calculus integration logarithms trigonometry limit induction derivative

Solution to [b]problem (5)[/b]: (Hope I am not wrong.)

$I = \int_{0}^{\infty}(\frac{\tan^{-1}x}{x})^{3}\,dx$

We will make the (somewhat obvious) substitution $\theta = \tan^{-1}x.$ So, we have ${I = \int_{0}^{\frac{\pi}{2}}(\frac{\theta}{\tan\theta})^{3}\sec^{2}\theta \,d{\theta}}.$

$\Rightarrow I =-\frac12 \int_{0}^{\frac{\pi}{2}}\theta^{3}\,d{(\frac{1}{\tan^{2}\theta})}=-\frac12 \left[ \frac{\theta^{3}}{\tan^{2}\theta}\Big|_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\frac{3\theta^{2}}{\tan^{2}\theta}\,d{\theta}\right]$

$\Rightarrow I =-\frac12\left[0-\lim_{\theta \rightarrow 0}(\frac{\theta}{\tan \theta})^{2}\cdot\theta\right]+\frac32 \int_{0}^{\frac{\pi}{2}}\theta^{2}\cot^{2}\theta \,d{\theta}$

$\Rightarrow I = 0+\frac32 \int_{0}^{\frac{\pi}{2}}\theta^{2}(\csc^{2}\theta-1)\, d{\theta}=-\frac32 \int_{0}^{\frac{\pi}{2}}\theta^{2}\,d{(\cot\theta)}-\frac32 \int_{0}^{\frac{\pi}{2}}\theta^{2}\, d{\theta}$

$\Rightarrow I =-\frac32 \left[ \theta^{2}\cot\theta\Big|_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}\theta\cot\theta \, d\theta \right]-\frac{\theta^{3}}{2}\Big|_{0}^{\frac{\pi}{2}}$

$\Rightarrow I =-\frac32 \left[ 0-\lim_{\theta \rightarrow 0}(\frac{\theta}{\sin\theta})\theta\cos\theta \right]+3\int_{0}^{\frac{\pi}{2}}\theta \,d{(\ln (\sin\theta))}-\frac{\pi^{3}}{16}$

$\Rightarrow I = 0+3\left[ \theta \ln (\sin\theta)\Big|_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\ln (\sin\theta) \,d\theta \right]-\frac{\pi^{3}}{16}$

$I = \boxed{\frac{3\pi}{2}\ln (2)-\frac{\pi^{3}}{16}}$

Solution to [b]problem (3)[/b]:

$I_{n}= \int_{0}^{\frac{\pi}{2}}\cos^{n}x\cos nx \, dx = \frac{1}{n}\int_{0}^{\frac{\pi}{2}}\cos^{n}x \,d(\sin nx)$

$\Rightarrow I_{n}= \frac{1}{n}\left[ \cos^{n}x \sin nx \Big|_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\sin nx \cdot n\cos^{n-1}x \sin x \, dx \right]$ ... (Using integration by parts.)

$\Rightarrow I_{n}= \int_{0}^{\frac{\pi}{2}}\cos^{n-1}x \sin x \sin nx \, dx$

$\Rightarrow 2I_{n}= \int_{0}^{\frac{\pi}{2}}\cos^{n-1}x (\cos x \cos nx+\sin x \sin nx)\, dx$

$\Rightarrow 2I_{n}= \int_{0}^{\frac{\pi}{2}}\cos^{n-1}x \cos (n-1)x \, dx$ ... (Using the identity $\cos (A-B) = \cos A \cos B+\sin A \sin B$)

$\Rightarrow I_{n}= \frac12 I_{n-1}$

A simple induction on $n$ shows that $I_{n}= \frac{1}{2^{n-1}}I_{1}$ (this is left as an exercise for the reader.)

Finally, $I_{1}= \int_{0}^{\frac{\pi}{2}}\cos^{2}x \, dx = \frac12 \int_{0}^{\frac{\pi}{2}}(1+\cos 2x)\, dx = \frac{\pi}{4}$

So, $I_{n}= \boxed{\frac{1}{2^{n-1}}\cdot \frac{\pi}{4}}$

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

Solution to [b]problem (4)[/b]: (The best solution seems to me the use of differentiation under the integral sign.)

I will use $a$ instead of $\alpha$ just to save myself the trouble of typing some extra characters.

Let $I(a) = \int_{0}^{\frac{\pi}{2}}\sec x \ln (1+\sin a \cos x)\, dx$

$\Rightarrow I'(a) = \int_{0}^{\frac{\pi}{2}}\sec x \left(\frac{\partial}{\partial a}\ln (1+\sin a \cos x)\right) \, dx$

$\Rightarrow I'(a) = \cos a \int_{0}^{\frac{\pi}{2}}\frac{dx}{1+\sin a \cos x}= \cos a \int_{0}^{\frac{\pi}{2}}\frac{dx}{1+\sin a \cdot \frac{(1-\tan^{2}\frac{x}{2})}{(1+\tan^{2}\frac{x}{2})}}$

$\Rightarrow I'(a) = 2\cos a \int_{0}^{\frac{\pi}{2}}\frac{\frac12 \sec^{2}\frac{x}{2}\, dx}{(1-\sin a)\tan^{2}\frac{x}{2}+1+\sin a}$

$\Rightarrow I'(a) = \frac{2\cos a}{(1-\sin a)}\int_{0}^{\frac{\pi}{2}}\frac{\frac12 \sec^{2}\frac{x}{2}\, dx}{\tan^{2}\frac{x}{2}+\frac{1+\sin a}{1-\sin a}}= \frac{2\cos a}{(1-\sin a)}\int_{0}^{\frac{\pi}{2}}\frac{\frac12 \sec^{2}\frac{x}{2}\, dx}{\tan^{2}\frac{x}{2}+(\frac{1+\sin a}{\cos a})^{2}}$

Now, we use the substitution $y = \tan \frac{x}{2}.$ So, we have

$I'(a) = \frac{2\cos a}{(1-\sin a)}\int_{0}^{1}\frac{dy}{y^{2}+(\frac{1+\sin a}{\cos a})^{2}}$

$I'(a) = \frac{2\cos a}{(1-\sin a)}\cdot \frac{\cos a}{1+\sin a}\tan^{-1}\left( \frac{y\cos a}{1+\sin a}\right)\Big|_{0}^{1}= 2\tan^{-1}\left( \frac{\cos a}{1+\sin a}\right)$

Now, note that $\frac{\cos a}{1+\sin a}= \frac{1-\tan^{2}\frac{a}{2}}{(1+\tan \frac{a}{2})^{2}}= \frac{1-\tan \frac{a}{2}}{1+\tan \frac{a}{2}}= \tan (\frac{\pi}{4}-\frac{a}{2})$

$\therefore I'(a) = 2(\frac{\pi}{4}-\frac{a}{2}) = \frac{\pi}{2}-a$
$\Rightarrow I(a) = \frac{\pi}{2}a-\frac{a^{2}}{2}+C$. But note that $I(0) = 0 \Rightarrow C = 0.$

Thus, $I(a) = \boxed{\frac{a}{2}(\pi-a)}$

Problem 1.

We must be brave and daring and break this up by partial fractions. Just going part-way, we split the integrand into

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

The integral of the second part is obviously $\ln(x^{4}-2x+1)$. For the first part, I just guessed it is the derivative of $\frac{P(x)}{x^{3}-x+1}$ so $-2x^{3}+1 =-(3x^{2}-1)P(x)+(x^{3}-x+1)P'(x)$ and by inspection $P(x) = x$ works. :P

So after we evaluate at the bounds, we get $\frac{3}{25}-\frac{2}{7}+\ln(76)-\ln(13)$ (unless I messed up my arithmetic).

This isn't very nice since there was a lot of guesswork. Does anyone have a better approach? (Especially $x^{3}-x+1$ part)

Thank you all for showing a great deal of interest in solving integrals. FieryHydra seems to be on fire when it comes to cracking integrals. We both have something in common, at least we hail from the same country I guess.

[quote]
2) Evaluate $I = \int\frac{\ln(\cos\theta+\sqrt{\cos 2\theta})}{\sin^{2}\theta}d\theta$
[/quote]
We use a method of integration in parts.   $\int u dv = vu-\int v du$  ,   $\left\{ \begin{array}{l}u = \ln(\cos\theta+\sqrt{\cos 2\theta}),\\ dv = \frac{1}{\sin^{2}\theta}d\theta,\\v =-ctg\theta, \end{array}\right.$

$I =-ctg \theta\ln(\cos\theta+\sqrt{\cos 2\theta})-\int\frac{-\sin\theta-\frac{\sin 2\theta}{\sqrt{\cos 2\theta}}}{\cos\theta+\sqrt{\cos 2\theta}}ctg \theta d\theta$

$A =-ctg \theta\ln(\cos\theta+\sqrt{\cos 2\theta}), B = \int\frac{-\sin\theta-\frac{\sin 2\theta}{\sqrt{\cos 2\theta}}}{\cos\theta+\sqrt{\cos 2\theta}}ctg \theta d\theta$   $\Rightarrow I = A-B$

Find   $B =-\int\frac{ ctg \theta(\sin\theta \sqrt{\cos 2\theta}+\sin 2\theta)}{\sqrt{\cos 2\theta}(\cos\theta+\sqrt{\cos 2\theta})}d\theta$ $\Rightarrow B =-\int\frac{\cos\theta(\sqrt{\cos 2\theta}+2\cos\theta)}{\sqrt{\cos 2\theta}(\cos\theta+\sqrt{\cos 2\theta})}d\theta$

$B =-\int\frac{\cos\theta}{\cos\theta+\sqrt{\cos 2\theta}}d\theta-2 \int\frac{\cos^{2}\theta}{\sqrt{\cos 2\theta}(\cos\theta+\sqrt{\cos 2\theta})}d\theta$

$B =-\int\frac{\cos\theta}{\cos\theta+\sqrt{\cos 2\theta}}d\theta-2\int\frac{\cos\theta}{\sqrt{\cos 2\theta}}d\theta+2\int\frac{\cos\theta}{\cos\theta+\sqrt{\cos 2\theta}}d\theta$$\Rightarrow B = \int\frac{\cos\theta}{\cos\theta+\sqrt{\cos 2\theta}}d\theta-2\int\frac{\cos\theta}{\sqrt{\cos 2\theta}}d\theta$

$X = \int\frac{\cos\theta}{\cos\theta+\sqrt{\cos 2\theta}}d\theta$ and $Y = \int\frac{\cos\theta}{\sqrt{\cos 2\theta}}d\theta$ $\Rightarrow B = X-2Y$

$X = \int\frac{\cos\theta(\cos\theta-\sqrt{\cos 2\theta})}{(\cos\theta+\sqrt{\cos 2\theta})(\cos\theta-\sqrt{\cos 2\theta})}d\theta = \int\frac{\cos\theta(\cos\theta-\sqrt{\cos 2\theta})}{\sin^{2}\theta}d\theta = \int ctg^{2}\theta d\theta-\int\frac{\cos\theta\sqrt{\cos 2\theta}}{\sin^{2}\theta}d\theta=-ctg\theta-\theta-\int\sin^{-2}\theta(1-2\sin^{2}\theta)^{\frac{1}{2}}d\sin\theta$ .
It is solved by means of theorem Chebyshev's.
${Y = \int\frac{d\sin\theta}{\sqrt{1-2\sin^{2}\theta}}=\frac{1}{\sqrt{2}}arcsin(\sqrt{2}\sin\theta})$


Let $v(n)$ denote the number of all $n$-digit perfect squares. For each $n\in \mathbb N$, find a formula for $v(n)$


I noticed that too, in the Arizona forum. Same guest, but they said "go arizona yeah message"
I haven't seen any other guests post...are the different state forums different?

It's weird, because I tried to see if I could post as a guest, but it didn't work.


In terms of difficulty, about what number on the AIME corresponds to the easiest problem on the USAMO? Or is the easiest problem on the USAMO simply beyond all questions on the AIME? Thanks.

(PS, even though this has nothing to do with the question above, but are the AIMEs from the 80s and 90s generally easier, harder or the same as current AIMEs?)

To your second question: Yes, earlier AIME's are generally easier (although you can't necessarily say that 2000 is harder than 1999, you can generally say that 2000 is harder than 1985).

AIME and USAMO test some pretty different things, so there's no way to compare an AIME problem to a USAMO one. Many AIME problems can be solved through cheap methods or assumptions, which would pretty much give you a 0 if that were a problem.

I know several people who got ~10 or higher on the AIME and solved less than one USAMO problem. On the other hand, I'm pretty sure there are several people who barely made USAMO and solved at least one problem.

Thanks. Ill clarify my question a bit. If you had to write full solutions for the AIME problems, are there any questions on the AIME which would be at par in difficulty with any USAMO questions?

In my opinion, on occasion, some AIME 13-15's are harder than some USAMO #1's or #4's. Like this year, on the first test. I found #15 to be quite difficult and #1 and #4 to be quite easy. That might just be me, though....

when I took this years USAMO the questions like 2 and 4 seemed really ez. Using Cauchy, AM-GM, and a tiny bit of induction you "could" prove them. But in USAMO more than knowing the answers you should have a clear and concise proof(neatness included). In that case to get like a full score on like number 2 was like get both 12(which I got :blush: ) and 13 on this years AIME even though questions seem easy.

In my opinion, AIME problems are never harder than USAMO problems. Not post-2002. AIME problems never require repeated, deep insight - maybe one or two realizations to solve a problem, but it usually works out fine. USAMO problems are always significantly more complex.

That said, the only way to really get a feel for it is to look at old tests, and especially to take them under timed conditions.

[quote="worthawholebean"]USAMO problems are always significantly more complex.[/quote]

This year, #1 was an easy 6 points for those who know radical axes, and #4 was a less-easy 7 points for those who know Cauchy-Schwarz.

I remember sending a PM about a AIME problem with a difficult rigorous proof, but now my small sentbox strikes. :(


In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$ appears $k$ times. It is a fact that there are integers $b$, $c$, and $d$ such that for all positive integers $n$,
\[ a_n=b\lfloor \sqrt{n+c} \rfloor +d,  \]
where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$. The sum $b+c+d$ equals

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

[hide]First, note that the second through fourth terms are all the same.  Therefore, $\lfloor\sqrt{n+c}\rfloor$ gives the same value for three consecutive integers $2+c$, $3+c$, and $4+c$.  These integers must be equal to 1, 2, and 3, so $c=-1$.

Next, note that the first term is 1.  We have $1=b\lfloor\sqrt{1-1}\rfloor+d\Rightarrow d=1$.

Finally, we see that the terms increase in increments of 2, so $b=2$ and the sum is $\fbox{2}$.[/hide]

Why must these be 1,2,3?
Also, is there a way to derive this formula (or prove it)?

it might help for you to look at $ \lfloor \sqrt {x} \rfloor$ and just try to shift and stretch this function.  starting with $ x \equal{} 0$, we get the sequence $ 0,1,1,1,2,2,2,2,2,3...$ etc.  observe that all numbers in this sequence appear an odd number of times, each two more than the last.  This happens because the sequence increases when it hits square numbers, and the difference between two consecutive squares is $ (x \plus{} 1)^2 \minus{} x^2 \equal{} 2x \plus{} 1$ - always odd.

It is easy to see $ b$ must be $ 2$ for the sequence to include only odds or only evens.  We want $ 1$ to appear once, as the first term, so we must map $ x \equal{} 0$, the only term that appears only once, to $ 1$.  Thus we must have $ n \plus{} c \equal{} 0$ for $ n \equal{} 1$, ie. $ c \equal{} \minus{} 1$, and then $ 2*0 \plus{} d \equal{} 1$, so $ d \equal{} 1$.  Thus $ a_n \equal{} 2\lfloor \sqrt{n \minus{} 1} \rfloor \plus{} 1$ and $ a \plus{} b \plus{} c \equal{} 2 \implies \boxed{C}$.


Consider $ \triangle ABC$ with point $ D$ on $ \overline{BC}$.  $ \triangle ABD$ is rotated about $ \overline{AD}$ to form a solid $ \mathcal{S}_1$ with volume $ V_1$.  $ \triangle ACD$ is rotated about $ \overline{AD}$ to form a solid $ \mathcal{S}_2$ with volume $ V_2$.  Prove that $ V_1\equal{}V_2$ if and only if $ \overline{AD}$ is a median.

[hide="Hint 1"]Express the volumes of $ \mathcal{S}_1$ and $ \mathcal{S}_2$ in terms of cones.[/hide]

[hide="Hint 2"]
[asy]pair A,B,C,D; A=(4,3); B=(0,0); C=(5,0); D=(2.5,0); draw(A--B--C--A--D); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,SE); draw(D--(2,-1)--B); draw(C--(3,1));[/asy]
[/hide]


algebra
polynomial
number theory
greatest common divisor
linear algebra
matrix
number theory unsolved

Let $n$ be a positive integer. Determine all positive integers $m \in \mathbb{N}$ for which there exists an integer polynomial
\[
f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \in \mathbb{Z}[x]
\]
such that $a_n \neq 0$, $\mathrm{gcd} \, (a_0,a_1,\ldots,a_n,m) = 1$, and $f(k)$ is divisible by $m$ for all integers $k \in \mathbb{N}$.

A beautiful problem!!!!!!!!! I think m must divide n!. In this case the polinomial $ f(X)=X(X-1)(X-2)...(X-n+1)$ solves the problem.

Now, suppose we have such a polynomial and take a prime p which divides m.
We have $ p| a_0+a_1\cdot k+...+a_n\cdot k^n $ for any $k=\overline{1,n+1}$. Consider this in $\mathbb{Z}_p$ as a linear system in $ a_0,...,a_n$. The determinant, which is a Vandermonde one must be 0 in $\mathbb{Z}_p$. This means that p must divide a number of the form i-j, so p is at most n.
  Now take any prime p smaller than n and suppose that its exponent in m is more than the exponent in n!. We have obviously $\frac{f(X)}{m}=b_0+b_1\cdot X+...+b_n\cdot\frac{x(x-1)...(x-n+1)}{n!} $ for certain $b_0,...,b_n $. We easily deduce now that $ b_i$ are integers, thus we can write $\frac{f(X)}{m}=\frac{c_0+...+c_n\cdot x^n}{n!} $ for certain $ c_i$ integers. But since $ a_i=\frac{m}{n!}\cdot c_i $ we must have $ p|a_i $ and of course p|m. This is a contradiction.

I don't understand $harazi's$ solution.can anyone explain it better?

I don't think $m|n$ for sure: $m=2 , n=3, f(x)=x^3+x^2$.

He meant $ m$ devides $ (n!)$. Very nice solution ! However, I dont think Vandermonde's matrix is acceptable in a TST. Therefore, how could you deduce $ p$ can't exceed $ n$ without resorting to it, Harazi!  :o


geometry perimeter trigonometry conics ellipse

Show that $ 2ph_a\gamma$ is a congruence criterion for triangles, being $ 2p$ it's perimeter, $ h_a$ its relative height of $ \overline{BC}$ and $ \gamma$ the internal angle in the vertice $ C$ .

I've seen the solution to this problem but I can't figure out how it's proving anything. Any help is appreciated.

If $ D$ is the foot of $ h_a$, then $ b \equal{} {h_a\over\sin\gamma}$ and $ CD \equal{} {h_a\over\cos\gamma}$ are uniquely determined by the given data, and $ \triangle ACD$ is easily constructible. Then $ DB \plus{} BA \equal{} 2p \minus{} b \minus{} CD$ is also uniquely determined. Therefore we can construct the ellipse with foci in $ A$ and $ D$ and with longer axis equal to $ 2p \minus{} b \minus{} CD$. Point $ B$ is then the intersection of the ellipse and the line $ CD$ (such that $ D$ is between $ C$ and $ B$).

Since the given data are sufficient for construction, they are also sufficient as a congruence criterion.


Say that we have a triangle ABC.  Now, take point P inside Triangle ABC such that $ \angle PAB \equal{}\angle PBA \equal{} 30$ and $ \angle PCB \equal{} 60$.  Draw S on AB such that BX=2AX and draw the midpoint of AC, we call this point T.  Prove that $ \angle PTS \equal{} 90$.
EDIT: Oh, sorry...


hello !! $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$ prove that :
$ \frac{ab}{1\plus{}c}\plus{}\frac{ac}{1\plus{}b}\plus{}\frac{bc}{1\plus{}a}\leq\frac{1}{4}$

Homogenize it, then normalize it, the rest is done with some computations...


If $x,y,z$ are real positive numbers with the property that $x+y+z=xyz$ and none of the numbers is $1/ \sqrt{3},$ prove that: $\frac{3x-x^{3}}{1-3x^{2}} + \frac{3y-y^{3}}{1-3y^{2}} + \frac{3z-z^{3}}{1-3z^{2}}= \frac{3x-x^{3}}{1-3x^{2}} \cdot \frac{3y-y^{3}}{1-3y^{2}} \cdot \frac{3z-z^{3}}{1-3z^{2}}$

Write $x = \tan{\alpha},\ y = \tan{\beta},\ z = \tan{\gamma}$ where $\alpha + \beta + \gamma = \pi$. Then we have \[ \tan{3\alpha} + \tan{3\beta} + \tan{3\gamma} = \tan{3\alpha}\tan{3\beta}\tan{3\gamma} \] and since \[ \tan{3x} = \frac{3\tan{x} - \tan^{3}{x}}{1 - 3\tan{x}^2} \] we get the result.


A regular octagon $ ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ ABEF$?
[asy]unitsize(8mm);
defaultpen(linewidth(.8pt)+fontsize(6pt));

pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5);

draw(A--B--C--D--E--F--G--H--cycle);
label("$A$",A,NNW);
label("$B$",B,NNE);
label("$C$",C,ENE);
label("$D$",D,ESE);
label("$E$",E,SSE);
label("$F$",F,SSW);
label("$G$",G,WSW);
label("$H$",H,WNW);[/asy]$ \textbf{(A)}\ 1\minus{}\frac{\sqrt2}{2} \qquad
\textbf{(B)}\ \frac{\sqrt2}{4} \qquad
\textbf{(C)}\ \sqrt2\minus{}1 \qquad
\textbf{(D)}\ \frac12 \qquad
\textbf{(E)}\ \frac{1\plus{}\sqrt2}{4}$

[hide="Click for solution"]
[asy]unitsize(12mm);
defaultpen(linewidth(.8pt)+fontsize(6pt));

pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5);

draw((B)--(F),dashed);
draw((A)--(E),dashed);
draw((A)--(F));
draw((B)--(E));

draw(A--B--C--D--E--F--G--H--cycle);
label("$A$",A,NNW);
label("$B$",B,NNE);
label("$C$",C,ENE);
label("$D$",D,ESE);
label("$E$",E,SSE);
label("$F$",F,SSW);
label("$G$",G,WSW);
label("$H$",H,WNW);[/asy]

Let $ O$ be the intersection of the diagonals of $ ABEF$.  Since the octagon is regular, $ \triangle AOB$ has area $ \frac{1}{8}$.  Since $ O$ is the midpoint of $ \overline{AE}$, $ \triangle OAB$ and $ \triangle BOE$ have the same area.  Thus, $ \triangle ABE$ has an area of $ \frac{1}{4}$, so $ ABEF$ has area $ \frac{1}{2}$, or answer choice $ \textbf{(D)}$.
[/hide]

D)1/2
the octagon has area $ 2s^2 \plus{} 2s^2 \sqrt {2}$
the rectangle has area $ s^2 \plus{} s^2 \sqrt {2}$
EDIT: beaten

An easy way to look at this:
Area of Octagon: $\frac {ap}{2}=1$
Area of Rectangle: $\frac {p}{8}\times 2a=\frac{ap}{4}$
You can see from this that the octagon's area is twice as large as the rectangle's area and voila! You get $\frac {1}{2}$.

[quote="Arrange your tan"][quote="AlphaMath1"]
Area of $ >> $ Hexagon $<<$: $\frac {ap}{2}=1$

You can see from this that the $>>$ hexagon's $ <<$area is twice as large as the
rectangle's area and voila! You get $\frac {1}{2}$.[/quote]

It is an octagon, not a hexagon.[/quote]

Woah oops. *fixed* I always get them mixed up somehow >.> I'm more used to working with hexagons because of their association with triangles and stuff.

[quote="Z-coli"]Another way to attack this problem would be to extend the four non-adjacent sides of the octagon to create a square.  Letting each equal leg of the four triangles created outside of the octagon have side length $ s$, we have that the side length of the octagon is $ s \sqrt{2}$.  Hence, the side length of the square is $ s\plus{}s \sqrt{2}\plus{}s\equal{}s(2\plus{}\sqrt{2})$.  The area of each little triangle is $ \frac{s^{2}}{2}$, so the area of all four is $ 2s^{2}$.  Because the area of the octagon is $ 1$, and is also the difference of the square and the four little triangles, we have an equation:
$ [s(2\plus{}\sqrt{2}]^{2}\minus{}2s^{2}\equal{}1$.
From this, we get that
$ s^{2}\equal{}\frac{1}{4\plus{}4 \sqrt{2}}$.
Notice that the area of rectangle $ ABEF$ is $ s \sqrt{2} \bullet s(2\plus{}\sqrt{2})\equal{}s^{2}(2 \sqrt{2}\plus{}2)$.  Multiplying
$ s^{2}\equal{}\frac{1}{4\plus{}4 \sqrt{2}}$ by $ (2 \sqrt{2}\plus{}2)$, we get that
$ [ABEF]\equal{}s^{2}(2 \sqrt{2}\plus{}2)\equal{}\frac{2 \sqrt{2}\plus{}2}{4\plus{}4 \sqrt{2}}\equal{}\frac{1}{2}$, or $ \boxed D$.[/quote]

I thought of doing the same but how do we know the triangles formed by creating a square are a 45 45 90 triangle or how do we know the legs are congruent?

Using the formula for the interior angles of a polygon: $\frac{180(8-6)}{8}=135$ The supplement of this angle, created by forming the square around the octagon is $180-135=45$.


topology
geometry
3D geometry
sphere
vector
inequalities
function

What are necessary and sufficient (geometric) conditions for a subset of $ \mathbb{R}^2$ to be the unit ball $ \{x \in \mathbb{R}^2 \ : \ \parallel{}x\parallel{} \leq 1 \}$ for some norm?

[quote="JRav"]I'm not sure what you mean.  Under different norms, the unit ball looks different.  You can probably cook up all sorts of norms that make different sets the unit ball.  For example, under the discrete metric, the unit ball is just the point in the middle.[/quote]

But the discrete metric is not given by a norm.

The question is: What conditions on $ K \subset R^2$ are necessary and sufficient for $ K$ to be the closed unit ball relative to some norm on $ R^2$? I believe the answer is that $ K$ must be convex, compact, symmetric (i.e., $ K \equal{} \minus{} K$) and $ 0\in int(K)$ (where "compact" and "int=interior" are relative to the usual topology on $ R^2$).

That's from memory. Look up the definition of the Minkowski functional; I'm sure it's somewhere online.

[quote="Phelpedo"]I may be misinterpreting http://en.wikipedia.org/wiki/Unit_sphere#Unit_balls_in_normed_vector_spaces, but it seems to imply that convexity is not necessary.[/quote]

A moment's thought shows convexity is necessary.

Phelpedo: the $ p$ norm for $ 0 < p < 1$ is not a norm - at least not one that satisfies the axiom of a normed vector space.

The key things a norm has to satisfy include scaling and the triangle inequality. (The $ p$ norm for $ p < 1$ scales but fails the triangle inequality; if you take its $ p$th power, you regain the triangle inequality but lose the scaling property).

A proof that the unit ball of a (real or complex) normed vector space must be convex, using only the scaling property and the triangle inequality:

Suppose $ \|x\| < 1,\|y\| < 1,$ and $ z_t \equal{} (1 \minus{} t)x \plus{} ty$ for $ 0\le t\le 1.$ Then $ \|z_t\| < 1.$

$ \|z_t\| \equal{} \|(1 \minus{} t)x \plus{} ty\|\le\|(1 \minus{} t)x\| \plus{} \|ty\| \equal{} (1 \minus{} t)\|x\| \plus{} t\|y\| < (1 \minus{} t)\cdot 1 \plus{} t\cdot 1 \equal{} 1.$

[hide="Preempted by previous post"]Phelpedo: You are misinterpreting, since the $ p$-norm isn't a norm for $ p<1$. Depending on which version we use, either the triangle inequality or the scaling property fails.

A quick proof that the unit ball in a normed vector space must be convex:

Suppose $ \|u\|$ and $ \|v\|$ are $ \le 1$. Choose an arbitrary linear combination $ w\equal{}tu\plus{}(1\minus{}t)v$ with $ 0\le t\le 1$. We have $ \|w\|\le \|tu\|\plus{}\|(1\minus{}t)v\|\equal{}t\|u\|\plus{}(1\minus{}t)\|v\|\le t\plus{}(1\minus{}t)\equal{}1$ by the triangle inequality and scaling respectively, so $ w$ is also in the unit ball. Repeating over all pairs, the ball is convex.[/hide]

The other properties WWW noted are similarly easy to prove necessary, and the Minkowski functional constructs a norm from such a ball.
I'll go through those proofs. Let $ B$ be the closed unit ball in some norm $ \|\cdot \|$ on $ \mathbb{R}^2$.
Compactness: First, we show that any norm must be continuous in the usual Euclidean topology. $ \|(a,b)\|\le \|(a,0)\|\plus{}\|(0,b)\|\equal{}|a|\|(1,0)\|\plus{}|b|\|(0,1)\|$; this goes to zero as $ a,b\to 0$ in any way. By the triangle inequality, continuity at the origin implies continuity elsewhere.
Now, $ B$ is the preimage of a closed set under a continuous function, and must be closed. Also, since the norm is continuous, it has a minimum $ c$ on the (compact) unit circle. Every nonzero element of $ \mathbb{R}^2$ is a positive multiple of something on the circle, so $ \|x\|\equal{}\|x\|_2\cdot \left\|\frac{x}{\|x\|_2}\right\|\ge c\cdot \|x\|_2.$ Thus, nothing outside the Euclidean ball of radius $ \frac1c$ is in $ B$, and $ B$ is bounded. A closed and bounded set in $ \mathbb{R}^2$ is compact.
0 in the interior: As we've already shown, $ \|\cdot\|$ is continuous. The open ball $ \{x: \|x\|<1\}$ is therefore an open (in the Euclidean sense) set contained in $ B$ containing $ 0$.
Symmetric: $ \|x\|\equal{}\|\minus{}x\|$, by the scaling property with coefficient $ \minus{}1$. Either both are in, or both are out.


Solve for $ x$:

$ \frac{\sqrt{a\plus{}x} \plus{} \sqrt{a\minus{}x}}{\sqrt{a\plus{}x} \minus{} \sqrt{a\minus{}x}}\equal{}\frac{a}{x}$

I use multiplied the left side by its conjugate and got $ \sqrt{(a\plus{}x)(a\minus{}x)} \equal{}a$, but the [hide="answer is"] $ x_1 \equal{} \minus{}a, x_2\equal{}a$.[/hide]

[quote="jeez123"]Solve for $ x$:

$ \frac {\sqrt {a \plus{} x} \plus{} \sqrt {a \minus{} x}}{\sqrt {a \plus{} x} \minus{} \sqrt {a \minus{} x}} \equal{} \frac {a}{x}$

I use multiplied the left side by its conjugate and got $ \sqrt {(a \plus{} x)(a \minus{} x)} \equal{} a$, but the [hide="answer is"] $ x_1 \equal{} \minus{} a, x_2 \equal{} a$.[/hide][/quote]

Let's see...

$ \frac {\sqrt {a \plus{} x} \plus{} \sqrt {a \minus{} x}}{\sqrt {a \plus{} x} \minus{} \sqrt {a \minus{} x}} \equal{} \frac {a}{x}$

$ \Rightarrow\frac {\sqrt {a \plus{} x} \plus{} \sqrt {a \minus{} x}}{\sqrt {a \plus{} x} \minus{} \sqrt {a \minus{} x}}\cdot\frac{\sqrt {a \plus{} x} \plus{} \sqrt {a \minus{} x}}{\sqrt {a \plus{} x} \plus{} \sqrt {a \minus{} x}} \equal{} \frac {a}{x}$

$ \Rightarrow\frac {2a\plus{}2\sqrt {(a \plus{} x)(a \minus{} x)}}{2x}\equal{} \frac {a}{x}$

$ \Rightarrow\frac {a\plus{}\sqrt {(a \plus{} x)(a \minus{} x)}}{x}\equal{} \frac {a}{x}$

So $ a\plus{}\sqrt {(a \plus{} x)(a \minus{} x)}\equal{}a\Rightarrow\sqrt{(a\plus{}x)(a\minus{}x)}\equal{}0$

So $ (a\plus{}x)(a\minus{}x)\equal{}0$, so the solutions are $ x\equal{}\pm{}a$

Remember, we were solving for $ x$, not $ a$...

[hide="here goes nothing"]
Call $ \sqrt{a\plus{}x}\equal{}c$ and $ \sqrt{a\minus{}x}\equal{}d$.
Convertify equation.
$ \frac{c\plus{}d}{c\minus{}d}\equal{}\frac{c^2\plus{}d^2}{c^2\minus{}d^2}$
Multiply LHS by a+b.
$ \frac{c^2\plus{}2cd\plus{}d^2}{c^2\minus{}d^2}\equal{}\frac{c^2\plus{}d^2}{c^2\minus{}d^2}$
Multiply both sides by $ c^2\minus{}d^2$.  Simplify.
$ cd\equal{}0$.
Whoa. That means either $ a\equal{}x$ or $ \minus{}a\equal{}x$.
 :D

[/hide]


[b]If f(x)=x^2+ax+b and the equation f(f(x)) has 4 real roots and 2 of its sum is
= -1. Then prove that b<=-1/4.[/b]


search
quadratics
algebra
difference of squares
special factorizations

Hi Guys and Girls I have this "funny" equation, but I don't have any idea of have to start. Can any of you please give me some kind of clue of how to get started?

(x+a)(x+b)(x+c)(x+d)+e = 0

[hide="Heres all I can think of"](let $a+b=p, \frac{ab+cd}{2}=q, r=\frac{ab-cd}{2}=q$)
\[(x+a)(x+b)(x+c)(x+d)+e=0\iff\\\iff-e=(x^{2}+px+ab)(x^{2}+px+cd)=(x^{2}+px+q)^{2}-r^{2}\]from which it is easily solved since we have two quadratics in terms or $a,b,c,d$.
[/hide]

Yes, that is essentially the solution...i think it is a bit easier to let $j=a+b=c+d$ and $k=x^{2}+jx$

then you have: $(x^{2}+(a+b)x+ab)(x^{2}+(c+d)x+cd)+e=0\implies (k+ab)(k+cd)-e=0$

This is a quadratic that can be solved for $k$. Then use those values to solve for $x$.

This solution is essentially the same. I just did not use those two other random substitutions that you used...

[quote="Altheman"]err...you are going to have to use the quartic formula...[search on google]

Here is a nicer problem: provided that $a+b=c+d$, solve $(x+a)(x+b)(x+c)(x+d)+e=0$[/quote]

Moreover, $ad-bc=0$ is also nice.

Well its exactly the same as in my solution except the difference of squares factorization has two squares each with an x in them.


modular arithmetic
number theory unsolved
number theory

Determine all solutions (x,k,n) of positive integers such that:
\[3^k-1=x^n\]

Suppose WLOG $n>1$.

[b]Case 1.[/b] $n$ is even. We have $3^k-1$ is even, so $x$ is even. It follows that $4\mid x^2$. As $4\mid 3^k-1$ we conclude $k$ is even: $k=2l$. We can write $x^2=3^k-1=(3^l-1)(3^l+1)$. Since $\gcd(3^l+1,3^l-1)=2$ we obtain $3^l-1=2u^2$ and $3^l+1=2v^2$. We see $v^2-u^2=1$ -- contradction.

[b]Case 2.[/b] $n$ is odd. Suppose $p\mid n$, $p\geq 3$ is a prime number. Consider new equation $3^k-1=y^p$. Then $3^k=(y+1)(y^{p-1}-y^{p-2}+...+1)$. Hence $y=3^l-1$, but it follows that $y^{p-1}-y^{p-2}+...+1=p\pmod{3}$. It means $p=3$. Thus we obtain $3^k=(y+1)((y+1)^2-3(y+1)+3)$, but $9\not\mid (y+1)^2-3(y+1)+3$. Therefore, $y^2-y+1=3$, i.e. $y=2$.

Answer: (2,2,3).


abstract algebra
induction
group theory
set theory
Ring Theory
superior algebra
superior algebra unsolved

hi

are there any fields with more elements than the reals

i really mean more in a cardinality sense, not in the sense of extensions like C or R( x)

my guess is yes, but I can't find any example

thx

Take a transcendental extension $R(x_\lambda)$ with a set $\{x_\lambda : \lambda \in \Lambda\}$ of  indeterminates indexed by as big a set $\Lambda$ as you like.

Edited to delete the stupidity I said :) (see fleeting-guest's comment below).

[quote="grobber"]I think $\Omega_p$, the metric completion of the algebraic closure of the metric completion of $\mathbb Q$ with respect to the $p$-adic norm also works (i.e. it has more elements than the reals).[/quote]

Algebraic closure of a field is quotient (by some equivalence relations) of the field formed by adjoining $n$ roots for each irreducible polynomial.  The number of roots adjoined is at most the number of elements of the field (if infinite), so the closure does not increase the cardinality of the field.
The (sequential) completion of a topological space is some quotient of the set of convergent sequences in that space, so does not raise the cardinality of an uncountable space.
So $\Omega_p$ has the same cardinality as $R$.

There is the question of "effective cardinality" --- whether there is any tame (e.g. measurable) bijection with $R$.   I am not sure about this because algebraic closure of an arbitrary field involves the axiom of choice, although for $Q_p$ there may be some more explicit approach.

[quote="kgbto"]i don't think that the surreals even form a set, rather a proper class but it's very likely for me to be wrong with that[/quote]

The surreals definitely do not form a set,  but I don't see any reason to require a field's elements to form a set.

If you don't like that, though, there's plenty of subsets of the surreals that work as well. Given a cardinal $C$, we can form the subset $S_C$ such that for all  $\{X_L|X_R\} \in S_C$:
$\|X_L\| < C$
$\|X_R\| < C$
$\forall x \in X_L: x \in S_C$
$\forall x \in X_R: x \in S_C$.

It's closed and a subset of a field, so it's a subfield. By choosing the right $C$, you should be able to get $\|S_C\|$ to be as large as you want.

what i meant was that the definitions for group, loop, ring, field (and categories like those) i have studied usually required a field for instance to be a "set, equipped with certain operations within it that satisfy certain conditions". Wolfram research website says the same:
http://mathworld.wolfram.com/Field.html
I think that when defining the elements of a field, one has to start from somewhere, to "gather" those elements from a somehow closed in an intuitive sense or already well-defined entity. If there is a definition that do not require that fields are sets, i am surely very interested to become familiar with it :) What does it say for the entity of the fields elements ? An arbitrary proper class ? It has to be somehow expressible in set theory terms.

Wikipedia seems to indicate that the surreals might be a field.
http://en.wikipedia.org/wiki/Surreal_number

Algebraic structure theory is just the study of the logical consequences of various properties and their combinations. So you could develop a version of, say, field theory where "is a field"  is simply replaced by "satisfies the field axioms". You could additionally specify that a structure "has elements that form a set", optionally. Otherwise, you leave the question of how the elements are contained to the structures themselves. Really, I just think of "field" as a very compact way of saying the property "is associative, commutative, closed, has inverses and identities under addition and multiplication.", or the fully written out version with the $\forall$s and the $\exists$s.
Does this make any sense?

And this isn't a problem with cardinality alone. What do you think of the abelian monoid of sets under unions? or the abelian monoid of abelian monoids under cartesian cross product? They contain themselves, but are of particular interest to abelian monoid theorists.

well, i have always thought of the elements of different abstract categories as taken from a particular set by default.  Along with that i have developed myself the belief that somehow those category theories are built within set theory, as the whole maths is supposed to be.
Nevertheless, the question of studying "pure" properties such as associativity, commutativity and others, starting from basic category axioms is sure of particular interest and has its own depth and value.
However, i doubt that one would be able to proove the same propositions when freeing himself form choosing elements from a set. For example, when we try to proove that every non-trivial commutative ring has a maximal ideal, we rely on the axiom of choice, which speaks only about sets. I'm not aware of a proof that doesn't use this axiom or any equivalent one. When prooving that every field can be extended to an algebraically closed one, we somehow use transfinite induction by adjoining some roots to the field and "waiting" for the moment when the construction becomes stable under this operation, and transfinite induction deals with ordinals - certain sets. So it seems like if we abandon the requirement that certain category theory lies within a set, we wouldn't be able to derive such variety of propositions and theorems. Maybe, what could be worth fighting for, would be certain logical implications that follow from the axiomatic ones. I just can't see how the existence of some ideals, subgroups, closures could be "set-free" prooved  :huh:


geometry
geometric transformation
reflection
trapezoid
parallelogram
rhombus
symmetry

Let $\Gamma{1}$ and $\Gamma{2}$ be two non-intersecting circles(with centers $O_1$ and $O_2$ respectively) and denote by $l_1$ and $l_2$ their external common tangents. Take a point $A$ on $l_1$(on the  segment determined by the points of tangency of the two circles with $l_1$)and consider points $B$ and $C$ on $l_2$ such that $AB$ is tangent to $\Gamma{1}$ and $AC$ is tangent to $\Gamma{2}$ . Let $D$ be the point of tangency of the ex-circle of $\triangle{ABC}$ with the side $BC$. If $X$ is the midpoint of $(O_1O_2)$ then prove that $\triangle{AXD}$ is isoceles.

Let $P_1, P_2$ be the tangency points of the circles $\Gamma_1, \Gamma_2$ with one of their common external tangent and $Q_1, Q_2$ the tangency points of these 2 circles with their other common external tangent. Let $R_1, R_2$ be the tangency points of the tangents from the point $A \in P_1P_2$ with the circles $\Gamma_1, \Gamma_2$ and $s$ the semiperimeter of the triangle $\triangle ABC$.

$2s = AB + BC + CA = (AR_1 + BR_1) + BC + (CR_2 + AR_2) =$

$= AP_1 + BQ_1 + BC + CQ_2 + AP_2 = (AP_1 + AP_2) + (BQ_1 + BC + CQ_2) =$

$= P_1P_2 + Q_1Q_2 = 2\ P_1P_2$

Hence, the semiperimeter of the triangle $\triangle ABC$ is constant and equal to the segment length of the common external tangent of the circles $\Gamma_1, \Gamma_2$ between the tangency points. The distances of the tangency point $D$ of the excircle $(E_A)$ against the vertex $A$ from the tangency points $Q_1, Q_2$ are equal to

$DQ_1 = BQ_1 + s - AB = BQ_1 + (AP_1 + AP_2) - (AR_1 + BR_1) =$

$= (BQ_1 - BR_1) + (AP_1 - AR_1) + AP_2 = AP_2$

$DQ_2 = CQ_2 + s - CA = CQ_2 + (AP_1 + AP_2) - (AR_2 + CR_2) =$

$= (CQ_2 - CR_2) + (AP_2 - AR_2) + AP_1 = AP_1$

Reflect the points $O_1, O_2, X$ in the tangent $P_1P_2$ into the points $O_1', O_2', X'$. The quadrilateral $O_1O_2O_2'O_1'$ is an isosceles trapezoid and $P_1, X, P_2, X'$ are the midpoints of its sides $O_1'O_1, O_1O_2, O_2O_2', O_2'O_1'$. Since the Varignon parallelogram $P_1XP_2X'$ of the isosceles trapezoid  $O_1O_2O_2'O_1'$ is a rhombus, $P_1X = P_2X$. Thus the triangle $\triangle P_1XP_2$ is isosceles and the angles $\angle XP_1P_2 = \angle XP_2P_1$ are equal. By symmetry of the tangency points $P_1, Q_1$ amd $P_2, Q_2$ with respect to the center line $O_1O_2$,

$Q_1X = P_1X = P_2X$

$\angle XQ_1D \equiv \angle XQ_1Q_2 = \angle XP_1P_2 = \angle XP_2P_1 \equiv \angle XP_2A$

Consequently, the triangles $\triangle AP_2X \cong \triangle DQ_1X$ are congruent by SAS, because $DQ_1 = AP_2, Q_1X = P_2X$ and $\angle XQ_1D = \angle XP_2A$. Hence, their remainig sides $XD = XA$ are also equal and the triangle $\triangle AXD$ is isosceles.


algorithm
calculus
number theory
greatest common divisor
Euclidean algorithm
number theory unsolved

i am a math teacher that is working on her master's. My teacher gave me this problem to work and other than going one by one to find a pattern, I am not sure how to do this. This is due tomorrow!  the problem is:

Find every positive integer (n) such that n^3+n^2+n^1+1 is square

thank you!!!

Already posted (and answered) in Intermediate Topics.

The solution in intermediate topics is wrong.  Write it as $n^3+n^2+n+1$.  Then it factors as $(n^2+1)(n+1)$.  We want to see what common factors $n^2+1$ and $n+1$ can have.  By the Euclidean algorithm, we find $(n^2+1)-(n-1)(n+1)=2$.  Hence the greatest common factor of $n^2+1$ and $n+1$ is $2$, and this occurs if and only if $n$ is odd.  Break it up into two cases.

1) $n$ is even.  Then we need $n^2+1$ and $n+1$ to both be perfect squares.  $n^2+1$ is never a perfect square unless $n=0$.  Hence $n=0$ is the only solution.

2) $n$ is odd.  Write $n=2k+1$.  Then $n^2+1=4k^2+4k+2$ and $n+1=2k$.  We can get rid of the factors of $2$, so that $2k^2+2k+1$ and $k$ are both perfect squares.  If $2k^2+2k+1=l^2$, then $(k+1)^2=(l-k)(l+k)$.  I'm kind of stuck here.  It appears the only solutions are $n=1$ and $n=7$.

This certainly doesn't belong in the calculus forum. Let's give it a better home.

[quote="blahblahblah"]2) ... Write $n=2k+1$.... $n+1=2k$...[/quote]

The second one should read $n+1=2k+2$

d'oh!  I was fiddling around with $n=2k-1$ and $n=2k+1$, and I think I changed one but not the other.  It doesn't change that I don't have a solution, unfortunately.

If $n$ is odd then see that $(n^2+1)(n+1)=4(2k^2+2k+1)(k+1)$.

Now look for the gcd of the numbers at the factorisation and notice that

$2k^2<2k^2+2k+1<2(k+1)^2$   ;)


Does anyone know who in the top 10 in MC STATE 08 used AoPS?
I know 2:
5.  Me
9. Rking


solve this please
[b]1.[/b] (y+1)^x = y! + 1[/b]
[b]2.[/b]If K is not equal to B and if K^n-A|K-b ,so a=b^n
   *If K^n-A divides K-b
[b]3.[/b]x^2 - y^3 = (2c)^3 - 1 solve this equation
Good Luck

Could you rewrite the equations using latex in a proper way and defining where your variables come from because like this nobody can understand. Maybe somebody will try to solve after you edit the post :maybe:


The numbers $a,b,a_{2},a_{3},.........a_{n-2}$ are real and $ab$ is not $0$. All the roots of the polynomial

$ax^{n}-ax^{n-1}+a_{2}x^{n-2}+a_{3}x^{n-3}+........+a_{n-2}x^{2}-n^{2}bx+b=0$

are real and positive. Prove that all the roots are mutually equal.

[hide]Let $x_{1}, x_{2}, ... x_{n}$ be positive roots.

Dividing by $a$ we have:

$x^{n}-x^{n-1}+\alpha_{2}x^{n-2}+...-n^{2}\beta x+\beta =$
$(x-x_{1})(x-x_{2})...(x-x_{n})=$
$x^{n}-\sigma_{1}x^{n-1}+...+(-1)^{n-2}\sigma_{n-2}x^{2}+(-1)^{n-1}\sigma_{n-1}x+(-1)^{n}\sigma_{n}$,
where $\sigma_{i}$ are elementary simmetrical polinoms. In particular the all of them are positive.

By comparison we obtain:
$\sigma_{1}=1, \ \ \sigma_{n-1}=(-1)^{n}n^{2}\beta, \ \ \sigma_{n}=(-1)^{n}\beta$
Two last give
$\sigma_{n-1}= n^{2}\sigma_{n}$

So we have
$x_{1}+x_{2}+...+x_{n}=1$
$\frac{1}{x_{1}}+...+\frac{1}{x_{n}}=n^{2}$
The LHS of the last equation has strictly minimum $n^{2}$ at $x_{1}=x_{2}=...=x_{n}= \frac{1}{n}$[/hide]


Determine the number of elements of the following set:
$M=\left \{ x\in Q|x=\frac{n^{2}+3}{n^{2}+n}, n\in \{ 1,2,...,50\}\right \}$

Let $f(n)=\frac{n^{2}+3}{n^{2}+n}$. we have $f(n)=f(m)\iff n=m$ or (simplifying) $(m-3)(n-3)=12$. since 12 has 6 disticnt divisors, we can have 3 distinct pairs (m,n), hence the ans is $47$. am i correct?

EDIT: calculation errors, sorry. :blush:

[quote="nayel"]Let $f(n)=\frac{n^{2}+3}{n^{2}+n}$. we have $f(n)=f(m)\iff n=m$ or (simplifying) $(n,m)=(5,6)$ or $(4,9)$. so the ans is $48$. am i correct?[/quote]

Huh? Neither of those ordered pairs work, and there are others that do...  :huh:
[hide]
\[f(m)=f(n)\iff \frac{n-3}{n^{2}+n}=\frac{m-3}{m^{2}+m}\]
after subtracting 1 from each side. Then multiplying out gives
\[(m^{2}+m)(n-3)=(n^{2}+n)(m-3)\iff (m-n)(mn-3m-3n-3)=0\]
If $m>n$, then $(m-3)(n-3)=12$ which gives the pairs $(15,4),\ (9,5),\ (7,6)$. So there are $\boxed{47}$ distinct values.
[/hide]


calculus
integration
analytic geometry
geometry
3D geometry
sphere
trigonometry

I am really not very good with polar coords and all its offshoots. So I don't even know where to start with this problem other than it's a triple integral: (I know it can be solved without cylindrical but it is asking specifically for cylindrical.)

Use [b]cylindrical coordinates[/b] to find the Volume of the solid enclosed by the sphere $x^{2}+y^{2}+z^{2}=9$, $z=0$ and the cylinder $x^{2}+y^{2}=3y$

I have so far that the cylinder is $x^{2}+(y+\frac{3}{2})^{2}=\frac{3}{2}^{2}$ so the cylinder is of radius $\frac{3}{2}$ centered about $(0, -\frac{3}{2}, z)$ Which I think means that is is $r=\frac{3}{2}$ and $0\leq\theta\leq2\pi$ and $z=z$. I have no idea how to do a sphere in cylindrical and what the $f(r,\theta,z)$ part of $\int\int\int f(r,\theta,z)drd\theta dz$ is.

Okay I just found this gem in my book: $\int\int\int_{E}f(x,y,z)dV=\int_{\alpha}^{\beta}\int_{h_{1}(\theta)}^{h_{2}(\theta)}\int_{u_{1}(r cos(\theta), r sin(\theta)}^{u_{2}(r cos(\theta), r sin(\theta)}f(r cos(\theta), r sin(\theta))r dz dr d\theta$

Therefore: $\int_{0}^{2\pi}\int_{0}^{\frac{3}{2}}\int_{0}^{\sqrt{r^{2}-9}}r(\sqrt{r^{2}-9}) dz dr d\theta$ ???

Unfortunately I haven't gotten to polar coordinates yet so I feel bad about answering this post in this way, but...
You don't have to use polar coordinates and instead could use a double integral.
I assumed you meant that the region intersects $z \ge 0$, but it doesn't matter if you say $z \le 0$
[hide="solution with double integral, but without polar coordinates"]
A few important things to note before integrating.
Drawing a picture is helpful.
$x^2+y^2=3y \rightarrow x^2+(y-\frac{3}{2})^2 = {\frac{3}{2}}^2$
Since $z \ge 0 \rightarrow z = \sqrt{9-y^2-x^2}$
This means the equation can be written:
$\int^{\sqrt{\frac{9}{4}-x^2}+\frac{3}{2}}_{-\sqrt{\frac{9}{4}-x^2}+\frac{3}{2}} {\int^{3/2}_{3/2} \sqrt{9-x^2-y^2}\,dx}\, dy$
Forget what I said.
[/hide]
Nevermind.

We just had our polar coordinate lecture and i'm pretty sure I can take on your problem now! Hooray.
Now, first thing first.  When you edit your post make sure you write correct stuff like $x^2+(y-\frac{3}{2})^2 = {\frac{3}{2}}^2$, which results in different answers on other parts of your post.

I used law of cosines to figure out that limits of integration of the radius for a given angle, they are $0$ and $\frac{3}{2} \sqrt{2+2\cos{2\theta}}$.
For a given angle, the area in terms of $\theta$ and $r$ is then $\int^{\frac{3}{2} \sqrt{2+2\cos{2\theta}}}_{0} \sqrt{9-r^2}*r\,dr$.
Thus the volume of the whole figure is, $\int^{\pi /2}_{-\pi /2} {\int^{\frac{3}{2} \sqrt{2+2\cos{2\theta}}}_{0} \sqrt{9-r^2}*r\,dr}\,d\theta$.
By letting $u = 9-r^2, du = -2rdr$, the innermost integral evaluates to: $\frac{1}{3}(27-(9-{\frac{3}{2}}^2(2+2\cos{2\theta}))^{3/2})$.
Thus the whole thing evaluates to $9 \pi - 9 \int^{\pi /2}_{-\pi /2} \sin{\theta}^3\,d\theta$, by a little factoring and using the fact that $1-\cos{2\theta} = 2 \sin{\theta}^2$.
Finally that just comes to $9 \pi$.  I think someone should check my work and another person should Maple or Mathematica this and another should do it a different way.

FMako, you make some good points but the Question specifically asks for [b]cylindrical coordinates[/b] and since cylindrical coords are 3D, I assume that also means a respective triple integral.

[quote="jmerry"]There's not much difference between cylindrical and polar coordinates for this problem. In this case, just replace the inner $\sqrt{9-r^2}$ with $\int_0^{\sqrt{9-r^2}}1\,dz$[/quote]

Seriously? Okay, will give that a shot...

Um the max volume of the cylinder in Question is $6.75\pi$ (due to $0\leq z\leq3$ from $z=0$ plane and z max of 3 based on the sphere) and the max volume of the sphere (since its cut in half by z=0 and half again due to the cylinder being only on the positive side of $x=0$) is $9\pi$ how can the answer be $9\pi$?

I must have made a mistake lajakahr.  I did it last night instead of homework on sticky notes so I assumed I was wrong in the first place.  I just wanted people to look at it and say if I was wrong or right, and find out what I did wrong or right, respectively.

[quote="jmerry"]It's $\int\int\int 1\cdot r\cdot dz\cdot dr\cdot d\theta$, but the limits of those integrals and the best order to do the integration aren't clear.
The boundary contains pieces with equations $z=0$, $z^2+r^2=9$, and $r=3\cos\theta$. In the last equation, we should restrict to $-\frac\pi2\le\theta\le\frac\pi2$, so we have
$\int_{-\frac\pi2}^{\frac\pi2}\int_0^{3\cos\theta}\int_0^{\sqrt{9-r^2}} 1\cdot r\cdot dz\cdot dr\cdot d\theta$[/quote]

Why is it: $-\frac\pi2\le\theta\le\frac\pi2$ and not $0\le\theta\le\pi$? Since the cylinder lies entirely on $x\ge0$?
$\theta$ describes the angle with respect to the xy plane correct?

The freaky part is that my calculator gives me the same answer either way....16.2743 and cranking it out by hand I get $9\pi$!!!!! (Is my calculator wrong?) Is this maybe the volume of both the portion above and below $z=0$???


[quote="Art of Owna"]Call an ordered pair of positive integers (m,n) "complex" if it requires carrying while adding m+n.  How many "complex" pairs of (m,n) are there that sum up to 2007[/quote]

[hide]The units digits sum to 7, the tens to 0, the hundreds to 0, and the thousands to 2, so $8\cdot 3=\boxed{24}$[/hide]

No offence, but your answer made no sense

[quote]hm i got 23[/quote]
wait, how can the answer be odd-if (m,n) works, then (n,m) works-and 2007 is odd, so there was be an even amount...[

You both need to reread the question-REQUIRES CARRYING-you both are solving when it does not require carrying-which is still wrong....
and yes- (986,1021) and (1021,986) are different

Whoops.

[hide]So then if you just use the previous number I got...there are 2006 possible (m,n) with no restrictions, 2006-24=1982, subtract two for the cases m=0 and n=0 to get 1980[/hide]

Close..check your number again.. why 2*8? and when you did that, you counted when m and n can be 0..

Blargh, edited, hopefully it's right now -.-

Alright- if the digits sum up to 7, there are 8 cases, the 0s dont matter, but if the digits sum up to 2, there are 3 cases, so 8*3=24, but we counted 0 as a choice for m and n, so 24-2=22, 2006-22=1984..
22 cases

1000-1007
1001-1006
1002-1005
1003-1004

reverse them

2000-7
2001-6
2002-5
2003-4
2004-3
2005-2
2006-1

reverse them

4*2+7*2=22

yeah i'm dumb, meant 22 >.>

did 8*3 and removed (0, 2007) but not (2007, 0)

man those hot ranch or w/e doritos are yummy

Is there a way to count them  up directly?  instead of doing total-not requring carrying

quick question.....is there a collection of mind bogglers published or are they only accessible through the actuarial digest.

and a comment on the footnote:  why doesn't anyone think about the water?  It's always about the glass....but I supposed that's for another thread


geometry
circumcircle
trigonometry
geometry proposed

Let $\triangle ABC$ be a triangle and denote $R,H_a$ the circumradius and the feed of the perpendicular from the vertex $A$ to the line $BC$ respectively. Show that:

$(1) \qquad \left( {AH_a } \right)^2  = BH_a  \cdot CH_a \Longleftrightarrow \measuredangle C - \measuredangle B = 90^\circ$

$(2) \qquad \measuredangle C - \measuredangle B = 90^\circ \Longrightarrow AB^2  + AC^2  = 4R^2$

[quote="Leon"]$(1) \qquad \left( {AH_a } \right)^2  = BH_a  \cdot CH_a \Longleftrightarrow \measuredangle C - \measuredangle B = 90^\circ$[/quote]
 :what?:  :what?: Clearly $\measuredangle C + \measuredangle B = 90^\circ$ if $H_a\in [BC]$.
I think you meant $H_a$ doesn't lie on segment $[BC]$. If it is true than we immediately obtain $\tan B\cdot\tan C=1$, so $|\measuredangle C - \measuredangle B| = 90^\circ$.

Your first assertion is false.
[quote="Myth"]  Clearly $\measuredangle C + \measuredangle B = 90^\circ$ if $H_a\in [BC]$.[/quote]

Your second comment
[quote="Myth"] I think you meant $H_a$ doesn't lie on segment $[BC]$. If it is true than we immediately obtain $\tan B\cdot\tan C=1$, so $|\measuredangle C - \measuredangle B| = 90^\circ$.[/quote]
me seems obvious, in fact $\measuredangle C - \measuredangle B = 90^\circ  \Rightarrow \measuredangle C = 90^\circ  + \measuredangle B \Rightarrow \measuredangle C$ is obtuse $\Rightarrow H_a$ doesn't lie on segment $BC$

[quote="Leon"]Your first assertion is false.
[quote="Myth"]  Clearly $\measuredangle C + \measuredangle B = 90^\circ$ if $H_a\in [BC]$.[/quote][/quote]
???
It is a well known property of right triangles.

Mikhail is right here. In fact both of you are right :) If the triangle is not right-angled then Leon is correct. However if the triangle is right-angled in $A$ Myth is correct :) So it's just a matter of cases ;)

[quote="Myth"]I am correct in all cases, because I wrote: "IF $H_a\in[BC]$ then ..." and "IF $H_a\not\in [BC]$ then ...". So I considered all cases.[/quote]Yes you are right, but I was trying to burry the hatch of war :P :) It is probably a small misunderstanding of geometrical notation ...

If <C - <B = 90 then <CAH = <ABH, so ACH ~ BAH and finally AH/BH = CH/AH.

Since <CAH = <ABH, HA is tangent at A to c(ABC). Then If M is the mid point of AC, OMA ~ AHB, and from this, as AC = 2AM, AB = 2OM. Therefore 4R^2 = 4AM^2 + 4OM^2 = AC^2 + AB^2.

If AH^2 = BH*CH [b]and H is external to segment BC [/b] then ACH ~ BAH then <C - <B = 90.

If AH^2 = BH*CH [b]and H is internal to segment BC [/b] then ACH ~ BAH then <C - <B = 90.

[quote="Valentin Vornicu"]... but I was trying to burry the hatch of war :P :)...[/quote]
Is it possible you mean "to bury the axe of war"? At least, it is an idiom in Russian. :?

[quote="Myth"][quote="sprmnt21"]If AH^2 = BH*CH [b]and H is external to segment BC [/b] then ACH ~ BAH then <C - <B = 90.
If AH^2 = BH*CH [b]and H is internal to segment BC [/b] then ACH ~ BAH then <C - <B = 90.[/quote]
???[/quote]

Sorry, I made a too quick cut and paste  :D.

Here it follows my real intention:

"If AH^2 = BH*CH [b]and H is internal to segment BC [/b] then ACH ~ BAH then <C + <B = 90"

I forgive you this time...  :lol:  :lol:  :lol:


what is the greatest possible number of points of intersection for 8 distinct lines in a plane?

THis is what the answer sheet gave me:
[hide]28[/hide]

I don't get how to do it.  pls show me

Mathcounts is next week for me, and i am very nervous.  :(  :maybe:
i feel really scared  :stretcher:

With every 2 lines, you have one intersection.  Therefore, the answer is simply $ \binom{8}{2}\equal{}28$

Or if you really feel like it, you can draw the diagram.  For the first two lines, there is only at most one intersection point.  To maximize the number, we wish to have each line intersect the previous ones drawn.  So the third line will create another two, etc.
Thus, our answer is $ 1\plus{}2\plus{}3\plus{}\hdots\plus{}7\equal{}\frac{7*8}{2}\equal{}\binom{8}{2}\equal{}28$ You can generalize this to $ n$ lines.


How to find out when the equality occurs in some inequality?

Inequalities are type such as $x_1\leq x_2 \leq \cdots \leq x_n$. Thus we have $x_1=x_n$ iff $x_i=x_{i+1}$ for every $1\leq i \leq n-1$. Now usually in competition you can use some well-known inequalities in every part of proving some inequality. Usually it is also known when the equality occurs is some well-known inequality. Of course you can directly try to solve the equation $x_1=x_n$. And in many cases the equality occurs if and only if all variables has the same value, but not always. Hope this helps.


[quote="4everwise"]If $\frac{4^x}{2^{x+y}}=8$ and $\frac{9^{x+y}}{3^{5y}}=243$, $x$ and $y$ are real numbers, then $xy$ equals:

$\text{(A)} \ \frac{12}{5} \qquad \text{(B)} \ 4 \qquad \text{(C)} \ 6 \qquad \text{(D)} \ 12 \qquad \text{(E)} \ -4$[/quote]
[hide]
Rearranging both:
$2^{x-y}=8$
$3^{2x-3y}=243$
$x-y=3$
$2x-3y=5$
$xy=4$[/hide]

[hide]
Set the first equation to base 2 and the second to have base 3.

$\frac{2^{2x}}{2^{x+y}}=2^3$

$\frac{3^{3x+3y}}{3^{5y}}=3^5$

By the rules of exponents:

$2x-x-y=3$ which simplifies to: $x-y=3$

$3x+3y-5y=5$ which simplifies to: $3x-2y=5$

Solving this system gives us:

$x=-1$
$y=-4$

So $xy=-1*-4=4$ So the answer is (B)
[/hide]

[hide="answer"]$2x-3y=5$
$x-y=3$

$y=1$ and $x=4$ so $xy=4$

Work:

$\frac{2^{2x}}{2^{x+y}}=2^3$

$\frac{3^{2x+2y}}{3^{5y}}=3^5$ Getting us:

$2x-3y=5$ and $x-y=3$. Find $x$ and $y$ by solving it by systems and multiply $x$ and $y$ together the get $xy=4$.

Edit: $\boxed{B}$ is the right answer.
[/hide]

[hide] From the first equation we have that $\frac{4^x}{2^{x+y}}=8 and \frac{9^{x+y}}{3^{5y}}=243$
Converting to bases 2 and 3,
Since $2x-x-y = 3$ and $3x+3y-5y = 5$, we use linear combinations to find that x = 4 and y = 1 so the answer is \boxed{B}[/hide]


I've tried a few examples, and it doesn't seem to be true.

Oh you are right, sorry for typo :blush:


[quote="mychrom"]Fie $ABCD$ un patruleter astfel incat $m(BAC)=m(BDC)$. Fie $O$ intersectia diagonalelor. Daca $OV \perp (ABCD)$ si $m(AVC)=90^{0}$ atunci si $m(BVD)=90^$.[/quote]
Din $VO\perp (ABCD)$ avem $VO\perp AC, BD$.
Din egalitatea unghiurilor, patrulaterul este ABCD inscriptibil, deci $AO\cdot CO$=$DO\cdot BO$. In trABC, putem aplica teorema inaltimii si avem $VO^{2}$=$AO\cdot CO$. Din egalitatea de mai sus (puterea punctului O fata de cerc), obtinem $VO^{2}$=$BO\cdot DO$ si conform reciprocei teoremei inaltimii, obtinem ca   $m(BVD)=90^{0}$


Show that you cant find an $ n$-digit number such that the $ n^{th}$ power of the sum of its digits is equal to the number for every positive base.

$ a_1a_2a_3......a_n \ne (a_1 \plus{}a_2 \plus{} a_3 \plus{} ...... \plus{} a_n)^n$
Show that statement holds for some $ n$ in every positive integer base.

I guess you mean for $ n\ge2$ because all one digit numbers have this property.

Also the part were you say for every positive base you mean:

1) You can't find $ k$ such that written in any base, the condition holds ??

or

2) You can't find $ k$ such that the condition holds at least in one base??

 :)

You have to show there exists at least one number n in every positive $ integer$ base k, such that in every base k there cannon exist some $ n$-digit number that is equal to the nth power of the sum of its digits.

$ a_1a_2a_3......a_n \ne (a_1 \plus{}a_2 \plus{} a_3 \plus{} ...... \plus{} a_n)^n$
Show that statement holds for some $ n$ in every positive integer base.


induction
number theory proposed
number theory

Show that there is a unique sequence $u_{0},u_{1},u_{2},\cdots$ of positive integers such that \[u_{n}^{2}= \sum_{r = 0}^{n}{n+r \choose r}u_{n-r}\] for all nonnegative integers $n$.

We have $u_{0}^{2}=u_{0}$, because $u_{0}$ positive number $u_{0}=1$.
Then $u_{1}^{2}=u_{1}+2$, because $u_{1}$ positive number $u_{1}=2$.
Let defined $u_{k}$ for all k<n. Then for $u_{n}$ is positive root of equation \[u_{n}^{2}-u_{n}=p, \ p=\sum_{r=1}^{n}{n+r \choose r}u_{n-r}.\] Because p is positive, this equation had only one positive solution. It proved, that exist unique positive sequence.
By induction we have $u_{n}=2^{n}$, because \[2^{n}=\sum_{k=0}^{n}{n\choose k}\] and \[\sum_{r=0}^{n}{n+r\choose r}\sum_{k=0}^{n-r}{n-r\choose k}=\sum_{k=0}^{n}\sum_{r=n-k}^{n}{n+r\choose r}{n-r\choose k}=2^{2n}.\]


Let $BE$ and $CF$ are two altitudes of a triangle $ABC$. Prove that $AB=AC$ iff $AB+BE=AC+CF$.

[hide="Solution"]
$AB\cdot CF= AC\cdot BE$ ($= 2\triangle ABC$)

$\Longleftrightarrow \frac{AB}{BE}= \frac{AC}{CF}$

Let $k=\frac{AB}{BE}= \frac{AC}{CF}$.

Then we have $AB=kBE$ and $AC=kCF$, so $AB+BE=kBE+BE=(k+1)BE$ and $AC+CF=kCF+CF=(k+1)CF$.

Thus $AB+BE=AC+CF\Longleftrightarrow (k+1)BE=(k+1)CF\Longleftrightarrow BE=CF$ ($k+1\ne 0$).

$kBE=kCF\Longleftrightarrow AB=AC$, and we are done.[/hide]


There are 12 persons sitting on a circular table. Each person shakes hand with just another one of the rest. How many ways can they shake without cross-arm? (any person must shake, and there's no one can shake with more than one person stimulatenously)

Really I have just found a solution, but it;s not well, and not general.

i don't see number theory in this problem.  :D

Isn't it just the ways to choose 2 people out of the total each time?
$\binom{12}{2}*\binom{10}{2}*\binom{8}{2}*\binom{6}{2}*\binom{4}{2}*\binom{2}{2}$ ways.

I don't think so:
when I understand correctly, when $A$ shakes with $B$ and $C$ shakes with $D$, then it's not allowed that they sit in the order $ACBD$ or similar in the circle.


Do you think the national competition should be at Detroit?

It should be in north carolina cuz i live there.

I just asked that 'cause I thought detroit seemed kinda boring.

The renaissance center is actually pretty nice.

I would rather it be in Washington D.C, but I think they pick the places to go by their sponsors.  Like GM is one of their sponsors, hence we're going to Detroit which is where their headquarters is.

It was in Washington D.C. 3 years ago.
It's been in Chicago for the past 2 years.

Actually, Nats was in DC last year, Chicago 2 years in a row before that, and DC every year before that.

Yeah, this is my first year going to nationals and I'm in eighth grade.

[quote="Secret Asian"]Yeah...Because my sister went to nationals twice once in Washington DC(2001) and once in Chicago(2002)[/quote]
wow, your sister went to nationals all three years? even in sixth grade? cool, she must be really smart

Im glad its in Detroit instead of DC because almost ALL national competitions are in DC, even MC most years.

I know one that isn't. The Reader's Digest Natioanl Word Power Challenge championship is in Orlando, at least for the 2 years I'm and I've gone. Before that it was in Williamsburg. I'm pretty sure it will stay in Orlando too.

Billy

Mobbed on the street by math-hating fanatics?  Beat 'em over the head with our calculators!

Or your wooden shirt! (Other forum joke)

Good idea.  While uncomfortable and subjects me to various bug attacks.  The defense/attack qualities of it are amost unsurpassable.  I'll just run into them.  If I run fast enough I can leave a "Secret Asian" imprint on their heads.

See if they mob you after that. . .
Also, Secret Asian should be imprinted backwards on your shirt so it's not a mirror image when you slam into them.  Then they'll be scarred for life.  Below Secret Asian, put Secret Asian in Binary.

No way!!  It should be in DC because I want to go to the capitol.  Anyhow, most kids want to make it to Nationals because it is usually in the capitol.  The capitol is the better spot, and I don't care that GM is a sponsor this year.

I don't think the competition should be in Detroit.  I'd rather go to DC because it was a trip I was going to take with my family and is just more appealing than Detroit is.  Plus, my mom would like it because it's warmer there.

[quote="mathgeek2006"]There are more of the interesting kinds of buildings in DC, and Detroit, well, what if we're mobbed on the street? (He-he) :D[/quote]Last time [b]I[/b] was in the DC with the team (in 2001 at the IMO) we got mobbed. We (the Romanian team) were walking with our guide (a Chigaco Univ. Junior student at that time) back from Lincon's Monument to the Potomac bay where a ship was supposed to wait for us to take a ride on the Potomac and let us watch nice fireworks (it was the 4th of July).

However at some point the guide didn't notice (well it was his first time in DC also :) ) that we are walking straight into an "unfriendly" neighborhood. Suddently some teenagers on bikes appear and start yelling at us (all besides me were wearing big math shirts) and a couple draw some guns. I got really really scared when they started shooting at us. It was not untill one of our team members was hit and then another one (ajorza on the site) that we realised that they were [hide="..."] paint bullets. Nevertheless we started running and they on their bikes after us. At some point they got bored with chasing us, and after ruining almost all our t-shirts we escaped :)[/hide]

[quote="09husbri"]That's kind of a funny story, even if it is, well, a bit disturbing. :)
But I don't see why you should be mobbed/painted when touring D.C. when everything to see there is in a pretty good neighborhood. Weird. :?[/quote]Being my (probably our would stand for everyone in the team at that time) first visit to US, when those teenagers drew out their guns (they looked and [b]sounded[/b] like real guns) we got really, but really scared :blush:. The guide wasn't from DC, it was his first time there, and he also did not bring a map with him, so we kinna got lost and ended up in that bad neighborhood  :?

[quote="AznChristmas"]I wish it were here in Nebraska. I hate flying.[/quote]
Well too bad, AznChristmas, because your flying to Nationals whether you like it or not.  Anyway, Nebraska is kinda boring.  (I know, I live in Lincoln.)  I wish it were still in DC though.

[quote="Valentin Vornicu"]Being my (probably our would stand for everyone in the team at that time) first visit to US, when those teenagers drew out their guns (they looked and [b]sounded[/b] like real guns) we got really, but really scared :blush:. The guide wasn't from DC, it was his first time there, and he also did not bring a map with him, so we kinna got lost and ended up in that bad neighborhood  :?[/quote]

 :) That's pretty funny - so did you get to see any of the good stuff there? My favorite place in DC is the FDR memorial - it's absolutely stunning.

I agree with whoever said that would be in Detroit since GM is a big sponsor, but I still think Detroit isn't as stimulating as DC or even Chicago.

how about we all go to
HAWAII!!!!!!!!
does hawaii have a MATHCOUNTS team???? how about alaska?

All 50 states and 7 territory/province-thingies of the US have MathCounts teams.

[quote="09husbri"]That's pretty funny - so did you get to see any of the good stuff there? My favorite place in DC is the FDR memorial - it's absolutely stunning.[/quote]It is funny now :) But wasn't funny at all then :D Anyway we pretty much visited everything imporant in DC. I didn't get to enter in the Congress building, nor in the FBI building since the waiting lines were toooo long. I however go to visit the Union Station :D :D


let a ,b,c,d be four naturel numbers st:ad-bc=1
show that there is no rational number between a/b and c/d whose denominator is less than b+d
show that there is one rational number whose denominator is b+d

Here is my solution:1/Supposing there exists m,n s.t [tex]gcd(m,n)=1[/tex] and :[tex]\frac{c}{d}<\frac{m}{n}<\frac{a}{b}[/tex]
Since [tex]ad-bc=1[/tex] we have: [tex]\frac{1}{bd}=(\frac{a}{b}-\frac{m}{n})+(\frac{m}{n}-\frac{c}{d})=\frac{an-bm}{bn}+\frac{md-nc}{nd}\geq \frac{1}{bn}+\frac{1}{nd}=\frac{b+d}{bdn}[/tex].Thus ,[tex]n\geq b+d[/tex].And we are done.
2/Obvios,[tex]\frac{a+c}{b+d}[/tex] satisfy the condition

Very good solution since the book ASKED FOUR QUESTIONS FOR THIS EXERCISE to reach the solution.


trigonometry
inequalities
geometry
circumcircle
geometry proposed

Prove that in any triangle $ ABC$ exists the identity $ \boxed {\ \left(2\cdot\sin\frac A2\minus{}1\right)^2\plus{}8\cdot\sin\frac A2\cdot\sin^2\frac {B\minus{}C}{4}\equal{}1\minus{}\frac {2r}{R}\ }$ .

$ 8\cdot\sin\frac A2\cdot\sin^2\frac {B - C}{4} = 4\cdot\sin\frac A2\left(1-\cos{\frac{B-C}{2}}\right)$ and so the identity become:

$ 2\sin\frac A2\cos{\frac{B-C}{2} - 2\sin^2\frac A2=\frac {r}{R}\ \Longleftrightarrow 2\sin\frac A2 \left(\cos{\frac{B-C}{2}-\sin\frac A2 \right) = \frac{r}{R} \ \Longleftrightarrow 4\sin\frac A2 \sin\frac B2 \sin\frac C2 = \frac{r}{R}}}$

wich is know

[quote="Virgil Nicula"]Prove that in any triangle $ ABC$ exists the identity $ \boxed {\ \left(2\cdot\sin\frac A2 \minus{} 1\right)^2 \plus{} 8\cdot\sin\frac A2\cdot\sin^2\frac {B \minus{} C}{4} \equal{} 1 \minus{} \frac {2r}{R}\ }$ .[/quote]
Here are some inequalities which can be obtained from this identity :

$ 1\blacktriangleright\ a\cdot\sin\frac A2\plus{}b\cdot\sin\frac B2\plus{}c\cdot\sin\frac C2\ge p$ , where $ 2p\equal{}a\plus{}b\plus{}c$ .

$ 2\blacktriangleright\ \frac pr\le \frac {a}{IA}\plus{}\frac {b}{IB}\plus{}\frac {c}{IC}\le\frac {p(3R\minus{}2r)}{2Rr}$ , where $ C(O,R)$ is circumcircle and $ C(I,r)$ is incircle of $ \triangle ABC$ .


Show that the arithmetic sum 1+2+3+4+....+n is a even square for infintie values n.
Reduce solution to suitable Pell Equation....

Big thanks in advance...

It's here: http://www.mathlinks.ro/Forum/viewtopic.php?t=62223
The even part comes from looking $\mod 4$ for $b$.


[quote="Davron"]Show that for all n element of Natural numbers A_n=5^n+2.3^n+1
are divisible by 8 n={1,2,3,...,n}.

Davron[/quote]
I don't understand(n=2k then $A_n \equiv 4(mad 8)$


Can 4-acetoxypyridine (pyridine with the group -OC(=O)CH3 attached in position 4) be prepared directly by esterification of 4-hydroxypyridine? If not, provide a synthesis of 4-acetoxypyridine from pyridine.

I think not bcos pyridine is an electron withdrawing molecule so it wont undergo esterification  :)
AS to the synthesis part i have no answer yet  :lol:

i think the reason for no estrification is the C-O bond is double bonded in character.
if we see the mech of esterification ROH---->R+
in this case since double bond character,so esterification will not take place.

Yes, that's right: the pyridone ressonance structure is somewhat more stable.


Find all sets of distinct integers 1 < a < b < c < d such that abcd−1 is exactly divisible by (a − 1)(b − 1)(c − 1)(d − 1).

Compare to IMO 1992/1. It's a pretty boring problem.


inequalities
integration
trigonometry
calculus
absolute value
real analysis
real analysis unsolved

Prove the inequality for any x positive:

$\int_{x}^{x+1}sint^2dt\leq \frac{1}{x}$

Integrate by parts.

$\int_x^{x+1}\sin t^2\,dt=\int_x^{x+1}\frac1{2t}\cdot2t\sin t^2\,dt$

$=\left.-\frac1{2t}\cos t^2\right|_x^{x+1}-\int_x^{x+1}\frac1{2t^2}\cos t^2\,dt$

Now estimate. In all cases, $|\cos t^2|\le 1.$ Using this, the absolute value of the integral is less than or equal to

$\frac1{2(x+1)}+\frac1{2x}+\int_x^{x+1}\frac1{2t^2}\,dt$

$=\frac1{2(x+1)}+\frac1{2x}+\frac1{2x}-\frac1{2(x+1)}=\frac1x.$

Hence we have proved that

$\left|\int_x^{x+1}\sin t^2\,dt\right|\le\frac1x.$

This is the estimate we would use for $x>1.$ For $0<x<1,$ we can simply say that the integral is $\le 1.$


In the xy-plane, what is the length of the shortest path from $(0, 0)$ to $(12, 16)$ that does not go inside the circle $(x - 6)^2 + (y - 8)^2 = 25$?

$\text{(A)}\ 10\sqrt 3 \qquad \text{(B)}\ 10\sqrt 5 \qquad \text{(C)}\ 10\sqrt 3 + \frac{ 5\pi}{3} \qquad \text{(D)}\ 40\frac{\sqrt{3}}3 \qquad \text{(E)}\ 10+5\pi$

[quote="Iversonfan2005"][hide="solution!!"] We have a circle centered at $(6,8)$ with radius of $5$ so we have two straight lines and a curve around the circle. Kind of hard to explain without a diagram, but the two straight lines are part of two $30,60,90$ triangles with base of $5$ which is the radius, so the two straight lines have length of $10\sqrt{3}$ and now we know that the part of the circle we want has $60^{\circ}$, so the circumference of the sector is $\frac{5\pi}{3}$, so the answer is $\boxed{10\sqrt{3}+\frac{5\pi}{3}}$
[/hide][/quote]Beat me to it.  :D

But is there any proof that the path is the shortest?
[quote=4everwise][hide="Solution"]Let $O=(0,0)$, $P=(6,8)$, and $Q=(12,16)$. Let the points of tangency from $Q$ and $O$ to circle $P$ be $M$ and $N$, respectively. The shortest path from $O$ to $Q$ consists of segments $ON$, $QM$, and minor arc $MN$.

By the distance formula, we know that $OP=10$. The radius of the circle is $5$. Thus, triangle $OPN$ is a $30^\circ, 60^\circ, 90^\circ$ right triangle. By the same reasoning, triangle $QPM$ is also a $30^\circ, 60^\circ, 90^\circ$ right triangle.

Thus, $ON=QM=5\sqrt{3}$. We slo know that $m\angle MPN=180^\circ-60^\circ-60^\circ=60^\circ$, because $O$, $P$, and $Q$ are collinear. Thus, minor arc $RT$ has length $\frac{5\pi}{3}$.

The answer is thus $10\sqrt{3}+\frac{5\pi}{3}$.[/hide][/quote]

The answer is [hide=yes]
Due to symetry we can look to half the path, say from $O$ to
the middle of the trip, with $OA=s$, where $A$ is the point
in the circle, where we change to the circular path till the
middle trip at $B$ equidistant from $O$ and $Q$.
Let $x=\angle APO$ so by law of cosines $s=5\sqrt{5-4\cos x}$
and the complete half path will be $s+5({{\pi}\over 2}-x)$
thus we can derive $y=\sqrt{5-4\cos x}+{{\pi}\over 2}-x$
getting $y'={{2\sin x}\over{\sqrt{5-4\cos x}}}-1$, negative
from $x=0$ to $x={{\pi}\over 3}$ where it becomes null;
but this is the point of tangency and the function doesn't
hold for farther inclination of $OA$. Anyhow, it is obvious
that this would result in an increase of the path.[/hide]


Sunny runs at a steady rate, and Moonbeam runs $m$ times as fast, where $m$ is a number greater than 1. If Moonbeam gives Sunny a head start of $h$ meters, how many meters must Moonbeam run to overtake Sunny?

$\text{(A)}\ hm \qquad \text{(B)}\ \frac{h}{h+m} \qquad \text{(C)}\ \frac{h}{m-1} \qquad \text{(D)}\ \frac{hm}{m-1} \qquad \text{(E)}\ \frac{h+m}{m-1}$

[quote="Silverfalcon"]Sunny runs at a steady rate, and Moonbeam runs $m$ times as fast, where $m$ is a number greater than 1. If Moonbeam gives Sunny a head start of $h$ meters, how many meters must Moonbeam run to overtake Sunny?

A. $hm$
B. $\frac{h}{h+m}$
C. $\frac{h}{m-1}$
D. $\frac{hm}{m-1}$
E. $\frac{h+m}{m-1}$[/quote]

[hide]
Let the rate at which Sunny is running be $r$ and the time be $t$.

$mr(t-\frac{h}{r})=rt$
$mrt-mh=rt$
$mrt-rt=mh$
$rt(m-1)=mh$
$rt=\frac{hm}{m-1}$
$d=rt$
$d=\frac{hm}{m-1}$

$\boxed{D}$
[/hide]

Edit: I think I forgot that moonbeam doesn't run during the head start and approached the problem incorrectly. :blush:  Fixed.

[hide="ccy,"]your solution is incorrect because we are seeking the distance Moonbeam runs, which is $mrt$, not $rt$.  So by multiplying the expression you got by $m$, you would get the correct answer.  (D)[/hide]

[quote="JesusFreak197"][hide="Yay for random numbers!"]Say $m=2$, $h=5$. The answer would then be $10$. However, both $A$ and $D$ give $10$. Looking at $A$, that obviously cannot be right for other values, so our answer is $\boxed{D}$. :lol:[/hide][/quote]

One word. LOL.  :rotfl:

I think the best way is to use "correct" method of what ccy did and check with Jesusfreak197's way. Jesusfreak197's way is quick but again, it's pretty risky.  :)

While being risky, it worked for me twice in one day  :D

I was also thinking what ccy wrote at first.  :blush:

[hide]
Problem here is I don't get why my solution works despite it 'working'. :P
So let's say Sunny runs $x$. Then Moonshine runs $x+h$. Sunny runs at a rate $s$ while Moon runs at a rate $m\times s=ms$.
We have the equation \[ms(x+h)=xs\] and solving we get that \[x(m-1)=hm\] \[x=\frac{hm}{m-1}\] This is $x$. But isn't that how much Sunny runs, not how long Moon needs to run to catch up? [/hide]

[hide]You should divide the distance by the speed (distance/time) to get the time taken: \[\frac{\text{distance}}{\text{speed}}=\frac{\text{distance}}{\text{distance/time}}=\text{time}\] the idea being that the time should be the same for both horses.

Your first equation should be $\frac{x+h}{ms}=\frac{x}{s}$, giving $x=\frac{h}{m-1}$. Then $x+h=\frac{h+h(m-1)}{m-1}$, etc.[/hide]


continuous function
algebra
four variables
IMO
IMO 1974

The variables $a,b,c,d,$ traverse, independently from each other, the set of positive real values. What are the values which the expression \[ S= \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d} \] takes?

Denote the given expression by $S(a, b, c, d)$. We have

\[ 1 = \sum_{\textrm{cyc}}\frac{a}{a+b+c+d} < S(a, b, c, d) < \frac{a}{a+b} + \frac{b}{a+b} + \frac{c}{c+d} + \frac{d}{c+d} =2.  \]

Also, $S(x,x,1,1) \rightarrow 1$ for $x\rightarrow \infty$ and $S(x,1,x,1) \rightarrow 2$ for $x\rightarrow \infty$.

Hence the range of $S$ is $]1,2[$.

If  $ a,b,c,d$ be positive real variables, let the expression

$ S(k) \equal{} \frac {a}{d\plus{}ka \plus{} b} \plus{} \frac {b}{a \plus{} kb \plus{} c} \plus{} \frac {c}{b \plus{} kc \plus{} d} \plus{} \frac {d}{ c \plus{} kd\plus{}a},$

then

$ S(k)>\frac{2}{k\plus{}1}$ holds if and only if $ 0\leq k\leq 1$;

$ S(k)<\frac{2}{k}$ holds if and only if $ 0\leq k\leq 2$.

[quote=Awesome_guy]Was on WOOT. I don't think previous solutions are rigorous enough...(one must show all numbers in the range work).

Note that
$$\sum_{cyc}\frac{a}{a+b+d}>\sum_{cyc}\frac{a}{a+b+c+d}=1$$
Note that $\lim_{(a, b, c, d)\to(\infty, 1, 0, 1)}\sum_{cyc}\frac{a}{a+b+d}=1$, thus the lower bound is $1$, exclusive.

Note that
$$\sum_{cyc}\frac{a}{a+b+d}<\frac{a}{a+b}+\frac{b}{b+a}+\frac{c}{c+d}+\frac{d}{d+c}=2.$$
Note that $\lim_{(a, b, c, d)\to(\infty, 1, \infty, 1)}\sum_{cyc}\frac{a}{a+b+d}=2$, thus the upper bound is $2$, exclusive.

Set $(a,b,d)=(\infty,1,1)$, and let $\sum_{cyc}\frac{a}{a+b+d}$ vary with $c$. Clearly our function in terms of $c$ is continuous, thus by the intermediate value theorem, all reals within $(1,2)$ can be attained. Thus $\sum_{cyc}\frac{a}{a+b+d}\in (1,2)$, as desired. $\blacksquare$[/quote]

I did not get your reasoning when you showed that 1 and 2 are not attained by the function. Could you clarify it, please?


AMC AIME USA(J)MO USAMO search function MATHCOUNTS

The best way to get better is to visit the forums often, try as many problems within your range as you can, and most importantly, keep track of the ones you can't do.  Come back to them later and see the solutions offered by the more advanced members, and learn from them.  Also, use the search function liberally if you see members referring to stuff you haven't heard of.  You'll be amazed how much you can learn in a month...

Also, the AoPS books are very helpful, and come with many challenging practice questions, and are a good start to advanced problem solving.

Just do as many AIME problems as you can.

When you can do 11-12+ on each AIME, take a look at USAMO questions.

[quote="The Sigma"]I am currently  a sophomore, and need some advice. I can answer upto 20 questions on the AMC 10. How does one prepare for the AIME, USAMO, and IMO. I mean, there has to be some way to crack the questions. If someone could give me some adivce, I would greatly appreciate it.[/quote]

This is the first year I've legitamately prepared for the AIMEs, as I'm a freshman. (I made them last year but didn't prepare at all and got a disastrous 4 :blush:.... but that's because I approached the problems like a Mathcounts test, and sort of got lost in most of the problems. In hindsight, I probably can get a 12 on that test now, both because of experience, and having learned a lot more math. AIME preparation goes a long way really.)

I would say to up your AMC score, and consequently get better at the AIMEs, definately practice the last 5 problems on the AMC10s/AMC12s, and then ease yourself into the AIMEs, starting with the first few (1-4), and eventually cracking the middle, and maybe end (this is where I could definately get better). The first few are definately approachable if you can attack some of the late AMC problems, and the middle just gets easier with practice. The end is quite difficult (here I'm referring to the last 3-5 problems), so don't get discouraged if you don't get them.

Then, on your weekends on breaks, sit yourself down occasionally for a full test (or half in the case of the AIMEs), and get yourself in the zone. Look at solutions, becuase they offer things to learn, and hint at what you should "look for", or so to speak, in AIME/AMC problems. And certianly spend time on these forums, because there's a wealth of knowledge here, and many (better I might add) people who all of us can learn from, from the greats like Mathew Crawford--

[size=75][color=green]SPELL HIS NAME CORRECTLY FOOL!!! don't worry, i fixed it for you.[/color][/size] :D

--to our peers on the forums. Eventually, you'll get better than most of your school, so this forum is useful to keep your level of math up. I'm hoping to make the USAMO this year taking the AIME II so I can't offer any advice there yet, but a great start is the AIMEs (I'm assuming you're taking the AIME II).

[quote="ShoeFactory"]When you can do 11-12+ on each AIME, take a look at USAMO questions.[/quote]

Actually, you should start looking at olympiad questions when you can solve 9-12 AIME, according to Mr. Crawford.


This question will be a cakewalk for most of the maths guys but still give it a try:-

According to one of the many Hindu mythologies, there are 3 towers and 64 disks(all of different sizes) are kept on one tower in such a manner that no disc rests above a disc of smaller size. The discs are to be removed one by one and transferred to another tower in such a way that no disc ever rests on another disc of bigger size. a [b]move[/b] is defined to be the removal of a disc from one tower and placing it one another tower.

Now the interesting part......

A Brahmin has started doing this work when the universe came into being and according to myth, the universe will end when he finishes doing the job. He takes 1 second to perform each [b]move[/b]. Assuming that the brahmin is very clever and will finish the task in the minimum possible time , find the lifetime of universe in years acc to this myth!!!

Well, this is the famous Edouard Lucas' "Tower of Hanoi" puzzle (he chose Hanoi for the exoticism of the name, not being acquainted with the Hindu legend).

For $ n$ disks, the minimal number of moves is $ 2^n \minus{} 1$. A famous recurrence relation $ h_{n} \equal{} 2h_{n\minus{}1} \plus{} 1$, with $ h_1 \equal{} 1$ yields this answer.

See [url]http://mathworld.wolfram.com/TowerofHanoi.html[/url]

also known as Tower of Brahma. [url=http://en.wikipedia.org/wiki/Tower_of_Hanoi]wikipedia link[/url] has a lot of detail on several solutions, a couple interesting applications include neuro-psychological testing of problem solving and/or frontal lobe deficits.


The game is Respond to someone without quoting them.

Rules:
The first three posts of this topic shall be as irrelevant to each other as possible. Thereafter:
Respond to any reply in the thread, but don't quote it! Respond normally, as if you did quote the post. Let's see how confusing this gets.

I hope we can get the troops out of Iraq soon.

People don't like staying there since it is such a dangerous hill.

Same-sex couples should be allowed to adopt children.

I don't see how Benny "Dangerous" Hill is relevant to this discussion!

They'd embarrass 'their' children while they grow up.

Hey whom do you think you are calling crazy?

Why is the hill dangerous anyway? Are there cliffs and rocks or something?

Sure, and they'd tell their grandchildren that they missed the biggest party in the world by sleeping through it.

I don't think he's referring to you. Relax.

Please don't use the word "whom" if you don't understand how to use it correctly.

Where do you learn how, though. In school ?

Oh, I see. I had better stay away from that hill then.

Do they tell them stories about how they were born ?

Purist eh?
Who do you think you are correcting?

Why? Then the whole world would come crashing down before us!

Probably not; they will find out sooner or later through some means anyway.

That crossed the line right there. :mad:

The lunch line is so long in middle school!

WHAT DO YOU MEAN YOU'RE ACTUALLY A PIKACHU

Cool upperclassmen with free blocks go [i]out[/i] for lunch.

People like that don't HAVE grandchildren

I'm not a pikachu. Stop yelling false things.

Iraq is certainly a bad place to go to war. Talking to yourself is fun by the way

I dare them to cross this line:
[u]                  START HERE               [/u]
                  END HERE

Some of them do, anyway. The rest stay at home.

There isn't a war there, I think. It just needs to get back onto it's feet.

Well don't dare [i]me[/i], or you'll make a fool of yourself.

oh hey i've gone out for lunch a couple times
even though i'm a freshman who should be in study hall :X

Iraqi's? In constipated penguin land?

-jorian

Bubka's right, the've crossed it a thousand times.


I have thought the following problem: I have given a positive integer $n_{10}$ in decimal system. Now convert it to binary number $n_{2}$. How can I calculate fast total number of ones of binary numbers $1,10,11,\cdots, n_2$?

I'd do it like this:

First convert $n_{10}$ to $n_2$ in logarithmic time. Let's say that $n_2=10001010$ - I'd calculate the number of ones from $1$ to $10000000-1$, then the number of ones from $10000000$ to $10001000-1$, then the number of ones from $10001000$ to $10001010-1$ and finally add 3 for $10001010$ itself.

The number of ones from $1$ to $10000000-1$ is exactly $\frac{1}{2}\times7\times2^7=k\times2^{k-1}$ if we're checking numbers $1$ to $2^k-1$.
The number of ones from $10000000$ to $10001000-1$ is $\frac{1}{2}\times3\times2^3+2^3$.
The number of ones from $10001000$ to $10001010-1$ is $\frac{1}{2}\times1\times2^1+2\times2^1$.

The number of ones from $101....0000....00$ to $101....0010....00-1$ would be $\frac{1}{2}\times k\times2^k+c\times2^k$ where $k$ is the number of zeros to the right of the digit we've changed and $c$ is the number of ones to the left of the digit we've changed.


Anyone else have trouble with the time limit of this problem?
For test case 6 it seems to me it would take more than a second just to read all of that data and process it into an array.  I've cut down the steps as far as I can, but I still can't get it to run in time.
Any suggestions?

Thanks

I don't think it takes more than a second to merely read the data.
Do you know dynamic programming? If not do a search on it and see if you can apply it to this problem.

If you're starting from the top to find the path, try starting from the bottom.  THat's what I did.

have an array to store values that have been computed already

Yea I ended up figuring out to start at the bottom and go up :D.  Thanks.


group theory
abstract algebra
function
linear algebra
matrix
LaTeX
binomial coefficients

I have searched the internet, the forum, and all my other books and cannot find anything about this.

Let G be the subgroup which permutes the sets {1,..,r} and {r+1,..,n} within themselves. Then each (distinct) left coset of G contains precisely one (n-r,r) shuffle. Let H denote the set of all such $ \left(\begin{array}{l}n\\ r\\ \end{array}\right)$ shuffles. Let K be a commutative ring with identity and let det be the determinant function on $ K^{nxn}$. If A is a matrix in $ K^{nxn}$,

Then $ \det A \equal{}\sum_{\sigma\in H}{({\mathop{\rm sgn}}\sigma )}\left|\begin{array}{l}A_{\sigma 1,1}...A_{\sigma r,r}\\ ...\\ A_{\sigma 1,1}...A_{\sigma r,r}\\ \end{array}\right|\left|\begin{array}{l}A_{\sigma 1,r\plus{}1}...A_{\sigma r,n}\\ ...\\ A_{\sigma 1,r\plus{}1}...A_{\sigma r,n}\\ \end{array}\right|$

(the vertical bars denote the determinant of the submatrix)

My book says other such Laplace expansions may be obtained by replacing the sets {1,...,r} and {r+1,...,n} by two different but complementary sets of indices. I understand the above expression but I'm not sure what complementary sets of indices are being referred to and how to prove equivalence. I am sure it is something simple. Also what are some of the uses of these expansions?

LaTeX hint: if you want to multiply, use \times.  $ n\times n$.  If you want binomial coefficients, use \binom{n}{r}.  $ \binom{n}{r}$.  Finally, in the process given, you split the set $ \{1,\ldots, n\}$ into the disjoint union of sets $ \{1,\ldots, r\}$ and $ \{r\plus{}1,\ldots, n\}$.  As for the content, I think the author is just pointing out that you could begin with any subset of $ \{1,\ldots, n\}$ in place of $ \{1,\ldots, r\}$ and the process would be identical, but the final result would be slightly different.

I'm guessing the difference will be a change of sign dependent on how the complementary subsets are chosen. How can we find the sign of each shuffle in terms of its elements?


Players 1, 2, 3, …, n are seated around a table and each has a single penny. Player 1 passes a penny to Player 2, who then passes two pennies to Player 3. Player 3 then passes one penny to Player 4, who passes two pennies to Player 5, and so on, players alternately passing one penny or two to the next player who still has some pennies. A player who runs out of pennies drops out of the game and leaves the table. Find an infinite set of numbers n for which some player ends up with all n pennies.

We will show that, if $n=2^{k}+1$, then the last player ends up with all  [b]n[/b] pennies

Let $0,1,2,..., 2^{k}, 2^{k}+1$ be the players

When someone runs out of pennies, we'll say that he is [u]dead[/u].

The coin of the player-[b]0[/b] will travel to all players. We'll call it the "gold coin"
The possesor of the gold coin is the person who has turn to play

[color=green]
[size=150][hide="Introduction"][/size]

[b][u]First move[/u]:[/b]
The player $0$ passes the gold coin.

After this move, we have $2^{k}$ players

[b][u]round 0[/u]:[/b]
The player $1$ passes his own coin and the gold coin
The player $2$ gets one coin and passes the gold coin
The player $3$ passes his own coin and the gold coin
The player $4$ gets one coin and passes the gold coin
.......................................................................
The player $2^{k}-1$ passes his own coin and the gold coin
The player $2^{k}$ gets one coin and passes the gold coin

After this procedure, the players $2,4,6,...,2^{k}$ have $2$ coins each.
The player $2^{k}$ has got also the gold coin.
The rest players are dead

The "alive" players are the positive multiples of $2$

[u][b]From this point we'll not count the "gold coin"[/b][/u]

[b][u]round 1.1[/u]:[/b]
The player $2$ gives one coin. $2-1=1$
The player $4$ gets one coin. $2+1=3$
.......................................................................
The player $2^{k}-2$ gives one coin. $2-1=1$
The player $2^{k}$ gets one coin. $2+1=3$

[b][u]round 1.2[/u]:[/b]
The player $2$ gives one coin. $1-1=0$
The player $4$ gets one coin. $3+1=4$
.......................................................................
The player $2^{k}-2$ gives one coin. $1-1=0$
The player $2^{k}$ gets one coin. $3+1=4$

The alive players are the positive multiples of $4$[/color]

[size=125][color=red]$\star$ We'll call "set 1" the set of {round 1.1, round 1.2}
Below, each set consists of some rounds. Only at the last round of each "set" some players die.[/color][/size]
[/hide]
$\hline$

[color=brown][size=150]Theorem[/size]
The set $m$ consists of  $2^{m}$ rounds.
At the begining of the round $(m,1)$ the alive players are all the multiples of $2^{m}$.
Each of them has got $2^{m}$ coins.
The last player $2^{k}$ has one coin more (the gold coin)
[/color]

[color=blue][size=150][hide="Proof"][/size]The theorem says that, in the end of the round $(m,2^{m})$ the alive players should be the multiples of $2^{m+1}$, with $2^{m+1}$ coins each of them.

Suppose that we start a set $m$, where the alive players are the multiples of $2^{m}$.
Then there are $\frac{2^{k}}{2^{m}}=2^{k-m}$ alive players

[b][u]round $(m,1)$[/u]:[/b]

The player $2^{m}$ gives on coin. Now he has $2^{m}-1$ coins
The player $2\cdot 2^{m}$ getss on coin. Now he has $2^{m}+1$ coins
.......................................................................
The player $(2a-1)\cdot 2^{m}$ gives on coin. Now he has $2^{m}-1$ coins
The player $(2a)\cdot 2^{m}$ getss on coin. Now he has $2^{m}+1$ coins

[b][u]round $(m,2)$[/u]:[/b]

The player $2^{m}$ gives on coin. Now he has $2^{m}-2$ coins
The player $2\cdot 2^{m}$ getss on coin. Now he has $2^{m}+2$ coins
.......................................................................
The player $(2a-1)\cdot 2^{m}$ gives on coin. Now he has $2^{m}-2$ coins
The player $(2a)\cdot 2^{m}$ getss on coin. Now he has $2^{m}+2$ coins

[b][u]round $(m,q)$[/u]:[/b]

The player $2^{m}$ gives on coin. Now he has $2^{m}-q$ coins
The player $2\cdot 2^{m}$ getss on coin. Now he has $2^{m}+q$ coins
.......................................................................
The player $(2a-1)\cdot 2^{m}$ gives on coin. Now he has $2^{m}-q$ coins
The player $(2a)\cdot 2^{m}$ getss on coin. Now he has $2^{m}+q$ coins

For $q=2^{m}\Rightarrow$ round $(m,2^{m})$
After this round, each of $(2a-1)\cdot 2^{m}$ is dead.
Each of $(2a)\cdot 2^{m}$ has $2^{m+1}$ coins

They are the players $a\cdot 2^{m+1}$. In other words, the multiples of $2^{m+1}$[/color][/hide]
$\hline$

Applying the theorem, we can see that if $n=2^{k}+1$ then the game will have end

[color=olive][size=150]Conjecture[/size]
Let $m=[\frac{n-1}{2}]$, where $[\ ]$ denotes the integer part
Let $p$ be the largest natural number such that $2^{p}\mid m$ and let $c=\frac{m}{2^{p}}$

Then, if $c=1$ then the game ends in a finite number of steps
else the game goes to a non-terminated loop with $c$ players
[/color]


Find all continuous functions $ f: (0,\infty) \rightarrow (0,\infty)$, such that for all positive reals $ x,y$ the following equation holds:

$ f(x)f(y) \equal{} f(xy) \plus{} f\left(\frac {x}{y}\right)$.

$ x\equal{}y\equal{}1\to f(1)\equal{}2$
$ y\equal{}x, x\equal{}y\to f(\frac xy )\equal{}f(\frac yx)$ or $ f(x)\equal{}f(x^{\minus{}1})$.
Let $ f(x)\equal{}x^a\plus{}x^{\minus{}a}, a\equal{}a(x)$, then $ f(x^2)\equal{}(x^a\plus{}x^{\minus{}a})^2\minus{}2\equal{}(x^2)^a\plus{}(x^2)^{\minus{}a}$. Therefore $ a(x^2)\equal{}a(x)$.
If $ a(x)\equal{}a(y)\equal{}a(x/y)$ then $ f(xy)\equal{}(x^a\plus{}x^{\minus{}a})(y^a\plus{}y^{\minus{}a})\minus{}x^ay^{\minus{}a}\minus{}x^{\minus{}a}y^a\equal{}(xy)^a\plus{}(xy)^{\minus{}a}$.
Therefore $ a(x^n)\equal{}a(x)$ and $ a(x^{m/n})\equal{}a(x^m)\equal{}a(x)$. By continiosly $ a(x)\equal{}const$.

[quote="Rust"]$ x \equal{} y \equal{} 1\to f(1) \equal{} 2$
$ y \equal{} x, x \equal{} y\to f(\frac xy ) \equal{} f(\frac yx)$ or $ f(x) \equal{} f(x^{ \minus{} 1})$.
Let $ f(x) \equal{} x^a \plus{} x^{ \minus{} a}, a \equal{} a(x)$, then $ f(x^2) \equal{} (x^a \plus{} x^{ \minus{} a})^2 \minus{} 2 \equal{} (x^2)^a \plus{} (x^2)^{ \minus{} a}$. Therefore $ a(x^2) \equal{} a(x)$.
If $ a(x) \equal{} a(y) \equal{} a(x/y)$ then $ f(xy) \equal{} (x^a \plus{} x^{ \minus{} a})(y^a \plus{} y^{ \minus{} a}) \minus{} x^ay^{ \minus{} a} \minus{} x^{ \minus{} a}y^a \equal{} (xy)^a \plus{} (xy)^{ \minus{} a}$.
Therefore $ a(x^n) \equal{} a(x)$ and $ a(x^{m/n}) \equal{} a(x^m) \equal{} a(x)$. By continiosly $ a(x) \equal{} const$.[/quote]
Why can you substitute $ f(x) \equal{} x^{a} \plus{} x^{ \minus{} a}$,even if $ a$ is a function?
Things you wrote seems to me very unclear and incorrect.

But then it immediately implies that $ f(x^2) \equal{} f(x)^2 \minus{} 2$, which can't be true for arbitrary function?

Sufficiently condition $ f: [1,\infty )\to (0,\infty)$ with $ f(x^2)\equal{}f(x)^2\minus{}f(1)$ (used only x=y).

[quote="Rust"][quote="Erken"]Why can you substitute $ f(x) \equal{} x^{a} \plus{} x^{ \minus{} a}$,even if $ a$ is a function?
Things you wrote seems to me very unclear and incorrect.[/quote]
For any $ f(x)$ we get $ f(x) \equal{} x^a \plus{} x^{ \minus{} a}$ were $ a \equal{} a(x) \equal{} \frac {ln(\frac {f(x) \plus{} \sqrt {f(x)^2 \minus{} 4}}{2})}{ln(x)}.$
If $ f(x) < 2$ a is complex and exist x, suth that $ f(x) < 0$.[/quote]
I dont agree the Rust's argument  when  $ f(x)<2$    :D So we have to proof that $ f(x) \geq 2$ .
$ f(x^{2^n})\equal{}f^2(x^{2^{n\minus{}1}})\minus{}2\geq 0 \Longrightarrow f(x^{2^{n\minus{}1}})\geq \sqrt 2 \Longrightarrow$$ f(x^{2^{n\minus{}2}})\geq \sqrt {2\plus{}\sqrt2}\Longrightarrow ... \Longrightarrow f(x)\geq \sqrt{2\plus{}\sqrt{2\plus{}\sqrt {2\plus{}..}}}, \forall n$
You make $ n\rightarrow \plus{}\infty$, so $ f(x)\geq 2$ because $ \sqrt{2\plus{}\sqrt{2\plus{}\sqrt {2\plus{}..}}}\rightarrow 2$ ,  And so on ... :cool:


For which $a$ is the solution set of the inequality $(x^{2}+ax+4)(x^{2}-5x+6)\leq 0$ an interval?

Too lazy to type this up, but

[hide]
We have
$(x^{2}+ax+4)(x-2)(x-3)\leq0$

Note that the values of $x^{2}+ax+4$ go from positive to negative to positive. The same is true with $(x-2)(x-3)$. Figure out/eliminate some cases.[/hide]


A circle is circumscribed about an equilateral triangle, and is inscribed in another equilateral triangle.

What is the ratio of the area of the outer triangle to the area of the inner triangle ?

Intuitively, the outer triangle is twice the area of the inner triangle. Could you provide a proof ?

[hide="Hint"]
Rotate the inner triangle.
[/hide]


calculus
integration
inequalities
complex numbers

A sequence consits of 4n complex numbers a_1, a_2, ..., a_{4n} which satisfy the following three properties:

(i) For each j, a_j = 1, -1, i, or -i, where i =Sqrt[-1].
(ii) For each value of k=1, ..., m, the real and imaginary parts of the sum a_1+a_2+...+a_k fall in the interval [0,3].
(iii) If 1<=j<k<4n, then \sum_{i=j}^k a_i \neq 0.

Find the maximum possible value for n for which such a sequence exists.

Well, if the sum (a_1 + a_2 + ... + a_k) equals the sum (a_1 + a_2 + ... + a_j), j < k, then the sum (a_(j + 1) + a_(j + 2) + ... + a_k) = 0.  So, each partial sum can occur at most once.  There are 16 possibilities for the partial sums with real and imaginary parts integral in [0, 3].  But we can't have a partial sum of 0.  So that gives us only 15 possibilities, so n is at most 3.  Now, we just need to show that we can do it with n = 3.

So, let's make this our sequence: 1, 1, 1, i, -1, -1, -1, i, 1, 1, 1, i.  It is evident that every desired property is satisfied.  All done.

Sorry, but I got n=4.

Here's my solution (what is incorrect?) This was from a math meet in January of a question that we never resolved. There is a strict inequality, k<4n.

http://68.38.137.202/dvml.pdf

Oh, if k < 4n, then it can be n = 4 because the last term of the sequence can bring you back to a previous sum.  Just add another -1, -1, -1, -i onto my sequence and it works fine.  But by the same reasoning, it can't be greater than 4.  (In fact, the sequence can't be of length greater than 16, even if it doesn't have to be a multiple of 4 in length.)


geometry
3D geometry
tetrahedron
geometry proposed

Show that for any tetrahedron, the condition that opposite edges have the same length is equivalent to the condition that the segment drawn between the midpoints of any two opposite edges is perpendicular to the two edges.

I don't know how to do these things.. can anyone post the solution?


Let
\[f = X^{n}+a_{n-1}X^{n-1}+\ldots+a_{1}X+a_{0}\]
be an integer polynomial of degree $n \geq 3$ such that $a_{k}+a_{n-k}$ is even for all $k \in \overline{1,n-1}$ and $a_{0}$ is even.
Suppose that $f = gh$, where $g,h$ are integer polynomials and $\deg g \leq \deg h$ and all the coefficients of $h$ are odd.
Prove that $f$ has an integer root.

You can prove g is linear by considering it in $\mathbb{Z}_{2}[x]$. The trick is that if h(x) divides f(x), then h(x) also divides $x^{n}f \left( \frac{1}{x}\right)$ (in $\mathbb{Z}_{2}[x]$).

Edit, ok... as perfect_radio requested, I will provide further details. For this solution we'll be working only in $\mathbb{Z}_{2}[x]$. We have $f(x) \equiv g(x)h(x)$ and since $h(x) \equiv h(1/x) \cdot x^{\deg h}$, we also have $h(x)g(1/x)x^{\deg g}\equiv f(1/x)x^{n}$. Notice that since $a_{k}+a_{n-k}\equiv 0$, we have $f(1/x)x^{n}+f(x) \equiv x^{n}+1$.

Now let $\deg h = d$, and we know $d \ge \frac{n}{2}$. By solving for the coefficients of g, we find that the first few are $g(x) = 1+x+x^{d+1}+x^{d+2}+\ldots$. However, if $\deg g \ge d+1$, then $\deg g+\deg h = 2d+1 > n$, so we must have $g(x) = 1+x$. So that pretty much shows that g is linear in $\mathbb{Z}[x]$, although we should probably also note that the leading coefficient on g must be odd (so that it doesn't vanish when reducing to $\mathbb{Z}_{2}[x]$).

[quote="probability1.01"]so we must have $ g(x) \equal{} 1 \plus{} x$. So that pretty much shows that g is linear in $ \mathbb{Z}[x]$, although we should probably also note that the leading coefficient on g must be odd (so that it doesn't vanish when reducing to $ \mathbb{Z}_{2}[x]$).[/quote]

Since we're working in $ \mathbb{Z}_{2}[x]$, don't you just know the coefficients are congruent to $ 1 \plus{} x (\mod 2)$?

Can anybody explain, or I'm misunderstanding something?  :maybe:

I think that's clear as the leading coefficient of g must be odd and divides 1 so it must no other way be 1.
However,solving the coefficients is really boring to me!!Could anyone kindly and have enough time to clarify it!  It remind me of failing to finish Problem 5 at the VMO 06!I wasnt able to end it!  :(

The cruelest solution is this:

In $ \mathbb Z \slash 2 \mathbb Z$ we have $ f \equal{} gh \equal{} \left( b_0 \plus{} b_1 x \plus{} \ldots \plus{} b_j x^j \right) \left( 1 \plus{} x \plus{} \ldots \plus{} x^k \right)$, where $ j \leq k$.

The coefficient of $ x^{j \minus{} 1}$ is $ a_{j \minus{} 1} \equal{} b_0 \plus{} \ldots \plus{} b_{j \minus{} 1}$ and the coefficient of $ x^{k \plus{} 1}$ is $ a_{k \plus{} 1} \equal{} b_1 \plus{} \ldots \plus{} b_j$. If $ j \minus{} 1 \geq 1$, then $ a_{j \minus{} 1} \equiv a_{k \plus{} 1} \left( \bmod \, 2 \right)$, so $ b_0 \equiv b_j \equiv 1 \left( \bmod \, 2 \right)$, contradiction with $ g(0) \equal{} 0$. Thus, $ j \equal{} 1$ and that's it.


let a,b,c>0. Prove that: $\sum\limits_{cycle}^{}(a+b)^4  \geq$ $\frac{4}{7}\sum\limits_{cycle}^{} a^4 $ :rotfl:  :D

this inequality is surely not at pre olympiad level! The problem should be true for all real a,b,c.

I don't know it at which level. But it's too hard to me!!! I hope everybody help me find solution for this problem!!!

Are you sure it's $\frac{4}{7}$ and not $\frac{7}{4}$?  We know that with a, b, c>0, the left side must be greater than the right side...