geometry
circumcircle
cyclic quadrilateral
projective geometry
geometry proposed

[quote="28121941"]I guess AK and AS are perpendicular, right? The symbol was not readable. Thanks[/quote]
You were right, why do I be able to read it?

Dear mr.danh, it is not your fault....I think is a problem of the compatibility between different processors.

This problem seems be forgotten by everybody....
If the circumcircles of KBD and KCE meet at A, then A would be the center of direct similarity which transforms ABC in ADE, if I am not wrong....I will still thinking about it

[quote="mr.danh"]Let a acute triangle ABC inscribed circle (O). SB,SC are two tangents of (O). K is located on line BC such that AK⊥AS. The circle centred by S passes through B,C meets line SK at D,E. Prove that $ \triangle ABC\sim\triangle ADE$.[/quote]
Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=982020#982020

1.Let KA meet Circle O at G  then G,C,E are collinear
2.A,B,D,K are concircle  => ∠ABC=∠ADE
3.∠BAS=∠BES=90-∠BDE=90-∠KAB A,B,S,E are concircle
4.∠SCA=∠ABK=∠ADK A,D,S,C are concircle  => ∠DAS=∠SAC
5.△DAE∽△BAC
DONE!

[quote="mr.danh"]Let a acute triangle ABC inscribed circle (O). SB,SC are two tangents of (O). K is located on line BC such that AK⊥AS. The circle centred by S passes through B,C meets line SK at D,E. Prove that $ \triangle ABC\sim\triangle ADE$.[/quote]
Thanks , dear 28121941.
But I guess triangle ABC need not to be an acute triangle. Is it right ? dear mr.danh  :)

[quote]Plane Geometry wrote:
But I guess triangle ABC need not to be an acute triangle. Is it right ? dear mr.danh[/quote]
I think mr. danh is right. Cause otherwise the problem is leading to two types of similarity, both of which resulting in sum of the angles of a triangle > 180

[quote="jayme"]Dear Plane Geometry,
How do you prove that G, C and E are collinear?
Sincerely
Jean-Louis[/quote]
oh， :blush: ，I didn't notice that before you mention it.I am still working on it now
Dear Jean-Louis , do you have any suggestions?Thank you in advance!

[quote="jayme"]Dear Plane Geometry,
let 0 the circumcircle of ABC, O the center of 0, G the second point of intersection of BD and 0, M the second point of intersection of BE and 0, G' the second point of intersection of CE and 0, M' the second point of intersection of CD and 0.
Let 1 the circle with center S passing through B.
By definition, 0 and 1 are orthogonal. It follows that DE is perpendicular GOM and G'OM'. Therefore, G=G' and M=M'.
First conclusion : M is the orthocenter of GDE.
Let 2 the circumcircle of BDK and A' the second point of intersection of 0 and 2.
By the pivot theorem, A', G and K are collinear.
Now we have to prove that A'=A.
Let Tc the tangente to 0 at C.
I think to applied the pascal theorem in a particular case : GA'MBC Tc G; it follows that S, M and A' are collinear. Then A'=A.
The rest follows...
I think that there is a better approach.
Sincerely
Jean-Louis[/quote]
Thank you my dear friend Jean-Louis.I learn from you a lot!
 :P

Let be the points $ A'\equiv (O)\cap AS,\ G\equiv (O)\cap A'O$ and so, we have that $ AG\perp AA'.$

Let be the points $ D\equiv BG\cap A'C,\ E\equiv CG\cap A'B$ and so, we have that the quadrilateral $ DBCE$ is cyclic with diameter the segment $ DE.$

Clearly the point $ A'$ is the orthocenter of the triangle $ \bigtriangleup GDE$ and so, the tangent lines of $ (O)$ at points $ B,\ C,$ pass through the midpoint of the segment $ DE,$ as well ( and easy to prove ).

Hence, we conclude that the point $ S,$ is the midpoint of the segment $ DE$ and consequently, the circumcenter of the cyclic quadrilateral $ DBCE.$

$ \bullet$ Let be the point $ K\equiv BC\cap DE$ and we will prove that the points $ K,\ A,\ G,$ are collinear.

From the complete quadrilateral $ GBA'CDE,$ we have that the points $ K,\ B,\ L\equiv BC\cap A'G,\ C,$ are in harmonic conjugation.

Based on the bellow [b][size=100]Lemma[/size][/b], we conclude that the point $ K'\equiv BC\cap AG,$ is the harmonic conjugate of $ L,$ with respect to the points $ B,\ C.$

Hence, we conclude that $ K'\equiv K$ and then, the points $ K,\ A,\ G,$ are collinear.

$ \bullet$ It is easy to show now, that the quadrilateral $ ABDK$ is cyclic, because of $ \angle KAB = \angle BCG = \angle BDE$ $ ,(1)$

So, we have $ \angle AKB = \angle ADB$ $ ,(2)$ and then, $ \angle ADE = \angle ADB + \angle BDE = \angle AKB + \angle KAB = \angle ABC$ $ \Longrightarrow$ $ \angle ADE = \angle ABC$ $ (3)$

Because of $ (KD)\cdot (KE) = (KB)\cdot (KC) = (KA)\cdot (KG),$ we conclude that the quadrilateral $ ADEG$ is cyclic and then we can easy to show now that $ \angle AED = \angle AGD\equiv AGB = \angle ACB$ $ ,(4)$

From $ (3),$ $ (4)$ we conclude that the triangles $ \bigtriangleup ABC,\ \bigtriangleup ADE,$ are similar and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ D$ be, the intersection point of its circumcircle $ (O),$ from the $ A$-symmedian. We denote as $ D',$ the antidiametric ( = diametric opposite ) point of $ D$ wrt $ (O)$ and let be the points $ K\equiv BC\cap AD'$ and $ L\equiv BC\cap DD'.$ Prove that the points $ K,\ B,\ L,\ C,$ are in harmonic conjugation.[/color][/size][/b]

Kostas Vittas.

PS. I will post here later the easy proof of the above Lemma have in mind.

[quote="plane geometry"]1.Let KA meet Circle O at G  then [b]∠GCB=∠BAK=∠BDE=180-∠BCE => G,C,E are collinear[/b]
2.A,B,D,K are concircle  => ∠ABC=∠ADE
3.∠BAS=∠BES=90-∠BDE=90-∠KAB A,B,S,E are concircle
4.∠SCA=∠ABK=∠ADK A,D,S,C are concircle  => ∠DAS=∠SAC
5.△DAE∽△BAC
DONE![/quote]
[b]EDIT.[/b] I'm sorry  :blush: I was looked at your image  :D

[quote="April"][quote="plane geometry"]1.Let KA meet Circle O at G  then [b]∠GCB=∠BAK=∠BDE=180-∠BCE => G,C,E are collinear[/b]
2.A,B,D,K are concircle  => ∠ABC=∠ADE
3.∠BAS=∠BES=90-∠BDE=90-∠KAB A,B,S,E are concircle
4.∠SCA=∠ABK=∠ADK A,D,S,C are concircle  => ∠DAS=∠SAC
5.△DAE∽△BAC
DONE![/quote][/quote]
Well,dear April,when you are in step 1,you don't have ABDK concircle,why ∠GCB=∠BAK=∠BDE=180-∠BCE ?

Dear Mathlinkers,
Dear Mathlinkers,
as I say, I wasn't satisfied with my proof. Kostas Vittas has opened the way of my new proof.
1. The begining and the notations of Kostas:
let 0 the circumcircle of ABC, O the center of 0, A' the second point of intersection of AS with 0, G the second point of intersection of OA' witn 0, 1 the circle centered at S and passing through B, D the second point of intersection of GB with 1, E the second point of intersection of GC with 1.
2. By définiton of 0 and 1, 1 is orthogonal to 0.
3. BA' BG ; G, C and E are collinear ; BG BE (1 is orthogonal to 0) ; it follows BA'//BE, BA'=BE ; B, A' and E are collinear.
Mutatis mutandis,	C, A' and D are collinear.
4. Let Tb the tangent to 0 at B, Tc the tangent to 0 at C.
5. Accordling to Pascal's theorem applied to TbGCTcA'B,	S, D and E are collinear.
6. Accordling to Boutin's theorem,			S is the midpoint of DE (see http://perso.orange.fr/jl.ayme  vol. 1 A propos du théorème de Boutin)
7. AG AA'S and AA'S AK; then AG//AK, AG=AK, A, G and K are collinear.
8. Accordling to Pascal's theorem applied to AGBTbCA'A,	K, D and S are collinear (with E).
9. Accordling to Pivot theorem,				K, A, B and D are concyclic.
10. Accordling to Monge's theorem,			A, D, E and G are concyclic.
11. The rest follows by an angle chasing.

Thank to Kostas
Sincerely
Jean-Louis

Dear Mathlinkers and Moderator,
I forgot "ortho" for perpendicular in my last message; perhaps the moderator can help me. If yes, you have the possibility to destroy my last message. Thanks in advance.

As I say, I wasn't satisfied with my proof. Kostas Vittas has opened the way of my new proof.
1. The begining and the notations of Kostas:
let 0 the circumcircle of ABC, O the center of 0, A' the second point of intersection of AS with 0, G the second point of intersection of OA' witn 0, 1 the circle centered at S and passing through B, D the second point of intersection of GB with 1, E the second point of intersection of GC with 1.
2. By définiton of 0 and 1, 1 is orthogonal to 0.
3. BA' ortho BG ; G, C and E are collinear ; BG ortho BE (1 is orthogonal to 0) ; it follows BA'//BE, BA'=BE ; B, A' and E are collinear.
Mutatis mutandis, C, A' and D are collinear.
4. Let Tb the tangent to 0 at B, Tc the tangent to 0 at C.
5. Accordling to Pascal's theorem applied to TbGCTcA'B, S, D and E are collinear.
6. Accordling to Boutin's theorem, S is the midpoint of DE (see http://perso.orange.fr/jl.ayme vol. 1 A propos du théorème de Boutin)
7. AG ortho AA'S and AA'S ortho AK; then AG//AK, AG=AK, A, G and K are collinear.
8. Accordling to Pascal's theorem applied to AGBTbCA'A, K, D and S are collinear (with E).
9. Accordling to Pivot theorem, K, A, B and D are concyclic.
10. Accordling to Monge's theorem, A, D, E and G are concyclic.
11. The rest follows by an angle chasing.

Thank to Kostas
Sincerely
Jean-Louis


I'm a novice in Asymptote, and I never really dealt with drawing solid shapes. What are the basics to drawing the shapes (just one line of code kind of stuff, like draw(unitsphere); maybe?)

I don't really know much about solids either but I did see:
[asy]size(200);
import solids;
sphere(O,1).filldraw(green,3,blue);[/asy]

Somebody should sticky [url=http://piprim.tuxfamily.org/asymptote/]this site[/url].. looking at the examples on that page answers a lot of the questions people are asking here.


Find all integers which are equal to the square of their number of positive divisors.

..oh.  :D

Then why not post the problem source?

I forgot, since in the BC there does not always appear a field for the problem source.

So did anyone of you think of a solution till now? C'mon, it's really simple!

Maybe a hint should help:
The decomposition of n in primes factors is
(1) n = prod_(k \geq 1)(p_k)^a_k
find the expression of d(n) in terms of a_k.
From the equation n=(d(n))^2, what can you say about the parity of
a_k in (1)
and ....


Suppose I want to set an image in the center of the page and then, write some text in it. How to make it? Besides, how to turn that image a bit transparent?

To make an image transparent you need to alter the image so you need an image manipulation program (eg Paint.NET or GIMP or a zillion others). Any such program will allow you to add text to the image.
If however you want a background image on the page then use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=wallpaper]wallpaper package[/url].


inequalities
geometry
circumcircle
geometry proposed

Given triangle $ABC$ let $M_{a},M_{b},M_{c}$ are the length of median extened and intersect with circumcenter,three median $m_{a},m_{b},m_{c}$ and three internal bisector $l_{a},l_{b},l_{c}$ prove that
$M_{a}+M_{b}+M_{c}\ge \frac{2}{3}(m_{a}+l_{a}+m_{b}+l_{b}+m_{c}+l_{c})$
 :D

I think you can strengthen the RHS by $\frac{4}{3}(m_{a}+m_{b}+m_{c})$ by using the fact that $M_{a}= \frac{b^{2}+c^{2}}{m_{a}}$


calculus
integration
derivative
logarithms
trigonometry
calculus computations

$\int cosec^{3}x$
$\int cosec^{4}x$

is this correct for (a)?
[hide]$\int cosec^{3}x.dx$=$\int cosec^{2}x.cosecx.dx$
=$\int (cot^{2}x+1)cosecx.dx$

$u=cotx$ $\frac{du}{dx}=-cosec^{2}x$

but how can i proceed from this[/hide]

That method works well for integrating $\csc^{4}x$, and in fact, $\csc^{n}x$ for even $n$. Finding an antiderivative of $\csc^{n}x$ for odd $n$ is a bit trickier. You can use integration by parts with $dv =\csc^{2}x\ dx$: for the particular case $n=3$, you can set $u=\csc x,\ dv=\csc^{2}x\ dx$, which gives
[hide="this"]\[\int_\frac{\pi}{2}^{a}\csc^{3}x\ dx=-\csc a\cot a-\int_\frac{\pi}{2}^{a}\csc x\cot^{2}x\ dx\]
(the derivatives of $\cot x,\ \csc x$ are $-\csc^{2}x,\-\csc x\cot x$). For that second integral, use $\cot^{2}x=\csc^{2}x-1$. Then the right hand side will have a $\int_\frac{\pi}{2}^{a}\csc x\ dx$ term and another $\int_\frac{\pi}{2}^{a}\csc^{3}x\ dx$. The first of these integrals is equal to $\ln |\csc a-\cot a|$, and it remains to isolate $\int_\frac{\pi}{2}^{a}\csc^{3}x\ dx$. You end up with
\[\int_\frac{\pi}{2}^{a}\csc^{3}x=\frac{1}{2}(\ln |\csc a-\cot a|-\csc a\cot a)\][/hide]

how do u integrate $cosecx$?

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

how would I go about solving this?

$\int cos^\frac{2}{3}xsin^{5}x$

thanks scorpious for the above explanation!

p.s are these qns considered difficult for [b]school[/b] standard?

$P = \int\cos^{\frac{2}{3}}x \sin^{5}x\,dx$.

The strategy here is to convert that product in another one(s) of the form

$\int [f(x)]^{a}\cdot f'(x)\,dx = \frac{[f(x)]^{a+1}}{a+1}$.

In our case, as we have an odd power of $\sin x$, let's isolate one $\sin x$ and the develop the $\sin^{4}x$, remembering that $(\cos x)' =-\sin x$:

$P = \int \cos^{\frac{2}{3}}x(\sin^{2}x)^{2}\sin x\,dx = \int \cos^{\frac{2}{3}}x(1-\cos^{2}x)^{2}\sin x\,dx$

$= \int \cos^{\frac{2}{3}}x(1-2 \cos^{2}x+\cos^{4}x) \sin x\,dx$

$=-\int \cos^{\frac{2}{3}}x(-\sin x)\,dx+2 \int\cos^{\frac{8}{3}}x(-\sin x)\,dx-\int \cos^{\frac{14}{3}}x(-\sin x)\,dx$

$=-\frac{3 \cos^{\frac{5}{3}}x}{5}+\frac{6 \cos^{\frac{11}{3}}x}{11}-\frac{3 \cos^{\frac{17}{3}}x}{17}+C$.

To get the primitive of $\csc x$ start by remembering that

$\frac{d}{dx}\ln[f(x)] = \frac{f'(x)}{f(x)}$.

If you now make $f(x) = \csc x-\cot x$ (or $f(x) = \csc x+\cot x$, which will do the same), then substituting in the equality above we get

$\frac{d}{dx}\ln|\csc x-\cot x| = \frac{-\csc x \cot x+\csc^{2}x}{\csc x-\cot x}= \frac{\csc x (\csc x-\cot x)}{\csc x-\cot x}= \csc x$

and so $\int \csc x\,dx = \ln|\csc x-\cot x|+C$. Using the formulas that relate $\sin x$ and $\cos x$ with $\tan \frac{x}{2}$ we can show that

$\ln|\csc x-\cot x| = \ln\left|\tan \frac{x}{2}\right|$

and so we also have $\int \csc x\,dx = \ln\left|\tan \frac{x}{2}\right|+C$.

This proof is not certainly the better, because why would we remember to consider $f(x) = \csc x-\cot x$? Perhaps someone else can provide a much better proof.

$\int \csc x \, dx = \int \frac{\sin x}{1-\cos^{2}x}\, dx$. Now you substitute $y = \cos x$ to get the integral of a rational function.
([i]Mynd you,[/i] the subsequent computations [i]Kan be pretti nasti...[/i])

kk i see how it works now. Appreciate the various soln.

bye


Ring Theory
group theory
abstract algebra
linear algebra
Units

Show that the group of units of a unital ring can't have order $ 5$.  Generalize?

If the ring is $ \mathbb{Z}_n$ the ring of units have even order for $ n\ge 3$. But I'm not able to go further, I don't know rings very well... But maybe it's true for almost all finite (and some infinite) unital rings...

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=374436

The proof does not seem to use commutativity.

I think that if $ p$ is an odd prime such that $ p\not\in\left\{2^n\minus{}1\mid n\in\mathbb{N}\right\}$, then there exists no unitary ring with exactly $ p$ units.
I hope my proof is correct... (it's rather easy, maybe too easy).

  darij

[quote="darij grinberg"]I think that if $ p$ is an odd prime such that $ p\not\in\left\{2^n \minus{} 1\mid n\in\mathbb{N}\right\}$, then there exists no unitary ring with exactly $ p$ units.
I hope my proof is correct... (it's rather easy, maybe too easy).

  darij[/quote]

I finally understand this comment, from my experience with units in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=292862]this thread[/url].  Yes, it's not so bad, as far as I can see--the characterstic $ k$ must be 2, (since $ \phi(k)$ divides $ p$) and since $ x^p \minus{} 1$ has no repeated roots, we are looking at a product of finite fields.

I was just asked to post a proof of this:

[quote="darij grinberg"]I think that if $ p$ is an odd prime such that $ p\not\in\left\{2^n-1\mid n\in\mathbb{N}\right\}$, then there exists no unitary ring with exactly $ p$ units.[/quote]

Since I don't remember my short proof of olden days, here is a long one:

Let $p$ be an odd prime such that $ p\not\in\left\{2^n-1\mid n\in\mathbb{N}\right\}$. Let $R$ be a unitary ring with exactly $p$ units. We must obtain a contradiction from this.

Let $Z$ be the subset of $R$ consisting of all integer multiples of $1_R$ (where $1_R$ is the unity of $R$). Then, $Z$ is a subring of $R$, so that $Z^{\times}$ is a subgroup of $R^{\times}$. Thus, $\left|Z^{\times}\right| \mid \left|R^{\times}\right|$ (by Lagrange's theorem). Since $\left|R^{\times}\right| = p$ (because $R$ has exactly $p$ units), this becomes $\left|Z^{\times}\right| \mid p$, so that $\left|Z^{\times}\right|$ must be either $1$ or $p$ (since $p$ is prime). We thus must be in one of the following two cases:

[i]Case 1:[/i] We have $\left|Z^{\times}\right| = 1$.
[i]Case 2:[/i] We have $\left|Z^{\times}\right| = p$.

Let us consider Case 1 first. In this case, the ring $Z$ has only one unit. Since $-1_R$ and $1_R$ are units of $Z$, this yields that $-1_R=1_R$, thus $2\cdot 1_R=0$. Hence, $R$ is a $\mathbb F_2$-algebra.
Since $\left|R^{\times}\right| = p$, the group $R^{\times}$ must be cyclic of order $p$ (since the only groups of order $p$ are the cyclic ones), so that there exists a $\zeta\in R^{\times}$ which generates $R^{\times}$. Consider this $\zeta$. Clearly, $\zeta^p=1$. Thus, $\mathbb F_2\left[\zeta\right]$ is a finite subring of $R$. Since $\mathbb F_2\left[\zeta\right]$ has exactly $p$ units (in fact, it cannot have more, since $R$ has only $p$ units, and it cannot have less, since $\zeta$ generates $R^{\times}$ and thus all units in $R^{\times}$ are also units in $\mathbb F_2\left[\zeta\right]$), we can thus replace $R$ by $\mathbb F_2\left[\zeta\right]$. As a consequence, we can WLOG assume that $R$ is a finite ring. If the Jacobson radical of $R$ was nonzero, then there would be a nonzero $j$ lying in the Jacobson radical of $R$, and thus the number of units of $R$ would be even (because for any unit $r$ of $R$, the element $r+j$ of $R$ would also be a unit of $R$), which contradicts the fact that the number of units of $R$ is odd (since this number is $p$, and $p$ is odd). Thus, the Jacobson radical of $R$ is zero. Hence, $R$ is a semisimple ring (since $R$ is finite, thus Artinian). Consequently, by the Artin-Wedderburn theorem, $R$ is a direct product of matrix rings over field extensions of $\mathbb F_2$. This direct product can have only one factor non-isomorphic to $\mathbb F_2$ (because any factor non-isomorphic to $\mathbb F_2$ is a ring with more than $1$ unit, and the number of units of a direct product is the product of the numbers of units of the factors, but $p$ is prime). Hence, we have $R\cong \mathbb \mathbb F_2^k \times \mathrm{M}_n\left(\mathbb F_{2^m}\right)$ for some integers $k$, $n$ and $m$. Now, $p$ is the number of units of this ring, i. e., the number of units of the ring $ \mathrm{M}_n\left(\mathbb F_{2^m}\right)$ (since the number of units of $\mathbb F_2$ is $1$), and this is clearly equal to

$\left|\mathrm{GL}_n\left(\mathbb F_{2^m}\right)\right| = \left(\left(2^m\right)^n-\left(2^m\right)^0\right)\left(\left(2^m\right)^n-\left(2^m\right)^1\right)...\left(\left(2^m\right)^n-\left(2^m\right)^{n-1}\right)$.

For this to be prime (and it must be prime, since $p$ is prime), we need $n$ to be $1$, and thus $p=\left(2^m\right)^n-\left(2^m\right)^0=2^{mn}-1$, contradicting $ p\not\in\left\{2^n-1\mid n\in\mathbb{N}\right\}$. Thus, we have got a contradiction in Case 1.

Now, let us consider Case 2. In this case, $\left|Z^{\times}\right| = p$, so that $Z$ has exactly $p$ units. But the ring $Z$ (by its definition) has the form $\mathbb Z/u\mathbb Z$ for some $u\in\mathbb N$. Thus, the number of units of $Z$ is $\phi\left(u\right)$, where we let $\phi\left(0\right)$ denote $2$. Thus, $\phi\left(u\right)=p$. But this is absurd, since $p$ is an odd prime, whereas values of the Euler phi function are always either even or $1$. So we got a contradiction in Case 2, too, and we are done.

A shortening of the Case 1 argument: we have that $R$ is a quotient of $\mathbb{F}_2[X]/(X^p-1)$, as we have assumed that $R$ is characteristic 2 and generated by a unit of order $p$.  It follows from the Chinese Remainder Theorem that $R$ is a product of finite fields (as $X^p-1$ does not have repeated roots), each of which has $2^n-1$ units for some $n$.

"we can thus replace R by $\mathbb F_2[\zeta]$"
If $R_1$ is a subring of $R_2$, and both $R_1$ and $R_2$ have the same set of units, that doesn't mean $R_1=R_2$.

[quote="dominicleejun"]If $R_1$ is a subring of $R_2$, and both $R_1$ and $R_2$ have the same set of units, that doesn't mean $R_1=R_2$.[/quote]  The units of $\mathbb{R}[x]$ are $\mathbb{R}^\times$, aren't they?

The point is that the units of the generated ring form a nontrivial subgroup, which must be the entire group since the order is prime.  We can thus assume that $R$ is the generated ring, without harming the argument (since we seek a contradiction, not any particular knowledge about $R$).


For what positive value of $ x$ does $ \frac{\frac12x}3 \equal{} \frac{12}{\frac12x}$?

cross multiply
 1/4x^2=36

make the x^2 alone

x^2=144
x=12

[hide="answer"]ill answer this in a series of steps

1.i cross multiplied because its easier

2.so when cross multiplying it looks like
 (x/2)(x/2)=12*3

3.so when we multiply to each side we get that looks something like
36=x^2/4

4.then we multiply each side by 4 and get
144=x^2

5.then we sqrt it and get

6.our answer is $\boxed{12}$[/hide]

[b][i][u]please rate my post thank you so much!![/u][/i][/b]

[size=50]please be fair and please PM me if there is a mistake thank you so much!![/size]

:ddr: :ddr:

 :ddr:

 :ddr:

Divide through on each sides:

$\frac{x}{6}={24}{x}$

$x^2=144$

$x=\boxed{12}$


Write each expression in the standard form a + bi.

(1) ((sqrt{3}/2) - (1/2)(i))^2

(2) i^14

$\left(\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)^{2}$

HINT: USE the FOIL method

[hide="solution"]

$\frac{3}{4}-\frac{\sqrt{3}}{4}i-\frac{\sqrt{3}}{4}i-\frac{1}{4}=$

$\frac{1}{2}-\frac{\sqrt{3}}{2}i$

wow i probably did that wrong its past dinner time and im getting hungry  :lol:

[/hide]

[hide="2"]
$i^{14}$

since $i^{2}=-1$ and the exponent is even, $i^{14}=-1$
[/hide]

editted for stupid typo :blush:

editted soo many times for stupid LaTeX and hide errors  :lol:

What is the correct answer for number one?

Try to FOIL with the complex binomial and see what you get...i'm 80% sure mine is right but I could be wrong

the demon's answer to nimber 1 is correct. i get the same thing.


inequalities
limit
inequalities unsolved

$ Let (X_n)$ be a sequence defined by:
$ x_1 = 1, x_{n1} = \frac {1}{1 + x_n}$ for all $ n\ge 1$.

Show that $ \mid x_{n + 1 + k} - x_{n + 1}\mid \le \frac {4}{9} \mid x_{n + k} - x_{n}\mid$ and hence determine $ \lim_{x\to\infty}x_n$.

Can someone please help. Thank you very much. :P

I hope this will help you, although in written Japanese.

1985 Aoyama University entrance exam.


Ok, so when I was doing math, I stumbled upon this question:

Show that there are infinitely many primes of the form 4n+3.

Yeah. I don't get it. Someone help me? :huh:

Assume not. Then there are only finitely many primes of that form. Let's say there are k of them, namely $ p_i \equal{} 4n_i \plus{} 3$ for $ i \equal{} 1,2,...,k$. Consider the number $ N \equal{} 4(p_1p_2\cdots p_k \plus{} 1) \plus{} 6$. Now notice that every prime number has either the form $ 4n \plus{} 1$ or $ 4n \plus{} 3$. Multiplying numbers of the form $ 4n \plus{} 1$ together gives a number of the form $ 4n \plus{} 1$, so if $ N$ is composite, it has to have a prime factor of the form $ 4n \plus{} 3$, namely $ p_i$ for some $ i$. But $ N \equiv 10 \pmod{p_i}$, which is not congruent to $ 0$ for any of possible value of $ p_i$. This contradicts the assumption, so there are infinitely primes of that form.

You could've just taken $ N \equal{} 4p_1 ... p_k \minus{} 1$  :huh:

Oh yeah... thanks! :rotfl:  :rotfl:

This is sorta the same kind of proof they use to prove that there are infinite regular primes.


From $x+y+z=0\Rightarrow \sum \frac{x^{2}}{y}=\sum \frac{(y+z)^{2}}{y}=\sum y+2\sum z+\sum \frac{z^{2}}{y}= \sum \frac{z^{2}}{y}$
So: \[\sum \frac{x^{2}}{y}= \sum \frac{z^{2}}{y}\Rightarrow\sum \frac{x^{2}}{y}=\frac{1}{2}(\sum \frac{x^{2}}{y}+\sum \frac{z^{2}}{y})=\frac{1}{2}(\sum \frac{x^{2}}{y}+\sum \frac{x^{2}}{z})=\frac{1}{2}\sum \frac{x^{2}(y+z)}{yz}=-\frac{1}{4}\sum x^{4}\]
In order to find the maximum ef $\sum \frac{x^{2}}{y}$ we must find the minimum of $\sum x^{4}$
From $\sum x=0$ and $xyz=2$ we can say that there are 2 negative numbers,and a pozitive one.
Let $x$ be the pozitive one and we can take $y,z$ pozitive with the condition that $x=y+z$ and of course $xyz=2$
We have that $\sum x^{4}=y^{4}+z^{4}+(y+z)^{4}$.
From $xyz=2\Rightarrow (y+z)yz=2$ but $yz\leq \frac{(y+z)^{2}}{2}$ so $y+z\geq 2$.Hence $(y+z)^{4}\geq 16$ and $y^{4}+z^{4}\geq \frac{(y+z)^{4}}{8}=2$.
In conclusion $min{ \sum x^{4}}=18$ for$y=z=1$ so $max{ \sum \frac{x^{2}}{y}}=-\frac{9}{2}$ for $(x,y,z)=(2,-1,-1)$


if $ a,b,c$ are positive number,then $ \sum_{cyc}\frac{4a^2\minus{}b^2\minus{}c^2}{a(b\plus{}c)} \leq 3$

[quote="arqady"][quote="apollo"]if $ a,b,c$ are positive number,then $ \sum_{cyc}\frac {4a^2 \minus{} b^2 \minus{} c^2}{a(b \plus{} c)} \leq 3$[/quote]
$ \sum_{cyc}\frac {4a^2 \minus{} b^2 \minus{} c^2}{a(b \plus{} c)} \leq 3\Leftrightarrow\sum_{cyc}(a^4b^2 \plus{} a^4c^2 \minus{} 2a^4bc \plus{} a^3b^2c \plus{} a^3c^2b \minus{} 2a^2b^2c^2)\geq0.$[/quote]
My solution is same :D


How does resonance help determine bond length? This was a question on a past local exam and the answer was NO2+.

The "true" structure of a molecule can be thought of as a (weighted) average of its resonance structures.  Hence, a bond that is single in one resonance structure and double in another will have length somewhere in between a single and double bond.  The exactly length will depend on the number of contributors and the total energy of each.


Let  $a,b,c>0$ . Prove that  $\dfrac{a+b+c}{1+a+b+c}\geq\dfrac{a}{1+3a}+\dfrac{b}{1+3b}+\dfrac{c}{1+3c}$

Rewrite as $f\left(\dfrac{a+b+c}{3}\right)\ge\dfrac{f(a)+f(b)+f(c)}{3}$ with $f(x)=\dfrac{1}{1+1/x}$, and the rest is history.

(And by history, I mean a simple application of Jensen's inequality).

Niko What is your solution? Could you post it ?

Another proof: the given ineq is a special case of the following
$\sum \dfrac{a_1}{1+ka_1} \le \dfrac{nS}{n+kS}$
where $k\ge 0$ and $S=\sum a_i$, which, in turn, can be deduced from $\sum x_i \sum y_i \ge \sum (x_i + y_i)\sum \dfrac{x_iy_i}{x_i+y_i}$ by letting $x_i=ka_i$ and $y_i=1$.


[quote="kunny"]For real number $ x$, denote the  greatest integer which is less than or equal to $ x$ by $ [x]$.

(1) For $ \frac {14}{3} < x < 5$, find $ [\frac {3}{7}x] \minus{} [\frac {3}{7}[x]]$.

(2) For any real number $ x$, prove that $ [\frac {1}{2}x] \minus{} [\frac {1}{2}[x]] \equal{} 0$.

(3) Let $ n$ be natural number. For real number $ x$, find $ [\frac {1}{n}x] \minus{} [\frac {1}{n}[x]]$.[/quote]

1. Well we can rewrite this as $ [\frac{3}{7}(x \minus{} [x])]$.

We know that $ x \minus{} [x]$ will be less than one but more than zero. So if we multiply this by $ \frac{3}{7}$ we will also get a number between zero and one. So that means that the greatest integer less than this product must be zero.

2.  This example is very similar to the last one except the fractions are different. $ x \minus{} [x]$ has to be between zero and one. So if we multiply this by one half we will get a number also between zero and one. The greatest integer less than any number between zero and one will be zero.

3. This will be zero as well.

One problem: I don't believe you can factor like that.

[quote="Flame"][hide="(2)"]
$ 2 < \frac {7}{3} < \frac {1}{2}x < \frac {5}{2} < 3 \implies [\frac {1}{2}x] \equal{} 2$

$ [x] \equal{} 4 \implies [\frac {1}{2}[x]] \equal{} 2$

$ [\frac {1}{2}x] \minus{} [\frac {1}{2}[x]] \equal{} 2 \minus{} 2 \equal{} \boxed{0}$[/hide][/quote]
For any real number $ x$.
But, this solution is $ \frac{14}{3}<x<5$.

Yes, you are right. I just rewrote the condition from (1) in my notebook  :blush:

[hide="2"]For positive $ x$
$ x\minus{}[x]<1$.

when we take $ \frac{1}{2}x$, then this differs from $ \frac{1}{2}[x]$ by less than $ \frac{1}{2}$

Since we know $ [x]$ is an integer, $ \frac{1}{2}[x]$ is of the form $ y.5$ where $ y$ is an integer, and so $ \frac{1}{2}x<y\plus{}1$
Thus, $ [\frac{1}{2}x]\equal{}y$ and $ [\frac{1}{2}[x]]\equal{}y$, so the difference is 0.

If $ x<0$, use the same reasoning.[/hide]

Three uses the same reasoning as two.

General Solution.
$ a, b \in \mathbb{N}, \ (a, b) \equal{} 1$

$ n \in \mathbb{Z}$
$ \left[\frac {b}{a}x\right] \equal{} n \ \Rightarrow \frac {a}{b}n \le x < \frac {a}{b}(n \plus{} 1)$
$ \left[\frac {b}{a}[x]\right] \equal{} n \ \Rightarrow \min m_n \le x < (\max m_n) \plus{} 1 \ \ \ \left(m_n \equal{} \left\{m \in \mathbb{Z}\left|\left[\frac {b}{a}\right.\right] \equal{} n\right\}\right)$

cycle:$ a$

$ \left[\frac {b}{a}x\right] \minus{} \left[\frac {b}{a}[x]\right] \equal{} 0$
$ \underbrace{\min m_0 \le x < \frac {a}{b}, \cdots , \min m_i \le x < \frac {(i \plus{} 1)a}{b}, \cdots, \min m_{b \minus{} 1} \le x < a}_b$

$ \left[\frac {b}{a}x\right] \minus{} \left[\frac {b}{a}[x]\right] \equal{} 1$
$ \underbrace{\frac {a}{b} \le x < \min m_1, \cdots , \frac {i a}{b} \le x < \min m_i, \cdots, \frac {(b \minus{} 1) a}{b} \le x < \min m_{b \minus{} 1}}_{b \minus{} 1}$

[b]For example[/b]
(1)
$ a \equal{} 7,b \equal{} 3$
$ \left[\frac {3}{7}x\right] \minus{} \left[\frac {3}{7}[x]\right] \equal{} 0: 0 \le x < \frac {7}{3}, \ 3 \le x < \frac {14}{3}, \ 5\le x < 7$
$ \left[\frac {b}{a}x\right] \minus{} \left[\frac {b}{a}[x]\right] \equal{} 1: \frac {7}{3} \le x < 3, \ \frac {14}{3} \le x < 5$

(2)
$ a \equal{} 2,b \equal{} 1$
$ \left[\frac {1}{2}x\right] \minus{} \left[\frac {1}{2}[x]\right] \equal{} 0: 0 \le x < 2$
$ \left[\frac {1}{2}x\right] \minus{} \left[\frac {1}{2}[x]\right] \equal{} 1: \text{nothing}$

(3)
$ a \equal{} n,b \equal{} 1$
$ \left[\frac {1}{n}x\right] \minus{} \left[\frac {1}{n}[x]\right] \equal{} 0: 0 \le x < n$
$ \left[\frac {1}{n}x\right] \minus{} \left[\frac {1}{n}[x]\right] \equal{} 1: \text{nothing}$


Point $ P$ is inside equilateral $ \triangle ABC$.  Points $ Q, R$ and $ S$ are the feet of the perpendiculars from $ P$ to $ \overline{AB}, \overline{BC}$, and $ \overline{CA}$, respectively.  Given that $ PQ \equal{} 1, PR \equal{} 2$, and $ PS \equal{} 3$, what is $ AB$?

$ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \sqrt {3}\qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 4 \sqrt {3}\qquad \textbf{(E)}\ 9$

[hide] ab=bc=ac thus s(1+2+3)/2=(s^2)*(sqrt3)/4 then s=4sqrt3
sorry i did not use the LaTeX [/hide]

[hide="Solution"]By a well known theorem, the altitude of the triangle is the sum of these distances.

Thus the height of the triangle is 6.
We have 30-60-90 right triangles, so we know that the sidelength of the triangle is $\frac{12}{\sqrt{3}}$

This becomes $4\sqrt{3}$, or Choice D
[/hide]

[quote="t0rajir0u"]Point $P$ is inside equilateral $\triangle ABC$.  Points $Q, R$ and $S$ are the feet of the perpendiculars from $P$ to $\overline{AB}, \overline{BC}$, and $\overline{CA}$, respectively.  Given that $PQ = 1, PR = 2$, and $PS = 3$, what is $AB$?

$\text{(A)}\ 4 \qquad \text{(B)}\ 3 \sqrt{3}\qquad \text{(C)}\ 6 \qquad \text{(D)}\ 4 \sqrt{3}\qquad \text{(E)}\ 9$[/quote]
[hide]Let $AB=s.$ Since the triangle is equilateral, $AB=BC=CA=s.$
We find the area of the triangle in 2 ways.
$[ABC]=[ABP]+[BCP]+[CAP]=\frac{PQ\cdot AB+PR\cdot BC+PS\cdot CA}{2}=\frac{s+2s+3s}{2}=3s.$
Now by heron's,
$[ABC]=\frac{s^{2}\sqrt3}{4}.$
Equating the two,
$\frac{s^{2}\sqrt3}{4}=3s,$
and it follows that $AB=S=4\sqrt3=D$[/hide]
edit:beaten to it. BY 2 PEOPLE OMG

Ummmm....

DiscreetFourierTransform

What well known Theorem are you talking about.... :huh:

Let the distance to the sides from any point P inside equilateral triangle ABC be a, b, and c. Use the fact that the sum of the areas of the three triangles formed by connecting P to two vertices at a time is the area of ABC.

The theorem is called [url=http://mathworld.wolfram.com/VivianisTheorem.html]Viviani's Theorem[/url].

Another approach could be placing triangle in a coordinate system if you lack ideas.

Can someone please present a proof of Viviani's Theorem?

Also I connected the medians and found the area of one of the smaller triangles. One can multiply by six and then work backwards to find the length of $ \overline{AB}$

[quote="mathemagician1729"]Can someone please present a proof of Viviani's Theorem?[/quote]

Express the area of triangle ABC in two different ways.

[hide="proof"]

Let $ s$ be the side length of the triangle. Let $ h$ be the height.

$ [ABC] \equal{} \frac 12 sh$

$ [ABC] \equal{} [ABP] \plus{} [BCP] \plus{} [CAP] \equal{} \frac 12 s (PQ) \plus{} \frac 12 s (PR) \plus{} \frac 12 s (PS)$

So we have

$ \frac 12 sh \equal{}  \frac 12 s (PQ) \plus{} \frac 12 s (PR) \plus{} \frac 12 s (PS) \implies h \equal{} PQ \plus{} PR \plus{} PS$
[/hide]

Here's a nice proof without words: http://www.cut-the-knot.org/Curriculum/Geometry/VivianiPWW.shtml


geometry
geometric transformation
homothety
geometry proposed

Let $ ABC$ be an acute triangle. Denote $ (C_1)$, $ (C_2)$, $ (C_3)$ circles that tangent to $ (AB,AC)$, $ (BC,BA)$, $ (CA,CB)$, respectively. Let $ (C_4)$ be the circle out-tagents to $ (C_1)$, $ (C_2)$, $ (C_3)$, respectively, by the point $ X$, $ Y$, $ Z$.
Prove that: $ AX$, $ BY$, $ CZ$ are concurent. :D

Nobody can solve this??? :D
Here is my solution: (Not really nice)
Denote $ I$ the center of the incircle wrt triangle $ ABC$. Denote $ M$ be the internal center of the homothety that turns $ (I)\mapsto (C_4)$. It is easy to see that $ A$ and $ X$, respectively, are the external and internal centers of the homotheties that turns $ (C_1)\mapsto (I)$, turns $ (C_4)\mapsto (C_1)$. Therefore, applying [b]The Monge-d'Alembert circle theorem[/b], we get, $ A,X,M$ are collinear. With the same argument, we also have, $ (B,Y,M)$ and $ (C,Z,M)$ are collinear.
Hence, $ AX,BY,CZ$ are concurent at $ M$ the internal center of the homothety of $ (C_4)$ and $ (I)$.
Which ends our proof. :blush:.


Let v,d, and e denote the number of vertices, diagonals, and edges of a triangular prism. FInd v+d+e.

Perhaps this should be in Middle School Classroom Maths?

Actually, so many problems here posted are actually a bit above real hsb level. This is definitely hsb level.

v+d+e = 6 + d + 9. a diagonal of a triangular prism. i have never heard of one.

A diagonal is typically a line connecting two vertices that does not coincide with the given shape at hand.


limit
logarithms
trigonometry
calculus
calculus computations

$ \lim_{x \to 0}\frac{ln(1\plus{}tan2x)\minus{}2x}{ln(cos4x)}$

hint::
AS the limit is in the form of 0/0 here we can apply L hospital Rule.
just differentiate the numerator and denominator and convert every expression in form of tan2x.
u will get the answer. :)

hello, applying the L'Hospital's rule we get
$ \lim_{x\to 0}\frac {\frac {2 \plus{} 2\tan(2x)^2}{1 \plus{} \tan(2x)} \minus{} 2}{\frac { \minus{} 4\sin(4x)}{cos(4x)}} \equal{} \lim_{x \to 0} \minus{} \frac {1}{4}\frac {2\sin(2x)\cos(2x) \minus{} 1}{\cos(2x)^2} \equal{} \frac {1}{4}$.
P.s. after a hint of kabi.
Sonnhard.

The limit is zero:

$ L \equal{} \lim_{x \to 0}\frac{\ln[1 \plus{} \tan(2x)] \minus{} 2x}{\ln[1 \minus{} 2 \sin^2(2x)]} \equal{} \lim_{x \to 0} \frac{\tan(2x) \minus{} 2x}{2 \sin^2(2x)} \equal{} \lim_{x \to 0} \frac{2x \plus{} \frac83 x^3 \minus{} 2x}{8x^2} \equal{} 0$.

Sorry about my previous post. My mistake.

$ L \equal{} \lim_{x \to 0}\frac{\ln[1 \plus{} \tan(2x)] \minus{} 2x}{\ln[1 \minus{} 2 \sin^2(2x)]} \equal{} \lim_{x \to 0} \frac{\ln[1 \plus{} (2x \plus{} O(x^3))] \minus{} 2x}{\ln[1 \minus{} 2(2x \plus{} O(x^3))^2]}$

$ \equal{} \lim_{x \to 0} \frac{[2x \plus{} O(x^3)] \minus{} \frac12[2x \plus{} O(x^3)]^2 \minus{} 2x}{\minus{} 2(2x \plus{} O(x^3))^2} \equal{} \lim_{x \to 0} \frac{2x \minus{} 2x^2 \plus{} O(x^3) \minus{} 2x}{\minus{}8x^2 \plus{} O(x^4)}$

$ \equal{} \lim_{x \to 0} \frac{1 \plus{} O(x)}{4 \plus{} O(x^2)} \equal{} \frac14$.

$ \lim_{x \to 0}\frac {ln(1 \plus{} tan2x) \minus{} 2x}{ln(cos4x)}\equal{}\frac{1}{4}$

Thank you very much.


subj :wink:

http://jarnik.osu.cz/index.php?page=year2008

Today the second announcement was sent to persons interested in.

There appeared online forms for registration.

http://jarnik.osu.cz/index.php?page=add-univ
http://jarnik.osu.cz/index.php?page=add-part

LMU Munich: Daniel Harrer (ZetaX), Alex Schreiber (he is on ML, but I forgot his nick), Christopher Wulff (Werwulff), me.

Bonn: Paul Jonas Hamacher (Pi-J) and two others (I am not sure whether they want me to disclose their names here, and I don't know who of them is on ML).
Peter Scholze finds conferences more interesting. :P

  darij

[quote="bus"]I will.... :). Is there a team from Ukraine that's coming this year???[/quote]

No, just me. :wink:
Hope next year.


algebra
polynomial
algorithm
superior algebra
superior algebra unsolved

Prove that if r has minimum polynomial f(t)= t^3+at+b over Q, that

discriminant [1, r, r^2]= -4a^3-27b^2.

I've been going at this for a while it's really getting to me so I've decided to ask for help. : /

I'm not sure what you mean by "discriminant [1, r, r^2]", but the discriminant is usually defined as $ (\alpha\minus{}\beta)^2(\beta\minus{}\gamma)^2(\gamma\minus{}\alpha)^2$, where $ \alpha$, $ \beta$, and $ \gamma$ are the roots of $ f$.

What you want to do is expand out that expression and apply the symmetric identities $ \alpha\plus{}\beta\plus{}\gamma\equal{}0$, $ \alpha\beta \plus{} \beta\gamma \plus{} \gamma\alpha \equal{} a$, and $ \alpha\beta\gamma\equal{}\minus{}b$.  There are "only" 27 terms in the product...

But you already know what the answer is supposed to be, so it's enough to show that $ (\alpha\minus{}\beta)^2(\beta\minus{}\gamma)^2(\gamma\minus{}\alpha)^2 \plus{} 4(\alpha\beta \plus{} \beta\gamma \plus{} \gamma\alpha)^3 \plus{} 27\alpha^2\beta^2\gamma^2$ is divisible by $ \alpha\plus{}\beta\plus{}\gamma$. (because then it's zero)

That sum, by the way, is equal to $ (\alpha\plus{}\beta\plus{}\gamma)((\alpha\plus{}\beta\plus{}\gamma)(\alpha\beta\plus{}\beta\gamma\plus{}\gamma\alpha)^2 \plus{}18\alpha\beta\gamma(\alpha\beta\plus{}\beta\gamma\plus{}\gamma\alpha) \minus{} 4\alpha\beta\gamma(\alpha\plus{}\beta\plus{}\gamma)^2)$.  Just multiply it out for a proof!  :roll:

there is a more systematic approach using the algorithm which represents a symmetric polynomial as a polynomial in the elementary symmetric functions. the discriminant may be computed without knowing it in advance.

Saying "a solution exists" is not the same thing as providing a solution, -oo-...

I see no problem in using knowledge of the answer as a computational tool if it gets me onto more interesting and less painful problems faster, but this is largely a matter of taste.  If you feel that it's somehow more instructive to use an algorithm, or write the coefficients as unknowns and plug in values, go for it.

I agree that it is important to know the result that any symmetric polynomial can be written as an algebraic combination of elementary symmetric polynomials.  This is fundamental in understanding these sorts of problems.


well folks it's the end of the season ... the IMO is avaiting so I won't bug you with anymore un-usefull details ... but I would like to know what did you think about the contest ... good or bad ...

... and advices of the next edition.

for the three winners congrats!

In my opinion, the problems may not be more difficult than IMO, but must be more technical than IMO

I'd like to participate to the next edition (even though I'm not romanian)

Do you plan to do a next edition?

yeah sure - this time i will announce the edition before

it's probably going to start in September this year, so in about 2 months ( plus/minus two weeks )

seven rounds - each containing 3 problems

more details you can find on the site and i will post some more here soon


hello.  i'd like to ask for an opinion.

We have to solve for $ x$ in $ log_x(4x\minus{}3)\equal{}2$.  We will get $ x\equal{}3$ and $ x\equal{}1$.  But we exclude $ x\equal{}1$ from the solution set (final answer) since the base of a log function cannot be 1 ( or any negative number ), right ?

I asked the Maple software to solve it but it didn't exclude 1, but I think $ x\equal{}1$ shouldn't be there.   :o
I want your opinion guys.   :)

Well, what is $ m\equal{}\log_1 1$? That's a number such that $ 1^m\equal{}1$, that can be any number. This is why $ \log$ doesn't have base 1, so $ x\equal{}1$ is not a solution.

thanks Bugi.

So simply, even though $ 1^2\equal{}1$, this doesn't imply that we can write it in the form $ log_{1}1\equal{}2$, since that would be erroneous.  I think the Maple software has a bug or the program code is not good enough since it gave $ x\equal{}1$ as a solution...    :o

[quote="maths486"]I think the Maple software has a bug or the program code is not good enough since it gave $ x \equal{} 1$ as a solution...    :o[/quote]
I wouldn't call it a bug.  Maple probably automatically converted the problem to $ \ln (4x \minus{} 3) \equal{} 2 \ln x$.  It's unnatural to state problems using nonconstant bases.


An unbiased coin is tossed $N$ times. What is the probability that no two consecutive heads appear?

The number of possible sequences is of course $2^N$
The number of sequences where no two consecutive heads occur in one toss is 2 (H, T)
The number of sequences where no two consecutive heads occur in two tosses is 3 (HT,TH,TT)
Now for N>3 note that the sequence either
1. starts with T and has an (N-1) sequence that works
or
2. starts with HT and has an (N-2) sequence that works
so
The number of sequences N long that work = the number of N-1 seq. + the number of N-2 seq.
This is of course the Fibonacci sequence, except first term is 2
so the answer is

$\text{(N+2)th Fibonacci number}/2^N$

Can you assert this pattern will continue?

[quote]
The number of sequences N long that work = the number of N-1 seq. + the number of N-2 seq.
[/quote]

I did assert that this will work since this is the Fibonacci sequence. You can rewrite it as
f(n)=f(n-1)+f(n-2)
where f(n) is the number of sequences n long where two consecutive heads appear.
The two cases that I brought up are the only two possible ways to increase the sequence by one toss.


function
calculus
Support
analytic geometry
calculus computations

Suppose we don't have a formula for the function f, but we know that f '(x) = 1/sqrt(2cos(x)+3) for all x and we know that f(2) = 8.

a. use linear approximation to estimate f(2.1).
b. Is your estimate from part a to large or too small? Use calculus to support your answer.

I got an answer of f(2.1) = 8.06792.  I am not sure how to do part b.

$f'$ is increasing -> $f$ is concave up -> the graph of $f$ is over the tangent line at $x=2$. The number 8.06792 that you got is not $f(2.1)$, but the $y$-coordinate of the point on the tangent line with $x=2.1$.


consider the sequence $ S_{1},S_{2},.....S_{n},.....$. let $ S_{1}=3/4$ and $ S_{n}=1+\frac{1}{S_{n-1}}$. what is $ S_{5}$ as a common fraction.

hint-write is as a complex fraction. eg. $ S_{3}$ would be $ 1+\cfrac{1}{1+\cfrac{1}{\frac{3}{4}}}$

variation-consider the sequence $ S_{1},S_{2},.....S_{n},.....$. let $ S_{1}=p/q$ and $ S_{n}=1+\frac{1}{S_{n-1}}$. what is $ S_{n}$ as a common fraction.(is there a way to do this like a formula?)

[quote="mountain.dew"]you might be able to find a formula, but i'd just go brute force on finding $ S_{5}$[/quote] so far i havent but yeah i was wondering because the kid got this in like 5 seconds at our chapter counddown round.


Find the largest possible value for $ k$ for which $ 3^{11}$ can be expressed as $ k$ consecutive positive integers.

[hide="solution"]$ n\plus{}(n\plus{}1)\plus{}...\plus{}(n\minus{}k\plus{}1)\equal{}\frac{(2n\plus{}k\minus{}1)k}{2}\equal{}3^{11}$

So $ 2 \times 3^{11}\equal{}(2n\plus{}k\minus{}1)k>k^2$ and $ k|(2 \times 3^{11})$

The largest factor of $ (2 \times 3^{11})$ smaller than $ \sqrt{2 \times 3^{11}}$ is $ 2 \times 3^5\equal{}486$

After checking , $ (n,k)\equal{}(122,486)$ satisfies the condition so the largest possible $ k$ is $ 486$[/hide]

[hide="source"]This problem is from the [url=http://www.artofproblemsolving.com/Wiki/index.php/1987_AIME_Problems/Problem_11]1987 AIME #11.[/url][/hide]


LaTeX
geometry
geometric transformation
algebra
difference of squares
special factorizations

yeah..I'm pretty much dumbfounded.

3x(to the fourth power) - 12 =

4c(to the second power) -4cd + d(to the second power)

Special thanks to anyone that helps.

Hint for 1: [hide]Factor out a 3 and solve as difference of squares.[/hide]

Solution for 1:
[hide]Factor out a 3 to get $ 3(x^{4} \minus{} 4)$

Now factor as a difference of squares so we get $ 3(x^{2}\minus{}2)(x^{2} \plus{} 2)$

from that we get $ 3(x \plus{} \sqrt{2})(x \minus{} \sqrt{2})(x \minus{} i\sqrt{2})(x \plus{} i\sqrt{2})$
[/hide]

2. [hide]$ (d \minus{} 2c)^{2}$[/hide]

for hint #1.. i get 3(x to the fourth power -4)..which isn't the right answer.

And i don't get your second hint..

hint for 2

[hide]It wasn't a hint[/hide]

When I mean difference of squares I mean $ a^{2} \minus{} b^{2} \equal{} (a \plus{} b)(a \minus{} b)$

[quote="gauss1181"]#2: (2c-d)^2

#1: 3(x^4-4)=3(x^2-2)(x^2+2)=3(x-sqrt2)(x+sqrt2)(x^2+2)[/quote]
latex translation because we all know latex is sooo much more better
#1 $ 3(x^{2}\minus{}4)\equal{}3(x^{2}\minus{}2)(x^{2}\plus{}2)\equal{}3(x\minus{}\sqrt{2})(x\plus{}\sqrt{2})(x^{2}\plus{}2)$
#2 $ (2c\minus{}d)^{2}$

Just use dollar signs $ \$ \text{text here} \$$

[quote="7h3.D3m0n.117"][quote="gauss1181"]OK. i know. but how do u get to use LaTEX text?[/quote]

AoPS has a good guide here:

[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url][/quote]

Outdated. [[LaTeX:About]] is the current reference.


rotation
conics
ellipse
hyperbola
parabola
geometry
3D geometry

The xy term "tilts" the graph.. there's a formula for the angle it gets tilted, which I can't recall at the moment (take a look in aops volume 2, if you have it).

As Caroline said, an xy term rotates the graph of a conic section.  First you want to identify the conic (by the descriminant).  Then you want to use the formula Caroline alluded to in order to find the angle that it is rotated at.  Next you want to do some substitution and some completing the square in order to get the equation into standard form w/o the xy term.  So you will graph this new equation, titled at the angle you found from the formula.

Okay I just checked, look on page 65 of AoPS vol 2, its a section called "that pesky xy term" which is exactly what you are looking for.

As for the other one, you can graph it on your TI if you want a quick way...but I don't think there is a trick...but I usually fail to see the shortcuts...so I can't answer that.

Hope this helped!

For the second one, there are only 3 cases to check:

x <= 4
4 <= x <= 13/2
13/2 <= x.

After that its just solving 3 easy linear equations. So no, there is no 'shortcut' that I know of, but the 'long' way isn't too bad.

[quote]Discriminant of a conic
For conic section defined by real polynomials of the form
Ax2+Bxy+Cy2+Dx+Ey+F=0,
the discriminant is equal to
B2-4AC,
and determines the shape of the conic section. If the discriminant is less than 0, the equation is of an ellipse or a circle. If the discriminant equals 0, the equation is that of a parabola. If the discriminant is greater than 0, the equation is that of a hyperbola. This formula will not work for degenerate cases (when the polynomial factorises). [/quote]

This is from http://www.sciencedaily.com/encyclopedia/discriminant

This might mean you just get meaningless results for degenerate conics.  I am not sure.

this isn't exactly an 'easy' way to do it, but it is a way nonetheless.

Where the equation of the conic is

Ax2 + Bxy + Cx2 + Dx + Ey + F = 0

A conic is degenerate if the determinant
[code]
| 2A     B      D  |

| B      2C     E  |    =  0

| D      E      2F |
[/code]

you probably don't want to memorize that though

Hmm... okay well if its a line its y=mx+b, but what equation does it come from?

Degenerate "circle"

x^2 + y^2 =0
x^2 = - y^2
Can't go anywhere.

Degenerate "ellipse":

2x^2 + y^2 = 0
Same deal as circle.

Degenerate "hyperbola":

x^2 - y^2 = 0
x^2 = y^2
y = +/- x
So this is the two intersecting lines.

Degenerate "parabola"

x^2 = 0
+/- x = 0
x=0
So here you get a vertical line.

Um I don't know if these are correct or not.  A point dosent have an equation, so I don't know how it would be algebraically derived from the equation of a conic section.  Also, isn't the rule <b> A </b> plane cutting into two cones?  So if ya only have 1 plane, how can you get 2 intersecting lines?  Also, I heard somewhere that a pair of parallel lines was also a degenerate conic....Don't know how this works either though...

I just found this reference...here's the link: http://mathforum.org/library/drmath/view/54756.html

[quote="JS1527"]Also, isn't the rule <b> A </b> plane cutting into two cones?  So if ya only have 1 plane, how can you get 2 intersecting lines?[/quote]

Imagine a plane passing through the vertex of the cone, cutting both the upper and lower halves of the cone.  You'll get two intersecting lines.

Also, a point does have defining equations: x = a, y = b.  Alternatively, the general formula for a circle is
[tex]\displaystyle{(x-a)^2+(y-b)^2=r^2}[/tex], right?  Well, if r = 0, that system has a unique solution (by the trivial inequality, [tex]\displaystyle{x^2\geq0}[/tex] for x real), namely x = a, y = b.  So we can call [tex]\displaystyle{(x-a)^2+(y-b)^2=0}[/tex] the equation of the point (a, b).


ratio MATHCOUNTS logarithms probability function inequalities geometry

[b]First Ever MATHCOUNTS Problem Solving Marathon![/b]

I got this idea from Silverfalcon's marathon in the Getting Started forum. You just answer the last question and make up your own. It has to be Mathcounts-level, though, and [b]you must provide solutions when answering.[/b]

[size=150]Problem 1[/size]

(MATHCOUNTS State 2005) When the digits in the number 2005 are reverses we obtain the number 5002, and $5002 = a \cdot b \cdot c$, such that $a$, $b$, and $c$ are three distinct primes. How many other positive integers are the products of exactly three distinct primes $p_{1}$, $p_{2}$, and $p_{3}$ such that $p_{1}+p_{2}+p_{3}= a+b+c$?

[quote="probability1.01"]I assume this was a target--I don't see how a student could finish this in under a minute.  Now nat mc, I think you're supposed to post another problem ;).[/quote]

It was team #7.

Oh.. ok umm...
(national 93-94 target #8)
A farmer has 60 meters of fencing with which to fence in her cows horses and pigs.  She wants each type of animal to have the same area.  What is the maximum number of square meters she can enclose.
______
l[color=white]gl[/color]l[color=white]gl[/color]l[color=white]gl[/color]l
-------

Three rectabgles side by side each with a common side next to it.

[quote="bob123"][quote="chess64"][quote="Xantos C. Guin"][hide]225 m^2
The farmer can enclose a 15 by 15 meter area.
15x15= [b]225 ANS.[/b][/hide][/quote]

I don't know if it's right, but go ahead and post problem 3.[/quote]

It's not right... Solution:

[hide]You want 3 squares. so you do 4x+6x=60, x=6. The side of each square is 6, 6^2=36.[/hide]

Too lazy to find another problem to post.[/quote]

I'll go ahead and post for you.

[b]Problem 3[/b]

What is the smallest positive whole number that ends with 1997 and divisible by 1999?

Both are wrong.
@Xantos there are three rectangles, each shring one side with the rectangle next to it. All sides need fencing.

[hide="solution"]call the horrizontals x and the verticals y

$6x+4y=60$

$3x+2y=30$

$x=10-\frac{2y}{3}$

Try to find the maximum value of $3xy$ which is the area

$3xy=-2y^2+30y$

$\text{maximum}=\frac{-b}{2a}=\frac{-30}{2\cdot{-2}}=\frac{15/2}$

$x=5$

$3xy=112.5$
[/hide]
I'll just post a different problem:
(nat 95-96 sprint #22)
On a 4 by 4 grid, the points are 1 unit apart horizontally and vertically.  Two points P and Q are randomly selected from this grid.  What is the probability that the distance between P and Q is less than 2.5?

The question was: [b]What is the maximum number of square meters she can enclose. [/b]
All three were enclosed, right? 37.5x3=112.5

Guys, post like this, so it's easier to tell when the next problem is posted:

[size=167]Problem 3[/size]

What is the smallest positive whole number that ends with 1997 and divisible by 1999?

[hide]
[b]8001997[/b]

$N=10000x+1997$, take the equation mod 1999. $5x +1997 \pmod {1999}$ is the remainder when $N$ is divided by $1999$. We want the remainder to be 0 or a multiply of 1999. Trial-and-error yields x=800, $N=8001997$.
[/hide]

[quote="SplashD"]I'll let you post another tricky one.  :D[/quote]
I don't have an infinite supply of tricky questions.  :P

Problem $\infty^\infty$:
[b]The perimeter of a regular octagon is numerically equal to its area. Find this numerical value. [/b]

[quote="SplashD"]Ddo you mean a regular octagon?[/quote]
Yes. :D Sorry.

[quote="ragnarok23"]Is it [hide]$4(\sqrt{2}-1)$[/hide][/quote]
That's correct. Solution?

An ant sitting on point A wants to get some tasty melt chocolate on point B. Point A and B are two opposite points of a cube. If the ant cannot travel on the side that it already traveled on, what is the largest number of sides the ant can go over?

[quote="Andrew Kim"]
       A --------------
      /                   /|
     /___________ /  |
    |                   |  |
    |                   |  |
    |                   |  |
    |                   |  /
    |____________|/
                         B

An ant sitting on point A wants to get some tasty melt chocolate on point B. Point A and B are two points of a cube. If the ant cannot travel on the side that it already traveled on, what is the largest number of sides the ant can go over?[/quote]
Nice diagram but the spaces don't show up. :D
[hide] I got 7. Just draw some of the longest possible paths, and you'll see that it's 7. [/hide]

[quote="Andrew Kim"]Actually there is more.
And there is a smarter way to prove it than just trying it millions of times.[/quote]
Well, if the ant can travel past B and come back again, the answer is [hide] 9 [/hide]

[quote="ragnarok23"]I derived a formula for an area of an octagon which was $2s^{2}(1+\sqrt{2})$ and you just solve $8s=2s^{2}(1+\sqrt{2})$ to get the answer.[/quote]
wow thats cool I derived that same formula.  I figured it out from problem 15-9 in AoPS vol.1

I know this was one of the largest and longest-running topics on AoPS, but this topic is just too big to moderate.  (And moderation became necessary due to some recent posts.)

I'm sorry, but I'm locking this topic.  Please start a new topic.


geometry
circumcircle
trigonometry
parallelogram
ratio
geometry proposed

Obviously $AD=AE$ and $AD \perp DK$. Let $A'$ a point on the circumcircle of $\triangle ABC$ such that $A'M \perp BC$. As $AH//A'M$, it is equivalent to prove that $\frac{A'M}{AH}=\frac{A'K}{AK}$. By angle chasing and trigo, it is simple to work out that:
\[ AH=2R \cos A; \]
\[ A'M=R-R \cos A; \]
\[ AA'=2R \sin \angle ABA'=2R \sin(\frac{A}{2}+B); \]
\[ AK = \frac{AD}{\cos \frac{A}{2}} = AH \frac{\sin \angle AHD}{\sin \angle ADH}\frac{1}{\cos \frac{A}{2}} = 2R \cos A \frac{\sin(\frac{A}{2}+B)}{\cos ^2 \frac{A}{2}}. \]
Now it is simple to work out that:
\[ \frac{A'K}{AK}=\frac{AA'-AK}{AK}= \frac{\cos ^2 \frac{A}{2} - \cos A}{\cos A} = \frac{1 - \cos A}{2 \cos A} = \frac{A'M}{AH}. \]

I can't find beautiful solution. Just we can calculate all that we need.
Note $ADE$ is isosceles, $AK$ is diameter - thus we can find
$CE$, $AE$ and $AK$ (I got $\displaystyle AK = \frac{(b+c)cos\alpha}{2cos^3\frac{\alpha}{2}}$)

Let $AH \cap BC = K, AI \cap BC = L$, all that we have to prove - Menelaos for triangle $AKL$ and secant $HM$.

$\frac{KH}{AH}=\frac{2Rcos\beta \cdot 2Rcos\gamma}{2Rcos\alpha}$

$\frac{ML}{MK}$ also easy to find, and $\frac{AK}{AL}$ is a bit hard, but also calculable $AL=\frac{2bc\cdot cos\frac{\alpha}{2} }{b+c}$.

Hence in such unbeautiful way we can solve the problem.

Mybe here is a nice solution :)
That is easy to see that $KE||BH$ and $KD||CH$.
Let $B_1\in{KD\cap{BH}}$,$C_1\in{KE\cap{CH}}$=>
$KB_1HC_1$ is paralelogram=>the midpoint of $B_1C_1$ lie on $HK$,but if $C'\in{{CH}\cap{AB}}$,$B'\in{{BH}\cap{AC}}$ then $\triangle{HBC'}$similar to$\triangle{HCB'}$=>$\frac{BD}{DC'}=\frac{CE}{EB'}$ and $\frac{BD}{DC'}=\frac{BB_1}{B_1H}$,$\frac{CE}{EB'}=\frac{CC_1}{C_1H}$=>$\frac{BB_1}{B_1H}=\frac{CC_1}{C_1H}$=>$B_1C_1||BC$,but as we have said the midpoint of $B_1C_1$ lie on $HK$=>the midpoint of $BC$ lie on $HK$ too. ;)

A relatively nice solution to the problem using areas is as follows:
Let BH intersect AC at L and CH intersect AB at M. It is easy to see with simple angle chasing that ADE is iscoceles.
To prove that HK is a median of triangle BHC, it suffices to prove |HKB|=|HKC|, i.e. |HKB|/|HKC|=1.
|HKB|/|HKC| =  (|HKL|*HB/HL) / (|HKM| * HC/HM)
                    =  (|HKL| / |HKM|) * (HB/HC)*(HM/HL)        (rearranging terms)
                    =  (|HEL| / |HDM|) * (HM^2 / HL^2 )           (since DK // MH, KE//HL, HB/HC=HM/HL since MLBC is concyclic)
                    =  (HL^2 / HM^2 ) * (HM^2 / HL^2 )           (since triangle HEL is similar to triangle HMD, with ratio equal to HL/HM)
                    =   1, as required.

Let $O$ be the center of $w$, the circumcircle of $ABC$, and $W,X$ be the intersections of $AH,AO$ with $w$.  Let $U,V$ be the midpoints of the minor and major arcs of $BC$, $M$ the midpoint of the line segment $BC$ and $S$ the intersection of $WV,BC$, and $\theta =\angle WAU =\angle UAX =\angle AWV$.

 $ADKE \sim VBUC$, and $\angle AHK =\angle SVM$, so $HK,SU$ intersect at angle $\theta$.  We know that $OM$ is parallel to and half the size of $AH$, so $M$ is the midpoint of $HX$.  It remains to be shown that $XM,SU$ intersect on $w$.  But project from $V$ to get that $(S,M;B,C)=BW/WC = XC/XB = (U,X;B,C)$, so this must be true.

Dear Mathlinkers,
a relation with
M.O. Russie (2000), 10th grade, problem 3
can be made...
Sincerely
Jean-Louis

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=519896]2005 ISL G5[/url].

Dear Mathlinkers,
an complete harmonic proof is possible by considering
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89098
Sincerely
Jean-Louis

Here is my solution for this problem
[b]Solution[/b]
Let $P$, $Q$ be orthogonal projections of $B$, $C$ on $AC$, $AB$; $N$ $\equiv$ ($APQ$) $\cap$ ($ABC$)
Since: $DE$ is external bisector of $\widehat{BHC}$, we have: $\dfrac{DB}{DQ}$ = $\dfrac{HB}{HQ}$ = $\dfrac{HC}{HP}$ = $\dfrac{EC}{EP}$
It's easy to see that: $\triangle$ $NBQ$ $\sim$ $\triangle$ $NCP$ and $\dfrac{DB}{DQ}$ = $\dfrac{EC}{EP}$, then: $\triangle$ $NBD$ $\sim$ $\triangle$ $NCE$
So: $\dfrac{ND}{NE}$ = $\dfrac{BD}{CE}$
But: $\widehat{DBH}$ = $\widehat{ECH}$, $\widehat{BHD}$ = $\widehat{PHE}$ = $\widehat{CHE}$ then: $\triangle$ $BHD$ $\sim$ $\triangle$ $CHE$
Hence: $\dfrac{DH}{EH}$ = $\dfrac{BD}{CE}$ = $\dfrac{ND}{NE}$ or $NH$ is internal bisector of $\widehat{DNE}$
But: $\triangle$ $NBD$ $\sim$ $\triangle$ $NCE$ so: $\widehat{BND}$ = $\widehat{CNE}$ or $\widehat{DNE}$ = $\widehat{BNC}$ = $\widehat{BAC}$
Therefore: $NH$ passes through $K$ or $HK$ passes through midpoint of $BC$

[quote=Vietnam TST 2006 P1]Given an acute angles triangle $ABC$, and $H$ is its orthocentre. The external bisector of the angle $\angle BHC$ meets the sides $AB$ and $AC$ at the points $D$ and $E$ respectively. The internal bisector of the angle $\angle BAC$ meets the circumcircle of the triangle $ADE$ again at the point $K$. Prove that $HK$ is through the midpoint of the side $BC$.[/quote]
[b]Solution:[/b] From [url=https://artofproblemsolving.com/community/c6h89098_hard_to_approach_it_]ISL 2005 G5[/url], We know, $AD=AE$ if $K$ $=$ $\odot (ADE)$ $\cap$ $\odot (ABC)$ And define $A-$ $\text{angle bisector} $ $\cap$ $ \odot (ABC)=L$, then, $KH$ bisects $BC$. Let $KH \cap \odot (ADE)=L'$ $\implies$ $\Delta ADL' \cong \Delta AEL'$ $\implies$ $L \equiv L'$ Hence Done!


[quote="aops_user"]Try the challenge presented by the characters at this link:

http://www.stripcreator.com/comics/aops_user/383214[/quote]

From AoPS moderator to AoPS user, a steep career!

  dg

It's Benford's Law, and it's $\log(1+\frac{1}{d})$ where $d$ is the string of leading digits (ie it works for numbers starting with the digits 33, or whatever, in addition to single digits). Thats a natural log, by the way.

Also, that's quite some length you went to to make an attack on someone who will probably just get additional enjoyment out of making fun of you for this and thus further prolonging this age-old feud.

[quote="Xevarion"]further prolonging this age-old feud[/quote]
The worst thread I've entered was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71819]this[/url]. I don't respond to this user's posts anymore, but I do watch them. The last username in that thread was clearly created as a personal insult to JesusFreak197, who has since left.

A PM I received at the time:
[quote="AOPS Moderator"]Subject: Regardless of your prime sum being correct, ....
Message: you initiated hostility in more than one user on the board. You need to leave out combative language that does that, and just state things matter of factly that need to change with a post. [/quote]


function
algebra
domain
real analysis
real analysis unsolved

Decide if the items are TRUE or FALSE.  Give a proof or counterexample

a.  Let D be a domain, and let f,g: D-->R be bounded, uniformly continuous on D.  Then fg is uniformly continuous on D.

b.  The function f(x)=1/(1+x^2) is uniformly continuous on R.

Read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85309]this topic.[/url]

If you understand what's written there, you should be able to figure out the answers.

I read it, but still a little bit confused.  Please continue to help me, give me more examples or step by step help me to see the solutions.  Thanks

Try to work through the definitions and examples in that topic yourself. They aren't terribly difficult. What is happening in each of those situations? Draw some pictures. What is the issue in each of Professor Merryfield's "easy to remember counterexamples"?

edit: If you don't know what some of the terminology means, look on Wikipedia or google it. That ought to help too.


geometry
3D geometry
sphere
calculus
integration
trigonometry

What is the surface area of the Earth between 23.5 degrees north (Tropic of Cancer) and 23.5 degrees south (Tropic of Capricorn) latitude, if Earth is a perfect sphere with radius 3963 miles and if the straight-line distance between the tropics is 3250 miles?

I tried to slove your problem but I have'n got any way. I think you shoud post it at forum [b]Geometry Unslove Problem[/b], at this have many body help you.
http://www.mathlinks.ro/index.php?f=46&sid=353ffd3e7f21d6a3cab18b63d02b5d24

i got it, you can use integral to solve this equation!

the equation is:
2 * 2 * $ {\pi}$ * radius of earth *$ \int^{23}_{0} \sin x\minus{}90\,dx$ <= note, that this is in degree

basically "2 * radius of earth *$ \int^{23}_{0} \sin x\minus{}90\,dx$" breaks the part of the earth down to its individual pieces along the arc with the new radius at each point.

the "* $ {\pi}$" part gives the circumference of each point along the arc. after adding them all up (the defination of an integral) you should get the area of the top section.

the last "*2" would double it, giving you the whole solution.

sorry, that this is a bit confusing


search
function
combinatorics unsolved
combinatorics

In an n x n array, each of the numbers 1, 2, .., n appears exactly n times. Show that there is a row or column in the array with at least $ \sqrt{n}$ distinct numbers.

let $ R_{1.k},R_{2,k},...,R_{n.k}$ and $ C_{1.k},C_{2.k},...,C_{n.k}$ be numbers of distinct numbers in rows and columns after writing n times 1,2,...,k in an array. At the begginning $ R_{i.0}\equal{}C_{i.0}\equal{}0$ $ i\equal{}1,2,...,n$. It's easy to prove that $ \sum_{}^{}C_{i.k} \plus{} \sum_{}^{}R_{i.k}\plus{}2 \sqrt{n} \le   \sum_{}^{}C_{i.k\plus{}1} \plus{} \sum_{}^{}R_{i.k\plus{}1}$.
so
$ max(C_{i.n},R_{i.n}) \ge  \frac{ \sum_{}^{} C_{i.n}\plus{}R_{i.n}}{2n} \ge  \sqrt{n}$

As a general tip, the search function is your friend here.  My search for "row column distinct" found this particular problem [url=http://www.mathlinks.ro/viewtopic.php?search_id=104878348&t=112667]here[/url], [url=http://www.mathlinks.ro/viewtopic.php?t=231717]here[/url], [url=http://www.mathlinks.ro/viewtopic.php?t=171494]here[/url],  and [url=http://www.mathlinks.ro/viewtopic.php?p=114481#114481]here[/url] (this last one contains a particularly nice solution by Fiachra).

[quote]this last one contains a particularly nice solution by Fiachra).[/quote]it's the same as mine  :) (it's hidden in part: "it's easy to prove")


show  :D

$ \sum_{k\equal{}1}^\infty\;\dfrac{k\cdot 2^k}{\dbinom{2k}{k}}\;\equal{}\;\boxed{3\plus{}\pi}$

[quote="misan"]show  :D

$ \sum_{k \equal{} 1}^\infty\;\dfrac{k\cdot 2^k}{\dbinom{2k}{k}}\; \equal{} \;\boxed{3 \plus{} \pi}$[/quote]
mm.. try to consider $ arcsin^{2} x$


Solve the equation for $ x\ge 0$ : $ x^3\minus{}7x\plus{}4\equal{}0$

With this equation you can use method :
$ x\equal{}ky$ where $ \frac{k^3}{4}\equal{}\frac{7k}{3}$ where $ k>0$
The equation equivalent : $ k^3y^3\minus{}7kx\plus{}4$
$ \Longleftrightarrow 4y^3\minus{}3y\plus{}\frac{16}{k^3}\equal{}0$
Then it is quite easy to solve .

[quote="Erken"]To quangpbc:
Could you please show your solution?
According to my soft,it has only two solutions. :(
Thank you.[/quote]

Ok, my friend .

Easy to show that $ x\le 3$ Let $ x\equal{}\sqrt{\frac{28}{3}}y$ then $ 0\le y<1\longrightarrow y\equal{}\cos t\longrightarrow x\equal{}\sqrt{\frac{28}{3}}\cos t$

The equation is equivalent to :

$ \frac{7\sqrt{7}}{3\sqrt{3}}\left(4\cos t^3\minus{}3\cos t\right)\equal{}\minus{}2\Leftrightarrow \frac{7\sqrt{7}}{3\sqrt{3}}\cos 3t\equal{}\minus{}2$....  :)


limit
combinatorics proposed
combinatorics

Do we take these triplets $(i,j,k)$ as ordered or as unordered sets?

I am going to bump this problem again I would like to see a full solution.

Hint for (a): Consider the generating function of $\frac{1}{\sqrt[3]{1-27x}}$.


suppose a function $ f$ satisfies $ f(0)\equal{}1$ and $ f(n)\equal{}nf(n\minus{}1)$ for all positive intergers $ n$
find $ f$

[quote="DIMITRIS T"]suppose a function $ f$ satisfies $ f(0) \equal{} 1$ and $ f(n) \equal{} nf(n \minus{} 1)$ for all positive intergers $ n$
find $ f$[/quote]

Clearly it's $ f(n)\equal{}n!$, if you really want a proof then use induction.


pigeonhole principle
modular arithmetic
number theory
relatively prime

Given a natural number $ k$, prove that there exists a prime $ p$ and a strictly increasing sequence $ a_1, a_2, a_3, ...$ of natural numbers such that $ p \plus{} ka_1, p \plus{} ka_2, p \plus{} ka_3, ...$ are all prime numbers.

Since there are an infinite number of primes, by the Pidgeonhole Principle for any k there must exist an integer i $ (0<i<k)$ such that there are an infinite number of primes congruent to i modulo k.  Let p be the first one, and choose the $ a_i$s accordingly, such that they generate the primes congruent to i mod k.

You could also use Dirichlet's theorem, choosing $ p$ as the greatest prime less than $ k$ which guarantees that it is relatively prime with $ k$.

[quote="bpgbcg"]Since there are an infinite number of primes, by the Pidgeonhole Principle for any k there must exist an integer i $ (0 < i < k)$ such that there are an infinite number of primes congruent to i modulo k.  Let p be the first one, and choose the $ a_i$s accordingly, such that they generate the primes congruent to i mod k.[/quote]
i am sorry but can you elaborate a little more? i find it a bit difficult to understand...

Given $ k$ is a positive integer, split the primes $ p_n$ into sets $ S_0,\dots,S_{k \minus{} 1}$ such that $ S_i \equal{} \{p_n|p_n\equiv i\pmod k\}$. Since $ \bigcup_{i\equal{}0}^{k\minus{}1} S_i$ is infinite, it follows that at least one of the $ S_i$ is infinite (I'm not sure if this is the Pigeonhole principle.)

Then, all of the elements of $ S_i$ are $ i$ mod k, or in other words, can be written as $ ka_n \plus{} p_0$.

[quote="Yongyi781"]Since $ \bigcup_{i \equal{} 0}^{k \minus{} 1} S_i$ is infinite, it follows that at least one of the $ S_i$ is infinite (I'm not sure if this is the Pigeonhole principle.)[/quote]

Yes, it is the Pigeonhole Principle: Infinitely many pigeons in finitely many pigeonholes $ \Rightarrow$ one of the holes has infinitely many pigeons.


calculus
inequalities unsolved
inequalities

Question: Maximize the expression
$a^{4}+(b+c)a^{2}+(b^{2}+bc+c^{2})$
subject to the constraints $a, b, c > 0$ and $a+b+c = 1$.

Again I have three variables and it's still symmetric in two of them. But I don't think assuming $b=c$ is the best approach here -- it will again just be a huge and ugly calculus problem. Are there nice ways to deal with this sort of thing?

Let $f(a,b,c) = a^{4}+(b+c) a^{2}+\left( b^{2}+bc+c^{2}\right)$.

Then,
\begin{eqnarray*}f(a,b,c) &=& a^{4}+(b+c) a^{2}+\left( b+c \right)^{2}-bc \\ \ &=& a^{4}+(1-a) a^{2}+\left( 1-a \right)^{2}-bc \\ \ &=& a^{4}+\left( a^{2}-a^{3}\right)+\left( 1-2a+a^{2}\right)-bc \\ \ &=& a^{4}-a^{3}+2 a^{2}-2a+1-bc \\ \ &\leq& a^{4}-a^{3}+2 a^{2}-2a+1 .\end{eqnarray*}

We have that $\frac{\partial^{2}}{\partial a^{2}}\left( a^{4}-a^{3}+2 a^{2}-2a+1 \right) = 12 a^{2}-6 a+4$, which has $\Delta < 0$, so it's always positive. Thus, $a \mapsto a^{4}-a^{3}+2 a^{2}-2a+1$ is convex. Hence, it reaches its maximum in one of its endpoints. Actually, it reaches its maximum in both endpoints, which is $1$.

Equality cases: $(a,b,c) = (0,1,0), \, (0,0,1), \, (1,0,0)$.

Note that I assumed $a,b,c \geq 0$, since otherwise the maximum wouldn't be reached.


Let $ x$, $ y$ and $ z$ are non-negative numbers. Prove that:
\[ \frac {x^3 \plus{} y^3 \plus{} z^3}{3}\ge xyz \plus{} \frac {3}{4}\,\lvert(x \minus{} y)(x \minus{} z)(y \minus{} z)\rvert
\]
The gift from me:
Prove that the maximal $ \alpha,$ such that the following inequality
\[ \frac {x^3 \plus{} y^3 \plus{} z^3}{3}\ge xyz \plus{} \alpha\lvert(x \minus{} y)(x \minus{} z)(y \minus{} z)\rvert
\]
holds for all non-negative $ x,$ $ y$ and $ z$ is
$ \alpha \equal{} 1.467889825... \equal{} f\left(\frac {1 \plus{} \sqrt3 \plus{} \sqrt {2\sqrt3}}{2}\right),$ where $ f(t) \equal{} \frac {t^3 \plus{} 1}{3(t^2 \minus{} t)}.$

Romanian JBTST V 2007, problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?p=858831#858831

[b]P.S.[/b] Sorry dear Michael, but is there any pure geometry problem?  :)  If so, could you post it? Thank you very much   :P

[quote="April"]Romanian JBTST V 2007, problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?p=858831#858831[/quote]
Wow! I forgot about it and I found another way for proof of the stronger inequality. :D
[quote="April"]
[b]P.S.[/b] Sorry dear Michael, but is there any pure geometry problem? [/quote]
Dear April!
My English is very bad in order to translate from Russian.
Maybe it will help:
http://lib.mexmat.ru/forum/viewtopic.php?t=9750
http://lib.mexmat.ru/forum/viewtopic.php?t=9732
http://lib.mexmat.ru/forum/viewtopic.php?t=9720
 :P


calculus
integration
logarithms
real analysis
real analysis unsolved

$ \int_0^1 \int_0^1 \frac{\ln(xy)\, x}{1\minus{}x^2y^2} \, dy\, dx$ is after substitution $ u\equal{}xy \quad \int_0^1 \int_0^x \frac{\ln u}{1\minus{}u^2}\, du \, dx$ .

Then reverse integration order $ \int_0^1 \int_u^1 \frac{\ln u}{1\minus{}u^2} \, dx \, du\equal{}\int_0^1 \frac{\ln u}{1\plus{}u}\, du$ and this is $ \minus{}\frac{\pi^2}{12}$ .


Source: AHSME 1973 #9

In $\triangle{ABC}$ with right angle at $C$, altitude $CH$ and median $CM$ trisect the right angle.  If the area of $\triangle{CHM}$ is $K$, then the area of $\triangle{ABC}$ is
$\text{(A)} \ 6K \qquad \text{(B)} \ 4\sqrt{3}K \qquad \text{(C)} \ 3\sqrt{3}K \qquad \text{(D)} \ 3K \qquad \text{(E)} \ 4K$

We have $BH=HM$ and $AM=2BH$
since area of triangle $CHM$ is $HM*CH/2$
and area of triangle $ABC$ is $AB*CH/2$
that means area of triangle $ABC$ is [b]four times[/b] the area of triangle $CHM$  :roll:


geometry
trapezoid
LaTeX
number theory
prime numbers

I am paraphrasing, but you can still get the general idea.

In trapezoid ABCD in which AB is parallel to CD (I know a line should go above "AB" and "CD", but I don't have LaTeX yet), point E is on AD and point F is on BC so that line segments AB, CD, and EF are parallel to each other.  The region above line segment EF has the same area as the region below it, i.e., the area of trapezoid ABFE is the same as EFCD.  The lengths of AB, EF, and CD are, respectively, [i]p[/i], [i]m[/i], and [i]q[/i], where [i]p[/i]<[i]m[/i]<[i]q[/i].  If [i]p[/i], [i]m[/i], and [i]q[/i] are all prime numbers, what is the lowest possible value of [i]p[/i]?

How do you do this?  Is it purely guess and check, or is there an actual method to this?

[hide="If you want the answer..."]...it's 7.[/hide]

I'm assuming you went to NYSML? :D

[hide]
The height, $h$, of the top trapezoid resulting from another parallel line drawn between $p$ and $q$ ranges from $h$ to $0$ while the base new trapezoid goes from $q$ to $p$. Using this relation, we can write the equation:
$\displaystyle \frac{p+m}{2}\times \displaystyle \frac{m-p}{q-p}=\displaystyle \frac{m+q}{2}\times\displaystyle \left(1- \displaystyle \frac{m-p}{q-p} \right)$

$m^2-p^2=q^2-m^2$

$2m^2=p^2+q^2$

We are told that $p$, $q$, and $m$ are all primes, so writing out the first squares of primes:

$4, 9, 25, 49, 121, 169, 289...$

$7^2+17^2=2(13^2)$
$p=\boxed{7}$

[/hide]

[quote="lingomaniac88"]And I'm assuming you went too.  Only someone who also went to NYSML would recognize that. :D[/quote]

And NYSML means ...  :?

New York State Math League... It's a competition very similar to ARML.

Edit: chess64 beat me... :D

Looking at the solution, I guess that you had to use guess and check eventually.  Like someone would have [hide]$7^2+17^2=2(13^2)$[/hide] in the of their head. :P

[quote="10000th User"][quote="lingomaniac88"]... but I don't have LaTeX yet)...[/quote]But you don't have to get LaTeX to use it on the forums. For example, if you want to type an overline above "AB" all you need to do is:

$\$$\overline{AB}$\$$

which produces $\overline{AB}$

Whenever you want to start a $\text{\LaTeX}$ formula, use dollar signs. This can also start a formula too but it will be centered format ${\backslash[\ldots\backslash]}$ like this

\[ \overline{AB}  \]

Hope this helps[/quote]

I figured that out an hour ago.


How many zeros do you write if you start from 1 to 10000 inclusive?

So if you were to count the zeros from 1-100, then you would find 11. So if you were to find 101-999, you would find 178. If you were to count the zeros from 1000-9999, then you would find 11143. Someone please tell me if that is WAY WAY WAY WAY OFF!!! I worked it out, but I'm pretty sure it's wrong. And 10000 is just 4. So 11+178+11143+4= 11336.

I really hope this helped... But the main key to this problem is figuring out how many zeros per "tens". For example: 20-29 had 1 "0". Since there are 9 tens in 1-99, then that means there are only 9 "0's" in 1-99.    :lol:

calculate the 0s in each digit (hundreds,tens,ones)-and take out 10000 for now
so 1/10 of the numbers would have a 0 in the ones digit, 1/10 of the two-digit and up numbers have a 0 in the tens place, 1/10 of the three-digit and up numbers have a 0 in the hundredth spot, and there is one 0 in the thousands place (10000)
so 1000+999+990+1=2990 zeroes
but that method includes the number 0 in it, so the answer is [b]2989[/b]
and yes u were WAY off

Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=263738]here[/url] for a similar problem. If we are counting the $ n$ digit numbers $ \overline{a_{n\minus{}1}a_{n\minus{}2}\ldots a_1a_0}$, then since by definition $ a_{n\minus{}1}\ne0$, there are $ n\minus{}1$ choices for the digit $ 0$ to be in and then $ 9\cdot10^{n\minus{}2}$ choices for the rest of the digits, so there are $ 9(n\minus{}1)10^{n\minus{}2}$ zeroes in the $ n$ digit numbers. Summing this from $ n\equal{}1$ to $ n\equal{}4$ gives $ 0\plus{}9\plus{}180\plus{}2700\equal{}2889$, but then we have to add the zeroes in $ 10000$ for a total of $ 2889\plus{}4\equal{}\boxed{2893}$.

3 different answers...but jonathan is WAY off
at least the others are close to each other (too lazy to do it)

I'm sorry. I redid the problem and got 2983. Hmm.....  I'm so sorry for my first stupid answer. :blush:

rofl every1 still has different answers :wink:

@math154: I don't understand where you got the $ 9 \cdot 10^{n \minus{} 2}$ choices for the other digits. Could you explain a little more? (Sorry brain not working today)

The answer is 2893. I confirmed it after doing it over again using a slower way.

yea math154, i dont get it either, could u explain it a little more.

Well....

Let's split this thing up into cases.

[b]2 digit[/b]: There are obviously 9 0's.

[b]3 digit[/b]: One 0: (9 ways to choose first digit)*(2 ways to choose the placing of the 0)*(9 ways to choose the other digit)=162.
Two 0's: 100,200,300,...,900 (9 ways)

[b]4 digit[/b]: One 0: (3 ways to choose the placing of the 0)*(9 ways to choose the first digit)*(9*9 ways to choose the other two digits besides 0 and first digit)=2187 ways
Two 0's: (3 ways to choose the placing of 0's)*(9 ways to choose first digit)*(9 ways to choose the other digit)=243 ways.
Three 0's: 1000,2000,3000,4000,...,9000 (9 ways)

[b]5 digit[/b]: WELL DUH, just 4 0's.

Now add them up.

9+162+9*2+2187+243*2+9*3+4=[b]2893[/b] ways!!!

Whew...


A rhombus is inscribed in triangle $ ABC$ in such a way that one of its vertices is $ A$ and two of its sides lie along $ AB$ and $ AC$.  If $ \overline{AC} \equal{} 6$ inches, $ \overline{AB} \equal{} 12$ inches, and $ \overline{BC} \equal{} 8$ inches, the side of the rhombus, in inches, is:

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 3 \frac {1}{2} \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

hello, let the $ x$ side length of the rhombus, then we have the relation $ \frac{6\minus{}x}{x}\equal{}\frac{1}{2}$, from here we get $ x\equal{}4$, so the right answer is $ \boxed{D}$.
Sonnhard.

[hide]
Let the vertex of the rhombus on $ BC$ be $ D$, and the vertex of the rhombus on $ AC$ be $ E$.
$ \bigtriangleup ABC \sim \bigtriangleup EDC$ by $ AAA$.

Let $ x$=$ AE$=$ ED$=the side of the rhombus.
$ EC \equal{} 6 \minus{} x$

Because the two triangles are similar, you set up the proportion:
$ \dfrac{EC}{AC} \equal{} \dfrac{ED}{AB}$

$ \dfrac{6 \minus{} x}{6} \equal{} \dfrac{x}{12}$

Cross-multiplying, you get

$ 72 \minus{} 12x \equal{} 6x$
$ 72 \equal{} 18x$
$ x \equal{} \boxed{4, (D)}$
[/hide]


geometry
3D geometry
sphere
calculus
symmetry
calculus computations

what is the volume of the solid that lies inside both of the spheres

x^2 + y^2 + z^2 + 4x - 2y + 4z +5 = 0

and

x^2 + y^2 + z^2 = 0

?????

[quote="math beginner"]what is the volume of the solid that lies inside both of the spheres

x^2 + y^2 + z^2 + 4x - 2y + 4z +5 = 0

and

x^2 + y^2 + z^2 = 0  :?

?????[/quote]

math beginner - please don't post the same problem twice. It has been moved here to the "Calculus computations and tutorials" section, which is where it belongs - if it belongs anywhere, since as a problem it has no substance.

Since I don't think you understood kunny's comment, let me spell it out more explicitly:

$x^2+y^2+z^2=0$ isn't a sphere with positive volume; instead, it is a single point with no interior and with volume zero.

oops sorry...I was unable to find my problem where I originally posted... so I thought it was missing or I didn't click submit....

yes...there is a mistake when I typed them...it should be

what is the volume of the solid that lies inside both of the spheres

x^2 + y^2 + z^2 + 4x - 2y + 4z +5 = 0

and

x^2 + y^2 + z^2 = [b]4[/b]

 :P

All right- that's two spheres of radius 2, with centers at distance 3. Rotating and translating to a nice position, consider the two spheres $(x-\frac32)^2+y^2+z^2=4$ and $(x-\frac32)^2+y^2+z^2=4$. This should be fairly easy to set up; you just need to find the area of the cross-section for each $x$ and integrate. The symmetry around $x=0$ is intended to help.
[hide="hint"]$\pi\cdot(4-(|x|+\frac32)^2)$[/hide]


function
real analysis
real analysis unsolved

Let f:]0,00[->R be a function with the property that f(x)<=f(nx) for all n in N* and all x>0.

a) if f is continuous, show tht there exist limit{x->0}f(x) and lim{x->00}f(x).
b) if f isn't continuous, show that the 2 limits above may or may not exist.

[i]Gheorghe Titeica contest/2000/11'th grade[/i]

Really nice and interesting problem.I think I have a solution for the first part. Let a such that there exists a sequence (x_n) which tends to zero and such that f(x_n) tends to a. We prove that f(x)>=a for all x.
  Indeed, let x>0. Since x_n tends to 0, the set {mx_n|m,n>=1} is dense in (0,infinity), so there is a sequence (m_kx_n_k)(k>=1) which thends to x. But f(m_k*x_n_k)>=f(x_n_k) and since f(x_n_k) tends to a, we have f(x)>=a.
  Now suppose that f doesn't have limit in 0. Thus, there are sequences a_n and b_n which tend to 0 and reals a and b (different) such that f(a_n) tends to a and f(b_n) tends to b. We may take a<b. Then we have seen that f(a_n)>=b for all n and so a>=b, false. Thus, f has limit in 0. For the limit in infinity, observe that f(x)>=f(x/n) and use the same argument (this time the dense set is {x_n/m|m,n>=1} if x_n tends to infinity).

Here's what I did:

Let's assume there exist sequences (xn) and (yn) which both tend to 0 and f(xn)->a and f(yn)->b<a. Then for some large enough n there exists an yn s.t. f(yn)0. Then there is an entire interval [u,v] s.t. f(x)<a-\varepsilon for all x in [u,v] (because f is continuous). This means that f(x)u/n, so the intervals [u/n,v/n] go to 0 and for large enough n's 2 consecutive such intervals aren' disjoint, so we find that there is a d>0 s.t. f(x)<a-\varepsilon for all 0<x<d, so there can't be a sequence xn->0 s.t. f(xn)->a, so we have a contradiction.


Hellooooooooooo!
I'm also interested in MO like Vivath. But I cannot find any useful website, difference from this one, about the way to solve geometry exercises.
I'm the worst at MO ....
Thanks in advance. :)

Find text books on plane geometry please !  :lol:

[i][color=blue] :?:  :(  well, thanks. But, the main problem is that there is no such a good book. That is the reson why I want to know the Web Pages. I hope the other will help me.[/color][/i]

I have some books on plane geometry, if you want then i will send them to you. If you need find a website then Mathlinks.ro is GOOD+est. :P


Can you save solution of the past ABACUS International Math Challenge?

isn't this supposed to be in another forum :wink:


I'll do #2 now, might do some more later.
[hide="#2"]
Note that $\cos{2x}=2\cos^{2}{x}-1$.
Then $2\cos^{2}{x}-\cos{x}-1=0$.
$\cos{x}=-\frac{1}{2},$ or $1$
$x=\frac{2\pi}{3}, \frac{4\pi}{3},$ or $0$.
[/hide]

Maybe this should go in HSB.

[quote="john1"]
These are due Monday.[/quote]

This sounds like homework. I don't think its right for us to be doing his homework.

Haha, yes this actually is homework. I have some 60 plus problems I have to do, so I only needed help with 4 of those.

And I put this in this forum because they are Pre-Calculus questions. Thanks for the help guys.

What about the other two though?

[hide="hint for 3"]
Use the identities

$\tan x=\frac{2\tan x}{1-\tan^{2}x}$
and
$\cot x=\frac{1}{\tan x}$
[/hide]

[hide="Hint for 4"]
$\sin{(4x)}=\sin{(2\cdot 2x)}$[/hide]

[quote="john1"][quote="rnwang2"][hide="1"]
$\sin{(2x)}=\frac{1}{2}$

$x=\frac{\pi}{12}, \frac{5\pi}{12}$
[/hide]
[hide="2"]
$2\cos^{2}{x}-1-\cos{x}=0$

$(2\cos{x}+1)(\cos{x}-1)=0$

$x=0, \frac{2\pi}{3}, \frac{4\pi}{3}$[/hide][/quote]

Wait, can you explain number one a bit more. If sin 2x = 2 sin x cos x, then how are you getting sin x = 1/2. Also do I subtract that 1 from the equation in order to get the equation to equal 0. For example:
   4 sin x cos x = 1
   4 sin x cos x - 1 = 0
   2 (2 sin x cos x) - 1 = 0

The back of my book lists the answers as 0, pi/12, 5pi/12, 13pi/12, and 17pi/12.

Oh and sorry I literally just signed up and don't know how to make actual pi symbols.[/quote]

$4\sin x\cos x=1$
$2\sin x\cos x=\frac{1}{2}$

There's no need to set equal to 0 unless you plan to factor.

As for making pi symbols and other mathematical expression, you a typesetting language called LaTeX. There's a guide on how to use (click the link at the left of the site). To use LaTeX on the site, put the code between dollar signs.

For example:

[code]
$5+2=\frac{14}{2}$
[/code]

will make $5+2=\frac{14}{2}$

$\frac{13\pi}{12}$ and $\frac{17\pi}{12}$ are solutions as well. Those are all of them between $0$ and $2\pi$, but you can keep on adding $2\pi$ to get more solutions.

Oops...I forgot about the last two solutions.

First, please don't post homework problems. Most of us have better things to do than to do other's homework problems.(yes, you can make up an excuse of not being able to do it, blahblahblah, but you say it's due monday. That's gives you 2 full days to try the problems.)
Second, "intermediate" does NOT mean "pre-calculus." The word intermediate isn't talking about the advanced-ness of the material involved, but the amount of creativity involved. And since these are easy problems(most school problems are), they should go in HSB.
And a tip, for when you're stuck on homework problems: skim the corresponding chapter of the book, and review the important identities, theorems, etc. Those are probably the key to solving the problem.

[quote="junggi"]First, please don't post homework problems. Most of us have better things to do than to do other's homework problems.(yes, you can make up an excuse of not being able to do it, blahblahblah, but you say it's due monday. That's gives you 2 full days to try the problems.)
Second, "intermediate" does NOT mean "pre-calculus." The word intermediate isn't talking about the advanced-ness of the material involved, but the amount of creativity involved. And since these are easy problems(most school problems are), they should go in HSB.[/quote]

I apologize, I literally just signed up to this forum a few moments ago today. Our teacher gave us these problems among others as an extra credit assignment and said that we had to teach ourselves how to do them. I'm not asking that people on these forums do them for me, but simply for help in doing them correctly. The problems I listed above have stumped me all day. I only need to know how to solve them by factoring, using the double angle formulas.

[quote="Altheman"][quote="john1"]I'm not asking that people on these forums do them for me, but simply for help in doing them correctly. [/quote]

what?

hmm....[/quote]

Likse guidance you know. Hints. If you guys are uncool about sharing your knowledge and helping somone solve a problem, then that's cool I guess, but at least helping me in someway shape or form to solve these problems would be nice. =]

[quote="tjhance"][quote="john1"]Hey for number 4, I subtracted the two terms from one another by getting common denominators and got the following result of:

3tan^2x - 1 / tan x - tan^3x

(I'm really sorry that isn't in "math" form by the way.)

Anyway what would I do now?[/quote]

I'm assuming you meant #3, cause it doesn't make sense for #4.

You've got a really ugly denominator. What can you do to get rid of it really easily?[/quote]

Oh, yes, I'm sorry I did mean number 3. Anyway to get rid of the denominator couldn't I multiply by the reciprocal?

Hey guys, number 3 is the only one I can't get. I got common denominators between the two terms and subtracted, but then I get a strange fraction I don't know what to do with.

Well there's a 0 on the other side of the equation, right? Just multiply out the denominator.

[quote="john1"]Ohhh, okay. I see now. So, 3tan^2 - 1 would be equal to 0 (tanx)(1 - tan^2x). But then 3tan^2 - 1 would just equal 0 after the two terms would be multiplied out on the one side. But the back of my book also says I should be getting pi/2 and 3pi/2 as answers. How would I get these as answers?

And then after that the only help I need is with number 4. Thanks alot.[/quote]
Dude look at my hint on the previous page.

[quote="john1"]Ohhh, okay. I see now. So, 3tan^2 - 1 would be equal to 0 (tanx)(1 - tan^2x). But then 3tan^2 - 1 would just equal 0 after the two terms would be multiplied out on the one side. But the back of my book also says I should be getting pi/2 and 3pi/2 as answers. How would I get these as answers?

And then after that the only help I need is with number 4. Thanks alot.[/quote]

Okay so you have
$3\tan^{2}x-1=0$
You should be able to solve for $\tan x$ yourself. ONce you find that $\tan x$ equals, consider a 30-60-90 degree triangle to figure out what angle will give u that value of the tangent.

alright thanks everyone who helped me with these.

[quote="john1"]In the following exercises, find the exact solutions of the equation in the interval [0, 2pi)

1. 4 sin x cos x = 1

2. cos 2x - cos x = 0

3. tan 2x - cot x = 0

4. sin 4x = -2 sin 2x

I am completely stumped and so any help would be extremely appreciated. These are due Monday. Thanks.[/quote]

I swear I did these problems..when I was going back reviewing this for AIME..and amazingly I actually used trig for number 9...


function
arithmetic sequence
Functional Equations
pen

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(m)+f(n))=m+n.\]

Plugging $n=m$ leads us to $(1)$  \[f(2f(m))=2m\]

We have then \[f(2f(m))=2m\]  \[f(2f(n))=2n\]
By adding: \[f(2f(m))+f(2f(n))=2m+2n\]

Taking both sides as arguments for $f$, we obtain $(2)$ \[2f(m)+2f(n)=f(2m+2n)\]
Now assume we have $f(1)$ and $f(2)$. Let $k \in N$ and $k>1$. Then from the condition $(2)$ we obtain: \[2f(k)+2f(k)=f(2k+2k)=f(2(k+1)+2(k-1))=2f(k+1)+2f(k-1)\]
which means that \[f(k+1)=2f(k)-f(k-1)\] so our function is an arithmetic sequence of form \[f(m)=am+b.\] Plugging this into our equation we get \[a^{2}(m+n)+2ab+b=m+n\] for every $m,n \in N$. Thus $a=1$ and $b=0$ and finally \[f(m)=m\]

Sorry to revive, I have a very short and nice solution.

Clearly $ f$ is injective. Take the smallest $ a$ for which $ f(a)\ne a$. If $ a>1$, $ f(a)\equal{}f(f(1)\plus{}f(a\minus{}1))\equal{}a$, contradiction. So $ a\equal{}1$. Let $ f(1)\equal{}n>1$. $ f(n\plus{}f(n\minus{}1)) \equal{} f(f(1)\plus{}f(n\minus{}1))\equal{}n\equal{}f(1)$. So $ n\plus{}f(n\minus{}1)\equal{}1$, which is impossible. Thus $ f(x)\equal{}x,\forall x\in\mathbb{N}$.

[quote="Johan Gunardi"]Sorry to revive, I have a very short and nice solution.

Clearly $ f$ is injective. Take the smallest $ a$ for which $ f(a)\ne a$. If $ a > 1$, $ f(a) \equal{} f(f(1) \plus{} f(a \minus{} 1)) \equal{} a$, contradiction. So $ a \equal{} 1$. Let $ f(1) \equal{} n > 1$. $ f(n \plus{} f(n \minus{} 1)) \equal{} f(f(1) \plus{} f(n \minus{} 1)) \equal{} n \equal{} f(1)$. So $ n \plus{} f(n \minus{} 1) \equal{} 1$, which is impossible. Thus $ f(x) \equal{} x,\forall x\in\mathbb{N}$.[/quote]
Very nice your solution,I like it :)

[quote=jastrzab]which means that \[f(k+1)=2f(k)-f(k-1)\] so our function is an arithmetic sequence of form \[f(m)=am+b.\] [/quote]

I don't really get this part...

soluiton; it's clear that $f$ is injective.
  we have $f(f(n+1)+f(n-1))=2n$ and $f(2f(n))=2n$
thus from injectivity $f(n+1)+f(n-1)=2f(n)$ ie $f(n+1)-f(n)=f(n)-f(n-1)$ ie $\frac{f(n+1)-f(n)}{(n+1)-n}=\frac{f(n)-f(n-1)}{n-(n+1)}=cte$ thus $f(n)=n+c$ replacing we get $c=0$

Oh yeah so next is $f(n+1)=f(n)+c$  thus
$f(n)=f(1)+(n-1)c\implies f(n)=nc+f(1)-c=an+b$
Where $a=c$ and $b=f(1)-c$

let $f(1)=c$ and we can easily see that $f$ is injective also $\forall i>1 \exists y : f(y)=i$ so we shall define for $i>1, a_i$

such as that $f(a_i)=i$ and then for $m>2, n>1 \implies$

$f(f(a_m)+f(a_n))=f(f(a_{m-1})+f(a_{n+1}))=f(m+n) \implies a_{n+1}-a_n=a_m-a_{m-1}$ now put $m$

a constant which will imply that $a_{n+1}-a_n = c_0, n \ge 2$ for some constant $c_0$. now combining $(1,1)$ gets us

$f(f(1)+f(1))=2 \implies f(2c)=2$ again combining $(2c,2c)$ get us $f(4)=4c$, combing $(4,4)$ would imply

$f(8c)=8$, now we see that $a_8=8c,a_2=2c, a_8-a_2=6c_0 \implies c_0=c$ and

now because $a_2=2c$ then $a_3=3c$ and then $\forall i>1 : a_i=ic$

assume that $c>1$ and $f(c-1)=k \implies c-1=a_k=kc$ which is a contradiction so $c=1$ and $\forall i \in N: f(i)=i$


Let I be the center of a circle inscribed in triangle ABC. AI inersect BC at D, intersects the circumcircle of triangle ABC at E. Prove AE/IE=IE/DE.  :?  :(

EDIT: trig is allowed

[quote="D.P.L"]Let I be the center of a circle inscribed in triangle ABC.[/quote]
is this circle the incircle or just a circle :?:

This has already been answered in the other topic.

Could you please post the link lotrgreengrapes7926, I'm curious to see what the solutions are compared to mine. :P


Prove that there do not exist solutions to $x^2y^2 = z^2(z^2-x^2-y^2)$ in positive integers.

Considering it as a cuadratic in z, let $a=x^2+y^2$, $b=2xy$ we have $a^2-b^2$ and $a^2+b^2$ (discriminant of the equation) are squares, so $a^4-b^4=t^2$ which is wellknown to be impossible!


There should be [b]more[/b] problem solving contests or competitions, even though there already is one.  Maybe a problem of the week EDIT, nevermind there's already a problem of the week.

[quote="anirudh"]maybe a competition (long one) by a moderator  :wink:[/quote]
There used to be a competition, but it didn't really work (people signed up and questions were posted, but then it just stopped--probably no one handed in answers.) Hopefully, if another one is started, it will continue longer.

[quote="SplashD"][quote="lotrgreengrapes7926"]I'm thinking of creating a competition. I'll probably post the information sometime in the next week.  :D[/quote]

be sure to include a countdown round  :wink:[/quote]
This is MOEMS, not MathCounts.  :P
I will probably have 2 written rounds and a countdown for the first few places.

another way is to create a problem of the less than or equal to hour thingy, where people just post and try to solve question

[quote="sirorange"]It is I, your friendly moderator-type figure sirorange.

Now, it is no secret that the MOEMS boards are slow, and get very little action. So, if you have any suggestions on how to make this board a more fun and active place for all, I would greatly appreciate if you posted them here, or on my blog.

Help me make AOPS a better place, thank you :D[/quote]

proof that the board is slow is that it took more than 8 months for someone to respond.  :P


Prove that there exist an infinite number of ordered pair (a,b) of integer such that: $ ab | a^{2007}\plus{}b^{2006}\plus{}1$

If $ (a,b)$ solution, then $ (a,\frac{a^{2007}\plus{}1}{b}), (\frac{b^{2006}\plus{}1}{a},b)$ are solutions too. Therefore we can build infinetely many solutions beginning from (1,1).

see more here
http://www.mathlinks.ro/Forum/viewtopic.php?t=167183


Let $ \minus{}2 \leq x\leq 5$.
Find the maximum and the minimum values of $ y\equal{}|3\minus{}5|4\minus{}|2x\plus{}4| | |$.

[quote="mathson"]The solution:
$ (0,47)$[/quote]

Please post the solution

We can distribute the 5 to get $ |3\minus{}|20\minus{}|10x\plus{}20\parallel{}|$

errr... since $ x \ge \minus{}2$, we get $ |3\minus{}|10x|$. If $ x>0$, we have $ |3\minus{}10x|$, Otherwise, we have $ |3\plus{}10x|$. The max and min of the first expression in $ [0,5]$ respectively are 47 and 0. Then max and min of the second in $ [\minus{}2,0]$ respectively is 17 and 0. Thus, the max is 47 and the min is 0.


I've got a continued fraction I would like to shrink in size.  How would I do that?

One way is to use \scriptstyle but because you need it at every fraction, redefine\cfrac to do so. Instead of [code]$1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\dotsb}}}}$[/code]use [code]\newcommand{\csfrac}{\scriptstyle \cfrac}
$1+\csfrac{1}{1+\csfrac{1}{1+\csfrac{1}{1+\csfrac{1}{1+\dotsb}}}}$[/code]and[code]\newcommand{\csfrac}{\scriptscriptstyle \cfrac}[/code]would make it smaller still.


Met het woordenboek , waar de university stuff apart moest, ben ik beetje verdwaald en weet ik niet meer waar wat moet.µ

Kan iemand me helpen deze termen naar het engels vertalen

stralenwaaier (alle rechten door een specifiek punt in 3d in een vast vlak door dat punt)
stralen schoof (alle rechten door een specifiek punt in 3d)
stralenveld (alle rechten in een specifiek vlak in 3d)

geraamte (een verzameling van n+2 punten in een ndimensionale projectieve ruimte, waarvan geen n+1 lineair afhankelijk zijn, in 2d zijn dat er dus vier waarvan geen drie collineair)  zou dit "frame" kunnen zijn?

schierveld (alle eigenschap van een deelring (zoals quaternionen) maar dan zonder de $(a+b)c=a c+b c$ eis, je hebt wel nog $a (b+c)=a b+a c$)  (nederlands/vlamingen opgelet : nederlanders zeggen deelring op quaternionen, vlamingen nietcommutatief lichaam , ik denk ook dat de nederlanders bijnalichaam zeggen op schierveld?)

Als je er een paar van zou weten zou ik al een hele stap vooruit zijn.

sorry als het verkeerd staat hoor

maar is er iemand die bijvoorbeeld al zou weten wat een stralenwaaier is in het engels, iets met ray of zo in is het blijkbaar niet
ik sta open voor alle ideeen

Ik denk eigenlijk dat je best eens een topic post met de engelse omschrijvingen in college section, en daar vraagt naar de correcte term.


[quote="236factorial"]The diameter of a circle is a whole number and the area of the circle is between 100 and 120 square units. What is the length of the circumference of this circle? Express your answer in terms of $\pi$[/quote]
[hide]
Using casework, we find that the perfect square, $36$ mulitplyed by $\pi$ is equal to a number between 100 and 120. (approximately $113$)

So the radius of the circle is $6$ and the diameter is $12$. So the circumference is $12\pi$[/hide]

You can do it without casework as follows.  Find the upper limit and the lower limit.  $\frac{120}{\pi}$ and $\frac{100}{\pi}$  They are about 40 and 32.  The only perfect square between 32 and 40 is 36.  The radius is therefore $\sqrt{36}=6$.  That would make the circumfrence of the circle $2\times6\times\pi=\mbox{12\pi}$.