatrix}\in{G}$. Let $ B\equal{}\begin{pmatrix}a&b\\c&d\end{pmatrix}\in{G}$ where $ a\plus{}d\equal{}2$. $ AB\equal{}\begin{pmatrix}a\plus{}c&b\plus{}d\\c&d\end{pmatrix}\in{G}$; therefore $ a\plus{}c\plus{}d\equal{}2$ and $ c\equal{}0$.  QED.


How can I keep a marble in motion for 2 minutes?

I've had 2 suggestions so far:

*A simple way would be to put a marble in a bucket and move the bucket so the marble spins along the walls.

*Another idea is to have a teeter-totter, using a long plastic tube. Have a string attached to each side. Put he marble in one one of the tube and pull the other one down. Then alternate pulling an end down to keep the marble moving. Have and caps on the tube so the marble can't escape.

But, if i can only touch it once and that's to start if off. how can i do that way?

Your thoughts?


Let sequence $ (a_n)$ such that
$ a_0\equal{}a_1\equal{}a_2\equal{}1$
$ a_{n\plus{}2}\equal{}\frac{a_na_{n\plus{}1}\plus{}1}{a_{n\minus{}1}}$
Prove that $ a_n\in N^*$ with $ \forall n\in N$

we have
$ a_{n \plus{} 2}a_{n \minus{} 1} \minus{} a_na_{n \plus{} 1} \equal{} 1$
$ a_{n \plus{} 1}a_{n \minus{} 2} \minus{} a_{n \minus{} 1}a_n \equal{} 1$
$ \implies a_{n \plus{} 2}a_{n \minus{} 1} \minus{} a_na_{n \plus{} 1} \equal{} a_{n \plus{} 1}a_{n \minus{} 2} \minus{} a_{n \minus{} 1}a_n$
$ \implies \frac {a_{n \plus{} 2} \plus{} a_n}{a_{n \plus{} 1}} \equal{} \frac {a_{n} \plus{} a_{n \minus{} 2}}{a_{n \minus{} 1}} \equal{} k$
$ \implies k \equal{} (a_2 \plus{} a_0)/a_1 \equal{} 2$ so we have $ a_{n \plus{} 2} \equal{} 2a_{n \plus{} 1} \minus{} a_{n}$
we r done!
[b][u]Edit:[/u][/b]
many thanks to latios, :oops:
the formula $ a_{n \plus{} 2} \equal{} 2a_{n \plus{} 1} \minus{} a_{n}$
just holds for even $ n$ and for odd $ n$ we have $ k\equal{}(a_3\plus{}a_1)/a_2\equal{}3$ so we get $ a_{n\plus{}2}\equal{}3a_{n\plus{}1}\minus{}a_{n}$ for odd $ n$.

that formula only holds for even n. for odd n we have An+2=3An+1-An

Hi;

Recurrence can be reduced to:

a[n+4] = 4 a[n+2] - a[n]

Easy to see that these are always integers.


[quote="i_like_pie"]5 cows eat 2 acres of grass in 10 days. 7 cows eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. How many days will it take 16 cows to eat 7 acres of grass?[/quote]

this sort of doesn't make sense because it says:

5 cows eat 2 acres in 10 days

7 cows eat 3 acres in 30 days
therefor 7 cows eat 1 acre in 10 days...

5 cows eat 2 acres and 7 cows eat 1 acre at the smae pace?? :maybe:

You can't simplify like that becuase you're forgetting the grass regrowth rate.

oh. ok :D
even though the grass won't grow that fast :)

I got [hide="(probably wrong...)"]7750 days?[/hide]

[quote="i_like_pie"]5 cows eat 2 acres of grass in 10 days. 7 cows eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. How many days will it take 16 cows to eat 7 acres of grass?[/quote]

[hide="I think I messed up somewhere, but oh well...."]Ok. In 10 days, 1 cow will eat $2/5$ of an acre. In 30 days, 1 cow will eat $3/7$ of an acre, but if the grass didn't grow, we would expect the cow to eat $6/5$ of an acre. This means that in 20 days, the grass grows $(6/5-3/7)$ or 4 acres, or the grass grows 1 acre every 5 days. In 10 days, the grass grew 2 acres, so the 1 cow ate $12/5$ acres. This means that in 1 day, a cow will eat $6/25$ acres, but $1/5$ acre will grow back, soooo, in 1 day, 16 cows will eat $96/25$ acres and $5/25$ acres will grow back, so each day, $91/25$ acres are eaten. We need 7 acres, or $175/25$ eaten, so the cows will take a total of $25/13$ days to eat the grass.[/hide]

edit: something weird happened there, hope I fixed it.

Both answers are incorrect.

Pianoforte - how did you get 7750 days?

[hide="New Answer- Is it... "]70?[/hide]
[hide="Solution"]Let x be the rate at which 1 cow eats in 1 day, y the rate at which 1 acre of grass grows in 1 day, A be the initial ammount of grass per acre, and D be the number of days we're looking for.

For each of 10 days, each of 5 cows eat x ammount of grass: 50x.
For each of 2 acres, there is A ammount of grass initially, and each are grows by y for 10 days: 2(A+10y)=2A+20y.
These are the same: $50x=2A+20y$

Similarly, we get $210x=3A+90y$ and
$16Dx=7A+7Dy$
$\implies D=\frac{7A}{16x-7y}$.

Solving for x and y in terms of A, $x=\frac{2A}{5}$, so that $y=\frac{9A}{10}$. Plugging these in, we get $D=\boxed{70}$?[/hide]

Correct! :D
Your solution and thinking is all right.


Hey all.
I need help with this one.
Given $ A$ is a real symmetric $ n*n$ matrix. Prove that there exist matrices $ B$ and $ C$ such that
$ BC = CB$
and
$ A = B^2 - C^2$

regards

A is symetric real it is diagonalisable

there exist P in O(n,R) such that $ P^{\minus{}1}AP\equal{}D$

$ D\equal{}D(1)\minus{}D(2)$

with D(1), D(2) are  diagonal matrix with positive eigenvalues,
 wich are square real numbers
you can finish now.


analytic geometry
combinatorics unsolved
combinatorics

An integer $ n$ is called good if there are $ n \geq 3$ lattice points $ P_1, P_2, \ldots, P_n$ in the coordinate plane satisfying the following conditions: If line segment $ P_iP_j$ has a rational length, then there is $ P_k$ such that both line segments $ P_iP_k$ and $ P_jP_k$ have irrational lengths; and if line segment $ P_iP_j$ has an irrational length, then there is $ P_k$ such that both line segments $ P_iP_k$ and $ P_jP_k$ have rational lengths.

(1) Determine the minimum good number.

(2) Determine if 2005 is a good number. (A point in the coordinate plane is a lattice point if both of its coordinate are integers.)

It is easy to see that $ 4$ is not good (without using proporties of rational and irrational numbers, but testing possibilities). If $ ABCDE$ is a pentagon with integer sides but $ ACEBD$ is a star with irrational sides, the conditions to $ n \equal{} 5$ be a good number are fulfilled. We can thus take

$ P_1 \equal{} (0, 0)$
$ P_2 \equal{} (5, 12)$
$ P_3 \equal{} (9, 9)$
$ P_4 \equal{} (15, 1)$
$ P_5 \equal{} (15, 0)$

[size=150]Solution of part (1) :-[/size]
It is easy to check that $n=3$ and $n=4$ does not work.
For $n=5$ we may take five points $(0,0),(1,0),(0,7),(5,3),(8,7)$ which clearly works.
So minimum good number is $\boxed 5$

Here's a potential solution for Part B:

Consider the sets
$A = { (1, 0), (2, 0), ..., (669, 0) }$ ,
$B={(1,1), (2, 1), ..., (668, 1)}$,
$C={(1, 2), (2, 2), ..., (668, 2)}$.

We claim that the $2005$ points in the set $S_{2005} = A\cup B\cup C$ satisfies the conditions of the problem. Consider a pair of distinct points $P_i = (x_i , y_i)$ and $P_j = (x_j , y_j)$ in the set $S_{2005}$. If either $x_i =x_j$ or $y_i = y_j$ , then clearly $P_iP_j$ is rational. Otherwise, either $|x_i-x_j|= 1$ or $2$, or $|y_i-y_j|= 1$ or 2, or both. Hence $|P_iP_j|^2$ can be written in the form of either $n^2+1$ or $n^2 + 4$ for some positive integer $n$. For a pair of positive integers $n$ and $m$ with $n > m$, we have $n^2 - m^2 > n^2\geq n^2- (n- 1)^2 = 2n - 1$. It follows that for a positive integer $n$ , $n^2 +1$ and $n^2 +4$ are not squares of an integer. We conclude that $(P_i , P_j)$ is rational iff line $P_iP_j$ is parallel to one of the axes.

If $P_iP_j$ is rational with $y_i = y_j$, then we set $P_k = (X_k , Y_k)$, with $x_k\neq x_i$ , $x_k\neq x_j$, and $y_k\neq y_i$. (We can do so because there are $668$ distinct $x$ values and three distinct $y$ values. ). Then $(P_i, P_j , P_k)$ is good. In exactly the same way, we can deal with the case when $P_iP_j$ is rational with $x_i = x_j$.

If $P_iP_j$ is irrational, then $x_i\neq x_j$ and $y_i\neq  yj$. We set $P_k =
(x_i,y_j)$. Then $(P_i , P_j , P_k)$ is good. Therefore, set $S_{2005}$ satisfies the conditions of the problem and $n=2005$ is good $\blacksquare$

[b]Note: [/b]
Let $S_6 = S_5\cup ({P_6 = (-24, 0)})$ and $S_7 = S_6\cup ({P7 = (- 24, 7) })$. We can now see both $6$ and $7$ are good. For every positive integer $n\geq  8 $, it is possible to generalize the construction shown above to show that $n$ is good. Therefore, all integers greater than $4$ are good.


I received a e-mail from "AoPS" saying they have a contest of who will send in the most MATHCOUNTS material, and the winner will get free tests.  Why I think this is a fraud:

The e-mail ID was aops.webmaster@gmail.com, and I'm positive AoPS would use their AoPS domain.

This was the original message:

[quote]The Art of Problem Solving is now accepting donations of MATHCOUNTS competitions.

The user who sends the most competitions will become a moderator and receive free AoPS texts. The deadline is March 21, 2009.

Thank you for your consideration.

--
AoPS - MathLinks Administrators[/quote]

If this is truly form AoPS, sorry.  I don't think it is though.

I think someone is trying to rip people off, and get some tests by doing nothing.  note something: it doesn't say how to turn in.

hm, yeah, that definitely sounds suspicious. i didn't get one...but yeah, thats weird.

The best place for this is probably Questions, Suggestions, and Announcements. You should also PM an admin.

[size=200][b]DO NOT SEND THIS PERSON ANY TESTS[/b][/size]

AoPS would NOT do this. Furthermore, isn't it funny how the deadline is a day when a lot of states are having the state competition?

The AoPS email is webmaster@artofproblemsolving.com or something like that.

Yes, this is fraudulent.  Do not send anything to that email address.  It has nothing to do with Art of Problem Solving.

Since DA MAN (Mr. Rusczyk) said it was a fraud, then just send an email back to that guy saying  :censored:  Lol. But that's kind of a stupid way to get the mathcounts competitions. You know something's a fraud just by it's server. I don't think aops wants a gmail account.  :rotfl:

Lol! Well... I heard someone tried (and failed) to impersonate some users from AoPS by emailing their coaches.  :P

Lol.. this is really amusing, actually.

woahhh... good thing I didn't get one of those  :lol:

If someone actually fell for that..... Wow... Just wow.... But since this is the only fraud so far reported.. I guess the mastermind behind this decided to quit in the middle of the operation.   :lol:

Track the IP. Usually you can view the IP of who sent it.

I've concluded that the IP Address was hidden when it came.  If you really want the header, I can give to you.


how would you solve any equation in the format x^x= something in this case x^x=24

A calculator.  Functions like this can't be inverted "algebraically" unless you really like the [url=http://en.wikipedia.org/wiki/Lambert%27s_W_function]Lambert W function[/url].

thanks but for the lambert w function it is for complex numbers and i am not talking about complex numbers

The real numbers are contained in the complex numbers.  For several reasons it makes more sense to develop the properties of the W function over the complex numbers first and then specialize to the real numbers.

thanks, but can you also explain how to do the W function


Euler
projective geometry
geometry unsolved
geometry

P is the intersection of diagonals of an inscribed quadrilateral ABCD. Prove that all 4 Euler lines of the triangles
PAB, PBC, PCD, PDA intersect in a only point.
----------------------------------------------------------
I have no solution for it!

This is a problem of Vittasko: http://www.mathlinks.ro/Forum/viewtopic.php?t=107997 .

  darij

This problem has been posted first time in Hyacinthos Forum at http://groups.yahoo.com/group/Hyacinthos/message/12112 and replied once by [b][size=100]Michael de Villiers[/size][/b], who informed me that probably it appears in the [b][size=100]I.M.Yaglom’s[/size][/b] book of Geometric Transformations.

(Translated in English from Russian by [b][size=100]Allen Shields[/size][/b] and [b][size=100]Abe Shenitzer[/size][/b] and Published by the Mathematical Association of America in three Volumes I, II, III).

Although I searched in this useful book, however I didn’t find something similar and I hope that I am not mistaken.

So, I am still searching if there is any reference about this problem, which has also been discovered by a friend of mine [b][size=100]Nikos Iosifides[/size][/b].

Kostas Vittas.

Hello, Mr [b]vittasko[/b]!
There are some PDF files in your website and one in them say of this problem! I don't see! Please send to me this file! Thank you!
Best regards!

Dear Mr vittasko and Mr darij grinberg!
Please explain me "What is the Desarques’s theorem?"
Thanks!

[quote="ricardokaka"]Please explain me "What is the Desarques’s theorem?"[/quote]

I think he means the Desargues theorem: http://www.mathlinks.ro/viewtopic.php?t=30323

  darij


MIT college AMC USA(J)MO USAMO geometry rectangle

Is anyone else here going to the January 13 Iowa GPML in Iowa Falls?

I haven't gone to any other local meets or practiced, so I'll probably suck the dirt at state...

I don't know what the exact time is, but I'll ask my coach.

I'll be going.  :lol:

You want to meet up somewhere there?

Since most of my math club's members have already qualified, so only of two of us are going (having missed the other meets via jazz band performances). My dad's taking us, so I can go wander wherever I want (without my coach yelling at me).

I don't know the exact time of the competition, but let's say meet at the front door of the school 30 minutes before registration, for starters.

Well, I hope I'm there by that time.

Uh... I guess I'll wear my MIT shirt so I'll stand out. It's the one with E/c^2, $\sqrt{-1}$, PV/nR on the front.

I hear Iowa City has a [i]really[/i] good team...

Last year Ames had two good seniors, but now they're gone.

Hehehehehe. I'm on that team.  :lol: This year has been very good to us. We have a lot of good people, a lot of them better than me. But yeah, Ames really did give us a hard time last year. Josh and Qi are quite the mathematicians! I'm looking forward to meeting you at the competition. Are you going to wear anything so I can tell who you are or anything like that?

Same here. Iowa City had 4 USAMO qualifiers, while we had Josh and Qi (aka Dave). I believe Josh is in Europe and Qi is at Iowa State University.

I don't have any good T-shirts that stand out a lot, and I usually wear something over it (given January weather). But not knowing what's in the laundry basket, I'll tell you by Friday. (Maybe I'll come up with a poster that says something like "Paul, come here!" like they do at the airport.)

It's unfortunate that we can't do a relay round without one more person, but I think Jim Strickland and I can pull a five on the team round.

Good luck to everyone who's going, wherever you are!

(And don't cram...we know what will happen :P )

Congratulations to everyone that qualified for state (although for AoPSers, that should be easy)!
I got 9th grade fourth place in both target and sprint  :( , but our two-man team got third.
Paul, you did quite nicely (it was expected of you...). If our regular team was there, Iowa City wouldn't have dominated as much.  :wink:

[b]GPML Qualifying Contest 10704, Alden High School (January 13, 2006) - Results for Top 3 Places[/b]

[u]Sprint[/u]:
1st (Tie) - Nicholas Allendorf (110), Iowa City West
        Paul Hlebowitsh (110), Iowa City West
        Lena Ye (110), Iowa City West
        Shuo Zhang (110), Iowa City West
5th - Curie Ahn (106), Iowa City West
6th - Jim Strickland (92), Ames High School

[u]Target[/u]:
1st - Nicholas Allendorf (80), Iowa City West
2nd (Tie) - Jim Strickland (70), Ames High School
                Adam Pottebaum (70), Atlantic
3rd (Tie) - Jim Anderson (60), Chariton High School
               Tyler Chenhall (60), JV Iowa City West
               Curie Ahn (60), Iowa City West
               Michelle Han (60), Iowa City West
               Shuo Zhang (60), Iowa City West
               Paul Hlebowitsh (60), Iowa City West

[u]Team - Large Schools[/u]:
1st - Iowa City West (80)
2nd - JV Iowa City West (50)
3rd - Ames High School (50)

[u]Team - Small Schools[/u]:
I'm missing the data for this division.

[u]Relay[/u]:
1st - Iowa City West: Allendorf, Hlebowitsh, Zhang (40)
2nd - Iowa City West: Ahn, Han, Ye (35)
3rd - JV Iowa City West: Gao, Hwang, Wei (25)

[u]Sweepstakes (Overall) - Large Schools[/u]:
1st - Iowa City West (322 5/6)
2nd - JV Iowa City West (196 1/6)
3rd - Kennedy High School (111 1/6)

[u]Sweepstakes - Small Schools[/u]:
1st - Home Schools of Eastern Iowa (215 2/3)
2nd - Waverly-Shell Rock (156 1/3)
3rd - Bishop Garrigan (150 1/3)

GPML target is like MC except you get 10 minutes for each pair. Here are the questions:

[b]January 13 - Round 10704 Target Round[/b]

[u]Pair 1[/u]
1. If Bob and Jane leave from their houses at the same time and walk towards each other each at their own constant rate, they will meet after 5 minutes . If Bob leaves 3 minutes later than Jane, they meet after Bob has walked 3 minutes. How long would it take Bob to walk from his house to Jane's? [hide="Answer"]15 minutes[/hide]
2. The diagram depicts three adjacent congruent squares. For angles $A$ and $B$ as shown, what is the radian measure of $A+B$? (diagram: rectangle ABCD with M and N trisecting AB, P and Q trisecting CD, and AM=MN=BN=BC=CP=PQ=QD=AD; angle A in the problem is angle APD, and angle B is ACD) [hide="Answer"]$\frac{\pi}{4}$[/hide]

[u]Pair 2[/u]
3. In how many distinct ways can you walk up a stairway which has 7 steps if you can take 1 or 2 steps up the stairs at a time? [hide="Answer"]21 ways[/hide]
4. A target consists of seven concentric circles of radii 1,2,...7. Determine the area of the largest of the three marked regions, A, B, and C, bounded as shown by the bold circles. (Diagram shows A containing rings of radii 1 through 4, B containing rings of radii 5 and 6, and C containing the ring of radii 7) [hide="Answer"]$20\pi$[/hide]

[u]Pair 3[/u]
5. Given that $a_{0}=1$, $a_{1}=3$, $a_{2}=7$ and $a_{n-3}=3a_{n-1}+a_{n}$ for $n\ge0$, we can find a general polynomial expression for $f(n)=a_{n}$. What is the sum of the degree and the coefficients of $f(n)$? [hide="Answer"]5[/hide]
6. IWC High School boys' basketball team has a 30% probability of winning Saturday's game while IWC High School girls' basketball team has a 40% probability of winning their game Saturday. If these two events are independent, what is the probability that IWC High School wins exactly one of the two games? [hide="Answer"]46%[/hide]

[u]Pair 4[/u]
7. Let $a$, $b$, and $c$ be the roots of the polynomial $x^{3}-17x+19$. What is the value of $a^{3}+b^{3}+c^{3}$? [hide="Answer"]-57[/hide]
8. Consider a grid with these 42 lines: 21 vertical lines for $x=i$ for $i=-10,-9, \ldots, 9, 10$; and 21 horizontal lines for $y=i$ for $i=-10,-9, \ldots, 9, 10$. On this grid are drawn 10 circles centered at the origin with radii: $k+0.1$ for $k=0, 1,\ldots, 9$. What is the total number of intersection points of the circles with the grid? [hide="Answer"]400 points[/hide]

I just looked for a pattern. I ended up messing up the function, but it's pretty clear what the function looks like after 2 or 3 terms.

Randomly came across this from school but I can give you the quick answer for number 5, you dont need to actually find the function to solve it.

Recall that the sum of the coefficients is $f(1)$ so $f(1) = a_{1}= 3$.  So now we need only to find the degree of the polynomial.  The difference of the differences in terms will always be 2 as a resut of the way the function is defined (dont have time to show that right now)
Since the second difference is constant, the polynomial is of second degree.  Thus the answer is 2+3=5.


Let a,b,c≥0 satisfying ab+bc+ca=1.Prove that:
$ \sum \frac{1}{\sqrt (a^2\plus{}bc)}$≥$ 2\sqrt2$


Let $ S$ be a convex set of points which contains at least three noncollinear points.

Suppose that the points in $ S$ are colored in $ p$ different colors. Prove that for each $ n\geq 3$ there exist infinitely many congruent $ n$-gons such that all vertices of those $ n$-gons are colored with the same color.

[quote="pbornsztein"]Consider a circle  $ C$ contained in $ S$ and an arbitrary origin $ w$ on $ C$ and use Van der Waerden theorem about monochromatic arithmetical progressions.

Pierre.[/quote]

Can you please explain in more detail?

Let $ k\equal{}W(n,p)$ the Van der Waerden number such that for each $ p$-colouring of $ \{1, \cdots, k \}$ there is a monochromatic arithmetical progression of length $ n$.
Now, choose $ e>0$ sufficientely small so that the points $ w\equal{}A_1,A_2, \cdots, A_k$ are on $ C$ with $ A_iA_{i\plus{}1} \equal{} e$ for each $ i<k$, and the arc $ A_1A_k$ does not overlap itself.
Any $ p$-coulouring of $ S$ induces a $ p$-colouring of $ C$ which induces a $ p$-colouring of $ \{1,\cdots, p \}$. Thus, there is a monochromatic $ n$-gon whose vertices are among the $ A_i$'s.
Now, do the same thing staring from any origin $ w' \in ]A_1,A_2[$. This gives an uncountable number of pairwise distinct monochromatic $ n$-gons. Such there are a finite number of arithmetical progressions of length $ n$ in $ \{1,\cdots,k \}$ and a finite number of colours, it follows that there are an uncountable number of these $ n$-gons which are congruent and all with the same colour.

Pierre.


On hypotenuse AB of right triangle ABC, D is the point for which CB = BD. If angle B measures 40 degrees, find measure of angle ACD.

[hide]Draw segment $\overline{CD}$.  Since $BC=BD$, $\angle BDC=\angle BCD=x$.  Then since the sum of angles in a triangle is $180^{\circ}$, we have $2x+40=180\Rightarrow x=70$.  Since $\angle ACD$ is complementary to $\angle BCD$, $\angle ACD=20^{\circ}$.[/hide]


Let me know if you guys figure this one out.

Signed,
An Adult.

That would be like reducing your life to school-related things, leisure/relaxation on afternoons/evenings/nights/weekends, and sleep. And that, too, is not at all a fair comparison, since adults have things to do [hide="such as"]
- Manage their life
- Make money and manage* finances
- Raise kids
- Do chores/Run errands
- Be accountable for their mistakes (to some extent)
- Plenty of other things that kid's don't really have to worry about in school, and sometimes in college

* -Saving up lunch money to buy Pokemon cards and such does [i]not[/i] count as managing your finances.[/hide]

Here's a better question -- what do adults [i]not[/i] do?

[quote="12markkram34"]It's a simple question. To me, it seems as though all they do is work, relax during nights and weekends, and sleep.[/quote] Do their hobbies? Is this even a question...

[quote="TZF"]Here's a better question -- what do adults [i]not[/i] do?[/quote]

They don't kid or horse around. Nor do they collect Pokemon cards, or look up random things on the internet.

I would guess that it would depend on the person. If you want it general, then they just live life.

[quote="12markkram34"]It's a simple question. To me, it seems as though all they do is work, relax during nights and weekends, and sleep.[/quote]

Heh. How old are you? Not criticizing, just wondering.

If you want to know what they do, think about what you do. Could your life be summarized as (in a general sense, not nessacarily specifically you) "Go to school, do homework, relax after homwork and weekends, and sleep"?

[quote="ZetaX"]At an age of 23, your brain will be replaced by a computer chip (the original one gets destroyed). That's all I found out so far... and I'm 22...[/quote]
I don't know which model of that chip I got. I sort of suspect that the software isn't supported any more.

I should complain to tech support. I think I got a defective one.

[quote="dragon96"]or look up random things on the internet.[/quote]

False. I know quite a few who do.

Don't worry about it.  It'll happens as it happens.

That's life.

It'll just happen, nothing can really change it. You get presented with decisions every second and need to make a choice about it, and that's basically life.

What a silly question... TZF posted a lot better question... If there were no adults, where would you live? How would you pay the bills? Who would keep up your house/flat? Who would give you 99.99% of what you have got?

Oh man, am I gonna get a bad chip? xD

[quote="Bugi"]What a silly question... TZF posted a lot better question... If there were no adults, where would you live? How would you pay the bills? Who would keep up your house/flat? Who would give you 99.99% of what you have got? [/quote]
In my opinion, the OP's question is pretty legitimate.
12markkram34, being 15 you have plenty of chances to never complain you "coulda been a contender.”  :D

<<If there were no adults, where would you live?>> this question and about the bills one too I guess is discussed in the following book:
From Wikipedia, the free encyclopedia
<<<<<
The Little Prince, published in 1943, is French aviator Antoine de Saint-Exupéry's most famous novella. Saint-Exupéry wrote it while living in the United States. It has been translated into more than 180 languages and sold more than 80 million copies making it one of the best selling books ever.
>>>>>

What do adults do? BOGTRO summed it up nicely. "They get presented with decisions every second and need to make a choice about it, and that's basically life."

Adults have more freedom than adolescents, which means they get to make more choices. Those choices affect the lives of themselves and other people around them. I think that this is what is daunting about the prospect of becoming an adult; the fact that you will need to make choices, and that they will play a heavier role in your life.

I'm 14.5, so I don't know if I got this completely accurate. :)

The main question-- What do adults do-- appears to be a probing one. Adults do a lot of things.:D I guess the "post" by [b] TZF [/b] gives us a highly accurate and reliable picture of "What adults mostly do."

[quote="12markkram34"] I'm 15 and I'm wondering what life will be like when I get a job, etc. I'm serious, not trolling. The prospect of becoming an adult is daunting.[/quote]

Yes, I would acquiesce (only to some extent) with your statement. When you become an adult, especially when you're working-- you can't be as flippant and carefree as you were as an adolescent. This was one of the "truths" which I had discovered after I became an adult. Also, as an adult you're expected to be more 'mature' and accountable.

On the flip side, you also get some privileges/benefits after you become an adult, for example, you get the privilege to exercise your franchise.

I guess life becomes really "tough" or "daunting" after one marries and starts a family. (I'm just guessing)

[quote="ZetaX"]At an age of 23, your brain will be replaced by a computer chip (the original one gets destroyed). That's all I found out so far... and I'm 22...[/quote]

I'm nearly of the same age as you're [b]ZetaX[/b]. But, I would like to know, what exactly did you mean by the above statement?

[quote="1=2"] I'm 14.5, so I don't know if I got this completely accurate. :)[/quote]

Wow! So you were a member of this forum right from the age of 10. (or 10 years 7 months maybe) Also, you appear (or were you feigning?) to be precocious.

[quote="12markkram34"]It's a simple question. To me, it seems as though all they do is work, relax during nights and weekends, and sleep.[/quote]When I was little (a long, long time ago, in a country far, far away), I looked at the adults around me, and saw them the same way. They seemed to want a mediocre job, to get a semi-decent salary, to be able to live dull lives. I wanted more from life, but was not sure what.

After that, my parents got a TV (I told you it was a long time ago), and I learned about other life options by watching a bunch of educational programs. That's how I decided (when I was 5) that I was going to have an exciting life, including:
1) climb Mount Everest
2) become a scientist
3) get the Nobel prize
4) emigrate to another country (maybe the US, or somewhere in Europe) and become a naturalized citizen there - I thought I would need 4) to be able to do the research needed to achieve 3).

Shoot for the stars, and you may get half of what you're aiming for.

Half a century later, I had achieved 2) and 4), decided that I do not want 1) (too cold), and realized that I will never achieve 3). Although my business card has the words "research scientist" on it, my job is interesting, but not as exciting as it sounds, or as I had imagined.

At some point before turning 30, I decided that I did not want to grow stale, and lose the ability of finding new and exciting things in life. I try to keep learning and growing. I often pretend I did not grow up, and have fun with the same activities much younger people enjoy. I like to solve problems, read (fiction and non-fiction), play minesweeper, listen to different points of view. I enjoyed Harry Potter, Eragon, The Little Prince, Mrs Frisby and the Rats of NIMH, and Johnatan Livingston Seagull, and can picture myself as any of the heroes in those books.

Other times I worry about the taxes, the mortgage, my children's education, the cost of their tuition, the price of gasoline, the cost of medicines, and a bunch of other boring stuff, like most adults do. At times I read/investigate/worry about education, civil rights, world hunger, wars, global warming, and a bunch of other not so boring (to me) stuff, like some adults do.

[b] @KMST: [/b] That was a really intriguing post sir! I guess the thing in [b]point no 3[/b]-- which you had mentioned in your list is conferred upon only a 'select' and 'lucky' few. And yes, Mount Everest sure is too cold-- certainly a place I wouldn't like to visit.


Let $a <0<c$ and we have that $9a^2=4b^2+36$ and $5c^2=7d^2+35$ .

Find the minimum price of $A=a^2+b^2+c^2+d^2-2(ac+bd)$ :)

i would say $11+4\sqrt 7$. if thats right, i will show my work ...

well since $a<0<c$: $a=-\sqrt{\frac49 b^2+4}$ and $c=\sqrt{\frac 75 d^2+7}$
plugging in $A=\frac49 b^2+\frac75 d^2 +11+(b-d)^2+2\sqrt{\left(\frac49 b^2+4\right)\left(\frac 75 d^2+7\right)}$, which is obviously the smallest when $b=d=0$, so $A_{min}=11+ 2\sqrt7$


How do you make a number that is divisible by $n$ but is $1mod(m)$ ?

[hide]Okay, first and foremost, GCD(m,n)=1, I believe.

Next, You have to find n(mod m).

Then you have to find the number <m, such that nk is congruent to 1(mod m).

I don't think I explained good.[/hide]

[hide="...an idea"]
Solve simultaneous linear congruences:
$x \equiv 1 (\text{mod m})$
$x \equiv 0 (\text{mod n})$

using the Chinese remainder Theorem
all x that satisfy the above would be ur numbers that are divisible by n and congruent to 1 mod m

[url=http://www.math.okstate.edu/~wrightd/crypt/lecnotes/node21.html]Link #1[/url]

[url=http://www.cut-the-knot.org/blue/chinese.shtml]Link #2[/url]

[url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Link #3[/url][/hide]

[quote="t0rajir0u"]http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity[/quote]
Um... I'm sorry, I don't seem to understand how this relates :oops:  :huh: . And we don't know the actual values of m and n, they are just variables.

[quote="laughinghead505"]Um... I'm sorry, I don't seem to understand how this relates :oops:  :huh: . And we don't know the actual values of m and n, they are just variables.[/quote]

Andre has already explained it, [hide="in fact"] We wish to find a $k$ such that

$k \equiv 1 \bmod m$
$k \equiv 0 \bmod n$

In other words,

$\exists x : k = 1-mx$
$\exists y : k = 0+ny$

Equating,

$1-mx = ny$

Or

$1 = mx+ny$

We wish to solve for $x$ and $y$ from which we may calculate $k$.  We do this with Bezout's identity:

http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity[/hide]
If you want an algebraic expression in terms of $m$ and $n$, there is none.

Edit:  From mathgenius, [hide="the other method"] We require that $gcd(m, n) = 1$.  Next, we claim that precisely one of the numbers

$n, 2n, 3n, ... (m-1)n$

Is congruent to $1 \bmod m$.  Consider their residues $\bmod m$.  If any two are equal, we have

$in \equiv jn \bmod m \Leftrightarrow (i-j)n \equiv 0 \bmod m$

Since $gcd(n, m) = 1$ it follows

$m | (i-j)$

But $i, j \in \{ 1, 2, 3, ... m-1 \}$, absurd.  Hence we have $m-1$ distinct remainders that are also nonzero (since $m$ has no common factors with $n$ and does not divide $1, 2, 3, ... m-1$).

It follows by the Pigeonhole Principle that one of these must have residue $1$.  [/hide]


geometry
3D geometry
modular arithmetic
algebra unsolved
algebra

hey isnt this imo 2002 shortlist problems.so solutions will be available on john schole's website

so what if solution is available there? most of the solutions there are not so nice or neat :) the problem is posted and still needs to be solved. :)

of course i dont mind posting the solution (especially since this is one of the few problems of the shortlist that i solved :D  :D ).
but i'll give a hint first (a major hint at that :) )
first prove that 2002^2002 can not be a sum of three cubes
(work modulo  9)
and than 2002 = 10^3 + 10^3 + 1^3 + 1^3
and 2002 = 667*3 + 1
well the solution is nearly finished
cheers!!

the sad part is that john scholes posted the same proof (believe me,i checked after making the previous post :(


[i][u]Problem :[/u]

  Let $ a \ , \ b \ , \ c$ be nonnegative real numbers

  Prove that :

 $ ( a^{2} \ \plus{} \ 3 )( b^{2} \ \plus{} \ 3 )(c^{2} \ \plus{} \ 3 ) \ \geq \ \frac{4}{27}( 3ab \ \plus{} \ 3bc \ \plus{} \ 3ca \ \plus{} \ abc)^{2}$[/i]

   [i]Nice problem :lol: [/i]

[quote="nguyenvuthanhha"][i][u]Problem :[/u]

  Let $ a \ , \ b \ , \ c$ be nonnegative real numbers

  Prove that :

 $ ( a^{2} \ \plus{} \ 3 )( b^{2} \ \plus{} \ 3 )(c^{2} \ \plus{} \ 3 ) \ \geq \ \frac {4}{27}( 3ab \ \plus{} \ 3bc \ \plus{} \ 3ca \ \plus{} \ abc)^{2}$[/i]

   [i]Nice problem :lol: [/i][/quote]
I only prove it by MV
It mean $ f(a,b,c)\ge f(\frac{a\plus{}b}{2},\frac{a\plus{}b}{2}.c)$There $ f\equal{}LHS\minus{}RHS$and $ a\ge b\ge c$
I think your proof is nicer than,any other solution

Allnames, can you post the full solution?


function
real analysis
real analysis unsolved

Let $ f: (0,\infty)\times (0,\infty)\rightarrow \mathbb{R}$ be such a differentiable function that \[ y\frac{\partial f}{\partial y}\equal{}2x\frac{\partial f}{\partial x}\] for all $ x>0,y>0$. Prove that there exists a function $ g: (0,\infty)\rightarrow\mathbb{R}$ such that $ f(x,y)\equal{}g(y^{2}/x)$.

Could you please give me some advice on how to solve the problem?

Note that
\[ y\frac {\partial f}{\partial y} \equal{} 2x\frac {\partial f}{\partial x} \iff \nabla f \cdot \left(\begin{array}{c} 2x \\
y \end{array} \right) \equal{} 0\]
Use this to find curves $ \gamma$ along which $ f$ is constant, i.e. $ f\circ \gamma \equiv const.$

This is closely related to the "method of characteristics" for solving first order PDEs.


Find $f(x): R \to R$ satisfy:
$f(x+y)f(xy)=xf(y^{2})+yf(x^{2})$

try this problem, it's much harder
Find all functions $f: R \to R$ such that: $f(x+y)f(xy)=x^{2}f(y)+y^{2}f(x)$ $\forall x,y \in R$


limit
search
calculus
calculus computations

Does anybody know proof for this :
$ \lim_{n \rightarrow \infty} \frac{e^{n}\cdot n!}{n^{n} \cdot \sqrt{n}}\equal{}\sqrt{2\pi}$
?  :)

It's basically Stirling approximation.

http://mathworld.wolfram.com/StirlingsApproximation.html
http://www.sosmath.com/calculus/sequence/stirling/stirling.html

Search for Stirling's formula. There is an argument [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=584310739&t=171251]here[/url].


:lol:  wafine
is the $\large LHS$ correct . :wink:

[quote="arkhammedos"]
is the $\large LHS$ correct .[/quote]

yes the LHS is correct !

nobody coud solve it  :?:

what about the third term on the RHS ?? :)

[hide="hint"]notice that the greatest value of the RHS is achieved when $b=\infty$[/hide]


[url=http://img22.imageshack.us/my.php?image=grida.png][img]http://img22.imageshack.us/img22/5967/grida.th.png[/img][/url]
How would I find the number of paths from the bottom left corner of the figure to the top right?

There was another thread on this forum that I read, but I didn't understand the explanation
any help would be greatly appreciated

So you start on the bottom right corner. (I'm assuming you can only travel along the black and that you can only move up and left.) Now choose the bottom corner adjacent to it. There is only 1 way to get to it. Now keep moving right, and you can see there is only one way to get to each corner. Now move up. There is only one way to get to all of them. Now choose The corner to the up and right of the original. Since you've marked out the bottom and left sides, add the two corners to the left and bottom of your new one. Do the same with every other spot.

Hard problem.. I'll do my best k? And don't trust my answer:
First of all, (I know dis for sure) let's say it was a regular $ 9 by 3$ rectangle. The equation for finding out how many paths for that is:
$ 12!/9!3!$

So if we evaluate that question, then we get an answer of $ 220$. So let's now look at the gap: There are 4 ways to use the gap (with imaginary lines) But since we can't use the gap, then we subtract: $ 220\minus{}4\equal{}216$ I hope that is the answer.  :roll:  :huh:

Other than the fact that $ 9\plus{}36\equal{}45\ne47$ :P , that is correct (you use the $ 45\plus{}2$ twice so you overcounted by $ 4$), but the answer should be $ \boxed{200}$.

Possibly faster is to note that all illegitimate paths go to (let the origin be the starting point) $ (3,3)$ eventually, and the rest is forced. There are thus $ \binom63\equal{}20$ bad cases; subtract this from $ \binom{12}3\equal{}220$ total cases to get $ \boxed{200}$.

Thanks so much! :D  :D
But I was wondering math, how did you get 6C3? :huh:

It's the same as where $ \binom{12}3\equal{}\binom{9\plus{}3}3$ came from. You have to go $ 3$ times up and $ 3$ times right, and there are $ \binom63\equal{}\binom{3\plus{}3}3$ ways to order those moves.


Prove that for any natural number $ n>1$

$ \frac{4^{n}}{n\plus{}1} < \frac{(2n)!}{(n!)^{2}}$.

hello, you can use induction to prove it.
Sonnhard.


HMMT
number theory
prime numbers
puzzles

This is one of the problems on an HMMT competition:
[quote="HMMT"]
Write down at least one, and up to ten, different 3-digit prime numbers. If you somehow fail
to do this, we will ignore your submission for this problem. Otherwise, you’re entered into a
game with other teams. In this game, you start with 10 points, and each number you write
down is like a bet: if no one else writes that number, you gain 1 point, but if anyone else
writes that number, you lose 1 point. [/quote]

I think we should start a game like this on AoPS. We'll start with [b]three-digit[/b] primes, then in each round increase the number of digits by one. The rules are the same; you can submit from one to ten prime numbers of the specified amount of digits in each round, in a PM to me. If you are the only one to submit a specific prime number, you get a point; if two or more people submit the same prime, they all lose one point each from their total score. Every person begins with 10 points.

[b]Anyone can join at any time, and will join everyone else, starting with 10 points. Everyone who sends me a PM [u]before[/u] Saturday May 19, will begin in the first round.[/b]

Please, DON'T use any outside resources, i.e. prime number lists or websites, on this.... I think the fourth or fifth round will serve as an end to the game.

Note that [u]any[/u] digit in any of the prime numbers you submit can be a zero, i.e. 007 is valid for the first round; but keep in mind that there are only four of these numbers (2,3,5,7) available, so they are risky to choose.

Enjoy the game!   :)

Edits: Start sending me your primes right away. No need to spend time for a sign-up period when the rules are already lain out

I'll try it out. :)

In our first PM, should we just tell you we will be participating, or should we give you a list of our primes right away?

3-digit primes are a bit to easy to find if you use a 32 by 32 grid(i wrote 1 out :D )

i'll join...so 1-10 primes?

the j person

shoot: klebian, can you PLEASE im me? it's really important and it's about pming.

hmm...
i'm in.
just for convenience is there like a list of all 3 digit primes?  :lol:

[quote="ra5249"]just for convenience is there like a list of all 3 digit primes?  :lol:[/quote]
[quote="Klebian"]Please, DON'T use any outside resources, i.e. prime number lists or websites, on this.... I think the fourth or fifth round will serve as an end to the game.[/quote]
Already answered for you. :wink:

It is very easy to find primes with some searching or writing a program (which I have...), but I hope nobody uses outside sources because that would make the whole thing unfair. If you think about it, the point is to see how many unusual primes you [i]know[/i], not how many you can [i]find[/i].

btw, i'll do different things instead of just 10 3-digit primes, 10 4-digit primes, etc.

Maybe something like 6 2-digit primes. We'll see.

Meh, I will extend this until I get more sign-ups. Or it will probably be scrapped. No point in playing otherwise.

Sign up, everyone!  I'm playing.  Bumped to get more people to look.

I'll play.  Very few people visit this forum; I suggest moving it to GFF to get more submissions.


limit
calculus
logarithms
real analysis
real analysis unsolved

(corrected)

1/j^(1+1/j) =
1/j*exp(-ln(j)/j) ~
1/j(1-ln(j)/j), which is not the general term of a convergent series.

Conclusion: your series diverges.

julien_santini is correct, of course. I'm going to repeat what he said just for clarity.

The key thing is that $\lim_{j\to\infty} j^{\frac1j}=1.$ This is a basic limit that every calculus student should just know, immediately. (This is especially true if you ever intend to make use of the root test for series convergence.)

Julien gave an argument: If $L=\lim_{j\to\infty} j^{\frac1j},$ then $\ln L=\lim_{j\to\infty}\frac{\ln j}j= 0$ because the logarithm grows more slowly than any postiive power. Since $\ln L=0,$ we have $L=1.$

Now, back to the original series. We have that the terms of the original series are $\frac1j$ times somthing that tends to $1$ as $j\to\infty.$ Hence the limit comparison test applies, and the series diverges by limit comparison to $\sum\frac1j,$ the harmonic series.


Ring Theory
algebra
polynomial
superior algebra
superior algebra unsolved

Find all the ideals of the ring Z[x]/(2,x^3+26)

I have no idea what the ring is look like....
or is there any other shortcut for this question??

R[x]/(p,f) = R/p[x]/f follows from the universal properties of both sides. in this case, we get the ring $ \mathbb{F}_2[x]/(x^3\plus{}26)\equal{}\mathbb{F}_2[x]/(x^3)$ whose ideals correspond (since $ \mathbb{F}_2[x]$ is a PID) to the divisors of $ x^3$.

What do you mean by universal properties of both sides.

polynomial algebras and quotients have universal properties, you can look them up and fill in easily the details.


Use the method of separation of variables to solve, at least formally, this diffusion problem (D and U are real numbers, D>0). Can't see how to deal with the inhomogeneous BC.  :(

$ u_t \minus{} Du_{xx} \equal{} 0,0 < x < \pi ,t > 0$
$ u(x,0) \equal{} g(x),0 \le x \le \pi$
$ u_x (0,t) \equal{} 0,t > 0$
$ u_x (0,t) \plus{} u(L,t) \equal{} U,t > 0$

are the boundary conditions really the ones you wrote, namely:
1: initial time: $ u(x, t\equal{}0)\equal{}g(x)$ for $ 0<x<L$
2: Neumann condition at $ x\equal{}0$: $ \partial_x u(x\equal{}0, t)\equal{}0$
3: Dirichlet condition at $ x\equal{}L$: $ u(x\equal{}L,t)\equal{}U$

and if so, do you assume $ g(L)\equal{}U$ ? Is $ U$ a constant ?

You are right alekk. The book has quite a few mistakes and imprecisions in it. Let's replace pi with L. U is a constant (I assume; not specified). There is a Robin condition at x=L, nakely: $ u_x (L,t) \plus{} u(L,t) \equal{} U,t > 0$. Everything else is correct.

I can no longer edit.


[i]Solve the following equation in $ \mathbb{R}$ :

   $ \left| \frac {6 \minus{} 2x}{\sqrt {5 \minus{} x}} \ \plus{} \ \frac {6 \plus{} 2x}{\sqrt {5 \plus{} x}} \ \right| \equal{} \ \frac {8}{3}$

   Ps Pco :

         [hide]  Have you try that problem :

  http://www.mathlinks.ro/viewtopic.php?t=321886
  [/hide][/i]

[quote="nguyenvuthanhha"][i]Solve the following equation in $ \mathbb{R}$ :

   $ \frac {6 \minus{} 2x}{\sqrt {5 \minus{} x}} \ \plus{} \ \frac {6 \plus{} 2x}{\sqrt {5 \plus{} x}} \ \equal{} \ \frac {8}{3}$

   Ps Pco :

         [hide]  Have you try that problem :

  http://www.mathlinks.ro/viewtopic.php?t=321886
  [/hide][/i][/quote]
Let $ a\equal{}5\plus{}x, b \equal{} 5\minus{}x$, $ f(x) \equal{} \frac{x\minus{}2}{\sqrt{x}} \equal{} x^{\frac{1}{2}} \minus{} 2 \cdot x^{\frac{\minus{}1}{2}}$. So we have to find $ a,b \ge 0, a\plus{}b\equal{}10$ such that:
$ f(a)\plus{}f(b) \equal{} \frac{4}{3} \equal{} f(9)\plus{}f(1)$
$ f''(x) \equal{} \minus{}\frac{1}{4} x^{\frac{\minus{}3}{2}} \minus{} \frac{3}{2} x^{\frac{\minus{}3}{2}}$, so $ f''(x)$ is concave. First assume $ a \ge b$. If $ a<9$ we find $ (9,1) \succ (a,b)$ so $ f(9)\plus{}f(1) < f(a)\plus{}f(b)$. If $ a>9$ we find $ (a,b) \succ (9,1)$ so $ f(a)\plus{}f(b) < f(9)\plus{}f(1)$. So there is only one solution when $ a \ge b$: $ (a,b)\equal{}(9,1)$. Similarily we also find $ (a,b)\equal{}(1,9)$, which gives the two solutions $ x \in \{4,\minus{}4\}$ which is indeed solutions to the original equation. Done :)


For what values of $ a$ and $ b$ does the function  $ f(x)\equal{}axe^{bx^2}$ have the maximum  value $ f(2) \equal{} 1$?

$ f(x) \equal{} axe^{bx^2}\Longrightarrow f'(x) \equal{} a(2bx^2 \plus{} 1)$, thus $ b < 0$ is needed for $ f(x)$ to have maximum at $ x \equal{} 2\Longleftrightarrow b < 0$.
That is to say, $ f'(2) \equal{} 0$ and $ f(2) \equal{} 1\Longleftrightarrow a(8b \plus{} 1) \equal{} 0$ and $ 2ae^{4b} \equal{} 1$, $ a \equal{} 0$ is trivial, so $ b \equal{} \minus{} \frac {1}{8}\Longrightarrow a \equal{} \frac {1}{2\sqrt {e}}$, then you need to confirm variation of $ f(x)$.

Because $ e^{bx^2}$ is a function, I only see $ f'(x) \equal{} a(2bx^2 \plus{} 1)$ being true when $ b\equal{}0$, please explain, thanks.

$ f(x) \equal{} a x e^{b x^2}\Longrightarrow f'(x) \equal{} a e^{b x^2} \plus{} a x e^{b x^2} \cdot 2 b x \equal{} a(2b x^2 \plus{} 1)e^{bx^2}$.

He forgot that exponent. But that's not gonna be zero anyway.


calculus
integration
function
floor function
algebra
calculus computations

Look at the interval from $k-1$ to $k.$ If $k-1\le x< k,$ then $\lfloor x\rfloor=k-1.$

$\int_{0}^{n}\lfloor x\rfloor\,dx=\sum_{k=1}^{n}\int_{k-1}^{k}\lfloor x\rfloor\,dx$

$=\sum_{k=1}^{n}\int_{k-1}^{k}(k-1)\,dx= \sum_{k=1}^{n}(k-1)=\frac{n(n-1)}2.$

[quote="myoplex11"]integrate f(x) from 0 to n. where n is a positive integer and f(x) is the greatest integer function.
I Know that i have to use the definiton of greatest integer function and break it into sum of the integrals how do i do this.[/quote]

how would you do this if we changed the problem so that n does not have to be an integer?

All you'd have to do is modify the very last term of the sum; everything else would remain the same. Draw a picture.


Sa  se rezolve in $\mathbb{Z}$  ecuatia: \[2x-3y+4z=5.\]

hai ca o rezolv eu, ca sunt dornic de "initiere"  :))   :blush:

deci avem de rezolvat ecuatzia : $2x-3y+4z = 5$ pe care o notam asa sa fim si noi eleganti cu $(\star)$  :blush:

lucram modulo 4 si anume : $2x-3y\equiv 1\pmod{4}$, deci $2x-3y=1+4s (\star\star)$ , cu $s \in \mathbb{Z}$

observam si noi ca o solutie particulara a lui $(\star\star)$ este :
$x_{0}=2+5s$
$y_{0}=1+2s$

acum conform faptului ca intr-o ecuatzie de forma $ax+by+c=0$, unde $(x_{0}, y_{0})$ este o solutie particulara a acestei ecuatzii, forma generala a tuturor solutilor $(x,y)$ este :
$x=x_{0}+bt$
$y=y_{0}-at$, cu $t \in \mathbb{Z}$

si sa revenim la oile noastre acum:deci forma generala a solutilor ecuatiei   $(\star)$ este :
$x=2+5s-3t$
$y=1+2s-2t$

acum inlocuind in $(\star)$ obtinem $z=1-s$

si acum sa tragem si o concluzie: $\Rightarrow (x;y;z) = (2+5s-3t ; 1+2s-2t ; 1-s)$ , cu $s,t \in \mathbb{Z}$ , of course :blush:


integration
function
calculus
real analysis
real analysis unsolved

Given $ J(y) \equal{} \int \limits_{0}^{\frac{\pi}{6}}\left[ (y'\minus{}\cos(x))^2\plus{}4y \right]dx$ find a stationary function, and use convexity to show its the unique minimizing function on the set

$ \mathcal {D} \equal{} \{ y \in C^1[0,\frac{\pi}{6}] : y(0)\equal{}0, y(\frac{\pi}{6})\equal{}\frac12 \}$

I get that $ f_y[y] \equal{} 4$ and $ \frac{d}{dx}f_z[y] \equal{} \frac{d}{dx}2(y'\minus{}cos(x))$ and then subtracting them and equating to 0 I solve for y and get $ y \equal{} x^2 \minus{} \sin(x) \plus{} c$. But this can't be right given the initial conditions of the set $ \mathcal{D}$ :( Help please.

Well, whatever the the title of the class is, this is a calculus of variations questions. For my tastes, that's too advanced for the "computations and tutorials" section, so I moved it.

You just forgot $ ax$ in your general solution. Then using boundary conditions one can find $ a$ and $ c$ :wink:

Ahhh haha thanks. I did indeed forget it.
Then for the boundary parts I got $ c \equal{} \frac {\pi}{6}, d \equal{} 0$.
Thanks. I feel dumb XD

This is for AMath456 at Waterloo


you don't need logs- just raise both sides of the equation to the 3/2 power.

You can't use logarithms because your base would be 'x'

x^(2/3) <= 27.6 is equivalent to log[base x]27.6 = 2/3

[quote="maisimplicity"]how would you solve an equation like this..

$ x^\frac{2}{3} \geq 27.6$[/quote]

Actually, if you really wanted to, you could do it, it would just be a waste.

Note that $ \log_a b \cdot \log_b a \equal{} 1$. So, $ \log_{x} 27.6 \equal{} \frac{2}{3} \iff \log_{27.6} x \equal{} \frac{3}{2}$.

Exponentiating with base $ 27.6$ gives $ x \equal{} 27.6 ^{\frac{2}{3}}$, which, gives the same result as the much simpler approach in post #2.

[quote="i_like_pie"][quote="maisimplicity"]how would you solve an equation like this..

x^(2/3) is greater than or equal to 27.6[/quote]
If $ x^{2/3}\ge27.6$, then $ x\ge27.6^{3/2}$ and $ x\le \minus{} 27.6^{3/2}$.

When taking the square root, don't forget about the negative root.[/quote]

Yeah i tried [code]\pom[/code] what is the latex code for it?

even simpler; \pm gives $ \pm$ and \mp gives $ \mp$


[hide]
$\tan12\tan72=\tan36\tan42$

$\implies \frac{-1}{4}(\cos84-\cos60)(\cos78+\cos6) = \frac{-1}{4}(\cos84+\cos60)(\cos78-\cos6)$

after simplification: $\implies \cos6\cdot \cos84 = \cos 60\cdot \cos78$

$\implies \cos 78 = \cos138 + \cos18$

$\cos138 + \cos18 = 2\cdot \cos78 \cdot \cos60$ :D [/hide]