geometry
trigonometry
analytic geometry
geometry proposed

Let be $ ABCD$ a square , $ M\in (BC)\ ,\ N\in (CD)$ so that $ MB \equal{} m \cdot BC$ and $ ND \equal{} n\cdot NC$ . Let be $ \{P\} \equal{} BN\cap MD$ . Prove that :

$ AP \equal{} AB\Longleftrightarrow mn \equal{} 2$

It doesn't seem right. What is the point of P?

[quote] dgreenb801 WROTE :I've computed this carefully with geogebra and the problem is wrong. Can anyone give a correct statement?[/quote]
I think this is true ,and we can use analytic geometry to solve it :
WE SHOULD KNOW THAT:$ {\boxed{\frac{BM}{MC}=m;\frac{DN}{NC}=n}}$
First : we prove :$ {AP=AB}$ $ {\rightarrow}$ $ {mn=2}$
Let O be vertex A  of square ABCD in $ {x-O-y}$ plane :let $ {D(r,0)}$;let $ {A(0,-r)}$; Circle O(A) with the equation $ {x^2+y^2=r^2}$; Chose a point $ {P}$ on circle O such that $ {P(xo,yo)}$; easy to get :$ {l1(MD)}$:$ {y=\frac{-yo}{r-xo}(x-r)}$;$ {l2(BN)}$:$ {y=\frac{yo+r}{xo}x-r}$; EASY to get :$ {BM=\frac{r(r-xo+yo)}{yo}}$;$ {DN=\frac{r(xo-yo-r)}{xo}}$;easy to find expression of $ {\frac{BM}{MC}*\frac{DN}{NC}[mn]}$ is $ {\frac{r-xo+yo}{xo-r}*\frac{xo-yo-r}{yo+r}}$;that is : with$ {[\boxed{xo^2+yo^2=r^2}]}$    $ {mn=\frac{-(r-xo+yo)^2}{(xo-r)(yo+r)}=\frac{r^2+2ryo-2rxo+\boxed{(xo^2+yo^2)}-2xoyo}{-r^2+xoyo-r(yo-xo)}}$  .$ {\rightarrow}$   $ {\frac{2(r^2-xoyo+r(yo-xo))}{r^2-xoyo+r(yo-xo)}=2}$ that is $ {mn=2}$ We have prooved the first part .
Second : we prove ：$ {mn=2}$ $ {\rightarrow}$ $ {AP=AB}$
Let $ {\frac{BM}{MC}=m}$;$ {\frac {DN}{CN}=n}$;easy to get that :$ {M(\frac{m}{m+1}r,-r)}$;and$ {N(r,-\frac{n}{n+1}r)}$ ;we get $ {l1(DM)}$:$ {y=(m+1)(x-r)}$;we also get $ {l2(BN)}$:$ {y=\frac{x}{n+1}-r}$; WITH equation of $ {l1}$ and $ {l2}$ WE GET with $ {[\boxed{mn=2}]}$ $ {P(x)=\frac{n+1}{mn+m+n}rm=\frac{rm(n+1)}{m+n+\boxed{mn}}=\frac{rm(n+1)}{m+n+2}}$ so $ {P(\frac{rm(n+1)}{m+n+2},\frac{-r(n+2)}{m+n+2})}$; so :AP=$ {\sqrt{\frac{r^2(\boxed{(n+2)^2+m^2(n+1)^2)}}{\boxed{(m+n+2)^2}}}}$ compare $ {(n+2)^2+m^2(n+1)^2})$ and $ {(m+n+2)^2}$
${ {(n+2)^2+m^2(n+1)^2=n^2+4n+4+(\boxed{mn})^2+2(\boxed{mn})}m+m^2=n^2+m^2+4(m+n)+8}$; and $ {(m+n+2)^2=m^2+n^2+2\boxed{mn}+4m+4n+4=m^2+n^2+4(m+n)+8}$;that is $ {(n+2)^2+m^2(n+1)^2=(m+n+2)^2}$; so $ {AP=\sqrt{r^2}=r=AB}$     ;That is :  $ {AB=AP}$   $ {\longleftrightarrow}$     $ {mn=2}$  ;  $ {Q.E.D}$

[geogebra]b248d0816bda18d579c818abf13d39ae5ee90a34[/geogebra]
It can't be true, see the diagram, we must have $ \frac {MB}{BC} \cdot \frac {ND}{NC} \equal{} 2$.

EDIT:O yes, I see, it is $ \frac{BM}{MC}$, not $ \frac{BM}{BC}$ as he typed.

Here is my solution:
Let s be the side length of ABCD. Draw the diagonal BD. Draw the circle center A radius s, passing through P. We have  $ \frac {MB}{BC} \cdot \frac {ND}{NC} \equal{}$
$ \frac{[BDM]}{[MDC]} \cdot \frac{[DBN]}{[NBC]}\equal{}$
$ \frac {BD \cdot sin DBN}{s \cdot sin NBC} \cdot \frac {BD \cdot sin BDM}{s \cdot sin MDC}$
But $ \frac {BD}{s} \equal{} \sqrt {2}$, and $ \angle DBN \equal{} \angle MDC$, as they intercept the same arc, similarly $ \angle NBC \equal{} \angle BDM$. Thus, $ \frac {MB}{BC} \cdot \frac {ND}{NC} \equal{} 2$
The converse is then easy to show.

[quote][color=darkred]Let $ ABCD$ be a square. For $ M\in (BC)$ , $ N\in (CD)$ denote  $ P\in BN\cap MD$ . Prove that  $ AP \equal{} AB\ \Longleftrightarrow\ \frac {MB}{MC}\cdot\frac {ND}{NC} \equal{} 2$[/color] .[/quote]

[color=darkblue][b][u]Proof[/u].[/b] Suppose w.l.o.g. $ AB \equal{} 1$ and denote $ \left\|\begin{array}{ccc} m(\widehat {NBD}) \equal{} x & ; & (\widehat {NBC}) \equal{} y \\
 \\
m(\widehat {MDB}) \equal{} y' & ; & (\widehat {MDC}) \equal{} x'\end{array}\right\|$ . Observe that $ x \plus{} y \equal{} x' \plus{} y' \equal{} 45^{\circ}\ (1)$ .

Using an well-known relation obtain $ \left\|\begin{array}{c} \frac {ND}{NC} \equal{} \frac {BD}{BC}\cdot\frac {\sin \widehat {NBD}}{\sin\widehat {NBC}} \equal{} \sqrt 2\cdot\frac {\sin x}{\sin y} \\
 \\
\frac {MB}{MC} \equal{} \frac {DB}{DC}\cdot\frac {\sin \widehat {MDB}}{\sin\widehat {MDC}} \equal{} \sqrt 2\cdot\frac {\sin y'}{\sin x'}\end{array}\right\|\ \bigodot\ \implies\ \frac {MB}{MC}\cdot\frac {ND}{NC} \equal{} \frac {2\sin x\sin y'}{\sin y\sin x'}$ .

Therefore, $ \boxed {\ \frac {MB}{MC}\cdot\frac {ND}{NC} \equal{} 2\ }\Longleftrightarrow \frac {\sin x\sin y'}{\sin y\sin x'} \equal{} 1$ $ \Longleftrightarrow$ $ \cos (x \minus{} y') \minus{} \cos (x \plus{} y') \equal{} \cos (y \minus{} x') \minus{} \cos (y \plus{} x')$ $ \stackrel{(1)}{\Longleftrightarrow}$

$ \cos (x \plus{} y') \equal{} \cos (y \plus{} x')$ $ \Longleftrightarrow$ $ x \plus{} y' \equal{} y \plus{} x'$ $ \stackrel{(1)}{\Longleftrightarrow} x \equal{} x'\ \wedge\ y \equal{} y'$ $ \Longleftrightarrow P\in C(A,1)$ $ \Longleftrightarrow AP \equal{} 1$ $ \Longleftrightarrow \boxed {\ AP \equal{} AB\ }$ .

[b][u]Remark[/u].[/b] I proposed this problem and it appears as the [u]exercise 2.40[/u] in my book

"[b]GEOMETRIE PLANA (sintetica, vectoriala, analitica). Culegere de probleme, Editura GIL, Zalau, 2002[/b]". [/color]


integration
real analysis
limit
number theory
prime numbers
number theory unsolved

Let $ \pi(x)$ denote the number of primes not exceeding $ x$. Show that

$ \sum_{p<\equal{}x} 1/p \equal{} \frac{\pi(x)}{x} \plus{} \int^x_2\pi(u)/u^2 du$

Then deduce that $ limsup_{x\minus{}>\infty} \frac{\pi(x)}{x/log x} >\equal{} 1$

If $ p_{1}, \ldots, p_{r}$ denote the prime numbers in the interval $ [2,x]$ and $ p_{r + 1}$ is initially set to $ x$ we have that

${{ \int_{2}^{x} \frac {\pi(u)}{u^{2}}du = \sum_{k = 1}^{r} \int_{p_{k}}}^{p_{k + 1}} \frac {\pi(u)}{u^{2}}du = \sum_{k = 1}^{r} \lim_{n \to \infty} \int_{p_{k}}}^{p_{k + 1} - \frac {1}{n + 1}} \frac {\pi(u)}{u^{2}}du = \ldots \\
\ldots = \sum_{k = 1}^{r} k\left(\frac {1}{p_{k}} - \frac {1}{p_{k + 1}}\right) = \left(\sum_{k = 1}^{r} \frac {1}{p_{k}}\right) - \frac {r}{p_{r + 1}} = \left(\sum_{k = 1}^{r} \frac {1}{p_{k}}\right) - \frac {\pi(x)}{x}$

and the veracity of violetcraze's first claim follows...


f is from positive reals to positive reals
for all x,y > 0 ; f(x+y)>f(x)(1+yf(x))
show that there isnt any function f which obtains this conditions.

Let $f(1)=a>0,g(x)=\frac{1}{a+1-x}$. From $\frac{f(x+y)-f(x)}{y}>f^{2}(x),f(1)=a=g(a),g'(x)=g^{2}(x)$
we have $f(x)\ge g(x)$ for all x>1. It give contradition.

i think g(a)=1 instead of a
and i cant understand anything from your solution (my algebra is very bad)

Here is an understable proof :)
1)$f(x+y)-f(x)>y(f(x))^{2}\implies$ $f$ is strictly increasing in $\mathbb{R^{+}}$
2)As $f\neq0$ $\implies$ $\exists a$ such that $f(a)\neq0$
$\forall y>0$we have $f(y+a)>f(a)+y(f(a))^{2}$ then  $\lim_{y\to\infty}f(y+a)=+\infty$ so $f$ is not bounded.then there exists $k$ such that $f(k)>0$
So $\forall x\geq k$ we have $f(x)>0$
For such $x$ take $y=\frac{1}{f(x)}$ then $f(x+\frac{1}{f(x)})>2f(x)$
Now take$x_{n}$ such that $x_{0}=k$ & $x_{n+1}=x_{n}+\frac{1}{f(x_{n})}$ and also take $A_{n}=f(x_{n})$
We have $A_{n+1}=f(x_{n+1})=f(x_{n}+\frac{1}{f(x_{n})})>2f(x_{n})=2A_{n}$
Then $A_{n}>2^{n}A_{0}$and then $A_{n}$is not bounded
But $x_{n+1}-x_{n}=\frac{1}{f(x_{n})}=\frac{1}{A_{n}}<\frac{1}{2^{n}A_{0}}$
Easy to deduce that $x_{n+1}<k+\frac{2}{A_{0}}$
But $f$ is inreasing then $A_{n+1}=f(x_{n+1})<f(k+\frac{2}{A_{0}})$ which gives a contradiction.


Let ABC.....E be a regular polygon of unit side 1.Consider the triangles at A that are determined by BE and the diagonals at A.Then,for each of these triangles,the length of one of the sides on BC is equal to th product of lengths of the other two sides.

This problem becomes almost trivial as soon as one understands the grammar ;) . Actually, here is how I would rewrite the problem:

[color=blue][b]Problem.[/b] Let $A_1A_2...A_n$ be a regular n-gon with sidelength $A_1A_2=A_2A_3=...=A_nA_1=1$. Let k be an integer such that $2\leq k\leq n-1$. Let the lines $A_1A_k$ and $A_1A_{k+1}$ meet the line $A_nA_2$ at the points $P_k$ and $P_{k+1}$. Prove that $A_1P_k\cdot A_1P_{k+1}=P_kP_{k+1}$.[/color]

[i]Solution.[/i] Since the n-gon $A_1A_2...A_n$ is regular, it has a circumcircle, and its equal sides $A_1A_2$, $A_2A_3$, ..., $A_nA_1$ are equal chords in this circumcircle. In particular, the chords $A_kA_{k+1}$ and $A_nA_1$ are equal. Since equal chords in a circle have equal chordal angles, the chordal angles $\measuredangle A_kA_1A_{k+1}$ and $\measuredangle A_nA_2A_1$ of the two equal chords $A_kA_{k+1}$ and $A_nA_1$ are equal. So we have shown that $\measuredangle A_kA_1A_{k+1}=\measuredangle A_nA_2A_1$. In other words, $\measuredangle P_kA_1P_{k+1}=\measuredangle A_1A_2P_{k+1}$. Also, trivially, $\measuredangle P_kP_{k+1}A_1=\measuredangle A_1P_{k+1}A_2$. Thus, the triangles $P_kA_1P_{k+1}$ and $A_1A_2P_{k+1}$ are similar. Consequently, $\frac{A_1P_k}{P_kP_{k+1}}=\frac{A_2A_1}{A_1P_{k+1}}$. In other words, $\frac{A_1P_k}{P_kP_{k+1}}=\frac{A_1A_2}{A_1P_{k+1}}$, so that $A_1P_k\cdot A_1P_{k+1}=P_kP_{k+1}\cdot A_1A_2$. Since $A_1A_2=1$, this becomes $A_1P_k\cdot A_1P_{k+1}=P_kP_{k+1}$. This solves the problem.

  Darij


conics
hyperbola
Asymptote
analytic geometry
graphing lines
slope
rotation

Find the values of $ a$ and $ b$ that make the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ congruent to the hyperbola $ xy=1$.

$ a = 1$ and $ b =1$

the slope of the asymptotes are a / b, and hence we have our asymptotes as

$ y = \pm x$

since the negative reciprocal of $ x$ is $ -x$, the asymptotes are perpendicular. Thus, the hyperbola

$ x^{2}=y^{2}= 1$ is a rotated version of $ xy = 1$

Also, since my other hyperbola thread has pretty much lost its attention, what would be the co-ordinates of the focii of a hyperbola whose formula is $ y = \frac{a}{x}$ ????

actually, $ a=b=2$

This is because the minimum distance from the origin must be $ \sqrt{1^{2}+1^{2}}$. And from the need to have asymptotes y=x, y=-x, you can derive a=b, but not a=b=1.

For your question about focii, consider $ \frac{x^{2}}{1}+\frac{y^{2}}{1}=1$. Here, a=b=1, so the eccentricity is:

$ b^{2}=a^{2}(e^{2}-1)$
$ e^{2}=2$
$ e=\sqrt{2}$

And so the focii lie at $ (\pm ae,0) = (\pm\sqrt{2},0)$

Which means the focii are $ \sqrt{2}$ * min distance from origin. Now, since $ y=\frac{a}{x}$ is [i]similar[/i] to $ \frac{x^{2}}{1}+\frac{y^{2}}{1}=1$, it's focii must also be $ \sqrt{2}$ * min distance from origin. Now, the points which are the minimum distance are $ (\sqrt{a},\sqrt{a}), (-\sqrt{a},-\sqrt{a})$, and this distance is $ \sqrt{\sqrt{a}^{2}+\sqrt{a}^{2}}= \sqrt{2a}$. This means the focii are $ \sqrt{2}*\sqrt{2a}= 2\sqrt{a}$ from the origin (and must like on y=x). This means each x, y coordinate must be: $ \sqrt{2}x=\pm2\sqrt{a}$, so $ x=y=\pm\sqrt{2}\sqrt{a}$, or rather:

$ (\sqrt{2a},\sqrt{2a}), (-\sqrt{2a},-\sqrt{2a})$

[quote="aspiringchemlete"]This might be a little tricky, so
[hide="Hint"]The asymptotes have to be perpendicular, and the distance from the center to the vertices must be the same as with $ xy=1$. [/hide][/quote]

When you mean "congruent", are you just saying that any rotation of the hyperbola is okay? Because there is no way to get that hyperbola exactly the same as $ xy=1$, since that would involve a rotation of axes, which produces the $ xy$ term in the General Quadratic Form $ Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F = 0$.

If we just want any old rotation, this is what we do:

[hide]From the hint, all we need to do is make the asymptotes perpendicular, and make the distance from the center to the vertices the same as with $ xy=1$.

To make the asymptotes perpendicular, note that the product of the slopes of the asymptotes must be $ -1$, and the slopes are $ \pm b/a$. Hence,
$ -b^{2}/a^{2}=-1$
$ b=\pm a$

Now we find the distance from the hyperbola to the vertex. The vertex is the closest point to the origin. Let $ (x,1/x)$ be the closest point to the origin. Then the distance to the origin is
$ \sqrt{x^{2}+\frac{1}{x^{2}}}$
which is greater than or equal to $ \sqrt{2}$ by AM-GM. Equality occurs when $ x^{2}= 1$, or $ x=\pm 1$. So the distance is then $ \pm \sqrt{2}$.

In a regular hyperbola, the distance from the center to the foci would be $ b$, so $ b=\pm \sqrt{2}$

In either case, the hyperbola becomes
$ \frac{x^{2}}{2}-\frac{y^{2}}{2}=1$[/hide]

vishalarul's answer is correct. By "congruent," I mean they have the same shape.

dammit... I've only started working on hyperbolas and briefly forgot that the divisors of the $ x^{2}$ and $ y^{2}$ terms were the squares of $ a$ and $ b$. I was still thinking of $ \frac{x^{2}}{2}+\frac{y^{2}}{2}= 1$ when I said $ a=b=2$.

[quote="t0rajir0u"]"Same shape" means similar. By congruent, I presume you mean that they can be mapped onto one another by translation and rotation only.[/quote]

you forgot flipped (I forget if there's a more mathematical word for "flipped")

It turns out my textbook (New Senior Mathematics, Four Unit Course For Year 12, J.B.Fitzpatrick), actually discusses the cases of hyperbolas in standard form being congruent to the $ xy=k^{2}$ form:

[quote="Textbook"]

If this rectangular hyperbola [$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{a^{2}}=1$] is rotated through $ \frac{\pi}{4}$ in an anticlockwise direction about the origin, its equation becomes $ xy=\frac{a^{2}}{2}$, i.e. $ xy=k^{2}$ where $ k^{2}=\frac{a^{2}}{2}$ and its asymptotes are the X-axis and Y-axis respectively.

[/quote]


Points $ A,B,C$ and $ D$ lie on a line, in that order, with $ AB\equal{}CD$ and $ BC\equal{}12$. Point $ E$ is not on the line, and $ BE\equal{}CE\equal{}10$. The perimeter of $ \triangle AED$ is twice the perimeter of $ \triangle BEC$. Find $ AB$.

$ \text{(A)}\ 15/2 \qquad
\text{(B)}\ 8 \qquad
\text{(C)}\ 17/2 \qquad
\text{(D)}\ 9 \qquad
\text{(E)}\ 19/2$

(D) =9 ...

let x be AB and
let y be AE (=ED)

nwo we have

32*2=64=12+2*(x+y)

x+y=26

let v be the hight od ADE(on AD)

then we have

v^2+36=100

(x+6)^2+v^2=(26-x)^2

now we only find x...


A student is playing computer. Computer shows randomly 2002 positive numbers. Game's rules let do the following operations
- to take 2 numbers from these, to double first one, to add the second one and to save the sum.
- to take another 2 numbers from the remainder numbers, to double the first one, to add the second one, to multiply this sum with previous and to save the result.
- to repeat this procedure, until all the 2002 numbers won't be used.
Student wins the game if final product is maximum possible.
Find the winning strategy and prove it.

Could someone explain what does "..to take 2 numbers from these, to double first one, [color=#f00]to add the second one and to save the sum.[/color]" mean?!

[quote=randomusername]The question can be translated to:
Given $a_1,a_2,\dots,a_{2n}$,
find the permutation $b_1,b_2,\dots,b_{2n}$ for which the product
$P=(2b_1+b_{n+1})(2b_2+b_{n+2})\dots(2b_{n}+b_{2n})$
is maximal.

We will use a switching algorithm. First, observe that if $b_k\le b_{k+n}$ ($k\in\{1,2,\dots,n\}$), then $2b_k+b_{k+n}\le 2b_{k+n}+b_k$, so switching the two yields a larger $P$. We can keep performing this operation until we get $x\ge y$ for all $x\in\{b_1,b_2,\dots,b_n\}$ and $y\in\{b_{n+1},b_{n+2},\dots,b_{2n}$. After this, we should investigate what happens when we switch $b_k$ for $b_\ell$ ($k,\ell\in\{1,2,\dots,n\}$,$k\neq \ell$). The product $P$ after this switch becomes $P'$, so $\frac{P'}P=\frac{(2b_\ell+b_{k+n})(2b_k+b_{\ell+n})}{(2b_k+b_{k+n})(2b_\ell+b_{\ell+n})}$. This increases $P$ iff $(2b_\ell+b_{k+n})(2b_k+b_{\ell+n})\ge (2b_k+b_{k+n})(2b_\ell+b_{\ell+n})$, which is equivalent to $b_\ell b_{\ell+n}+b_{k+n} b_k\ge b_k b_{\ell+n}+b_{k+n}b_\ell$, factoring to $(b_\ell-b_k)(b_{\ell+n}-b_{k+n})\ge0$. From here we can see that the maximal $P$ is taken when sequences $b_1,b_2,\dots,b_n$ and $b_{1+n},b_{2+n},\dots,b_{n+n}$ are oppositely sorted, and the first sequence is larger than the latter. Indeed, if this doesn't hold, we can always conduct a switch described above.[/quote]

Do we also switch $b_i$ with $b_{n+j}$ where $i$ is not equal to $j$?

[quote=randomusername]The question can be translated to:
Given $a_1,a_2,\dots,a_{2n}$,
find the permutation $b_1,b_2,\dots,b_{2n}$ for which the product
$P=(2b_1+b_{n+1})(2b_2+b_{n+2})\dots(2b_{n}+b_{2n})$
is maximal.

We will use a switching algorithm. First, observe that if $b_k\le b_{k+n}$ ($k\in\{1,2,\dots,n\}$), then $2b_k+b_{k+n}\le 2b_{k+n}+b_k$, so switching the two yields a larger $P$. We can keep performing this operation until we get $x\ge y$ for all $x\in\{b_1,b_2,\dots,b_n\}$ and $y\in\{b_{n+1},b_{n+2},\dots,b_{2n}$. After this, we should investigate what happens when we switch $b_k$ for $b_\ell$ ($k,\ell\in\{1,2,\dots,n\}$,$k\neq \ell$). The product $P$ after this switch becomes $P'$, so $\frac{P'}P=\frac{(2b_\ell+b_{k+n})(2b_k+b_{\ell+n})}{(2b_k+b_{k+n})(2b_\ell+b_{\ell+n})}$. This increases $P$ iff $(2b_\ell+b_{k+n})(2b_k+b_{\ell+n})\ge (2b_k+b_{k+n})(2b_\ell+b_{\ell+n})$, which is equivalent to $b_\ell b_{\ell+n}+b_{k+n} b_k\ge b_k b_{\ell+n}+b_{k+n}b_\ell$, factoring to $(b_\ell-b_k)(b_{\ell+n}-b_{k+n})\ge0$. From here we can see that the maximal $P$ is taken when sequences $b_1,b_2,\dots,b_n$ and $b_{1+n},b_{2+n},\dots,b_{n+n}$ are oppositely sorted, and the first sequence is larger than the latter. Indeed, if this doesn't hold, we can always conduct a switch described above.[/quote]

Moreover, what do you mean by "oppositely sorted'? One of them increasing the other one decreasing, or vice versa?

@Arslan, yes that is what "oppositely sorted" means. In this question, it should be clear that the "vice versa" part does not matter.

WLOG suppose that this set of numbers $\{a_n\}$ obeys the following: $a_1 \geq a_2 \geq ... \geq a_{2002}$.

Claim: The numbers that are doubled are precisely the largest 1001 numbers in $\{a_n\}$.
Proof: Suppose by contradiction that the maximal product $P$ has a doubled number $a_i$ not in $\{a_1, a_2, ..., a_{1001}\}$. In $P$, there exists two factors $(2a_i+a_j)$ and $(2a_k+a_l)$. For the sake of maximality, $a_j$ must be less than or equal to $a_i$, otherwise switching the two numbers would yield a larger product. By the pigeonhole principle, there exist numbers $a_k, a_l$ where $a_k, a_l \in \{a_1, a_2, ..., a_{1001}\}$. So $P = (2a_i+a_j)(2a_k+a_l)K = (4a_i a_k + 2a_i a_l + 2a_j a_k + a_j a_l)K$. Define $Q = (2a_l+a_j)(2a_k+a_i)K = (4a_k a_l + 2a_i a_l + 2a_j a_k + a_i a_j)K$. It is clear that $Q > P$ and $P$ is not the largest product; contradiction. Thus, the largest product cannot have any doubled numbers that are not from the set $\{a_1, a_2, ..., a_{1001}\}$, and the claim is proven.

Claim: For the maximal product $P=(2a_1+b_1)(2a_2+b_2)...(2a_{1001}+b_{1001})$, we have that $b_1\leq b_2 \leq ... \leq b_{1001}$. That is, $a_1, a_2,...,a_{1001}$ is oppositely ordered from the sequence $b_1,b_2,...,b_{1001}$.
Proof: Suppose that there exists two indices $i, j$ such that $a_i > a_j$ and $b_i > b_j$. Rewrite $P$ as $(2a_i+b_i)(2a_j+b_j)K=(4a_i a_j + 2a_i b_j + 2a_j b_i + b_i b_j)K$. Define $Q = (2a_i+b_j)(2a_j+b_i)K = (4a_i a_j + 2a_i b_i + 2a_j b_j + b_i b_j)K$. Clearly, similarly sorted sequences imply that $Q > P$, so there is once again a contradiction.

Therefore the maximal product must not have $b_i > b_j$ and $i > j$ for any pairs $(i,j)$. This forces $b_1 \leq b_2 \leq ... \leq b_{1001}$, and so the largest possible answer is $P=(2a_1+a_{2002})(2a_2+a_{2001})...(2a_{1001}+a_{1002})$.


calculus
integration
function
limit
real analysis
real analysis unsolved

let $ g$ be continious function defined on $ [0,1]$ such that $ g(1)\equal{}0$ , let $ n$ be natural number and
$ L_n\equal{}n \int_{1\minus{}\frac{1}{\sqrt{n}}}^1 t^ng(t) dt$ prove that $ \lim_{n\to\infty} L_n\equal{}0$.

Take $ 0 < a \leq 1$ . It's easy to see (with the mean theorem) that there is $ x_n\in [1\minus{}\frac 1{n^{a}},1]$ such that
$ L_n\equal{}n\int_{1\minus{}\frac 1{n^a}}^1t^ng(t)dt\equal{} ng(x_n)\int_{1\minus{}\frac 1{n^a}}^1 t^ndt \equal{} g(x_n)\frac n{n\plus{}1}(1\minus{}(1\minus{}\frac 1{n^a})^{n\plus{}1})$
For $ n\rightarrow \infty$ we can write $ \frac n{n\plus{}1}(1\minus{}\frac 1{n^a})^{n\plus{}1}\rightarrow finite\_limit , \ x_n\rightarrow 1,\ g(x_n)\rightarrow g(1)\equal{}0$ and finally $ L_n\rightarrow 0$
 :cool:


In how many ways can 2004 be written as a sum of two or more consecutive positive integers written in increasing order?

after 5 seconds of thought [hide]there are 500 ways to write it as a sum of 2 numbers, in increasing order, at least, i think so[/hide]

You didn't read all the requirements.  You missed "consecutive" and "or more."

After a bit more than 5 seconds thought ...

[hide]Sum of consecutive integers from m to n is (n+m)(n-m+1)/2, so we seek distinct positive n and m such that (n+m)(n-m+1)=4008=8x3x167. Noting that n+m and n-m+1 have opposite parity, and that n+m>n-m+1 (since m>0) we have :-

4008x1 : n=m=2004 - but n and m not distinct so not valid solution

1336x3 : n=669, m=667

501x8 : n=254, m=247

167x24 : n=95, m=72

so we have 3 ways.

[/hide]

I think the answer is [hide]3[/hide], how I did it: [hide]The sum of consecutive positive integers equals (a+a+n-1)n/2, where a=the first member of the sequence, n=number of consecutive integers. So (2a+n-1)n=4008. n has to be a divisor of 4008 and odd because 4008/n is an even, which plus 1-n is also even, so 1-n is even and n is odd.there are only 3 odd divisor of 4008, 3 and 167 and 501. So the answer is tres.[/hide]

I've been lurking for a few weeks, and this is my first post anywhere on the board. Go easy on me, since I have a gut feeling that my answer is wrong .

[hide]

In general, n consecutive terms can be written as

x + (x+1) + (x+2) + (x+3) + ... + (x + n - 1) = 2004

This can be rewritten as

nx + n(n-1)/2 = 2004

n(2x) + n^2 - n = 4008

n^2 - n(1-2x) = 4008

n(n + 2x - 1) = 4008

Factors of 4008 are:

1-4008

2-2004

3-1336

4-1002

6-668

8-501

12-334

24-167

n(n + 2x - 1) can be either even * odd or odd * even. This cancels out 2-2004, 4-1002, 6-668, and 12-334.

We are left with 1-4008, 3-1336, 8-501, 12-334, 24-167 for n(n + 2x - 1).

(n + 2x -1) will always be greated than n, so obviously the first numbers of that series correspond to n. 1-4008 doesn't fit the directions. Dividing 4008 from the other n, we see that 4008/12 is an even number, and thus cannot be the answer (334 would be the median, but there is no clear median among even numbers).

Thus, we conclude that only n=3, 8, and 24 work. Thus, 3 sets.

[/hide]

Nuts, beta beat me to it while I was writing out my incredibly long solution (compared to his). Haha, oh well.

Good job all of you who've gotten it so far.

And don't worry rlee. It's not a race to see who can post the first solution. It's just an exercise for everyone to try. Welcome to the board.


trigonometry
limit
percent
rotation
geometry
analytic geometry
graphing lines

The 5 should be flipped as it should be the same way the 8 is.

It's an 8. How can you tell which way it's flipped?

2 more:

Did you get these off of Nonie's website?

I think he did.

Eitherway, I have a bad joke.

Why don't mathematicians go to the beach?

Cuz all they need is sine and cosine to get tan. :P  :P  :P

haha tat is corny

my math teacher has showed me the "find x" before

The one with "Expand $(a+b)^{n}$" is soooo hilarious  :D

"Peter" is probably the unfortunate guy who did the homework innocently only to come back later and find it posted on someone's website and eventually making it into a "Math Jokes" thread on a math forum.

I don't get the "math4.jpg" one... the one with all the weird stuff under the radical...

50c that "peter" is peter zhang.

the 4th picture: the guy blotted out all the stuff under the radical and hung himself :rotfl:

[quote="Ignite168"]Phelpedo is Tony Feng I believe, not "Peter" whoever that is.[/quote]

on the $expand$ $(a+b)^{n}$

it says (in red) very funny, peter

I liked the "Find x" one the best. The rest I had seen before. They are okay. The other new one was the hanging, and it was quite good too!

As for the 8 and 5, here is something that happened in my class five years ago, when i was in eighth:

Ma'am was teaching mensuration. The example we took had radius $7$ something. We took $\pi=\frac{22}7$ and it nicely got cancelled. Then she gave us a sum with radius $9$ something. One boy in solving it actually took $\pi=\frac{22}9$!!!

Ok I found a better version of the limit joke

$\lim_{x\rightarrow8}\frac{1}{x-8}=\infty$
$\lim_{x\rightarrow3}\frac{1}{x-3}=\omega$

bubka, I believe your answer is incorrect, I believe the answer should be "m" not "w"

I see no "w." What is this "w" of which you speak?

No, the $\omega$ is an omega, which signifies Cantor's infinite ordinal numbers just like the alephs signify the infinite cardinals.

Don't ask.

[quote="Xantos C. Guin"]One of my friends actually did the "find x" thing on his test. Needless to say, it was marked wrong.[/quote]
Its always nice when peopel make math jokes threads. This is hilarious. :rotfl:
btw, are there more? :P

Here are some more random ones--

A circle is a round straight line with a hole in the middle.

Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.

"The number you have dialed is imaginary. Please rotate your phone 90 degrees and try again."

Q: how many times can you subtract 7 from 83, and what is left afterwards?
A: I can subtract it as many times as I want, and it leaves 76 every time.

Old mathematicians never die; they just lose some of their functions.

Q: What is the area of a circle?
A: pi R^2?
R: Pie are not square. Pie are round. Cornbread are square.

How do you prove in three steps that a sheet of paper is a lazy dog?
1. A sheet of paper is an ink-lined plane.
2. An inclined plane is a slope up.
3. A slow pup is a lazy dog.

The combination of the Einstein and Pythagoras discoveries:
E= m c^2 = m (a^2 + b^2)

[size=75]source:some website out there[/size]

My old signature was something like this:

Mathcounts: 50% practice, 50% speed, 50% avoiding stupid mistakes

How many quantum physicists does it take to change a lightbulb?

Wow these jokes are so hilarious.
I particularly liked the expand joke and the find $x$ joke. :rotfl:

oh that IS funny.
but i loved the ones by Phelpdo. :)

Here is a good joke posted long ago in Canadian forum(it's a bit immature):
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42012

Another one of my favorites:

Several scientists were asked to prove that all odd integers higher than 2 are prime.

 Mathematician: 3 is a prime, 5 is a prime, 7 is a prime, and by induction - every odd integer higher than 2 is a prime.

Physicist: 3 is a prime, 5 is a prime, 7 is a prime, 9 is an experimental error, 11 is a prime. Just to be sure, try several randomly chosen numbers: 17 is a prime, 23 is a prime...

Engineer: 3 is a prime, 5 is a prime, 7 is a prime, 9 is an approximation to a prime, 11 is a prime,...

Programmer (reading the output on the screen): 3 is a prime, 3 is a prime, 3 a is prime, 3 is a prime....

Biologist: 3 is a prime, 5 is a prime, 7 is a prime, 9 -- results have not arrived yet,...

Psychologist: 3 is a prime, 5 is a prime, 7 is a prime, 9 is a prime but tries to suppress it,...

Chemist: What's a prime?

Politician: "Some numbers are prime.. but the goal is to create a kinder, gentler society where all numbers are prime... "

Programmer: "Wait a minute, I think I have an algorithm from Knuth on finding prime numbers... just a little bit longer, I've found the last bug... no, that's not it... ya know, I think there may be a compiler bug here - oh, did you want IEEE-998.0334 rounding or not? - was that in the spec? - hold on, I've almost got it - I was up all night working on this program, ya know... now if management would just get me that new workstation that just came out, I'd be done by now... etc., etc. ..."

Two passengers are talking in a plane.

- Are you afraid that one of the passengers has the bomb?
- Well, the probability is about 1/1000, but I'm not afraid.
- Why?
- Well, the probability that two passengers have the bomb is much smaller, it's (1/1000)^2 = 1/1000000!
- I don't understand. Why do you say that?
- Well, (don't tell anybody), I have one bomb in my bag!

-- Have you heard that the pope has settled the continuum hypothesis?
-- ???
-- He has declared that cardinals above 80 have no powers.

Two male mathematicians are in a bar. The first one says to the second that the average person knows very little about basic mathematics. The second one disagrees, and claims that most people can cope with a reasonable amount of math.

The first mathematician goes off to the washroom, and in his absence the second calls over the waitress. He tells her that in a few minutes, after his friend has returned, he will call her over and ask her a question. All she has to do is answer one third x
cubed.

She repeats "one thir -- dex cue"?
He repeats "one third x cubed".
Her: `one thir dex cuebd'? Yes, that's right, he says.
So she agrees, and goes off mumbling to herself, "one thir dex cuebd...".

The first guy returns and the second proposes a bet to prove his point, that most people do know something about basic math. He says he will ask the blonde waitress an integral, and the first laughingly agrees.

The second man calls over the waitress and asks "what is the integral of x squared?".

The waitress says "one third x cubed" and while walking away, turns back and says over her shoulder "plus a constant!"

One day a farmer called up an engineer, a physicist, and a mathematician and asked them to fence of the largest possible area with the least amount of fence.

The engineer made the fence in a circle and proclaimed that he had the most efficient design.

The physicist made a long, straight line and proclaimed "We can assume the length is infinite..." and pointed out that fencing off half of the Earth was certainly a more efficient way to do it.

The Mathematician just laughed at them. He built a tiny fence around himself and said "I declare myself to be on the outside."

[quote]50c that "peter" is peter zhang.[/quote]

As a Peter Zhang, I feel it is my duty to dispute that fact.

Tony, you still going to PA next year?

Wow, this forum hasn't been too active recently...

well, here's a weird little joke, courtesy of my math teacher

Ok, so there's a $45^\circ$ angle.

$135^\circ$ [b]angle[/b]: Look, theres acute angle! I should go ask her out.

$90^\circ$ [b]angle[/b]: Don't be so obtuse. You know she's taken.

$135^\circ$ [b]angle[/b]: Yeah, you're always right.

 :roll:

i would say this is the most intresting ct forum going on..:)

I saw the Women one as $money=\sqrt{evil}$.


geometry
3D geometry
pyramid
tetrahedron

Consider the 4 3D points (-2,3,5),(-4,1,9),(1,2,3), and (5,11,6).  Each point is joined to the 3 other points by line segments.  Find the volume of the figure bound by these line segments.

I know an easy way to do this, but I'm interested to see how one would go about finding this without the method I know.  But if you too know the easy way, go ahead and post it, I guess.  (Also, I randomly chose those numbers so I doubt it'll work out to a nice number.  I'll change them to make them work well if nobody gets this by tomorrow.)

You can calculate the lenghts of the piramid. Then you need the base area and the pyramid height(which is not so hard to find it ;) ).

so whats the 'easy' way youre referring to?

Let $A(-2,3,5),B(-4,1,9),C(1,2,3)$, and $D(5,11,6)$.The equation of the plane passing through three points $A,B$ and $C$ is

$\alpha:x+y+z=6$, $\overrightarrow{AB}=(-2,-2,4),\overrightarrow{AC}=(3,-1,-2)$,

therefore the area of triangle $ABC$ is $\triangle ABC=\frac12 \sqrt{|\overrightarrow{AB}|^2|\overrightarrow{AC}|^2-(\overrightarrow{AB}\cdot \overrightarrow{AC})^2}=4\sqrt{3}$.

Using the distance between a point and a plane,the hight of the tetrahedron equals to the distance between point $A$ and plane $\alpha$,

$\frac{|-2+3+5|}{\sqrt{1^2+1^2+1^2}}=\frac{6}{\sqrt{3}}$.

The desired volume is $\frac13\cdot 4\sqrt{3}\cdot \frac{6}{\sqrt{3}}=8$

[quote="Palytoxin"]i doubt the coordinates were chosen randomly... and i doubt he knows the easy way, waiting for someone to answer problem for him...[/quote]

Wow, brilliant you are, you giant  :!: , especially since school isn't even going on.  But before you make any more stupid assumptions, I'll tell you that I wrote this problem myself after learning about scalar triple products.  However, when drawing it the other day I realized that my method doesn't work (I orginally thought it would simply be half the scalar triple product of any three matrices from one point).  I'm trying to tweak my scalar triple product method, but assuming 8 is the correct answer, I haven't had luck yet.  Still working tho, but apparently people get accused if they wait too long, so I had to post something...

[quote="seamusoboyle"]Could someone please elaborate on how to get an area with matrices? Ive never seen the method before.... :?[/quote]
$A=$area of triangle with vertices at $(x_1,y_1),(x_2,y_2),(x_3,y_3)$
\[A=\frac1{1\cdot 2}\left|
 \begin{array}{ccc}
  x_1 & y_1 & 1\\
  x_2 & y_2 & 1\\
  x_3 & y_3 & 1\\
 \end{array}\right|\]

$A>0$ if the vertices are placed in a counterclockwise direction
$A=0$ if they are collinear
$A<0$ if they are placed in a clockwise direction

And knowing maths, this generalises :P
Giving us what ye used. Thanks!


A circle of radius 1 is tangent to a circle of radius 2. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $\overline{AB}$ and $\overline{AC}$ are congruent. What is the area of $\triangle ABC$?

$\text{(A) }\frac{35}2 \qquad \text{(B) }15\sqrt{2}\qquad \text{(C) }\frac{64}3 \qquad \text{(D) }16\sqrt{2}\qquad \text{(E) }24$

[img]http://farrago.hopto.org/images/triangleratio.gif[/img]

kinda hard to find the area without having the drawing...

yeah I don't completely see the picture lol . just vaguely

I can't see the picture, but whatever. I remember this problem vividly. The two triangles are similar. Enough said.

I don't know what happened. I created an image, but I couldn't actually see the image. I hope this is somewhat what it looked like:

[img]http://img120.imageshack.us/img120/7775/triangleratioze2.gif[/img]

Draw in the lines from the tangent points of the circles to the radii. This makes similar right triangles. Also, if you draw the line parallel to the base and tangent to both circles, you get two similar isosceles triangles.,

Oops made a fatal mistake in my solution.

I remember this problem too. What was the answer by the way?

Ah got it now. The solution is fairly simple. Just smash and bash using the fact that tangents are perpendicular to radius.

16sqroot2 is the answer.

or you could just form similar triangles. maybe a little nicer?


calculus
derivative
analytic geometry
graphing lines
slope
geometry
function

I am just wondering, if the derivative is the slope of a graph at a certain point, then how come that the derivative is also used in finding the maximum area? Can somebody explain it.

I want to understand its concept.

Generally in order to find the maximum area of a function you take the deriative and set it equal to zero.

Now, what does it mean (geometrically) when the derivative is zero? Well, it means that slope of the tangent line is zero. So where in the graph can the slope of the tangent line be zero? You can easily convince yourself that for any graph the slope of the tangent line can be zero only where the graph has a maximum or a minimum!

Now, to answer your question, in order to find the maximum area, you plot whatever relationship you have between your area and some variable x and then you look at the graph for points that will maximize the area (i.e. a maximum). Although, this could be done without taking derivatives and just by looking at the graph, you save your self some time by using derivatives to find that same maximum! :lol:

This is a nice explanation. Now I got it. Thanks.

[quote="swapnillium"]You can easily convince yourself that for any graph the slope of the tangent line can be zero only where the graph has a maximum or a minimum![/quote]

You can easily convince yourself of this, but unfortunately it is not true -- consider, for instance, the function $f(x) = x^3$ which has derivative 0 at $x = 0$ but does not have either a maximum or a minimum at this point.

However, the converse is true: if you have a function which is differentiable on an open interval, then all the extrema on the interval (if they exist) are at points where the derivative is 0.


trigonometry
geometry
geometric transformation
rotation
calculus
calculus computations

So the question is:

Find all the values of the following.

[img]http://img190.imageshack.us/img190/7148/bdd5e504719da10a0e20802.png[/img]

[img]http://img513.imageshack.us/img513/6757/2e8ab53030f19df0bc1950b.png[/img]

[img]http://img29.imageshack.us/img29/8966/ad3aefe0ca1b758e28144ca.png[/img]

This is complex number question. I really don't know how to start answering the question. Please help me.

thank you

[hide="First Problem"]Assume $ x \equal{} (\minus{}256)^{1/4}$. So, $ x^4 \plus{} 256 \equal{} 0$, so $ (x^2 \plus{} 16i)(x^2 \minus{} 16i) \equal{} 0$. $ x^2 \equal{} \pm 16i$ has solutions $ x \equal{} 4\sqrt{\pm i}$. $ \sqrt{i} \equal{} \pm \frac{\sqrt{2}}{2} \pm i\frac{\sqrt{2}}{2}$. $ \sqrt{\minus{}i} \equal{} \pm \frac{\sqrt{2}}{2} \mp i\frac{\sqrt{2}}{2}$.

So the four solutions are:

$ \boxed{x \equal{} 2\sqrt{2} \plus{} 2i\sqrt{2}}$

$ \boxed{x \equal{} 2\sqrt{2} \minus{} 2i\sqrt{2}}$

$ \boxed{x \equal{} \minus{}2\sqrt{2} \plus{} 2i\sqrt{2}}$

$ \boxed{x \equal{} \minus{}2\sqrt{2} \minus{} 2i\sqrt{2}}$

[/hide]

thank you so much darkdieuguerre. would you also go over the second and the third one. I've just tried to slove them following your steps in the first one but I couldn't find the answer.

thanks

Use the relationship

$ re^{i\theta} \equal{} r(\cos\theta \plus{} i \sin\theta)$

and

$ (re^{i\theta})^n \equal{} r^n e^{in\theta}$.

For instance,

$ i \equal{} e^{i \pi/2} \equal{} e^{i (\pi/2 \plus{} 2\pi)} \equal{} e^{i (\pi/2 \plus{} 4\pi)} \equal{} e^{i (\pi/2 \plus{} 6\pi)}$

so

$ i^{1/4} \equal{} \{e^{i \pi/8}, e^{i(\pi/8 \plus{} \pi/2)}, e^{i(\pi/8 \plus{} \pi)}, e^{i(\pi/8 \plus{} 3\pi/2)}\}$.

The third one is essentially asking for $ z$ such that $ z^5=1$, also known as the fifth roots of unity.

Those happen to lie on the vertices of a regular pentagon (think about why; the fifth roots of unity will be of the form $ e^{2\pi ki/5}$ for integers $ k$. Why is this?). The first is clearly $ 1$, and the rest are $ 72^\circ$ rotations of $ 1$ around the origin in the complex plane, so we can get each one by multiplying by $ e^{2\pi i/5}$.

$ \boxed{1,e^{\pm 2\pi i/5},e^{\pm 4\pi i/5}}$


Find all solutions of $ x^2\minus{}10y^2\equal{}\minus{}1$ in natural numbers.

[quote="ZetaX"]Consider powers of $ (3 \plus{} \sqrt {10})$  :wink:[/quote]

Can u be a bit more specific please? :oops:

The method of solving this equation is in no point different from Pell equations with a $ \plus{} 1$. In fact all solutions of $ x^2 \minus{} ny^2 \equal{} \pm 1$ are archieved by $ x \plus{} \sqrt n y \equal{} \pm (x_0 \plus{} \sqrt ny_0)^k$, where $ k$ is an integer and $ x_0 \plus{} \sqrt ny_0$ is the "smallest" solution (i.e. $ x_0,y_0 > 0$, $ x_0^2 \minus{} ny_0^2 \equal{} \pm 1$ and they are the smallest such pair).
If there is a solution to $ x^2\minus{}ny^2\equal{}\minus{}1$, then there is a smallest one (defined as above) and all solutions are of the form $ \pm (x_0\plus{}\sqrt n y_0)^{2k\plus{}1}$.


Let $ y_{n}= (-1)^{n}x_{n}$. Then $ y_{n}=-\frac{1}{2004}y_{n-1}+1$ and by simple observation, $ y_{n}$ converges.

Applying $ n\to\infty$ to the recursive formula, the limit of $ y_{n}$ is $ \frac{2004}{2005}$.

So $ \lim x_{n}^{2}= (\lim y_{n})^{2}= \left(\frac{2004}{2005}\right)^{2}$.


Posts after posts, I've seen the word: circumradius. Does anybody know what it is?  :huh:

The circumradius of a polygon (if it is cyclic) is the radius of its circumcircle (the circle it can be inscribed in).


It costs 2.5 cents to copy a page. How many pages can you copy for $ \$20$?


I recently encountered a problem that I know is solvable, but seems to be a problem only solvable by luck.

Here it is:
'The sum of all positive integer factors of a number is 91.  What is the number?'

[hide="The solution is:"]36.[/hide]

I tried about 10 numbers on this problem, but all failed.  I would've tried more, but there were 19 other problems to solve and to start with I only had 35 minutes to solve each problem.  It would have really been nice to know a fast way to solve such a problem as this.

Any helpful hints?

Thanks in advance.[/hide]

Well...

The sum of all positive factors of a particular number $ p_1^n*p_2^k*p_3^l*\hdots$

Is $ (1\plus{}p_1\plus{}p_1^2\plus{}\hdots)(1\plus{}p_2\plus{}\hdots)\hdots$

This works because all possible combinations of factors are found.

So we want a number such that the above expression is 91.

However, factoring $ 91$ gives us $ 7*13$

$ 7\equal{}2^3\minus{}1$, which gives us reason to believe that there is a $ 2^2$ factor.

Now $ 13\equal{}1\plus{}3\plus{}3^2$, and thus the number is $ 2^2*3^2\equal{}36$

In order to see these possibilities, you should have a knowledge of geometric sequence sums.


Prove that the sum of the squares of two odd integers cannot be a perfect square.

[hide="Solution"]
In $ \pmod{4}$ the quadratic residues are $ 0,1$, they are $ 1$ if the number being squared is congruent to $ 1$ or $ 3$ in $ \pmod{4}$ or in other words they are odd, hence their addition should be congruent to $ 2\pmod{4}$ which is not a quadratic residue in $ \pmod{4}$ and we are done. :ninja:
[/hide]


[img]http://img149.imageshack.us/img149/9210/kklup0co8.gif[/img]
thanks and best regards

It might help you to write
$ 16^{1/3} \plus{} 4^{1/6} \minus{}8^{1/9}$

Yeah i know this.i solved this way but i found 6 is the power sgrt 2^5  :(
where is wrong i dont know.  :maybe:


linear algebra
matrix
vector
inequalities
complex numbers
linear algebra solved

Let $ A \equal{} \left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ and $ B \equal{} \left(b_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ be two $ n\times n$ positive definite matrices. Show that the matrix $ C \equal{} \left(c_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ defined by $ c_{i,j} \equal{} a_{i,j}b_{i,j}$ is also positive definite.

[i]Remark (inserted by moderator).[/i] Under a [i]positive definite matrix[/i], we understand a symmetric (or Hermitean, if we work over the complex numbers) matrix $ U$ such that $ x^TUx > 0$ (or $ x^*Ux > 0$, respectively) for any nonzero vector $ x$.

is that correct?
the product of two positive definite matrices isn't even always a symmetric (or hermitian if you work complex)

take A = [ 1  1/2
               1/2  1

B =  3   1
      1    2

the product is not symmetric!

? he just claimed it's positive definite, not symmetric...

:? whenever i think i know something i make a fool of myself
indead, i am so used at working with symmetric positive definite that i thought other positive definite matrices
weren't considered
sorry!

[quote="harazi"]Then use the fact that the sum of positive semidefinite matrices is semidefinite.[/quote]
Hi Harazi, could you explain little more your proof. I don't understand how you conclude with the sum of $x\cdot x^{*}$. Thanx  :)

[quote="harazi"]Use Schur decomposition theorem to prove that any positive semidefinite matrix can be written as the sum of k matrices of rank 1 of the form $ x\cdot x^{*}$ where k is the  rank of A and the vectors x are ortogonal. Then use the fact that the sum of positive semidefinite matrices is semidefinite.[/quote]
What is Schur decomposition theorem?

I think here that to prove the statement of Harazi you just have to diagonalize the matrices in an orthogonal basis.
Conclusion follows.

I still don't understand this...
What is the next step after obtaining the spectrum decomposition of the positive definite matrices?

Well, it suffices to write that $A=U^{*}DU$ with D diagonal and to identify the coefficients in the equality.

Can you give a detailed solution...
I still don't understand.

I will post a complete solution tomorow. Now I have a discussion... guess with whom.

[quote="harazi"]So, any positive semidefinite matrix A of rank k can be written in the form $ v_1 v_1^{*}+...+v_k v_k^{*}$ with $v_i$ ortogonal vectors. Indeed, write by spectral theorem $A=UDU^{*}$ with U ortogonal and D diagonal with entries $d_i\geq 0$ and take $v_i$ the vector $\sqrt{d_i}\cdot M_i$ where $M_i$ is the column i vector in the matrix U.[/quote]
I don't you think understood the question as stated.  For the case of 2x2 the following is a positive definite matrix.
A= [1 -6 ; 7 2].  Taking the specified product above of A with itself.
cij = aij^2 which is equals C = [1 36 ; 49 4] which is not positive definite.   Thus the above conjecture that "the kronecker product of two positive definite matrices is positive definite" is not true.

I don't think  it's true.
The positive defination is symmetric?

hi ,wath are you talking about , a matrix witch is defined positivly isn't necesserly symetric ,take for exemple  (1  -1)  ( of course in IR-space )
                                            (0   1)

Do you know what I find neat about this topic? The matrix $ C$ defined above is [i]not[/i] the Kronecker product of the matrices $ A$ and $ B$, but throwing in the words "Kronecker product" points to another (in my opinion, better) solution of the problem:

The matrix $ C$ is a principal minor of the Kronecker product of the matrices $ A$ and $ B$ (in fact, we can think of the rows of the Kronecker product indexed by pairs of numbers $ \left(i,j\right)\in\left\{1,2,...,n\right\}^2$, and of the columns indexed the same way; now, the matrix $ C$ is the principal minor of this Kronecker product obtained by taking only the rows $ \left(i,i\right)$ and the columns $ \left(k,k\right)$). Now, the Kronecker product of the matrices $ A$ and $ B$ is positive definite (says the Wikipedia, and I guess this is not hard to prove), and a principal minor of a positive definite matrix is positive definite again (trivial by the definition). Thus we are done.

By the way, for me, a positive definite matrix is supposed to be symmetric by the definition of "positive definite". The Wikipedia agrees with me...

  darij

the proper phrase is 'hadamard product', i believe.

as Darij says, it is not hard to see that if $ A,B$ are positive, then so is their Kronecker product. For the simple case, $ dim (A) \equal{} dim(B)$ : the Kronecker product of $ A,B$ can be seen as the matrix of the application $ \phi_{A,B}: M_n \to M_n$ defined by $ \phi_{A,B}(M) \equal{} AMB$. One has to show that for every matrix $ M$, the following holds:  $ Tr(M^tAMB) \equal{} Tr(AMBM^t) \geq 0$. Write $ A \equal{} O^t D O$ where $ D$ is diagonal with positive elements to see that $ Tr(M^t AMB) \equal{} Tr(D (OM) B (OM)^t)$. The conclusion then follows since $ [(OM) B (OM)^t]_{i,i} \geq 0$ (because $ B$ is positive).

here is a nice application of the result , prove that if A is symetrique positif matrix then so is the matrix (exp(ai,j))

[quote="alekk"]Let $ A = \left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ and $ B = \left(b_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ be two $ n\times n$ positive definite matrices. Show that the matrix $ C = \left(c_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ defined by $ c_{i,j} = a_{i,j}b_{i,j}$ is also positive definite.
[i]Remark (inserted by moderator).[/i] Under a [i]positive definite matrix[/i], we understand a symmetric (or Hermitean, if we work over the complex numbers) matrix $ U$ such that $ x^TUx > 0$ (or $ x^*Ux > 0$, respectively) for any nonzero vector $ x$.[/quote]
  By $ A \in S_n^{ + }(\mathbb{R})$ then the exist $ \omega \in O_n{\mathbb{(R)}}$ such that : $ A = \omega D \omega^{T}$ ("*")
 Here $ \omega = [\omega_{ij}]_{1 \leq i,j \leq n}$ and  $ \omega^T = [\omega_{ji}]_{1 \leq i,j \leq n}$ , $ D = diag(\lambda_1,\lambda_2,..,\lambda_n)$ ,$ \lambda_k \geq 0 ,\forall k \in \{1,2,..,n\}$
Then easy see from  "*" : $ a_{ij} = \sum_{k = 1}^{n}\sum_{m = 1}^{n}{\omega_{ik}\lambda_{ik}\omega_{jm}} = \sum_{k = 1}^{n}{\lambda_k\omega_{ik}\omega_{jk}}$ (1)
 You known that $ A \in S_n^{ + }(\mathbb{R})$  then $ \forall x \in M_{n1}{\mathbb{(R)}}$ then $ x^TAx \geq 0$
Now put $ y_{ik} = \omega_{ik}x_i,1 \leq i ,j \leq n$ and $ y_k = \begin{bmatrix}\omega_{1k}x_1 \\
\omega_{2k}x_2 \\
.... \\
\omega_{nk}x_n \end{bmatrix}=\begin{bmatrix}y_{1k}\\y_{2k}\\...\\y_{nk}\end{bmatrix}$ , $ 1 \leq k \leq n$
We have :
$ x^TCx = \sum_{i = 1}^{n}\sum_{j = 1}^{n}{c_{ij}x_ix_j} = \sum_{i = 1}^{n}\sum_{j = 1}^{n}{b_{ij}(a_{ij}x_ix_j)} = \sum_{i = 1}^{n}\sum_{j = 1}^{n}\sum_{k = 1}^{n}{b_{ij}(\lambda_k (\omega_{ik}x_i)(\omega_{jk}x_j))} = \sum_{k = 1}^{n}{\lambda_k(\sum_{i = 1}^{n}\sum_{k = 1}^{n}{b_{ij} (\omega_{ik}x_i)(\omega_{jk}x_j))}) = \sum_{k = 1}^{n}{\lambda_k(\sum_{i = 1}^{n}\sum_{k = 1}^{n}{b_{ij} y_{ik}y_{jk}}) = \sum_{k = 1}^{n}{\lambda_k}{y_k^TBy_k} \geq 0}}$
 By  $ B \in S_n^{ + }(\mathbb{R}),y_k \in M_{n1}{\mathbb{(R)}}$.....
 You can read a different proof of [b] Elizabeth Million [/b] about it but I don't find proof [b]Oppenheim's Inequality[/b] can post it  :oops:
 With $ \forall A,B,C \in M_n{\mathbb{(C)}}$ , $ C = [a_{ij}b_{ij}]_{1 \leq i,j \leq n}$ then $ rank(C) \leq rank(A \otimes{B})$ (easy see small matrix)
 About [b]Hadamard's Inequality [/b] ,$ A \in S_n^{ + }{\mathbb{(R)}}$
Then $ \det(A) \leq a_{11}a_{22}...a_{nn}$ You can use lemma $ A = M^TM,M \in M_n{\mathbb{(R)}}$ then :
 $ |\det(M)| = |\det(m_1,..,m_n)| \leq \parallel{}m_1\parallel{}...\parallel{}m_n\parallel{}$ ([b]Hadamard's Inequality [/b]) then easy $ a_{ii} = \sqrt {\parallel{}m_i\parallel{}} , 1 \leq i \leq n$ ....
 You can talk more  ?  :rotfl:

Solution:
Well known that ($A$, $B$ is positive definite) exist $C$, such that $C^{T}AC=E$, $C^{T}BC=diag(\lambda_1,\lambda_2,..,\lambda_n)$, so $A=(C^{-1})^{T}C^{-1}$, $B=\sum_{i=1}^n(C^{-1})^{T}diag(0,0,...,\lambda_i,0,...,0)C^{T}$, denote $A\circ B=^{def}\left(a_{i,j}b_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$. Let $D=C^{-1}=\left(d_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}=\sum_{i=1}^n \left(\delta_{i,k}* d_{k,j}\right)_{1\leq k\leq n,\ 1\leq j\leq n}=\sum_{i=1}^n D_i$, so $D^{T}=\sum_{i=1}^nD_i^{T}$ ($\delta_{i,k}$- Kronecker symbol). $A\circ B=\sum_{1\leq i\leq n, 1\leq j\leq n}\lambda_j *D_i^{T}D_i\circ D_j^{T}D_j$, so for every vector $v$, $v(A\circ B)v^{T}=\sum_{1\leq i\leq n, 1\leq j\leq n}\lambda_j *v(D_i^{T}D_i\circ D_j^{T}D_j)v^{T}$. Easy to see that for some matrix $N$, $v(D_i^{T}D_i\circ D_j^{T}D_j))v^{T}=vN^{T}Nv^{T}\geq 0$, so $v(A\circ B)v^{T}=\sum_{1\leq i\leq n, 1\leq j\leq n}\lambda_j *v(D_i^{T}D_i\circ D_j^{T}D_j)v^{T}\geq 0$, if $\sum_{1\leq i\leq n, 1\leq j\leq n}\lambda_j *v(D_i^{T}D_i\circ D_j^{T}D_j)v^{T}=0$, then $\sum_{1\leq i\leq n, 1\leq j\leq n}v(D_i^{T}D_i\circ D_j^{T}D_j)v^{T}=0=v((C^{-1})^{T}C^{-1}\circ (C^{-1})^{T}C^{-1})v^{T}$, but $v((C^{-1})^{T}C^{-1}\circ (C^{-1})^{T}C^{-1})v^{T}=vv^{T}>0$. done


[quote="math92"]Which sum is the largest?
A. –6.2 + 8.07
B. –8.41 + 10.26
C. 0.99 + 0.75
D. –0.4 + 2
E. –11 + 12.86[/quote]

[hide="with simple addition"]
[i][u][b][size=200][color=blue]A[/color][/size][/b][/u][/i][/hide]


I am in the Intro to Algebra class and no longer see the class forum.  It seems to have disappeared.  Is anyone else having problems?


[quote="kyyuanmathcount"]Somewhat reworded:

Arnav walks out of the his house sometime between 12:00 and 1:00.  When he gets back, Zhai, who is always right, tells him that it is still between 12:00 and 1:00.  Then, it takes the rest of the IMO team to realize that both when Arnav left and when he returned had times where the difference between the measures of the hour and minute hands were 110 degrees.  How long was Arnav gone on his little journey?

I found a nice countdown solution  :lol: .[/quote]

Let's see... I think I thought of a way to solve it easily when I was checking the 5/6 set at the competition.

[hide]The angle goes from 110 to 360-110=250  for a difference of 140 minutes. For each minute, the minute hand's position increases by 6 degrees and the hour hand's position also increases by 1/2 of a degree, for a total difference of 5.5 degrees per minute. So the answer is 140/5.5 = 280/11 minutes, and this problem is different from the one in the actual target round, right?[/hide]

You could also do it this way, but it's much slower.

[hide]Minute hand moves 6 degrees / minute
Hour hand moves 1/2 degree / minute.

So one 110 degree angle will be when the minute is less than 30. 6x - x/2 = 110 so x = 20.

The other 110 degree angle will occur when the minute is greater than 30. 6(60-y) + y/2 = 110 --> y = 500/11

(500-220)/11 = 280/11.[/hide]

I think the times were 6 and 7 instead of 12 and 1 on the test, because the answer was an integer (40 if I remember correctly)

[quote="kyyuanmathcount"]Oh, how about we do the problem with the 6 and 7?  I had forgotten what the numbers actually were  :( .[/quote]

I don't really think that the times (6 and 7) matter... just the angle, as long as the time elapsed is maintained as less than an hour. I think a general solution (as displayed in my earlier solution) would be:

[hide] (360-2A)/5.5 or 2(360-2A)/11, where A is the angle.[/hide]

In the original problem, I think that the angle was 70 degrees, hence:

[hide] (360-2(70))/5.5 = 220/5.5 = 40 [/hide]

[b]Edit:[/b] No, no, actually it was 110 (see post below)-- it was a different case. The general solution here is for a Case II problem, like the original one here.

[quote="vector2101"][quote="kyyuanmathcount"]Oh, how about we do the problem with the 6 and 7?  I had forgotten what the numbers actually were  :( .[/quote]

[b]I don't really think that the times (6 and 7) matter... [/b]just the angle, as long as the time elapsed is maintained as less than an hour. I think a general solution (as displayed in my earlier solution) would be:

[hide] (360-2A)/5.5 or 2(360-2A)/11, where A is the angle.[/hide]

In the original problem, I think that the angle was 70 degrees, hence:

[hide] (360-2(70))/5.5 = 220/5.5 = 40 [/hide][/quote]
actually it matters alot, just imagine a clock, remember they have to be in the period of one hour, so the clocks would go differently.

Like if it was between 6-7, the original minute hand would be behind the hour hand and the second time, the minute hand would be ahead of the hour hand

(if you draw it and compare minute hand relative to hour hand, it will be -110 degrees and +110 degrees)

if it was 12-1, the original minute hand would be ahead, not behind the hour, and the second minute hand would even be more above the hour (so 250 degrees)
(if you draw it and compare minute hand relative to hour hand, it will be +110 degrees, and then another +140 degrees-because it will be 110 from the hour hand in the other direction so +250 degrees)

im not really good at solving the problem, but when i imagined this the clcok hands were positioned differently (i need a picture to explain really well)

Oh, okay... maybe it does, but there are only two cases. The first is when the first time's minute hand is behind the hour hand and the second's is ahead, for which the general solution is:

[hide] 2A/5.5 [/hide]

And the second is with the first ahead and the second behind, for the case I showed earlier.

The original problem was a Case I, this one was a Case II.

so the weird thing is that I don't think the IMO team was announced when you posted this

GASP

YOU MUST BE CLAIRVOYANT

A man leaves his house between 6 and 7 and notes the clock hands make a 110 degree angle.  When he returns, more than 5 minutes later, the hands are in a 110 degree angle again.  How long was he gone?

My logic:  This answer is wrong.
At 6:00 there are 180-110=70/5.5=about 12 minutes until the first time the clock reached 110 degrees.  It takes 20 minutes for the angle to collapse, and 20 minutes for it to set back up again.  So it would be 40 minutes.

[hide="Solution"]It is easy to find a formula for the angle between two hands of a clock.  This is given by $ |30H \minus{} 5.5M|$.

Since $ H \equal{} 6$, we have $ |180 \minus{} 5.5M| \equal{} 110$
$ 5.5M \equal{} 70$ or $ 290$
Thus the difference in times will be $ \frac {220}{5.5} \equal{} 40$ minutes.
[/hide]

Edit: he's supposed to return again in between 6 and 7.

[hide]

so it starts at 180º, 70º later, or 12 8/11 minutes, (since it is 5.5 degrees a minute) it is 110º, so he left at 7:12 8/11

220º later, it will be 110º again, so 220/5.5=40 minutes later. Are you sure that you have the wrong answer?

I got 40 too. But my coach said it was wrong. Maybe I should ask her again. Or you should Alex :D.


geometry
circumcircle
angle bisector
geometry proposed

Let $ ABC$ be a triangle inscribed in a circle $ (O)$.$ AO$ meet $ (O)$ at the second point $ P$.Let $ H$ be the orthocenter of $ ABC$.The perpendicur line from $ H$ to the angle bisector of $ \hat{A}$ meet $ AB,AC$ at $ M,N$.The circumcircle of $ AMN$ meet the angle bisector of $ \hat{A}$ at $ Q$.$ M'$ is the midpoint of $ BC$.Prove that $ H,M',P,Q$ are collinear.

Dear Mathlinkers,
this problem is a nice variation of     M.O. Russie (2000), 10th grade, problem 3.
The second point of intersection of the circumcircle of ABC and the circle with diameter AH is also on this line.
Sincerely
Jean-Louis


calculus integration trigonometry LaTeX logarithms algebra polynomial

Evaluate

(1) $ \int \frac{\left( 1\plus{}x^{2}\right) \left( 2\plus{}x^{2}\right) }{\left( x\cos x\plus{}\sin x\right) ^{4}}\allowbreak dx$

(2) $ \int_{\pi }^{2\pi }\frac{\left( 1\plus{}x^{2}\right) \left( 2\plus{}x^{2}\right) }{\left( x\cos x\plus{}\sin x\right) ^{4}}\allowbreak dx$

Ans to (2)   [hide]$ \frac{7\plus{}12\pi ^{2}}{24\pi ^{3}}$[/hide]
:D

Am in a hurry, as usual, but have an idea:

$ (x\cos x \plus{} \sin x)\prime \equal{} 2\cos x \minus{} x\cos x$
$ (x\sin x \plus{} \cos x)\prime \equal{} x \cos x$
$ (x\cos x \minus{} \sin x)\prime \equal{} \minus{}x\sin x$
$ (x\sin x \minus{}\cos x)\prime \equal{} x\cos x \plus{} 2\sin x$

Then $ \frac{d(\frac{u}{v})}{dx} \equal{} \frac{v\frac{du}{dx} \minus{} u\frac{dv}{dx}}{v^{2}}$

$ \equal{} \frac{(u\cos x\plus{} \sin x)\frac{du}{dx} \minus{} u\times 2(x\cos x\plus{} \sin x)(2\cos x \minus{} x\sin x)}{(x\cos x\plus{} \sin x)^{4}}$

Perhaps we can figure out $ u$ and $ v$, from the given integrand, using

$ x^{2} \equal{} x^{2}(\cos^{2} x \plus{} \sin ^{2} x)$.

and then the answer should be $ \frac{u}{v}$.

[quote="AndrewTom"]Am in a hurry, as usual, but have an idea:

$ (x\cos x \plus{} \sin x)\prime \equal{} 2\cos x \minus{} x\cos x$
$ (x\sin x \plus{} \cos x)\prime \equal{} x \cos x$
$ (x\cos x \minus{} \sin x)\prime \equal{} \minus{} x\sin x$
$ (x\sin x \minus{} \cos x)\prime \equal{} x\cos x \plus{} 2\sin x$

Then $ \frac {d(\frac {u}{v})}{dx} \equal{} \frac {v\frac {du}{dx} \minus{} u\frac {dv}{dx}}{v^{2}}$

$ \equal{} \frac {(u\cos x \plus{} \sin x)\frac {du}{dx} \minus{} u\times 2(x\cos x \plus{} \sin x)(2\cos x \minus{} x\sin x)}{(x\cos x \plus{} \sin x)^{4}}$

Perhaps we can figure out $ u$ and $ v$, from the given integrand, using

$ x^{2} \equal{} x^{2}(\cos^{2} x \plus{} \sin ^{2} x)$.

and then the answer should be $ \frac {u}{v}$.[/quote]

Interesting idea, but perhaps you can share with the forumers in greater details when you are not in a hurry... :ninja:

Yes, IndoChina, perhaps tomorrow, but I have a life and I'm going out for the evening.

All the best,

AndrewTom.

I forgot what I did yesterday evening and started again this morning:

The answer must be expressible in the form $ \frac{u}{(x\cos x \plus{} \sin x)^{3}}$ where $ u \equal{} u(x)$.

Differentiating, we have

$ (x\cos x \plus{} \sin x)^{\minus{}3}\times u\prime \minus{} 3u(x\cos x\plus{}\sin x)^{\minus{}4}(2\cos x\minus{}x\sin x)$

$ \equal{}\frac{(x\cos x \plus{} \sin x)u\prime \plus{}(3x\sin x\minus{}6\cos x)u}{(x\cos x\plus{}\sin x)^{4}}$

$ \Rightarrow\,(x \cos x \plus{} \sin x)u\prime \plus{} (3x\sin x\minus{}6\cos x)u \equal{} (1\plus{}x^{2})(2\plus{}x^{2}) \equal{} 1 \plus{} 3x^{2} \plus{} x^4$

$ \Rightarrow\, u\prime \plus{} \frac{3(x\sin x \minus{} 2\cos x)}{(x \cos x\plus{}\sin x)}u \equal{} \frac{1\plus{}3x^{2} \plus{} x^{4}}{(x\cos x\plus{}\sin x)}$

We can solve this differential equation, using an integrating factor, as follows:

The integrating factor is $ e^{\int \frac{3(x \sin x \minus{} 2\cos x)}{x\cos x\plus{}\sin x}}\, dx \equal{} e^{3\ln(x\cos x\plus{}\sin x)} \equal{} (x\cos x\plus{}\sin x)^{3}$

Multiplying both sides of the differential equation by this, we get

$ (x\cos x \plus{} \sin x)^{3}u\prime \plus{} 3(x\cos x\plus{}\sin x)^{2}(x\sin x \minus{} 2\cos x)u$

$ \equal{} (1\plus{}3x^{2} \plus{} x^{4})(x\cos x\plus{}\sin x)^{2}$

$ \Rightarrow\, (x\cos x\plus{}\sin x)^{3} \equal{} \int(1\plus{}3x^{2}\plus{}x^{4})(x\cos x\plus{}\sin x)^{2}\,dx$

We can now use $ \cos (2x) \equal{} 2\cos^{2} x \minus{} 1 \equal{} 1 \minus{} 2\sin^{2} x$, integrate by parts (which is easy but will take a while to do) and finally divide both sides by $ (x\cos x\plus{}\sin x)^{3}$.

I hope this is OK for you.

actually i think i have got a better idea . put x= tanx in the integral so we get $ \int\frac {\left( 1 \plus{} tan^{2}x\right)\left( 2 \plus{} tan^{2}x\right) }{\left( tanx\cos x \plus{} \sin x\right)^{4}}\allowbreak {dx}cos^{2}x$  so after that the i + tan ^(2)  x  and the cos^(2) x in the numerator cancel leaving us with 2 + tan^(2) x / 16*sin^(4)  x .now separate the the integrals and integrate ...

i hope i am correct please mention if i am  wrong... :)  :)  please pardon me[b] i dont know latex[/b] is there any way to learn ??? do put  more interesting questions like these

but there is a problem this substitution is not valid for 2) as tanx is not single valued..

Hi AndrewTom,

Your initial assumption aside, how did the result
$ \int \frac{3\left( x\sin x\minus{}2\cos x\right) }{x\cos x\plus{}\sin x}dx\equal{}e^{3\ln \left( x\cos x\plus{}\sin x\right) }$ come about? It make a lot of difference if you miss out a sign. :wink:

Hi vijaymenon.

Surely $ x = \tan x$ is an equation with three real solutions ($ 0$ and approximately $ \pm 4.5$) and not a valid substitution.

$ x = \tan y$ gives something which starts out looking nice but contains a term like $ \sin(x + y)$ in the denominator; it was the first thing I tried yesterday evening.

I have picked up a little $ \text{\LaTeX}$ from the Latex Help link at the top of the page and also by clicking on what others have written.

Hi IndoChina.

Just seen your post and realised I got the sign wrong!

sorry i ment x = tany  who ya i am sorry made a mess of it extremely sorry.. :oops_sign:

I think my initial assumption is okay as $ u$ could contain a factor $ (x\cos x \plus{} \sin x)^{n}$.

But I feel intuitively that the answer should have $ (x\cos x \plus{} \sin x)^{3}$ as the denominator and that my first method will give a cumbersome expression for the numerator involving products of low-degree polynomials with sines and cosines. This looks tedious and, rather than pursue this, I think it might be easier to try to find another way of solving this problem. I cannot think of a suitable substitution at the moment.

I'm guessing that as people like farenhajt, Sonnhard, kunny etc. haven't posted any replies to this that it (like your other one (Simple if you know (6)?) is quite tricky.

P.S. Have just visited Wolfram Alpha where I found the following horrendous expression for the integral:

$ \frac { \minus{} 3(x^{2} \plus{} 1)\cos (x) \plus{} (1 \minus{} 3x^{2})\cos (3x) \plus{} 2x\sin(x)(2x^{2} \plus{} (x^{2} \minus{} 3)\cos (2x))}{6(\sin (x) \plus{} x \cos (x))^{3}}$,

among others.

P.P.S. It timed out for your Simple if you know (6)!

IS this non integrable or something??  :(  :(

[quote="IndoChina"]Evaluate

(1) $ \int \frac {\left( 1 \plus{} x^{2}\right) \left( 2 \plus{} x^{2}\right) }{\left( x\cos x \plus{} \sin x\right) ^{4}}\allowbreak dx$
[/quote]

Here is a solution!

Let $ x \equal{} \tan \alpha$
$ \Longrightarrow dx \equal{} \sec^2 \alpha d\alpha, 1 \plus{} x^2 \equal{} \sec^2\alpha$ and $ x\cos x \plus{} \sin x \equal{} \sec \alpha \sin(x \plus{} \alpha)$
So
$ \int \frac {\left( 1 \plus{} x^{2}\right) \left( 2 \plus{} x^{2}\right) }{\left( x\cos x \plus{} \sin x\right) ^{4}}\allowbreak dx \equal{} \int \frac {\sec^2 \alpha(1 \plus{} \sec^2 \alpha)}{\sec^{4}\alpha \sin ^{4}(x \plus{} \alpha)}\allowbreak \sec^2 \alpha d\alpha \equal{} \int \frac { 1 \plus{} \sec^2 \alpha}{\sin ^{4}(x \plus{} \alpha)}\allowbreak d\alpha \equal{} \int \frac { 1 \plus{} \sec^2 \alpha}{\sin ^{4}(\tan \alpha \plus{} \alpha)}\allowbreak d\alpha$

Now let $ u \equal{} \alpha \plus{} \tan \alpha \Longrightarrow du \equal{} (1 \plus{} \sec^2 \alpha)d\alpha$
$ \Longrightarrow\int \frac { 1 \plus{} \sec^2 \alpha}{\sin ^{4}(\tan \alpha \plus{} \alpha)}\allowbreak d\alpha \equal{} \int \frac {1}{\sin ^{4}u}\allowbreak du \equal{} \int cosec ^{4}u\allowbreak du \equal{} \minus{} \frac {(\cot u)^3}{3} \minus{} \cot u \plus{} c$
$ \Longrightarrow\int \frac { 1 \plus{} \sec^2 \alpha}{\sin ^{4}(\tan \alpha \plus{} \alpha)}\allowbreak d\alpha \equal{} \minus{} \frac {(\cot(\alpha \plus{} \tan \alpha) )^3}{3} \minus{} \cot (\alpha \plus{} \tan \alpha) \plus{} c$

So
$ \int \frac {\left( 1 \plus{} x^{2}\right) \left( 2 \plus{} x^{2}\right) }{\left( x\cos x \plus{} \sin x\right) ^{4}}\allowbreak dx \equal{} \minus{} \frac {1}{3}(\frac {1 \minus{} xtanx }{x \plus{} tanx})^3 \minus{} \frac {1 \minus{} xtanx }{x \plus{} tanx} \plus{} c$ :D

Hope all calculations are correct.
[hide]I realised that there is a problem when $ x \equal{} \frac {\pi}{2} \plus{} k\pi$ because in this case there is no such $ \alpha$. But it does not affect the above solution[/hide]

but i think $ x \equal{} tan\alpha$ is not a valid substitution as any substitution in definite integrals should be a continous within the limits and $ x\equal{}tan\alpha$ is not continous at $ \frac{3\pi}{2}$

[quote="vijaymenon"]but i think $ x \equal{} tan\alpha$ is not a valid substitution as any substitution in definite integrals should be a continous within the limits and $ x \equal{} tan\alpha$ is not continous at $ \frac {3\pi}{2}$[/quote]

I also thought about this but as I remember if there is only finitely many discontinuity points then you can still do this. I mean you can divide $ (\pi,2\pi)$ as $ (\pi,\frac{3\pi}{2})$ and $ (\frac{3\pi}{2}, \pi)$. Then do above calculations.

is it but consider $ \int _ {\frac {1}{ \minus{} \sqrt3}}^{\frac {1}{\sqrt3}} \frac {dx}{1 \plus{} x^{2}}$
here we get 2 different answers one by the normal method that is considering $ \frac {1}{1 \plus{} x^{2}}$ as the derivative of $ \arctan x$ .  in the same problem try substituting $ x \equal{} \frac {1}{t}$ we see that we get a different answer although $ \frac {1}{t}$ in not continuous only at 0.

[quote="IndoChina"]Well done, cefer :)[/quote]

Thanks.
 But I think the way I used may not be correct. Because $ x\equal{}tan \alpha$ is not valid for $ x\equal{}\frac{\pi}{2}\plus{}k\pi$ and the example that [b]vijaymenon[/b] gave is very good to understand where is the problem in my solution.

The only thing that I am sure is the things that I wrote solves part $ (b)$

Take $ A(x)\equal{}\frac{\left( 1\plus{}x^{2}\right)\left( 2\plus{}x^{2}\right) }{\left( x\cos x\plus{}\sin x\right)^{4}}$ and $ B(x)\equal{}\minus{}\frac{1}{3}(\frac{1\minus{}xtanx }{x\plus{}tanx})^{3}\minus{}\frac{1\minus{}xtanx }{x\plus{}tanx}$

$ B'(x)\equal{}A(x) \Longrightarrow B(x)$ is an antiderivative of $ A(x)$. And $ A(x)$ is continuous on $ [\pi,2\pi]$

So by Fundamental Theorem of Calculus

$ \int_{\pi }^{2\pi }\frac{\left( 1\plus{}x^{2}\right)\left( 2\plus{}x^{2}\right) }{\left( x\cos x\plus{}\sin x\right)^{4}}\allowbreak dx\equal{}\int_{\pi }^{2\pi }A(x)\allowbreak dx \equal{}B(2\pi)\minus{}B(\pi)\equal{}\frac{7\plus{}12\pi^{2}}{24\pi^{3}}$


How 4 boxes can contain 21 beans such that each box has a odd number of beans ?

[hide="maybe"]
7 in the first box, 7 in the second box, 7 in the third box, and the first box in the fourth box.
[/hide]


function
limit
floor function
algebra
calculus
calculus computations

[x^2]-[x]^2

discuss the continuity of the function where [  ] denotes greatest integer function\

Q2  http://www.sosmath.com/CBB/viewtopic.php?p=187990&highlight=#187990 PL SEE THESE TWO QUESTIONS ALSO .IS THERE ANY OTHER WAY OF DOING THESE QUESTIONS .ie  WITHOUT L HOSPITALS RULE(q1 )

I think it is not continuous at least in $ x\equal{}0$ because $ \lim_{x \to 0^\minus{}} f(x) \equal{} \minus{}1$ and $ \lim_{x \to 0^\plus{}} f(x) \equal{} 0$.

You meant the floor function, right?

[quote="uglysolutions"]I think it is not continuous at least in $ x \equal{} 0$ because $ \lim_{x \to 0^ \minus{} } f(x) \equal{} \minus{} 1$ and $ \lim_{x \to 0^ \plus{} } f(x) \equal{} 0$.

You meant the floor function, right?[/quote]yes

$ \lfloor x^2 \rfloor$ has jump discontinuities at the x-values where $ x^2$ has an integer value. That would be at $ \{\pm \sqrt{k} | k \in \mathbb{Z}, k > 0 \}$. These jumps are by a value of $ \minus{}1$ at negative x-values, and $ \plus{}1$ at positive x-values.

$ \lfloor x \rfloor^2$ has jump discontinuities at $ \{k | k \in \mathbb{Z}, k \neq 0 \}$. These jumps are also by $ \minus{}1$ at negative x-values and $ \plus{}1$ at positive x-values.

Therefore, $ \lfloor x^2 \rfloor \minus{} \lfloor x \rfloor^2$ has discontinuities at the x-values where $ \lfloor x^2 \rfloor$ jumps, but $ \lfloor x \rfloor^2$ does not. This would be at the points $ \{\pm \sqrt{k} | k \in \mathbb{Z}, k > 0, and \sqrt{k} \notin \mathbb{Z}\}$, i.e. at $ \pm \sqrt{2}, \pm \sqrt{3}, \pm \sqrt{5}, \pm \sqrt{6}, \ldots$

[quote="gauss202"]$ \lfloor x \rfloor^2$ has jump discontinuities at $ \{k | k \in \mathbb{Z}, k \neq 0 \}$. These jumps are also by $ \minus{} 1$ at negative x-values and $ \plus{} 1$ at positive x-values.[/quote]

try again ...

Ah, that was indeed a mistake. I'll leave it to someone else to correct it though. Hopefully what's there is enough to get someone started if they were stuck.


[quote="Ji Chen"]Let $ ABCD$ be a quadrilateral with $ \angle B = 90^{\circ}$.

$ DM\parallel BC$ and $ DM\cap AC = M$. Prove that

$ AC \times (BD^2 - DA^2) = AB^2 \times (MC - AM)$.
[asy]size(6.8cm,0);
import geometry_dev;

pair a=(-1.6,0);
pair b=(0.3,-1.1);
pair c=(0.9,0);
pair d=(0.7,2.2);
pair m=(-0.7,0);

point A=a, B=b,C=c, D=d,M=m;

dot("$A$",A);
dot("$B$",B);
dot("$C$",C);
dot("$D$",D);
dot("$M$",M);
draw(A--B);
draw(B--C);
draw(C--D);
draw(D--A);
draw(A--C);
draw(D--M,red);
draw(D--B);[/asy][/quote]

Let's $ DM\cap AB = K$(so we have $ AB \perp DK$)
Then we have that:
$ BD^2-AD^2=BK^2-AK^2=AB(BK-AK)$
It's left to prove that:
$ AC\cdot AB \cdot (BK-AK)=AB^2\cdot (MC-AM)$
or
$ \frac{AK-BK}{AB}=\frac{AM-MC}{AC}$
or
$ \frac{AK}{AB}=\frac{AM}{AC}$
which is right.


[hide]
$x^2+x-1=0 \Rightarrow x=\frac{\sqrt{5}-1}{2}$[/hide]

[quote="danne89"]Can you post an image istead. I'm sitting here under linux without any possibility to open *.doc-files.[/quote]
hello danne89, i know how that feels without the proper software/OS to view it, below is the picture riddler is referring to:

[quote="riddler"]palytoxin, can you tell me how to get the picture onto the workspace?[/quote]
copy and paste it into mspaint. save it as .gif. when you post, click "Add an Attachment" just right below when you Post a reply (after the Preview, Submit buttons). saving as .gif is good because it keeps simple image files as small as possible. if your image has colors... well i suggest the lower quality .jpeg because if you save it as .gif in mspaint you will lose colors thou

[quote="shobber"]This is the golden ratio. This 36-72-72 triangle is very beautiful just for that.[/quote]

Actually, this is the negative of the negative Golden Ratio. What is normally considered the Golden Ratio is $\frac{1\pm \sqrt{5}}{2}$; the square of the positive root is one more than the positive root, and the square of the negative root is one less than the positive root.

I wonder if they got golden ratio from this triangle or use golden ratio to find this triagnle, but after all, this 36-72-72 follows the definition of golden ratio.

[quote="shobber"]This is the golden ratio. This 36-72-72 triangle is very beautiful just for that.[/quote]
I never knew that.  :roll:    Form that you can find the exact value of

$\sin{\frac{\pi}{60}}=\frac{\sqrt{30}+\sqrt{10}-\sqrt{6}-\sqrt{2}-\sqrt{12\sqrt{5}+60}+\sqrt{4\sqrt{5}+20}}{16}$

:D


ratio
trigonometry
geometry
angle bisector

In triangle $ABC$, points $D$ and $E$ are on $AB$ and $AC$, and angle-bisector $AT$ intersects $DE$ at $F$. If $AD=1$, $DB=3$, $AE=2$, and $EC=4$, compute the ratio $AF: AT$.

[hide="answer"]Since $BT: TC=AB: AC=4: 6=2: 3,$ use Menelaos' theorem we have $\frac{1}{3}\cdots \frac{5}{3}\cdots \frac{TF}{FA}=1\Longleftrightarrow TF: FA=9: 5,$ yielding $AF: FT=5: 14.$[/hide]

[quote="Iversonfan2005"][quote="4everwise"]The answer is actually [hide]5/18[/hide].

I'm not sure how to get that.  :([/quote]

Use $\frac{1}{2}ab\sin{c}$, for the small triangle and the whole triangle and note that the small triangle is equal to the two smaller triangles it's composed of. By equating the two together, you can find a general relationship between the sides and will get you the ratio. The answer is $\frac{5}{18}$

If you don't have any idea what i just said, i will post a solution.[/quote]oooooh. That's clever.  :D They had that in AOPS 2 and I understood it, but I still haven't used it enough to recognize when it'll be useful.

I should be able to figure it out now. :yup:

[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=763&sid=99231af7f9e517b6b003f3439f0cfc5e[/img]

In the diagram shown above, let $AF=x$ and $AT=y$. Because we are only given two sides of the triangle and a pair of congruent angles, we think to apply the triangle area formula $\text{area}=\frac12 ab\sin C$.

We know that $[\triangle ADE]=[\triangle ADF]+[\triangle AEF]$, so $\frac12 (1)(2)(2\sin \theta \cos \theta)=\frac12 (1)(x)(\sin \theta)+\frac12 (2)(x)(\sin \theta)\implies x=\frac43 \cos \theta$.

Similarly, $[\triangle ABC]=[\triangle ABT]+[\triangle ACT]$ and $\frac12(4)(6)(2\sin \theta \cos \theta)=\frac12(4)(y)(\sin \theta)+\frac12(6)(y)(\sin \theta)\implies y=\frac{24}{5} \cos \theta$. Therefore, our answer is $AF: AT=\frac43: \frac{24}{5} =\boxed{18: 5}$.  [/hide]


geometry
3D geometry
tetrahedron
trigonometry
trig identities
Law of Sines
geometry unsolved

Let $ABCD$ be a tetrahedron with $\angle BAD = 60^{\cdot}$, $\angle BAC = 40^{\cdot}$, $\angle ABD = 80^{\cdot}$, $\angle ABC = 70^{\cdot}$. Prove that the lines $AB$ and $CD$ are perpendicular.

Are you sure the problem statement is correct? $\angle{ABD} > \angle{ABC}$?

Yes, it's correct I suppose (otherwise I cannot imagine how those two segments can be perpendicular)

Strange. I don't manage to draw the quadrilateral - at least, not a convex one.

Trig Ceva gives that $\sin{\angle{ADC}}=\dfrac{1}{2}$.

Sorry, after all there was a mistake but not in angles  :blush: it should be tetrahedron instead of quadriteral, everything else is correct.

Let [Φ] denote the plane containing shape Φ. Lay triangle ADB on [ΔABC] as shown on the next page. Point D is now D’. From D’ draw the altitude D’I to AB and extend it to meet AC at C’. Since ∠BAD’ = 60°, ∠BAC = 40° and ∠ABC = 70°, we have ∠AD’I = 30°, ∠AC’I = 50°.

Now let AD = a. It’s easily seen that AI = a/2 and D’I = a×sqrt(3)/2.

Applying the law of sines, we get

D’B = a×sqrt(3)/(2×sin80°),  BI = a×sqrt(3)×sin10°/(2×sin80°),
AB = AI + BI = a/2 + a×sqrt(3)×sin10°/(2×sin80°), and
AC’ = a/(2×sin∠AC’I) = a/(2×sin50°).

Now let’s prove that
1/sin50° = 1 + sqrt(3)×sin10°/sin80°						(i)
1/sin50° = 1 + sqrt(3)×sin10°/cos10°, or
1/[sqrt(3)×sin50°] = 1/sqrt(3) + tan10°, or
1/[sqrt(3)×sin50°] = tan30° + tan10°, or
1/[sqrt(3)×sin50°] = sin(30° + 10°)/cos30°×cos10° = sin40°/(cos30°×cos10°), or

sqrt(3)×sin50°×sin40° = cos30°×cos10°, or
sin50°×sin40° = cos30°×cos10°/sqrt(3), or
sin50°×sin40° = cos10°/2, or

-½(cos90° – cos10°) = ½cos10°, and finally this is obvious because cos90° = 0.

Now multiplying both sides of equation (i) by a/2, we get
a/(2×sin50°) = a/2 + a×sqrt(3)×sin10°/(2×sin80°).

Therefore, AC’ = AB and ∠AC’B = ∠ABC = 70°, or C’ coincides C, and the three points D, I and C are collinear, and DC ∈[ΔIDC]. This confirms that AB perpendiculars CD.


Suppose a computer can generate a whole number from 1 to k in one instance randomly.

What is the expected number of instances, n required for each given k such that a complete set of 1 to k has been produced? Ie, the first time where you get all of the numbers from 1 to k.

How do I calculate Prob(set first completed in exactly nth instance) for all n?

Thanks all!

Can you plz explain the problema bit more. it doesn't seem clear to me!!! :?


A Density is a very complicated matter and therefore i would like to bring thi =s matter up. so please help on my homeworks.

A density curve looks like an inverted letter V. The first segment goes from point (0, 0.6) to the point (0.5, 1.4). The second segment goes from (.5, 1.4), to (1, 0.6)

What percent of the observations lie belo 0.3?
What percent of the observations lie between 0.3 and 0.7?

This has to do with z-scores, etc. You should look it up in your book.

yea but it's not in the book. The teacher said it was a problem that's going to be on the quiz similar, and I don't understand it at all.

[quote="Ravi B"]I don't think this problem has to do with z-scores.  That's a concept for normal distributions.

Anyway, look at the region below the density curve but above the x-axis.  That region consists of two trapezoids.  Each trapezoid has area 1/2, which means the area of the region is 1.  That's good -- that's what's supposed to happen with a density curve.

To solve the first problem (percent of observations below 0.3), look at the portion of the region in the previous paragraph that is to the left of x = 0.3.  That subregion happens to be another trapezoid.  The area of that subregion will be the probability (which you can then convert to a percent).

You can solve the second problem similarly.[/quote]

Ah okay thanks, I'm pretty sure I get it now


Your school is ordering computer equipment. Let c represent the cost of one personal computer and p represent the cost of one printer.

a. Write an expression for the total cost of 15 personal computers and 8 printers.

b. The school discovers that it needs 8 additional computers and 4 additional printers. Write an expression for the total cost of this additional equipment.

c. Write an expression that will represent the total amount of money spent on computer equipment.

[hide="a"]

$15c+8p$

[/hide]

[hide="b"]

$23c+12p$

[/hide]

[hide="c"]

$23c+12p$

[/hide]

it seems too easy?

I believe b. is $8c+4p$ as it is only asking for the cost of the additional equipment... either way, it's just semantics  :P.

It is great to have the answers but how did you get the answers?

I want to know what you did to get the answers.

Thanks!

[hide]
This is the way he got the answer to a:
Since the cost of one computer is c, we can say the cost of 15 computers is 15c.  Since the cost of one printer is p, we can say that the cost of 8 printers is 8p.  Therefore, the total amount payed is 15c+8p.
This is how he got the answer to b:
Since 8 additional computers are needed, this will cost 8c.  We also know that 4 additional printers are needed, therefore 4p.  When you add these up, you get 8c+4p.
This is how he got the answer to c:
The answer to b. is the [b]additional[/b] amount, so the total amount is previous cost+additional cost, which is 8c+4p+15c+8p=23c+12p.  [/hide]

I appreciate all the help given but the last math tutor went into detail, which is what I mainly seek by posting questions here.

Thanks!


function
real analysis
real analysis solved

Show that for each f:QxQ->R there is a function g:Q-.R s.t. f(x,y)<=g(x)+g(y) for any x,y\in Q.

There was a point (b) asking to construct f:RxR->R s.t. there is no function g:R->R with the same propertie as in the statement above, but I couldn't post it here because I haven't solved it, but I'm still trying :).

Go to
http://www.ucl.ac.uk/~ucahjej/imc/imc2003/day2_solutions.pdf
......

Oh, I already have the solutions to both problems, but I still want to think about the second one, because it just looks so nice :).


Atasat este un program freeware de facut figuri geometrice.

Eu am un alt program, [url=http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/doc_en/]C.a.R.[/url] [url=http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/Download/car.exe](Compass and Ruler)[/url]. (Cu mentiunea ca foloseste JAVA)

[quote="enescu"][quote="Marius Damian"]Corel Draw 12 si multa rabdare :) .[/quote]
In nici un caz...:) Recomand [url=http://www.mit.edu/~ibaran/kseg-0.401.zip]programul acesta[/url] . Nu e de desenat, e pentru geometrie.
Altfel, mai exista resurse, dar nu freeware: Geometer's Sketchpad, Cabri, etc[/quote]
Multumesc pentru recomandare. Voi folosi si eu acest program.

[quote="Valentin Vornicu"][quote="Marius Damian"]Corel Draw 12 si multa rabdare :) .[/quote]Eu credeam ca postul asta se vrea sa fie o gluma initial :)[/quote]
De ce ai crezut asta?
E chiar asa nepotrivit Corel pentru realizarea desenelor de geometrie?  :?

[quote="Valentin Vornicu"][quote="Marius Damian"]
De ce ai crezut asta?
E chiar asa nepotrivit Corel pentru realizarea desenelor de geometrie?  :?[/quote]In comparatie cu Kseg sau Geometer's Sketchpad, da este.[/quote]
Cand am inceput sa fac desene pe calculator nu stiam de programele pe care le amintesti. Aveam doar un CD de la PC Magazin cu un COREL demo. Asa ca... :)

salut.....
ma chema iurie si particip deasemenea la olimpiade de matematica ......dar nu in tara romina(dar sunt romin)
dar nu am suficiente materiale pentru a invata ......poti sami dai telefonul ori adresa de  la vreun professor de matematica ce face intrenari pentru olimpiade .....
                          bine ....... iti multumesc pentru rabdare si astept un curind raspuns.......
                                                                                ciao

Pentru Cimabue: Cum poti sa copiezi un desen din "kseg" in Word?

Il exportezi ca imagine (jpg, png, ce vrei tu) din kseg (sau oricare alt program), apoi, in word, insert $\mapsto$ image (sau ceva de genul).

[quote="enescu"]E forte bun si [url=http://www.geogebra.at/]programul asta[/url], nu numai pt. geometrie, dar si pt. grafice, analiza etc. In plus, e freeware!
Iar pt. geometrie in spatiu, [url=http://ns31710.ovh.net/data/cabri3d_1.1_eval_win.exe]Cabri 3D[/url].[/quote]
Nu pot sa descarc Cabri 3d..mai aveti si altele?

Poate cineva sa posteze programul Cabri 3D versiunea pe care o poti folosi nelimitat???Va rog frumos :!: Am cautat pe internet si nu am gasit decat versiunea de 30 de zile.  :(

andrei, cred ca ar fi bine sa iti editezi postul de mai sus. eu nu sunt admin sau ceva, dar ce zici tu acolo tine de warez, care nu sunt prea legale...si..intelegi tu :)
ontopic :
intr`adevar e misto kseg. si ies desenele mult mai misto decat cu paint :)

Eu am programul Geogebra care mi se pare mai misto decat kseg

texgraph http://perso.orange.fr/Fradin.Patrick/TeXgraph/accueil.html
care e destul de bun,spre deosebire de altele stie sa exporte in LaTeX
pentru a putea folosi respectiva figura in vreun document.

Eu stiu unul foarte usor de manuit, pe care-l folosesc de citiva ani! [b][color=red][b]Se numeste: Euklides (e facut de maghiari).[/b][/color] [/b]Un desen la o figura destul de complicata pt. o problema cu acest program nu-mi ia mai mult de 2-3 min. Desenul poate fi salvat apoi in format *.wmf si importat intr-un document word. Poate desena pana si conice, poate gasi si locuri geometrice cu conditii impuse (le deseneaza, demonstratia trebuie gasita insa individual)!
In attachment gasiti acest program pentru incercare!
In alt attachment va anexez un fiser word care contine un astfel de desen facut cu acest program.

[quote="mihai miculita"]Eu stiu unul foarte usor de manuit, pe care-l folosesc de citiva ani! [b]Se numeste: Euklides (e facut de maghiari). [/b]Un desen la o figura destul de complicata pt. o problema cu acest program nu-mi ia mai mult de 2-3 min. Desenul poate fi salvat apoi in format *.wmf si importat intr-un document word. Poate desena pana si conice, poate gasi si locuri geometrice cu conditii impuse (le deseneaza, demonstratia trebuie gasita insa individual)!
In attachment gasiti acest program pentru incercare!
In alt attachment va anexez un fiser word care contine un astfel de desen facut cu acest program.[/quote]


Let m,a,b be positive integers such that

[b]m^2 < a < b.[/b]

If a*b is a square integer,

prove that [b] b> (m+1)^2[/b]

sorry N.T.TUAN, i didn't see you had posted before.. :)

[quote="EUCLA"]The square which follows $m^{2}$ is $(m+1)^{2}$

or $a=b$ which can't be since $a<b$

or $a=1$ and $b$ is a perfect square whuch can't be since $m^{2}<a$ and $m,a$ are positive integers

or $a=x , b= x^{3}$

then it can't be $m^{2}< x^{2}< (m+1)^{2}$
hence $x^{3}>x^{2}>(m+1)^{2}$[/quote]

i fail to see the significance of your statements.can you elaborate more?

why did you take just three cases,
(a=b) or (a=1, b is perfect square) or (a=x ,b=x^3)

c^2 = a*b > m^4

[quote="N.T.TUAN"]Same idea for  http://www.mathlinks.ro/Forum/viewtopic.php?t=127497  :wink:[/quote]

can you explain in more clarified form

how that fact applies here.

Your problem is case particular of following problem: Let $k$ and $n$ be positive integers. Let $I(k,n)=\{j\in\mathbb{N}|k^{n}<j<(k+1)^{n}\}$.
a)For $n=2$ and all $k$, prove that there do not exist distinct $a,b\in I(n,k)$ such that $ab$ is a square.
b)For each $n>2$, prove that when $k$ is sufficiently large there exist $n$ distinct integers in $I(k,n)$ whose product is the $n-$ th power of an integer.
[i]H. A. ShahAli, AMM.[/i]

At this time, i know only prove of a). I will post later if you want :)

You are right SOUMYASHANT NAYAK, i need to explain better what i did..thank you to make me notice that! :)

For $ab$ as a square we have three cases,

$(1)$ $ab< m^{2}$ which can't be since $a, b > m^{2}$ and $a,b\geq 1$
$(2)$ $ab=m^{2}$  which can't be for the same reason as above

then we have $(3)$ $ab> m^{2}$

$ab$ should be a square, then the exponents of its divisors in its factorization should be even

The simplest cases are
$(1)$ $a=b$ which can't be since $a<b$

$(2)$ $a=1$ and $b$ is a perfect square which can't be, since $m^{2}<a$ and $m, a$ are positive integers

For the other cases, just take the simplest. Take a factor of $ab$ which is contained
in $a$ a number $x$ of times and in $b$a number $y$ of times, such that $y>x$.
There should be one with this property, if not then $a>b$.

Now, $x+y=2k$
$(1)$ both $x, y$ are even.  Then the smallest $(x,y)$ is such that $x=2$, $y=4$
$(2)$ both $x, y$ are odd.  Then the smallest $(x,y)$ is such that $x=1$, $y=3$

Coming back to the $b$, we have the least $b$ when there is only the factor whose exponents are $x, y$.
Call that factor $z$, then the least $b$ is $z^{y}$.

The following square of $m^{2}$ is $(m+1)^{2}$
then we have $y\geq 3$. But $z^{y}\geq z^{3}>z^{2}\geq (m+1)^{2}$ $[b>m^{2}]$
$b\geq z^{y}>(m+1)^{2}$

If there is still something which is not clear, please ask me  :)


$ S\equal{}1^{2007}\plus{}2^{2007}\plus{}3^{2007}\plus{}\cdots \,\plus{}2006^{2007}$
prove that S is divisible by 2007.
PS: sorry for my english i am moroccan member  :D

0 = (1+2005)^2007 = 1^2007 + 2005^2007 mod 2007

all the middle terms of the expansion cancel out
Just use a similar process for other terms

How do I make it so it looks like Latex

on the last post i meant to say :
2006^2007 = 2006 mod 2007

Here's an interesting twist on the problem:
Determine, with proof, all values of n such that $ 1^{n}\plus{}2^{n}\plus{}...\plus{}2003^{n}$ isn't a multiple of 2003.


trigonometry
inequalities
algebra unsolved
algebra

Given sequence $ x_n$ such that:$ x_n \in R;x_{n \plus{} 1} \equal{} x_n \plus{} \frac {1}{2}(c{\rm{os}}\,x_n \plus{} \sin \,x_n )(\forall n \in N^* )$.Find limited of sequence depending on $ x_1$.

$ (x + \frac {sinx + cosx}{2})' = 1 + \frac {cosx - sinx}{2} > 0$

Thus, for $ \frac {3\pi}{4} > x_n > - \frac {\pi}{4}$, we have

$ x_{n + 1} = x_n + \frac {sinx_n + cosx_n}{2} < \frac {3\pi}{4} + \frac {sin\frac {3\pi}{4} + cos\frac {3\pi}{4}}{2} = \frac {3\pi}{4}$

So $ x_n < x_{n + 1} < \frac {3\pi}{4}$ (the first inequality comes from $ sinx + cosx > 0 \forall x \in ( - \frac {\pi}{4}, \frac {3\pi}{4})$). Then, $ x_n$ is an increasing and bounded sequence case $ x_1 \in ( - \frac {\pi}{4}, \frac {3\pi}{4})$. This sequence must have a limit $ L$ for which

$ L = L + \frac {sinL + cosL}{2} \Rightarrow sinL + cosL = 0 \Rightarrow L = \frac {3\pi}^{4}$.

The case $ \frac {3\pi}{4} < x_1 < \frac {7\pi}{4}$ is similar. The cases $ x_1 = \frac {3\pi}{4}$ and $ x_1 = \frac {7\pi}{4}$ give constant sequences. The other cases are obtained from these cases by adding $ 2\pi$ to $ x_1$, which just translates the sequence.


vector
analytic geometry
calculus
calculus computations

What is the best way to prove identities such as the following:

(1) $ div(\nabla f\times\nabla g)\equal{}0$
(2) $ curl(curl\,$ [b]F[/b]) $ \equal{} grad(div\,$ [b]F[/b]$ ) \minus{} \nabla ^2$[b]F[/b]

Is this specifically for $ \mathbb{R}^3$? If so, then you could brute force it by working it out component-wise.

This must be specifically $ \mathbb{R}^3.$ The cross product is specifically 3-dimensional, and while we can talk about something like the curl in all dimensions, only in $ \mathbb{R}^3$ is the curl of a vector field also a vector field; hence only in $ \mathbb{R}^3$ does "the curl of the curl" makes sense.

Some proofs of properties I've seen are related to properties of the dot and cross product themselves such as:

$ curl(\nabla f) \equal{} \nabla\times (\nabla f)$, which is analogous to $ a\times a \equal{} 0$.

and

$ div(curl F) \equal{} \nabla\cdot(\nabla\times F)$, which is analogous to $ a\cdot (a\times b) \equal{} (a\times a)\cdot b\equal{}0$.

Does this work in general?

A proof for #2 may be found here. It simply utilizes properties of Cartesian coordinates and expands each expression.

http://en.wikipedia.org/wiki/Vector_Laplacian/Proofs


geometry
trapezoid
trigonometry
power of a point
radical axis

[color=darkblue]Let $ABCD$ be a right trapezoid $(AD\parallel BC$ , $AB\perp BC$ , $BC>AD)\ .$Denote : the middlepoint $M$ of the side $[AB]$ ; the projection $E$ of the point $D$ to the side $[BC]$ ; the point $F\in (DE)$ so that $DF=CE\ .$ Prove that $MF\perp CD$ $\Longleftrightarrow$ $BF=BC\ .$[/color]

The first equivalence follows from the fact that both angles are acute.
\[MF \perp CD \iff \tan{\widehat{DCE}}=\tan{\widehat{AMF}}\iff \frac{AB}{EC}= \frac{AD}{\frac{AB}{2}-EC}\]
\[\iff EC.(AD+AB) = \frac{AB^{2}}{2}\iff (AB-CE)^{2}+AD^{2}= (AD+CE)^{2}\]

So conclusion follows from Pythagoras, because $(AB-CE)^{2}+AD^{2}=BF^{2}$

[color=darkblue][b]Another enunciation.[/b] Let $ABCD$ be a right trapezoid $(AD\parallel BC$ , $AB\perp BC$ , $BC>AD)\ .$Denote : the middlepoint $M$ of the side $[AB]$ ; the projection $E$ of the point $D$ to the side $[BC]$ ; a point $F\in (DE)\ .$ Prove that if two from the following three statements are truly, then and the another is truly :
\[\begin{array}{ccccc}1.\ BF=BC\ . & \maltese & 2.\ DF=CE\ . & \maltese & 3.\ MF\perp CD\ . \end{array}\]
[b]Remark.[/b] As far as possible, a synthetical proof ![/color]

[hide]
Construct circles $\omega_{1}$ and $\omega_{2}$ centered at $C$ and $D$, respectively, such that both are tangent to $\overline{AB}$, let $P$ and $Q$ be the intersection points of the circles, and $M^\prime$ be the intersection of $\overline{PQ}$ and $\overline{AB}$. By Power of a Point,
\[M^\prime A^{2}=M^\prime P\cdot M^\prime Q=M^\prime B^{2}\]
so $M^\prime=M$ and $M$ is on the radical axis of the two circles, making (3) equivalent to "$F$ is on the radical axis of the two circles", which is in turn equivalent to "$DA^{2}-DF^{2}=CB^{2}-CF^{2}$" (applying Power of a Point to $F$ on the lines through the centers of $\omega_{1}$ and $\omega_{2}$ and setting them equal).

So any two of the given statements will imply the third if \[AD^{2}-CE^{2}=BF^{2}-CF^{2}\]
But $AD=BE$, $BF^{2}=BE^{2}+EF^{2}$, and $CF^{2}=CE^{2}+EF^{2}$. Substituting, the above equation reduces to $BE^{2}-CE^{2}=BE^{2}-CE^{2}$, a true statement.
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trigonometry
geometry
function
ratio
trig identities
Law of Sines

The opposite is $ c$.  Which gives that $ \sin{C}\equal{}\frac{c}{c}\equal{}1$.

I don't think that is correct.

Does anybody else think something different.

I don't think hunter is correct because there is a formula to find the area of a right triangle that involves multiplying by SinC and I don't think that would be part of the formula because that would just be multiplying by 1.

Oh and another quick question.

If you have something like $ sin^3 A$ is that basically $ (sinA)^3$ or something else?

Why isn't it correct? How is the opposite not $ C$?

And the formula is $ \frac12ab\sin C$, which yields $ \frac12ab$ for right triangles (this is clearly correct).

And yes. What you should be careful about is $ \sin^{\minus{}1}$, which is the inverse sine function (not really a function), not the multiplicative inverse.

[quote="Wickedestjr"]I don't think hunter is correct because there is a formula to find the area of a right triangle that involves multiplying by SinC and I don't think that would be part of the formula because that would just be multiplying by 1.
[/quote]

The formula you're talking about is $ \text{Area}\equal{}\frac{1}{2}ab\sin C\equal{}\frac{1}{2}bc\sin A\equal{}\frac{1}{2}ca\sin B$

Thus, the area of the triangle would be $ \frac{1}{2}ab\sin C\equal{}\frac{1}{2}ab$, which is correct.

And yes, $ \sin^3A$ is $ (\sin A)^3$.

Basically, think as if you have a triangle, and you were finding the O/H ratio of the right angle. In that case, we have degenerate triangle as a line segment, in which O and H are the same, meaning they're congruent, and if they're congruent, the ratio of one to another must be $ 1$.

$ C\equal{}90$ then $ sinC\equal{}sin90\equal{}1$

[quote="Wickedestjr"]Okay, sorry hunter. I just thought it was wierd to add sinC to that formula when it was basically multiplying by one.[/quote]

That's there because C is not nessacarily 90. It could be 45 for example, or 60.

Oh now i understand. Thanks for clarifying that BOGTRO. I always assume that C is a right angle.

[quote="Wickedestjr"]I think I may have asked this before, but if you have right triangle $ ABC$, and $ C$ is the right angle, then how do you find $ sinC$. It should be opposite/hypotenuse, and the hypotenuse is $ c$, but what would the opposite be? Is there something I'm missing?[/quote]
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There are several ways you can prove that the sine of a right angle equals $ 1$. The easiest way is using the identity:
\[ \cos^2x \plus{}\sin^2x \equal{} 1\]

For your case, $ \cos^2\frac{\pi}{2} \plus{}\sin^2\frac{\pi}{2} \equal{} 1$.

You can see by inspection that when $ x \equal{}\frac{\pi}{2}$ the length of the adjacent side is $ 0$.

Consequently, $ \cos\frac{\pi}{2} \equal{} 0 \implies \cos^2\frac{\pi}{2} \plus{}\sin^2\frac{\pi}{2} \equal{}\sin^2\frac{\pi}{2} \equal{} 1$.

\[ \therefore \sin\frac{\pi}{2} \equal{} \pm 1\].

Considering the geometry, the solution $ \sin\frac{\pi}{2} \equal{} \minus{}1$ is absurd.

Thus
\[ \boxed{\sin\frac{\pi}{2} \equal{} 1}\]

Another way to prove it is using the Law of Sines
\[ \frac{\sin A}{a} \equal{}\frac{\sin B}{b} \equal{}\frac{\sin C}{c}\]

for any $ \triangle ABC$ whatsoever, right, isosceles, or scalene.

$ a$ is the length of the side opposite $ A$, $ b$ is the length of the side opposite $ B$, and $ c$ is the length of the side opposite $ C$.

We'll pick two angles, the right angle and one other, say $ A$ then
\[ \frac{\sin A}{a} \equal{}\frac{\sin C}{c} \\ \\ \implies \sin C \equal{}\frac{c}{a}\sin A\]

Since $ \triangle ABC$ is a right triangle $ \sin A \equal{}\frac{a}{c}$  therefore,
\[ \sin C \equal{}\frac{c}{a}\sin A \equal{}\frac{c}{a}\cdot\frac{a}{c} \equal{}1\][/hide]


[quote="ashrafmod"]tan(1\2)   +tan( 1\3) +tan ( 1\4) ................[/quote]
I don't see the question. Dou you want to evaluate $\tan\left(\frac12\right)+\tan\left(\frac13\right)+\tan\left(\frac14\right)+\ldots$? If you do, then what is
[quote="ashrafmod"]-1           -1             -1[/quote]
 :?:

I think he meant $\sum_{i=2}^{\infty}\tan^{-1}\left(\frac{1}{i}\right)$, which diverges :roll:

EDIT: 'converges' $\Rightarrow$ 'diverges' of course...

Use $\LaTeX$ and don't double post ( http://www.mathlinks.ro/Forum/viewtopic.php?t=112232 ).


The student council has 11 members who signed up to be on committees for the school dance. The committees are listed below. There is only one time slot open for meetings each day and each committee needs one full time slot to get its work done. What is the least number of days needed so that each member can attend the committee meetings without conflicts?
\begin{tabular}{|l|c|}\hline Band&Sean, Kirsten, Mike\\\hline Decorations&Kirsten, Lien, Kim, Angela\\\hline Tickets&Kim, David, Dan, Lee\\\hline Food&Shannon, Elise, Mike, David\\\hline Chaperones&Lien, Sean\\\hline\end{tabular}

[quote="GameBot"]The student council has 11 members who signed up to be on committees for the school dance. The committees are listed below. There is only one time slot open for meetings each day and each committee needs one full time slot to get its work done. What is the least number of days needed so that each member can attend the committee meetings without conflicts?
$ \begin{tabular}{|l|c|}\hline Band&Sean, Kirsten, Mike\\\hline Decorations&Kirsten, Lien, Kim, Angela\\\hline Tickets&Kim, David, Dan, Lee\\\hline Food&Shannon, Elise, Mike, David\\\hline Chaperones&Lien, Sean\\\hline\end{tabular}$[/quote]

We go in an organized manner.

Assume that the Band group meets on Day $ 1$. The Tickets group has no members that are in the Band group, so we schedule both the Band and Tickets groups on Day $ 1$.

Assume that the Decorations group goes on Day $ 2$. We can also schedule the Food group on that day.

The final day, Day $ \boxed3$, is for the Chaperones group only.


When doing questions concerning Gibbs Energy:

$ \Delta G ^{\circ} \equal{} \minus{}RT \ln{K}$

How do we know if $ K \equal{} K_c$ or $ K\equal{} K _p$.

K is the equilibrium constant expressed with concentrations. If the partial pressures for the system in study are relevant (gases) then we can express the concentrations as functions of pressures by using the equation of state.

The problem is that I was doing the Prepatory problems for 2007 IChO. I got to this question about Gibbs energy. Because I was given the reaction with a pure liquid and a solid making the constant:

$ K_c \equal{} \frac{1}{[O_2]}$ so I decided to make it into $ K_p$ with $ PV\equal{}nRT$ but the solution simply assumed $ K_c \equal{} K_p$!

1. Post the problem as given to you, in order to make a better judgement.

2. What is the relation between Kc and Kp? Under what conditions are they equal?

http://www.icho39.chem.msu.ru/preparatory-problems/Problems_theoretical.pdf

It is problem 6.