geometry
rectangle
real analysis
real analysis unsolved

Consider the measure space $ [0,1]$ with the Borel sigma field, and let $ \mu$ be the Leb. measure and $ \nu$ be the counting measure on $ [0,1]$. Let $ \mu \times\nu$ be the product measure on the product sigma field. Can someone help me to compute $ (\mu \times\nu)(E)$ where $ E\equal{}\{(x,x)/x\in [0,1]\}$.

Thanks.

what is your definition of the product measure. Usually, the product measure $ \mu \times \nu$ is defined when both $ \mu$ and $ \nu$ are $ \sigma$-finite. Remember that this hypothesis implies the unicity of any measure $ m$ on the product space (with product sigma field) that satisfies $ m(A \times B) \equal{} \mu(A)\nu(B)$.

So what product measure are you talking about, since the counting measure is not $ \sigma$-finite on $ [0;1]$ ?

Yes alekk, I understand that when $ \mu$ and $ \nu$ are not $ \sigma$-finite then $ \mu \times \nu$ is not unique. But (according to Folland, Real Analysis) if there are two such measures $ \lambda_1$ and $ \lambda_2$(i.e. $ \lambda_i(A\times B) \equal{} \mu(A).\nu(B)$ for all measurable rectangles $ A\times B$) then $ \lambda_1(E)\le \lambda_2(E)$ for all $ E$ in the product sigma field, with equality when $ \lambda_2(E) < \infty$.

Is it right?Can you make this concept more clear, please?

alekk, just neglect my previous post. Actually according to Folland, for non $ \sigma$-finite measures $ \mu$ and $ \nu$, by $ \mu \times \nu$, I mean the outer measure which is obtained from the premeasure $ \mu_0$ on the field of finite disjoint unions of measurable rectangles where $ \mu_0(\cup_{i\equal{}1}^{n}(A_i\times B_i))\equal{}\sum_{i\equal{}1}^{n}\mu(A_i).\nu(B_i)$

So now my problem reduces to find the outer measure of the diagonal $ E$.
So for any covering of $ E$, say $ E\subseteq \cup_{i}(A_i\times B_i)$, I should somehow be able to conclude that $ \mu(A_i).\nu(B_i)\equal{}\infty$ for some $ i$. Then it would follow that $ (\mu \times \nu)(E)\equal{}\infty$
Can you help me in proving this?

perhaps some sort of cardinality argument on the diagonal would be a good idea.


A telephone pole is supported by a steel cable which extends from the top of the pole to a point on the ground 3 meters from its base. When Leah walks 2.5 meters from the base of the pole toward the point where the cable is attached to the ground, her head just touches the cable. Leah is 1.5 meters tall. How many meters tall is the pole?

You have similar triangle. $ \frac {.5}{1.5} \equal{} \frac{3}{x}$ which yields $ x \equal{} \boxed{9 meters}$


Edit's grandmother's great grandmother's age was 1/31  of her own birth year when she died.
(Count her age in full years.) How old was she in 1900?

Since she obviously lived before and after $1900$, we can put $1900-x$ as the year of birth and $1900+y$ as the year of death, giving

$x+y={1\over 31}(1900-x)\iff 32x+31y=1900$

By any of the known ways of solving linear diophantics in two variables, we get $(x,y)=(-31n+9,32n+52),n\in\mathbb{Z}$. Since both $x$ and $y$ must be non-negative, we get $n\in\{-1, 0\}$, yielding $(x,y)\in\{(9,52),(40,20)\}$. Therefore we have two possibilites:

1. She was born in $1891$ and died in $1952$, meaning that in $1900$ she was $9$ years old;

2. She was born in $1860$ and died in $1920$, meaning that in $1900$ she was $40$ years old.

(By an abundance of words "great" in the problem, I guess the creator opted for the second solution.)

Another problem that I can do in less than 30 minutes :o move it fast!

[hide]Great granny was born in a year that was divisible by 31. Using my refined slacker techniques, I discover that $31*60= 1860$. From this, I quickly find that $31*61=1891$ and $31*59=1829. 1829+59<1900$, so 1829 isn't a possibility for Great Grannies birth year. $1860+60=1920$, so that is a definite possibility. In 1900, Granny would have been 40. If 1891 was her birth year, than Granny would have been 9. My guess for the answer moldlee is looking for is 40, because I like the number 40 better than the number 9.[/hide]


inequalities function algebra domain limit induction calculus

I've been working through old qualifying exam problems, and for the most part, problems that end with "show that $ f\equiv 0$" are pretty easy. Just a quick application of one of the theorems that say a function must be constant.

For this one, the trick seems to elude me, though. Suppose $ U$ is a domain containing $ 0$, and $ f$ is analytic on $ U$. Suppose also that $ |f(1/n)|<e^{\minus{}n}$ for sufficiently large $ n$. Show that $ f\equiv 0$.

Any hints?

Consider the functions $ f_k(z)\equal{}z^{\minus{}k}f(z)$ for $ k\ge 0$. If $ \lim_{z\to 0}f_j\equal{}0$ for $ j<k$, then $ f_k(0)\equal{}\frac1{k!}f^{(k)}(0)$.
Now, look at the values of $ f_k$ at $ \frac1n$. $ \left|n^kf\left(\frac1n\right)\right|<n^ke^{\minus{}n}\to 0$ for every $ k\ge 0$. By induction, $ f^{(k)}(0)\equal{}0$ for every $ k\ge 0$. If all of the derivatives of an analytic function are zero at a point, the function is identically zero in a neighborhood of that point.

Incidentally, that "analytic" can be in either the real or complex sense.

Another way (almost the same) is based on the considering k-th finite difference of the function $ f$ at a point $ 0$. :wink:


A super ball dropped from a height of 36 inches rebounds three-fourths of the distance of the previous bounce. Write an expression to represent the distance traveled by the ball in n bounces. Justify your answer.

Thank you

after the first bounce the ball has traveled 36+36*3/4.
second bounce 36+36*3/4+36*(3/4)^2
now using the geometric progression formula
a(1 - r^n )
------------
  1 – r

where a=36 r=3/4 and n is the variable

hope this helps

Ok what if I asked how far the ball has traveled if it keeps on bouncing (taking it to infinity)

[hide]Well, this will obviously be a series. So let's take a look at the first few terms.

After 0 bounces, when the ball first strikes the ground, it will have travelled $36$ inches. This is obvious.

After 1 bounce, when the ball comes back to the ground and is about to hit again, the ball had rebounded to a height of $36 * \frac{3}{4}$ inches, so it travelled a grand total of $36+2*36* \frac{3}{4}$ inches.

Using similar reasoning, we see that after 2 bounces, it has travelled $36+2*36* \frac{3}{4}+2*36* (\frac{3}{4})^{2}$ inches.

Extrapolating, we see that after $n$ bounces, the ball has travelled $36+2*36* \frac{3}{4}+...+2*36*(\frac{3}{4})^{n}$ inches. In closed form, that's:

$36+\sum^{n}_{i = 1}2*36*(\frac{3}{4})^{i}$.

Since it's geometric, for an infinite number of bounces, you can use the formula previously mentioned.[/hide]

umm i think mathamatrix forgot to multiply the terms by two (except the first one) because the ball also comes back down that same amount..

sorry about that, yes all of the terms after the 36 must be multiplied by 2


I must solve the following equation in $(\mathbb{R},+\cdot,\leq)$

$x=\sqrt{19}+\cfrac{92}{\sqrt{19}+\cfrac{92}{\sqrt{19}+\cfrac{92}{\sqrt{19}+\cfrac{92}{\sqrt{19}+\cfrac{92}{x^2}}}}}$

It must have at least one real root (since is a third grade equation) but I can't find it!

Don't triple post. It's a little overkill  :P

Sorry, I didnt want to... :blush:  but I cant delete them (I'm not administrator)...

Do you mean in the set $\mathbb{R^+}$?

And on a solution.... i'm workin' on it  :)

[hide="Method and resulting polynomial"]
To get rid of the huge denominator, what is did was subtracted the $\sqrt{19}$ over, multiply the denominator over to the other side, move the variable expression over, repeat until done. The resulting 3rd degree polynomial is (if i didn't accidentally forget any of the radicals) $14069x^3-(25031\sqrt{19})x^2+(18676\sqrt{19})x-906292=0$.  I don't know what to do next with those gigantic coefficients, as i can't find any factors easily. Anyone want to tackle this?[/hide]

[quote="JBL"]Cincodemayo, Mathematica doesn't think your polynomial is correct.  It also doesn't think that either your polynomial or the original question has a "nice" solution.  The original problem has one real solution with numerical value about 10.4355....[/quote]

:rotfl: yay mathematica!
Well then i probably did one too many layers or something. If there's no pretty solution, then solving for it must be really hard/tedious.. eh.

I used Derive and I also found an awful polynomial. It has one real root x = 10.43548985, so I quit this way...

I changed the problem a bit (I switch $x^2$ to $x$), ie,

$x=\sqrt{19}+\cfrac{92}{\sqrt{19}+\cfrac{92}{\sqrt{19}+\cfrac{92}{\sqrt{19}+\cfrac{92}{\sqrt{19}+\cfrac{92}{x}}}}}$

and I found a nice solution with no much effort. I wonder if there is a similar trick...

I'm solving the modified problem....

Consider the equation $x^2-\sqrt{19}x-92=0$, thus $x=\sqrt{19}+\frac{92}{x}$. Then, $x=\sqrt{19}+\cfrac{92}{\sqrt{19}+\cfrac{92}{x}}$, and going on this way...

So the solutions are the roots of the equation: $x^2-\sqrt{19}x-92=0$

What do you think? I still have to solve the original problem!


farz konid n adadi [b]morakab[/b] bashad. mo'adeleye hamneheshti
x^n=x  (mod n)
ra dar nazar begirid ke be ezaye tamame x haye sahih bargharar ast.
saabet konid binahayat n ba khavaase bala vojoud darad.

fekr mikonam in masale opene , intor nist ?
be in adada migan avvalnamaye motlagh , faghat 43 ta adade avvalnamaye motlagh kamtar az $ 1000000$ hastesh . koochiktarinesh $ 561$ hast hamchenin $ 10585$ , $ 6601$ , $ 1729$ niz avvalnamaye motlaghan :)

vali ye seri soale moshabehe in hast ba ye seri taghirate kheili jozi ke hal shodan va soalhaye ghashangi ham hastand...


ok who took the amc this year and how did you do/what was your score.

I got first in my school with 22 I think?  Anyway not too great considering some people regularly get a perfect.  It's just stupid mistakes really and careless calculations.  I already know all the material.  Considering for a 8th- math test I am in a lot higher math then almost anyone taking the test.

Search in this section of the forum for past threads on this topic. Back during February, we had polls and such of the results.


geometry
3D geometry
algebra
polynomial
induction
factorization
sum of cubes

(1^3+2^3+..........to the n^3)'s formula is (1+2+3+........+n)^2(squared).So, this problem is just simply a problem that you can cancel out the bottom with. ((1+2+.....blah)^2)/(1+2+.....blah)+36. 1+2+....+n=36. n=8.

i think you could easily find a proof by googling "formula for sum of cubes" or something lik that. Wolfram is bound to have something like that. :read:

[quote="archimedes1"]I'm refreshing like crazy...[/quote]
http://usamts.org/Gallery/G_Gallery.php
Methods for proof:
Assume it's a quartic polynomial (of the form $ax^{4}+bx^{3}+cx^{2}+dx+e$)
Use hockeystick

You can prove the formula

$1^{3}+2^{3}+3^{3}+...+n^{3}= (1+2+3+...+n)^{2}$

by induction.

I've always disliked induction though... It never really intuitively makes sense as opposed to other kinds of proofs.

[quote=davidlizeng][quote="i_like_pie"][hide]$\frac{1^{3}+2^{3}+3^{3}+\ldots+n^{3}}{1+2+3+\ldots+n}=36$

$\frac{\left(1+2+3+\ldots+n\right)^{2}}{1+2+3+\ldots+n}=36$

$1+2+3+\ldots+n=36$

$n=\boxed8$[/hide][/quote]

how do you prove that the blah^3 = (blah)^2 ??
(i feel really dumb asking this)[/quote]

use induction

[quote=davidli]For what value of $n$ is $\frac{1^{3}+2^{3}+3^{3}+...+n^{3}}{1+2+3+...+n}=36$?[/quote]

See this [url=https://artofproblemsolving.com/community/c3h1930773_sum_of_cubes]post[/url]


$ x, y, z$ are real numbers that is not negative.

Prove that

$ 8(x^3\plus{}y^3\plus{}z^3)^2\ge9(x^2\plus{}yz)(y^2\plus{}zx)(z^2\plus{}xy)$.

[quote="Stephen"]$ x, y, z$ are real numbers that is not negative.

Prove that

$ 8(x^3 \plus{} y^3 \plus{} z^3)^2\ge9(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} xy)$.[/quote]
It 's weak,we can solve by Holder inequality
$ RHS \le \frac {(x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx)^3}{3}$
$ \le \frac {8(x^2 \plus{} y^2 \plus{} z^2)^3}{3}$
So,we only need to prove:
$ 3(x^3 \plus{} y^3 \plus{} z^3)^2 \ge (x^2 \plus{} y^2 \plus{} z^2)^3$
Which 's obvious true.

Expand gives $ 8\sum_{cyc} {a^6}\plus{}7\sum_{cyc} {a^3b^3}\ge 18a^2b^2c^2\plus{}9\sum_{cyc} {a^4bc}$ which is true using $ AM\minus{}GM$ because :

 $ a^6\plus{}a^6\plus{}b^3c^3\ge 3\sqrt[3]{a^{12}b^3c^3}\equal{}3a^4bc$

 $ \sum {a^3b^3}\ge 3a^2b^2c^2$

 and :

 $ \sum {a^6}\ge \frac{1}{3}(a^3\plus{}b^3\plus{}c^3)^2\ge abc(a^3\plus{}b^3\plus{}c^3)$


algebra
linear equation
real analysis
real analysis unsolved

How can one solve
$ \phi_t\equal{}a^2\phi_{xx}\plus{}1$
$ 0\le x \le L$
if the boundary conditions are of the form
$ \phi(0,t)\equal{}1$ and $ \phi_x(L,t)\equal{}0$
and the initial condition is $ \phi(x,0)\equal{}1\minus{}x$

With a nonhomogeneous linear equation, be on the lookout for a particular solution. In this case, notice that $ \phi(x,t)\equal{}t$ satisfies the equation and the boundary conditions.

So let $ u(x,t)\equal{}\phi(x,t)\minus{}t.$ We have that $ u_t\equal{}a^2u_{xx}$ with boundary conditions $ u(0,t)\equal{}0,\ u_x(L,t)\equal{}0$ and initial condition $ u(x,0)\equal{}1\minus{}x.$

Now, can you solve that for $ u?$ It's a pretty standard separation of variables argument.

Nice! thanks, that was a nice trick!

In the book it is asked to be solved with finite transforms. How could i apply that given these boundary conditions?

actually wouldn't the initial condition be $ u(0,t)\equal{}1\minus{}t$ after the substitution? :)

Oops, I misread that - I thought you had said $ \phi(0,t)\equal{}t.$ This makes things a little trickier, since it's not that easy to bring it to a homogeneous form.


Since there have been quite a few mock tests lately, I was wondering for the sake of future mocktestwriters when people are most free to do these mock tests. At first I thought it would probably be on some holiday like Thanksgiving or Christmas break, but at the same time some people would be on vacation. So when are you most free to take these tests? What time frames would work best for you?

[quote="probability1.01"]What time frames would work best for you?[/quote]
Long ones. Might as well include holidays, and not holidays, if you can. If people care they'll fit it in.

I can write a mock contest preferably on holidays, or week-ends sometimes. :)

Im usually free on a school night...seeing that I'm still in middle school and don't have much homework. I probably can't make it once or twice a week next year cuz of aops classes though.

Holidays are still the best. Even if people are on vacation, they can still print the test out and think about it from time to time.

wait are u making a mock usamo/IMO or something? cuz if so that'd be cool and i'd like to do it if i have time. I think thanksgiving would be good cuz all our early decision apps will be sent and u don't have to do anything for that :wink:

Holidays are fine.

But any day works since I would be doing a mock test.

I'd prefer the holidays. Gotta study during school days


vector
linear algebra
matrix
linear algebra unsolved

Let $ W, U$ and $ V$ be such subspaces of the vector space $ X$, that $ X\equal{}U\oplus V\equal{}U\oplus W$. Show, that there exists such isomorphism $ T: X \rightarrow X$, that we have $ Tx\equal{}x$ for all $ x\in U$ and $ T(V)\equal{}W$. Is it unique?

There's no way we can get uniqueness. The most we might hope for is something like uniqueness modulo automorphisms of $ V.$

Here's a cheap example: let $ X\equal{}\mathbb{F}^3,U\equal{}\text{span}\{(1,0,0)^T\},$ and $ V\equal{}W\equal{}\text{span}\{(0,1,0)^T,(0,0,1)^T\}.$

Let $ T_1\equal{}I_3$ and let $ T_2\equal{}\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}.$ Both $ T_1$ and $ T_2$ satisfy your condition.

Here is my solution.

1. $ \dim{U \cap V}=0$:

Since $ X=U\oplus V \Leftrightarrow U \cap V=\{0\} \Rightarrow \quad \dim{U \cap V}=0$.

2. $ \dim{V}=\dim{W}$, $ V=W$:
\[ \dim{U\oplus V}+\dim{U\cap V}=\dim{U}+\dim{V}\Rightarrow\\
\dim{X}=\dim{U\oplus V}=\dim{U}+\dim{V}\Rightarrow\\
 \dim{X}-\dim{U}=\dim{V}\]
By the analog reasoning we get $ \dim{W}=\dim{V}$ and since bases of $ W$ and $ V$ added to the base of $ U$ give us the base of $ X$, $ W$ and $ V$ must be the same.

3. Now we define the operator:
\[ T: X \rightarrow X\\
T(x_{1}+x_{2})=T(x_{1})+T(x_{2}) \quad (1)\\
T(\alpha x)=\alpha T{x} \quad (2)\]
$ T(u_{i})=u_{i}$, where $ u_{i}$ is the base vector of $ U$
$ T(v_{j})=w_{j}$, where $ v_{j}$ is the base vector of $ V$ and $ w_{j}$ is the base vector of $ W$

4. We must just check that $ T$ is well defined, $ T(V)=W$ and $ T$ is bijective (hence isomorphism):

[b]i.) well defined:[/b]

$ T(u_{i})=u_{i}, T(v_{j})=w_{j}$ for all $ i$ and $ j$. Since all vectors in the vector space have unique representation in the given bases, those vectors cannot come into contradiction with (1), (2).

[b]ii.)[/b] $ T(V)=W$

Since the base of $ V$ maps into base of $ W$, $ T(V)=W$.

[b]iii.)[/b] $ T$ [b]injective:[/b]

Since $ \dim{ \imT}=\dim{X}\Rightarrow \dim{%Error. "kerT" is a bad command.
}=0 \Rightarrow T$ injective.

[b]iv.) [/b]$ T$ [b]surjective:[/b]

All base vectors of $ X$ are the image of some vectors from $ X$, hence $ T$ surjective.

In your step 2, you assumed that all of the vector spaces in this problem were finite dimensional.

What would happen if you dropped that assumption?

(Note that both the existence of a Hamel basis and some statement about the cardinality of such a basis may appear.)

Thank you for your kindness, professor Kent.

But the problem is, that I do not have any experience with infinite vector spaces. So before answering on your question I must study some theory.  :)


analytic geometry
trigonometry
logarithms

Suppose that beyond our known world there is another similar world the "other world". The two are connected by a wormhole through which brave astronauts can travel safely.  A possible solution for this is the following space time.
Let the coordinates be $ (t,l,\Theta ,\phi)$, with $ l>0$ in our known world and $ l<0$ in the other one, and $ 0\leq \Theta \leq \pi$ and $ 0 \leq \phi < \2\pi$ the usual spherical angles. Let the square of the infinitesimal interval be given by:

$ ds^2=e^{2\Phi (l)}dt^2+dl^2+r(l)^2(d\Theta^2+\sin^2 \Theta d\phi^2)$

Where $ \Phi (l)=-2M/|l|$, and $ r=|l|-M \ln (|l|/r_0)$.Here $ M>0$ is the mass of the wormhole, and $ r_0>0$ is the "radius" of its throat. This spacetime is indeed Minkowskian in the $ |l| \to \infty$ limit.

How should $ M$ and $ r_0$ be chosen so that the observer descending slowly and radially into the wormhole    should not feel more than the terrestrial acceleration $ g$?

For more challenging problems like this check [url=http://ortvay.mafihe.hu/main.html]here[/url]

Wow very nice problems.  :) Yet another competition I have not heard.

i am not sure i know the correct solution
maybe shreyas can help u!

Hehe, an undergraduate physics competition. But I didn't expect them to be asking about wormholes just yet.   :huh:

Well, this is out of my league. Anyone want to take a stab at it?


algebra
polynomial
superior algebra
superior algebra solved

prove that $X^{72}-X+47$ is irreducible over $Z$

This is a variation on an old chestnut.  Since 47 is prime, one of the factors must have constant term 1 and so a root at most 1 in magnitude.


side $\bar {AB}$ of square ABCD is $\sqrt{50}$, E is between B and H, and BE = 1.  What is the area of the inner square EFGH?
(my drawing is really off...: the inner part is "supposed" to be a square) :(
This problem is from AMC 10, 2005
And:
I couldn't get this one:
An equiangular octagon has four sides of length 1 and four sides of length $\sqrt{2}/2$ , arranged so that no two consecutive sides have the same length.  What is the area of the octagon?

[hide="2nd problem"]
Let the vertices of the octagon be $A$, $B$,...$H$, such that $AB=CD=EF=GH=1$ and $BC=DE=FG=HA=\frac{\sqrt{2}}{2}$.

Extend the sides $BC$, $DE$, $FG$ and $HA$ to the both directions. The 4 lines intersect at 4 different points, forming a square.

Let $P$ be the intersection of (the extensions of) $BC$ and $ED$.
$\triangle PCD$ is a 45-45-90 triangle, so $PC=\frac{\sqrt{2}}{2}$.

Then the side length of the large square is $\frac{\sqrt{2}}{2}+2\cdot \frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}$.
Therefore, the area of the octagon is $\text{(Area of the square)}-4\times \text{(Area of the small 45-45-90 triangle)}$
$=(\frac{3\sqrt{2}}{2})^2-4\cdot \frac{1}{4}=\boxed{\frac{7}{2}}$[/hide]

EDIT- Fixed.

Wow that's a really cool approach...
Yes, nice try, but sorry that is not in the answer choices.  :(
I think you went wrong at $P = sqrt{2)/2}$; if CD is 1, then it is the square root of 2 side; so it would be x(sqrt(2)) on that side and x*1 on the CP and PD sides.  X is then $sqrt{2}/2$, so CP and PD would be $sqrt{2}/2$, maybe?  I hope i'm correct... I did the rest out and got 6sqrt(2) - 1; but that also is not correct
The answer choices are:
a. 7/2 b. (7sqrt(2))/2 c. (5+4sqrt(2))/2 d. (4+5sqrt(2))/2 e. 7

EDIT:
Oh I see what I got wrong - i forgot to square the 2 underneath the fraction... lol, nice job

Oops, you're right, I somehow thought the side lengths were $1$ and $2\sqrt{2}$, not $\frac{\sqrt{2}}{2}$ :oops: I fixed my solution.

[hide="answer"]
36, i got it using magic
[/hide]
OR
[hide="hint"]
BH = ?
AB = sqrt(50)
AH = 1
EH = ?
[/hide]

[hide="1"]
Denote length EH as x. Using the pythagorean theorem:

$\sqrt{50-1}=7=x+1\\ x=6$

Our desired area is $x^2$, which is $\boxed{36}$.
[/hide]

-BLt

Because ABCD is a square, $AD=\sqrt{50}$. Since BE=1, DG=1 and AH=1 as well. Using Pythagorean theorem, AG=7. HG=7-1=6. The area of the inner square is 6.6=24.

[hide="1"]BH by pythagorean theorem = sqrt{50-1}=7
HE = 7-1=6
Area of HEFG=36[/hide]

Oh yeah. What a miscalculation! :)  :blush:


How many $ \frac12$-inch cubes are needed to make 1 cubic foot?

One cubic foot is equal to $ 12^3$ cubic inches. $ \frac {1}{2}$ inch cubes have a volume of $ \frac {1}{8}$ cubic inches. $ \frac {12^3}{1/8}$ equals $ 13824$. Thus, it takes $ 13824$ half-inch cubes to make one cubic foot.


trigonometry
AMC
AIME
geometry
trig identities
Law of Sines
Law of Cosines

Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ$.  Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ$.  Find the number of degrees in $\angle CMB$.

[hide]

WLOG let $AC=BC=1$.  Then, using Law of Sines in triangle $AMC$, we get $\frac{1}{\sin 150}=\frac{MC}{\sin 7}$, and using the sine addition formula to evaluate $\sin 150=\sin (90+60)$, we get $MC=2 \sin 7$.

Then, using Law of Cosines in triangle $MCB$, we get $MB^2=4\sin^2 7 + 1 - 4\sin 7(\cos 83) = 1$, since $\cos 83 = \sin 7$. So triangle $MCB$ is isosceles, and $\angle CMB = 83^{\circ}$

[/hide]

[hide]
Take point $N$ inside $\triangle ABC$ such that $\angle CBN=7^\circ$ and $\angle BCN=23^\circ$.

$\angle MCN=106^\circ-2\cdot 23^\circ=60^\circ$.
Also, since $\triangle AMC$ and $\triangle BNC$ are congruent (by ASA), $CM=CN$.
Hence $\triangle CMN$ is an equilateral triangle, so $\angle CNM=60^\circ$.

Then $\angle MNB=360^\circ-\angle CNM-\angle CNB=360^\circ-60^\circ-150^\circ =150^\circ$
We now see that $\triangle MNB$ and $\triangle CNB$ are congruent.
Therefore, $CB=MB$, so $\angle CMB=\angle MCB=\boxed{83^\circ}$

[/hide]

[quote="the future"]lol this problem reminds me of 2008 AMC 10B #24...

just use a protractor and measure the angle, its AIME you get 3 hours, so take like 15 minutes to make a perfect diagram.[/quote]
Or use pro sketching skills and do it in 2. I guess and checked this one correctly. I somehow have the most godly sketching skills ever for geo stuff, but I suck at geo. One of the angles looked to be in the 30s range, so I tested every single number until I found the right one. Yea I admit its a bad strategy, but it does get you pretty close sometimes, even correct :D

[hide="Another solution"]Let $O$ be the circumcenter of $\triangle AMC$. It is easy to see that $\angle AMC = 150^{\circ}$ which implies $\angle OAC = 60^{\circ}$. So, $\triangle AOC$ is an equilateral. Notice that $\angle OCM = 60^{\circ} + 23^{\circ} = 83^{\circ}$ and $\angle BCM = \angle 106^{\circ}-23^{\circ} = 83^{\circ} = \angle OCM$. Combining this with the fact that $OC = AC = CB$, we conclude that $OB$ is the perpendicular bisector of $CM$. Since $OC = OM$, then $OC = OM = BM = BC$ and thus $\angle CMB = \angle MCB = 83^{\circ}$[/hide]

@above please don't bump 10-year-old threads...

@above ... I don't think he has done anything wrong doing it. If you see the solution, its quite different from the ones which have been given by the others. [b]Alternate solutions help the users to get a new dimension over solving a problem. There is nothing wrong if one is posting a new approach for the same question.[/b] After all, it is only helping the others to get new solving strategies to the question. I am not trying to be offensive. It is just my advice to you for the future. I hope you get my point ...

Take $P$ reflection of $C$ across $AM$, see that $\triangle CMP$ is equilateral and $m(\widehat{ACP})=m(\widehat{BCM})=83^\circ$, thus triangles $CBM$ and $CAP$ are congruent, s.a.s. criterion, consequently $\triangle CBM$ is isosceles, as $CAP$, and $m(\widehat{CMB})=m(\widehat{MCB})=83^\circ$.

Best regards,
sunken rock

[hide=Solution]Denote $\angle ABM = x$. From Trig Ceva, we have $\frac{\sin 30}{\sin 7} \cdot \frac{\sin{(37-x)}}{\sin x} \cdot \frac{\sin 23}{\sin 83}=1$, or $\frac{1}{2}\sin{(37-x)}\sin 23 = \sin 7 \sin 83 \sin x$. We have $\sin 7 \sin 83 = \frac{1}{2}\cos 76$ from Product-to-Sum, so $\sin{(37-x)}\sin 23 = \sin x \cos 76$. This is equivalent to $\frac{\sin{(37-x)}}{\sin x}=\frac{\sin 14}{\sin 23}$, meaning that $x=23$. We have $\angle CMB = x+60=\boxed{83}$.[/hide]

Generalisation of the problem at [url]https://stanfulger.blogspot.com/2021/08/aops-own-generalisation-of.html[/url]

Best regards,
sunken rock

[quote=Bluesoul]angular Ceva theorem
which is sin83/sin23 * sin7/sin30 * sinx/sin(37-x)=1
which is 2cos7sin7/sin23 * sinx/sin(37-x)=1
sin14/sin23 * sinx/sin(37-x)=1
hence x=23 in degrees
the answer is 30+30+23=83 degrees we are done[/quote]

where can i find the proof of this theorem?


[quote="sylow_theory"]Why isnt my signature showing?
Is it because latex is not allowed?[/quote]

 go to profile --> Preferences --> posting a message

and hit "always attach my signature"


calculus
integration
function
calculus computations

$\int \frac{x^{3}}{1+x^{8}}dx$. By substitution ($t=x^{4}$), I get $\frac14\arctan(x^{4})+c$. Is it right?

EDIT: forget it, I know it's right.

well we can always differentiate to check

$\frac{d}{dx}\frac{1}{4}\arctan(x^{4})+C$

$\frac{1}{4}\frac{4x^{3}}{1+(x^{4})^{2}}$

$\frac{x^{3}}{1+x^{8}}$


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Why is there no option for over 50? Also, the options overlap.


Given a triangle $ OAB$ with area $ 1$. Let $ C$ be the point such that $ AC: CB \equal{} 2: 1$ and $ D$ be the mid point of $ OB$. Suppose that moving points $ P,\ Q$ are on the sides $ OA,\ OB$ respectively and $ R$ of the barycentre of triangle $ CPQ$ are on the line segment $ AD$.Find the area of the region swept by line segment $ OR$.


Assume a regular 8x8 chessboard. Join the top and bottom, and right and left to form a torus. What is the maximum number of knights that can be placed on the board so they do not attack each other?

[hide="hint"]Think about parity on a chessboard[/hide]

[quote="JBL"][hide]Just as with a conventional torus, the answer is 32: place one on each white square to attain this result.  We cannot do better: The $n$ knights on white squares each attack 8 black squares (as it is clear that no knight attacks the same square twice, since the attacked squares are at most 5 units apart vertically or horizontally) and each black square is attacked by at most 8 knights, so the $n$ knights necessarily block $n$ possible knights on the white squares.  Thus there cannot be a total of 33 or more knights on the board.  Indeed, it seems that making the board a torus could not possible allow [i]extra[/i] knights to fit (since each knight just gets to attack more squares) while it doesn't make it any harder for them to fit, either (since they still only attack squares of the opposite color).[/hide]

I have several related questions:
For which I do know the answer:
1) How many knights at most can fit on a (regular) $n \times n$ chessboard?
2) For even $n$, how many knights can fit on a toroidal $n \times n$ chessboard?
For which I don't know the answer:
3) A cylindrical board would be one in which just one pair of edges are identified (joined).  For odd $n$, how many knights can fit on a cylindrical $n \times n$ chessboard?
4) For odd $n$, how many knights can fit on a toroidal $n \times n$ chessboard?[/quote]

Wow, that proof is the exact same as mine, except for minor phrasing issues.

Good job, I'll get working on your problems. Thanks for the problems, new ones are always fun.

I have developed a hypothesis for my question 3, but I wouldn't be terribly surprised if it can be beaten sometimes in larger cases.  Certainly, though, we can set a lower bound:
[hide]If we instead think of just any $m \times n$ cylinder, $n$ odd, where $n$ is the "circumference" measure and $m$ is the "height," we can get $n \cdot \lceil \frac m3 \rceil$ by filling in every 3rd row up the height of the cylinder.[/hide]

Three years later, some solutions:

The answer to question 1 is given by [url=http://www.research.att.com/~njas/sequences/A030978]A030978[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url]: for $ n$ odd, it's $ \frac {n^2 \plus{} 1}{2}$, for $ n \equal{} 2$ it's $ 4$, and for $ n > 2$ even it's $ \frac {n^2}{2}$.  For $ n \neq 2$, consider a knight's tour (on the odd board, a path, on the even, a circuit).*  Then the positions of our knights form an independent subset of an $ n^2$-path (for $ n$ odd) or an $ n^2$-cycle (for $ n$ even), also subject to some other conditions, and so have size at most the claimed values.  In both cases, we achieve this bound by putting knights on all the squares of the same color (in the odd case, including the corner squares).

For the even case, this shows that the answer is the same for the cylindrical and toroidal boards, since it is still true that a knight on a square of one color attacks only squares of the opposite color.  However, this is no longer true for boards of odd dimension (a knight on the boundary now attacks some squares of the same color as its own).

For completeness, it's easy to see that for a usual $ 2 \times 2$ chessboard we can fit four knights, but if we make it toroidal or cylindrical then we can only fit the expected two.

For odd-size cylindrical and toroidal boards, the answer must be asymptotic to $ \frac {n^2}{2}$ (since we can put a knight on every black square except near the edges, which include a vanishingly small fraction of the whole board), but I'm still unsure what the true value should be.  Neither problem appears to be associated to a sequence in the OEIS, unfortunately.

* Apparently, I was wrong that knight's tours are known to exist for all board sizes.  Instead, start pairing off squares that are a knights-move apart.  We can divide the board into disjoint pairs of this sort (with one square left over in the odd case) when $ n \neq 2$ and this gives us the claimed result.

This is a problem that's actually even prettier than the Iranian problem (courtesy of K. Cordwell):

Find, with proof, the maximal number of knights that can fit on a $ 5 \times 5$ toroidal chessboard.

Further remarks on torus chessboards with at least one odd edge: we've shown that a $ 3 \times n$ board allows at most $ n$ knights, that a $ 4 \times n$ board allows at most $ 2n$ knights if $ n$ is even or $ 2n \minus{} 4$ knights if $ n$ is odd, and that a $ 5 \times 3n$ board allows at most $ 5n$ knights.  We also conjecture that for every board size, the maximum is achieved either by "pinstriping" (filling several rows or columns completely with knights and leaving the rest of the board blank) or by trimming off some number of edges (zero on an even-by-even board, two on an even-by-odd board or four on an odd-by-odd board) and filling the rest in the usual chessboard pattern.  We have upper bounds that are upper bounds for a cylindrical board with one odd edge (based on decomposing the board into a union of disjoint odd cycles) and which are better than what you get by pretending the board is flat, but still a ways from the truth.


pigeonhole principle
combinatorics unsolved
combinatorics

A square of side $ n$ is divided into $ n^2$ unit squares each of which is colored red, yellow or green. Find the smallest value of $ n$ such that, for any coloring, there exist a row and a column having at least three equally colored unit squares.  :(

by pigeonhole principle it always works for $ n\equal{}7$ for each row and column
and for $ n\equal{}6$  alternate the three colors

1	2	3	1	2	3
3	1	2	3	1	2
2	3	1	2	3	1
1	2	3	1	2	3
3	1	2	3	1	2
2	3	1	2	3	1

with no nice coloring.


geometry
circumcircle
projective geometry

Quadrilateral $ ABCD$ is cyclic. The tangents to the circumcircle of $ ABCD$ at $ A, B, C, D$ are, respectively, $ t_A, t_B, t_C, t_D$.

$ t_A \cap t_C \equal{} P_1$

$ t_B \cap t_D \equal{} P_2$

$ AB \cap CD \equal{} P_3$ (the lines)

$ BC \cap DA \equal{} P_4$ (the lines)

Prove $ P_1, P_2, P_3, P_4$ are collinear.

It's not too tough, but I thought it was a pretty nice property.

[hide="Solution"]

Lemma: Consider points $ LMNPQR$ on a circle. By Pascal's theorem, $ LM \cap PQ$, $ MN \cap QR$, $ NP \cap RL$ are collinear. If we force $ L$ and $ M$ to coincide and call the point $ S$, and we force $ P$ and $ Q$ to coincide and call that point $ V$, then we get $ t_S \cap t_V$, $ SN \cap VR$, $ NV \cap RS$ are collinear.

If we apply the lemma, we find that $ t_A \cap t_C, AB \cap CD, BC \cap DA$, or $ P_1, P_3, P_4$, are collinear.

Applying the lemma again, we find that $ t_B \cap t_D, AB \cap CD, BC \cap DA$, or $ P_2, P_3, P_4$, are collinear.

Since two points determine a line, $ P_1, P_2, P_3, P_4$ are all collinear, and we are done.

[/hide]

nah, i rediscovered it during my speech class.  :lol:


[color=blue] [size=150]FRANCIA 1997[/size][/color]

[color=blue][i]Enunciado[/i][/color]

Dado un poligono de 1997 lados, .. se etiquetan sus vertices con numeros enteros, de tal modo que la suma de estos enteros sea 1.
Comenzamos el etiquetado por algun vertice del poligono, siguiendo el orden antihorario.

Podremos elegir un vertice de inicio, tal que la suma de los $k$ primeros enteros escritos sea positivo (para $k = 1, \cdots, 1997$) ?

(Sugerencia)
[hide="Hint"] Definiamos $b_k$: suma de los primeros $k$ enteros.. analicemos $b_1, b_2, \cdots, b_{1997}$[/hide]

Carlos Bravo :)
Lima- PERU


Will writing my solutions by hand make me less likely to get a commended solution? This is an example of my handwriting:
[img]http://img90.imageshack.us/img90/5663/handwritingsf1.png[/img]

I don't think it should.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166536

[quote="Wikipedian1337"]Will writing my solutions by hand make me less likely to get a commended solution?[/quote]No.


geometry
circumcircle
symmetry
vector
perpendicular bisector
power of a point
radical axis

Two points $ A$ and $ C$ are chosen on a circle $ \Gamma$. Let $ O$ be an arbitrary point on the perpendicular bisector of $ \overline{AC}$. A circle centered at $ O$ cuts $ \overline{OA}$ and $ \overline{OC}$ at $ A',C'$ respectively. The tangents to this circle at $ A'$ and $ C'$ meet $ \Gamma$ at points $ B$ and $ D$ respectively such that $ A,B,C,D$ lie on $ \Gamma$ in this order. $ AB$ and $ CD$ meet at $ P$. Show that the segment adjoining the midpoints of $ \overline{AB}$ and $ \overline{CD}$ is perpendicular to $ OP$.

$ M, N$ are midpoints of $ AB, CD.$ $ (M), (N)$ are circles with diameters $ AB, CD.$  $ \angle AA'B, \angle CC'D$ are right $ \Longrightarrow$ $ A' \in (M), C' \in (N).$ $ \overline {PA} \cdot \overline {PB} \equal{} \overline {PC} \cdot \overline {PD}$ because $ ABCD$ is cyclic with circumcircle $ \Gamma.$ $ \overline {OA'} \cdot \overline {OA} \equal{} \overline {OC'} \cdot \overline{OC}$ by symmetry. $ PO$ is radical axis of $ (M), (N)$ perpendicular to their ceter line $ MN.$

Almost same as the previous problem ([url]http://www.mathlinks.ro/viewtopic.php?p=1176173#1176173[/url])
I am also completing like the previous one.
We just need to prove that, $ OB^2 \minus{} OD^2 \equal{} AB^2 \minus{} CD^2$
$ OB^2 \minus{} OD^2 \equal{} BA'^2 \minus{} C'D^2 \equal{} (B'A^2 \plus{} AA'^2) \minus{} (C'D^2 \plus{} CC'^2) \equal{} AB^2 \minus{} CD^2$
Here, from symmetry $ AA' \equal{} CC'$
And we are done!  :)

after seeing yetti"s solution i am.....speechless :o (oooo...) :D

in my solution i just showed the following
$ OM^2\minus{}ON^2\equal{}(AB^2\minus{}CD^2)/4\equal{}PM^2\minus{}PN^2$
then the result follows by the lemma.......(well known)
i am really impressed by yettis solution
nayel wats ur solution?????moon said urs is by vector
can u send it here??? :)

Yetti's solution is just awesome. :alien:

yetti's idea was simply brilliant! :D  I wanted to post another similar problem:

[hide]$ ABCD$ is a cyclic quadrilateral and $ \Gamma$ is any circle passing through $ A$ and $ D$. The line perpendicular to $ AB$ at $ B$ meets $ \Gamma$ at $ B'$ and the line perpendicular to $ CD$ at $ C$ meets $ \Gamma$ at $ C'$. Let $ AB$ and $ CD$ meet at $ P$ and let $ AB'$ and $ C'D$ meet at $ P'$. Show that the segment adjoining the midpoints of $ AB'$ and $ C'D$ is perpendicular to $ PP'$.[/hide]
But after yetti's ingenious solution this doesn't get much place.. :oops:

[quote="nayel"]$ ABCD$ is a cyclic quadrilateral and $ \Gamma$ is any circle passing through $ A$ and $ D$. The line perpendicular to $ AB$ at $ B$ meets $ \Gamma$ at $ B'$ and the line perpendicular to $ CD$ at $ C$ meets $ \Gamma$ at $ C'$. Let $ AB$ and $ CD$ meet at $ P$ and let $ AB'$ and $ C'D$ meet at $ P'$. Show that the segment adjoining the midpoints of $ AB'$ and $ C'D$ is perpendicular to $ PP'$.
[/quote]Anyway, After the solution of Mr. yetti, when I got this Problem yesterday I sent this diagram to Nayel as a proof (almost) without word.  :D
[u][b]Few Words:[/b][/u] The radical center of $ \omega_1,\omega_3,\omega_4\equal{}P$ and of $ \omega_2,\omega_3,\omega_4\equal{}P'$

[quote="yetti"]
Thank you guys. The fact that radical axis is perpendicular to the center line is so well known that one sometimes forgets, how is it proved. It is proved in exactly the same way the other posted proofs are done.[/quote]Actually I always forget this property and it is not always that easy (for me) to find it out. :blush:


conics
ratio
trigonometry
geometry
trig identities
Law of Sines
geometry proposed

in a triangle 3 concurrent cevians form 6 angles and the angle bisectors of these meets the side lines on 6 point. Prove that these 6 points are on a conic.

I think that the problem is not correct, generally.

As the six points $ D',$ $ D'',$ and $ E',$ $ E'',$ and $ F',$ $ F'',$ on the side lines $ BC,$ $ AC,$ $ AB$ respectively, of a given triangle $ \bigtriangleup ABC,$ $ ($ let $ P$ be, the concurrency point of the cevians $ AD,$ $ BE,$ $ CF$ and suppose the points $ D',$ $ E',$ $ F',$ as adjoined to $ B,$ $ C,$ $ A,$ respectively $ ),$  form the convex hexagon $ D'D''E'E''F'F'',$ I see by the drawing that the points $ Q\equiv D'D''\cap E''F',$ and $ R\equiv E'E''\cap D'F''$ and $ S\equiv F'F''\cap D''E',$ are not collinear, indispensable by [b][size=100]Pascal’s theorem.[/size][/b]

Kostas vittas.

ehm really they are collinear, i prove the problem so... :D

[img]http://img513.imageshack.us/img513/1482/geometria24ze4.png[/img]

I apologize because I misunderstood your problem.

I thought the 6 angles in which divided the angles of the given triangle, from the concurrent cevians. :blush:

$ ($ $ \angle BAX \equal{}\angle XAX'',$ $ \angle X''AX' \equal{}\angle X'AC,$ .... $ ).$

Sorry again, Kostas Vittas.

no problem, i have exspleined bad (i'm not good in english  :blush: )

At first, we present a [b][size=100]Lemma[/size][/b], based on which, we will prove the proposed problem.

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ P$ be, an arbitrary point on the side segment $ BC.$ WE define two points $ D,$ $ E,$ on the side segments $ AB,$ $ AC$ respectively, such that $ \angle BPD =\angle CPE$ and we denote as $ D',$ $ E',$ the intersection points of $ AB,$ $ AC,$ from the angle bisectors of $ \angle APD$ and $ \angle APE,$ respectively. Prove that the line segments $ BC,$ $ DE,$ $ D'E',$ are concurrent at one point.[/color][/size][/b]

This problem has already been posted in [b][size=100]Hellenic Forum[/size][/b], at http://www.mathlinks.ro/Forum/viewtopic.php?t=166326

includes two alternative proofs based on [b][size=100]Double Ratio Theory[/size][/b] and one, by [b][size=100]Menelaus theorem.[/size][/b]

$ \bullet$ In our configuration now (see the [b][size=100]schema t=165703[/size][/b]), we can easy to see that the triangles $ \bigtriangleup DEF,$ $ \bigtriangleup PLL'$ are perspective, because of $ A\equiv DP\cap EL\cap FL'$ and so, by [b][size=100]Desarques’s theorem[/size][/b], we have that the points $ Q\equiv DE\cap PL,$ $ R\equiv EF\cap LL',$ $ S\equiv DF\cap PL',$ are collinear.

Based on the above [b][size=100]Lemma[/size][/b], we have that the line segments $ KK',$ $ DE,$ $ LM',$ are concurrent at one point and so, we conclude that the line segment $ KK',$ passes through the point $ Q,$ as the intersection point of $ DE,$ $ LM'.$

Similarly, the line segments $ LL'$ and $ MM',$ pass through the points $ R,$ $ S,$ respectively.

$ \bullet$ We will prove now, that the point $ C'\equiv LL'\cap MM',$ lies on the line segment $ CP.$

We denote the points $ C'\equiv CP\cap LL',$ $ C''\equiv CP\cap MM'$ and it is enough to prove that $ C'\equiv C''.$

The pencil $ R.CK'EL,$ is intersected from the line segments $ AC,$ $ CP$ and so, we have $ (C,K',E,L) = (C,P,F,C')$ $ ,(1)$

The pencil $ S.CKDM',$ is intersected from the line segments $ BC,$ $ CP$ and so, we have $ (C,K,D,M') = (C,P,F,C'')$ $ ,(2)$

Because of the concurrency of $ KK',$ $ DE,$ $ LM',$ we conclude that $ (C,K',E,L) = (C,K,D,M')$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ (C,P,F,C') = (C,P,F,C'')$ $ \Longrightarrow$ $ C''\equiv C'.$

By the same way we can prove that the points $ A'\equiv KK'\cap MM'$ and $ B'\equiv KK'\cap LL',$ lie on the line segments $ AP,$ $ BP,$ respectively.

$ \bullet$ So, because of the concurrency of $ AA',$ $ BB',$ $ CC',$ we have that the triangles $ \bigtriangleup ABC,$ $ \bigtriangleup A'B'C',$ are perspective and then by [b][size=100]Desarques’s theorem[/size][/b] again, we conclude that the points $ X\equiv AB\cap A'B',$ $ Y\equiv BC\cap B'C',$ $ Z\equiv AC\cap A'C',$ are collinear.

But, we can also see these points, as the intersection points of the opposite sidelines of the convex hexagon $ KK'LL'MM'$

$ ($ $ X\equiv KK'\cap L'M,$ $ Y\equiv LL'\cap KM',$ $ Z\equiv MM'\cap K'L$ $ ).$

Hence, by [b][size=100]Pascal’s theorem[/size][/b], we conclude that the six points as the problem states, lie on a conic and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Son Hong Ta.[/size][/b]

Kostas Vittas.

thank you vittasko, this is my solution:

[img]http://img472.imageshack.us/img472/3290/geometria26bb8.png[/img]

[b]Carnot's Theorem[/b]

[i]if a conic intersect the sides BC at X, X', CA at Y, Y' and AB at Z, Z' then[/i]

$ \frac{BX}{XC}\cdot\frac{BX'}{X'C}\cdot\frac{CY}{YA}\cdot\frac{CY'}{Y'A}\cdot\frac{AZ}{ZB}\cdot\frac{AZ'}{Z'B}\equal{}1$

[i]and if the relation is true then the 6 points are on a conic.[/i]

[b]Proof[/b]

for the law of sines:

$ \frac{BX}{XC}\equal{}\frac{PX\cdot\sin 1}{\sin b}\cdot\frac{\sin c'}{PX\sin{(1\plus{}2\plus{}2)}}\equal{}\frac{\sin 1\cdot\sin c'}{\sin{(1\plus{}2\plus{}2)}\cdot\sin b}$

$ \frac{BX'}{X'C}\equal{}\frac{PX'\cdot\sin{(2\plus{}1\plus{}1)}}{\sin b}\cdot\frac{\sin c'}{PX'\sin2}\equal{}\frac{\sin{(2\plus{}1\plus{}1)}\cdot\sin c'}{\sin2\cdot\sin b}$

$ \frac{CY}{YA}\equal{}\frac{PY\cdot\sin 3}{\sin c}\cdot\frac{\sin a'}{PY\sin{(3\plus{}1\plus{}1)}}\equal{}\frac{\sin 3\cdot\sin a'}{\sin{(3\plus{}1\plus{}1)}\cdot\sin c}$

$ \frac{CY'}{Y'A}\equal{}\frac{PY'\cdot\sin{(1\plus{}3\plus{}3)}}{\sin c}\cdot\frac{\sin a'}{PY'\sin1}\equal{}\frac{\sin{(1\plus{}3\plus{}3)}\cdot\sin a'}{\sin1\cdot\sin c}$

$ \frac{AZ}{ZB}\equal{}\frac{PZ\cdot\sin 2}{\sin a}\cdot\frac{\sin b'}{PZ\sin{(2\plus{}3\plus{}3)}}\equal{}\frac{\sin 2\cdot\sin b'}{\sin{(2\plus{}3\plus{}3)}\cdot\sin a}$

$ \frac{AZ'}{Z'B}\equal{}\frac{PZ'\cdot\sin{(3\plus{}2\plus{}2)}}{\sin a}\cdot\frac{\sin b'}{PZ'\sin3}\equal{}\frac{\sin{(3\plus{}2\plus{}2)}\cdot\sin b'}{\sin3\cdot\sin a}$

so with ceva and $ \sin(2\plus{}1\plus{}1)\equal{}\sin(2\plus{}3\plus{}3)$ and cyclyc we have:

$ \left (\frac{\sin{a'}\cdot\sin{b'}\cdot\sin{c'}}{\sin{a}\cdot\sin{b}\cdot\sin{c}}\right )^{2}\cdot\frac{\sin{1}\cdot\sin{2}\cdot\sin{3}}{\sin{1}\cdot\sin{2}\cdot\sin{3}}\cdot$$ \frac{\sin{(2\plus{}1\plus{}1)}}{\sin{(2\plus{}3\plus{}3)}}\cdot\frac{\sin{(1\plus{}3\plus{}3)}}{\sin{(1\plus{}2\plus{}2)}}\cdot\frac{\sin{(3\plus{}2\plus{}2)}}{\sin{(3\plus{}1\plus{}1)}}\equal{} 1$


Disclaimer: if you are a Titanic fanatic, I ask you to leave this topic or if you insist on checking out the link, I have NO full responsibility for the production of it, NOT EVEN PART of it.

Check out [url]http://www.angryalien.com/0604/titanicbunnies.html[/url]
 :D  :D  :D  :rotfl:  :D  :D  :D

So someone actually reads my littl' old blog  :blush:  :P

THose bunny animations are FUNNY. I laughed my ass off with the one about The Shining


pigeonhole principle
number theory solved
number theory

Show that for every natural m,n m not divisible by 2 or 5 there exists a multiple of m formed by putiing together the digits of n a number of times-for example if n is 235 the multiple must be of the form 235235...235 .

This is clearly Pigeonhole Principle.
Let xt be the number obtained by repeating the digits of n exactly t times.
[For example, if n = 217 then x3 = 217217217.]
Consider the numbers x1, x2, ..., xm + 1.
We have m + 1 numbers of this form. There are only m residue classes modulo m.
So some two of them, say xa and xb, are congruent modulo m.
This means that m divides xa - xb.
But xa - xb = 10r xs for some r and s. As m is coprime to 10, m divides xs.

:)


Okay this is a really wierd problem I discovered in this book of puzzles and brainteasers. There is this snake. It starts eating its tail and starts eating itself. The question is, what happens at the end. There is a snake eating itself starting at the tail. Does it;

*Die in the process of eating itself.
*Disappear in thin air.
*Lose the ability to eat any more at a certain point.

I don't know if the answer has been found yet.

I think it will die in the process of eating itself, thereby losing the ability of eating more of itself at a certain point.

But animal's brains do not work like human's brains, they care highly about survivalism and would not "try" to suicide.

I agree with james4l. But even if for some reason it began to eat itself, due to starvation or something, it would probably faint from pain and then die eventually.

Would if it didn't die though, then what do you think would happen.

Okay, we assume it would die.

New Problem;

The same snake is eating itself again starting at the tail, but the snake is also immortal, so it does not die. What happens now?

I have doubts as to if it can reach after a certain extent.

Well, even if it could open its mouth as wide as it could, it wouldn't be able to twist its head around 360 degrees, it would just be stuck like that for the rest of its life.

It would probably stop very soon after starting due to large amounts of pain, too.

Yeah true. I wonder how the people that wrote about this question came up with it.

I think once it got to its stomach it wouldn't exactly be able to do much...

At some point the snake would not be able to proceed any further because its own body would be in the way.  Picture the snake eating itself as a torus.  The "hole" enclosed within the torus (analogous to the hole in a donut) decreases in area, so eventually it will reach zero.  Once that happens, the snake cannot continue eating itself because it just wouldn't have the room.

out of curiosity, where does a snake discharge waste??
also, i don't think a snake can eat itself. most snakes swallow prey whole, and then the prey digests in the snake's stomach. If you can imagine it, i think the tail end of the snake would enter the mouth, keep moving down, and "coil inside" the snake.Draw a picture-you'll see what i mean. Eventally, the snake would not be able to move and be stuck that way for the rest of his life. This is the weird part: if the tail is [i] outside[/i] of the body, how can it be both [i]outside[/i] and [i] inside[/i] at the same time??! I thought the tail would begin to be digested, but how can the STOMACH digest itself (or a part connected to itself)?? wow this is actually interesting...

[quote="herefishyfishy1"]If the snake was immortal, then it would become smaller and smaller by volume but its mass would be conserved. So it would turn into a sort of mini black hole. But only if it was able to twist its body around many times.[/quote]

Ingenious. We don't even need an LHC, just a snake!

There are a lot of different posibilities. I don't even think scientists have figured it out yet. When I read about this in the book I got it from, it said that they weren't too sure that it would die.

"repeat experiment for all n in the set of integers, which we use n snakes, each eating each other."

Hmmm. Good thought. It MIGHT be possible: well, one snake is mad at the other snake so they bite the other snake's tail. Then in retaliation they bite the next snake, and the more pain, the more retaliation. Now that's giving an incentive to the snakes to make a black hole!  :D

SNAKES CAN'T BITE A CHUNCK OUT OF SOMETHING

[quote="pacman2812"]Or, maybe a better way would be with two snakes, each one eating the other's tail. Interesting  :|[/quote]

Any solutions for this problem? This one is pretty interesting too.

I've heard a similar version of this before.
There are 4 snakes eating each other. The result is that they become so fat after a while, that they can't finish.

violin321 has a good point. They don't take a huge bite out of something; they swallow it whole.

Hmm, here's a good one, would if there are 5 snakes. All of which immortal, can open their mouth as wide as they wish, and can stretch their heads to any extent.

They would probably choke and be forced to stop.

all the snakes would like to eat each other, so, seeing each other they would open their mouths. each snake would see that others' mouth size is increasing so, they will keep increasing their size to infinite sizes and nobody would be able to eat anything :D
very sad  :D

Or, maybe, they successfully eat eachother, and they all disappear in thin air, and create a black hole, that sucks up all matter. It could be a possible lee. lee - life extinguishing event (like Yellowstone erupting)

this is assuming the snakes arent poisonous, right?

Um, please don't revive. We have already established that the snakes don't kill each other. If you are interested in this thread, then start a new one. The problem has already been solved.


Having trouble with this, please help

The natural numbers greater than 1 are arranged this way

       2    3    4     5    6    7
 13  12   11   10   9    8
      14   15   16  17   18   19
 25  24   23   22  21   20

In which column and row will 150 appear?  what about 1001 and 5237?  When N >1, how can we determine what row and column N will be in?

The numbers do not match up as I had entered them 2,3,4,5,6,7,are in the right place but the 8 should line up under the 6 so 13 hands on its own.  When the numbers wrap they continue in this manner.  This must have an impact on the answer

HELP PLEASE

meh i try, i fail
$ \ 150\div\6=25$
so ans is
$ \ 2nd column and 25th row$

[hide="hint"]look at the columns (each column) and its terms mod 12 [/hide]

hmmm ok heres what i did if what you meant was that the order would be like this:
      2    3    4    5    6   7
13  12   11  10   9    8
     14   15  16   17  18 19...

 ok for 150 what i did was: (150-1)/6= 24 with a residue of 5. so it would be in row # 25. then with the residue i just positioned it on the column the way the uneven rows are meaning it would be on column #7

 I did the same with the other and my results were: 1001= row # 167, column #5.
5237= row #873, column #5. and well i would find the row of n>1 by using this equation:
  n-1/6= row-1(if there is a residue)
  n-1/6= row(if tehre is no residue)

 first answer i post so please notify me if i am wrong   :blush:

Using Nikeballa's hint, wouldn't it be column 5(#6), row 25?

ohh now i wrote the 150 one compleatly and it landed on column #6 im sry i meant to write column 6 not 7 on the 150


[quote]Find the number of n digit even and odd numbers with the property that no two adjoining digits are the same
Even if u guessed the answer please do it without induction,it kills the problem.[/quote]

Hmm..@Pardesi: Do you want me to separately solve for even and odd???

Total ans is obvious 9^n.

if you want separately:  i can think of symmetry between even and odd.. either no of even=no of odd omitted no's or they differ by a constant factor proportional to n. Anyway, tell me.

yes the total is obviuosly $ 9^{n}$
ok let me tell u the answer it is $ \frac {9^{n} \minus{} ( \minus{} 1)^{n}}{2}$
just prove it  :)

Just use Recursion ( Maybe this wasnt too easy as i thought )  :D


Putnam algorithm function algebra domain calculus integration

I was reading the solution to a Putnam problem when it was stated that $(1+x)^{1/2} = 1 + x/2 + O(x^2)$ when $-1 < x < 1$. I know the very basics of big-O notation, but I don't understand why the statement is true. Could someone please explain?

Thanks.

I actually had to look this one up.  I've seen big-O notation a lot in regards to computing algorithms, but one use I did not know was

Big O can also be used to describe the error term in an approximation to a mathematical function. For example,

    $e^x=1+x+x^2/2+\hbox{O}(x^3)\qquad\hbox{as}\ x\to 0$

expresses the fact that the error is smaller in absolute value than some constant times $x^3$ if $x$ is close enough to 0.

For small $x$, by the binomial theorem,
\[(1 + x)^r = 1 + rx + \frac{r(r - 1)}{2!} x^2 + \frac{r(r - 1)(r - 2)}{3!} x^3 + \dots.\]
Taking $r = 1/2$ gives
\[(1 + x)^{1/2} = 1 + \frac{x}{2} + O(x^2),\]
where the $O(x^2)$ stands for terms containing $x^2$ or higher powers of $x$.

Thanks for the explanations guys. I understand now.

So in the solution I was reading, I saw the statement

$x^{1/2} (\sqrt{1+a/x} -1) = x^{1/2} (1 + a/2x + O(x^{-2}))$.

Shouldn't the right equality be

$x^{1/2} (\sqrt{1+a/x} -1) = x^{1/2} (a/2x + O(x^{-2}))$

since the 1's cancel, or I am doing something wrong (I got this from the official Putnam solutions for a particular year, but maybe there was a typo?)?

Yeah, that's a typo. I've seen miswritten questions as well.

Ok.

By the way, in the equation I wrote in my last post, the solution equated $\sqrt{1+a/x}$ and $1 + a/2x + O(x^{-2})$, but how do we know that $|a/x|<1$ is true so that we can use the extension of the binomial theorem jhaussmann5 provided?

Note that x is just a real variable and a is a positive real number (constant). By the way, my questions are regarding problem A2 here: http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1995.pdf .

[quote="blahblahblah"]$|a/x|<1$ is true for all sufficiently large $x$ - the integrand is bounded everywhere else so it gives some finite integral.[/quote]

It is true for large $x$, but the solution is expressing the function to be integrated using big-O notation and THEN integrating this function. Thus, the use of the big-O notation should express the function properly over all x (in the domain) and not just for large x (as in the upper bound of integration), right? Furthermore, the lower bound of integration is a constant $b$, and $|a/x|<1$ may not be true when x is around $b$.

No.  We don't care about the value of the integral on a bounded interval, as long as it doesn't blow up there.  There are really only two ways you can get a divergent integral (that is, one that goes off to infinity)

The function can blow up, or the measure of the set we're integrating over can blow up.  We already can eliminate the first case, so we can ignore the integral of the function over bounded (compact) intervals.

Thanks. I see.

I just have one more, erm, slight problem...so now, due to the solution typo mentioned above, I have to find $(a/2x + O(x^{-2}))^{1/2}$ in terms of a sum of terms of the form $cx^k$ and/or $O(...)$. Using the binomial theorem to express $(a/2x + O(x^{-2}))^{1/2}$ does not appear to yield anything too nice, or at least anything that would give me a bunch of $cx^k$ terms and an $O(x^{-2})$ anyways. How would you suggest to work out this dilemma?

I did post a classroom handout of mine on the subject [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31517]here[/url].


Let $ a,b,c$ be non-negative real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that

1/ $ \frac a{a\plus{}2bc}\plus{}\frac b{b\plus{}2ca}\plus{}\frac c{c\plus{}2ab}\ge1$

2/ $ \frac a{2a\plus{}bc}\plus{}\frac b{2b\plus{}ca}\plus{}\frac c{2c\plus{}ab}\le1$

Ps: A well known's Vasc :  $ \frac a{a\plus{}bc}\plus{}\frac b{b\plus{}ca}\plus{}\frac c{c\plus{}ab}\ge\frac3{2}$

Problem 2.
\[ \sum_{cyclic}\frac{a}{2a\plus{}bc} \leq 1 \longleftrightarrow \sum_{cyclic}\frac{2a}{2a\plus{}bc} \leq 2\]
\[ \longleftrightarrow \sum_{cyclic}(1\minus{}\frac{2a}{2a\plus{}bc}) \geq 1\]
by Cauchy-Schwarz ; \[ \sum_{cyclic}\frac{bc}{2a\plus{}bc} \equal{}  \sum_{cyclic}\frac{(bc)^2}{2abc\plus{}b^2c^2} \geq \frac{(ab\plus{}bc\plus{}ca)^2}{6abc\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2}\]
\[ \equal{} \frac{a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}2abc(a\plus{}b\plus{}c)}{6abc\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2}\equal{}1\]

Problem 1. \[ \frac a{a\plus{}2bc}\plus{}\frac b{b\plus{}2ca}\plus{}\frac c{c\plus{}2ab} \equal{} \sum_{cyclic}\frac{a^2}{a^2\plus{}2abc} \geq \frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2\plus{}6abc}\]
from am-gm ; $ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \geq 9abc$ ,so $ ab\plus{}bc\plus{}ca \geq 3abc$
\[ \therefore \frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2\plus{}6abc} \geq \frac{a^2\plus{}b^2\plus{}c^2\plus{}6abc}{a^2\plus{}b^2\plus{}c^2\plus{}6abc} \equal{} 1\]

[quote="Vasc"][quote="dduclam"]Ps: A well known's Vasc :  $ \frac a{a \plus{} bc} \plus{} \frac b{b \plus{} ca} \plus{} \frac c{c \plus{} ab}\ge\frac3{2}$[/quote]
This inequality is nice. But where I posted it?[/quote]
It is equivalent to the following of you, VasC
If $ a,b,c \ge 0,ab\plus{}bc\plus{}ca\equal{}3$, then
\[ \frac{1}{a^2\plus{}1}\plus{}\frac{1}{b^2\plus{}1}\plus{}\frac{1}{c^2\plus{}1} \ge \frac{3}{2}\]
:)

[quote="can_hang2007"][quote="Vasc"][quote="dduclam"]Ps: A well known's Vasc :  $ \frac a{a \plus{} bc} \plus{} \frac b{b \plus{} ca} \plus{} \frac c{c \plus{} ab}\ge\frac3{2}$[/quote]
This inequality is nice. But where I posted it?[/quote]
It is equivalent to the following of you, VasC
If $ a,b,c \ge 0,ab \plus{} bc \plus{} ca \equal{} 3$, then
\[ \frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} 1} \ge \frac {3}{2}
\]
:)[/quote]
You are right, Can (by replacing a,b,c with 1/a, 1/b, 1/c). Thanks. And this inequality is in Algebraic Inequalities (page 7 and 27). :P

[quote="Allnames"][quote="Vasc"][quote="dduclam"]Ps: A well known's Vasc :  $ \frac {a}{a \plus{} bc} \plus{} \frac b{b \plus{} ca} \plus{} \frac c{c \plus{} ab}\ge\frac3{2}$[/quote]
This inequality is nice. But where I posted it?[/quote]
I think it must be$ \frac a{1 \plus{} bc} \plus{} \frac b{1 \plus{} ca} \plus{} \frac c{1 \plus{} ab}\ge\frac3{2}$
I know it from Hungkhtn's book(volume 1) and he wrote that  its author was you[/quote]
This is an other inequality (see page 69 and 97 in Algebraic Inequalities).

[quote="dduclam"][quote="Vasc"][quote="dduclam"]Ps: A well known's Vasc :  $ \frac a{a \plus{} bc} \plus{} \frac b{b \plus{} ca} \plus{} \frac c{c \plus{} ab}\ge\frac3{2}$[/quote]
This inequality is nice. But where I posted it?[/quote]

It was posted in this forum, but I haven't found the links :(

And I have at least 8 solutions for it.  :P[/quote]
And for this below have you a solution?

Let $ a,b,c,d$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}4$. Prove that

$ \frac a{a\plus{}bc}\plus{}\frac b{b\plus{}cd}\plus{}\frac c{c\plus{}da}\plus{}\frac d{d\plus{}ab} \ge 2$.

[quote="Vasc"] Let $ a,b,c,d$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$. Prove that

$ \frac a{a \plus{} bc} \plus{} \frac b{b \plus{} cd} \plus{} \frac c{c \plus{} da} \plus{} \frac d{d \plus{} ab} \ge 2$.[/quote]
This is equivalent to

$ \frac {ac(1 \minus{} bd)}{(a \plus{} bc)(c \plus{} da)} \plus{} \frac {bd(1 \minus{} ac)}{(b \plus{} cd)(d \plus{} ab)} \ge 0$,

$ \frac {1 \minus{} bd}{(1 \plus{} \frac{bc}{a})(1 \plus{}\frac{da}{c})} \plus{} \frac {1 \minus{} ac}{(1 \plus{} \frac{cd}{b})(1 \plus{} \frac {ab}{d})} \ge 0$,

$ 2(1 \minus{} abcd) \plus{} (1 \minus{} bd)(\frac {ab}{d} \plus{} \frac {cd}{b}) \plus{} (1 \minus{} ac)(\frac {bc}{a} \plus{} \frac {da}{c}) \ge 0$.

It suffices to show that

$ (1 \minus{} bd)(\frac {ab}{d} \plus{} \frac {cd}{b}) \plus{} (1 \minus{} ac)(\frac {bc}{a} \plus{} \frac {da}{c}) \ge 0$.

Can somebody prove the last inequality ?

[quote="Vasc"] Let $ a,b,c,d$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$. Prove that

 $ \frac a{a \plus{} bc} \plus{} \frac b{b \plus{} cd} \plus{} \frac c{c \plus{} da} \plus{} \frac d{d \plus{} ab} \ge 2$;

$ (1 \minus{} bd)(\frac {ab}{d} \plus{} \frac {cd}{b}) \plus{} (1 \minus{} ac)(\frac {bc}{a} \plus{} \frac {da}{c}) \ge 0$. [/quote]
I think the following inequality is also true:

[i]Let $ a,b,c,d$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$. Prove that[/i]
$ \frac a{2a \plus{} bcd} \plus{} \frac b{2b \plus{} cda} \plus{} \frac c{2c \plus{} dab} \plus{} \frac d{2d \plus{} abc} \ge \frac 4{3}$.

[quote="dima ukraine"] $ a \equal{} 0,b \equal{} c \equal{} d \equal{} \frac {4}{3}$ :wink:[/quote]
What means this ?


geometry
parallelogram
LaTeX
geometry solved

Let $RSTU$ be a cyclic quadriteral with $ST>RU$ and $TU>RS$. Take points $X,Y$ on sides $ST,TU$ so that $SX=RU$ and $UY=RS$. Let $M$ be the midpoint of $XY$. Pove that $\angle SMU$ is a right angle.

The wording is easy , so I do not use TATEX.

 Let $N$  the symmetric of  $S$with respect to $M$ .Then $SXNY$ is a parallelogram and the triangles  $RSU$ and $UYN$ are equal( $RU = SX = NY , RS = UY$and $\angle SRU = \angle UYN = 180 - \angle T$ ( $NY // TS$)). Thus $US = UN$.

 In isoskeles $\triangle USN$, $UM$ is a median , so it is an altitude as well.

[u] Babis[/u]

Yes  , ;)  , finally , I have not used TATEX , but a little LATEX !

  Sorry - Babis

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=16158 .

  darij


Let $ ABC $ be a triangle . $ AD , BE , CF $ respectively are internal bisectors of triangle $ ABC $ . $ H $ is the orthocenter of triangle $ ABC $ . Let $ A' , B' , C' $ be the midpoints of the segments $ AH , BH , CH $ respectively . Prove that : $ DA' , EB' , CF' $ are concurrent .

I don't think they are. My dynamic sketch says otherwise :).


solve the system of equation
 $ x\plus{}y^2\plus{}z^4\equal{}0\\y\plus{}z^2\plus{}x^4\equal{}0\\z\plus{}x^2\plus{}y^4\equal{}0$

obviously no real solution other than (0,0,0).
because if x , y OR z is >= 0 , equation 1 , 2 or 3 will imply that x = y = z = 0.
edit : i was asleep or .. sorry ...

what is S set ?
and how do you prove S is empty set ?

sorry for the second case , what i said is false .. :blush: maybe  i was sleepy

Note that if $ x = y = z$, the system collapses into $ x^4 + x^2 + x = 0$.  Note that for $ x = - 1$, the LHS is positive, but for $ x = - \frac {1}{2}$, the LHS is negative, so there is at least one solution with $ x = y = z\in( - 1, - \frac {1}{2})$...

Hint:
[hide]
If there is a solution where one of $ x,y,z$ is nonzero (wlog $ x$), then $ y = - x^4 - z^2 < 0$, $ z = - x^2 - y^4 < 0$, and consequently $ x = - y^2 - z^4 < 0$, or calling $ a = - x$, $ b = - y$, $ c = - z$ we have system
\[ a = b^2 + c^4,\qquad    b = c^2 + a^4,\qquad c = a^2 + b^4,\]
where now $ a,b,c$ are non-negative reals...
[/hide]

OK.but what we do then ? Could you post a full solution plz ?
Thanks you very much

Necessarily $ x,y,z\leq0$. For example $ x\leq{y}\leq{z}\leq0$.
$ x^2\plus{}y^4\geq{y^2\plus{}z^4}$. Thus $ \minus{}z\geq{\minus{}x}$ and $ z\equal{}x\equal{}y$.
$ x^4\plus{}x^2\plus{}x\equal{}0$ implies that $ x\equal{}0$ or $ x$ is the real root of $ x^3\plus{}x\plus{}1\equal{}0$ ($ \approx{\minus{}0.68}$).

I think we can't assume $ x \leq y \leq z$

we can assume $ x\ge y\ge z$ because of the simetry of the equations.
It doesn't matter if $ x\ge y\ge z$ or $ y\le x\ge z$, in other words, you can "transform" each variable in another and the equations stay  :)

if we replace $ x \rightarrow y$ and $ y \rightarrow x$ then the equation don't stay
if we replace $ x \rightarrow y$, $ y \rightarrow z$ and $ z \rightarrow x$ then the equation stay.
so we can't assume $ x \leq y \leq z$ we only can assume that $ x \leq y$ and $ x \leq z$
what I said is right? I don't know

No, the system is not symmetric. It is cyclic. dulldrake's example is correct (switch $ x$ and $ y$). For this reason we can't assume that $ x\geq y \geq z$. We can, however, assume $ x\equal{}max(x,y,z)$, I believe.


$ ABC$ is triangle such that $ AB\equal{}AC$ and $ \angle BAC \equal{} 20^\circ$ .The point $ X$ on $ AB$ is such that $ \angle XCB \equal{} 50^\circ$ . The point $ Y$ on $ AC$ is such that $ \angle YBC \equal{} 60^\circ$ .Find $ \angle AXY$

Dear Mathlinkers,
this problem is well known. It is call "adventitious angle" . This name was given by Langley in 1922 after my sources.
The result here is 50°.
A nice way of a prove can be seen in my web site : http://perso.orange.fr/jl.ayme  vol. 2 In Memoriam "Juan Carlos Salazar"
Sincerely
Jean-Louis

[url=http://pagesperso-orange.fr/jl.ayme/Docs/In%20memoriam%20Juan%20Carlos%20Salazar.pdf]Here (pdf)[/url] is the direct link. Jean-Louis, your page will be more useful for many people if you were using English!  :)

it's a very old problem, se also here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=179738[/url]


These guidelines should be followed in the new Fun Factory subforum of Games. They are generally similar to those that were in place when we had games and fun factory.

1) Absolutely no personal insults will be tolerated.

2) There are users of all ages and backgrounds on this site. Anything that may be deemed offense will not be tolerated.

3) Spam will be edited or removed at the discretion of Moderators and Administrators. If we feel a topic is still too 'frivolous' it is up to our discretion to lock or delete it (more often the former).

4) Treat each other with respect.

5) Moderators are here to help you. Should you have any questions, ask one of us.

6) The discussion and promotion of illegal activities is not allowed. (This includes filesharing)

7) Do not advertise the products or services of any organization. There is a difference between an advertisement and a recommendation.

8) Use your discretion when bumping an 'inactive' thread (generally a thread that has not been posted in for a month or so). If you feel your comment is relevant to discussion posted then you can bump it but it is ok to start a new discussion.

9) This forum was created because moderators and administrators felt that Games and Fun Factory was getting overloaded with spam and flooding out the more serious discussions that don't quite fit in Round Table. This doesn't give you permission to spam. You should still refrain from making one word or irrelevant posts.

Thanks.

[quote="7h3.D3m0n.117"][quote="Brut3Forc3"][quote="7h3.D3m0n.117"]Another Hurt and Heal? Jesus - I might ban all Hurt and Heals in the Fun Factory if this keeps up...[/quote]
Could you? Please?[/quote]

Hmm, I'm having a bad day actually, so from here on out, I decree that any more Hurt and Heal variations will be deleted on the spot.[/quote]
No more hurt and heals!

How do you delete your own post? I want to delete the post I made in the "Witty quotes".

If you're referring to your first post in the thread (post #48), you can't delete it.  The only time you can fully delete one of your posts is when it is the most recent post in a particular thread.  You can edit out the contents of one of your posts that is not the most recent, but there is a time limit for editing posts, which has expired for the post in question.  (I think the time limit is 36 hours, but I'm not sure).

[quote="7h3.D3m0n.117"][quote="1=2"]Read my idea in the same topic: [quote="I"]P.S. Idea: Contain all the hurt+heal into one topic and do one at a time.[/quote] Consider.[/quote]

Wait,wait. Into one [i]thread[/i]?[/quote]

Yes.

I posted a thread, but people did not follow directions, so we had to start over. I started a new thread. Could you please lock the old thread? It is the one with 3 pages, do not lock the new one by mistake. The name of the game is "NOT hurt and heal"

I tried to crack a joke, but nobody thought it was funny. Could you please delete or lock it?

Can you please stop making spam posts and topics. The mod doesn't need to lock it or delete it. Just don't post on that thread anymore.

like yah. My 'Fan Club' fail club was basically a call to arms.

lol you can't deny spam's got style.
however deceiving he might look now, old klebby was an out and out spammer three years back.
of course, they say responsibilities change people.... :rotfl:

Why dont these posts count toward your post counts? :huh:

Are boys vs. girls banned?

From James4l's post, it says he is fine with it.

What about the mods?

(I'm not planning to start one...I think :P)

[quote="eagle2"]why did you lock talk about anything?[/quote]

The topic itself was a bit too broad, so it's considered "spam" for a Fun Factory thread.

[quote="Joe10112"]Are boys vs. girls banned?[/quote]

No. Just don't go overboard with them.

[quote="Joe10112"]What about the mods?[/quote]

Yeah. All of the mods are banned.

This topic is very amusing.

The slang term "mod" is a shortened version of "moderator". A "moderator" is a person that has special powers on a forum, such as locking threads, deleting threads, and moving threads. They do these actions so that order is maintained in the forum. If all of the moderators were banned, then chaos would ensue, which would be bad. This means that dragon96 was being sarcastic when he made the comment "Yes, all of the mods are banned." Obviously, none of the moderators are banned.

well if there were no mods... wouldnt we all be mods???

yes very funny

Nononononono.

I know what mods are.

I was asking for the mods opinion.

What's the point of locking a thread when you can delete it? I mean, locking a thread will still allow others to see it and post stuff about the thread and spam all over again, while deleting a thread on the spot will limit the number of people who see it and therefore reduce spam comments relating to the deleted thread.

If the topic is locked, it can be used in the future as an example of what's spam/unacceptable.

Actually it's the other way around. Deleting a thread makes people wonder where the thread is, and simply make new threads about it. Locking a thread serves as a reminder (like Tungsten said).

just stumbled here. what's with fun factory and threads?

I want to ask a question.
I saw a game in the fun factory or the games forum. I want to see it again now but I cannot see it. Should I post another topic with the same rules(same game)?

Have you tried a search? Do so; you should get the thread. (If you don't, you can tell some moderator.) If the game has indeed finished/dead (no post for a long time), you can try making a new one.

I tried searching, I could'nt get it.I suppose I should make a new one.


[quote="hedeng123"][quote="Ligouras"]Prove that

$ \frac {m_a}{h_a}\leq\frac {R}{2r}$[/quote]

$ \frac {m_a}{h_a}\leq\frac {R}{2r}\Longleftrightarrow \frac {R^2}{4r^2} \minus{} \frac {m_a^2}{h_a^2}\geq 0$

$ \Longleftrightarrow 4b^3c^2 \plus{} 4b^2c^3 \minus{} c^2a^2b \plus{} 4c^3ab \plus{} 3b^2a^3 \minus{} 3b^3a^2 \minus{} b^2a^2c \plus{} 3c^2a^3 \minus{} 3c^3a^2 \plus{} a^4b \minus{} 2a^3bc$

$ \minus{} 2ab^4 \plus{} 4ab^3c \plus{} 2b^5 \minus{} 2b^4c \minus{} 2c^4a \minus{} 2c^4b \plus{} 2c^5 \minus{} a^5 \plus{} a^4c\geq 0$

$ \Longleftrightarrow 56y^2x^2z \plus{} 24z^2x^2y \plus{} 8x^3yz \plus{} 8xy^3z \plus{} 24z^2y^2x \plus{} 32z^3xy \plus{} 8z^5 \plus{} 8y^2x^3 \plus{} 8y^3x^2$

$ \plus{} 8y^3z^2 \plus{} 24y^2z^3 \plus{} 8z^2x^3 \plus{} 24z^3x^2 \plus{} 24z^4x \plus{} 24z^4y\geq 0$[/quote]

I think this prove must have something wrong, because the last ineq can not get "=", but the original inequality can get "="(when a=b=c)...

[size=150]yes,

$ (x\minus{}z)^2y^3\plus{}(13z^2x\plus{}z^3\plus{}3x^3\minus{}9zx^2)y^2\plus{}x(z\plus{}3x)(\minus{}2z^2\minus{}3zx\plus{}x^2)y\plus{}x^2(z\plus{}x)^3 \geq 0$;

For  the ineq,xue zhi yang have good proof.

BQ[/size]


Prove that there are infinite prime numbers

Assume there are finitely many primes.

Let M = p1 * p2 * p3 * ... * pn
Where p1..pn are the set of all primes.

Consider M+1.
M+1 must be prime.
If M+1 weren't prime, then some prime pk | M+1, which means (M+1)/pk is integral - or M/pk + 1/pk is integral. But this cannot be so because the smallest prime is two, and 1/2 is no longer integral.

This question was already asked on the forum and a whole lot of solutions were posted ;)


Okay. Imagine that you have a 4x4 grid which may look something like:
[asy]int i, j;
for(i=0; i<4; i=i+1)
for(j=0; j<4; j=j+1)
{
draw((i,j)--(i,j+1), red);
draw((i,j)--(i+1,j), red);
}
draw((0,4)--(4,4),red);
draw((4,0)--(4,4),red);[/asy]

In which you can fill up with numbers. I have 1 dice, from 0-9, which I will announce one at a time. The point of the game is to have the highest score. Thinking of your stategy? Great. Sign up now!!!

Okay sorry guys:
The rules:
I will post a number, you put it in anywhere you want to make the higest sum. It stays there. Then finally, after 16 rolls, you sum up collums.

Then each square is counted once, so we get the same score no matter what. Unless you weight each square differently.

So would:
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
Score: 4936?

yes.

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

I'll play, but i still don't totally understand scoring...

There is no strategy to this game. It's random.  :rotfl:

[quote="Dojo"]Okay AIME, here are the rules, just for you again: :)
[kindergarden teacher voice]
I roll a dice numbered 0-9.
You put it in one of the red little squares okay?
You put it in to get the highest number possible, which is added up my collum.
[/kindergarden teacher voice][/quote]
Wait so wouldn't it be the same score no matter what? Since each column is counted once?

And how did James4l get 4936? 1+1+1+1+2+2+2+2+3+3+3+3+4+4+4+4=40.

edit-OH i get it now. It's 1234+1234+1234+1234=4936

thanks.  what protestant said is what i was lookin for...

no need for the kindegarten teacher voice.

BTW dojo i beat brian in some math thing today!!

whats the signup list?  when will we start?

Brute there is some strategy actually. figure it out on your own.

1. pacman
2. AIME15
3. chensi
4. ernie

Use:

____
____
____
____

And fill in your blank. Please play honest, and don't share answers!!!
First number: 3

Oh, I misread the rules. I thought someone picked a square, then you rolled for them individually. Sorry.

So we PM you our choices?  i'm pming now...

\in

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

Okay sign ups are open until I post the second number.

1. pacman
2. AIME15
3. chenhsi
4. ernie
5. dragon96

You could choose a square that's on the main diagonal since it won't matter, then find out later... I think the numbers are added up as rows based on james4l's example.

cough cough collums (horizontal)
not rows (vertical)

yes

rofl dojo....i hope that was sarcastic :P or else i feel reallly bad for you.

for those not sarcastic today:

1 2 3 4
1 2 3 4
1 3 6 7

The first [b]row[/b]is 1234, the second is 1234, the third is 1367.

the first [b]column[/b] is 111, second 223, third 336, last 447.

ahem

rows = horizontal
columns = vertical

So its horizontal?

(you spelled "collum" wrong too)

-,-
grrr

look at AIME15's example

*sigh*

collum[b]n[/b]s

talking to dojo not me right?  because last time i checked, my rows are going horizontal... :|

*slaps forehead*

You spelled "collumns" wrong
Do you realize the correct spelling is in my previous post?

OMG

okay co[b]l[/b]um[b]n[/b]s

Any more sign ups?

we pm to you right?  i sure hope so........

ERNIE AND PACMAN I WILL WAIT ONE MORE DAY. IF YOU DO NOT RESPOND IN THAT TIME YOU WILL BE DQED.


Given an integer $a$ and a positive integer $n$, show that

$a \equiv 1 \bmod p^{n}\Leftrightarrow a^{p}\equiv 1 \bmod p^{n+1}$

[hide="Hint"] The right direction is very easy.  Subtract $1$ from both equations. [/hide]

[hide]I assume $(a, p) = 1$
First $\Rightarrow$: $a^{p}-1 \equiv (a-1)(a^{p-1}+a^{p-2}+...+1) \bmod p^{n+1}$. The first factor is divisible by $p^{n}$, while the second is $\equiv p \bmod p^{n}$.

Now $\Leftarrow$: Suppose $ord_{p}(a) = d$. $a^{1}+a^{2}+...+a^{d}\equiv 0 \bmod p$. The sequence $a^{0}, a^{1}, ...$ is periodic with period $d$. On the other hand $p-1 = k \cdot ord_{p}(a)$ with $k$ integer. Hence $a^{p-1}+a^{p-2}+...+a^{1}+1 \equiv p \cdot \frac{p-1}{ord_{p}(a)}\equiv 0 \bmod p$. Hence $a-1 \equiv 0 \bmod p^{n}$.[/hide]

[hide="another way"]The $p$-element group of $1\mod{p^{n}}$ elements of $\mathbb{Z}_{p^{n+1}}$[/hide]


function
quadratics
floor function
modular arithmetic
AMC

Let $\lfloor{x}\rfloor$ be the greatest integer less than or equal to $x$. Then the number of real solutions to $4x^2-40 \lfloor {x}\rfloor + 51 = 0$ is

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

Clearly x is positive.

Case 1:  x is an integer - then $4x^2 - 40x + 51 = 0$ becoming $(2x-17)(2x-3)$ which doesn't work.

Case 2:  x is not an integer - since $4x^2$ must be an integer, it follows $x = y + 1/2$. ($y \in \mathbb{Z}$)  It becomes $(2y+1)^2 - 40y + 51 = 0$ or $4y^2 - 36y + 50 = 0$.  The discriminant is not an integer.

Answer (A).

[hide]$\lfloor x \rfloor = \frac{1}{40} (4x^2+51) \implies 4x^2+51 \equiv 0 \pmod {40}$, $4x^2\equiv 29 \pmod {40}$. The LHS is always even, but the RHS is always odd. $\boxed {(A)}$.[/hide]

Is that right?

EDIT: Oh nvm $x$ isn't an integer :wallbash:

[quote="nsato"][quote="Singular"]Case 2:  x is not an integer - since $4x^2$ must be an integer, it follows $x = y + 1/2$. ($y \in \mathbb{Z}$).[/quote]

Umm, that part doesn't look right to me.[/quote]

you are right, i am wrong.  sorry i was thinking x rational

[hide="hmm"]
We know that $\lfloor x \rfloor>1$ because $4x^2>0$.  Suppose $\lfloor x \rfloor=2$.  Then $4x^2 = 80-51 = 29 \rightarrow x = \frac{\sqrt{29}}{2} = 2.693$, satisfying the original condition that $\lfloor x \rfloor=2$.  That is the only value that will work.  Answer: B
[/hide]

The [hide="answer"]There are four roots: one with integer part 2 ($\frac{\sqrt{29}}{2}$), one with integer part 6 ($\frac{\sqrt{189}}{2}$), one with integer part 7 ($\frac{\sqrt{229}}2$), and one with integer part 8 ($\frac{\sqrt{269}}{2}$).[/hide]

yea, actually you're right.  If you plot $x$ as a function of $\lfloor x \rfloor$, it's all the values on that curve between $x<\lfloor x \rfloor +1$ and $x>\lfloor x \rfloor$.


Hey guys,

I'm wondering what sort of stuff you need to know for USACO. I'm not a particular expert with C++, C, or Java, but I have been programming for a few years on the web development side. While web languages are not as strongly typed for instance and generally looser and "easier", if USACO is more algorithm based then I could attempt to start trying some of the training stuff on the site.

Do I need to know file/IO or is the example they gave us on the first page enough? Of course some math functions will probably be needed. Anything in particular that is pretty specific towards a certain language that would be important to know?

Thanks

Well you need algorithomic knowledge. and it will be ok if you don't know much about coding (like me :lol: )

You can make it in USACO with only basic programming knowledge (but knowing a little more is always better ;) )...

Reading a simple "Learn C/C++/Java in 24 hours book" is enough. :D

And, yes, you will need to know how to read or write plain text files. Like, using iostream in C++ or scanf/printf for C...

Math functions is good to know, but only useful in very few tasks (mostly geometry ones)...

USACO (and nearly all other programming contests) focus on algorithms. The training page is an excellent source to learn that.

Btw, what language are you going to use?

Although you won't need it right away, something that might come in handy if you're using C++ is the STL, or Standard Template Library.  This is a library of header files that includes all kinds of neat stuff like sort functions, queues, vectors, stacks, deques, and many other useful functions and data structures that you might want to use.  Find more information at [url]http://www.sgi.com/tech/stl/[/url].

Thanks, I just wanted to know if I needed extensive knowledge of C++, Java, or C. I'm probably going to be using C++ since it seems interesting to me, and I've always wanted to learn it.

I'll pay special attention to the file i/o sections, and thanks for the links countersixte. I'll probably have to read about pointers and stuff since I'm more familiar with object oriented programming even though it tends to be a bit slower usually.

I mainly wanted to know if I was wasting my time by attempting some of the USACO training challenges since I usually have to look up functions on Google when I need them, but thanks for the feedback :)

well, if u want to practise contest programming like IOI or ACM then USACO training page is the suitable place. But if u r looking for some other things like software development or game making ;)  or some thing else then i don't think USACO is for u :P

Yeah :) Game programming interests me, but algorithms are an important part of AI -> games. Since I only have my pathetic knowledge of web programming, I modeled a semi-OOP pathfinding A* algorithm with binary heaps and everything in Javascript (yes, you can actually write stuff that works in that language lol) that worked pretty well. But I'm interested in things like DP which I don't know too much about so I'll check out the training pages :)

I started on USACOs when I was only a month into a high school course on VB (earlier this year--my teacher soon gave me the basic rundown of C++ so I could submit of course ).  So all in all, those training pages basically walked me through learning how to program.  I'm only in the third chapter in the training pages, but you should totally go for it.  Especially since you have a knowledge basis of programming to begin with (yeah, I'm still working on that).  When you get a little ways into the Training pages, they'll give you some algorithm explanations, too.  And if you dont get something, running here to ask for algorithm help or explanations is always good.  :lol: My only warning is that they can be quite addicting.  :P

I am interested to what degree of optimization is needed? Are programs specifically judged based on the speed of algorithms or also on how the program is structured? If it's specifically based on speed, of course that discourages OOP programming somewhat, but how [i]much[/i] optimization is needed? I'm about 200 pages into the C++ book I'm reading and it's mentioned some performance tips that may be a bit extreme.

For instance, everyone knows that function calls have overhead, but if a function is called enough, should we not use functions? Or should we use inline functions even if they increase program size?

Since the program that is to be created is tailor made for a particular program, do we have to worry about portability and extensibility?

[quote="Zyloch"]I am interested to what degree of optimization is needed?[/quote]
For the first three chapters or so, you don't need to worry about optimization too much.  There might be a key observation that will reduce runtime, such as in checkers, but nothing too major.

[quote="Zyloch"]Are programs specifically judged based on the speed of algorithms or also on how the program is structured?[/quote]
Structure your programs in a way that suits you.  The only thing that counts is that the program runs in time (usually .5 seconds or less) and it provides correct answers for all the test cases.

[quote="Zyloch"] If it's specifically based on speed, of course that discourages OOP programming somewhat, but how [i]much[/i] optimization is needed? I'm about 200 pages into the C++ book I'm reading and it's mentioned some performance tips that may be a bit extreme.[/quote]
You most likely won't need OOP too much in the training pages, unless it's STL.  Remember that these problems are contest problems that were originally done in a timed setting (i.e. 3-5 hrs to code several problems), and OOP is generally too difficult to debug in such a short time.  Using structs when necessary is probably a better idea.

I don't think you'll need to use any extreme performance tips.  The pages are testing to see if your algorithm is correct, not how many extra little time-savers you can cram into your code.

[quote="Zyloch"]For instance, everyone knows that function calls have overhead, but if a function is called enough, should we not use functions? Or should we use inline functions even if they increase program size?[/quote]
Again, you probably won't have to use inline functions at all in the training pages.  There are certain programs in the training pages  (cowcycles and fence rails come to mind) where you [i]might[/i] want to try some inlining, but generally such optimizations make little difference in the runtime of the program anyway.

[quote="Zyloch"]Since the program that is to be created is tailor made for a particular program, do we have to worry about portability and extensibility?[/quote]
No.  As long as the program works on the online grader when you submit it, no portablility or extensibility is necessary.

My advice is to just try some of the problems on the training pages.  They start off easy so you get the hang of it, and gradually increase in difficulty.  Remember that the main object of the pages is to test algorithms and data structures.


I'm curious as to how you think all life on Earth might be destroyed. I do not believe that this will happen in the near future, but maybe one day it will. Which one of these options do you think is most likely to happen?

Well, if you're saying ALL life, then only something really extreme.

But if its human life--- ehhh, a much smaller rock is needed.  Maybe we'd do it ourselves- it seems quite possible at this time.

Seriously, I think Nukes might destroy it the soonest. A war has to happen in the next few centuries.

Nuclear War will not destroy the world, assuming that it starts with a small attack and ont the russians or us unleashing our entire arsenal.  Even if someone nuked us there would still be opposition to retaliating with nukes and we could simply conventionally bomb the country back to the stone age.  One or two nukes would not destroy life they would make it uncomfortable for a few years but even humans would survive.
Also of note Switzerland has superb bunkers which might survive most disasters that dont destroy the world.

I think the survey needs a new category labled "me"

No matter what happens, I think extinguishing all deep-sea life would be next to impossible.

On an unrelated note: Melissa, does your signature come from [i]Going Postal[/i]?  It reminds me of that book...

All life on Earth?  I think we'll need to wait for the sun to start to expand (as it will eventually).

Just humans?  When fossil fuels run out, that'll be a huge blow.  I'm not sure if it would finish us off, but it would certainly start the process.

The Earth will somehow accelerate in spinning speed and then it will throw everybody out to the space.

[quote="dts"]
On an unrelated note: Melissa, does your signature come from [i]Going Postal[/i]?  It reminds me of that book...[/quote]
Yes it does  :)

I say global warming: we've already had like 10 hurricanes hit the southwest in less than a year...

Edit: yes I do mean southwest. Sorry about that...

i say that people will get to greedy and want to conquer the earth so then WW3 begins and some weirdo ends up blowing up the earth.  Or the earth will end by global warming.  THe polar ice caps will melt and the whole earth will be flooded so we are forced to leave to another planet.

[quote="ashpotter"]why would there be a war?[/quote]

lol, i don't think we need to worry about the why so much as the when...

im not sure the war would destroy the earth, but i wouldn't think the earth is done with wars.

and btw whoever voted for the fossil fuels forgot that  humans did not start to use fossil fuels until the 17th century (burn trees!...whale blubber oil anyone?)

& as far as the world ending, i bet the bible is true... i.e. Jesus will come back and take over after people start getting really oppressive of christians & other bad stuff.

Global warming will just make our lives miserable
There will always be a lot of humans

Nuclear war is what I voted, although I don't think everyone will die in that either.
But think of it, it is so hard to maintain the balance (during the Cuba crisis, it seems a Russian order for launch was actually issued, but that the commander of a submarine simply refused...)
I really don't see humanity holding out for ten millennia like this.  It only takes one freak country to ignite it.


[quote="pda"]Thee exterior angles of a triangle are in the ratio 4:5:6. Determine its three interior angles.[/quote]

36,60,84...equations.

[hide="Solution"]
The sum of the exterior angles of any polygon is always $ 360$.
So we have $ (4+5+6)x = 360$, making $ x = 24$

Thus the EXTERIOR angles measure 96, 120, 144 degrees.
Because the interior angles are $ 180-\text{exterior}$, we can get that the interior angles are
$ \boxed{84, 60, 36}$ degrees.
[/hide]

[quote="MCPC"]Question though... if the triangle were equilateral, wouldn't each exterior angle be 300 degrees?  I don't get why the sum of the exterior angles are always 360 degrees.[/quote]
I think you misunderstood the definition of an exterior angle.
An exteriror angle is the non-reflex angle formed when one side of the polygon is extended.  Basically, if $ \theta$ is an interior angle, then $ 180^\circ-\theta$ would be the measure of the corresponding exterior angle.