trigonometry
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relatively prime

Let $ p_{n}$ denote the $ nth$ prime number and let $ \alpha\in [0,2\pi)$.

For which values of $ \alpha$ is the following series

$ \sum\limits_{n\equal{}1}^{\infty}\frac{\cos (p_{n}\alpha)}{p_{n}}$

 convergent?

There's at least one $ \alpha$ for which it diverges, namely $ \alpha\equal{}0.$

There's at least one $ \alpha$ for which it converges. The cheapest example I can think of is $ \alpha\equal{}\frac{\pi}2,$ for which all but one term of the sum is zero.

Two points settled, uncountably many to go.

I believe it should be possible to show that the series converges at every rational multiple of $ \pi$ using the stronger statement of Dirichlet's theorem (primes are equidistributed $ \bmod n$) and something like monotone or dominated convergence, although I'm not familiar with such arguments.

This problem is hard. I suggest solving this first:

Find all values of $ \alpha \in [0,2\pi)$ such that the series $ \sum_{n\equal{}1}^{\infty} {\frac{\cos(n \alpha)}{n}}$ is convergent.

(We ignore the "prime" condition).

My bad! I cannot come up with the solution for this new problem also.

mathangel: that's fairly easy, and the answer is that that series converges for all $ \alpha\in(0,2\pi).$ (In other words, it diverges only for $ \alpha\equal{}0.$)

The key is a summation by parts argument, summing the cosines and differencing the $ \frac1n.$

Here's a more general theorem: If $ b_n$ is (for sufficiently large $ n$) a positive, monotone nonincreasing sequence such that $ \lim b_n\equal{}0,$ then the power series $ \sum_{n\equal{}0}^{\infty}b_nz^n$ converges for all $ z$ such that $ |z|\equal{}1,z\ne1.$

That theorem includes the alternating series test as a special case (let $ z\equal{}\minus{}1.$) It also includes mathangel's example, by letting $ b_n\equal{}\frac1n$ and taking the real part.

No, didilica's problem is far different than (and presumably more difficult than) that problem. It seems likely to involve some nontrivial aspects of number theory.

[quote="t0rajir0u"]I believe it should be possible to show that the series converges at every rational multiple of $ \pi$[/quote]It certainly doesn't converge for $ \alpha \equal{} \frac {k\pi}{3}$ for any integer $ k$, right?  What kind of argument did you have in mind?

I actually kind of have the opposite feeling - that this sum ought to converge for 'most' [i]irrational[/i] multiples of $ \pi$.  When $ \alpha$ is rational the set $ \{\cos \left(p_n \alpha \pi \right): n\in \mathbb{N}\}$ is finite, while when $ \alpha$ is irrational the same set is dense in $ (0,1)$ and looks a lot like samples of $ \cos X$ where $ X$ is a random variable uniformly distributed in $ (0,2\pi)$.  Intuitively this would suggest that when $ \alpha$ is irrational the noise tends to cancel out, in the same way that limit theorems hold for i.i.d. collections of random variables.  Of course, this kind of reasoning is utterly fuzzy, and will fail for certain carefully chosen $ \alpha$.

The series converges in $ L^2$, so we expect it to converge almost everywhere. The same could be said of any trigonometric series with $ O(n^{ \minus{} 1})$ coefficients. The hard part is nailing down any specific points, because we don't have a nice function to match it to.

[quote="Kent Merryfield"]mathangel: that's fairly easy, and the answer is that that series converges for all $ \alpha\in(0,2\pi).$ (In other words, it diverges only for $ \alpha \equal{} 0.$)

The key is a summation by parts argument, summing the cosines and differencing the $ \frac1n.$

Here's a more general theorem: If $ b_n$ is (for sufficiently large $ n$) a positive, monotone nonincreasing sequence such that $ \lim b_n \equal{} 0,$ then the power series $ \sum_{n \equal{} 0}^{\infty}b_nz^n$ converges for all $ z$ such that $ |z| \equal{} 1,z\ne1.$
[/quote]
Sir, I couldn't get your proof for mathangel's sum as well as the proof of the theorem mentioned. Could you elaborate further please or provide a suitable link?

Following t0r's suggestion, I will try to show that the series converges when $ \frac{\alpha}{\pi}$ is a rational number with even denominator.  Specifically, suppose that $ \alpha \equal{} \frac{a}{b} \pi$ where $ b$ is even and $ a$ is relatively prime to $ b$.

Let $ C(m) \equal{} \sum_{n \equal{} 1}^m \cos(p_n \alpha)$.  By summation of parts, it suffices to show that $ \sum_{m \equal{} 1}^\infty C(m) \left( \frac{1}{p_m} \minus{} \frac{1}{p_{m \plus{} 1}} \right)$ converges.  We will show that this sum converges absolutely.

First, we need an estimate on $ C(m)$.  Note that a cosine term corresponding to $ p_n \equiv x \mod 2b$ cancels out with a cosine term corresponding to $ p_n \equiv (x \plus{} a^{\minus{}1}b) \mod 2b$.  According to results about primes in arithmetic progression, there are about the same number of primes in each residue class $ x \mod 2b$, where $ x$ is relatively prime to $ 2b$.  Specifically, according to the [url=http://en.wikipedia.org/wiki/Siegel-Walfisz_theorem]Siegel-Walfisz theorem[/url], the difference between the numbers is $ O\left(\frac{m}{\log m}\right)$.  So $ \left|C(m)\right|$ is $ O\left(\frac{m}{\log m}\right)$.

So we have to show that $ \sum_{m \equal{} 2}^\infty \frac{m}{\log m} \left( \frac{1}{p_m} \minus{} \frac{1}{p_{m\plus{}1}} \right)$ converges.  Using summation in parts in reverse, it suffices to show that $ \sum_{n \equal{} 2}^\infty \left( \frac{n \plus{} 1}{\log(n \plus{} 1)} \minus{} \frac{n}{\log n} \right) \frac{1}{p_n}$ converges.  But that follows from $ p_n \equal{} \Omega(n \log n)$.  So we are done.

Indeed, as Ravi B observed, the convergence at $ \frac ab 2\pi$ is governed by the number $ \sum_{x: (x,b) \equal{} 1}e^{2\pi i\frac xb}$. If the real part of this sum is $ 0$, we have convergence; otherwise, we have unboundedness. Note that this sum is a multiplicative function of $ b$ that equals $ \minus{} 1$ at all primes and $ 0$ at all higher powers of primes, so it is always real and not zero if and only if $ b$ is squarefree (Ravi did the case when $ 4|b$). Now we know that the set of rationals where the series is unbounded is dense. This implies unboundedness at some specially chosen irrationals (those that can be approximated by the "divergent rationals" sufficiently well), so we have a divergence set of cardinality continuum. Simple $ L^2$ estimates allow one to show convergence a.e., as jmerry noticed. This makes the problem of the full description seem transcendentally hard. I do not even know if we can choose $ \alpha$ such that the partial sums are bounded but the series does not converge :(.


function
modular arithmetic
induction
algebra proposed
algebra

Find all functions $ f: \mathbb r N \rightarrow \mathbb r N$ such that for all positive integers $ m$ and $ n$
following equation holds:

$ f(f(m+n)+f(m-n))=8m$

May be equation holds for all $ m,n\geq 0$?

$ f(m) \equal{} f(n)\Rightarrow 8m \equal{} f(2f(m)) \equal{} f(2f(n)) \equal{} 8n\Rightarrow m \equal{} n$
Now we have
$ f(f(m \plus{} n) \plus{} f(m \minus{} n)) \equal{} 8m \equal{} f(f(m \plus{} n') \plus{} f(m \minus{} n'))$
$ \Rightarrow f(m \plus{} n) \plus{} f(m \minus{} n) \equal{} f(m \plus{} n') \plus{} f(m \minus{} n')$
$ \Rightarrow f(2m \plus{} 1) \equal{} f(2m) \plus{} f(2) \minus{} f(1)$ and $ f(2m \plus{} 2) \equal{} f(2m \plus{} 1) \plus{} f(1) \minus{} f(0)$
From two this conditions we have $ f(n \plus{} 1) \equal{} f(n) \plus{} a$, $ a \equal{} f(2) \minus{} f(1) \equal{} f(1) \minus{} f(0)$
Then $ f(n) \equal{} an \plus{} b$, $ b \equal{} f(0)$
Now our equation become $ 2a^2m \plus{} 2ab \plus{} b \equal{} 8m\Leftrightarrow a \equal{} 2,b \equal{} 0\Leftrightarrow f(n) \equal{} 2n$

well, the solution is the same even if you assume $ m,n>0$  :wink:

[quote="Albanian Eagle"]well, the solution is the same even if you assume $ m,n > 0$  :wink:[/quote]

If $ m,n > 0$ then we must use a bit different way (not hard) to prove that $ f(m) \equal{} f(n)\Leftrightarrow m \equal{} n$ because in this we used $ 8m \equal{} f(f(m \plus{} 0) \plus{} f(m \minus{} 0))$
But a more important thing is may be $ f(2) \minus{} f(1)\neq f(1) \minus{} f(0)$ because we proved it by $ f(1 \plus{} 1) \plus{} f(1 \minus{} 1) \equal{} f(1 \plus{} 0) \plus{} f(1 \minus{} 0))$

I found it hard to complete  :oops: , but it's basically all your idea, tdl.   :)

$ f: \mathbb{N}\to\mathbb{N}$ satisfies
$ (*)$ $ f(f(m) \plus{} f(n)) \equal{} 4m \plus{} 4n$ for all $ m,n\in\mathbb{N}$ with $ 0\not \equal{} m \minus{} n$ even.

Then
$ 1)$
$ f(m) \equal{} f(n)\Rightarrow m \equal{} n$ for all $ m,n\in\mathbb{N}$ with $ m \minus{} n$ even.
PF:
Take $ k > m,n$ with $ k\equiv m\equiv n\pmod{2}$.
Then $ 4k \plus{} 4m \equal{} f(f(k) \plus{} f(m)) \equal{} f(f(k) \plus{} f(n)) \equal{} 4k \plus{} 4n\Rightarrow m \equal{} n$.

$ 2)$
$ f(n)\to \infty$ for $ n\to \infty$.
PF:
Otherwise we find $ N\in\mathbb{N}$ with $ f(n)\le N$ for infinitely many $ n\in\mathbb{N}$.
We then find $ 3$ different $ k,m,n\in\mathbb{N}$ with $ f(k) \equal{} f(m) \equal{} f(n)\le N$,
and $ 1)$ gives the contradiction $ k \minus{} m$ and $ m \minus{} n$ and $ n \minus{} k$ odd.

$ 3)$
We find some $ N\in\mathbb{N}$ with $ f(m) \equal{} f(n)\Rightarrow m \equal{} n$ for all $ m,n\in\mathbb{N},m > N$.
PF:
We need to show, that the set $ S: \equal{} \{m\in\mathbb{N}: f(m) \equal{} f(n)\text{ for some }n\in\mathbb{N},n\not \equal{} m\}$ is finite.
Assume $ S$ contains $ 3$ different odd elements $ m_1,m_2,m_3$
and take $ n_i\in\mathbb{N},n_i\not \equal{} m_i$ with $ f(m_i) \equal{} f(n_i)$ for $ i\in\{1,2,3\}$.
Then using $ 1)$ we have $ n_1,n_2,n_3$ even with
$ f(n_i) \equal{} f(m_i)\not \equal{} f(m_j) \equal{} f(n_j)\Rightarrow n_i\not \equal{} n_j$ for all $ i,j\in\{1,2,3\},i\not \equal{} j$.
And now using $ (*)$ we get
$ 4m_i \plus{} 4m_j \equal{} f(f(m_i) \plus{} f(m_j)) \equal{} f(f(n_i) \plus{} f(n_j)) \equal{} 4n_i \plus{} 4n_j$
$ \Rightarrow m_i \minus{} n_i \equal{} \minus{} (m_j \minus{} n_j)$ for all $ i,j\in\{1,2,3\},i\not \equal{} j$.
This gives $ m_1 \minus{} n_1 \equal{} \minus{} (m_2 \minus{} n_2)$ and $ m_2 \minus{} n_2 \equal{} \minus{} (m_3 \minus{} n_3)$ and $ m_3 \minus{} n_3 \equal{} \minus{} (m_1 \minus{} n_1)$
and therefore $ \sum_{i \equal{} 1,2,3}(m_i \minus{} n_i) \equal{} \minus{} \sum_{i \equal{} 1,2,3}(m_i \minus{} n_i)$
and $ 0 \equal{} \sum_{i \equal{} 1,2,3}(m_i \minus{} n_i)$ is odd. Contradiction!
So $ S$ contains at most $ 2$ odd elements and analog at most $ 2$ even elements.

$ 4)$
We find some even $ L\in\mathbb{N}$ with $ f(m) \plus{} f(n) \equal{} f(k) \plus{} f(l)$
for all $ m,n,k,l\in\mathbb{N}$ with $ L < m \plus{} n \equal{} k \plus{} l$ even.
PF:
Using $ 2)$ we find $ M\in\mathbb{N},M\ge 2$ with $ f(m) > N$ for all $ m > M$.
Then for all $ m,n,k,l\in\mathbb{N}$ with $ m\not \equal{} n,k\not \equal{} l$ and $ m \plus{} n \equal{} k \plus{} l$ even
we have
$ 2M < m \plus{} n$
$ \Rightarrow M < m\text{ or } M < n$
$ \Rightarrow N < f(m)\text{ or } N < f(n)$
$ \Rightarrow N < f(m) \plus{} f(n)$.
Then $ (*)$ gives
$ f(f(m) \plus{} f(n)) \equal{} 4m \plus{} 4n \equal{} 4k \plus{} 4l \equal{} f(f(k) \plus{} f(l))$
and using $ 3)$ we get $ f(m) \plus{} f(n) \equal{} f(k) \plus{} f(l)$
for all $ m,n,k,l\in\mathbb{N}$ with $ m\not \equal{} n,k\not \equal{} l$ and $ 2M < m \plus{} n \equal{} k \plus{} l$ even.

So we can complete $ 4)$ with $ L: \equal{} 2M \plus{} 2$ just by showing
$ 2f(m) \equal{} f(m \minus{} 1) \plus{} f(m \plus{} 1)$ for all $ m > M \plus{} 1$.
But now we already have
$ f(m) \plus{} f(m \plus{} 2) \equal{} f(m \minus{} 1) \plus{} f(m \plus{} 3)$
$ f(m) \plus{} f(m \minus{} 2) \equal{} f(m \plus{} 1) \plus{} f(m \minus{} 3)$
$ f(m \minus{} 2) \plus{} f(m \plus{} 2) \equal{} f(m \minus{} 3) \plus{} f(m \plus{} 3)$
for all $ m > M \plus{} 1\ge 3$, which gives $ 2f(m) \equal{} f(m \minus{} 1) \plus{} f(m \plus{} 1)$ immediately.

$ 5)$
We find some $ a\in\mathbb{Z}$ with $ f(n) \equal{} f(n \minus{} 1) \plus{} a$ for all $ n\in\mathbb{N},n > 1$.
PF:
$ 4)$ gives $ 2f(L) \equal{} f(L \minus{} 1) \plus{} f(L \plus{} 1)$, and we take $ a: \equal{} f(L \plus{} 1) \minus{} f(L) \equal{} f(L) \minus{} f(L \minus{} 1)$.
Again using $ 4)$ we get for $ n$ even
$ f(n) \plus{} f(L) \equal{} f(n \minus{} 1) \plus{} f(L \plus{} 1)\Rightarrow f(n) \minus{} f(n \minus{} 1) \equal{} f(L \plus{} 1) \minus{} f(L) \equal{} a$
and for $ n$ odd
$ f(n) \plus{} f(L \minus{} 1) \equal{} f(n \minus{} 1) \plus{} f(L)\Rightarrow f(n) \minus{} f(n \minus{} 1) \equal{} f(L) \minus{} f(L \minus{} 1) \equal{} a$.

And finally for $ b: \equal{} f(1) \minus{} a$ we get $ f(n) \equal{} an \plus{} b$ for all $ n\in\mathbb{N}$ by induction
and then $ (*)$ gives $ (a,b) \equal{} (2,0)$.


limit
real analysis
real analysis unsolved

[quote="hien"]Calculate:
$\lim_{n\to\infty}\frac{1+2^{2}+3^{3}+...+n^{n}}{n^{n}}$[/quote]
answer is 1.
Sulution: there is many way to solve it  but I choos this way:
$\sum_{1\leq k\leq n}k^{k}$~$n^{n}$ $\Rightarrow $ $\lim\frac{\sum_{1\leq k\leq n}k^{k}}{n^{n}}=1$


Trouble with Sons: A woman had two sons who were born on the same hour of the same day of the same year. But they were not twins. How could this be so?

They were triplets (or quadruplets, or quintuplets, etc.).


If $ x\equal{}2$ and $ x\equal{}\minus{}3$ are the only real roots of some 4th degree polynomial, which of the following CANNOT possibly be that polynomial?
$ A: x^4\plus{}(k \plus{} 2)x^3\plus{}(k^2\plus{}k\minus{}5)x^2\plus{}(k^2\minus{}6k\minus{}6)x\minus{}6k^2$
$ B: x^4\plus{}(2k\plus{}3)x^3\plus{}(4k\minus{}4)x^2\minus{}(10k\plus{}12)x\minus{}12k$
$ C: 2x^4\plus{}(k\plus{}3)x^3\plus{}(2k\minus{}10)x^2\minus{}5(k\plus{}1)x\minus{}6(k\plus{}1)$
$ D: 2x^4\plus{}(k\plus{}3)x^3\plus{}(k\minus{}9)x^2\minus{}(6k \plus{} 4)x\minus{}12$
$ E: \text{none of the above}$

The answer is B but i don't know how

[quote="funmath"]If $ x \equal{} 2$ and $ x \equal{} \minus{} 3$ are the only real roots of some 4th degree polynomial, which of the following CANNOT possibly be that polynomial?
$ A: x^4 \plus{} (k \plus{} 2)x^3 \plus{} (k^2 \plus{} k \minus{} 5)x^2 \plus{} (k^2 \minus{} 6k \minus{} 6)x \minus{} 6k^2$
$ B: x^4 \plus{} (2k \plus{} 3)x^3 \plus{} (4k \minus{} 4)x^2 \minus{} (10k \plus{} 12)x \minus{} 12k$
$ C: 2x^4 \plus{} (k \plus{} 3)x^3 \plus{} (2k \minus{} 10)x^2 \minus{} 5(k \plus{} 1)x \minus{} 6(k \plus{} 1)$
$ D: 2x^4 \plus{} (k \plus{} 3)x^3 \plus{} (k \minus{} 9)x^2 \minus{} (6k \plus{} 4)x \minus{} 12$
$ E: \text{none of the above}$

The answer is B but i don't know how[/quote]
Plug in x=2 into the second equation, and we get $ 16\plus{}16k\plus{}24\plus{}16k\minus{}16\minus{}20k\minus{}24\minus{}12k\equal{}\minus{}16$, but no matter what k is this can't possibly be 0.


function
trigonometry
algebra
domain
trig identities

What is the minimum value of the function

$f(x)=\frac{\sin{x}}{\sqrt{1-\cos^2{x}}}+\frac{\cos{x}}{\sqrt{1-\sin^2{x}}}+\frac{\tan{x}}{\sqrt{\sec^2{x}-1}}+\frac{\cot{x}}{\sqrt{\csc^2{x}-1}}$

as $x$ varies over all numbers in the largest possible domain of $f$, is

(a) -4
(b) -2
(c) 0
(d) 2
(e) 4

What does that last statement mean, and how do you solve general "minimum" type problems (or maximum)?

[hide]$\sqrt{1-\cos^2 x}=|\sin x|$, $\sqrt{1-\sin^2 x}=|\cos x|$, $\sqrt{\sec^2 x-1}=|\tan x|$, $\sqrt{\csc^2 x-1}=|\cot x|$

$\Rightarrow f(x)=\frac{\sin x}{|\sin x|}+\frac{\cos x}{|\cos x|}+\frac{\tan x}{|\tan x}+\frac{\cot x}{|\cot x|}$.

Each of these terms is 1 or -1 depending on whether the numerator is + or -. Not all 4 numerators can be negative, for $\sin x<0$, $\cos x<0\Rightarrow \tan x>0$.

But it is possible to get 3 to be negative. At an angle of $91^\circ$, the first fraction will be 1, but the other three will be -1, so the sum is -2.

Then $\min f=-2\Rightarrow \boxed{B}$.[/hide]

[hide="Answer"]Using a few trig identities, the equation simplifies to $\frac{\sin x}{|\sin x|}+\frac{\cos x}{|\cos x|}+\frac{\tan x}{|\tan x|}+\frac{|\cos x|}{\cos x}$. If both $\sin x$ and $\cos x$ are negative, then $\tan x$ is positive. If $\tan x$ is negative, then one of the others is positive. Therefore, the lowest possible value the funciton can take is $-2\Rightarrow \boxed{B}$.[/hide]

[quote="JesusFreak197"][hide="Answer"]Using a few trig identities, the equation simplifies to $1+1+1+1=4\Rightarrow \boxed{E}$.[/hide][/quote]

That's what I thought at first, but the square root can be negative.  ;)

look above your post.  :P

Answered too quickly. Be a little more patient and you'll see me fix it, since I always check other's answers when I'm done with mine. ;)


In a plane, 10 lines intersect such that no 3 lines meet at the same point. What is the maximum number of regions created by the lines?

THe first line creates 1 new region, the second makes 2, the third makes 3, etc.  Then we have 1+2+3+...+10=55. but add one b/c we started wiht 1, so we have 55+1=56.

what we know that for every one that makes a line there will be 2 more so therefore we get the equation

1+2+3...+55 = using the gauss formula we can use and get the equation

so we get the answer of 56

[quote="JayZz428xX"]1+2+3, and 55.
So the answer would be 56.[/quote]

I don't get that. Exactly what do you mean by that?

Look, at Markoboy's post He explained it better than I did :D.

it's recursively defined that $f(n) = f(n-1)+n$, where $f(n)$ is the maximum number of regions $n$ lines can divide a plane into.

It can be derived that $f(n) = \frac{n(n+1)}2+C$ where $C$ is a constant. plugging in $f(0)=1$ or $f(1)=2$, we find $C=1$

the function is therefor $f(n)= \frac{n(n-1)}2+1$

it's easy to evaluate $f(10)=10\cdot11\cdot\frac12+1=55+1=56$


calculus
integration
function
trigonometry
calculus computations

I'm in a class in which Fourier analysis rears its ugly head every now and then (a physics class). What is a good way to deal with integrals in which you don't have the beloved orthogonal functions making everything zero. Such as

$ \int_0^L x\sin\left(\frac{m\pi x}{L}\right)\cos\left(\frac{n\pi x}{L}\right)dx$

I know I can use $ u\equal{}x$ and $ dv\equal{}\text{other stuff}dx$ and integrate by parts with the aid of an integral table, but it seems like there should be a nicer way consider how frequently these come up.

Fourier analysis is not ugly  :mad: and neither is integration by part. Just use a [url=http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities]product-to-sum identity[/url] before integration.

Thanks...I completely forgot about that.

I didn't mean to call it ugly. I was speaking out of anger.  :lol:

I'm taking this class because I wanted to learn some interesting new concepts where mathematics is used. Instead, it is eating all of my time by performing long integrations, and I'm not really learning concepts. They should rename the class to: Learn to Evaluate Definite Integrals that Appear in Physics Classes.

Use product to sum as mlok suggested. And then use integration by parts.

The integral is $ 0$ when $ L \equal{} \pi$
Otherwise I got $ \frac {L^{2}}{2}( \minus{} 1)^{m \plus{} 1}[\frac {1}{(m \minus{} n)( \minus{} 1)^{n}} \minus{} \frac {1}{m \plus{} n}]$
I might have made some errors, but I do not have time to double check it.


geometry
incenter
trigonometry
analytic geometry
cyclic quadrilateral
projective geometry
geometry proposed

:?: [b]P1.[/b] Let $\triangle ABC$, the midlepoint [b]M[/b] of the side BC, the orthocentre [b]H[/b] what is projected on the sides of the triangle in the points $D\in BC,\ E\in CA,\ F\in AB$ and intersections $L\in BC\cap EF,\ S\in AM\cap LH$.
Prove that: $AM\perp LH;\ \widehat{ASB}\equiv\widehat{ASC}\Longleftrightarrow AB=AC$.

[b]P2.[/b] Let $\triangle ABC$, the incentre [b]I[/b] what is projected on the sides of the triangle in the points $X\in BC,\ Y\in CA,\ Z\in AB$ and the intersections $T\in BC\cap YZ,\ V\in TI\cap AX$. Prove that $TI\perp AX$ and $\widehat{AVB}\equiv\widehat{AVC}$.

[b]P3.[/b] Let $\triangle ABC$, the circumcentre [b]O[/b], the orthocentre [b]H[/b], the intersection $D\in BC\cap AH$, the symmetrical point E of the point C with respect to the point D $\ [D\in (EC) ,\ DE=DC]$ and the intersection $F\in AB\cap HE$. Prove that $DF\perp DO$.

[b]P4.[/b] Let $\triangle ABC,\ AB\ne AC$, the centroid $G$ and the point $D\in BC,\ GD\perp BC$. Ascertain the positions of the points $X\in AB,\ Y\in AC$ for which $G\in XY$ and $\widehat{GDX}\equiv\widehat{GDY}$. :!:

To the first one, you can have a  following generalization
   If $ABCD$ is a cyclic quadrilateral and $M, N, P$ are the intersections of $AB$ and $CD$ , $BC$ and $AD$, $AD$ and $BC$. Then $O$ is the orthocenter of $MNP$

Anyway, here is the proof to the second problem
   We notice that $A$ and $X$ lie on the polar of $T$ wrt $(I)$. This implies that $TI$ is perpendicular to $AX$. Also, we have $(TXBC) = -1$ and $\measuredangle{TVX} = 90^0$. Hence $VX$ is the bisector of $\measuredangle{BLC}$, and the second question follows.

\[\underline{\overline{|<>Lemma<>.|}}\] Let the line [b]AB[/b] and the points $M,N\in AB$, the middlepoints [b]S,T[/b] for the segments [b](AB), (MN)[/b] respectively. Prove that the following assertions are equivalently:

[b]A1.[/b] There is $m\ne 1$, such that $\overline{MA}=m\cdot \overline{MB},\ \overline{NA}=-m\cdot \overline{NB}$, i.e. [b](A,B,M,N)[/b] is a harmonious division;

[b]A2.[/b] $AB^2+MN^2=4\cdot ST^2$;

[b]A3.[/b] $TM^2=\overline{TA}\cdot \overline{TB}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( SA^2=\overline{SM}\cdot \overline{SN}\right) ;$

[b]A4.[/b] $\overline{BM}\cdot \overline{BN}=\overline{BA}\cdot \overline{BT}\ \ \ \ \ \ \ \ \ \left( \overline{AM}\cdot \overline{AN}=\overline{AB}\cdot \overline{AT}\right);$

[b]A5.[/b] $\overline{TA}\cdot \overline{TB}+\overline{TM}\cdot \overline{TN}=0\ \ \ \ \ \left( \overline{SM}\cdot \overline{SN}+\overline{SA}\cdot \overline{SB}=0\right);$

[b]A6.[/b] $\overline{NA}\cdot \overline{NB}=\overline{NM}\cdot \overline{NS}\ \ \ \ \ \ \ \ \left( \overline{MA}\cdot \overline{MB}=\overline{MN}\cdot \overline{MS}\right);$

[b]A7.[/b] $\frac{\overline{TA}}{\overline{TB}}=\left(\frac{\overline{MA}}{\overline{MB}}\right)^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \ \frac{\overline{SM}}{\overline{SN}}=\left(\ \frac{\overline{BM}}{\overline{BN}}\right)^2\ \right];$

[b]A8.[/b] $\frac{2}{\overline{AB}}=\frac{1}{\overline{AM}}+\frac{1}{\overline{AN}}\ \ \ \ \ \ \ \ \ \ \  \ \left(\frac{2}{\overline{MN}}=\frac{1}{\overline{AN}}+\frac{1}{\overline{BN}}\right);$

[b]A9.[/b] $\frac{2}{\overline{AB}}=\frac{1}{\overline{MB}}+\frac{1}{\overline{NB}}\ \ \ \ \ \ \ \ \ \ \ \left(\frac{2}{\overline{MN}}=\frac{1}{\overline{MA}}+\frac{1}{\overline{MB}}\right).$

[u][i]Remark.[/i][/u] With this lemma we will solve soon many problems.

The my [b]solution for the first problem.[/b]
Because $(L,D,B,C)$ is harmonious division $(h.d.)$ , from the [b]equivalence A4[/b] results that $DB.DC=DL.DM$ . Thus,

$(L,D,B,C)-h.d.\Longleftrightarrow DB\cdot DC=DL\cdot DM\Longleftrightarrow DA\cdot DH=DL\cdot DM$ $\Longleftrightarrow \frac{DA}{DM}=\frac{DL}{DH}\Longleftrightarrow$
$\tan \widehat {AMD}=\tan \widehat{LHD}\Longleftrightarrow \widehat{AMD}\equiv\widehat{LHD}\Longleftrightarrow HL\perp AM\ .$

[b](Continuation of solution of P1).[/b]

$\widehat{ASB}\equiv\widehat{ASC}\Longleftrightarrow \widehat{BSM}\equiv\widehat{CSM}\Longleftrightarrow (L,M,B,C)-h.d.$ $\Longleftrightarrow LS\parallel BC\Longleftrightarrow AB=AC.$

The my [b]solution for the second problem.[/b]

$TI\perp AX\Longleftrightarrow TA^2-TX^2=IA^2-IX^2\Longleftrightarrow TA^2-(TI^2-r^2)=$ $IA^2-r^2(=AY^2)\Longleftrightarrow$
$TA^2-TI^2=YA^2-YI^2\Longleftrightarrow TY\perp AI,\ what \ is\ true.$

Because $(T,X,B,C)$ is harmonious division and $TV\perp VX$ results that $\widehat{BVX}\equiv\widehat{CVX}$.

The my [b]solution of the third problem[/b]:
I note the symmetrical point $L$ of the point $H$ w.r.t. the point $D$ and $S\in FD\cap LC$. From the relation $DF=DS$ and the problem of butterfly results $OD\perp FD\ .$

The my [b]solution of the fourth problem[/b]:
I note the point $E\in BC$ for that $(B,C,D,E)-h.d.$ Since $GD\perp BC$ results $\widehat{GDX}\equiv\widehat{GDY}\Longleftrightarrow E\in XY$ .

Here is other problem whom I solve with the equivalence A4:

[b]P5.[/b] Let $C_k=C(O_k,r_k),\ k\in\overline{1,2}$ be two circles for which $O_1\in C_2,\ A,B\in C_1\cap C_2$. A some line $d$ meets $C_1$ in the points $M,N$ and (second time) the circle $C_2$ in the point $P$. I note the intersection $R\in AB\cap d$. Prove that $\widehat{PAM}\equiv\widehat{RAM}.$
[i][u]Solution[/u].[/i] From equivalence $A4$ results that $RM\cdot RN=RP\cdot RO_1\Longleftrightarrow (P,R,M,N)-h.d.$. Because $AM\perp AN$ results the conclusion of the proposed problem.

[quote="Virgil Nicula"] \[\underline{\overline{|<>Lemma<>.|}}\] Let the line [b]AB[/b] and the points $M,N\in AB$, the middlepoints [b]S,T[/b] for the segments [b](AB), (MN)[/b] respectively. Prove that the following assertions are equivalently:

[b]A1.[/b] There is $m\ne 1$, such that $\overline{MA}=m\cdot \overline{MB},\ \overline{NA}=-m\cdot \overline{NB}$, i.e. [b](A,B,M,N)[/b] is a harmonious division;

[b]A2.[/b] $AB^{2}+MN^{2}=4\cdot ST^{2}$;

[b]A3.[/b] $TM^{2}=\overline{TA}\cdot \overline{TB}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( SA^{2}=\overline{SM}\cdot \overline{SN}\right) ;$

[b]A4.[/b] $\overline{BM}\cdot \overline{BN}=\overline{BA}\cdot \overline{BT}\ \ \ \ \ \ \ \ \ \left( \overline{AM}\cdot \overline{AN}=\overline{AB}\cdot \overline{AT}\right);$

[b]A5.[/b] $\overline{TA}\cdot \overline{TB}+\overline{TM}\cdot \overline{TN}=0\ \ \ \ \ \left( \overline{SM}\cdot \overline{SN}+\overline{SA}\cdot \overline{SB}=0\right);$

[b]A6.[/b] $\overline{NA}\cdot \overline{NB}=\overline{NM}\cdot \overline{NS}\ \ \ \ \ \ \ \ \left( \overline{MA}\cdot \overline{MB}=\overline{MN}\cdot \overline{MS}\right);$

[b]A7.[/b] $\frac{\overline{TA}}{\overline{TB}}=\left(\frac{\overline{MA}}{\overline{MB}}\right)^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \ \frac{\overline{SM}}{\overline{SN}}=\left(\ \frac{\overline{BM}}{\overline{BN}}\right)^{2}\ \right];$

[b]A8.[/b] $\frac{2}{\overline{AB}}=\frac{1}{\overline{AM}}+\frac{1}{\overline{AN}}\ \ \ \ \ \ \ \ \ \ \ \ \left(\frac{2}{\overline{MN}}=\frac{1}{\overline{AN}}+\frac{1}{\overline{BN}}\right);$

[b]A9.[/b] $\frac{2}{\overline{AB}}=\frac{1}{\overline{MB}}+\frac{1}{\overline{NB}}\ \ \ \ \ \ \ \ \ \ \ \left(\frac{2}{\overline{MN}}=\frac{1}{\overline{MA}}+\frac{1}{\overline{MB}}\right).$

[u][i]Remark.[/i][/u] With this lemma we will solve soon many problems.[/quote]

Could you post a proof for every of this to make a beautiful <<harmonic>> packet mr Nicula ?

Silouan, Mr. Nicula's lemma is easy to prove! You can use coordinates for example (or other equivalent ways): $A(0), B(1), M(p), N(q)$; then note that $\left( A,B,M,N \right)$ is a harmonic division if and only if $p+q = 2pq$; using simple algebra, you can prove that $A.1, \ldots, A.9$ are equivalent to the same $p+q = 2pq$.
I think that the applications are more interesting.

can anyone tell me how A,B,M,N,S,T are placed on a straight line?

Because
[quote="Virgil Nicula"]... [b]Let the line $ AB$ and the points $ M,N\in AB$, the middlepoints $ S$ , $ T$ for the segments $ (AB)$ , $ (MN)$ respectively. ... [/b][/quote]Is it O.K. now ?!

first of all. 'M,N belongs to AB' does it mean that M,N can be on the extension of AB, ie. A-M-B-N?

A-M-S-B-T-N
A-M-T-N-S-B,
A-M-T-S-N-B,
A-M-S-T-N-B,

my question is the lemmas are all suitable for the above arrangement? :?:

A lot of those lemmata can be proven once we show that the circles with diameters $ AB$ and $ MN$ are orthogonal (ie, that harmonic division and inversion are equivalent).

[quote="Virgil Nicula"][b]P4.[/b] Let $\triangle ABC,\ AB\ne AC$, the centroid $G$ and the point $D\in BC,\ GD\perp BC$. Ascertain the positions of the points $X\in AB,\ Y\in AC$ for which $G\in XY$ and $\widehat{GDX}\equiv\widehat{GDY}$. :!:[/quote]

$XY\parallel BC$.


What is the cosine of the vertical angle?
In an isosceles triangle, if the altitudes intersect on the inscribed circle, then the cosine of the vertical angle A is?

A)1/9
B)1/3
C)2/3
D)None

[quote="fishythefish"]Define the vertical angle.

Generally, vertical angles are any pair of two opposite angles when two lines intersect.[/quote]
In general, in an isosceles triangle, the vertical angle denotes the angle opposite the unequal side !

[quote="implex"]In general, in an isosceles triangle, the vertical angle denotes the angle opposite the unequal side ![/quote]
Wrong in two ways.
First of all, generally, [b]vertex[/b] angles are in isosceles triangles.  Vertical angles are defined the way fishythefish said so, so your definition fails in this first way.
Second of all, "opposite the unequal side" doesn't make sense, because isosceles triangles are defined to be triangles with at least two congruent sides.  All equilateral triangles are isosceles.  In those cases, there is no "unequal side," so your definition fails in this second way.

But, I think that MNrule was referring to a vertex angle of an isosceles triangle, not a vertical angle.

Sorry for the typo. I meant vertex angle of isosceles triangle ABC. Anyone got the answer?

Whats the result of the orthocentre lying on the inscribed circle?


MATHCOUNTS
geometry
geometric transformation
reflection

From Elias Saab's mathcounts drills website: http://mathcounts.saab.org I found what seemed to be a particularly interesting and confusing problem:

Consider the expression  123571113 . If you are allowed to put as many parentheses as you want, how many different values can be obtained for this exprssion?

The answer the website gave was this:
[hide]
Here is one way of doing this problem:
A moment of reflection reveals that one can obtain 2^(7-2)=32 different values.
The best way to prove this is by induction.
This is true if m=2. If we have 1  2, there is only one value 1/2.
The formula is true for m=2, 2^(2-2)=1.
If m=3, we have the following situation
1  2  3.
We can have here two values
1  (2  3 )= 3/2 and
(1  2)  3 = 1/6.
The formula is true for m=3, 2^(3-2)=2.
Notice that the addition of the third number 3 resulted into the two values
3/2 = (1/2) * 3 and
1/6= (1/2) / 3 which are the previous value 1/2 multiplied by 3 and divided by 3.
Suppose that the formula is true for m=n. We need to prove it for m=n+1. The same reasoning above applies here. The addition of the (n+1)th. prime number p will result in double the values we get for m=n. In fact we get all the previuos values, then we multiply them by p and divide them by p. The number of different values will be
2* 2^(m-2)= 2^(m-1)=2^((m+1) -2).[/hide]

However, I don't really get how his "reflection" got the answer and how he knew for sure adding a [hide] prime will double the number of values.[/hide] Could someone explain the reasoning behind Mr. Saab's answer? Thanks.

Basically what he is saying is that the problem is equivalent to finding the number of subsets of a set with 5 distinct elements.  Let's look at finding the number of subsets of a 5 element set.  Each element can either be in the subset or not be in it.  Thus each element has two choices and contributes a factor of 2 to the total number of subsets.  Thus there are $2^5$ total subsets (you still count the original set and the null set in this count).

Now let's look at that problem.  The value of the expression depends on where you place each set of parentheses.  And really it only matters on where you place the close parenteses i.e. the ")" s.  Putting a parenthesis after the one won't change anything so we don't worry about it.  What does matter is if there is a parenthesis after 2, 3, 5, 7 or 11.  Once again it doesn't matter whether there is one after 13 or not.  Thus each parenthesis can either be there or not yielding $2^5=32$ possible values.

Thanks alot for your help :), but I still don't get one part of what you said. [quote]And really it only matters on where you place the close parenteses i.e. the ")" s[/quote]
Suppose I put a close parentheses only after the seven. There are then many possibilities for where I could put the opening parentheses, among them:

1/2/3/(5/7)/11/13=7/4290
1/2/(3/5/7)/11/13=35/858

The two give different results, but they are defined by the same subset of the five numbers {2,3,5,7,11}

Could you explain this part a bit more? Thanks.


algebra
polynomial
function
number theory
number theory unsolved

Hello ! I am looking for an elementary solution to this problem :

Does exist a polynom P with integer coefficients such that  $ P(n)$ is prime for any $ n \in \mathbb{N}$

I have a solution but it uses hard theorems of number theory

[hide]Use dirichlet theorem on arithmetic progressions ![/hide]

[hide="Hint"]For any polynomial with a constant term other than $ \pm 1$, we can plug in the constant term to get a multiple of that. Therefore assume that the constant term of $ p(n)$ is $ \pm 1$. Now try proving that if $ p(n) \equal{} p$, a prime number, then $ p(n \plus{} p)$ is composite. You don't need Dirichlet's theorem for this.[/hide]

Stupid comment:Yes,there exist a such polynomial,take any constant polynomial with prime value.
Assuming that you forgot to say about degree of $ P$:
Simpler approach works here:
Assume the contrary,i.e there exist a such polynomial.
Suppose that $ P(n)\equal{}q$,where $ q$ is some prime number.
Claim:If $ m\equal{}n\plus{}k\cdot q$,then $ P(m)\equal{}p$.
Proof:As we know $ P(m)\minus{}P(n)$ is divisible by $ k\cdot q$,therefore,$ P(m)$ is divisible by $ q$,it means that polynomial $ Q(x)\equal{}P(x)\minus{}x$,has the value equal to $ 0$ in each point $ n,q\plus{}n,2q\plus{}n,\dots$,therefore $ Q(x)\equiv 0$,hence $ P(x)\equal{}x$,but $ P(6)\equal{}6$ is not the prime number,contradiction.

[quote="popolux"]I apologize for this but i made a mistake while posting :

the exact question is find all polynoms of $ \mathbb{Z}[X]$ mapping $ \underline{primes \, \, numbers}$ on primes number.

Of course it implies the precedent question, but in this case i did not succeed to answer without dirichlet theorem..But i would prefer to find  an elementary proof like yours..Could you help me?[/quote]
I guess this is your solution as it uses Dirichlet:
[hide="Solution"]
Let the function be $ f(n)$ and let $ f(n)\equiv x\bmod n$.  If $ x \equal{} 0$, then let $ f(n) \equal{} mn$.  If $ m\ne \pm 1$, then, clearly, $ f(n)$ is composite for all $ n$.  Then, $ m \equal{} \pm 1$, so we obtain solutions $ f(n) \equal{} \pm n$.  Now, if $ x\ne 0$, then plug in $ n \equal{} p$, where $ p$ is a prime that is coprime with $ x$.  This gives that $ f(p)$ is coprime with $ p$.  If $ f(p) \equal{} k$, then we consider the numbers $ p \plus{} ka$ for integers $ a$.  There are infinitely many primes among these by Dirichlet's Theorem.  Let the $ a$'s that let $ p \plus{} ka$ be a prime be $ a_1$, $ a_2$, $ \cdots$, so $ p \plus{} a_ik$ is prime for natural $ i$.  It is well-known (and can be proven with binomial expansion) that $ f(ka \plus{} p)$ is a multiple of $ k$ when $ f(p) \equal{} k$.  Since $ f(ka \plus{} p) \plus{} p \equal{} 0$ and $ f(ka \plus{} p) \minus{} p \equal{} 0$ both have a finite number of roots, we see that $ f(a_ik \plus{} p) \equal{} p$ and $ f(a_ik \plus{} p) \equal{} \minus{} p$ both have a finite number of roots.  Then, there is a $ a_j$ (actually, infinitely many $ a_j$) so that $ f(a_jk \plus{} p)$ is a multiple of $ k$, but not equal to $ \pm k$, so is composite.  Then, we have found a prime, namely $ a_jk \plus{} p$ (let it equal $ q$), so that $ f(q)$ is not prime, which is a contradiction.  It follows that the only existing functions are $ f(x) \equal{} \pm x$. [/hide]

EDIT: Dear, [b]Allnames[/b], can you please explain?  $ x\equal{}4$ gives that $ x$ is not prime, so $ f(x)$ does not have to be prime either.  I don't quite understand...

[quote="The QuattoMaster 6000"] It follows that the only existing functions are $ f(x) \equal{} \pm x$. [/hide][/quote]
How about $ x \equal{} 4$
The result is not exist ,but with a polynom with more degree and more variables,it is exist(I  remember the degree is 24,and the number of variables is 16,I am not sure)
But if we solve it in $ \mathbb P$ the set which contains all primes then your proof is trur and it is old

Thank you for the link!
Damn , it seems it is hard to prove it by an elementary way...

[quote="popolux"]Thank you for the link!
- , it seems it is hard to prove it by an elementary way...[/quote]
No an other ways is consider series :
$ \sum_{p \in {P}} \frac {1}{f(p)}$
This series tend to infinite so degree of $ f$ must be equal 1 . Then easy check that there is only a function satisfy condition is $ f(x) \equal{} x$

Yes but why $ \sum_{\mathcal{P}} \frac{1}{f(p)}$ is infinite, we don't know that f is surjective on prime,  there may be few enough primes of the form $ f(p)$ such that $ \sum_{\mathcal{P}} \frac{1}{f(p)}$ is finite.

[quote="popolux"]Yes but why $ \sum_{\mathcal{P}} \frac {1}{f(p)}$ is infinite, we don't know that f is surjective on prime,  there may be few enough primes of the form $ f(p)$ such that $ \sum_{\mathcal{P}} \frac {1}{f(p)}$ is finite.[/quote]
Obvious the leading coefficient of P is a positive number ,therefore exist $ n_0$ such that $ f'(x)>0$ for all x is larger than $ n_0$ ,it means that f is increasing from this . Therefore the series must tend to infinity.

Yes but it not involves that the serie tend to infinity , f is not bijective from primes to primes .
You may be right but could you detail more, number theory is not my cup of tea :D


Where can u get old AHSMEs?

P.S. Do my contest!!! :P

This question has been asked many times before.  Please use the search feature.


geometry trapezoid symmetry circumcircle trigonometry algebra polynomial

I came up with this yesterday messing around with drawings.

Say we have a triangle ABC in a circle, and say we have point P inside the triangle. Extend CP to AB, so that they intersect at D, and do the same for the other vertices so that AP meets CB at E, and BP meets AC at F. Now extend FE so that it meets the circle at L_1 and L_2, extend ED so that it meets the circle at L_3 and L_4, and extend DF so that it meets the circle at L_5 and L_6. What are necessary and sufficient conditions so that the 9-gon formed by A,B,C, and the L_k is regular?

Just for reference.
[img]http://www.geocities.com/habukii/nona_circle.JPG[/img]
[hide]If the nonagon is regular, then any angle between consecutive vertices must equal 140.  First, examine quadrilateral $CBL_1L_3$.  If the nonagon is regular, then this will be a trapezoid such that $m\angle BCL_3=40$ and similarly $m\angle ACL_5=40$ Therefore, $m\angle ACB=140-80=60$ By symmetry, $\triangle ABC$ must be equilateral. Again by symmetry $FD=DE=EF$ giving $\triangle DEF$ is equilateral.  Since $\triangle ABC\sim\triangle DEF$, D, F, and E are the midpoints of $\triangle ABC$. Giving P is the intersection of the medians.
In conclusion, to form a regular nonagon using this method, $\triangle ABC$ must be equilateral and P must be the intersection of its medians.[/hide]

Also note that the intersection of the medians of an equilateral triangle is also the circumcenter, the center of the circle circumscribed about the triangle.  For this particular case, that means that P is the center of the circle, as well!

[quote="jelyman"]Excellent, very close to my solution. A question... how long did it take for u to "see" the solution?[/quote]
Oh, about three seconds. :D   But then again, I cheated.  :blush:  This figure above was constructed by dragging points around using geometer's sketchpad in which it was pretty simple to see the solution.  If I had done it by hand, it probably would have been somewhere under ten minutes.

http://www.keypress.com/sketchpad/

From what I hear it is the best geometry program around.  You have to buy it however.

[quote="seamusoboyle"]Where did you get the geometry software? Sounds interesting.[/quote]
Personally, I got it for free.  My school owns a server edition which may be installed on as many computers as you like.  I am sure that this would be a rather expensive purchase, so ask around your school and see if any of the geometry teaachers or computer labs already have the software.

Unfortunately, it does not work - no condition is sufficient (see the 1st attachment). You cannot construct a regular nonagon with compass and straightedge, because that would allow you to construct the angle of $40^o$, which in turn, would effectively trisect the angle of $60^o$.

$\cos{3 \alpha} = \cos  \alpha \cos{2 \alpha} - \sin  \alpha \sin{2 \alpha} = \cos  \alpha (\cos^2{ \alpha} - \sin^2{ \alpha}) - 2 \sin^2{ \alpha} \cos  \alpha = $

$= \cos \alpha (2 \cos^2{ \alpha} - 1) - 2 (1 - \cos^2{ \alpha}) \cos \alpha =  4 \cos^3{ \alpha} -3 \cos \alpha$

Applying this to the angle $ \alpha = 20^o$, we get

$\cos{60^o} = 4 \cos^3{20^o} -3 \cos{20^o}$

Hence, $x = \cos{20^o}$ is a root of the cubic equation

$\frac 1 2 = 4x^3 - 3x$

$8x^3 - 6x - 1 = 0$

By the rational root theorem for polynomials with integer coefficients, the only rational roots this cubic equation can have are $x = \frac p q$, where p is a divider of the absolute term (1) and q is a divider of the coefficient at the highest power (8), i.e., $x = \pm 1,\ \pm \frac 1 2,\ \pm \frac 1 4,\ \pm \frac 1 8$. By inspection, none of these numbers is a root. Therefore, we cannot factor this cubic equation into $x - \frac p q$ and a quadratic equation with rational coefficients. But the only numbers constructible with compass and straightedge are roots of quadratic equations with rational coefficients or coefficients, which are themselves constructible numbers. Hence, $x = \cos{20^o}$ cannot be constructed with compass and straightedge. It can be constructed exactly only by methods that allow to find roots of cubic equations, such as constructions with compass and marked ruler or paper-folding (origami). There are various approximation methods to trisect (or n-sect) angles and consequently, to construct approximately arbitrary regular n-gons. One is based on a rather accidental fact that the function

$f(x) = \frac{\pi}{2} \cdot \frac{\sqrt 3 \sin x}{\sqrt 3 + \cos x} \doteq x$

is almost linear in the interval $\left(-\frac \pi 2, +\frac \pi 2 \right)$. This allows the following approximate constructrion of any regular n-gon: Divide the segment $A_0A_n$ into n equal parts, such that $A_0A_1 = A_1A_2 = ... = A_{n-1}A_n = \frac{A_0A_n}{n}$. Construct the circle $(O)$ on the diameter $A_0A_n$ and the vertices $B, C$ of the equilateral triangles $\triangle A_0A_nB, \triangle A_0A_nC$ above and below the segment $A_0A_n$. Select either the even index points $A_0, A_2, A_4, ...$ or the odd index points $A_1, A_3, A_5, ...$. Let's make the first choice. Draw the lines $BA_0, BA_2, BA_4, ...$ intersecting the circle $(O)$ at points $A_0 \equiv D_0, D_2, D_4, ...$ on the opposite side of the line $A_0A_n$ than the point $B$. Similarly, draw the lines $CA_0, CA_2, CA_4, ...$ intersecting the circle $(O)$ at points $A_0 \equiv E_0, E_2, E_4, ...$ on the opposite side of the line $A_0A_n$ than the point $C$. The set of points $A_0, D_2, D_4, ..., E_2, E_4, ...$ approximately forms a regular n-gon (see the 2nd attachment).


My school is already registered for mathcounts as a 2-person team. I know this is a disadvantage. I am captain of my team and kids in my school are so disinterested in math. How can I get them involved? Force them?

Also, my team partner doesn't seem to get that many problems on the team round. How can I help her? Any strategies for working together on team round?

I know it's rather obvious, even if I get a 46 ind. score at chapter, I can't drag the team up to first place alone!  So what should I do?

I don't think forcing is a very good method. I attend a school with some students that are forced to be on math team by their parents, and are forced to try out for MathCounts as well. Although a few of them would do as much as their parents make them do regardless, most of them highly dislike it and become less interested than ever before, and it's hard to succeed in something you don't find very interesting. Even if you did manage to force people onto the team, this effect would most likely take place.

However, there are always opposite cases, in which kids are not interested in MathCounts, but would be if they knew a little more about it. For instance, I had no idea what MathCounts was until our math team coach told us about it, and I became very interested and worked hard at it. My parents wouldn't have known anything about it, I'm the last of 4 and the only one to be on math team.

Maybe you could try to show your fellow classmates the rewarding elements of the MC competition. It worked for me, and it may for them too.

I have a slightly similar situation on my team, although there are 4 people. On our last practice team round I contributed 5 of the 7 total problems we solved, and there was someone that didn't solve any problems correctly.

I have no idea how to get people interested in mathcounts. In my opinion, either they're interested or they aren't, and you can't change that. However, if there are people that might be interested, be sure they know the opportunity exists.

What you might want to do on the test is: give your teammate the easier problems that she is more likely to be able to solve, and do the harder problems yourself. Also, if there are any problems that are similar to problems your teammate might have done in class or be able to do (i.e. basic algebra problems, arithmetic problems, combinatorics where one can list out all of the cases, logic problems, calculator bash problems...basically anything that she can do), you should let your teammate solve them and work on the harder problems yourself. Sometimes, someone unversed in more advanced mathcounts techniques can still check some of the harder questions' answers by brute force or by checking reasonability.


trigonometry
logarithms
function
calculus
calculus computations

Answer the following question.

[1] $\frac{d}{dx} \sqrt{1-x}\ (x<1)$

[2] $\frac{d}{d\theta} \frac{\cos \theta}{(1-\sin \theta)^2}$

[3] $\frac{d}{dx} \ln \frac{1+\sin x}{\cos x}$

[4] If $x=2t-\sin t,y=2-\cos t\ (0<t<2\pi)$,we have $\frac{dy}{dx}=0$ when $(x,y)=\boxed{\ \ }$.

[5] For a function $f(x)=\frac{1}{2}\left(e^x+\frac{1}{e^x}\right)$, let $g(x)=\frac{d}{dx} f(x)$ and set $h(x)=g^{-1}(x)$.

   Find $(x^2+1)\frac{d^3}{dx^3} h(x)+3x\frac{d^2}{dx^2} h(x)$

1
$\frac{-1}{2\sqrt{1-x}}$
2
$\frac{2+\sin^3{\theta}-\sin{\theta}(1+2\cos{\theta})}{1-\sin{\theta})^4}$
3
its equilvalant to $\ln{|sec{x}+\tan{x}|}$
so the answer is: $\sec{x}$
4
i dont understand.
$\frac{dy}{dx}=\frac{\sin{t}}{2-\cos{t}}$ so if you set i equl zero, you get $t=k\pi$. but you said. $t\in (0,\pi)$

[quote="amirhtlusa"]1
$\frac{-1}{2\sqrt{1-x}}$ :)
2
$\frac{2+\sin^3{\theta}-\sin{\theta}(1+2\cos{\theta})}{1-\sin{\theta})^4}$[color=blue] Please express in terms of sine[/color].
3
its equilvalant to $\ln{|sec{x}+\tan{x}|}$
so the answer is: $\sec{x}$ :)
4
i dont understand.[color=blue]Sorry that I was made a typo. I will edit.[/color]:blush:
$\frac{dy}{dx}=\frac{\sin{t}}{2-\cos{t}}$ so if you set i equl zero, you get $t=k\pi$. but you said. $t\in (0,\pi)$[/quote]

kunny

2
$\frac{2+\sin^3{\theta}-\sin{\theta}+\sin{(2\theta)}}{(1-\sin{\theta})^4}$
4
$t=\pi\Rightarrow (x,y)=(2\pi,1)$
5
i dont get this one.
whay do you mean by that $g(x)=\frac{d}{dx}$??

[quote="amirhtlusa"]2
$\frac{2+\sin^3{\theta}-\sin{\theta}+\sin{(2\theta)}}{(1-\sin{\theta})^4}$ :(
4
$t=\pi\Rightarrow (x,y)=(2\pi,1)$ :(
5
i dont get this one.[color=blue]Sorry for again, amirhtlusa. I have just edited.[/color] :blush:  :blush:
whay do you mean by that $g(x)=\frac{d}{dx}$??[/quote]

kunny

[quote="amirhtlusa"]4
$(2\pi,3)$ :blush:[/quote] :)


How many solutions has the following equation:
                 $cos\pi{x}=\frac{x^{2}}{16}$

[hide]Evidently we may consider positive solutions and then duble their number. For  $x>0$ we have $x\le 4$ since $\cos \pi x \le 1$. Try to finish it - it's very easy now.[/hide]

the graphs are in this picture:
it comes out that we have 8 solutions..


vector
topology
advanced fields
advanced fields unsolved

$M,N,N_{1}$ are differential manifold, $E$ is a vector bundle over $M$. $f: N\to M$ is a $C^{\infty}$ mapping.
$u: N_{1}\to N$ is a $C^{\infty}$ mapping. If $G_{1}: N_{1}\to E$ is any lifting of $f\circ u$, and if $z_{1}\in N_{1}$,
Prove:there is a lifting $G$ of $f$ to $E$ such that $G(u(z_{1}))=G_{1}(z_{1})$.

my thinking is $u$ may not be onto, how do we even define a $G$ (which must be $C^{\infty}$) on whole $N$ and not just on $u(N_{1})$?

[quote="mlok"]The statement looks strange, because $N_{1}$ is not really involved in it. Maybe you were asked for $G$ such that $G(u(z_{1}))=G_{1}(z_{1})$ for [b]all [/b]$z_{1}\in N_{1}$?[/quote]

yes, for all $z_{1}\in N_{1}$, otherwise trivialy true.


limit logarithms algebra function domain integration inequalities

Why is
\[\lim_{x \rightarrow \infty} (1+\frac{1}{x})^x = e\]
true?

Start by giving your limit a name, say, $L$, and then taking the logarithm. This what you do with all tricky limits that involve exponents. Since the logarithm is continuous on its domain, we can interchange the function and taking limits.

$\ln L = \lim_{x\to\infty}x\ln\left(1+\frac1x\right).$

For what I'm about to do, it would make sense to substitute $y=\frac1x.$

$\ln L = \lim_{y\to0^+}\frac1y\ln(1+y).$

Note that $\ln(1+y)=\int_1^{1+y}\frac1t\,dt.$ Note that if $1\le t\le1+y$, then

$\frac1{1+y}\le\frac1t\le1,$ thus $\frac{y}{1+y}\le\int_1^{1+y}\frac{dt}t\le y.$

$\frac1{1+y}\le\frac1y\ln(1+y)\le1.$

Take the limit as $y\to0$: by the "squeeze theorem," since the two outside quantities of this chain of inequalities both tend to one, the quantity in the middle must also tend to 1.

So $\ln L=1,$ which means that $L=e.$

I think it's easier to say:
$\lim_{x \to 0+} \frac{\ln \left( 1+x \right) }{x}= 1$ since it's standart derivation of $\ln(x)$ at $1$ and thus $=1$

$f(x)=x\ln\left(1+\frac1x\right).$

Use asymptotic expansion

$f(x)=x ( 1/x -1/(2x^2) +o(1/x^2))= 1 -1/(2x) + o(1/x)$

$f$ tend to 1 when x tends to +oo

and $e^1=e$


inequalities function algebra domain geometry perimeter calculus

This is one of the problems I know about and know that are true, but never tried to prove it  :D (It is "clear" from continuity :D).

Some things to note:

The inequalities are chosen to permit the formation of some quadrilateral.

Among all quadrilaterals with the same sides, there exists one with maximum area. (This is really the matter of a continuous function on a compact domain assuming its maximum.) It doesn't matter what order you put the sides - the maximum area is the same.

For given sides, the maximum are is acheived when the quadrilateral is cyclic. Can you prove that?

In fact, any set of $n$ lengths that satisfy the inqualities needed to form an $n$-gon can form a cyclic $n$-gon. The argument is the same.

Well, to prove the initial problem:

Since we can find $a,b$ and $c,d$ (this is, in fact, a permutation of the $a,b,c,d$ we were initially given) s.t. $a+b\le c+d$, we can construct first a triangle (it can even be degenerate) with sides $a+b,c,d$. Mark a point on the side of size $a+b$ dividing it into two pieces of sizes $a,b$.

We can now play around with this quadrilateral. We may assume, for instance, that $a+d\le b+c$. This means that we can deform the quadrilateral until it becomes a triangle with sides of lengths $a+d,b+c$. In the first position the sum of two opposite angles was larger than $\pi$, and in the other position, the sum of the same two angles was less than $\pi$. By continuity arguments, at some point that sum must have been $\pi$ at some time during the deformation.

[quote="Kent Merryfield"]For given sides, the maximum are is acheived when the quadrilateral is cyclic. Can you prove that?[/quote]

It follows immediately from Bramhagupta's formula.  The other proof I know involves calculus.

Thank you very much, Grobber, for your proof, but I would like also to see a constuctive proof which doesn't use the continuity principle of analysis, that is an elementary proof in the sense of synthetic geometry  :)

Regarding the problem posted by Kent, I remember posting a proof for the general case: given $n$ lengths which can form a convex $n$-gon, we can find a cyclic $n$-gon with sides of these lengths, and its area is maximal among all the areas of such $n$-gons.

The fact that a cyclic $n$-gon exists can be proved by continuity like in the case of a quadrilateral (above). As for the area, consider we have constructed a cyclic $n$-gon with those sides, inscribed in a circle $\mathcal C$. For each side, construct the arc subtended by that side when the $n$-gon is cyclic. We get a figure with perimeter equal to that of $\mathcal C$, and its area is the largest when it's a circle (the isoperimetric inequality: among all the figures with the same perimeter, the one with the largest area is the circle), and since the areas of the regions between the sides of the $n$-gon and the arcs are always the same, it means that making the $n$-gon cyclic also maximizes its area.

Isn't it nice? :) Is this what you had in mind, Kent?

We can also prove the general case with the quadrilateral case: if the polygon is not cyclic, we construct one with greater area.

Just choose four consecutive vertices that are not concyclic. Draw the diagonal that cuts these off from the rest of the polygon. Now, the three sides and the diagonal form a quadrilateral which can be enlarged by making it cyclic with the same side lengths.
Since the maximum exists, it must come when the polygon is cyclic.

The quadrilateral case can be done pretty cleanly by calculus:
Suppose $AB=a,BC=b,CD=c,DA=d$ and $\angle ABC=\alpha$, $\angle CDA=\beta$.
The area is $\frac12(ab\sin \alpha+cd\sin\beta)$. The constraint is derived from $AC$; $a^2+b^2-2ab\cos\alpha=c^2+d^2-2cd\cos\beta$
Using Lagrange multipliers, $ab\cos\alpha=\lambda\cdot ab\sin\alpha$ and $cd\cos\beta=-\lambda\cdot cd\sin\beta$. This gives $\lambda=\cot\alpha=-\cot\beta$, so $\alpha$ and $\beta$ are supplementary and the area-maximizing quadrilateral is cyclic.
With more work, we can establish Brahmagupta's formula from here.

[quote="manlio"]Suppose that there are 4 positive numbers $a,b,c,d$ that satisfy

$a<b+c+d,   b<c+d+a,  c<d+a+b,   d<a+b+c$

Prove that $a,b,c,d$ can be the sides of a cyclic quadrilateral[/quote]

Here is a synthetic proof:

We will construct the required cyclic quadrilateral with sidelengths a, b, c, d. Then, the determination will show that the construction always works, provided the inequalities a < b + c + d, b < c + d + a, c < d + a + b and d < a + b + c hold. This will clearly solve the problem.

[b]Construction of the cyclic quadrilateral with sidelengths a, b, c, d:[/b]

[I have first seen this method of construction in [url=http://groups.yahoo.com/group/Hyacinthos/message/9190?expand=1]Hyacinthos message #9190[/url].]

Let C, D and E be three points lying on a line in this order and satisfying CD = c and $DE=\frac{bd}{a}$. Now, the locus of all points W such that $\frac{CW}{EW}=\frac{a}{d}$ is a circle, the so-called [b]Apollonius circle[/b] of the segment CE with the ratio $\frac{a}{d}$. We call this circle k, and we denote by A one of the two points of intersection of this circle k with the circle centered at D and having radius d. Finally, let B be the point such that the triangle ABC is directly similar to the triangle ADE (clearly, there is one and only one such point B in the plane). Then, the quadrilateral ABCD is our required cyclic quadrilateral with sidelengths a, b, c, d.

[b]Proof:[/b]

We have to prove that AB = a, BC = b, CD = c and DA = d and that the quadrilateral ABCD is convex and cyclic.

First, DA = d is clear because the point A lies on the circle centered at D and having radius d. Furthermore, CD = c follows from the definition of the points C and D.

Since the point A lies on the circle k which is the locus of all points W such that $\frac{CW}{EW}=\frac{a}{d}$, we have $\frac{CA}{EA}=\frac{a}{d}$. Since the triangle ABC is similar to the triangle ADE, we have $\frac{AB}{AD}=\frac{CA}{EA}$, and thus

$AB=AD\cdot\frac{CA}{EA}=AD\cdot\frac{a}{d}=DA\cdot\frac{a}{d}=d\cdot\frac{a}{d}=a$.

Also, since the triangle ABC is similar to the triangle ADE, we have $\frac{BC}{DE}=\frac{CA}{EA}$, and thus

$BC=DE\cdot\frac{CA}{EA}=DE\cdot\frac{a}{d}=\frac{bd}{a}\cdot\frac{a}{d}=b$.

Since the triangle ABC is directly similar to the triangle ADE, we have < ABC = < ADE with directed angles modulo 360° (not modulo 180°). But, with directed angles modulo 360°, we have < ADE = 180° - < CDA. Thus, < ABC = 180° - < CDA, and it follows that the quadrilateral ABCD is convex and cyclic.

This completes the proof of our construction.

[b]Determination:[/b]

Now we are going to show the really important thing, namely that the construction above always yields a quadrilateral ABCD, as long as the input values a, b, c, d satisfy the inequalities a < b + c + d, b < c + d + a, c < d + a + b and d < a + b + c. In fact, the only point where the construction above could fail is our definition of the point A as of the two points of intersection of the circle k with the circle centered at D and having radius d. In order to show that the construction always works, we have to prove that the circle k and the circle centered at D and having radius d always have two points of intersection. In the following, we will prove this statement, provided the conditions a < b + c + d, b < c + d + a, c < d + a + b and d < a + b + c hold.

Let u be the circle centered at D and having radius d. We have to prove that the circle k and the circle u always have two points of intersection.

As in the above, we won't use directed segments. We will also assume that $a\neq d$ (in fact, the case a = d is left to the reader - it is rather simple).

Let S and T be the two points dividing the segment CE in the ratio $\frac{CS}{ES}=\frac{CT}{ET}=\frac{a}{d}$ internally and externally, respectively (i. e., the point S lies inside the segment CE, and the point T lies outside the segment CE; the point T is a finite point since $a\neq d$). Since the circle k is the locus of all points W such that $\frac{CW}{EW}=\frac{a}{d}$, this circle must pass through the points S and T. Now, if we succeed to show that the point S lies inside the circle u, while the point T lies outside the circle u, it will follow that the circle k, passing through the points S and T, must have two points of intersection with the circle u. And the proof will be complete.

Hence, what we have to prove is that the point S lies inside the circle u, while the point T lies outside the circle u.

The circle u has center D and radius d; thus, in order to prove that the point S lies inside the circle u, it is enough to show DS < d, and in order to prove that the point T lies outside the circle u, it is enough to show DT > d.

So it remains to show that DS < d and DT > d.

Since the points C, D and E lie on a line in this order, we have CE = CD + DE. In other words, $CE=c+\frac{bd}{a}=\frac{ac+bd}{a}$.

Since $\frac{CS}{ES}=\frac{CT}{ET}=\frac{a}{d}$, we have $\frac{ES}{CS}=\frac{ET}{CT}=\frac{d}{a}$.

Since the point S lies inside the segment CE, we have CE = CS + ES. Thus,

$\frac{CE}{CS}=\frac{CS+ES}{CS}=1+\frac{ES}{CS}=1+\frac{d}{a}=\frac{a+d}{a}$.

Consequently,

$CS=\frac{CE}{\left(\frac{a+d}{a}\right)}=\frac{\left(\frac{ac+bd}{a}\right)}{\left(\frac{a+d}{a}\right)}=\frac{ac+bd}{a+d}$.

Hence,

$DS=\left|CS-CD\right|=\left|CS-c\right|=\left|\frac{ac+bd}{a+d}-c\right|=\left|\frac{ac+bd-c\left(a+d\right)}{a+d}\right|$
$=\left|\frac{\left(b-c\right)d}{a+d}\right|=\frac{\left|b-c\right|d}{a+d}$.

But we have to show that DS < d. Thus, we have to show $\frac{\left|b-c\right|d}{a+d}<d$. This is clearly equivalent to $\frac{\left|b-c\right|}{a+d}<1$, or, in other words, to $\left|b-c\right|<a+d$. But this is clearly true: If $b\geq c$, then $\left|b-c\right|=b-c$, and $\left|b-c\right|<a+d$ is equivalent to b - c < a + d, what follows from b < c + d + a; if $b\leq c$, then $\left|b-c\right|=c-b$, and $\left|b-c\right|<a+d$ is equivalent to c - b < a + d, what follows from c < d + a + b. Hence, DS < d is proven.

Now remains to show DT > d. In fact, since the point T lies outside the segment CE, we have two possibilities: Either the point T lies on the extension of the segment CE beyound its endpoint C, or the point T lies on the extension of the segment CE beyound its endpoint E.

Consider the first case, when the point T lies on the extension of the segment CE beyound its endpoint C. Then, CE = ET - CT. Thus,

$\frac{CE}{CT}=\frac{ET-CT}{CT}=\frac{ET}{CT}-1=\frac{d}{a}-1=\frac{d-a}{a}$.

This actually shows that d - a > 0 (in fact, CE and CT are non-directed segments, hence $\frac{CE}{CT}$ must be positive, and since $\frac{CE}{CT}=\frac{d-a}{a}$, it follows that d - a is positive).

From $\frac{CE}{CT}=\frac{d-a}{a}$, we conclude

$CT=\frac{CE}{\left(\frac{d-a}{a}\right)}=\frac{\left(\frac{ac+bd}{a}\right)}{\left(\frac{d-a}{a}\right)}=\frac{ac+bd}{d-a}$.

Hence,

$DT=CT+CD=CT+c=\frac{ac+bd}{d-a}+c=\frac{ac+bd+c\left(d-a\right)}{d-a}=\frac{\left(b+c\right)d}{d-a}$.

But we have to show that DT > d. Thus, we have to show $\frac{\left(b+c\right)d}{d-a}>d$. This is clearly equivalent to $\frac{b+c}{d-a}>1$, or, in other words, to b + c > d - a (we have multiplied by d - a, what is unproblematic as we know that d - a > 0). But this follows from a + b + c > d, what is simply our condition d < a + b + c, slightly rewritten.

The second case, when the point T lies on the extension of the segment CE beyound its endpoint E, can be handled similarly using the condition a < b + c + d (just believe me - or check this for yourself).

This completely solves the problem.

  Darij

Here is a proof of existence somewhat simpler than Darij's one.

Drawing the diagonal shows that it must hold \[ a^2\plus{}b^2\minus{}2ab\cos\beta\equal{}c^2\plus{}d^2\plus{}2cd\cos\beta,\quad(*)\] where $ \beta$ denotes the angle between $ a$ and $ b$ (and then $ \pi\minus{}\beta$ equals the angle between the other two sides). Since it must hold $ |\cos\beta|<1$, from the above condition it follows \[ \left|\frac{a^2\plus{}b^2\minus{}c^2\minus{}d^2}{2(ab\plus{}cd)}\right|<1,\quad(**)\] and it is not hard to check that this is actually equivalent to the asertion that each side is less than the sum of the other three. On the other hand, having $ (**)$ given, one could construct a diagonal accordingly to $ (*)$, and everything is over then.


If $ a^2 \equal{} 6$ and $ a^4 \equal{} 4b^3$ and $ a$ and $ b$ are positive, what is the simplified value of $ a^2b^3$?

We have that $ (a^2)^2\equal{}a^4\equal{}36$. So $ 36\equal{}4b^3 \implies b^3\equal{}9$. Combining information,
                                                     $ a^2b^3\equal{}6\times{9}\equal{}\boxed{54}$.


On the circumference of a circle there are red and blue points. One may add a red point and change the colour of both its neighbours (to the other colour) or remove a red point and change the colour of both its previous neighbours. Initially there are two red points. Prove that there is no sequence of allowed operations which leads to the configuration consisting of two blue points.

Name the operation: "add a red point and change the colour of both its neighbours" as $ O_1$ and the other one as $ O_2$. Let $ n_0$ be the initial number of points. If some point is colored red, assign the number 1, if it's blue, assign 0. Let $ S_k$ be the sum of the numbers in the circle at the $ kth$ operation and let $ n_k$ be the total number of points at that time. Then, $ n_k \minus{} S_k$ is the number of blue points at the $ kth$ operation. Clearly, after each operation, $ n_k$ changes its parity, as well as $ S_k$ does. Then the parity of $ n_k \minus{} S_k$ remains invariant. If $ n_0$ is odd, then $ n_k \minus{} S_k$ is always odd, then it is impossible to have 2 blue points.

If $ n_0$ is even, it is possible to have such a configuration. If the initial configuration is

RB|BRB|BB|BBB and you apply $ O_1$ where there is a |, you will get RRRRRRRRRRRBB and if you have

RB|BRB|BB|BB|....|BB|BBB you will have RRRRRRR....RRRBB.

I don't know, maybe something is missing.

^ This solution is not valid because $n_0$ is even.


Given acute angle triangle $ ABC$. Let $ CD$be the altitude , $ H$ be the orthocenter and $ O$ be the circumcenter of $ \triangle ABC$ The line through point $ D$ and perpendicular with $ OD$ , is intersect $ BC$ at $ E$. Prove that $ \angle DHE \equal{} \angle ABC$.

Call F the symmetric of A w.r.t. D, from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=203653]Balkan[/url] $ E \in DF$, so $ \angle DHE \equal{} \angle AHD \equal{} \angle ABC$

[quote="stergiu"][quote="Virgil Nicula"][quote="Litlle 1000t"] [color=darkred]Let $ ABC$ be an acute triangle with circumcircle $ w = C(O)$ and orthocenter $ H$ .

Denote $ \{\begin{array}{c} D\in AB\ ,\ CD\perp AB \\
\ E\in BC\ ,\ DE\perp DO\end{array}$ . Prove that $ \widehat{DHE}\equiv\widehat{ABC}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] The circle $ w$ cut again the line $ CD$ in the point $ F$ . Denote $ K\in AF\cap DE$ . Apply the "[b]property of butterfly[/b]"

 (see[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=73560][color=red][b]here[/b][/color][/url] ) [color=darkblue]to the interior point $ D\in CF\cap AB\cap EK$ of the circle $ w$ : $ \boxed {\ OD\perp \overline {EDK}\ \Longleftrightarrow\ DE = DK\ }$ .

Thus, $ \{\begin{array}{c} DE = DK \\
\ DH = DF\end{array}\|$ $ \implies$ $ HE\parallel FA$ $ \implies$ $ \widehat {DHE}\equiv\widehat {DFA}\equiv\widehat {CBA}$ , i.e. $ \widehat {DHE} = \widehat {ABC}$ .[/color][/quote]

 Dear Virgil , to which form of the buterfly theorem are you reffering to ? The form you give at the link , is the classic form . Do you use any generalisation of the theorem ?

 [u]Babis[/u][/quote]

 Well , I had the extension of  side BC and not side BC. Everything is OK now.
  Virgil , sorry !
 Babis

Sorry,Why
$ \boxed {\ OD\perp \overline {EDK}\ \Longleftrightarrow\ DE \equal{} DK\ }$ ? :blush:
Thank you

Can anyone,please,explain to me why $ OD\perp \overline {EDK}\ \Longleftrightarrow\ DE \equal{} DK$ Thanks :omighty:

Let $F= (BHC) \cap \overline{AB}$ and let $G_1,G_2$ be the points of intersection $\overline{ED}$ with the circumcircle.

Throw the configuration onto the complex plane and scale down the circumcircle to the unit circle and let $b=\frac{1}{a}$.

Using the fact that $CA=CF$ we have that $a+f=2d$, where $d=\frac{1}{2}\left(a+\frac{1}{a}+c-\frac{1}{c}\right)$, this implies that $f=\frac{1}{a}+c-\frac{1}{c}$.

By definition we have that:
$$\frac{g_1-g_2}{\overline{g_1-g_2}}=-\frac{d}{\overline{d}}$$
$$g_1=\frac{d}{g_2.\overline{d}}$$

Plugging in what we have we get that $HFG_1G_2$ is cylic.

Now by the theory of the radical center on the circles $(ABC),(BCHF)$ and $(HFG_2G_1)$ we have that the radical center is $E$, this implies that $\angle EHD = \angle FHD = \angle ABC$

Here is a solution by [b]mueller.25[/b]

Switch to $A$ labelling.

Observe that $O,H$ are isogonal in angles $A,C,D$ and so $H$ has an isogonal conjugate in $ACDE$. So, $\angle EHD + \angle AHC = 180 \implies \angle DHE = \angle ABC$. $\blacksquare$

Sorry I am stupid but can someone post the problem wording in $A$ labelling? I do not know which side the line through point $ D$ and perpendicular with $OD$ meets?

Oops, should probably have been specific.

[quote=Rephrased Problem]
Let $ABC$ be a triangle with $H$ orthocenter and $O$ circumcenter. Let $AH$ meet $BC$ at $D$ and let $E$ be a point on $AB$ such that $\angle EDO = 90^\circ$. Show that $\angle DHE = \angle ABC$
[/quote]

[quote=Shishkin]It is shortlisted problem. (1996, G3)[/quote]

Indeed, see [url = https://artofproblemsolving.com/community/c6h1133p3555] ISL 1996 G3 [/url].


the angles are x, 3x-18, and 168-4x

if we set up inequalities, we see that x must be between 20 and 39 inclusive. These are the values of x.

[quote="Trisha"]HI,
I need help with the following problem.  Having a problem setting it up.  Please show steps to take.  Thanks. Trish

[color=green]In an obtuse triangle, the measure of one acute angle is 18* less than three times the measure of the other acute angle.  What are the possible measures of the smaller acute angle?[/color][/quote]

[hide="answer"]
0° < x < 27°
[/hide]


group theory
abstract algebra
invariant
superior algebra
superior algebra solved

Let G be a finite group and H is a subgroup of G containing all elements of the form $x^2$. where x belongs to G.
 What can you say about H?

Hmmm... Would it be wrong to say that $\left|G\right|=2\left|H\right|$?

well what do you mean
THE subgroup generated by all squares or just some group containing all squares

that equality certainly is not correct think of $(\mathbb{Z}_3,+)$ where every element is square

Let $S=<x^2|x\in G>$. Then $S$ is a characteristic subgroup of $G$ and $G/S$ is abelian, of exponent $2$. Since $S\leq H\leq G$, we have $H/S\leq G/S$, and since $G/S$ is abelian, $H/S$ is a normal subgroup of $G/S$. Hence $H$ is a normal subgroup of $G$, and $G/H$ is an abelian group of exponent $2$.

How do you say that S is characteristic.

A subgroup of $G$ is called characteristic iff it remains the same under all automorphisms of $G$. The trivial subgroups $\{e\}$ and $G$ itself are always characteristic for example, and so are the squares. The reason is that being a square of some element isn't influenced by automorphisms.

A characteristic subgroup is a subgroup invariant under all automorphisms. Any automorphism must fix the set of squares ($a^2\to (\phi(a))^2$), so it fixes the group generated by them.

A characteristic group is also normal, so the quotient makes sense. The other fact used here is that if all elements of a group have order 1 or 2, the group is abelian.


Find x in this equation:
$ 3^x$ $ \plus{}$ $ 4^x$ $ \equal{}$ $ 5^x$

By FLT, the only solution is
$ x\equal{}2$

[quote="behdad.math.math"]What is FLT?[/quote]
[hide]fermat's little thm.[/hide]

Actually, TFPNH referred to Fermat's Last Theorem, which eliminates any positive integer solution larger than 2 ($ a^n \plus{} b^n \equal{} c^n$ never holds for $ n \in N, n > 2, a,b,c \in N$). But the problem is about real $ x$:
$ 3^x \plus{} 4^x \equal{} 5^x \implies (\frac {3}{5})^x \plus{} (\frac {4}{5})^x \equal{} 1$. The LHS is decreasing for positive $ x$, and since $ x \equal{} 2$ is a solution, it must be the only positive real solution. If $ x\le 0$, then $ (\frac {3}{5})^x \plus{} (\frac {4}{5})^x \ge 2 > 1$. Thus, $ x \equal{} 2$ is also the only real solution.

More easy:

$ \left(\frac {3}{4}\right)^x \plus{} 1 \equal{} \left(\frac {5}{4}\right)^x$

Left side is monotonous decreesing function, right - increesing. This means that number of roots $ \leq 1$. But $ x \equal{} 2$ is a root, so it is unique.  :D

x should be an even number.
consider x=2k then $ 9^k$ $ \plus{}$ $ 16^k$ $ \equal{}$ $ 25^K$
we ca write $ 25^k$ = $ (16\plus{}9)^K$
and now K>1 is wrong.


Let $ n$ be integer, and let $ a,b,c$ be real numbers such that $ a\plus{}b\plus{}c\equal{}0$. Then when $ n\geq1$,

$ abc(a^{2n\minus{}1}\plus{}b^{2n\minus{}1}\plus{}c^{2n\minus{}1}) \ge 0$;

when $ n\leq 1$, the inequality sign is reversed.

2 cases are possible for $ n \geq 1$
C1) $ abc \geq 0$ : 1 positive number, 2 negative numbers . For ex. $ b,c \leq 0 , a \equal{} \minus{}b\minus{}c$  :  $ a^{2n\minus{}1}\plus{} b^{2n\minus{}1}\plus{} c^{2n\minus{}1} \equal{} \sum_{k\equal{}0}^{2n}C_{2n\minus{}1}^k (\minus{}b)^k(\minus{}c)^{2n\minus{}1\minus{}k }\geq 0$
C2) $ abc<0$ : 1 negative number, 2 positive number . For ex . $ b,c >0 , a\equal{}\minus{}b\minus{}c$  : $ \minus{}a^{2n\minus{}1}\minus{} b^{2n\minus{}1}\minus{} c^{2n\minus{}1} \equal{} \sum_{k\equal{}0}^{2n}C_{2n\minus{}1}^k b^kc^{2n\minus{}1\minus{}k } > 0$
.....
For $ n \equal{} 1\minus{}m <1 , m\geq 1$ the inequality is equivalent as $ a^{2m\minus{}1}b^{2m\minus{}1}\plus{}b^{2m\minus{}1}c^{2m\minus{}1}\plus{}c^{2m\minus{}1}a^{2m\minus{}1} \leq 0$
1) 1 positive number, 2 negative numbers , $ b,c <0 , a>0$ . From $ C1$ we obtain : $ a^{2m\minus{}1}\geq \minus{} b^{2m\minus{}1}\minus{}c^{2m\minus{}1}$, so
$ a^{2m\minus{}1}b^{2m\minus{}1}\plus{}b^{2m\minus{}1}c^{2m\minus{}1}\plus{}c^{2m\minus{}1}a^{2m\minus{}1} \leq \minus{}(b^{2m\minus{}1}\plus{}c^{2m\minus{}1})^2$ $ \plus{} b^{2m\minus{}1}c^{2m\minus{}1}$ $ \equal{} \minus{}b^{4m\minus{}2}\minus{}b^{2m\minus{}1}c^{2m\minus{}1}\minus{}c^{4m\minus{}2} \leq 0$
2)1 negative number, 2 positive number . For ex . $ b,c >0 , a<0$ . From $ C2$ we obtain $ a^{2m\minus{}1}\leq \minus{} b^{2m\minus{}1}\minus{}c^{2m\minus{}1}$, so
$ a^{2m\minus{}1}b^{2m\minus{}1}\plus{}b^{2m\minus{}1}c^{2m\minus{}1}\plus{}c^{2m\minus{}1}a^{2m\minus{}1} \leq \minus{}(b^{2m\minus{}1}\plus{}c^{2m\minus{}1})^2 \plus{} b^{2m\minus{}1}c^{2m\minus{}1}$$ \equal{} \minus{}b^{4m\minus{}2}\minus{}b^{2m\minus{}1}c^{2m\minus{}1}\minus{}c^{4m\minus{}2} \leq 0$
 :cool:


here's what i think
i think we should finish the first tourney, since it's almost over
the team tourney, however, will not get finished in time, so i think that should be suspended
CTC should keep going in the background

maybe CTC could be like a mod in FTWOS
others can make tournies, but if there are too many, CTC can cancell some
in june, CTC can assume it's original position as the head of tourney creation

[quote="Dojo"]CTC seems to be losing interest.
The most active forum CTC related seems to be the board member discussion...

The school year has started and there is not much time for FTW tourneys anyways. CTC was very helpful during the summer but the school year is preeeeety busy anyway. (I'm writing this post even though my school hasn't started)

One more thing. The president(/supreme dictator) is not in a comfortable position, at least from my point of view. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223856]Here[/url] (Sorry mewto)

Maybe CTC should be started up again in June.[/quote]

It is not losing interest, there just haven't been any recent elections. Also, plenty of administrations have conitnued operating through scandals, whether real or imagined. Vallon, why are you trying to close CTC down? You seem to be the most vocal opponent of CTC yet you are part of its government.

i do not wish to close down CTC, but it just seems less necessary now that school has started
not many people go on AoPS during school
however, i am gonna stop talking to prevent getting into this deeper.... :D

you guys can do what you want
i'll be checking AoPS once in a while, but i won't be on as often, seeing as school is starting

yeah, after the team tourney is done. i'm also gonna leave ctc. wel it doesn't really matter cause i'm not really gonna come on.

If you volunteer to make a tourney, and people vote yes, you will be able to make a tourney. In any case, if you drop CTC and make a tourney, who will join?

well i'm not gonna really start any tourneys. i'm just not really gonna haev time to join any tournies.

[quote="jain_harshi"]Well, I am quitting! I hate to not make tourneys, so yeah. I am leaving CTC![/quote]

Silly jainjain. JUst pm me or another CTC board member asking to make a tournament and we'll see what we can do!

mewto, tahts what jain wants. he wants to start tournies without having to ask somone for permission.

as i said before, we should make CTC more like a mod of tourneys, instead of a government
let members make the tourneys, and then CTC will just certify it
if there's too many tourneys going on, CTC can shut down a few

and no, mewto, i'm not trying to shut down CTC
i'm just giving you my opinion on what has been said the last few posts


summer program
Mathcamp
ARML
USAMTS
AMC
AIME

How'd Round 4 go for everyone?

I'm so happy :D  this week- I got an invitation for Mathcamp, made the MN ARML team, and, now, got a perfect score on this round!  Yay!

No commendations- you can only get so much  :lol:

Hurrah!  Gold Winner! :first:

Alright alright now Mr. Conceited...  :lol: lol just kidding. I dont blame you. If I am ever that good I would brag all the way to China. Anyway, I did ok in this round 17, 84 total. Hope to do better next year.

Congrats btw

Heh, I'm pretty happy too.  I got a 25, first 4 commended!  :D  Congrats everyone!

25, two commended.  I'm happy about the number of commendations, though on one problem (that wasn't commended), the grader got really nitpicky and said "you didn't prove it was the ONLY solution", though I'm pretty sure I did. :P  98 total, so I'm pretty happy with my performance.

25 this round, 100 total.  Commendations requirements have been getting much stricter...zero this round again...but I think that's good.

Congratulations everyone!

18 this round, 81 total.  I think it's funny how the two scores form a palindrome. :P

Overall, I'm pleased with my score.  I made a few stupid mistakes here and there, but I'm still a "Silver Winner" so I can't complain, right?  Last year, my score was about 8-10 points higher, but I believe the grading was a bit tougher this year... not that I mind, it just means I have to work harder, which isn't necessarily a bad thing.

Next year, I'm aiming to be a Gold Winner.

I didn't attend USAMTS this year but it seems that people who post here always have high score (BTW, congrats to everyone for their scores and wow, mathfanatic, your score is perfect..).

How do you practice for USAMTS? :blush:

I'm just hoping to make AIME through it in case I mess up on the AMC again. :(

I turned in nothing for #4, but I got a 5 on all other problems (20 points) and also commended on 2, 3, and 5 (funny how round 1 and 4 are the only ones where I got any solution commended!). 93 total for the year. I'll glady take it.  :)

Just wondering, how many points do you need to be a Gold Winner?

YAY!!! I got a total score of 54!!! in the 8th grade. 13 this last round!

[quote="litoxe2718281828"]Just wondering, how many points do you need to be a Gold Winner?[/quote]96 or higher.  There were 35 Gold Level winners (out of 662 first round participants).

[quote="JGeneson"]I got 25 w/ 1 commended. I am happy with this result.[/quote]
Same here :D

[quote="likoucalvinhao"]Congrats everyone,

Mine was really nit-picky, I got a 4 on the second one because I didn't "show" that a+b+c=219, overall was a 19, first time.[/quote]
I also got a point docked from my score on problem #2 for a trivial "error." Basically the grader said "you lose a point because you didn'y plug your solutions back in to see if they work." I thought this was totally bogus -- the grader can take off points because I didn't double-check my work?? I'll post the solution later today, because I think it is ridiculous that a correct, fairly well-written solution should recieve less than 5 points.  :huh: Enough complaining though. :P

I'm happy though, because I got an 88 overall (7- or 8-point improvement over last year) with a 19 on this round (didn't solve #3) + a commendation on problem #1. :) Congratulations everyone!

[quote="ZennyK"]
I also got a point docked from my score on problem #2 for a trivial "error." Basically the grader said "you lose a point because you didn'y plug your solutions back in to see if they work." I thought this was totally bogus -- the grader can take off points because I didn't double-check my work?? I'll post the solution later today, because I think it is ridiculous that a correct, fairly well-written solution should recieve less than 5 points.  :huh: Enough complaining though. :P [/quote]
I didn't think much about it when I wrote my solution either, but I'm guessing you squared both sides of the given equations at some point? You could get extra roots this way (like $\sqrt{x}=-1\Rightarrow x=(-1)^2=1$)...that's probably why the grader deducted a point.

This competition sounds really fun. I'll be participating next year.


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ F$ be a finite field.
$ F[x]$ contains infinitely many irreducibles.
Why so?

Assume there are only finitely many $ p_1,...,p_n$, and consider $ f(x)\equal{}1 \plus{} \prod p_i(x)$

[quote="Jacobson"]Assume there are only finitely many $ p_1,...,p_n$, and consider $ f(x) \equal{} 1 \plus{} \prod p_i(x)$[/quote]

In $ Z_5[x]$, $ x \minus{} 1$, $ x \minus{} 2$, $ x \minus{} 3$, $ x \minus{} 4$ are irreducibles.
But $ f(x) \equal{} 1 \plus{} (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} 4)$ is also irreducible because $ f(0) \equal{} 0$.
How can I prove my question?

Jacobson's comment is a generalization of Euclid's proof that there are infinitely many primes.  If $ p_1, p_2, ... p_n$ is an enumeration of all irreducible polynomials and they have degree $ d_1, d_2, ... d_n$, then $ f(x)$ has degree greater than any of these (so it can't be irreducible) but is not divisible by any of $ p_1, p_2, ... p_n$ (so it can't be reducible); contradiction.  Hence there are infinitely many irreducible polynomials.

In fact, it's possible to count the number of irreducible polynomials of degree $ n$.  Count the number of monic polynomials of degree $ n$ (of course, $ |F|^{n \minus{} 1}$) and then the number of monic reducible polynomials of degree $ n$ (which are a product of some irreducible polynomial and some polynomial, both of lower degree than $ n$).


If $ a$, $ b$ and $ c$ are the sidelengths of a triangle, then prove the inequality

$ \frac {\left(b \plus{} c\right)^{2}}{a^{2} \plus{} bc} \plus{} \frac {\left(c \plus{} a\right)^{2}}{b^{2} \plus{} ca} \plus{} \frac {\left(a \plus{} b\right)^{2}}{c^{2} \plus{} ab}\geq 6$.

Now that the deadline has passed, I will post the beautiful solution to this inequality:
$ \frac{(b\plus{}c)^{2}}{a^{2}\plus{}bc}\plus{}\frac{(c\plus{}a)^{2}}{b^{2}\plus{}ca}\plus{}\frac{(a\plus{}b)^{2}}{c^{2}\plus{}ab}\geq 6$
$ \Leftrightarrow \sum_{cyc}(\frac{(b\plus{}c)^{2}}{a^{2}\plus{}bc}\minus{}2)\ge 0$
$ \Leftrightarrow \sum_{cyc}\frac{(b\plus{}c)^{2}\minus{}2a^2\minus{}2bc}{a^{2}\plus{}bc}\ge 0$
$ \Leftrightarrow \sum_{cyc}\frac{b^2\plus{}c^2\minus{}2a^2}{a^2\plus{}bc}\ge 0$
$ \Leftrightarrow \sum_{cyc}(\frac{b^2\minus{}a^2}{a^2\plus{}bc}\plus{}\frac{a^2\minus{}b^2}{b^2\plus{}ac})\ge 0$
$ \Leftrightarrow \sum_{cyc}\frac{(b^2\minus{}a^2)((b^2\plus{}ac)\minus{}(a^2\plus{}bc))}{(a^2\plus{}bc)(b^2\plus{}ac)}\ge 0$
$ \Leftrightarrow \sum_{cyc}\frac{(b^2\minus{}a^2)((b\minus{}a)(b\plus{}a)\minus{}c(b\minus{}a))}{(a^2\plus{}bc)(b^2\plus{}ac)}\ge 0$
$ \Leftrightarrow \sum_{cyc}\frac{(b\minus{}a)(b\plus{}a)(b\minus{}a)(a\plus{}b\minus{}c)}{(a^2\plus{}bc)(b^2\plus{}ac)}\ge 0$
$ \Leftrightarrow \sum_{cyc}\frac{(b\minus{}a)^{2}(b\plus{}a)(a\plus{}b\minus{}c)}{(a^2\plus{}bc)(b^2\plus{}ac)}\ge 0$
which is true because $ (b\minus{}a)^2\ge 0, (b\plus{}a),(a^2\plus{}bc),(b^2\plus{}ac)\ge 0$ as $ a,b,c$ are positive integers (as they are sides of a triangle), and $ a\plus{}b\minus{}c>0$ as $ a,b,c$ are sides of a triangle. Then, the original inequality follows immediately. :lol:

[quote="canada"]If $ a$ $ b$ and $ c$ are side lengths of triangle prove the inequality:

$ \frac {(b \plus{} c)^{2}}{a^{2} \plus{} bc} \plus{} \frac {(c \plus{} a)^{2}}{b^{2} \plus{} ca} \plus{} \frac {(a \plus{} b)^{2}}{c^{2} \plus{} ab}\geq 6$.[/quote]
It's true for all positive $ a,$ $ b$ and $ c.$ :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60128

Another solution: WLOG suppose that $ a\ge b\ge c$. Then $ a^2\plus{}bc\ge b^2\plus{}ca\iff (a\minus{}b)(a\plus{}b\minus{}c)\ge 0$, which is true. Thus we get $ a^2\plus{}bc\ge b^2\plus{}ca\ge c^2\plus{}ab$. Combining this with $ (b\plus{}c)^2\le (c\plus{}a)^2\le (a\plus{}b)^2$ we have from chebyshev's inequality,
\[ \sum\frac{(b\plus{}c)^2}{a^2\plus{}bc}\ge 3\left(\frac{(b\plus{}c)^2\plus{}(c\plus{}a)^2\plus{}(a\plus{}b)^2}{a^2\plus{}bc\plus{}b^2\plus{}ca\plus{}c^2\plus{}ab}\right)\equal{}6\]
With equality iff $ a\equal{}b\equal{}c$.


some convex poligons are colored red or blue, every poligons of different dolors intersect, and there are 2 red disjoint poligons, and 2 blue disjoint poligons, prove there is a line that intersects all the poligons.

I don't really know why this is here: "there are 2 red disjoint poligons, and 2 blue disjoint poligons". It's the most difficult case of the problem obtained without considering it, but we don't need it, do we? :?

Take any line in the plane, and project the polygons on it. We get blue segments and red segments. We are told that at all times, all the red segments intersect all the blue segments. We have to show that there is a line s.t. when we project the whole thing on it, every segment intersects every other segment.

When there are two disjoint red segments, all the blue segments contain the segment between the two red segments, so there are no disjoint blue sgements. The same is true if we reverse colors.

Now take a line and start rotating it. The situation changes continuously, so if there's a case when there are two disjoint red segments and another case when there are two disjoint blues segments, then between those positions of the line, there must be a position when there are no disjoint segments, because at no time can we have both two disjoint red segments and two disjoint blue segments.


calculus
integration
trigonometry
calculus computations

Calculate the following indefinite integrals.

[1] $\int (4-5\tan x)\cos x dx$

[2]  $\int \frac{dx}{\sqrt[3]{(1-3x)^2}}dx$

[3] $\int x^3\sqrt{4-x^2}dx$

[4] $\int e^{-x}\sin \left(x+\frac{\pi}{4}\right)dx$

[5] $\int (3x-4)^2 dx$

For 1,2 and 5, You are right, amirhtlusa.

How about 3 and 4?

[4]
$\sin (x+\frac{\pi}{4}) = \frac{1}{\sqrt2}(\sin x + \cos x)$.
$\int e^{-x}\sin(x+\frac{\pi}{4}) = \frac{1}{\sqrt2}\left(\int e^{-x}\sin xdx + \int e^{-x}\cos xdx\right)$.

$\int e^{-x}\sin xdx + \int e^{-x}\cos xdx = \int e^{-x}d(-\cos x) + \int e^{-x}\cos xdx = -e^{-x}\cos x - \int (-e^{-x})(-\cos x)dx + \int e^{-x}\cos xdx = -e^{-x}\cos x + c$

Hence $\int e^{-x}\sin(x+\frac{\pi}{4}) = -\frac{e^{-x}\cos x}{\sqrt 2}+\mbox{const}$