algebra
polynomial
inequalities
complex numbers
algebra open

1.Without         removing       brackets       at       any       stage         factorize

   $2y(y+z)-(x+y)(x+z)$

2.Factorize        the        polynomial        in     x

  ($t^2$+1)$x^3$-($t^2$-1)$x^2$-($t^2$+1)$x$+($t^2$-1)

3.Factorize      $a^5$(b-c)  +   $b^5$(c-a)   +   $c^5$(a-b)

2)  First, try to find a value of $x$ that is a root of the expression, regardless of $t$.  Notice that the $x^3$ coefficient is the negative of the $x$ coefficient and that the $x^2$ coeffcient is the neagitve of the constant.  So, if we could make $x^3$, $x^2$, and $x$ all be 1, we could reduce the entire expression down to zero, regardless of $t$.  $x=1$ does precisely this.  Therefore, 1 is a root of the expression that is independent of $t$.
So, use synthetic (or long) division to divide the expression by $x-1$.  You will obtain $(t^2+1)x^2+(2)x+(-t^2+1)$.  Set it to zero and solve for $x$.
$x=\frac{-2\pm\sqrt{4-4(-t^4+1)}}{2t^2+2}=\frac{-1\pm\sqrt{t^4}}{t^2+1}=\frac{-1\pm\sqrt{t^4}}{t^2+1}=\frac{-1\pm t^2}{t^2+1}$
So, $\frac{t^2-1}{t^2+1}$ and $-1$ are also roots of the expression.

So, factoring yields: $(t^2+1)(x-1)(x+1)(x-\frac{t^2-1}{t^2+1})=(x-1)(x+1)((t^2+1)x-(t^2-1))$

(t^1+1 was added because the that was the leading coefficient.)

1) Just look at $(x+y)(x+z)$:
If $x=y$, $(x+y)(x+z)=2y(y+z)$.
If $x=-2y-z$, $(x+y)(x+z)=(-y-z)(-2y)=2y(y+z)$.
The two roots were easy to identify! We just have to be careful with the coefficient:
$2y(y+z)-(x+y)(x+z)=-(x-y)(x+2y+z)$.

[quote="draperp"]Would one always include $t^2+1$ as a separate factor?  I'd combine it with the $x-\frac{t^2-1}{t^2+1}$ to eliminate the fraction.  Or is it just a matter of personal preference.[/quote]
I was taught to factorize in this form $k(x-x_1)(x-x_2)...(x-x_n)$.

[quote="lordWings"]3) If we want to factorize in $a$, we identify easily the roots $a=b$ and $a=c$, and the coefficient is $(b-c)$, so the factorization begins with $(b-c)(a-b)(a-c)...$[/quote]
If $a^5(b-c)+b^5(c-a)+c^5(a-b)=(b-c)(a-b)(a-c)P(a,b,c)$
then $P(a,b,c)=\frac{a^5}{(a-b)(a-c)}+\frac{b^5}{(b-a)(b-c)}+\frac{c^5}{(c-a)(c-b)}$.
Nice! It looks like an expression that would appear in an inequality, doesn't it?
Maybe the factorization would help to solve such inequality...

[quote="lordWings"]$a^5(b-c)+b^5(c-a)+c^5(a-b)=(b-c)(a-b)(a-c)P(a,b,c)$
$P(a,b,c)=\frac{a^5}{(a-b)(a-c)}+\frac{b^5}{(b-a)(b-c)}+\frac{c^5}{(c-a)(c-b)}$.[/quote].
I just tried with software to find the other three roots (call me cheater only if you've spent hours trying to factorize it manually).
I get three "monster" expressions (probably the software used Cardano), but perhaps they could be simplified.
There's one real root, $r-\frac{1}{3}(c+b)$, and then there's a pair of complex roots $re^{\pm i\frac{2\pi}{3}}-\frac{1}{3}(c+b)$, (so they're the three values of $\sqrt [3]{r^3}-\frac{1}{3}(c+b)$, with $r$ real).
If you're brave enough, you may want to watch the [hide="monster"]
$r=\frac{1}{6}\sqrt [3]{-80c^3-60bc^2-60b^2c-80b^3+12\sqrt{48c^6+72c^5b+105c^4b^2+150b^3c^3+105c^2b^4+72b^5c+48b^6}}-6{\frac{2/9c^2+1/9bc+2/9b^2}{\sqrt [3]{-80c^3-60bc^2-60b^2c-80b^3+12\sqrt{48c^6+72c^5b+105c^4b^2+150b^3c^3+105c^2b^4+72b^5c+48b^6}}}}$
.[/hide]
Please tell me there's a beautiful solution. I'm starting to think $r=0$ would be wonderful. I'll try later.

[quote="lordWings"]That's the easy part, Ramanujan. :P What about the next part?[/quote]

What part ??

$\\ ...=(a-b)(a-c)(b-c)[a^3+b^3+c^3+a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+abc]$

That is what we are looking for ... i think .  :roll:

Lots of questions pop to my mind:
Why do you stop factoring at that point? Aren't there any more factors? How do we know when we're done factoring? Hasn't a polynomial of degree 3 at least one real root? Shouldn't we finish with polynomials of degree 1 (real root) or at most 2 (couple of complex conjugate roots)?
Maybe I'm assuming something I shouldn't because of the multiple variables (I'm not quite used to it). Please correct me if I do.


geometry
geometric transformation
reflection

Equilateral triangle ABC has AB=1. For each vertex, one of the other two vertices are chosen. A reflection of the former vertex is done with respect to the latter. Suppose the image is triangle PQR. How many possible areas of PQR are possible?

[hide]Let $ P$ be the reflection of $ A$ that is closest to $ C$. This is our "fixed" point. There are two possible locations for $ Q$, and two for $ R$. If $ Q$ is chosen as the reflection of $ B$ closest to $ P$, then both possible locations for $ R$ give $ PQR$ the same area. If $ Q$ is chosen as the reflection of $ B$ farthest from $ P$, then one location of $ R$ gives $ PQR$ the same area as before, but the other gives a different, larger area. There are no other possible arrangements of the reflected vertices, so in all, there are $ 2$ possible areas for $ PQR$.[/hide]


A double is a positive integer that is obtained by writing the digits of another positive integer twice, for example, the numbers 156156 and 320320 are doubles but the number 36036 is not a double. Prove that infinitely many doubles are perfect squares.

I guess that the silence says it all! :D :D

cheers!


Consider the set $ A\equal{}\{ 1,2,3,4,...,2010,2011 \}$. We replace each two elements with the difference between the largest and the smallest until the set contains only one element. Find the parity of this element.

[hide="Solution"]The sum of the elements, mod 2, before doing anything is $ 1$.

After we remove $ 2011$ and $ 1$ and add $ 2010$, our original sum of $ 1$ gets two subtracted from it (2 odd numbers lost), but is still $ 1\bmod 2$.

After we remove $ 2010$ and $ 2$, and add in $ 2008$, we subtract 0 and add 0, so it's still $ 1$.

After we remove $ 2010$ and $ 3$, and add in $ 2007$, we subtract 1 and add 1, so it's still $ 1$.

Upon seeing this, we now claim that the sum of the elements of $ A$, mod 2, is invariant.
We prove this by casework:
Case 1. The largest and smallest elements are even. Then their difference is even, so we're adding and subtracting 0s mod 2, so nothing happens.
Case 2. The largest is odd and the smallest is even. Their difference is -1, and we're removing an odd element, so we subtract 2, which again makes no difference.
Case 3. The largest is even and the smallest is odd. By the same argument as Case 2, the sum remains the same.
Case 4. They're both odd. Their difference is then even, so we're adding 0 from their difference and subtracting -2, which again makes no difference.

Now we've shown that the sum of the elements is invariant mod 2 after changing the set around. Then, after every number has been replaced and the set has one element left, that element must also be 1 mod 2.
Then the parity of the remaining element is $ \boxed{\mathrm{odd}}$.[/hide]


AMC
AIME
geometry
3D geometry
pyramid
analytic geometry
graphing lines

Consider the paper triangle whose vertices are $(0,0)$, $(34,0)$, and $(16,24)$. The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?

w00t i was able to solve a 3D geometry problem  :lol:  (assuming my answer is correct)

[hide]
Let the points be A(0,0); B(34,0); C(16,24)
First we establish the midpoints: D(17,0); E(8,12); F(25,12)

When you picture the side's folding up in your head, you should see that the vertex (ignoring the the fact that it rises) moves perpendicular to the side connecting the midpoints; for example, the point (0,0) would move perpendicular to the line DE. We can see the slope of DE is $\frac{0-12}{17-8}=\frac{-4}{3}$. So the line perpendicular to that through point A is $y=\frac{3}{4}x$. If we draw the perpendicular from C, we see that it is completely vertical, so the x-coordinate is 16. Plugging that in, $y=\frac{3}{4}(16)=12$ So we know that the vertex is above the point (16,12)

Let the vertex of the pyramid above the plane have coordinates (16,12,z). By the distance formula, we calculate $AE=\sqrt{208}$. The vertext must then be that same distance from point E, so we use the 3-dimensional distance formula:
$\sqrt{(16-8)^{2}+(12-12)^{2}+z^{2}}=\sqrt{208}$
Solve for z to get 12. We now have the height and we can calculate the area of the base to be $\frac{1}{2}(17)(12)=102$. Then the volume is $\frac{1}{3}(102)(12)=\boxed{408}$
[/hide]

[hide]Or you can plot on the 3D coordinate plane and solve for z.

We have the points (17,0,0), (8,12,0), and (25,12,0).  And let the 4th point be (x,y,z).  Then:
$(x-17)^{2}+y^{2}+z^{2}=289$
$(x-8)^{2}+(y-12)^{2}+z^{2}=208$
$(x-25)^{2}+(y-12)^{2}+z^{2}=225$
And solve for z (first of all, the (y-12)^2 nicely cancel out in the bottom two equations, and you don't have to worry about anything more than a linear system as x^2+y^2+z^2 cancels out of all the equations).  When you get z, you see that z=12, the area of the base is 12*17/2=102.  And the volume is 102*12/3=408.[/hide]


geometry
floor function
inradius
incenter
number theory
prime numbers

Source: My head
1) What is the sum of the multiples of seven between 0 and 1000?

2) Find the sum of the coefficents of (a+b+c+1)^5.

3) A circle is inscribed in a 3,4,5 right triangle. what is the area of the circle.

4) What is the sum of the first 101 prime numbers?

5) What is 0!

 [hide="challenge"]5+X=10 Find X[/hide]
[hide="hint"] X=10-5[/hide]

[hide="1"]
There are $\lfloor \frac{1000}{7} \rfloor = 142$ multiples of $7$ less than $1000$.  Hence the sum is $7 \frac{142(143)}{2}$.
[/hide]

[hide="2"]
$a = b = c = 1$ gives the sum, which is $4^5 = 1024$.
[/hide]

[hide="3"]
Denote $3 = a + b, 4 = b + c, 5 = c + a$ where $a, b, c$ are the lengths that the inradius of the circle cuts the circle up into.  Consider the square formed by point $C$ and the incenter - it has side length $r = b$.  Hence $r = b = \frac{(a + b) + (b + c) - (a + c)}{2} = \frac{3 + 4 - 5}{2} = 1$, so the area is $\pi$.
[/hide]

[hide="5"]
$\frac{n!}{(n-1)!} = n$, then substitute $n = 1$ and we get $\frac{1!}{0!} = 1 \implies 0! = 1$.  (This does not technically constitute a proof - it's more of a definition.)
[/hide]

$4$ is no fun.   :(

[hide="1"]There are 142 multiple of 7s below 10000.

$7*(\frac{142*143}{2})=71071$[/hide]

[hide="3"]Area is 6.
6=sr=6r, where s is semiperimeter, r is inradius
r=1
Area of circle $=\pi$[/hide]


Find all real functions f such that for all x and y we have xy+f(x^2+y^2)=f(x)f(y)+xf(x)+yf(y).

A try for this one :

f(x)=g(x)+x

equation becomes :

g(x²+y²) = g(x)g(y) + (x+y)(g(x)+g(y)) (*)

y=0 : g(x²) = g(x)g(0) + x(g(x)+g(0))
y=0 : g(x²) = g(-x)g(0) - x(g(-x)+g(0))
x=y : g(2x²) = g(x)² + 4xg(x)
x=-y : g(2x²) = g(x)g(-x)

so if g(x)!=0, g(-x)=g(x)+4x which gives for x!=0 g(x)=g(0)-2*x which is not a solution of (*).

So g(x)=0 at least for x!=0. Looking at initial equation gives g(0)=0.

So only solution is f=Id.

What about f(x)=-x for all x? Isn't that a solution? And there are others!

Digging up the old...
If we let x=y=0, f(0)=f2(0) ==> either f(0)=0 or f(0)=1.
Keeping x fixed for the moment and letting y=0:
Case 1: If f(0)=0, then f(x2)=xf(x). Then f(x4)=x2f(x2)=x2*xf(x)=x3f(x). Note that f(x) is an odd function in this case, so since inductively we can get f(x2n)=x2n-1f(x), then f(x)= \pm x.

Case 2: If f(0)=1, then f(x2)=(1+x)f(x). Also f(1)=0. Then f(x4)=(1+x2)f(x2)=(1+x2)*(1+x)f(x). Again inductively, we get
f(x2n)=(1+x2n-1)(1+x2n-2)...(1+x2)(1+x)f(x).

This is similar to a well-known identity that the difference of two squares can be factored into 2 factors: a2-b2=(a+b)(a-b).
So we can supply the function f(x)=1-x satisfying the above properties.

f(x)= \pm x, 1-x. Note that I have assumed that f is continuous and that the monotonicity of f is kept intact. I believe these are the only ones. Are there any others?


[hide]add all together to get 3x+3y+3z+3w=198, so x+y+z+w=66, then subtract this and 4th equation to get x=81[/hide]

[hide]i got 85. I ll probably explain later even though i may be wrong[/hide]

kheg, did u try and solve for 4 equations.. that will take too long, its a mathcounts problem, but i got a different answer doing something differnet

[hide]Add the four equations together and then divide by 3[/hide]


Let $F(x)$ be such that $F\left(\frac{1-x}{1+x}\right) = x$ for all $x \not= -1$.  Which of the following statements is true?

(A) $F(-2-x) = -2 - F(x)$
(B) $F(-x) = F\left(\frac{1+x}{1-x}\right), x \not= 1$.
(C) $F\left(\frac{1}{x}\right) = F(x), x \not= 0$.
(D) $F(F(x)) = -x$.

(A).

F(x) = (1-x)/(1+x)

A is the only choice that fits


Fie $a,b,c$ strict pozitive. Aratati ca: \[\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\geq\frac{3}{4}\eqno{(I)}\]  \[\frac{a}{3a+b}+\frac{b}{3b+c}+\frac{c}{3c+a}\leq\frac{3}{4}\eqno{(II)}\]  \[\sqrt{\frac{a}{3a+b}}+\sqrt{\frac{b}{3b+c}}+\sqrt{\frac{c}{3c+a}}\leq\frac{3}{2}\eqno{(III)}.\]

$\sqrt{\frac a{a+3b}}+\sqrt{\frac b{b+3c}}+\sqrt{\frac c{c+3a}}\leq 3\sqrt{\frac{\frac a{a+3b}+\frac b{b+3c}+\frac c{c+3a}}3}\leq 3\sqrt{\frac{\frac{3}{4}}{3}}=\frac{3}{2}$[Am folosit inegalitatea dintre media aritmetica si media patratica]

Dupa numitor comun si eliminarea termenilor comuni,inegalitatea este echivalenta cu:
$3(a^{2}b+b^{2}c+c^{2}a)+5(ab^{2}+bc^{2}+ca^{2})\geq 24abc$,inegalitate ce reiese din medii(AM-GM) pentru toate numerele din membrul stang.Prin urmare problema s-a demonstrat.Insa acum ma intreb care este solutia mai frumoasa a acestei inegalitati..

Cred ca merge cu a teorema a III-a de maximizare.Mie mi-a dat ceva, dar inca nu e 100% completa solutia...

[quote="stancioiu sorin"]Cred ca merge cu a teorema a III-a de maximizare.[/quote]

M-ai spart, pustiule  :roll:

Oare de ce?Inca nu am obtinut o solutie completa.. :(

Putem scrie:
$\frac{a}{3a+b}+\frac{b}{3b+c}+\frac{c}{3c+a}\leq\frac{3}{4}\Leftrightarrow \frac{3a}{3a+b}+\frac{3b}{3b+c}+\frac{3c}{3c+a}\leq\frac{9}{4}\Leftrightarrow$
$\Leftrightarrow3+\frac{3a}{3a+b}+\frac{3b}{3b+c}+\frac{3c}{3c+a}\leq 3+\frac{9}{4}\Leftrightarrow$
$\Leftrightarrow 3-\frac{9}{4}\leq \frac{b}{3a+b}+\frac{c}{3b+c}+\frac{a}{3c+a}\Leftrightarrow$
$\Leftrightarrow \frac{b}{3a+b}+\frac{c}{3b+c}+\frac{a}{3c+a}\geq\frac{3}{4}$, evidenta, daca tinem cont de (I).

Folosim inegalitatea evidenta $(\sqrt{x}-\frac{1}{2})^{2}\geq 0$, $(\forall) x\geq 0$, scrisa echivalent $\sqrt{x}\leq x+\frac{1}{4}$ si ajungem imediat la inegalitatea (II).


Let $ABC$ be an equilateral triangle. Points $A_1$,$B_1$,$C_1$ are chosen inside the triangle in such a way that $B_1 \in AA_1$, $A_1 \in CC_1$, $C_1 \in BB_1$ and $AB_1=B_1A_1$, $B_1C_1=C_1B$,$CA_1=A_1C_1$. Prove that the triangle $A_1B_1C_1$ is also equilateral.

[hide]Please tell me if it works  :) [/hide]
Draw segments from $A$ to $C_{1}$, from $B$ to $A_{1}$ and from $C$ to $B_{1}$.

Now consider the seven triangles $AC_{1}B$, $AB_{1}C_{1}$, $AB_{1}C$, $BC_{1}A_{1}$, $BA_{1}C$, $CA_{1} B_{1}$ and $A_{1}B_{1}C_{1}$. They have all the same area.

Now consider the triangles with side $l$ (where $l$ is the side of $\Delta ABC$).

Since $\left[AC_{1}B_{1}\right] = \left[BA_{1}C_{1}\right] = \left[AB_{1}C_{1}\right]$, and they have a common side $l$, then their altitudes are equal. Let $h$ be this altitude.

The same for triangles $AB_{1}B$, $BC_{1}C$ and $CA_{1}A$. Hence their altitude is $h_{1} = 2h$.

Let $\angle CAB_{1} = \alpha$. Then $\frac{h}{\sin \alpha} = \frac{2h}{\sin (60 - \alpha)}$.

We get $\tan \alpha = \frac{\sqrt{3}}{2}$. Do the same for other angles. We get $\angle CAB_{1} = \angle ABC_{1} = \angle BCC_{1}$.

Also $\angle B_{1}AB = \angle C_{1}BC = \angle A_{1}CA$.

So they are similar. But for ASA $AB_{1}B \equiv BC_{1}C \equiv CA_{1}A$.

Hence $AB_{1} = BC_{1} = CA_{1}$, and $\Delta A_{1}B_{1}C_{1}$ is equilateral.


LaTeX
geometry
trapezoid
geometric transformation
homothety
rotation
ratio

Let $ ABC$ be an isocele triangle at A, let $ M \in [AB]$ and $ N \in [AC]$ and $ P \in BN \cap CM$ prove that:

$ \widehat {APM} \equal{} \widehat {APN}$  $ \iff$ $ (BC)  \parallel  (MN)$

EDITED.

[quote="FOURRIER"]Let $ ABC$ be an isocele triangle at A, let $ M \in [AB]$ and $ N \in [AC]$ prove that:

$ \widehat {APM} \equal{} \widehat {APN}$  $ \iff$ $ (BC)$ is parallel to $ (MN)$

what's the latex code for "parallel" please ?[/quote]

First of all, please make your statement correct.

For latex code of parallel just use \parallel :P $ \implies BC\parallel MN$
for more info: http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols

I forgot to mention:
$ P \in BN \cap CM$
ok.

well it is easy
first if BC is parallel to MN
then BCMN is a isoscales trapezium and hence cyclic
thou art done
if AMP=ANP
then p lies on aangle bisector A i.e. median
then use cevas
and finish it

[quote="gauravpatil"]
if [u][b]AMP=ANP[/b][/u]
then p lies on aangle bisector A i.e. median
then use cevas
and finish it[/quote]

it's not AMP=ANP it's $ \widehat{APM}\equal{}\widehat{APN}$

actually i meant the same so i think it does not make much of a difference as the solutiongets completed with cevas

Since $ ABC$ is isosceles, note that $ \angle APM \equal{} \angle APN$ implies $ AE \equal{} AF$, where $ AE\perp MP,AF\perp PN$. Then $ \triangle ACE\cong\triangle ABF$ and thus $ \angle EAM \equal{} \angle FAN$. Therefore $ \triangle AEM\cong\triangle AFN$. Thus $ AM \equal{} AN$ and by homothety we are done. The opposite direction is proved trivially.

Let us prove a more general result:

Let $ ABC$ be a triangle and let $ E$ and $ F$ be points on $ AC$ and $ AB$ respectively. Let $ BE$ and $ CF$ meet at $ P$. Then $ \overrightarrow{AP}$ bisects segment $ EF$ iff $ EF\parallel BC$.

Proof: First let $ EF\parallel BC$ and let $ \overrightarrow{AP}$ meet $ BC$ at $ D$. Then applying Ceva's theorem on the concurrent lines $ AD,BE,CF$ we get $ \frac {BD}{CD}\cdot\frac {CE}{AE}\cdot\frac {AF}{BF} \equal{} 1$. But $ \frac {AF}{BF} \equal{} \frac {AE}{CE}$, which implies $ BD \equal{} CD$. Now triangles $ AFE$ and $ ABC$ are similar and hence the composition of the rotation about $ 180^{\circ}$ at $ P$ and the homothety with ratio $ \frac {EF}{BC}$ sends $ BC$ to $ EF$. And this also sends $ D$ to the midpoint of $ EF$. Hence we conclude that $ AP$ bisects $ EF$.

Now suppose that $ D$ is the midpoint of $ EF$. Then from hypothesis, $ A,P,D$ are collinear. Thus again from Ceva's theorem we have $ \frac {ED}{FD}\cdot\frac {FC}{AF}\cdot\frac {AE}{EB} \equal{} 1$, which implies $ \frac {AF}{CF} \equal{} \frac {AE}{BE}$. Hence $ EF\parallel BC$. This completes the proof.


how many grids of triangles are there in a triangle of 15units on the sides.
calculate 7^99 without using calculator.
A rectangular plot of measures 12m by 5m.A path of constant with runs along one side and the path is 120m^2,find the width of the path.

Can't understand the first two.

For the third, we have:
$ 4 \left( \frac{1}{4} \pi w^2 \right) \plus{} 2(5w) \plus{} 2 (12w) \equal{} 120$
$ \pi w^2 \plus{} 34 w \equal{} 120$
$ w \equal{} \frac{\minus{}17\plus{} \sqrt{289\plus{}120 \pi}}{ \pi} \approx 2.80329 \text{m}$

I can't understand first two either.

For the second one:

If you meant the unit digit, find the pattern.

If you meant anything more than the unit digit, use Binomial Expansion although I'm not sure if that'll work out nicely (the best I can think of is $ (10 \minus{} 3)^{99}$ but that's not nice either.. )

he could mean the last two digits :P.  It'd be a mathcounts questions after that.

Yeah, I don't really see how to do 7^99...
without a calculator
and even then...
for the 1st
do we draw lines every inch or something?


geometry
geometric transformation
reflection
circumcircle
modular arithmetic
geometry unsolved

Let $ A'$ be the reflection of $ A$ in line $ BC$, and define $ B'$ and $ C'$ cyclically. Let $ A_b$ be the reflection of $ A'$ in $ AB$, and define $ A_c, B_c, B_a, C_a, C_b$ cyclically. Let

       $ A_1 \equal{} BA_b\cap CA_c$, and define $ B_1$ and $ C_1$ cyclically,
       $ A_2 \equal{} BA_c\cap CA_b$, and define $ B_2$ and $ C_2$ cyclically,
       $ A_3 \equal{} BB_a\cap CC_a$, and define $ B_3$ and $ C_3$ cyclically,
       $ A_4 \equal{} BB_c\cap CC_b$, and define $ B_4$ and $ C_4$ cyclically,
       $ A_5 \equal{} BC_a\cap CB_a$, and define $ B_5$ and $ C_5$ cyclically,
       $ A_6 \equal{} BC_b\cap CB_c$, and define $ B_6$ and $ C_6$ cyclically.

Then $ AA_n,BB_n,CC_n$ are concurent, for $ n \equal{} 1,2,3,4,5,6$.

For $ n \equal{} 1$
By angle chase, we can proof that $ (BCA_1), (CAB_1), (ABC_1)$ have a common point $ S_1$.
And by angle chase again, we can proof that $ S_1 \in AA_1, S_1 \in BB_1, S_1 \in CC_1$
For $ n \equal{} 3$
$ BB_a \perp AC', CC_a \perp AB'$. By Carnot's theorem, we have $ AA_3 \perp B'C'$. Similarly: $ BB_3 \perp C'A', CC_3 \perp A'B'$.
[hide="So $ AA_3, BB_3, CC_3$ are concurrent at the Kosnita point"]
Let $ O,S$ be the circumcenter of $ (ABC), (BOC)$
$ BC' \cap CB' \equal{} G$, we can proof $ G \in (BOC), G \in (ABB'), G \in (ACC')$ easily by angle chase
$ B'A \cap (ACC') \equal{} \{ A, M \} , C'A \cap (ABB') \equal{} \{ A, N \}$ , by angle chase we can proof $ M,N,B',C'$ are concyclic
Then $ \mathcal{P}_{B'/(S)} \equal{} \overline{B'C}.\overline{B'G} \equal{} \overline{B'A}.\overline{B'M}$
So $ \mathcal{P}_{B'/(S)} \minus{} B'A^2 \equal{} \overline{B'A}. \overline{AM} \equal{} \minus{} \mathcal{P}_{A/(MNB'C')}$
Similarly, we have $ \mathcal{P}_{C'/(S)} \minus{} C'A^2 \equal{} \minus{} \mathcal{P}_{A/(MNB'C')}$
So $ \mathcal{P}_{B'/(S)} \minus{} B'A^2 \equal{} \mathcal{P}_{C'/(S)} \minus{} C'A^2$, or $ B'S^2 \minus{} B'A^2 \equal{} C'S^2 \minus{} C'A^2$
Then $ B'C' \perp AS$, so $ AA_3$ passes through the circumcenter of $ \triangle{BOC}$...
[/hide]
Or we can proof $ AA_3, BB_3, CC_3$ are concurrent by Carnot's theorem
For $ n \equal{} 4$
Let $ H$ be the orthocenter of $ \triangle{ABC}$. It's easy to see that $ A_4$ is the reflection of $ H$ through $ BC$. Then $ AA_4, BB_4, CC_4$ are concurrent at $ H$
For $ n \equal{} 5$
We can proof $ \triangle{AB_aC} \equal{} \triangle{ABC_a}$ easily, then we can have $ A_5 \in (ABB_a), A_5 \in (ACC_a)$
Let $ f$ be the inversion has $ A$ to pole and $ AB.AC$ to power. Denote $ f(X) \equal{} X^*$ for every point $ X$
Let $ At$ be the bisector of $ \angle{BAC}$
We have $ A_5 \in (ABB_a) \cap (ACC_a)$, then $ A_5^* \equal{} B^* B_a^* \cap C^* C_a^*$
We have $ B^*, C^*, B_a ^*, C_a ^*$ are the symmetric point of $ C,B, C_a, B_a$ wrt $ At$
Then $ A_5^* \equal{} B^* B_a^* \cap C^* C_a^*$ is the symmetric point of $ A_3 \equal{} CC_a \cap BB_a$ wrt $ At$
Note that $ A, A_5, A_5^*$ are collinear, we have $ AA_5$ is the isogonal lines of $ AA_3$ wrt $ \angle BAC$
Then $ AA_5,BB_5, CC_5$ are concurrent at the isogonal conjugate of Kosnita point, which is the nine-point center
For $ n \equal{} 6$
By angle chase, we can proof that $ (BCA_6), (CAB_6), (ABC_6)$ have a common point $ S_6$
And by angle chase again, we can proof that $ S_6 \in AA_6, S_6 \in BB_6, S_6 \in CC_6$

Sorry, in the above post I forgot the case $ n \equal{} 2$, and I can't edit that post now  :blush:
Let $ O$ be the circumcenter, and $ Q$ be the inversive image of the Kosnita point in the circumcircle of $ \triangle{ABC}$
$ Q_a, Q_b, Q_c$ are the symmetric points $ Q$ wrt $ BC,CA,AB$
By angle chase, we can prove that $ (BA_c, CA_b) \equiv (A'C', A'B') \plus{} 2(AC,AB) \ \pmod{\pi}$ and $ (QB, QC) \equiv (A'B', A'C') \plus{} 2(AB,AC) \ \pmod{\pi}$
So $ (Q_aB, Q_aC) \equiv \minus{} (QB, QC) \equiv (BA_c, CA_b) \equiv (A_2B, A_2C) \ \pmod{\pi}$, which means $ Q_a,A_2,B,C$ are concyclic.
Note that $ Q \in (BA'C')$, so $ Q_a \in (BAC_b)$
We have:
$ (Q_aA_2, Q_aB) \equiv (CA_2,CB) \equiv(CA_b, CB) \equiv (C'B, C'A') \ \pmod{\pi}$
And:
$ (Q_aB,Q_aA) \equiv (C_bB, C_bA) \equiv (C'A', C'B) \ \pmod{\pi}$
So, $ A, A_2, Q_a$ are collinear. Similarly, we have $ B,B_2, Q_b$ are collinear, and $ C,C_2,Q_c$ are collinear.
We can prove that $ AQ_a, BQ_b, CQ_c$ are concurrent at a point lies on $ (ABC)$, it's easy by angle chase.
So, $ AA_2, BB_2, CC_2$ are concurrent, as desired


inequalities
search
inequalities unsolved

See http://www.mathlinks.ro/Forum/viewtopic.php?search_id=902617529&t=1566, which generalizes your problem.

Hi, Arne, I have the book "Bulgarian Math Olympiads" 1951-1990 I took a look at the problem's solution. I also looked at the materials of the link but I still cannot understand your idea :-( sometimes I'm too stupid. Please throw some light.

[i]After edition of the topic:[/i] I understood your idea. I think if we should prove the inequality in this way - our inequality is some kind of "ugly" and "difficult" inequality. Even the original solution in the book looks complicated and with some kinds of errors. Thank you for the very good idea.

[quote="borislav_mirchev"]Given positive numbers p, q with p+q=1. Prove that for all positive integers m, n the following inequality holds: (1-p^m)^n+(1-q^n)^m>=1.[/quote]
Given positive numbers $p, q $ with $p+q=1$. Prove that for all positive integers$ m, n $the following inequality holds:$ (1-p^m)^n+(1-q^n)^m\ge 1.$


What is the difference between the two largest prime factors
of 159,137?

First you find the prime factorization of the number, and from there it is straight forward.

[quote="GameBot"]What is the difference between the two [size=200]largest[/size] prime factors
of 159,137?[/quote]

The answer is 14

[quote="isabella2296"]ernie? read? I've never seen those two words in the same sentence!!!

It's REALLY important to read, ernie! With this powerful tool of education, you can one day reach 1500! This non-reading thing of yours is very detrimental to your FTW Health!  :D

$ 37 \minus{} 23 \equal{} \boxed{14}$.

Oh, whoops, didn't see your post, Dragon, sorry about that![/quote]

Unless the sentence is:

[b]ernie doesn't read.[/b]

That is an exception.  :P

Anyway, we're getting off-topic, so let's let this thread die.


calculus
derivative
logarithms
calculus computations

For the equation $ V(t) \equal{} 27(0.87)^t$ I have determined $ V(4)$ and $ {V}'(t)$ however I do not know how to evaluate $ {V}'(4)$? Should it just be plugging in $ 4$ for $ t$ in the derivative?
[hide]
$ V(4) \equal{} 15.468$
$ {V}'(t) \equal{} \minus{} 3.760(.879)^t$
[/hide]

$ \boxed{\frac{d}{dt} a^t\ (a\neq 1, a>0)\equal{}a^t\ln a}$

$ V'(t)\equal{}27\cdot .87^t\cdot\ln .87\implies \boxed{V'(4)\equal{}\minus{}2.154}$.


y = x * |x|

Why  is the inflection point 0,0 but f"(x) does not exist?
Mathematical proof would be apprecaited.

Hints:
1) Plot the function.
2) Compute $f'(x)$. What happens at point $x=0$?

May you do the question for me please. This is supposed to be a challenge question for my high school level. But I really want to do this.

V4Vendetta, first answer these two questions:
  i. What is the value of $f'(x)$ when $x>0$ ?
  ii. What is the value of $f'(x)$ when $x<0$ ?

[hide="Hint"]An inflection point occurs at a point on the graph of $f(x)$ when the sign of $f'(x)$ changes at that point.  So, $f''(x) = 0$ or $f''(x)$ does not exist at that point.  This question basically asks if you know what the definition of an inflection point is.[/hide]


geometry
circumcircle
symmetry
geometric transformation
homothety
Euler
ratio

The triangle $ A_0 B_0 C_0$ is the image of triangle $ ABC$ via symmetry with origin $ O$. Hence, their orthocenters are image of each other via this transformation. This means: $ O$ is the midpoint of $ HH_0$.

[quote="sinankaral53"]Given an acute angled triangle $ ABC$ , $ O$ is the circumcenter and $ H$ is the orthocenter.Let $ A_1$,$ B_1$,$ C_1$ be the midpoints of the sides $ BC$,$ AC$ and $ AB$ respectively. Rays $ [HA_1$,$ [HB_1$,$ [HC_1$ cut the circumcircle of $ ABC$ at $ A_0$,$ B_0$ and $ C_0$ respectively.Prove that $ O$,$ H$ and $ H_0$ are collinear if $ H_0$ is the orthocenter of $ A_0B_0C_0$[/quote]
Turkish solution

Denote $ G$ the centroid of triangle $ ABC$. Consider the homothety with center $ G$ which turns $ A_1\mapsto A$, $ B_1\mapsto B$, $ C_1\mapsto C$. Therefore, it maps triangle $ A_1B_1C_1$ to triangle $ ABC$. But, according to a very well- known result, we get that $ H, O, G$ are collinear (The Euler line), further, $ \overline {GO} \equal{} \minus{}\frac {1}{2}\overline {GH}$. Hence it also turns $ O\mapsto H$, which means $ O$ is also the orthocenter of triangle $ A_1B_1C_1$. Now, consider the homothety with center $ H$, ratio $ 2$, which obviously turns $ A_1\mapsto A_0$, $ B_1\mapsto B_0$, $ C_1\mapsto C_0$. Therefore, triangle $ A_1B_1C_1$ is mapped into triangle $ A_0B_0C_0$, also. Hence it turns $ O\mapsto H_0$, which leads to the result that $ H,O,H_0$ are collinear.
Our proof is completed :D .


[quote="GameBot"]The Gothic arch shown is constructed on the sides of an equilateral triangle. A side of the triangle is also a radius of each arc. The side length of the triangle is 6 feet. How many square feet are in the area outside the triangle but inside the arch? Express your answer as a decimal to the nearest tenth.
[asy]draw((0,0)--(1,0)--dir(60)--(0,0)); draw(Arc((0,0),1,0,60));
draw(Arc(dir(0),1,120,180));[/asy][/quote]

The circle has raiuds 6, so an area of $ 36\pi$.

Thus, each of the small arcs have area $ 6\pi\minus{}9sqrt{3}$.

$ 12\pi\minus{}18\sqrt{3}\plus{}9\sqrt{3}\equal{}12\pi\minus{}9\sqrt{3}$.

So 22.1

They actually want the area outside of the triangle, so the answer should be $ 12\pi\minus{}18\sqrt{3}\approx\boxed{6.5}$


[quote="redcomet46"]The digits 1,2,3,4,5,6,7 and 9 are used to form four two-digit prime numbers, with each digit used exactly once. What is the sum of these four primes?

$ \text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190$[/quote]
[hide]29,41,67,53 by brute forcing. $ 29 \plus{} 41 \plus{} 67 \plus{} 53 \equal{} 190$ or E [/hide]

[quote="redcomet46"]The digits 1,2,3,4,5,6,7 and 9 are used to form four two-digit prime numbers, with each digit used exactly once. What is the sum of these four primes?

$ \text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190$[/quote]

[hide="fast without brute force or finding the primes"]
The units digits have to be 1, 3, 7, and 9, because 5 would make the number nonprime.
Add up the tens digits and multiply by 10 $ (2\plus{}4\plus{}5\plus{}6)/cdot 10\equal{}170$
Add up the units digits $ 1\plus{}3\plus{}7\plus{}9\equal{}20$
so it's E $ 170\plus{}20\equal{}190$

[/hide]


modular arithmetic
induction
algebra
difference of squares
special factorizations

Prove that $ \forall n\in\mathbb{N}$,there exist a natural number $ k$,such that
$ 19^k\minus{}97$ is divisible by $ 2^n$.

i dont get the $ k$ part
do you mean ? prove that $ 19^{n}\minus{}97\equiv0\mod2^n$ or in other words prove that $ 2^n|19^{n}\minus{}97$

No, it means exactly what it says: prove that for each $ n$, there exists a $ k$ such that $ 2^n$ divides $ 19^k \minus{} 97$.

This problem is not so easy,though i know its solution,and i think that someone must move this topic to the olympiad section.

[hide]
Induct on $ n$. If $ n\leq 3$, we can just choose $ k\equal{}2$. For the inductive step, consider $ k$ such that $ 19^k\equiv 97\pmod{2^n}$, where $ n\geq 3$. If $ 19^k\equiv 97\pmod{2^{n\plus{}1}}$, we're done.

Otherwise, we must have $ 19^k\equiv 2^n\plus{}97\pmod{2^{n\plus{}1}}$. In this case,
\[ 19^{k\plus{}2^{n\minus{}2}}\equiv (2^n\plus{}97)19^{2^{n\minus{}2}}\pmod{2^{n\plus{}1}}\]
Now show $ 19^{2^{n\minus{}2}}\equiv 2^n\plus{}1\pmod{2^{n\plus{}1}}$ by repeatedly factoring difference of squares:
\[ 19^{2^{n\minus{}2}}\minus{}1\equal{}18\prod_{k\equal{}0}^{n\minus{}3} (1\plus{}19^{2^k})\]
which contains exactly $ n$ factors of 2. So this gives
\[ (2^n\plus{}97)19^{2^{n\minus{}2}}\equiv (2^n\plus{}97)(2^n\plus{}1)\equiv 97\pmod{2^{n\plus{}1}}\]
so we can choose $ k\plus{}2^{n\minus{}2}$ as the exponent.
[/hide]

Not so nice as scorpius119's solution,but this proof shows how to generalize the problem,and it is very useful in solving problems of this type
Problem:Let $ a\in N$ be a number of the form $ 8m\pm 3$.Prove that for all $ n\in\mathbb{N}$,there exist number $ k\in\mathbb{N}$,such that
$ a^{k} \minus{} 97$ is divisible by $ 2^n$.
         Solution:
Let $ \mathbb{Z}_{k}^{*} \equal{} \{i,$ such that $ (i,k) \equal{} 1$ and $ 1\leq i\leq k\}$.
All the numbers will be considered in $ \mathbb{Z}_{2^n}$.
It is obvious that $ a\in\mathbb{Z}_{2^n}^{*}$.
Let $ G \equal{} \{1,a,a^{2},a^{3},...,a^{k}\}$,such that $ k \plus{} 1$ is the smallest natural number,such that $ a^{k \plus{} 1}\equiv 1$.
        [b]Lemma 1:[/b]
$ |G| \equal{} 2^{n \minus{} 2}$.
Proof:
It is easy to prove by induction that
$ 2^{n}\parallel{}b \equal{} a^{2^{n \minus{} 2}} \minus{} 1$.     $ (1)$
Suppose that there exist $ m \equal{} 2^{t}r$,such that
$ a^m\equiv 1$ and $ m < 2^{n \minus{} 2}$.
Since $ (2^{n \minus{} 2},m) \equal{} 2^{t}$,then
$ a^{2^{t}} \minus{} 1$ is divisble by $ 2^{n}$,but from $ (1)$,it is not divisble by $ 2^{t \plus{} 1}$,hence $ t\geq n$.Contradiction.
So the lemma 1 is proved.
        [b]Lemma 2:[/b]
There exist exaclty $ 4$ numbers $ x$ in $ \mathbb{Z}_{2^{n}}^{*}$,such that
$ x^2\equiv 1$.
Proof:
It is easy to understand from the equation
$ 2^n|(x \minus{} 1)(x \plus{} 1)$ and $ (x \minus{} 1,x \plus{} 1)\leq 2$.
         They are $ x_1 \equal{} 1,x_2 \equal{} 2^n \minus{} 1,x_3 \equal{} 2^{n \minus{} 1} \plus{} 1,x_4 \equal{} 2^{n \minus{} 1} \minus{} 1$.
Since $ x_2,x_4\equiv 3(mod 4)$,then the following equation
$ y^2\equiv x_2,x_4$ has no solutions.
[b]Lemma 3:[/b]
Every natural number $ k\in\mathbb{Z}_{2^{n}}^{*}$ is either in $ G$,or can be represented as $ b\cdot c$,where $ b\in\mathbb{Z}_{2^{n}}^{*}/G$,$ c\in G$.
Or with another words for every $ b\in\mathbb{Z}_{2^{n}}^{*}/G$,the set
$ b\cdot G \equal{} \mathbb{Z}_{2^{n}}^{*}/G$.
Proof:
1.$ |b\cdot G| \equal{} |G| \equal{} 2^{n \minus{} 2}$.
Indeed,let there be $ a_1,a_2\in G$,such that $ a_1\neq a_2$ and $ ba_1 \equal{} ba_2$,but then
either $ 2^n|b$,or $ a_1 \equal{} a_2$.Both of the previous relations are impossible,hence
we've just proved the first part of the 3rd lemma.
2.$ b\cdot G\cap G \equal{} 0$
Let there be some numbers $ a_1,a_2\in G$,such that $ ba_1 \equal{} a_2$.
But for every $ a\in G$,there exist $ a'$,such that $ a\cdot a' \equal{} 1$.(All of the elemtents in $ G$,can reprensented as $ a^{i}$,then we take
$ a' \equal{} a^{p \minus{} 1 \minus{} i}$.
So let $ a_1'\in G$,be the natural number,such that $ a_1\cdot a_1' \equal{} 1$.
Hence $ b \equal{} a_1'a_2\in G$,contradiction.
[b]Lemma 4:[/b]
$ x_1,x_3\in G$ and $ x_2,x_4\in\mathbb{Z}_{2^{n}}^{*}/G$.
Proof:
Well-known and easy to prove.
[b]Proof of the problem:[/b]
Suppose for the contrary,that
$ 97\in\mathbb{Z}_{2^{n}}^{*}/G$.
Then from lemma 3 and lemma 4,there exist a natural number
$ c$,such that $ c\in G$.
$ 97 \equal{} x_4c$.But $ x_4 \equal{} \minus{} 1$,hence
$ 97 \equal{} \minus{} a^{s}$,where $ s\in N$.
Then it is easy to understand that $ s$ is an even number,so we must consider only one case,when $ s$ is odd.
It is easy to prove by induction that
$ 2^{n}\parallel{}97^{2^{n \minus{} 5}} \minus{} 1$,so since $ s$ is odd,then
$ 2^{n}|(a^s)^{2^{n \minus{} 5}} \minus{} 1$,or
$ 2^n|a^{2^{n \minus{} 5}} \minus{} 1$,contradiction to the first lemma.
Q.E.D
Remark:
During reading my solution,some of you will understand that $ 97$ could be replaced by some other number,i'm waiting from you some good generalizations of this generalization,(i mean replacing $ 97$ by some other number).Good luck.


function
search
combinatorics proposed
combinatorics

Find the number of subsets of the set $ \{1,2,3,...,5n\}$ such that the sum of the elements in each subset are divisible by 5.

This is a standard problem for a generating function approach with "roots of unity filter."  See for example:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1596684844&t=211616
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1596684844&t=258033
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1596684844&t=243014
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1596684844&t=224993
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1596684844&t=204359


Suppose that $n\geq 2$ be a natural number. Prove that there exist $A\subset \{1,2,...,n\}$ s.t.

a) $\mid A \mid \leq 2[\sqrt{n}]+1$(where [$x$] denote the greatest integer less than or equal to $x$)
b) $\{\mid x-y \mid \mid x,y\in A ,x$ not equal$y\}=\{1,2,...,n-1\}$

It was also used in the QEDMO, see http://www.mathlinks.ro/Forum/viewtopic.php?t=60235 , but nobody gave a correct answer in that other topic :(
(but the problem is easy, just try)

This one is not hard. Here is my approach.

Now, denote $k=[\sqrt{n}]$ and let $S=\{n;n-k;n-2k;...;n-k^2;k;k-1;...;1\}$.

We have $|S|\le 2k+1=2[\sqrt{n}]+1$ and $n-k^2-k\le k$.

besides: for any $i\in [1;n-1]$, We have $n-i=ak+b$,thus, $i=(n-ak)-b$.  Therefore:

$\{|x-y||x;y\in S, x \neq y\}=\{1;2;...;n\}$ and we're done


quadratics
search
algebra
polynomial
Rational Root Theorem

Can somebody tell me how many ways of solving quadratic equations there are, and name (and briefly explain) all of them? Thanks :D

I use the quadratic formula or factor the equation out.

quatratic formula: $x=\frac{-b(+ or -)\sqrt{b^2-4ac}}{2a}$

Yeah those are the only ones I know, just asking if there are any [i]other[/i] ways to solve one. :D

the way i solve quadratic equations:
I. use Viet formulas :$x_1+x_2=-\frac ba$ and $c=\frac ca$
Example $x^2-20x+51=0$ so $x_1*x_2=51$ and $x_1+x_2=20$ so it i easy to see that $x_1=3$ and $x_2=17$
II. next check with x=1 or x=-1
Example: $2006x^2-1999x-7=0$ we see $x=1$ and again by Viet $x=\frac{-7}{2006}$
1. This way its more fun  :D
2. Sometimes its much easier (try to solve the second equatin with the formula :oops:
III. If u are stuck [color=red]Use the formula[/color] :ninja:

[quote="fireemblem13"]completing the square. i don't really use it as i prefer the quadratic formula but it's there. search it if you want to know.[/quote]

They're equivalent.

It depends on if you want to solve them generally or in a specific case.  Generally, the only sensible method is the quadratic formula.

Now, if the coefficients are integers and you know the solutions are rational, then there is a better way:  the Rational Root Theorem.

Basically, what the Rational Root Theorem says is that if a polynomial (in our case, a qudaratic) with integer coefficients has a rational root, then the rational root must be of the form

$\frac{p}{q}$

Where $gcd(p, q) = 1$, $p |c, q | a$.

If this seems a little confusing, let's use a very simple example.  Consider the quadratic equation

$2x^2 - 5x + 2 = 0$

Then the Rational Root Theorem then tells us the only possible roots are $\pm 1, \pm 2, \pm \frac{1}{2}$.  By inspection, we find that $2$ is a root (this is actually another important technique: you can plug in numbers that you suspect are roots, and if you get $0$ then you can factor it out).  Therefore,

$2x^2 - 5x + 2 = (x + 2)(2x - 1)$

So the other root is $\frac{1}{2}$.

You can also use the "splitting the middle term" technique. Basically you multiply A and C. You find two numbers that multiply up to A*C and add up to B. You write these two new linears in place of the original B.  Then you do "factor by grouping", undo the distributive property, and use zero-product property.

For example--

6x^2-11x-7=0

[u]Prod[/u]  -42           [u]Sum[/u] -11
the two numbers are 3 and -14.

rewrite quadratic as this--
6x^2+3x-14x-7=0

factor by grouping...
3x(2x+1) -7(2x+1)=0
(3x-7)(2x+1)=0
x=7/3 or -1/2

that's the way my dad first taught me how to solve a quadratic equation... i find it a bit more difficult though than factoring or the quadratic formula.


It's been awhile since I've had decimals, but I have a pretty good question.
This relates to part of my game coding I am working on. All I really need is a number to input into my script.  In a way, you would be helping me create the game. :D

Anyway I'm trying to make the projectiles faster and the script uses decimals for timing.

The current time is: 0.075075

 The strange thing about that is if I raise it to: 0.099999 it goes alot slower and if I change it 0.022222 it goes slower. Why is that?

What is a decimal that would be faster than 0.075075?

Thanks.

$0.099999>0.075075>0.022222$

well you would obviously know that but you should specify.
So what does the number do?
Is it suppose to increase the speed?

Your question is $0.099999$ is slow and $0.022222$ is slower.
But your saying the speed works differently and $0.075075$ is the fastest.

So not by increasing numbers, but speed wise

$0.075075>0.099999>0.022222$

Maybe because it a trig equation?
for example.

$sin(100)>sin(150)$

Something like this maybe?

If you specify your program, I could help by analyzing it and giving a more complex equation, but considering the forum this is as far as I can explain here.

// ############################################################################
// Settings

::PB::DelayFire	= "0.075075";	// Delay between shots

Yes, it is supposed to increase the firing speed and the delay between shots. All I am asking for is a number that is would be bigger than 0.075075 which would in turn increase the firing rate. Also, the .0999999 and 0.0222222 both have the same rate of fire.

So to make it faster than 0.075075, choose a decimal value greater than 0.075075 and your result should be greater. What is a number greater than this?


At the beginning of a study 24 fish are caught, tagged and released back into a large pond containing fish. A few days later, 19 fish are caught from the pond and three of them have tags. Using this ratio, how many total fish would you expect to be living in the pond?

so every 19 fish, we expect 3 to be taggeed. therefore for 24 tagged fish, we need 19*(24/3)=152 fish


An 8.5-meter ladder is leaning against a vertical wall. How many meters is its base from the wall if the ladder reaches 7.5 meters up the wall?

The wall and the ground form a right angle so all we have to do is set up an equation using the Pythagorean theorem. In this case, the length of the ladder is the hypotenuse so:

$ 8.5^2 \equal{} 7.5^2 \plus{} x^2$

$ x \equal{} 8.5^2 \minus{} 7.5^2 \equal{} (8.5 \plus{} 7.5)(8.5 \minus{} 7.5) \equal{} 16$

$ \boxed{x \equal{} 4}$


Consider the statements:

$\text{(I)}~~\sqrt{a^2+b^2}=0$
$\text{(II)}~~\sqrt{a^2+b^2}=ab$
$\text{(III)}~~\sqrt{a^2+b^2}=a+b$
$\text{(IV)}~~\sqrt{a^2+b^2}=a-b$,

where we allow $a$ and $b$ to be real or complex numbers. Those statements for which there exist solutions other than $a=0$ and $b=0$ are:

$\text{(A)} \ \text{(I)},\text{(II)},\text{(III)},\text{(IV)} \qquad \text{(B)} \ \text{(II)},\text{(III)},\text{(IV)} \qquad \text{(C)} \ \text{(I)},\text{(III)},\text{(IV)} \qquad \text{(D)} \ \text{(III)},\text{(IV)} \qquad \text{(E)} \ \text{(I)}$

[hide]$I$ is true when $a = n$ and $b = ni$ so they add to zero.

For $II$, square each side and obtain $a^2 + b^2 = a^2b^2$

Let $x = a^2$, and $y = b^2$

$x + y = xy$

This works for 2 because $2*2 = 2+2$

so $x = y = \sqrt{2}$

No other choices have I and II so its $\boxed{A}$[/hide]

[quote="4everwise"]Consider the statements:

$\text{(I)}~~\sqrt{a^2+b^2}=0$
$\text{(II)}~~\sqrt{a^2+b^2}=ab$
$\text{(III)}~~\sqrt{a^2+b^2}=a+b$
$\text{(IV)}~~\sqrt{a^2+b^2}=a-b$,

where we allow $a$ and $b$ to be real or complex numbers. Those statements for which there exist solutions ofter than $a=0$ and $b=0$ are:

$\text{(A)} \ \text{(I)},\text{(II)},\text{(III)},\text{(IV)} \qquad \text{(B)} \ \text{(II)},\text{(III)},\text{(IV)} \qquad \text{(C)} \ \text{(I)},\text{(III)}\text{(IV)} \qquad \text{(D)} \ \text{(III)},\text{(IV)} \qquad \text{(E)} \ \text{(I)}$[/quote]

[hide]
I think it's A.

I obviously works you may just pick $a = x$ and $b = xi$;
II works because you may factor it into $(1 - a^2)(1 + b^2)=0$ which has an infinite number of solutions in complex numbers
III works because it can be simplified as $-2ab = 0$ which also has an infinite number of solutions if either a or b is 0
IV works for the same reason as 3.[/hide]


advanced fields
advanced fields unsolved

I am finding the general solution of (1 - 4x^2)y"+ 6xy' - 4y = 0
I got stuck later in the general solution...

I got an expression involving series... c_0 (1 + 2x^2) +

(x - (2/6)x^3 - (20/120)x^5 - (1080/5040)x^7 - (140400/362880)x^9 - ... )c_1

I separate x from my summation expression.. but

Right now, I can express as a summation from n = 1 to infinity of

(??) x^(2n + 1) / (2n + 1)!

That is.. what is my pattern in this: 2, 20, 1080, 140400,... ?

 I cannot see any pattern except for a factor of 2^n.

Hint: Your expression makes it clear that $1+2x^2$ is a solution. Find the other one by reduction of order.


factorial
logarithms
number theory unsolved
number theory

How to find the first digit of N! when N is big enough? Is there any recurrence equation? The Stirling Formula is not good for computing with hand :roll: .

\[\ln{N!}=\frac{1}{2}\ln{(2\pi N)}+N(\ln N-1)+\sum_{k=1}^{2n}\frac{B_{2k}}{2k(2k-1)N^{2k-1}}+\delta\] , were $n=[\pi N]$ and $|\delta|<\frac{2\sqrt N }{(2\pi N-1)e^{2\pi N}}$. This formul give first $[N\frac{2\pi}{\ln{10}}]$ digits and all digits correct if $N\le 1455$.


the lengths in orfer of four consecutive sides of an equiangular hexagon are 1, 7, 2, 4 respectively what is the sum of the lengths of the two remaining sides?

ahh can someone help me?is there like some special properties to a hexxagon

any help would be appreciated

thanks

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=733790984&t=204658]Searching helps quite a bit. :)[/url]

If you extend teh sides of the hexagons to meet, you form an equilateral triangle.

Try to work from there.

i remember this... try drawing the hexagon. Then draw a bunch of assistant lines to help you. I think you have to use 30-60-90 properties a lot of times.
BTW, the answer is 9. Good Luck!

uhh thanks guys...yes i got the thing with the equilateral triangle...will remember

and uhh to those of u going to states, Good Luck and to those of u going to nats, congrats.


Let $u$ and $v$ be real numbers such that
\[(u+u^{2}+u^{3}+\cdots+u^{8})+10u^{9}= (v+v^{2}+v^{3}+\cdots+v^{10})+10v^{11}= 8. \]
Determine, with proof, which of the two numbers, $u$ or $v$, is larger.


Έστω το πολυώνυμο $ P(x)$ βαθμού $ n$ με $ n\geq 2$ και με συντελεστές ακέραιους αριθμούς. Να αποδείξετε ότι το σύνολο $ A\equal{}\left\{ x\in Z: P\left(P(x)\right)\equal{}x\right\}$ έχει το πολύ $ n$ στοιχεία.

Ειναι ειδικη περιπτωση προβληματος IMO (2006 νομιζω) του οποιου η λυση εχει μπει εδω:

http://www.mathlinks.ro/viewtopic.php?t=137646


An academic dept with five faculty members Anderson , Box, Cox, Cramer, and Fisher-- must select 2 of its members to serve on a personnel review committee. Because the work  will be time consuming, no one is anxious to serve, so it decided that the representative will be selected by putting five slips of paper in a box, mixing them and selecting two.

a) What is the probability that both Anderson and Box will be selected?

b) What is the probability that at least one of the two memmbers whose name begins with a C is selected?

c) If the five faculty members have taught for 3, 6,7,10, and 14 years, respectively, at the university, what is the probability that the two chosen representatives have at least 15 years of teaching experience at the university

there are 5 ways to choose the first person, and 4 ways to choose the second person BUT there is always 2 orders to pick a pair of people (for example fisher box and box fisher)
so there are 10 possibilities for the chosen pair
that being said
1) there is 1 pair that satisfies the problem (anderson/box) and 10 total pairs so the probability is $ \frac{1}{10}$
2) there are 7 pairs (there are many ways to get this-- one way is shown here) that satisfy the problem
this is because there are 3 pairs that do not satisfy the problem (a-b b-f a-f)
and 10 total pairs so the probability is $ \frac{7}{10}$
3) there are 6 pairs that satisfy the problem (3+14, 6+10, 6+14, 7+10, 7+14, 10+14)
and there are total 10 pairs so the probability is $ \frac{6}{10}\equal{}\frac35$


[quote="nayel"]Let $a,b,c>0$. Prove that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{\sqrt[3]{abc}}{a+b+c}\ge 2 \]
[/quote]

I think there must be something wrong here?

Even without the $\frac{\sqrt[3]{abc}}{a+b+c}$ term, we have \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{abc}{bca}}=3,\] so the inequality is true, but far from optimal.

I have the following inequality with a very nice proof

\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c}\ge 4.\]

Alternatively, we have the following inequality

\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\ge 2,\]

with the equality for $a=b=c$ or $a=b,c=0$. I wait for a nice solution for it.

(deleted) yes it was wrong. :blush: i'll think about it for a while.

[quote="nayel"][quote="nayel"]well i'm very sorry, i edited it. before, equality couldn't occur. but now can. :lol:[/quote]

someone rated this post as spam. why? please show your reason.[/quote]

Don't worry  :lol: I've rated you "Very good". Rating is not the main matter. Could you solve my problem, since your previous is not true. The hint is that "classical method " :lol:

[quote="hungkhtn"]Alternatively, we have the following inequality
\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\ge 2, \]
with the equality for $a=b=c$ or $a=b,c=0$. I wait for a nice solution for it.[/quote]

thank you very much hungkhtn, you're very kind. :)

what about this solution:

we need to prove that \[\frac{x+y+z}{\sqrt[3]{(x+y)(y+z)(z+x)}}+\frac{3\sqrt[3]{xyz}}{2(x+y+z)}\ge 2\]so we'll be done if we can show that \[\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{x+y+z}\ge \frac{3\sqrt[3]{xyz}}{2(x+y+z)}\]which is true since \[\sqrt[3]{(x+y)(y+z)(z+x)}\ge 2\sqrt[3]{xyz}\ge \frac{3}{2}\sqrt[3]{xyz}\]

[quote="nayel"][quote="hungkhtn"]Alternatively, we have the following inequality
\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\ge 2, \]
with the equality for $a=b=c$ or $a=b,c=0$. I wait for a nice solution for it.[/quote]

thank you very much hungkhtn, you're very kind. :)

what about this solution:

we need to prove that
\[\frac{x+y+z}{\sqrt[3]{(x+y)(y+z)(z+x)}}+\frac{3\sqrt[3]{xyz}}{2(x+y+z)}\ge 2 \]
so we'll be done if we can show that
\[\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{x+y+z}\ge \frac{3\sqrt[3]{xyz}}{2(x+y+z)}\]
which is true since
\[\sqrt[3]{(x+y)(y+z)(z+x)}\ge 2\sqrt[3]{xyz}\ge \frac{3}{2}\sqrt[3]{xyz}\]
[/quote]

It is still not true, nayel. I think it should be

[quote]so we'll be done if we can show that[/quote]
\[\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{x+y+z}\le \frac{3\sqrt[3]{xyz}}{2(x+y+z)}, \]
which is wrong.

[quote="hungkhtn"]
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c}\ge 4.$
[/quote]
Are you sure hungkhtn ?
With AM-GM we have :
$2\frac{a}{b}+\frac{b}{c}\geq\ 3\frac{a}{\sqrt[3]{abc}}$
So that we have :
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\ \frac{a+b+c}{\sqrt[3]{abc}}$
Apply AM-GM again we have :
$3[\frac{a+b+c}{3\sqrt[3]{abc}}]+\frac{3\sqrt[3]{abc}}{a+b+c}\geq\ 4\sqrt[2]{\frac{a+b+c}{3\sqrt[3]{abc}}}\geq\ 4$
Done !

[quote="nayel"]Let $a,b,c>0$. Prove that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{9\sqrt[3]{abc}}{a+b+c}\ge 6 \]
[/quote]
And apply AM-GM :
$3[\frac{a+b+c}{3\sqrt[3]{abc}}]+3[\frac{3\sqrt[3]{abc}}{a+b+c}] \geq\ 6$
Done !

[quote="hungkhtn"]Alternatively, we have the following inequality
\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\ge 2, \]
with the equality for $a=b=c$ or $a=b,c=0$. I wait for a nice solution for it.[/quote]

What about this one, Tan?

[quote="hungkhtn"]Alternatively, we have the following inequality
\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\ge 2, \]
with the equality for $a=b=c$ or $a=b,c=0$. I wait for a nice solution for it.[/quote]
Here my solution for this problem:
Let’s $p=a+b+c;q=ab+bc+ca;r=abc$.
The inequality is equavalent to:
$\frac{p^{3}-2pq+3r}{pq-r}+\frac{3\sqrt[3]{r}}{2p}\ge 2$.
So we must prove:
$2p^{4}-4p^{2}q+6pr+3pq\sqrt[3]{r}-3r \sqrt[3]{r}\ge 4p^{2}q-4pr$
$2p^{4}-8p^{2}q+10pr+3pq\sqrt[3]{r}-3r \sqrt[3]{r}\ge 0$.
But we have:
$2p^{4}-10p^{2}q+8q^{2}+12pr \ge 0$( easy to prove).
So we have to prove:
$2p^{2}q-8q^{2}-2pr+3pq \sqrt[3]{r}-3r \sqrt[3]{r}\ge 0$.
It’s equivalent to:
$2(a+b+c)^{2}(ab+bc+ca)-8(ab+bc+ca)^{2}-2(a+b+c)abc+3(a+b+c)(ab+bc+ca)\sqrt[3]{abc}-3abc \sqrt[3]{abc}$.
Normalize $abc=1$.
We need to prove:
$2(a+b+c)^{2}(ab+bc+ca)-8(ab+bc+ca)^{2}-2(a+b+c)+3(a+b+c)(ab+bc+ca)-3$.
We put:
$f(a,b,c)= 2(a+b+c)^{2}(ab+bc+ca)-8(ab+bc+ca)^{2}-2(a+b+c)+3(a+b+c)(ab+bc+ca)-3$.
Suppose that $c \ge b \ge a \ge 0$.
Consider difference:
$f(a,b,c)f(a,\sqrt{bc},\sqrt{bc}) =(\sqrt{b}-\sqrt{c})^{2}[2a^{3}+2bc(\sqrt{b}+\sqrt{c})^{2}+2a(b+c+\sqrt{bc})^{2}-4a(ab+ac+2bc+2a\sqrt{bc})+3a^{2}+3bc+3a(\sqrt{b}+\sqrt{c})^{2}]$.
(It’s true because:
$3ab(1-a)+3ac(1-a)+ab(b-a)+ac(c-a)+(2a \sqrt{bc}-2abc)+(4a \sqrt{bc}-4a^{2}\sqrt{bc})+(4abc-4a^{2}\sqrt{bc}) \ge 0 )$.
So we need to prove:
$f(a,\sqrt{bc},\sqrt{bc})\ge 0$.
It’s equivalent to:
$(x-1)^{2}(4x^{6}+8x^{5}+18x^{4}+12x^{3}+6x^{2}+12x+6) \ge0$
with $x=\sqrt{a}$.
We are done! :)
Am I right,hungkhtn? :blush:

[quote="The soul of rock"]Let $x^{3}=a, y^{3}=b,z^{3}=c$
We have to prove that
$\sum_{cyc}\frac{(x^{3}-y^{3})^{2}}{(x^{3}+z^{3})(y^{3}+z^{3})}\ge \frac{(x+y+z)\left(\sum_{cyc}(x-y)^{2}\right)}{x^{3}+y^{3}+z^{3}}$
Or
$\sum_{cyc}(x-y)^{2}\left(\frac{(x^{2}+y^{2}+xy)^{2}}{(x^{3}+z^{3})(y^{3}+z^{3})}-\frac{x+y+z}{x^{3}+y^{3}+z^{3}}\right)\ge 0$
I think the rest is easy![/quote]
But how do you prove the rest?Can you post your solution completely?I think it isn't simple! :wink:

[quote="hungkhtn"]Alternatively, we have the following inequality
\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\ge 2, \]
with the equality for $a=b=c$ or $a=b,c=0$. I wait for a nice solution for it.[/quote]

i have a question: does the following inequality hold \[\frac1p+\frac1q+\frac1r+\frac{3}{p+q+r}\ge 4\]where $p=\frac{x}{y+z-x},q=\frac{y}{z+x-y},r=\frac{z}{x+y-z}$ and $x,y,z$ are the side lengths of a triangle?

or the simpler
\[\frac1p+\frac1q+\frac1r+\frac{3}{p+q+r}\ge 4 \]
where $p=\frac{a+b}{c},q=\frac{b+c}{a},r=\frac{c+a}{b}$. does this hold?

[quote="VIETNAM"][quote="The soul of rock"][quote="VIETNAM"]
Normalize $abc=1$.
[/quote]
You lost one case of equality[/quote]
I agree with you about it!But this case is trivial! :wink:[/quote]

Very good solution Vietnam. But still a better solution exists. Using Schur and Cauchy-Schwarz.

[quote="hungkhtn"][/quote][quote="VIETNAM"][quote="The soul of rock"][quote="VIETNAM"]
Normalize $abc=1$.
[/quote]
You lost one case of equality[/quote]
I agree with you about it!But this case is trivial! :wink:[/quote][quote="hungkhtn"]

Very good solution Vietnam. But still a better solution exists. Using Schur and Cauchy-Schwarz.[/quote]
I have a nice solution using Schur and Cauchy-Schwarz ...
Aplly Cauchy-Schwarz inequality we have:

\[\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+a}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\\ \ge \frac{(a+b+c)^{2}}{2(ab+bc+ca)}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\]

Now we will prove that:
\[\frac{(a+b+c)^{2}}{ab+bc+ca}+\frac{3\sqrt[3]{abc}}{a+b+c}\ge 4\]

We can put: $p=a+b+c,q=ab+bc+ca,r=abc$ they are Viet's expression
We have: $p^{3}-4pq+9r\ge 0$. This is well-known. Schur's inequality with degree $3$.
WLOG, we can assume $r=abc=1$. Then the inequality become:

\[\frac{p^{2}}{q}+\frac{3}{p}\ge 4 \iff p^{3}+3q\ge 4pq \iff (p^{3}-4qp+9r)+(3q-9r)\ge 0\]

It is true because :
$p^{3}-4pq+9r\ge 0$ and $3q\ge 9r$
It means $3q=3(ab+bc+ca)\ge 9\sqrt[3]{(abc)^{2}}=9=9r \iff 3q\ge 9r$

We are done.  :D

Yes, it is exactly my solution, thank you, Hoang.  :lol:


How can be found general term of Fibonacci sequence?For example what is the 50.term of that sequence?

The result is pretty well known though it's complicated. I think it is something like:
(((1+ :sqrt: 5)/2)^n+((1- :sqrt: 5)/2)^n)/4 (that's not right but it's something like that.)

I'm sure you could find the answer if you google it.

Correct version (Binet's formula):
$f_n = \frac{1}{\sqrt 5} \left( \left( \frac{1+\sqrt 5}{2} \right) ^n - \left( \frac{1-\sqrt 5}{2} \right) ^n \right)$

There are several possibilities to get this formula (to prove it you can use simple induction), for example the characteristic equation of the recurrence / the power series expansion.


I was surfing the net for something when I saw this beautiful explanation for the mechanism of swern oxidation.Click on it and then see.

Wow what a cool way to share mechanism. :D Maybe one of us could have taken it as project. Given a reaction mechanism, show it in an animated fashion.

[quote="madness"]You could have done it for compsci da rock  :D[/quote]
Ya I did periodic table no! :D  :D

[quote="madness"]no i meant such a project wherein you animate all reactions you know (The project would have gone for 20000 lines then)  :D[/quote]
actually our project with all the header files lines compiles for more than 15000 lines(I think 15980 approx)

Now here's one more of those reactions "VON RICHTER SYNTHESIS"

i meant only the main prog with your inbuilt fns. Anyways here is a another site full of practical teaching
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animationsindex.htm.  :D
There are many more with such stuff but organic rns are generally rare  :maybe:

The phosphorus pentoxide formed exhibits a strong phosphorescence after illumination

Want one more.
Then here it is artificial chemical rainbow


In the two circles shown, the radius of the inner circle is four units, and the number of square units in the area of the inner circle equals the number of square units in the area of the shaded region. How many units are in the length of the radius of the largest circle? Express your answer in simplest radical form.
[asy]fill((1,0)..(0,1)..(-1,0)..(0,-1)..cycle,gray(.5));
unfill((.707,0)..(0,.707)..(-.707,0)..(0,-.707)..cycle);[/asy]

We know the area of the inner circle is $ 16\pi$. Suppose the length from the outside of the inner circle to the outside of the larger circle is $ x$. We could then express the area of the shaded region as $ (4 \plus{} x)^2\pi \minus{} 16\pi$, and we know that this is equal to $ 16\pi$. Equating these gives us $ x^2 \plus{} 8x \minus{} 16 \equal{} 0$, so $ x \equal{} 2\sqrt{6} \minus{} 4$ and $ \minus{}2\sqrt{6} \minus{} 4$. Since $ x$ obviously can't be negative, it is $ 2\sqrt{6} \minus{} 4$, and thus the radius of the larger circle is $ \boxed{2\sqrt{6}}$.

I got something a little different.  The area of the large circle is $ 16\pi\plus{}16\pi\equal{}32\pi$.  The area of the large circle is $ \pi r^2$, for $ r$ being the radius.  So, if we set them equal, we get \[ \pi r^2\equal{}32\pi\implies r^2\equal{}32\implies r\equal{}\boxed{4\sqrt2}\]

I agree with jxl28.  The answer should be $ \sqrt {32} \equal{} \boxed{4\sqrt2}$.

isabella2296, your solution was fine until you got $ x \equal{} 2\sqrt {6} \minus{} 4$ and $ \minus{} 2\sqrt {6} \minus{} 4$.  It should be $ x \equal{} \frac { \minus{} 8 \plus{} \sqrt {64 \plus{} 64}}{2}$ which simplifies to $ \frac { \minus{} 8 \plus{} 8\sqrt {2}}{2}$ which also is $ \minus{} 4 \plus{} 4\sqrt {2}$.  The final answer is $ 4\sqrt {2}$.


Given that $ 591,3d8$ is divisible by 12, what is the sum of all of the
digits that could replace $ d$?

$ 590,300$ is divisible by 12 so the $ d8$ part is the only part that matters.  The only way to produce a multiple of 12 ending in 8 is to either multiply 12 by 4 or 9 (or anything congruent mod 10).  But $ 12\times 9\equal{}108$, which is too big.  So 4 is the only number that works.


We first count the edges on the hexagonal prism, subtract $ 6$, and add the other ones of the hexagonal pyramid to get $ \boxed{30}$.

[quote="GameBot"]A right hexagonal prism and a right hexagonal pyramid with congruent bases are glued together at their bases, as seen here. How many edges does the resulting polyhedron have?

[asy]draw((-10,8.7)--(-5,0)--(5,0)--(10,8.7));
draw((-10,8.7)--(-5,17.3)--(5,17.3)--(10,8.7),dashed);
draw((-10,18.7)--(-5,10)--(5,10)--(10,18.7));
draw((-10,18.7)--(-5,27.3)--(5,27.3)--(10,18.7),dashed);
draw((-10,8.7)--(-10,18.7));
draw((-5,0)--(-5,10));
draw((5,0)--(5,10));
draw((10,8.7));
draw((10,18.7));
draw((5,17.3)--(5,27.3),dashed);
draw((-5,17.3)--(-5,27.3),dashed);
draw((-10,18.7)--(0,44));
draw((10,18.7)--(0,44));
draw((-5,27.3)--(0,44),dashed);
draw((5,27.3)--(0,44),dashed);
draw((5,10)--(0,44));
draw((-5,10)--(0,44));
draw((10,8.7)--(10,18.7));[/asy][/quote]

I used a similar approach (using Exclusion-Inclusion principle) as professordad, but arrived at a different answer:

(Edges of Hexagonal Prism)+(Edges of Hexagonal Pyramid)-(Overlap edges)
(18)+(12)-6
24

[quote="cgyao15"]Why subtract 6?[/quote]
I think this is to correct for overcounting the same edges twice.

You count the edges of the hexagonal prism (18) (edges are the ends of the faces for example a square would have 4 edges). After that you add the edges of the hexagonal pyramid (12). Because they are glued together, the bottom hexagon of the pyramid and the top hexagon of the prism share an edge, thus making you subtract 6. I think professordad forgot to subtract 6 because you would end up with 24.


A container containing a liquid is moving down a fixed  inclined plane of inclination $\theta$  with an acceleration $a$.If $\beta$ is the angle made by the water surface with with the inclined plane when equilibrium is attained then show that
     $tan({\theta}+{\beta})$ = $\frac{a cos{\theta}}{g-a sin{\theta}}$

i think the question needs some correction :maybe:

consider one of the  extreme cases when say $\theta$ is $0$
then it is a well known result that the free surface of the liquid makes an angle $\beta$ with the horizontal that satisfies the relation

$tan{\beta}=\frac{a}{g}$

hence i suppose $cos{\theta}$ and $sin{\theta}$ have to change places.

anyways here is an intro to the proof
let my frame of refrence be the inclined plane
take any two points A and B in the liquid consider an area s(very small)
consider the pressure there .consider the liquid bounded by the cylinder containing the points. then consider the weight of the liquid inside the cylinder .equate the result to ma.consider the atm pressure
consider the ht of the points from free surface
get an expression for the right hand side equation from there

Ye syou are right .....I have corrected the problem..


what are the P vs. Np and ? above my avatar mean?

Designates you as a heavy posting person.  At a certain level of posts, you get a little question mark and a title.  At another level, you get another question mark and another title.  (Though don't go posting nothing messages to try gather question marks - we'll zap clutter, and the levels get farther and farther apart.)

Now...  What is 'P versus NP'?

akatookey
[b]P versus NP[/b]

Joined: 26 May 2003
Posts: 26
Location: Kentucky...Hicville USA

Its some famous unsolved problem, something about being easy to count or easy to find or something that doesn't really seem like a proper maths problem to me! Is there a simple way of stating it?

I figure out what the P vs NP problem was!!!

[url]http://www.claymath.org/Millennium_Prize_Problems/P_vs_NP/url] Its here[/url][/url]

EDIT : Sigh.... I must give credit where it is due...thanx andy17null


Suppose $ S\subset \mathbb{R}^n$ is compact, $ f: S\rightarrow \mathbb{R}$ is continuous and $ f(x)>0$ for all $ x\in S$. Show that there is a number $ c>0$ such that $ f(x)\geq c$ for every $ x\in S$.

Suppose not, then $ 0 = \mathop {\inf }\limits_S f\left( x \right)$. Therefore, there exists a sequence$ \{ x_n \}_n \subset S$ such that
\[ 0 < f\left( {x_n } \right) < \frac{1}{n}.
\]
Since $ S$ is compact in $ \mathbb R^N$, then passing to a subsequence, still denote by $ \{ x_n \}_n \subset S$, $ \{ x_n \}_n \to x \in S$. As $ n \to \infty$, using the continuity $ f$ we deduce that $ f(x)=0$. That's a contradiction.

This is also true evenly $ f$ is a lower semicontinous function  :)


just for clarification:

how many roots of unity does $x^{n} - 1 = 0$ have

and how many roots of unity does $x^{n-1} + x^{n-2} + x^{n-3} + \ldots + 1 = 0$ have?

The roots to the first one are $1$ and all the complex ones ($n$ in total).

The roots to the second are just all the complex ones ($n-1$ in total).

aite thanks just wanted to make sure of that


[url]http://ktj.in/woodstock[/url]
[url]http://ktj.in/woodstockUS[/url]

It is part of the IIT Kgp techno-fest Kshitij 2007 and will probably end when the fest ends on February 4 2007.

[quote="Koda"][url]http://ktj.in/woodstock[/url]
[url]http://ktj.in/woodstockUS[/url]

It is part of the IIT Kgp techno-fest - 2007 and will probably end when the fest ends on February 4 2007.[/quote]

oooh

nice! I like it  :lol:

i thought this was about the music festival

[quote="Koda"]:huh:  :huh:
what music festival?[/quote]

the really big one 20 or so years ago where every big rock band came for a 3 day festival.


Fie $ABC$ un triunghi inscris in cercul $w$ si $I$ centrul cercului inscris. Fie $B' \in BI \cap AC$, $C' \in CI \cap AB$. Acum fie $X,Y$ punctele de intersectie ale dreptei $B'C'$ cu cercul $w$. Sa se arate ca raza cercului circumscris triunghiului $IXY$ este de $2$ ori mai mare decat raza cercului $w$.

Exista o solutie sintetica si "luminoasa" la pb asta? Cred ca da, dar nu ma prind cum din cauza atator functii trigonometrice :)

Uite o solutie sintetica... Nu stiu bine latex, asa ca voi schita doar demonstratia  :blush:
Fie M si N intersectia dreptelor CI si BI cu cercul. Se arata usor ca raza cercului circumscris triunghiului IMN este egala cu raza lui w. Apoi, consideram R si S simetricele lui I fata de M, respectiv N. Pentru a demonstra problema e suficient sa aratam ca punctele I, X, Y, R, S sunt conciclice. Vom arata pe rand pentru (I, X, S, Y) si (I, X, R, Y).
C'X * C'Y = C'A * C'B ( din puterea punctului C' fata de w). (1)
R, I, A si B sunt pe un cerc cu centrul in M si diametru RI ( R este si centrul cercului exinscris in ABC corespunzator laturii AB) => C'I * C'R = C'A * C'B. (2)
Din (1) si (2) => I, X, R, Y conciclice.
Analog se demonstreaza pentru I, X, S, Y. => gata problema!  :D

da, asta este si solutia mea, ideea de baza era ca centrele exinscrise ale lui B si C sunt pe cercul $IXY$.


A movie place charges 9 dollars for an adult ticket and 6 dollars for a child ticket. On a recent night, the sales of children tickets were 5 times the sale of adult tickets. If the total amount collected for the tickte sales WAS NOT greater than 3000 dollars, what is the greatest number of adults who could have purchased the tickets.

[hide]Let $A$ be the # of adults. Money: $9A$

Let $C$ be the # of children. Money: $6C$

TOTAL: $9A+6C$

$C=5A$

Substituting into the total, total = $39A$. But we know that total<3000

Thus,

$39A<3000$

$13A<1000$

$13A<1001$

$13A<7\cdot 11\cdot 13$

$A<77$

$76$ adults.[/hide]

[quote="sharkman"]Where did number 13 come from?[/quote]

I divided both sides by 3...

[quote="sharkman"]Why divide both sides ONLY BY 3?

Why not divide both sides by the coefficient 39?[/quote]
3000 is not divisible by 39.  Vishalarul tried to first simplify the equation $3000=39A$ as much as possible before he found what A equals.


quadratics
function
algebra
calculus
calculus computations

Find the general solution to:

$ \frac{dy}{dx}\equal{}\frac{y}{x}\frac{x^{3}\minus{}y}{x^{3}\plus{}y}$

If you substitute $ y \equal{}\frac{z}{x^{3}}$, you actually end up with something separable.  After that, you end up solving a quadratic equation to find $ z$.  Unfortunately, I don't know the formal justifications one needs in order to apply these steps.

At the heart of differential equations lies a single theorem: the existence-uniqueness theorem. And part of the power of this theorem is that justifications can be after the fact, not before.

For this equation, there are some curves in $ \mathbb{R}^{2}$ along which is it ill-behaved (just look at where that denominator is zero.) These curves divide the plane into several open regions. In asking about the "general solution," we should really be making this a separate question for each of those regions of validity.

OK, so we do that. What then? We get clever. We apply some manipulation that seems to work. (No need for me to make any suggestions - JBL has already done that.) From this, we arrive at one or several formulas - that is, you have a family of functions. At this point I ask you one question: within your family of solutions, can you solve for all allowable initial conditions? If the answer is "yes," then you're done. You've found everything.

And if the answer is "no," then the fix is often easy. Here's a cheap example: your equation is $ \frac{dy}{dx}\equal{}3\minus{}y,$ and for whatever reason you decide that you want to use separation of variables. You wind up with $ \minus{}\ln(3\minus{}y)\equal{}x\plus{}C,$ which you can rearrange to $ y\equal{}3\pm e^{\minus{}(x\plus{}C)}\equal{}3\pm e^{\minus{}C}e^{\minus{}x}.$ Now, can we solve for every possible initial condition? No, this formula cannot cope with an initial condition for which $ y\equal{}3.$ But we also make separate note that $ y\equal{}3$ is a solution to the differential equation, and we "lost" this solution when we divided by $ 3\minus{}y$ as part of the separation of variables. Include that function and we can redefine $ C$ so we can write $ y\equal{}3\plus{}Ce^{\minus{}x}.$ You can solve for all initial conditions with that, so we've found everything.

Essentially, this is the plan:

Step 1: do something clever. (OK, that's not much of a "plan".)
Step 2: check that the functions you've found are solutions to the DE.
Step 3: check that you can solve for all allowable initial conditions.
If step 3 is "yes," you've found them all. If "no," see if you can fill the holes while still taking advantage of what you've already don.

hello, i have derived the following solutions
$ y(x)\equal{}\frac{Cx^3}{C\pm\sqrt{C^2\plus{}x^4}}$
Sonnhard.


probability
combinatorics unsolved
combinatorics

A stick of length 10 is marked with 9 evenly spaced marks (so each is one unit apart). An ant is placed at every mark and at the endpoints, randomly facing either right or left. Suddenly, all the ants start walking simultaneously at a rate of 1 unit per second. If two ants collide head-on, they immediately reverse direction (assume that turning takes no time). Ants fall off the stick as soon as they walk past the endpoints (so the two on the end don’t fall off immediately unless they are facing outwards). On average, how long (in seconds) will it take until all of the ants fall off of the stick?

Solution is given on page 19 of http://www.mathlinks.ro/viewtopic.php?mode=attach&id=6257.

I only understand the ants passing each other part after that I don't understand it was derived . Can someone please care to explain ?

E(maxTwait) = n×(P(either endpoint faces inwards)+
(n−1)×P(both endpoints face outwards + either of 2nd closest points face inwards)+...

E(maxTwait) =$ \left(\sum_{i \equal{} 0}^{\frac{n}{2} \minus{}1}\frac{3}{4} \left(\frac{1}{4}\right)^{i}(n\minus{}i)\right) \plus{} \left(\frac{1}{2}\right)^{n}\frac{n}{2}$

Since the intersections of the ants don't matter, the amount of time it takes for all of the ants to fall off is the time it takes for the ant that has to walk the farthest to an end. If either of the ants on the ends of the ruler face inwards, it takes them $ n$ seconds to fall off the other end. This occurs with probability $ 1\minus{}\left(\frac{1}{2}\right)^2 \equal{} \frac 34$.

If both of them face outwards (with probability $ \left( \frac 12\right)^2 \equal{} \frac 14$), then the next longest possible time occurs with the ants that are one unit into the ruler. If either of those face inwards, it takes them $ n\minus{}1$ seconds to fall off the end, again with $ 3/4$ probability. In general, for the ant that is $ k$ units away from the end of the ruler, there is a $ \left( \frac 12\right)^{k\minus{}1}$ chance that all of the other ants closer to the ends face outwards, so

$ \overline{\text{average\ wait}} \equal{} \frac{3}{4}n \plus{} \frac{1}{4} \cdot \frac{3}{4}(n\minus{}1) \plus{} \left(\frac{1}{4}\right)^2 \cdot \frac{3}{4} (n\minus{}2) \plus{} \cdots \equal{} \sum \text{the\ sum\ in\ your\ post}$

The $ \left(\frac 12\right)^n \cdot \frac n2$ term comes from considering the ant in the middle, which does not have a paired ant and so always takes $ n/2$ seconds to fall off.


superior algebra
superior algebra theorems

let x,y be elements of a group G such that xy(x^-1)=y^5, x has order 3, and y is diffrent than 1 has odd order. Find the order of y.


limit
analytic geometry
graphing lines
slope
search
trigonometry
function

I wasn't sure where to post this question but so i decided here.

I am currently in an argument to prove to my friend that n/0 is not infinity. I heard there are a few paradoxes that happen if n/0 is treated as infinity.

How would you explain why n/0 is NOT infinity?

$ \frac n 0$ is simply undefined.  Also, infinity is not a number.  If somebody tried to tell me $ \frac n 0 \equal{} \infty$, I would say, "what do you mean by '$ \frac n 0$', and what do you mean by '$ \infty$'?"

If somebody did try to define "$ \infty$" and "$ \frac n 0$" in such a way that $ \frac n 0 \equal{} \infty$ for all $ n > 0$, then the standard rules of arithmetic would no longer hold true.  For example, multiplying both sides by $ 0$ and assuming $ 0\cdot \infty \equal{} 0$ would presumably show you that $ n \equal{} 0$, a contradiction.  Perhaps somebody might say, "Oh, well $ 0 \cdot \infty$ is not $ 0$, it is $ \infty$."  But then you have found that $ n \equal{} \infty$, another contradiction.  Somebody else may try to say, "$ 0 \cdot \infty$ is not $ 0$, and it is not $ \infty$, in fact it is some finite positive number."  But which one?  This still leads to contradictions.

If $ \frac10$ and $ \frac20$ were both equal to "$ \infty$", then presumably we would have $ \frac10 \equal{} \frac20$, and multiplying both sides by $ 0$ would show that $ 1 \equal{} 2$, a contradiction.

Mathematicians deal with all this by simply leaving $ \frac n 0$ undefined.

He tried explaining that in the fraction n/d, as d gets smaller, and n is unchanged, then the fraction gets bigger. So, therefore, if d is the smallest possible non-negative number, then, the value of the fraction is the greatest possible number. Since the smallest non-negative is 0, then the value of n/0 must equal infinity.

I'm thinking of explaining that if your statement is true, then if the denominator is the largest possible number (infinty in his case) then the fractions value would be the smallest possible: 0. so if you have the equation 2/0=infinity, then you have 2=infinity*0, which says 2=0. So it's false.

As far as I know, Infinity is not a set number of only one value. There are infinities of different sizes.

His argument is that is $ f(x)\equal{}1/x^2$, then $ \lim_{x\to 0} f(x)\equal{}f(0)$ which isn't true. You should try searching for this topic. These are all over the place and you'll get a lot more information that way.

Also, because multiplication is repeated addition, and multiplication is the inverse of division, some people just ask "how many times do we have to add 0 'til we get n?" Because you can add forever, some people say "infinity". (Not true) And if n=0, then there are infinitely many solutions.

$ \frac {n}{0}$ is undefined. $ \lim_{x\to 0}\frac {n}{x} \equal{} \infty$ where $ n>0$

[quote]He tried explaining that in the fraction n/d, as d gets smaller, and n is unchanged, then the fraction gets bigger. So, therefore, if d is the smallest possible non-negative number, then, the value of the fraction is the greatest possible number. Since the smallest non-negative is 0, then the value of n/0 must equal infinity. [/quote]

This would be the graphical, slope argument. As has been mentioned quite a few times, division by zero is just contradictory, so it is undefined.

What i'm asking is how is it contradictory

Yes i've looked around the forum and did a search and all i got was random unrelated stuff or really complacated stuff.

unless n is 0 right? so does that mean that 0/0 is infinity?

The reason division by zero is not defined over most number systems is that doing so would cause the system to lose the desireable properties of a field. However, there are many examples of cases where division by zero is defined (the Real Projective Line for example).

$ \frac00$ is undefined as well.

$ \frac a b$ is supposed to be the number that you multiply by $ b$ to get $ a$.  But any number multiplied by $ 0$ gives you $ 0$, so there is not a unique value that can be assigned to $ \frac00$, and mathematicians leave $ \frac00$ undefined.

Essentially, what Singularitarian is saying, is that:

Let $ \frac{a}{b}$ be the solution to the equation $ bx\equal{}a$ when such a value exists and is unique.

If $ b\equal{}0$, then $ 0x\equal{}a$ or $ 0\equal{}a$. In the case where $ a\ne 0$ the equation has no solution, and in the case where $ a\equal{}0$ the solution set is infinite. In both cases, a unique value for $ \frac{a}{b}$ does not exist.

$ \frac {n}{0} \equal{} \frac {n}{1(0) \minus{} 2(0)} \equal{} \frac {n}{ \minus{} 0} \equal{} \minus{} (\frac {n}{0})$
You can do the 2nd step in this because the denominator is still equal to $ 0$
If $ \frac {n}{0} \equal{} \infty$, then $ \infty \equal{} \minus{} \infty$ by substitution. The 2nd statement ($ \infty \equal{} \minus{} \infty$) is not true because $ \infty$ is infinitely large, and $ \minus{} \infty$ is infinitely small, or infinitely large in the negative direction.
Since $ \infty \equal{} \minus{} \infty$ is not true, $ \frac {n}{0} \equal{} \infty$ cannot be true.

basically what i did here was simply take advantage of the fact that $ 0$ is absolute and signless ($ 0 \equal{} \minus{} 0$), but $ \infty$ usually has to have a sign.

He assumed that $ \frac {n}{0}$ is $ \infty$ because he was approaching $ 0$ from the positive side. However, $ 0$ is neither positive nor negative.

A clearer example you can do is to use actual numbers.
Take $ \frac {1}{0}$ for example.
$ \frac {1}{0} \equal{} \frac {1}{ \minus{} 0} \equal{} \minus{} (\frac {1}{0})$
if that was $ \infty$, it'd mean $ \infty \equal{} \minus{} \infty$

ProtestanT, what about the case of the Real Projective Line where $ \infty\equal{}\minus{}\infty$? Just something to think about.

i know that $ \tan{90} \equal{} \frac{3}{0}$  :rotfl:

Today, he made a new statement that n/0= +/- infinity.

Jezz, how am I gonna put up with this guy. He says that n/0 is undifined in algebra. But it is defined in calculus. I tried to explain to him that in the function f(x)=n/x for a constant n, the value of the function gets larger as x decreases, as long as x is greater then 0. How do i prove this to him?

I told him what you said singularitarian. [quote="Singularitarian"]I derived several contradictions.

Suppose there were such a number as $ \frac n 0$.  Then we would have $ 0 \cdot \frac n 0 \equal{} n$, but on the other hand $ 0$ times anything is $ 0$, so we would have $ n \equal{} 0$, a contradiction.[/quote]

He says n/0 can not be algebraicly worked with, and must be done with calculus. I asked him to solve 2/0=infinity with calculus and he says this "by the axiom that n/0 = infinity, 2/0 = infinity.

I'm not smart enough math wise to formulate a proof for him. I'll try ProstanT's thing next

[quote="ashpotter"]He says that n/0 is undifined in algebra. But it is defined in calculus. [/quote]
Does he know calculus?  The relevant correct statements are that $ \lim_{x \to 0^{ \plus{} }} \frac {1}{x} \equal{} \plus{} \infty, \lim_{x \to 0^{ \minus{} }} \frac {1}{x} \equal{} \minus{} \infty$, and $ \lim_{x \to 0} \frac {1}{x}$ [b]does not exist.[/b]  None of these statements can be translated into a meaningful interpretation of the symbol $ \frac{n}{0}$.

This guy sounds like a fuzzy thinker.  What is he, a liberal arts major?  $ \frac n 0$ is either defined or it isn't.  And in fact, it isn't.

[quote="t0rajir0u"][quote="ashpotter"]He says that n/0 is undifined in algebra. But it is defined in calculus. [/quote]
Does he know calculus?  The relevant correct statements are that $ \lim_{x \to 0^{ \plus{} }} \frac {1}{x} \equal{} \plus{} \infty, \lim_{x \to 0^{ \minus{} }} \frac {1}{x} \equal{} \minus{} \infty$, and $ \lim_{x \to 0} \frac {1}{x}$ [b]does not exist.[/b]  None of these statements can be translated into a meaningful interpretation of the symbol $ \frac {n}{0}$.[/quote]

Somehow my teacher says that $ \lim_{x \to \infty} \frac {1}{x} \equal{} \infty$
And on my limits test we had a question like this and I put DNE and got it wrong.

[quote="iniquitus"][quote="moogra"]Somehow my teacher says that $ \lim_{x \to \infty} \frac {1}{x} \equal{} \infty$
And on my limits test we had a question like this and I put DNE and got it wrong.[/quote]

That's ridiculous.

First of all, if you allow a limit to infinity, the limit yields two values.

Secondly, if the limit goes to infinity, (I think) you are allowed to put that it Does Not Exist since it is undefined (or at least in BC m-c questions)[/quote]How is that DNE? $ \lim_{x\rightarrow\infty}\frac1x\equal{}0$  :huh:

However, if you meant $ \lim_{x\rightarrow0}\frac1x$, this is DNE because the two one-sided limits do not agree.

You two should know that $ \infty$ is not a number, it's a concept and it means 'larger than any given real number'. I've said this tons of times. Perhaps our high school teachers fail to convey this message to the newer students.

I've also noticed that among many students, (not just math students), that they consider infinity to be "the largest number in the world"...

Most people, unless they have extensive math backgrounds, are oblivious to the characteristics of infinity.

[quote="10000th User"][quote="iniquitus"][quote="moogra"]Somehow my teacher says that $ \lim_{x \to \infty} \frac {1}{x} \equal{} \infty$
And on my limits test we had a question like this and I put DNE and got it wrong.[/quote]

That's ridiculous.

First of all, if you allow a limit to infinity, the limit yields two values.

Secondly, if the limit goes to infinity, (I think) you are allowed to put that it Does Not Exist since it is undefined (or at least in BC m-c questions)[/quote]How is that DNE? $ \lim_{x\rightarrow\infty}\frac1x \equal{} 0$  :huh:

However, if you meant $ \lim_{x\rightarrow0}\frac1x$, this is DNE because the two one-sided limits do not agree.

You two should know that $ \infty$ is not a number, it's a concept and it means 'larger than any given real number'. I've said this tons of times. Perhaps our high school teachers fail to convey this message to the newer students.[/quote]

I meant $ \to 0$. If it was $ \to \infty$, it would be 0. I don't see how this is using $ \infty$ as a number. There are plenty of problems that have limits that go to infinity (approaches might be a better word).

The two notations $ \lim_{x \to \infty} f(x) \equal{} L$ and $ \lim_{x \to a} f(x) \equal{} \infty$ are just that, notations.  What they actually mean is

$ \forall \epsilon > 0 \exists N : x > N \implies |f(x) \minus{} L| < \epsilon$

and

$ \forall N \exists \delta : |a \minus{} x| < \delta \implies f(x) > N$.

See the [url=http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html]epsilon-delta definition of a limit[/url].  The symbol $ \infty$ is just convenient notation that approximates intuitively a precise concept.

Newest update on the argument!

He is now basing his thought that n/0 is infinity off the fact that informaly
"1/infinity=0"
so multiplying by infinity/0, you get 1/infinity=0

http://mathworld.wolfram.com/Infinity.html

I can't understand your expressions, t0rajir0u and other people.

Well, that's because when you start messing around with things like this, the math can get pretty complex. (I don't understand half of it either...)

i.e. This is really high-level material...

[quote="ashpotter"]informaly[/quote]
Which is exactly the problem.  You can't claim a formal result and base it off of an informal argument.

[quote="moogra"] I don't see how this is using $ \infty$ as a number. There are plenty of problems that have limits that go to infinity (approaches might be a better word).[/quote]

I think this is one of the main problems about the concept of the limit that I know, by experience in my high school calculus class, many of my peers were confused with. I am by no means an expert, but I'd like to clarify one thing I think might be hazy, and I hope t0rajir0u and others will brutally correct me if I am wrong:

When you say, for example, that $ \lim_{x \to \infty} \frac{1}{x}\equal{}0$, you are [i]not[/i] saying that the "limit of $ \frac{1}{x}$ as $ x \to \infty$ [i]approaches[/i] $ 0$". Rather, you are saying that the "limit of $ \frac{1}{x}$ as $ x \to \infty$ [u]is[/u] $ 0$". As t0rajir0u posted, this is just notation, which formally means that, for any finitely small positive number $ \epsilon$, I can find you a very large number $ N$ such that, for all $ x>N$, the difference between $ f(x)$ and $ 0$ is less than $ \epsilon$. The limit is not "approaching" zero, but is defined as zero.

Similarly, when you say that $ \lim_{x \to 0} \frac{1}{x^2}\equal{}\infty$, you are not saying that infinity is a number, but rather that, for any large number $ N$, I can find you a finitely small positive number $ \delta$ such that for all $ x$ in $ (\minus{}\delta, \delta)$, $ f(x)>N$, which is essentially what is meant when we say something is "approaching infinity".

The limit, however, does not "approach" in either case.

[quote="jvenezuela716"][quote="moogra"] I don't see how this is using $ \infty$ as a number. There are plenty of problems that have limits that go to infinity (approaches might be a better word).[/quote]

I think this is one of the main problems about the concept of the limit that I know, by experience in my high school calculus class, many of my peers were confused with. I am by no means an expert, but I'd like to clarify one thing I think might be hazy, and I hope t0rajir0u and others will brutally correct me if I am wrong:

When you say, for example, that $ \lim_{x \to \infty} \frac {1}{x} \equal{} 0$, you are [i]not[/i] saying that the "limit of $ \frac {1}{x}$ as $ x \to \infty$ [i]approaches[/i] $ 0$". Rather, you are saying that the "limit of $ \frac {1}{x}$ as $ x \to \infty$ [u]is[/u] $ 0$". As t0rajir0u posted, this is just notation, which formally means that, for any finitely small positive number $ \epsilon$, I can find you a very large number $ N$ such that, for all $ x > N$, the difference between $ f(x)$ and $ 0$ is less than $ \epsilon$. The limit is not "approaching" zero, but is defined as zero.

Similarly, when you say that $ \lim_{x \to 0} \frac {1}{x^2} \equal{} \infty$, you are not saying that infinity is a number, but rather that, for any large number $ N$, I can find you a finitely small positive number $ \delta$ such that for all $ x$ in $ ( \minus{} \delta, \delta)$, $ f(x) > N$, which is essentially what is meant when we say something is "approaching infinity".

The limit, however, does not "approach" in either case.[/quote]You got much of the ideas but like you said, a bit 'hazy' so let me correct your thoughts and hopefully you can finally grasp this:

'The limit is "approaching" zero' is a statement that we intuitively understand from the verb "approach". However, the word "approach" is not rigorous enough, hence came the concept of the delta-epsilon definitions. The definition of a limit was established to clarify this distinction and to bring a more rigorous concept of the limit. Saying $ x\rightarrow0$ as $ f(x)\rightarrow\infty$ where the arrow stands for "approach", does make sense in an intuitive sense. Only that we do not use that verb when we are dealing with rigorous proofs. So the limit does "approach" (approach does not mean equal, that's why this verb is used only in an intuitive sense). From here arises the fact that by writing equality, many students believe it would eventually equal. It is not a hard concept but being able to see the difference here does take a bit of abstraction.

EDIT: can we let this topic die? As JBL said, it's getting pointless and discussed countless of times...

If the disscusion gets closed, can you not delete it, cause i want this disscusion for future reference.

He finally agreed with me that dividing by 0 in almost all cases yeilds no solution. I worded to him wrong, i said null the first time.

Thanks to you guys, I learned something, and proved him wrong. So you can quit posting on here if you'd like

Now we got into a disscusion about how time only travels in one direction in this dimension and how theres a theory that things will always get farther away from where they are supposed to be. PLEASE DON'T REPLY TO THIS!

correction: he is wierd.

And also, aren't we all weird if you go on aops. I mean, there is not really a "normal"

Ok this is getting off topic. Please close!


number theory
prime numbers
number theory proposed

Find the smallest possible natural number$n$
such that $n^2 + n + 1$ can be written as a product of four prime numbers.

[quote="jin"]Do the four prime need to be different?
If yes,then $n=14$.[/quote]
How said Megus your answer is wrong and without different prims,but I want to say,that
in my solution I have let that prime numbers should be different! ;)

[quote="Rust"]$What \ is \ problem? \ Let \ f(n)=n(n+1)+1. \ If \ p|f(n), \ then \ p \ is \ odd, \ and \ p|(k^2+3).$
$Therefore \ Lejander \ symbol \ (-3/p)=1 \Longrightarrow p=3 \ or \ p=1(mod \ 6).$
$3|f(n) \Longrightarrow n=1(mod \ 3), \ 9 \not|f(n).$
$7|f(n) \Longrightarrow 2n+1=\pm 2, \ n=2,4 (mod \ 7),$
$13|f(n) \Longrightarrow n=3,9 (mod \ 13),$
$19|f(n) \Longrightarrow n=7,11 (mod \ 19), \dots$
$Small \ primes \ are \ 3,7,13,19,31,37,43,61,67,73,79,97.$
$We \ can \ find \ 2n+1 \ satisfied \ for \ 4 \ primes.$
$If \ 3*7|f(n) %Error. "Longarrow" is a bad command.
\ n=4,16 (mod \ 21).$
$If \ 7*13|f(n) \Longrightarrow \ n=9,16,74,81 (mod \ 91).$
$If \ 7*19|f(n) |longrightarrow \ n=11,30,102,121 (mod \ 133).$
$Therefore \ f(100),f(121) \ had \ 4 \ primes \ divisors, \frac{f(100)}{3*7*13}=37.$
$f(100)=3*7*13*37, \ and \ it \ is \ minimal \ n.$[/quote]
Congratulation!your solution is rigth,and it is also my solution ;) .Nice answer,isn't it?


induction
number theory unsolved
number theory

 Suppose a number N which equals to a product of many numbers.Prove that if N is divisible by a prime number p then at least one of the numbers is divisible by p.
Please try to write a solution like writing during the IMO ... :lol:
Davron

Well, for some people this is the definition of prime...

If p divides ab, then p divides a or p divides b (Euclid's Lemma).  You can then use induction to extend this to a product of n numbers.


Let be $ a,b,x,y>0$ such that $ 4a>b^2$ . Prove that :

$ (1\plus{}bx\plus{}ax^2)(1\minus{}by\plus{}ay^2)\ge xy(4a\minus{}b^2)$

[quote="Farenhajt"][hide]$ \left(ax \plus{} b \plus{} {1\over x}\right)\left(ay \minus{} b \plus{} {1\over y}\right)\stackrel{\mathrm{AM \minus{} GM}}{\geqslant}(2\sqrt {a} \plus{} b)(2\sqrt {a} \minus{} b) \equal{} 4a \minus{} b^2$[/hide][/quote]

Fast...  :)


Hey all, recently i was encountered with a question, after extensive searching on the net i was still unable to find a way to do it. Then i was lead to here. Please help me find what method to use etc etc. The problem seems to be quite simple and perhaps is, but i just can't seem to wrap my head round it:S

Let X have a binomial distribution with 28 trials and p=0.347.
What is the probability it takes less than 5 trials until the first success?

It sounds soo simple, its killing me, but i just dont understand how to solve this. Please help me
Thank you

Well the probability that it takes less than 5 is equal to the probability that it takes 1,2,3, or 4. So you can calculate each of these individually and you'll get it I think.

sounds like a negative binomial to me. fixed number of successes, and number of failures until a success. you might want to look into that.  just a formula to plug in


Βρείτε όλους τους θετικούς ακεραίους $ m$ ώστε ο αριθμός

$ A \equal{} \frac {2^ma^m \minus{} (a \plus{} b)^m \minus{} (a \minus{} b)^m}{3a^2 \plus{} b^2}$ να είναι ακέραιος για κάθε ζεύγος $ (a,b)$ μη μηδενικών ακεραίων .

Δεν υπαρχει m με αυτη την ιδιοτητα....Παρατηρησε οτι στον αριθμητη υπαρχει και ο ορος $ 2mxy^{m \minus{} 1}$. :wink:

Αρχημήδης 6 ... αν  m=1 τότε ο αριθμητής είναι ίσος με μηδέν, οπότε διαιρείται απο οποιαδήποτε τιμή της παράστασης $ 3a^2 \plus{} b^2$ , :)

επομένως σίγουρα υπάρχει η τιμή m=1 :)

Σωστα...Αυτες ειναι οι "τετριμμενενες λυσεις"....Μπορεις να μου βρεις και αλλη? (Αρχημηδης 6=Δημητρης) :)

Πολύ καλή άσκηση  :) . Βάζω τη λύση μου.

Παίρνουμε τα $ a,b$ έτσι ώστε $ gcd(a,b)=1$ και έστω $ p$ ένας πρώτος διαιρέτης του $ 3a^2+b^2$. Τότε $ b^2 \equiv -3a^2 \pmod {p}$. Άρα έχουμε \begin{eqnarray*}(a+b)^m+(a-b)^m&=&2\sum_{0 \le i \le m \atop i \equiv 0 \pmod {2}} \binom {m}{i}a^{m-i}b^{i}\\&=&2\sum_{0 \le i \le m \atop i \equiv 0 \pmod {2}} \binom {m}{i}a^{m-i}(b^2)^{\frac {i}{2}} \\&\equiv& 2\sum_{0 \le i \le m \atop i \equiv 0 \pmod {2}} \binom {m}{i}a^{m-i}a^{i}(\sqrt{-3})^{i}\\&=&2a^{m}\sum_{0 \le i \le m \atop i \equiv 0 \pmod {2}} \binom {m}{i}(\sqrt{-3})^{i}\\&=&a^{m}[(1+\sqrt{-3})^{m}+(1-\sqrt{-3})^{m}] \pmod {p} \end{eqnarray*} και συνεπώς  \[ 2^{m}a^{m}-(a+b)^m-(a-b)^m \equiv a^{m}[2^{m}-(1+\sqrt{-3})^{m}-(1-\sqrt{-3})^{m}] \equiv 0 \pmod {p}\] δηλαδή αφού $ gcd(a,b)=1$, οπότε και $ gcd(a,p)=1$ \[ 2^{m} \equiv (1+\sqrt{-3})^{m}+(1-\sqrt{-3})^{m} \pmod {p}\] Όμως μπορούμε να πάρουμε τα $ a,b$ και τον $ p$ αρκετά μεγάλο ώστε $ p > |2^{m}-(1+\sqrt{-3})^{m}-(1-\sqrt{-3})^{m}|$.
Επομένως θα πρέπει $ 2^{m}=(1+\sqrt{-3})^{m}+(1-\sqrt{-3})^{m}$, που εύκολα μπορούμε να δούμε ότι ισχύει μόνο για $ m=1$ (για τον αριθμό $ \omega = \frac {1+\sqrt{-3}}{2}$ ισχύει $ {\omega}^3=-1$  :wink: ).

Ωραια λυση...Εγω προτιμησα εναν αλλο τροπο...Παρατηρουμε οτι Αν για δυο πολυωνυμα $ P_1$ και $ P_2$ με $ P_1(a,b) \equal{} (3a^2 \plus{} b^2)t_xSa^xb^y$ και $ P_2(a,b) \equal{} 2mab^{m \minus{} 1} \plus{} t_xSa^xy^{m \minus{} x}$ S=αθροισμα τοτε ειναι αδυνατον για καθε ζευγος ακεραιων $ (a,b)$ να ισχυει :$ P_1 \equal{} P_2$  Αρα το προβλημα μας δεν εχει λυση αλλη λυση μονο την m=1 :)


Can any body give me the main romanian olympiads problems??
I`ve search it but I didnt get anything usefull so any body could help me??
I need the recent 7-8 years in english
thanx

Search on the forum and/or look at http://www.mathlinks.ro/Forum/resources.php?c=142


$Question 1.$ In an Equilateral Triangle $OAB$, $O$ and $B$ are $(0, 0)$ and $(8/\sqrt(3), 0)$ respectvely. $P$ is a point on $AB$ from where perpendiculars $PL$ and $PM$ are drawn to $OA$ and $OB$ respectively such that $PL=2$. With $PM$ as diameter a circle is drawn meeting $AB$ in $N$. Determine the co-ordinates of point $N$.

$Question 2.$ What will be the sum if you have a number with $n$ digits and if we allow procedure of permutation of these n digits to form $n!$ numbers each of $n$digits (leading zeros permitted) and then add all these numbers together?

$Question 3.$ Find a number $N = abcdef$ (concatenated) such that:

$abcdef \times 2 = cdefab$
$abcdef \times 3 = bcdefa$
$abcdef \times 4 = efabcd$
$abcdef \times 5 = fabcde$
$abcdef \times 6 = defabc$

Enjoy !

[hide="3"]
$\frac{999999}{7}=142857$

It's also the first 6 digits of the demical expression of $\frac{1}{7}$.

($\frac{1}{7}=.\overline{142857}$)
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[hide="1"] We can find that A is at $(\frac{4\sqrt3}{3}, 4)$. Since it is halfway on the x-axis and the altitude is $\frac{8}{\sqrt3}\cdot\frac{\sqrt3}{2}=4$. The next step is to find P. Since PL is 2, or half of the altitude, P is the midpoint of AB or $(2\sqrt3,2)$. (Is this correct?) Then, we can write an equation for the circle with diameter PM and the equation for the line AB. The circle has a center of $(2\sqrt3,1)$ and a radius of $1$. The line has a slope of $-\sqrt3$ and a y-intercept of $8$.
$(x-2\sqrt3)^2+(y-1)^2=1$.
$y={-\sqrt3}x+8$.
We can now solve this system.
$x^2-4\sqrt3x+12+3x^2-14\sqrt3x+49=1$.
$4x^2-18\sqrt3x+60=0$
$2(2x^2-9\sqrt3x+30)=0$.
$(x-2\sqrt3)(2x-5\sqrt3)=0$.
$x=2\sqrt3$ or $x=2.5\sqrt3$.
Drawing the diagram, we can see that $x=\frac{5\sqrt3}{2}$.
Therefore, $y=-\sqrt3(\frac{5\sqrt3}{2})+8=\boxed{\frac{1}{2}}$.
$\boxed{\boxed{N(\frac{5\sqrt3}{2}, \frac{1}{2})}}$. :D The circle also meets AB at point $\boxed{P(2\sqrt3,2)}$. [/hide]
EDIT: OK, I did something wrong :( Can anyone see what it is??
EDIT: I fixed it now. :D

[quote="xxxyyyy"]any takers for $Question 2.$ ??[/quote]
[hide="Only answer"]If that number is $N=d_1d_2d_3\dots d_n$, then the sum $S$ is
\[ S=\frac{10^n-1}9\cdot (n-1)!\sum_{k=1}^nd_k  \][/hide]
Any takers for explaining my answer? :D

I will, but from the question's wording I think that $13_13_22$ is different from $13_23_12$

[hide]
consider the unit digit.  any digit $a_k$ will appear $(n-1)!$ times because you have that many ways to permutate the digits before it.  Therefore, the sum in the unit digit will be $(n-1)! * \sum {a_k}$.  But this is the same in other places, except it is multiplied by 10, 100, ...  Adding them all up, we get $(n-1)! * \sum{a_k} * \frac{10^n-1}{9}$
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frt and pianoforte, can you guys explain how the $\frac{1}{7}$ thing works??


Simplify the radical:

\[ \sqrt{\frac{2\minus{}\sqrt{3}}{4}}\]

[hide="hint"] This is equivalent to $ \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$. Remember those half-angle identities?[/hide]

Right, I just wanted to know how to simplify the radical because the book is giving me:

$ \frac {\sqrt {6} \minus{} \sqrt {2}}{4}$

[hide="Another hint"]
Let $ \sqrt {\frac {2 \minus{} \sqrt{3}}{4}} \equal{} (\sqrt {a} \minus{} \sqrt {b})^2$ for $ a,b \in \mathbb{Q}$. You can solve for $ a$ and $ b$ by either substitution or polynomial construction.
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