I don't have time for many questions but I feel like posting some problems so here they are. Enjoy!

1. Let O be the center of base ABCD of the square pyramid. The top point, V, of the pyramid such that VO is perpendicular to base ABCD. Find the distance from O to triangle VBC.

2. Show that the triangle with side $2xy, x^2-y^2, x^2+y^2$ is always a right triangle.

3. Given that points A,B,C, and D are the consecutive vertices of a regular decagon with sides 20 cm long. When ray AB and ray CD are drawn, their intersection is X. Calculate the length of BX nearest to integer.

4. From the stage of the famous theater AoPS ( :P ), the angle of elevation of the first balcony is 19 degrees. The angle of elevation of the second balcony, 6.3 m directly above the first, is 29 degrees. How high above stage level is the first balcony of AoPS?

Note to #4: I have no right on the creation of AoPS theater. :P

5. Easy but here it is:

Without calculator, what's:

111,111,111 x 111,111,111?

5 the easy one!
[hide]12345678987654321[/hide]
2
[hide](x^2 - y^2)^2 + 4x^2y^2 = (x^2 + y^2)^2[/hide]
3
[hide]It easy to see that the angle formed is 120 degrees and that the triangle must be isoscoles so the base angles are each 30. If we draw the altitude from X, there are 30-60-90 triangles. 10 is half the base, the height is 10/(sqrt3) and the hypotenuse is 20/(sqrt3) which is [20*sqrt3]/3 or approximately 12.[/hide]
4
[hide]We can see that there are a bunch of right triangles. The ratio of the height, x, is from the ground to y, the other leg of the right triangle is = to TAN19. So y = x/(TAN19). y also = (6.3 + x)/TAN29. By substitution, x/(TAN19) = (6.3 + x)/TAN29. Solving this out, x = approximately 10.33.[/hide]

[quote="Silverfalcon"]2. Show that the triangle with side $2xy, x^2-y^2, x^2+y^2$ is always a right triangle.[/quote][hide]$(2xy)^2+(x^2-y^2)^2=(x^2+y^2)^2,$ and $a^2+b^2=c^2$ only for right triangles where $a,b,c$ are the side lengths.[/hide]

[hide="4"]

Looking at the uploaded image, it is easy to see that $tan 29 = \frac{6.3+x}{a}$
which also means
$a=\frac{6.3+x}{tan 29}$

Another equation we may obtain is

$\frac{x}{a}=tan 19$

$a=\frac{x}{tan 19}$

Set the two $a$ values equal to one another.

Solve for $x$

$x = 10.33$ meters

[/hide]

For number 2:

Can a triangle have negative sides? What if x < y :P ?

A triangle cannot have negative sides, at least, not in our universe :). It also wouldn't matter if $x<y$. In fact, we can say that $x \le y$ because we can always rearrange the two numbers for $x$ and $y$ we pick.

As for the problem with the decagon, a minor issue, but it would be more appropriate to say ray DC (or am I missing something)?

Finally, for #1, are you sure the question is phrased correctly? Do you want us to find the exact distance, or the distance relative to the side of the square base, or what? And also, by distance, you mean the distance of $OZ$ where $Z$ is a point in $\triangle VBC$ and $OZ$ is perpendicular to $VBC$?

1
[hide]
We can look at this from the side of the pyramid.  We see that the height, a line from V to O, and half the base, create a right angle with hypotenuse being the slant height of the triangle $l$.  Since we are given no values, everything will be in ratios.  We can express $l$ as  $\sqrt{h^2+(s/2)^2}$ where $h$ is the height (VO), and $s$ is the length of the side of the base.
To find the distance from O to the triangle, we can draw an altitude from the right angle in the triangle to the hypotenuse.  We'll lable this altitude $x$ as it is the solution the problem.  We can set up an equation of the areas.  Since the area is always the same, we can express it as $bh/2$ two different ways, and set them equal.

$hs/4 = xl/2$
$hs = 2xl$

$x = hs/2l$
[/hide]


symmetry
geometry
geometric transformation
reflection
calculus
calculus computations

(See Attachment)
I don't quite understand what i am supposed to optimize, and what my restriction formula is. Is QT constant? But in that case, how could i optimize PRS?
I tried the following:
l = PR + RS

 PR^2 = PQ^2 + QR^2

cos theta1= {QR} / {PR}

 PR = {QR} / {cos theta1}

Similarly, RS = {TR} / {cos theta2}

So l = {TR} / {cos theta2} + {QR}  / {cos theta1}

But i don't see where this could go.
Thank you.

Let $S'$ be the point of symmetry with respect to the line $QT.$ since $\triangle{RSS'}$ is isocesceles triangle, we have $\overline{PR}+\overline{RS}=\overline{PR}+\overline{RS'}\geq \overline{PS'}.$
The equality holds $P,R,S'$ are colinear, which completes the proof.

basically what he is saying is that if we reflect one of the points over the line, then that reflection and the original have the same length, so if we want the least rope length, it is obviously a straight line, and that makes the 2 angles equal due to them being a pair of vertical angles

Thanks... although I knew that.

It's an optimization problem so I am supposed to use optimization calculations-- not just pure reasoning.

It's okay though, I've figured it out already.

$Lisa + Alex$


A circle with center $ A$ and radius three inches is tangent at $ C$ to a circle with center $ B$, as shown. If point $ B$ is on the small circle, what is the area of the shaded region? Express your answer in terms of $ \pi$.

[asy]filldraw(circle((0,0),6),gray,linewidth(2));
filldraw(circle(3dir(-30),3),white,linewidth(2));

dot((0,0));
dot(3dir(-30));
dot(6dir(-30));

label("B",(0,0),NW);
label("A",3dir(-30),NE);
label("C",6dir(-30),SE);[/asy]

Well, if the radius of the small circle is $ 3$, and the radius of the large circle is $ 3\times2\equal{}6$, then the area of the shaded region is $ \pi(6^2)\minus{}\pi(3^2)\equal{}36pi\minus{}9pi\equal{}27pi$.

[hide="alternate"]
The unshaded region is $ \frac{1}{4}$ of the whole circle, so the area of the shaded region is $ \frac{3}{4}\times6^2\pi\equal{}\frac{3}{4}\times36\pi\equal{}\boxed{27\pi}.$[/hide]


inequalities
rearrangement inequality
combinatorics unsolved
combinatorics

The fields of the $8\times 8$ chessboard are numbered from $1$ to $64$ in the following manner: For $i=1,2,\cdots,63$ the field numbered by $i+1$ can be reached from the field numbered by $i$ by one move of the knight. Let us choose positive real numbers $x_{1},x_{2},\cdots,x_{64}$. For each white field numbered by $i$ define the number $y_{i}=1+x_{i}^{2}-\sqrt[3]{x_{i-1}^{2}x_{i+1}}$ and for each black field numbered by $j$ define the number $y_{j}=1+x_{j}^{2}-\sqrt[3]{x_{j-1}x_{j+1}^{2}}$ where $x_{0}=x_{64}$ and $x_{1}=x_{65}$. Prove that \[\sum_{i=1}^{64}y_{i}\geq 48\]

Note that the $i$ corresponding to white fields are either the even or odd integers;  WLOG assume it is the even integers.  By the rearrangement inequality, we have

$\sum_{k=1}^{32}\sqrt[3]{x^{2}_{2k-1}x_{2k+1}}\leq \sum_{k=1}^{32}x_{2k-1}$

and similarly,

$\sum_{k=1}^{32}\sqrt[3]{x_{2k}x^{2}_{2k+2}}\leq \sum_{k=1}^{32}x_{2k}$.

So

$\sum_{i=1}^{64}y_{i}\geq \sum_{i=1}^{64}1+\sum_{i=1}^{64}x_{i}^{2}-\sum_{i=1}^{64}x_{i}= 48+\sum_{i=1}^{64}( x-\frac{1}{2})^{2}\geq 48$.


calculus
integration
trigonometry
function
limit
real analysis
real analysis unsolved

For $x>0$  find the improper integral

\[ G(x)=\int_0^{\infty} e^{-xt}\ \frac{\sin t}{t}\ dt.  \]

$x \to 1/x$, $G(x) = f(1/x)$ with $f(x) = \int_0^{\infty} e^{-t}\ \frac{\sin t*x}{t}\ dt = \int_0^{\infty} f(x,t) dt.$

Function is normally convergent on $\mathbb{R}^*$, $\int_0^{\infty} \partial f(x,t)/\partial x dt$ is normally convergent on every compact interval. We can swap $\int$ and $\partial$  :roll:

On $\mathbb{R}^{+*}$, $f'(x) = -1/(1+x^2)$.

$f'(x)$ hase a limit when $x \rightarrow 0$ so classical theorem gives $f'(0)$ exists. $f$ is thus $C^1$ on $\mathbb{R}^{+}$ and $f(x) = Cte - \arctan x$.

$f$ is continuous and $f(0) = \pi/2$ so $f(x) = \pi/2 - \arctan x$

that integral is a kind of laplace transformation
that is $\mathbb{L} \left (\frac{\sin{t}}{t} \right )$
with the same argument as tµtµ's, we can derive that $\mathbb{L} \left (-tf(t) \right ) = F'(x)$ provided that $\mathbb{L} (f(t)) = F(x)$.
I got $G(x)= -arctan(x)$
I don't know where tµtµ got $\frac{\pi}{2}$

Beginner:

I believe you actually showed that $G'(x)=-\frac1{1+x^2}.$

From this you get that $G(x)=-\arctan x+C.$

But don't forget the constant of integration!

We now look at the integral to show that $\lim_{x\to\infty}G(x)=0.$

That's where the $\frac{\pi}2$ comes from.

tµtµ actually got his constant from the other end, from knowing that

$\int_0^{\infty}\frac{\sin t}t\,dt=\frac{\pi}2.$

This begs two questions: the first is, how do we know that? The problem at hand is one of the ways of computing that in the first place. The other is this: how do we know that

$\lim_{x\to0^+}G(x)=G(0)?$

This is true, but it's not exactly obvious. It's can be considered a continuous analogue of Abel's theorm on the convergence of power series.

Oh  :o  I see it. I've just study Abel's theorem. But My mistake is I ignore the constant of integration, since I do the computation without paper (I thinks that really bad to do this).


Let $ R^2$ be the Euclidean plane in which the distance of $ x\equal{}(x1, x2)\in R^2$ to 0 is defined as $ \parallel{}x\parallel{}\equal{}\sqrt{(x1)^2\plus{}(x2)^2}$.

If we define the inner product $ <x, y>\equal{}(x1)(y1)\plus{}(x2)(y2)$ for $ x\equal{}(x1, x2), y\equal{}(y1, y2)$, then $ (\parallel{}x\parallel{})^2\equal{}<x, x>$.

We say that $ x$ is orthogonal to $ y$ if $ <x, y>\equal{}0$.

If $ x$ is orthogonal to $ y$, $ (\parallel{}x\plus{}y\parallel{})^2\equal{}(\parallel{}x\parallel{})^2\plus{}(\parallel{}y\parallel{})^2$.

Prove that if $ f: R^2\rightarrow R$ satisfies $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ whenever $ x$ is orthogonal to $ y$ and $ f(x)\ge0$ for all $ x\in R^2$, then $ f(x)\equal{}c\parallel{}x\parallel{}^2$ for some $ c\ge0$.

Pleaseeeeee...I need help.

[quote="Stephen"]Let $ R^2$ be the Euclidean plane in which the distance of $ x \equal{} (x1, x2)\in R^2$ to 0 is defined as $ \parallel{}x\parallel{} \equal{} \sqrt {(x1)^2 \plus{} (x2)^2}$.

If we define the inner product $ < x, y > \equal{} (x1)(y1) \plus{} (x2)(y2)$ for $ x \equal{} (x1, x2), y \equal{} (y1, y2)$, then $ (\parallel{}x\parallel{})^2 \equal{} < x, x >$.

We say that $ x$ is orthogonal to $ y$ if $ < x, y > \equal{} 0$.

If $ x$ is orthogonal to $ y$, $ (\parallel{}x \plus{} y\parallel{})^2 \equal{} (\parallel{}x\parallel{})^2 \plus{} (\parallel{}y\parallel{})^2$.

Prove that if $ f: R^2\rightarrow R$ satisfies $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ whenever $ x$ is orthogonal to $ y$ and $ f(x)\ge0$ for all $ x\in R^2$, then $ f(x) \equal{} c\parallel{}x\parallel{}^2$ for some $ c\ge0$.

Pleaseeeeee...I need help.[/quote]

Nobody's interested in it?

Is it very hard?

Is $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ ONLY when $ x$ is orthogonal to $ y$ or, is it just when $ x$ is orthogonal to $ y$. Because, every linear map from $ R^2$ to $ R$ satisfies $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ always.


Given $ a, b, c \geq\ 0$ satisfy $ a\plus{}b\plus{}c\equal{}3.$
Prove that: $ (a^3\plus{}b^3\plus{}c^3)(ab\plus{}bc\plus{}ca)^2 \leq\ 36$
 :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3.$
Prove that: $ (a^3 \plus{} b^3 \plus{} c^3)(ab \plus{} bc \plus{} ca)^2 \leq\ 36$
 :)[/quote]
But it is wrong Try $ a\equal{}1,9.b\equal{}1,c\equal{}0,1$
If you meant $ (a^3 \plus{} b^3 \plus{} c^3)(ab \plus{} bc \plus{} ca)^2 \geq\ 36$
Try $ a\equal{}b\equal{}c\equal{}1$ :)

Sorry for my mistakes. We can prove the weaker easily with pqr:  :)
$ (a^3 \plus{} b^3 \plus{} c^3)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \leq\ 36$
with $ a, b, c \geq\ 0$ and $ a\plus{}b\plus{}c\equal{}3$

[quote="nguoivn"]Sorry for my mistakes. We can prove the weaker easily with pqr:  :)
$ (a^3 \plus{} b^3 \plus{} c^3)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \leq\ 36$
with $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$[/quote]
Try $ a\equal{}b\equal{}1.5$ and $ c\equal{}0.$


geometry
circumcircle
inequalities
function
area of a triangle
Heron\u0027s formula
inequalities unsolved

Let a,b,c be three given positive real numbers. Prove that
1) (a+b+c) 3 (a+b-c)(b+c-a)(c+a-b) <= 27a 2 b 2 c 2
2) (a 2 +b 2 +c 2 )(a+b-c)(b+c-a)(c+a-b) <= abc(ab+bc+ca)
3) (a+b+c)[2(ab+bc+ca)-(a 2 +b 2 +c 2 )] <= 9abc

All three are simple geometric interpretations of the fact that in a triangle the circumradius in very big. In all cases the sharp estimate $ a^2+b^2+c^2\leq 9R^2$ works very well. By the way, the third one is also Schurs inequality. I remember to have already solved the second one on the forum. Anyway, the idea was that with the circumradius. Now you have solutions for all three.

Thanks Harazi!
But I still don't understand your idea that how the inequality a 2 +b 2 +c 2  <= 9R 2 could work here
Can you explain more clearly?

Now I can prove the inequality (2) and (3) by Muihead and Schur's inequality
(2) <--> [(5,0,0)+(3,1,1)-2(4,1,0)]+[(2,1,1)-(3,2,1)] >= 0
 ...........              (Schur for r=3)  ................               (Muihead)
(3) <--> (3,0,0)+(1,1,1)-2(2,1,0) >= 0 which is Schur's inequality for r=1
 :)
But the first inequality is more harder and I can't solve it! :(

Come on! You know how to express the circumradius in function of the sides of the triangle. :D

Hello Minkkoa
Here is a solution to your first inequality

Left side is( (a+b+c)^2)[ (a+b+c)((a+b-c)(b+c-a)(a+c-b)]

or ((a+b+c)^2)[16F^2] where F is the area of a triangle with sides a,b and c.                                 (Using Heron's formula for triangle area)

The inequality to be proved can be written ( a+b+c)^2 <= 27(a^2).(b^2).(c^2)/(16F^2)  =27R^2  (Using formula R=abc/4F). R is circuradius

Now assuming the result 9R^2 >= a^2 +b^2 +c^2 *

We get 27R^2  >=3(a^2 +b^2+c^2) >= (a+b+c)^2 which is is what we have to show

The last inequality can be solved without using Schurs Inequality.Take a triangle sides a,b and c .Then the square roots of these lengths are the sides of another triangle. Use the inequality * on the sides ,circumradius. of this new triangle. And  write its area using the Heron formula .This is the easiest of the three inequalities.

Just let me make some remarks on this problem, for the sake of completeness.

[color=blue][b]Original problem.[/b] Let a, b, c be three positive real numbers. Prove the following inequalities:

[b]First inequality:[/b] $\left(a+b+c\right)^3\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\leq 27a^2b^2c^2$.

[b]Second inequality:[/b] $\left(a^2+b^2+c^2\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\leq abc\left(bc+ca+ab\right)$.

[b]Third inequality:[/b] $\left(a+b+c\right)\left(2\left(bc+ca+ab\right)-\left(a^2+b^2+c^2\right)\right)\leq 9abc$.[/color]

What concerns the third inequality, everything what can be said about it has been said: It is actually equivalent to Schur's inequality $a\left(a-b\right)\left(a-c\right)+b\left(b-c\right)\left(b-a\right)+c\left(c-a\right)\left(c-b\right)\geq 0$, since

$9abc-\left(a+b+c\right)\left(2\left(bc+ca+ab\right)-\left(a^2+b^2+c^2\right)\right)$
$=a\left(a-b\right)\left(a-c\right)+b\left(b-c\right)\left(b-a\right)+c\left(c-a\right)\left(c-b\right)$.

As for the second inequality, it has been discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=4274 before.

Now I have to add some remarks about the first inequality. At first, it holds for [i]any reals[/i] a, b, c, not only for positive reals (although the case of some of a, b, c being negative is an easy consequence of the positive case).

Here is a purely algebraic proof of the first inequality, which works for any reals a, b, c (without distinguishing cases whether they are positive or some of them are negative):

If $\left(a+b+c\right)^3\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\leq 0$, then the first inequality is trivial, so in the following, we will only consider the case $\left(a+b+c\right)^3\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\geq 0$.

The numbers $a^2$, $b^2$, $c^2$ are nonnegative, and thus we can apply to them the third inequality:

$\left(a^2+b^2+c^2\right)\left(2\left(b^2c^2+c^2a^2+a^2b^2\right)-\left(\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2\right)\right)\leq 9a^2b^2c^2$.

Some straightforward algebra shows that

$2\left(b^2c^2+c^2a^2+a^2b^2\right)-\left(\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2\right)=\left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)$;

thus we get

$\left(a^2+b^2+c^2\right)\left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\leq 9a^2b^2c^2$.

But by Cauchy-Schwarz, we have $3\left(a^2+b^2+c^2\right)=\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\geq\left(1a+1b+1c\right)^2=\left(a+b+c\right)^2$, so that $\left(a+b+c\right)^2\leq 3\left(a^2+b^2+c^2\right)$, and thus

$\left(a+b+c\right)^3\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)$
$=\left(a+b+c\right)^2\left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)$
$\leq 3\left(a^2+b^2+c^2\right)\left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)$
$\leq 3\cdot 9a^2b^2c^2=27a^2b^2c^2$.

Thus, the first inequality is proven.

  darij


i got 900 and wrote it done in my answer sheet.
afterward i found the official answer was 899.
im pretty sure the answer is a perfect square.
did any1 get the same answer as me? or as the official answer?
Ps:the so-called official site seems personal but it has the stat of previous years

can u post the solution?
cuz notice it's n-1 above the ∑ thing but not n

You should have gotten something like $2^{n-1}(n+1)$ or something, that's what my friend said anyway. So n is one less than a perfect square and is odd.

Also, check the solution here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78380

now i get it.
all my exponents are one less than the right one bcz of the 2k thing...


geometry
geometric transformation
homothety
LaTeX
function

Two circles are externally tangent at A.  A line passing through the intersection point of their common external tangents meets the circles at B, C, D, and E in that order, where B is the point closest to the intersection point of their tangents.  Prove that BA is perpendicular to AD. (from AoPS Olympiad Geometry Course)

[hide="My Solution"] :

Let the circles be $C_{1},C_{2},$ with centers $O_{1},O_{2}$ and radiuses $r_{1},r_{2}$ .

Because their common external tangents meet in a point (lets name it $O$) so $r_{1}\neq r_{2}$. Let $r_{1}< r_{2}$, So $B \in C_{1}$.

Let $BA$ meets $C_{2}$ in $K,K \neq A$.

$C_{1}$ has been transformed to $C_{2}$ by a homothecy transformation by the center $O$ ,thus \[BO_{1}\parallel DO_{2}\] By a similar homothecy by the center $A$ ,we have \[BO_{1}\parallel KO_{2}\] Thus $D,O_{2},K$ are Collinear thus $\angle DAK=90$ .Hence the result.[/hide]

Nice.  You can also let A' be the point diametrically opposite to A in C1.  Construct the tangents to C1 at A' and A.  Let them meet the common external tangent which B is closest to at X' and X respectively.  Under the homothecy centered at O that takes the C1 to C2, A' goes to A, X' to X, and B goes to D.  Hence angle X'A'B= angle XAD.  But angle X'A'B=angle A'AB. Thus angle BAD= angle A'AX= 90 degrees.  (I should really learn Latex!)

This problem (2001? INMO) also has an elegant solution:

Find all R to R functions such that f(x+y)=f(x)f(y)f(xy) holds for all x,y.


ai are  positive reals , i=1,...,5  and  Σ[1/(1+ai)] \geq 4 . Prove  that
Πai \leq (1/4)^5

Take 1/(1+a_i)=1-u_i. Then we have  \sum u_i<=1 and we must prove that prod (1-u_i)>=4^5*prod u_i. This follows from prod (1-u_i)>=prod (u_2+u_3+u_4+u_5) and 5 applications of AM-GM.


calculus
integration
function
inequalities
limit
real analysis
real analysis solved

Let $f:\mathbb R\to\mathbb R$ be a continuous function with the property that $e^{f(x)}+f(x)\ge x+1$ for any real $x$.
Prove that $\int_0^ef(x)dx\ge 2$.

The problem is false as written. It would be true with $\frac32$ on the right hand side of the inequality instead of 2.

Let $\varphi(t)=e^t+t-1.$ This is an increasing, hence invertible, function. The given condition is that $\varphi(f(x))\ge x$, or $f(x)\ge\varphi^{-1}(x).$

Hence, $\int_0^ef(x)\,dx\ge\int_0^e\varphi^{-1}(x)\,dx.$

But $\int_0^e\varphi^{-1}(x)\,dx=e\cdot1-\int_0^1e^y+y-1\,dy=\frac32.$

Since we can set $f(x)=\varphi^{-1}(x),$ this minimum is realized.

Sorry for my mistake  :blush:  ... You are right Kent.

[quote="Kent Merryfield"]

$\int_0^e\varphi^{-1}(x)\,dx=e\cdot1-\int_0^1e^y+y-1\,dy=\frac32.$

[/quote]

This is an application of Young inequality

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21574

when $b=f(a)$ it becomes equality.

Here the way how manufacture other exercice like this one

$f:R+\rightarrow R+$ continous, for all $x\in R+$ , $g(f(x))\geq x$
where $g:R+\rightarrow R+, C^{1}$ $g(0)=0, g'>0, \lim_{+\infty}g=+\infty$
Then for a >0 $\int_{0}^{g(a)}f(x)dx \geq ag(a)-\int_{0}^{a}g(x)dx$

Idea of the proof:
$g$  is bijective $f(x)\geq g^{-1}(x)$

$\int_{0}^{g(a)}f(x)dx \geq \int_{0}^{g(a)}g^{-1}(x)dx$

conditions on $g$ invite us to use  Young equality


geometry
3D geometry
tetrahedron
limit
Euler
geometry proposed

[u][b]The author of this posting is : kunny[/b][/u]
____________________________________________________________________

Given the tetrahedron $ABCD$ whose faces are all congruent. The vertices $A$, $B$, $C$ lie in the positive part of $x$-axis, $y$-axis, and $z$-axis, respectively, and $AB=2l-1$, $BC=2l$, $CA=2l+1$, where $l>2$. Let the volume of tetrahedron ABCD be $V\left(l\right)$.

Evaluate \[\lim_{l\to 2} \frac{V\left(l\right)}{\sqrt{l-2}}\].

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

Are you sure? Definitely, $V\left(l\right) \not\to 0$ when $l\to 2$.

Maybe you mean $D$ is the origin and not all faces are congruent?

Myth, is it possible that you overlooked this here:

[quote="kunny"]Given the tetrahedron $ABCD$ [b]whose faces are all congruent[/b].[/quote]

In other words, our tetrahedron is an isosceles one. An isosceles tetrahedron with edgelengths a, b, c has volume $\sqrt{\frac{1}{72} \left(b^2+c^2-a^2\right)\left(c^2+a^2-b^2\right)\left(a^2+b^2-c^2\right)}$ (see [1]). We plug in our values and get

$\lim_{l\to 2} \frac{V\left(l\right)}{\sqrt{l-2}} = 8$.

[b]References[/b]

[1] [url=http://www.math.fau.edu/yiu/Geometry.html]Paul Yiu, [i]The volume of an isosceles tetrahedron and the Euler line[/i][/url]. See the file attached to this posting in case the server doesn't work properly.

Of course, the following sentence is completely unnecessary:

[quote="kunny"]The vertices $A$, $B$, $C$ lie in the positive part of $x$-axis, $y$-axis, and $z$-axis, respectively,[/quote]

I guess this is what irritated Myth.

  Darij

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

[quote="darij grinberg"]Myth, is it possible that you overlooked this here:

[quote="kunny"]Given the tetrahedron $ABCD$ [b]whose faces are all congruent[/b].[/quote][/quote]

No, it is impossible.

[quote="darij grinberg"]In other words, our tetrahedron is an isosceles one.[/quote]

Kakogo...

[quote="Myth"][quote="darij grinberg"]In other words, our tetrahedron is an isosceles one.[/quote]

Kakogo...[/quote]

In russian, "isosceles tetahedron" is called "равногранный тетраедр". See problem 304 in the book

  И. Ф. Шарыгин, [i]Задачи по Геометрии, Стереометрия[/i], Москва 1984.

  Darij


group theory
abstract algebra
advanced fields
advanced fields unsolved

Consider the covering spaces $f: (X,x_{0})\to (\mathbb{S}^{1},1),$ where $(X,x_{0})=(\mathbb{R},0), p(x)=e^{2\pi i x}$ or $(X,x_{0})=(\mathbb{S}^{1},1), p(z)=z^{n}$
$a)$ What are $p_{*}\pi_{1}(X,x_{0})$?
$b)$ What are the subgroups of $(\mathbb{Z},+)$ and related conjugate classes?
$c)$ What can you conclude?

a) in the first case $\pi_{1}(X,x_{0})$ is trivial, and so is its image under $p_{*}$. In the second case a generator $[\gamma]$ of $\pi(X,x_{0})$ is mapped into $[\gamma]^{n}$, and you get $n\mathbb Z$ sitting inside of $\pi_{1}(\mathbb S^{1},1)=\mathbb Z$.
b) and c) are beyond my comprehension.

maybe b/ c/ means that one has found all the connected covering spaces of $\mathbb{S}^{1}$, up to isomorphisms ?


Let $ a,b,c$ be nun-negative real numbers. Prove that
$ \frac{a^3\plus{}b^3\plus{}c^3}{(a\plus{}b\plus{}c)^3}\plus{}\frac{5\sqrt[3]{abc}}{12(a\plus{}b\plus{}c)}\ge\frac1{4}$

[quote="kunny"]When $ a \equal{} 0$, the inequality can be written as $ \frac {b^3 \plus{} c^3}{2} \geq \left(\frac {b \plus{} c}{2}\right)^3$ for $ b,\ c\geq 0$, which is true by Jensen's inequality.

Let $ \frac {b \plus{} c}{a} \equal{} x,\ \frac {bc}{a^2} \equal{} y$, Under the condiotion that $ x\geq 0,\ y\geq 0,\ y\leq \frac {1}{4}x^2$, we can rewrite the given inequality  as follows.

$ \frac {1 \plus{} x^3 \minus{} 3xy}{(1 \plus{} x)^3} \plus{} \frac {5}{12}\cdot \frac {\sqrt [3]{y}}{1 \plus{} x}\geq \frac {1}{4}$.[/quote]

Could you to complete your proof,kunny?
Thank you!

[quote="dduclam"]Let $ a,b,c$ be nun-negative real numbers. Prove that
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b \plus{} c)^3} \plus{} \frac {5\sqrt [3]{abc}}{12(a \plus{} b \plus{} c)}\ge\frac1{4}$[/quote]
It's very easy, ddlam:D!
The inequality is equivalent to
$ 12(a^3\plus{}b^3\plus{}c^3)\plus{}5\sqrt[3]{abc}(a\plus{}b\plus{}c)^2 \ge 3(a\plus{}b\plus{}c)^3$
By AM-Gm inequality, we have:
$ 5\sqrt[3]{abc}(a\plus{}b\plus{}c)^2 \ge 5\sqrt[3]{abc}\cdot 9\sqrt[3]{a^2b^2c^2}\equal{}45abc$
So we need to prove that:
$ 12(a^3\plus{}b^3\plus{}c^3)\plus{}45abc \ge 3(a\plus{}b\plus{}c)^3$
$ \Leftrightarrow a^3\plus{}b^3\plus{}c^3\plus{}3abc \ge \sum ab(a\plus{}b)$
Which is clearly true by Schur inequality.


Find the positive integers $x_1, x_2, \ldots, x_{29}$ at least one of which is greater that 1988 so that

\[ x^2_1 + x^2_2 + \ldots x^2_{29} = 29 \cdot x_1 \cdot x_2 \ldots x_{29}. \]

Start with the trivial solution $(1,1,\dots,1)$.

Then repeat the following process until $x_1$ exceeds 1988:

-Let $(x_1,x_2,\dots,x_{29})$ be a given solution.
-Replace $x_1$ by $29x_2\cdots x_{29} - x_1$ to get a new solution $(x_1',x_2',\dots,x_{29}').$
-Interchange $x_1'$ and $x_2'$ to get a new solution $(x_1,x_2,\dots,x_{29})$.
-Start over again :)

It's obvious that we can generate "arbitrarily large" solutions this way.

This problem belongs to so called Markov's equations.


Let $ A,B \in M_{n}(C)$ Such that : $ AB\minus{}BA\equal{}B$ Prove that : $ B^{n}\equal{}0$


LaTeX geometry trigonometry Asymptote Showcase

This thread is your chance to show off clever things you've found you can do in LaTeX so that they inspire others to learn more about LaTeX.

I'll start it off with a couple of things that I've already posted in another thread:
$\setlength{\unitlength}{0.5cm}\begin{picture}(5,2) \put(5,0){\line(-1,0){5}}\put(0,0){\line(1,1){2}}\put(2,2){\line(3,-2){3}}\put(-0.5,-0.1){\small A}\put(1.9,2.1){\small B}\put(5.1,-0.1){\small C}\put(4,0.1){\tiny 30}\qbezier(3.8,0)(3.9,0.55)(4.1,0.55) \put(0.6,1){\footnotesize 2}\put(3.6,1){\footnotesize 3}\put(2.3,-0.6){\footnotesize 5}\put(-1,-0.2){\tiny \shortstack{i\\m\\p\\o\\s\\s\\i\\b\\l\\e}}\put(6,-0.2){\tiny \shortstack{t\\r\\i\\a\\n\\g\\l\\e}}\end{picture}$ and $\setlength{\unitlength}{2mm}\begin{picture}(30, 20) \linethickness{0.075mm}\multiput(0, 0)(1, 0){31}{\line(0, 1){20}}\multiput(0, 0)(0, 1){21}{\line(1, 0){30}}\linethickness{0.15mm}\multiput(0, 0)(5, 0){7}{\line(0, 1){20}}\multiput(0, 0)(0, 5){5}{\line(1, 0){30}}\linethickness{0.3mm}\multiput(5, 0)(10, 0){3}{\line(0, 1){20}}\multiput(0, 5)(0, 10){2}{\line(1, 0){30}}\end{picture}$

OK, over to you. Even if you've already posted something good elsewhere then do repeat it here (hint to i_like_pie :))

[quote="i_like_pie"]Should I do some complicated stuff?[/quote]Yes, if you think it will be helpful to others and perhaps inspire them.

Lots of reading - see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117360]LaTeX How-Tos[/url] for where to find the information

[quote="anirudh"]How did you guys learn that stuff?[/quote]
[url]http://tobi.oetiker.ch/lshort/lshort.pdf[/url]

Graphics is Chapter 5 (page 105).

[quote="roadnottaken"]Note that if you hold the cursor over a $\LaTeX$ image, the source code will be displayed in a box.  I realize that this is not quite as easy to read, but it is something to keep in mind.[/quote]
I noticed that, too, but it only displays the first line or so.

To get the entire code, you can do one of two things:

1. Quote the post and copy the code.
2. If the topic is locked (you can't quote anymore), right-click the image and click Properties. The code is in Miscellaneous Properties.

[quote="chess64"][quote="anirudh"]Doesn't it take a looooong time to type that all up...

Isn' it faster to do with Geometer's sketchpad? :huh:[/quote]

Eww, geometer's sketchpad is so ugly. Besides you can't even export it to TeX or ps.[/quote]

If you're going to use a geometry program to do that stuff try Kig (http://edu.kde.org/kig/) I think with enough work you can make KDE and thus kig run under Windows.

Here are some [url=http://zoonek.free.fr/LaTeX/LaTeX_samples_title/0.html]cool title pages[/url].

There's some stunning stuff at [url=http://www.tug.org/texshowcase/]The TeX showcase[/url]

Another way is the hover your mouse onto the code, which allows you to see shorter codes. :)

Does this aslo include asymptote or just $\LaTeX$?
I have some interesting stuff in asympote.
I can't make an image in $\LaTeX$ yet.

[quote="kangrui (on the Test-Forum)"]$\text{I am writing this with\ \LaTeX. Cool!}$[/quote]   :coolspeak:  :coolspeak: :coolspeak:

This makes me wonder if you can make latex images bigger.

Not quite too clever I suppose but pretty nice:

$\left (\frac{^{^{\text{The}}}\text{ \LaTeX y}}{\underline{|| Kitchen ||}}\right )^ {\Uparrow \sim} - \ \text{A Concoction from the \ \LaTeX y Kitchen.}$

[code]$\left (\frac{^{^{\text{The}}}\text{ \LaTeX y}}{\underline{|| Kitchen ||}}\right )^ {\Uparrow \sim} - \ \text{A Concoction from the \ \LaTeX y Kitchen.}$[/code]

[asy]
import graph;
size(300);
real f(real t) {return 89cos(10*t) - 8sin(10*t) + 2cos(t);}
path g=polargraph(f,-10pi,10pi,10000, operator--);
draw(rotate(1)*g,yellow);
draw(rotate(7)*g,purple);
draw(rotate(6)*g,indigo);
draw(rotate(7)*g,green);
draw(rotate(9)*g,blue);
draw(rotate(11)*g,black);
draw(rotate(13)*g,darkgreen);
draw(rotate(15)*g,orange);
draw(rotate(17)*g,pink);[/asy]
 Try it.

That's asymptote. Is there a way to do that using latex?

I want to do sth special as $\sqrt[\sqrt{m}]{m}$

but doing with \sqrt[\sqrt[4]{m}]{m} , why this doesn't go?

$\sqrt[\sqrt[4]{m}]{m}$
$\sqrt[\sqrt[4]{m}]{m}$$\sqrt[\sqrt[4]{m}]{m}$
$\sqrt[\sqrt[4]{m}]{m}$

You need to group the inner \sqrt using {} so it is treated as one character \sqrt[{\sqrt[4]{m}}]{m} $\sqrt[{\sqrt[4]{m}}]{m}$


function
induction
calculus
integration
quadratics
modular arithmetic
algebra

Find all function $ f : \mathbb{Z} \to   \mathbb{Z}$ such that :

   $ f(x\plus{}y\plus{}f(y)) \ \equal{} \ f(x) \plus{}2y \ \forall \ x;y \ \in \  \mathbb{Z}$

hi!
for $ x\equal{}0$ we get that f is bijective,so for $ x\equal{}y\equal{}0$ ==> $ f(0)\equal{}0$
then we put $ a\equal{}y\plus{}f(y)$ we get : $ f(x\plus{}a)\equal{}f(x)\plus{}f(a)$ and since $ f: \mathbb{Z}\to\mathbb{Z}$
we conclude that: $ f(x)\equal{}x$.

Not hard at all :)

  $ f(x \plus{} y \plus{} f(y)) \equal{} f(x) \plus{} 2y$  (*)
  Assume $ f(x) \equal{} f(y)$.  Than,   $ f(x) \plus{} 2y \equal{} f(x \plus{} y \plus{} f(y)) \equal{} f(y \plus{} x \plus{} f(x)) \equal{} f(y) \plus{} 2x \equal{} f(x) \plus{} 2x$, from which $ x \equal{} y$. So, $ f$ is injective.
Take $ y \equal{} 0$ in (*). Using the injectivity and $ f(x \plus{} f(0)) \equal{} f(x)$, we conclude $ x \plus{} f(0) \equal{} x$, which yields $ f(0) \equal{} 0$. Hence, when $ x \equal{} 0$ in (*), we get:  $ f(y \plus{} f(y)) \equal{} 2y$.  (1)
   Plug $ a \plus{} f(a)$ instead of $ x$ and $ b$ instead of $ y$ in (*):
     Using (1), (*), we get:   $ 2(a \plus{} b) \equal{} f(a \plus{} f(a)) \plus{} 2b \equal{} f(a \plus{} b \plus{} f(a) \plus{} f(b))$, but from (1), $ 2(a \plus{} b) \equal{} f(a \plus{} b \plus{} f(a \plus{} b))$. The last $ 2$ equations together yield  $ f(a \plus{} b \plus{} f(a \plus{} b)) \equal{} f(a \plus{} b \plus{} f(a) \plus{} f(b))$, but as $ f$ is injective,   $ a \plus{} b \plus{} f(a \plus{} b) \equal{} a \plus{} b \plus{} f(a) \plus{} f(b)$ and  $ f(a \plus{} b) \equal{} f(a) \plus{} f(b)$  (**);     (**) means that  $ f(n \plus{} 1) \equal{} f(n) \plus{} 1$ and $ f(n \minus{} 1) \equal{} f(n) \minus{} 1$, for any integer $ n$. From this, using $ f(0) \equal{} 0$, the straightforward induction is enough to prove that $ f(x) \equal{} x$ holds for any $ x$ in Z. Easy to verify that the solution indeed works.  :)

Unfortunately, the solutions posted are both incorrect. The correct solution is as follows:

$ f(x \plus{} y \plus{} f(y)) \equal{} f(x) \plus{} 2y \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}(*)$
Setting $ y \equal{} 0$ in $ (*)$, we get
$ f(x \plus{} f(0)) \equal{} f(x)$
If $ f(0) \neq 0$, then $ f$ is a periodic function and hence bounded. This would imply $ f(x) \minus{} f(y)$ is also bounded.
But from $ (*)$, we have
$ f(x \plus{} y \plus{} f(y)) \minus{} f(x) \equal{}  2y$ and $ 2y$ is unbounded. Hence, $ f(0) \equal{} 0$.
Suppose $ f(\alpha) \equal{} 0$.
Setting $ x \equal{} 0$ and $ y \equal{} \alpha$ in $ (*)$, we get
$ f(\alpha \plus{} f(\alpha)) \equal{} 2\alpha \implies 0 \equal{} 2\alpha \implies \alpha \equal{} 0$.
In other words, $ f(x) \equal{} 0 \iff x \equal{} 0$.
Setting $ x \equal{} 0$ in $ (*)$, we get
$ f(y \plus{} f(y)) \equal{} 2y$
Setting $ x \equal{} u \plus{} f(u)$ in $ (*)$, we get
$ f(u \plus{} f(u) \plus{} y \plus{} f(y)) \equal{} f(u \plus{}f(u)) \plus{} 2y \implies f(u \plus{} f(u) \plus{} y \plus{} f(y)) \equal{} 2u \plus{} 2y$.
Setting  $ u \equal{} \minus{}y$ in the above equation, we get
$ f(f(\minus{}y) \plus{} f(y)) \equal{} 0 \implies f(\minus{}y) \plus{} f(y) \equal{} 0 \implies f(\minus{}y) \equal{} \minus{}f(y)$.
Setting $ x \equal{} \minus{}y$ in $ (*)$, we get
$ f(f(y)) \equal{} f(\minus{}y) \plus{} 2y \implies f(f(y)) \equal{} \minus{}f(y) \plus{} 2y \implies f(f(y)) \plus{} f(y) \equal{} 2y$.
Setting $ y \equal{} f(y)$ in $ (*)$, we get
$ f(x \plus{} f(y) \plus{} f(f(y))) \equal{} f(x) \plus{} 2f(y) \implies f(x \plus{} 2y) \equal{} f(x) \plus{} 2f(y)$.
From here, one can easily conclude by simple induction that $ f(n) \equal{} cn \;\;\forall n \in \mathbb{Z}$ where $ c \equal{} f(1)$.
Substituting, $ f(x) \equal{} cx$ in $ (*)$, we get
$ cx \plus{} cy \plus{} c^2y \equal{} cx \plus{} 2y \implies c^2 \plus{}c \equal{} 2 \implies c \equal{} 1$ or $ c \equal{} \minus{}2$.
Hence, the solutions to the given functional equation are $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{}2x$.

I think my solution is correct,since $ f(x\plus{}a)\equal{}f(x)\plus{}f(a)$ with $ a\equal{}f(y)\plus{}y$ and since f is surjective so $ f(y)$ can take any value in $ \mathbb{Z}$ then $ f(x)\equal{}cx$ / $ c\in{\mathbb{Z}}$
then we put $ f(x)\equal{}cx$ in our f.e we get: $ c^{2}\plus{}c\equal{}2$ ==> $ c\equal{}\minus{}2$ or $ c\equal{}1$.

When $ f(x) \equal{} \minus{} 2x$ , $ f$ is not a bijection. I do not understand , how you were able to conclude that $ f$ is bijective in the first place.  Your solution is not correct. In fact, it is not even clear how you conclude $ f(0) \equal{} 0$. I would like to point out that we are working in integers and not real numbers. Hence, setting  $ x \equal{} 0$ gives you $ f(y \plus{} f(y)) \equal{} 2y$ which only shows that $ f$ assumes even integers and does not say anything about odd integers.

[quote="nguyenvuthanhha"][quote="nguyenvuthanhha"]Find all function $ f : \mathbb{Z} \to \mathbb{Z}$ such that :

   $ f(x \plus{} y \plus{} f(y)) \ \equal{} \ f(x) \plus{} 2y \ \forall \ x;y \ \in \ \mathbb{Z}$[/quote]

 [i] Yes , Although the problem is very easy , there is only one correct solution which belongs to nicetry007 , Congrats

    The official solution is a bit shorter than nicetry007's one   :wink:

   I can post it if someone want to see[/i][/quote]

Please post the official solution. I would like ot see it. Nice solution by nicetry007.

Could someone please tell me what's wrong in my solution? I can't find anything wrong in it. If the problem is that $ a\plus{}f(a)$ doesn't take any integral value, I say that as the given equation is true for any pairs $ x,y$, I can clearly take $ x\equal{}a\plus{}f(a)$ for any integer $ a$, because $ a\plus{}f(a)$ will be an integer and I plug this value in the original equation. It's sometimes hard to understand. Is there anything other?

Anyway, the mistake in Reason's solution is indeed obvious because you get $ f(x\plus{}a)\equal{}f(x)\plus{}f(a)$ where $ a$ is of the form $ x\plus{}f(x)$, where $ x$ is an integer and you don't know that the set of all values of $ x\plus{}f(x)$ is exactly the set of all integers. :)

[i]nguyenvuthanhha has found mistake in Lasha 's solution ,

  Actually ,  To prove $ f(n\plus{}1) \equal{} f(n) \plus{}1$ , you must prove $ f(1) \equal{}1$

  and that fact hasn't been mentioned in your post , Lasha  :) [/i]

[quote="lasha"]Anyway, the mistake in Reason's solution is indeed obvious because you get $ f(x \plus{} a) \equal{} f(x) \plus{} f(a)$ where $ a$ is of the form $ x \plus{} f(x)$, where $ x$ is an integer and you don't know that the set of all values of $ x \plus{} f(x)$ is exactly the set of all integers. :)[/quote]

yes you're right  :) , I didn't pay attention.

Yes but after $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ in $ Z$ it's well-known fact that the only solution is $ f(x) \equal{} kx$ where $ k \equal{} f(1)$ is a integer constant.
Easy to get that $ f(1) \equal{} 1$ after plugging it back to the condition. Hence, $ f(x) \equal{} x$. If it's unclear for you, I'll prove that but iot's too boring to post the well-known easy things. :) I forgot to mention that $ k\equal{}\minus{}2$ is also the solution and hence, $ f(x)\equal{}\minus{}2x$ is solution as well.

[i]In order to avoid spamming , I will post the solution and if Lasha read it carefully , he can easily see that he had made a very silly  mistake :

   We have $ f(x \plus{} y \plus{} f(y)) \equal{} f(x) \plus{} 2y \ \forall \ x;y \in \mathbb{Z} \ \ (1)$

   Setting $ x \equal{} 0$ in $ (1)$ :$ f(y \plus{} f(y)) \equal{} 2y \ \forall \ y \in \mathbb{Z} \ \ (2)$

  From $ (1) \ \rightarrow f ( x \plus{} y \plus{} f(y) ) \plus{} x \plus{} y \plus{} f(y) \equal{} f(x) \plus{} 3y \plus{} f(y) \plus{} x \ \forall \ x;y \in \mathbb{Z}$

 $ \rightarrow f( f ( x \plus{} y \plus{} f(y) ) \plus{} (x \plus{} y \plus{} f(y)) ) \equal{} f( (3y \plus{} f(y)) \plus{} x \plus{} f(x)) \ \forall \ x;y \in \mathbb{Z}$

  $ \rightarrow 2(x \plus{} y \plus{} f(y)) \equal{} f(f(y) \plus{} 3y) \plus{} 2x \ \forall \ x;y \in \mathbb{Z}$ ( using $ (2)$ )

   $ \rightarrow 2 (y \plus{} f(y)) \equal{} f( 2y \plus{} y \plus{} f(y)) \ \forall \ y \in \mathbb{Z}$

   $ \rightarrow 2 (y \plus{} f(y)) \equal{} f(2y) \plus{} 2y \ \forall \ y \in \mathbb{Z} \ \rightarrow \boxed{ f(2y) \equal{} 2f(y) \ \ (3) }$

  Change $ y$ to $ y \plus{} f(y)$ in $ (1)$ , we have :

$ f(x \plus{} (y \plus{} f(y)) \plus{} f(y \plus{} f(y))) \equal{} f(x) \plus{} 2(y \plus{} f(y)) \ \forall \ x;y \in \mathbb{Z}$

 $ \rightarrow f( (x \plus{} 2y) \plus{} y \plus{} f(y)) \equal{} f(x) \plus{} 2(y \plus{} f(y)) \ \forall \ x;y \in \mathbb{Z}$

  $ \rightarrow f(x \plus{} 2y) \plus{} 2y \equal{} f(x) \plus{} 2(y \plus{} f(y)) \ \forall \ x;y \in \mathbb{Z}$

   $ \rightarrow f(x \plus{} 2y) \equal{} f(x) \plus{} 2f(y) \ \forall \ x;y \in \mathbb{Z} \rightarrow f(2x \plus{} 2y) \equal{} f(2x) \plus{} 2f(y) \ \forall \ x;y \in \mathbb{Z}$

    $ \rightarrow f(x \plus{} y) \ \equal{} \ f(x) \plus{} f(y) \ \forall \ x;y \in \mathbb{Z}$ ( using $ (3)$ )      $ \rightarrow f(x) \equal{} ax \ \forall \ x \  \in \ \mathbb{Z}$  ( $ a \equal{} f(1) \ \in \ \mathbb{Z}$ )

 So , in $ (2)$ , change $ y$ to $ x$ we have $ f(x \plus{} ax ) \equal{} 2x \ \forall \ x \in \mathbb{Z} \rightarrow a(a \plus{} 1)x \equal{} 2x \ \forall \ x \in \mathbb{Z}$

   Thus , $ a(a \plus{} 1) \equal{} 2 \iff a^2 \plus{} a \minus{} 2 \equal{} 0$

 $ a \equal{} 1$ and $ a \equal{} \minus{} 2$ are two root of this quadratic equation . So $ f(x) \equal{} x \ \forall \ x \in \mathbb{Z}$ or $ f(x) \equal{} \minus{} 2x \ \forall \ x \in \mathbb{Z}$

  An easy check can guarantee that those are all solution of our functional equation , Done :lol: [/i]

Is it the official solution? My solution seems to be much little. By the way, my mistake is too easy to be corrected. I think it's not the ain difficulty of that problem.  :lol: Cauchy equation is well-known and it should have been clear for you the rest even if I hadn't mentioned anything after that. :)

Lasha's solution is indeed very nice. It was my fault. I didn't read it carefully. My apologies. Except for the last bit, everything else is perfect.


Let $ A_1,A_2,\dots$ be the points sequence on the plane,such that for all $ n\in\mathbb{N}$,point $ A_{n\plus{}3}$ is circumcenter of the triangle $ A_nA_{n\plus{}1}A_{n\plus{}2}$.Find all $ k\in\mathbb{N}$,such that $ A_{n\plus{}k}\equal{}A_{n}$ for all $ n\in\mathbb{N}$.

Does anybody want to see a solution?Or must i wait a little bit more?


Find all functions $ f,g : \mathbb R\to\mathbb R$ such that $ f(g(x))=x^3$ and $ g(f(x))=x^4$ for all real $x$.

[quote="MJ GEO"]Find all couples $ f(x),g(x)$ of functions from $ \mathbb R\to\mathbb R$ such that $ f(g(x))\equal{}x^3$ and $ g(f(x))\equal{}x^4$ $ \forall x$[/quote]

From $ f(g(x))\equal{}x^3$, we get that $ g(x)$ is injective and that $ f(x)$ is surjective
From $ g(f(x))\equal{}x^4$ and $ f(x)$ surjective, we get that $ g(\mathbb R)\equal{}\mathbb R^\plus{}_0$
From $ g(f(x))\equal{}x^4$ and $ g(f(\minus{}x))\equal{}x^4$ and $ g(x)$ injective, we get that $ f(x)\equal{}f(\minus{}x)$ $ \forall x$
From $ g(f(x))\equal{}x^4$, we get that $ f(a)\equal{}f(b)$ $ \implies$ $ a^4\equal{}b^4$ and so $ a\equal{}b$ or $ a\equal{}\minus{}b$

So $ g(x)$ is a bijection from $ \mathbb R\to\mathbb R^\plus{}_0$
And the restriction $ f_r(x)$ of $ f(x)$ to $ \mathbb R^\plus{}_0$ is a bijection from $ \mathbb R^\plus{}_0\to\mathbb R$

Let then $ h(x)$ from $ \mathbb R\to\mathbb R^\plus{}_0$ the inverse of $ f_r(x)$ (so a bijection too)

From $ f_r(g(x))\equal{}x^3$, we get that $ g(x)\equal{}h(x^3)$
From $ g(f_r(x))\equal{}x^4$, we get that $ g(x)\equal{}h(x)^4$

So $ h(x^3)\equal{}h(x)^4$ $ \forall x\in\mathbb R$ and so :
$ h(\minus{}1)\equal{}h(\minus{}1)^4$ and $ h(\minus{}1)\in\{0,1\}$ (remember that $ h(\mathbb R)\equal{}\mathbb R^\plus{}_0$)
$ h(0)\equal{}h(0)^4$ and $ h(0)\in\{0,1\}$ (remember that $ h(\mathbb R)\equal{}\mathbb R^\plus{}_0$)
$ h(1)\equal{}h(1)^4$ and $ h(1)\in\{0,1\}$ (remember that $ h(\mathbb R)\equal{}\mathbb R^\plus{}_0$)

And so at least two of the three values $ h(\minus{}1),h(0)$ and $ h(1)$ must be equal but this is impossible since $ h(x)$ is a bijection.

So no solution to this equation.


find all $ a,b,m,n\in\mathbb{N}$, with $ (m,n)\equal{}1$ such that \[ (a^2\plus{}b^2)^m\equal{}(ab)^n\]

He's just saying that that's one set of numbers that works.

[hide="Hint"]Suppose $ p$ divides $ (ab)^n$. Then $ p$ must divide $ ab$ as well.

Using a similar argument, $ p$ must divide $ a^2 \plus{} b^2$. Therefore, $ a^2 \plus{} b^2$ and $ ab$ have identical lists of prime factors.

Continue this argument.[/hide]

I think this [hide="Lemma"] $ a,b,x,y \in \mathbb Z$ and $ a,b,x,y>1$ $ (a,b)\equal{}1$ If $ x^a\equal{}y^b$ then there exist integer $ n>1$ such that $ x\equal{}n^b$ and $ y\equal{}n^a$  [/hide] can help for the solution...

[quote="TZF"]Suppose $ p$ divides $ (ab)^n$. Then $ p$ must divide $ ab$ as well.

[/quote] I disagree with this.  Take $ n>2$ and $ p\equal{}(ab)^2$, but clearly $ (ab)^2$ does not divide $ ab$ with $ ab\neq 1$.

Good call, I forgot to explicitly mention that $ p$ is prime.


function
induction
number theory proposed
number theory

Show that for every $n \in \mathbb{N}$ we have:
$2^n | 1+[(3+\sqrt{5})^n]$

Is that greatest integer function on the exponential part???


Sa se arate ca un tetraedru in care muchiile opuse sunt congruente si formeaza unghiuri congruente este regulat.

exprima volumul ca $\frac{1}{6}sin (AB, CD)*AB*CD*d(AB;CD)$. apoi obtii ca $AB^{2}*d(AB;CD)=BC^{2}*d(BC;AD)$. dar cum tetraedrul este echifacial distanta dintre doua drepte necoplanare va fi chiar bimediana lor. se stie ca bimediana se calculeaza ca $MP^{2}=AD^{2}+BC^{2}+AC^{2}+BD^{2}-AB^{2}-CD^{2}$, unde MP este bimediana laturilor $AB$ si $CD$. si de aici cred ca iese cu calcule.

iese usor si vectorial... se tine cont de:

         $cos(\vec{AB},\vec{CD})= \frac{AD^{2}+BC^{2}-AC^{2}-BD^{2}}{2AB \cdot CD}$


modular arithmetic
number theory
relatively prime
algebra
system of equations

[hide="1"]Umm, simply divide and get $ \frac{2007}{9}\equal{}223$, so the units and tens digits are 3 and 2, respectively.[/hide]

[hide="2"]We want to know $ 2^{19}\pmod{100}$, or, $ 2^{19}\pmod{25}$ and $ 2^{19}\pmod4$.

$ 2^{19}\equiv0\pmod4$

$ 2^{\phi25}\equiv1\pmod{25}$

$ 2^{20}\equiv1\pmod{25}$

$ 2^{19}\equiv13\pmod{25}$

Combining these, we have that $ 2^{19}\equiv88\pmod{100}$.[/hide]

Not really sure what you want on #3...

[quote="binomial_4eva"][color=blue]1[/color]:what is the unit and tenth digits when $ 2007$ is divided by $ 9$?

[color=red]2[/color]:what is the unit and tenth digits of $ 2^{19}$?

[color=green]3[/color]:express the unit and tenth digit when $ a\plus{}b\plus{}c\plus{}d$ is divided by $ 2$[/quote]

[hide="1"]$ 2007/9 \equal{} 223$

So $ 3$ and $ 2$? I really don't see the point of this...[/hide]
[hide="2"]$ 2^{9}\equal{} 512\equiv 12\pmod{100}$
$ 2^{18}\equiv 144\equiv 44\pmod{100}$
$ 2^{19}\equiv 88\pmod{100}$

So, the last two digits are $ \boxed{88}$.

But, I'd rather just compute $ 2^{19}$:
$ 2^{19}\equal{} 524288$[/hide]
3 makes no sense

[quote="binomial_4eva"][quote][hide="2"]We want to know $ 2^{19}\pmod{100}$, or, $ 2^{19}\pmod{25}$ and $ 2^{19}\pmod4$.

$ 2^{19}\equiv0\pmod4$

$ 2^{\phi25}\equiv1\pmod{25}$

$ 2^{20}\equiv1\pmod{25}$

$ 2^{19}\equiv13\pmod{25}$

Combining these, we have that $ 2^{19}\equiv88\pmod{100}$.[/hide][/quote]

why $ \pmod{100}$ or $ \pmod{25}$ or $ \pmod{4}$ can i choose any number instead of 100 there?

and where did get $ 0$ and $ 1$ from in $ 2^{19}\equiv0\pmod4$ and $ 2^{20}\equiv1\pmod{25}$

can you explain in abit more detail please.[/quote]

[hide="Explanation"]b-flat used the Chinese Remainder Theorem (I just learned this recently, so please correct me if this explanation has any errors in it), abbreviated as CRT. Since we need to find $ \pmod{100}$ of that, we can consider $ \pmod{25}$ and $ \pmod{4}$ (since $ 25\cdot 4 \equal{} 100$).

Since $ 2^{19}\equiv 13\pmod{25}$, we have that $ 2^{19}\equal{} 13\plus{}25j$ for some integer $ j$. And since $ 2^{19}\equiv 0\pmod{4}$,

$ 13\plus{}25 j\equiv 0\pmod{4}$

To find the possible residues for $ j$, just plug in some really small numbers. We get that $ 3$ works. So $ j\equiv 3\pmod{4}$. So let $ j \equal{} 3\plus{}4k$:

$ 2^{19}\equal{} 13\plus{}25 j\equiv 13\plus{}25\cdot (3\plus{}4k) \equal{} 88\plus{}100k\equiv 88\pmod{100}$.[/hide]

EDIT: What's wrong with me today?

@vishalarul:  don't you mean $ j\equiv3\pmod4$?

CRT says that if we have the system of equations:

$ N\equiv a_{1}\pmod{b_{1}}$
$ N\equiv a_{2}\pmod{b_{2}}$
$ \vdots$

And the $ b$'s are pairwise relatively prime, then there exists such a positive integer, $ N$ (in fact, there are infinitely many $ N$, spaced out at intervals of $ b_{1}\cdot b_{2}\cdots$

In the case of this problem, we have that $ b_{1}\equal{}4$ and $ b_{2}\equal{}25$.

As for the choice of 100, taking any number mod 100 is the same as taking the last two digits, which is what we wanted.

$ 2^{19}\equiv2^{2}\cdot2^{17}\equiv4\cdot2^{17}\equiv0\pmod4$

Euler's Theorem states that for any $ a$ and $ b$ such that $ GCD(a,b)\equal{}1$, we have that:

$ a^{\phi b}\equiv1\pmod b$

Any thing else I need to explain?


Let $ a,b,c$ be positive numbers.
(a) If $ a\le b\le c$, then

$ \sqrt\frac{3a}{a\plus{}2b}\plus{}\sqrt\frac{3b}{b\plus{}2c}\plus{}\sqrt\frac{3c}{c\plus{}2a}\le 3$;

(b) If $ a\ge b\ge c$, then

$ \sqrt\frac{a}{a\plus{}3b}\plus{}\sqrt\frac{b}{b\plus{}3c}\plus{}\sqrt\frac{c}{c\plus{}3a}\ge\frac {3}{2}$.

(a) Let $ \frac{b}{a}\equal{}x,\ \frac{c}{b}\equal{}y\Longrightarrow\frac{a}{c}\equal{}\frac{1}{xy}\ (x\geq 1,\ y\geq 1)$, we can write the given inequality as

$ \frac{1}{\sqrt{1\plus{}2x}}\plus{}\frac{1}{\sqrt{1\plus{}2y}}\plus{}\frac{1}{\sqrt{1\plus{}2\cdot\frac{1}{xy}}}\leq\sqrt{3}$.


I've been working through the amc10s... almost out of them.. I can't find the 2007B anywhere.. does anyone have it that they're willing to share for the good of others?  :lol: thanks in advance

Buy a copy. I have spent a lot of time searching for it but haven't been able to find it.

i bought the book of AMC 10's- its very good

I bought a book, but it only had 2000-2005 tests. Thanks for the link

I f you purchase the 2008-2009 AMC 10/AMC 12 Math Club package you get a CD that has all ( 2000-2008 ) of the AMC 10 contests and the solutions, as well as the all the AMC 8s from 1998-2007, all the AMC 12s from 1999-2008 plus all the AIMEs 2001-2008, and the USAMOs 2001-2008.  As a bonus, there is some of the study material and tests from MOSP 2007 and 2008.  In the Math Club Package booklet are two different indexes of problems categorized by NCTM Standard and also by content using the MathWorld.com orgnaization scheme.  There are also about 10 Topic quizzes that collect problems from many years on a single topic such as Circle Geometry.

We have also just today completed two new CDs that have all the old AHSME tests.  Volume I has all the years 1950 - 1974 and Volume II has all the years 1975-1999.  Included are both the contests and the solutions.

You can order these by using the order blank at:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/2007-2008Pub,Revision.pdf
or by calling the AMC office.  All contact information is on our web page of course, and also on all of our printed materials.

The two new CDs are so new that we may not have them listed on the order form, so if you don't find a space to order them, just call or email.

Steve Dunbar
MAA Director, American Mathematics Competitions

Mr. Dunbar:

Do you have any resources that are sold that includes old AIME exams other than Contest Problem Book V?

Yes, in about 2 weeks or so, we will have a CD that will include all the AIMEs from the beginning to 1999 together with all the USAMOs from the beginning to 1999.  We are busy preparing that disk now.  The current Math Club Package disk will have all the AIMEs 2000-2008 together with all the USAMOs 2000-2008 and some additional material.

Steve Dunbar
MAA Director, American Mathematics Competitions.

To support AMC?  Also  you said "almost all".  YOU WANT ALL!!!!

I'm going to buy the AHSME vol.1 and 2 on monday.

Here are some more previous tests if you do not have them.

[url]http://www.math.ksu.edu/main/handbook/ProblemSets[/url]

[quote="mathemonster"]...almost all the aimes are on aops- why buy them????[/quote]There are a bunch of advantages: You want to support the AMC, or you want them in the actual form they are given on test day, or you don't want to deal with bad/incomplete diagrams, etc.

[quote="annoyingpi"]I found this using Google:
[url]http://www.unl.edu/amc/mathclub/4-3,contests/2-AMC10pdfs/2007AMC10B.pdf[/url]

Going up a few directories, I found this:
[url]http://www.unl.edu/amc/mathclub/4-3,contests/[/url]

I hope that's helpful.[/quote]

How the heck did you find that?

its in your quote...

I found this using Google.

ROFL

btw, the links are screweed up: theres a bug

Something's weird with the links. Its including the < character which mysteriously appeared, and I can't make it go away. Let's try this:

[url=http://www.unl.edu/amc/mathclub/4-3,contests/2-AMC10pdfs/2007AMC10B.pdf]Link for file[/url]
[url=http://www.unl.edu/amc/mathclub/4-3,contests/]Link for the whole directory[/url]

I searched for "amc 10 2007b filetype:pdf"

Apparently access to it is denied now lol

Thanks a lot, annoyingpi! Now that the AMC people found out about the link, it is no longer open to all.  :P


Please everyone have a look at a proposed proof of Riemann Hypothesis I found in the web.Please provide some feedback and forward it to anyone interested..

http://www.aronpalmer.com/riemann_proof.htm

Some remarks:

Not sure you need to know much about the area, the last bit about the zeta function being expressable in an even positive decreasing
function is pretty definitely correct, and the first bit about not being able to have zeros for s> 0 is pretty definately correct, the bit I am worried
about is going from the one sided transform to the two sided transform and being able to show that Odd(s) + Even(s) is a convergent polynomial.

Thanks,
Nick

Wait, so you expect us to seriously critique a proposed proof of an extremely deep result in number theory that has so far resisted all such effort?  I think you might be in the wrong forum... also, it seems like if there were a serious contender for a proof, then it would be widespread knowledge already.

Sorry nfrer, but I don't understand: is that link a "non-official proof" or Riemann Hypothesis? Or just a really sumerized way of how to solve it???? ... A proof in just 7 pages????? It took 200 to Wiles to proof Fermat...

How on earth can someone prove Riemann in 7 pages? I mean I don't understand 3/4 of that proof but it seems like it would/should be longer.. if mathematicians have struggled for so long.

There was, supposedly, a French mathematician who recently (in the past year or two) proved the Riemann hypothesis.  However, I heard his proof was being verified, and considering no big news was heard, there was probably a flaw in it.

This proof seems flawless to me.  However, I am obviously not qualified to make any such judgements.  Try submitting it somewhere else...

The current header of the site:

[i]Apologies to all, but this proof is flawed. Right at the end we need to show that we can express the one sided transform in a product form. In fact we can prove that this cannot be the case. I think this may not be fatal, but it is not trivially fixed. Please forgive optimism in remarks section/thanks etc since I am hoping not to have to retype.[/i]

LOL.  Of course, I warned you I wasn't an expert.

[quote="JBL"]Here's a 2-line proof of the Riemann Hypothesis (courtesy of my logic professor):

Let Larry be the name of the sentence:
"If Larry is true, then the Riemann Hypothesis is true."

If Larry is true, then by Larry, the Riemann Hypothesis is true.  However, this is a proof of Larry.  Thus Larry is true, and so the Riemann Hypothesis is true as well.[/quote]

Jajaja lol

:lol: "If I am not mistaken   the proof is  valid " :lol:

[quote="mathnerd314"]I believe the same person tried to prove that the power set of a countably infinite set is also countably infinite. I don't trust them anymore.[/quote]

Wasn't me honest, never posted any other maths ever anywhere.

Sorry Sean for letting you down, it took me ages to spot the "inevitable" mistake and that was looking at it every day.  Still optimistic that it can be fixed,  though it means the proof is going to be a lot longer which should please those of you who thought it was too short, but means the chances of anyone looking at in online must be close to zero.


function
algebra
domain
algebra unsolved

very simple.
for 0<=x<=1, define f(x) = g(x)  to be an arbitrary function from [0,1] such that f(1)=1.
then for every x>1, f(x)=g(1/x)
we are done.

[quote="dheardie"]Find the most general solution to f(x)=f(1/x) for x > 0 where f is continious, and has a single function over its domain (ie. no hybrid functions)[/quote]

General solution to this equation is $ f(x)\equal{}h(x)\plus{}h(\frac{1}{x})$, for any continuous function $ h(x)$ defined for $ x>0$

Demo :
$ 1)$ $ f(x)\equal{}h(x)\plus{}h(\frac{1}{x})$ with $ h(x)$ continuous implies $ f(\frac{1}{x})\equal{}h(\frac{1}{x})\plus{}h(x)\equal{}f(x)$ and $ f(x)$ continuous.

$ 2)$ $ f(x)\equal{}f(\frac{1}{x})$ an $ f(x)$ continuous implies $ f(x)\equal{}h(x)\plus{}h(\frac{1}{x})$ for $ h(x)\equal{}\frac{f(x)}{2}$ and $ h(x)$ is continuous.


[quote="Allan Z"][quote="bpms"][quote="Ignite168"]Given $log(25)=W$, find $log_{100}(160)$ in terms of W.[/quote]
[hide]
$\log 25=2\log 5$
$\log_{100}160=\log_{100}32+\log_{100}5$
$5\log_{100}2+\frac{W}{4}$
Now I'm stuck[/hide][/quote]

if you used change base formula, you should realize that $log_{100}{160}=\frac{log_{10}{160}}{log_{10}{100}}=\frac{log_{10}{160}}{2}=\frac{1+4log_{10}{2}}{2}=\frac{1+4(1-log_{10}{5})}{2}=\frac{5-4log_{10}{5}}{2}$

and you know $log5=\frac{W}{2}$
so..

$log_{100}{160}=\boxed{\frac{5-2W}{2}}$[/quote]
I didn't use change of base, but I should have remembered to do that.


Let $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}3$
Prove that
 $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c} \geq ab\plus{}bc\plus{}ca$

[quote="8826"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$
Prove that
 $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \geq ab \plus{} bc \plus{} ca$[/quote]
It very old :wink:
It is equivalent to: $ a^2\plus{}b^2\plus{}c^2\plus{}2(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}) \ge 9$
But $ a^2\plus{}2\sqrt{a} \ge 3a$
Similar,we have Q.E.D
With the same condition,we have the inequality harder and old:
$ \sqrt[3]{a}\plus{}\sqrt[3]{b}\plus{}\sqrt[3]{c} \ge ab\plus{}bc\plus{}ca$ :)

[quote="enndb0x"][quote="quykhtn-qa1"][quote="8826"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$
Prove that
 $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \geq ab \plus{} bc \plus{} ca$[/quote]
It very old :wink:
It is equivalent to: $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}) \ge 9$
But $ a^2 \plus{} 2\sqrt {a} \ge 3a$
Similar,we have Q.E.D
With the same condition,we have the inequality harder and old:
$ \sqrt [3]{a} \plus{} \sqrt [3]{b} \plus{} \sqrt [3]{c} \ge ab \plus{} bc \plus{} ca$ :)[/quote]
I can not see why they're equivalent.And also I can not see why holds this $ a^2 \plus{} 2\sqrt {a} \ge 3a$[/quote]

I think the solution is right.
The first step:mutiply 2 in the both sides,then add $ a^2 \plus{} b^2 \plus{} c^2$ both sides too,you can get  $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}) \ge 9$
The last step:Just use AM-GM, $ a^2 \plus{} 2\sqrt {a}$ = $ a^2 \plus{} \sqrt {a} \plus{} \sqrt {a} \ge ....$


geometry
geometric transformation
reflection
rotation
rectangle
perimeter
function

What is the smallest region in $ \mathbb{R}^2$ (in terms of area) that can fit any closed curve of length $ 1$ (after some displacement, reflection, and rotation)?

for some reason i want to WLOG it to rectangular curves, but i can't quite justify that.  maybe triangles?

A circle of radius 1, for example, does not fit inside any rectangle or triangle with perimeter 1.

well a circle of radius 1 has length $ 2\pi$, so its not considered..?

this problem seems hard to find a starting point

You're right, I should have written a circle with perimeter $ 1$ doesn't fit inside any rectangle or triangle with perimeter $ 1$ (circles have the largest area of all shapes with fixed perimeter).

Here's what seems like a good starting point to me: I [i]think[/i] we can consider without loss of generality only convex sets.  If we take the diameter of the set to be the $ x$-axis, then the halves of the curve above and below the axis can be seen as graphs of two functions of $ x$, which we can maximize at each point in the space of continuous functions with the perimeter constraint.  Unfortunately, the answer this approach yields isn't minimal, since it's not always true that the optimal positioning of a given shape is the one that places its diameter along the axis.


[quote="yhh3002"]Can you post your score in Michigan and other people can look at this too.
Tell me what chapter you were in...

So I can kind of get an estimation of where I stand at States
 :D[/quote]
This should be in the Michigan forum.

yeah, but in the Michigan Forum, there are only one middle schooler that joined the Michigan (private) forum, so I'm asking the whole entire mathcounters...

This means there is only one michigan middle schooler on AoPS :D


$ \begin {cases} x\plus{}y\plus{}z\equal{}6\\ x^2\plus{}y^2\plus{}z^2\equal{}18\\\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z}\equal{}4\end {cases}$

Replace $ x,y,z$ with their squareroots to get

$ \begin {cases} x^2 \plus{} y^2 \plus{} z^2 \equal{} 6 \\
x^4 \plus{} y^4 \plus{} z^4 \equal{} 18 \\
x \plus{} y \plus{} z \equal{} 4\end {cases}$

Let $ p \equal{} x \plus{} y \plus{} z \equal{} 6, q \equal{} xy \plus{} yz \plus{} zx, r \equal{} xyz$. By the identity $ x^2 \plus{} y^2 \plus{} z^2 \equal{} p^2 \minus{} 2q$, we get $ q \equal{} 5$. And by the identity $ x^4 \plus{} y^4 \plus{} z^4 \equal{} p^4 \minus{} 4p^2q \plus{} 2q^2 \plus{} 4pr$ we get $ r \equal{} 2$.

Hence $ x,y,z$ are roots of the equation $ x^3 \minus{} px^2 \plus{} qx \minus{} r \equal{} x^3 \plus{} 4x^2 \plus{} 5x \plus{} 2 \equal{} (x \minus{} 1)^2(x \minus{} 2)$. That is: $ (x,y,z)$ is a permutation of $ (1,1,2)$. Taking squares again we solve the system, and get

$ (x,y,z)\in\{ (1,1,4), (1,4,1),(4,1,1)\}$

I can done it by other way. I will post it later.


How many integers from 1 through 9999, inclusive, do not
contain any of the digits 2, 3, 4 or 5?

Wait...where did the $ 6^4$ come from again?

Personally, I prefer casework counting to complementary counting  :lol:

Gosh dang it I wrote 1296 the first time

[quote=AIME15]This is $ 6^4\equal{}1296$.[/quote]

Subtract 0000 so it should be 1295

[hide]A number $n$ has an odd number of divisors if and only if it is a perfect square. To see this, notice that the divisors $d$ and $n/d$ pair up, except for when $d=n/d$, or $n=d^2$. :-D  :blush:

Therefore, the only integers not counted are the perfect squares. Since $45^2=2025$ and $44^2=1936$, there are 44 positive integers less than 2008 with an odd number of divisors, leaving $2007-44=\boxed{1963}$ positive integers less than 2008 with an even number of divisors. :-D [/hide]

Thanks! I understand the problem so much better now!!!  :D

[quote=r15s11z55y89w21]I don't think the solution provided is the easiest. Isn't this way easiest?

In all numbers:
First digit can be $ 1$, $ 6$, $ 7$, $ 8$, $ 9$ ($ 5$ choices); following digits can be those plus $ 0$ ($ 6$ choices)

Case #1: 4-digit number: $ 5\times6\times6\times6\equal{}1080$
Case #2: 3-digit number: $ 5\times6\times6\equal{}180$
Case #3: 2-digit number: $ 5\times6\equal{}30$
Case #4: 1-digit number: $ 5\equal{}5$

$ 1080\plus{}180\plus{}30\plus{}5\equal{}\fbox{1295}$

I don't know, maybe it's just me. I think this way is much easier to think through.[/quote]

It isn't just you, I got it the same way.

[quote=Bestestsocks]why is this problem in "complementary counting"?[/quote]

because you can complementary count it?

I would just like to say that this problem taught me that it was a lot easier to have the leading zeroes be allowed for things like this  :love:

thanks for coming to my ted talk <3

I don’t get why we need to calculate $6^4$. It seems really big.

No, it's really not. It's just 6^2*6^2, which is 36^2, which is (40-4)(40-4) which is 1600-240-240+16 which is just 1296. It's possible to do in your head.

[quote=AIME15]Ugh, I forgot to subtract $ 1$.  But if you remember to do that, mine is way easier.

$ 6^4\minus{}1\equal{}\boxed{1295}$.[/quote]

me too...

[hide=Hint]Do casework for 1/2/3/4 digit integers[/hide]

Can I get an f for 1296. [hide=Sol] 6 options for each digit, so $6^4=1296$. BUTTTTTTTTTTTTTTTTTTTTTTTTTTT SUBTRACT 0! $1296-1=\boxed{1295}$[/hide]

I did this digit by digit

There are 5 numbers with 1 digit. 30 numbers with 2 digit's. 180 numbers with 3 digits, and 1080 with 4 digit.

5+30+180+1080 = 1295

In my opinion, this way is easier since you don't need to remember to minus 1 for 0000


[hide="Idea..."]
$\cos x(8\cos^{2}x+\sqrt3)=7\sin x \rightarrow$$8\cos^{2}x+\sqrt3=7\tan x$.
Now, $\cos^{2}x=\frac{1}{1+\tan^{2}x}$, so we get
$\frac{8}{1+\tan^{2}x}-7\tan x+\sqrt3=0$. Taking away the denominators and defining $z=\tan x$, it becomes
$7z^{3}-\sqrt3z^{2}+7z-(8+\sqrt3)=0$, which can be factored in $(7z-\sqrt3)(z^{2}+1)-8=0$... even if I'm not sure that this helps... but I think that the idea of getting an equation with $\tan$ does...
[/hide]

[quote="ashrafmod"]$8\cos^{3}x-7 \sin x+\sqrt{3}\cos x=0$[/quote]

[hide="Solution"]$8\cos^{3}x-7 \sin x+\sqrt{3}\cos x=0$
$8\cos^{2}x-7\tan x+\sqrt{3}= 0$
$\frac{8}{\tan^{2}x+1}-7\tan x+\sqrt{3}= 0$
$8-7\tan^{3}x-7\tan x+\sqrt{3}\tan^{2}x+\sqrt{3}= 0$

Let $y = tan x$

$7y^{3}-y^{2}\sqrt{3}+7y-(8+\sqrt{3}) = 0$

This thing has no rational roots... However, it is possible to find that
$y \approx 0.887035$

So $x = 41.5742 \textrm{ OR }221.5742$[/hide]

well, then my solution is complete, except for the values of the unknown! Thanks, vishalarul!...  :)

why are you looking for rational roots?

Common values of tangent include $\sqrt{3}$ and $\frac{\sqrt{3}}{3}$ which aren't rational either

there must be a more exact way of going about this, because if you are just going to use the calculator, you might as well use it right at the beginning


Find  $ Lim x^{x^x \minus{} 1}$  (x->0+)

It is well known that $ e^t \equal{} 1 \plus{} t \plus{} \frac {t^2}{2!} \plus{} ....$ by Taylor expansion.
Let $ t \equal{} xlnx$. Then when $ x\rightarrow 0^ \plus{}$, then $ t\rightarrow 0^ \minus{}$ by L'Hospital's theorem.  So, when $ x\rightarrow 0^ \plus{}$, $ e^{xlnx} \equal{} 1 \plus{} xlnx \plus{} \frac {(xlnx)^2}{2!} \plus{} ......$.
 So, when $ x\rightarrow 0^ \plus{}$, $ e^{xlnx}\approx 1 \plus{} xlnx$.


Yes, I know these problems are stupid, but hey, I was the only one on our math team that got them within two seconds...

1.  What is the sum of the last two digits of 5^2000 ?
2.  What is the last digit of 6^2000 ?

 :icecream: have fun!!

(I still know these problems are dumb  )

[hide]1. 7
2. 6[/hide]
Yes, I agree with you, they are really easy...  :P

The answers to those 2 qusetions and then right them in order of tens digit for number 1 and ones digit for number 2 will get you 76, which are the last two digits of 6^2000.

wow...
i didn't know that..  :coolspeak:
i'm only going into eighth grade...
thanks..
did you actually put 6^2000 to get the last two digits 76??
i dont think my TI-89 could do that..
oh well.
whatever..

TI-89 might be able too.... I'm not sure....

Dude no, you dont use caculators...

just observe 6^5 is 76 mod 100
so is 6^10 is  76 mod 100

infact every 6^{5n} is 76 mod 100 so is 6^2000 then.

for the first one: it's not hard because the tens and ones digit of 5 raised to any power will always be _ _ ....25
what is the sum of the last three digits of 5^2000?
this is only 1% harder. maybe not even that much :P

[quote="mihail911"]what is the sum of the last three digits of 5^2000? [/quote]
[hide]It would be 13, because for $ 5^{2n}$, the last 3 digits are 625[/hide]

5^1 is obviously just 5. mihail911 was talking about 5^2 and above.

[hide="Answers"]1. 7
2. 6[/hide]
which book did u get these from?

right...what hm29168 said...and you're right vamathletes..good job! :)

actually i think this was like a countdown round it went something like: find the sum of the units and tens digit of $ (2+3)^{7}3$.....or to some outrageously large exponent..

wow... u guys are expanding this to more than i thought... hmmm... :clap2:
oh well... this year i'm aiming for state!!!  :lol:
(yes, I know that might be lame, but i might have a chance!! i was 19th in chapter this year... and nationals might be a bit of a stretch...) :juggle:

oh... and i got these problems from this one booklet that my mathcounts coach uses...
i don't know what because she usually uses the overhead...
and i DID try to plug in 6^2000 into my TI-89, but it didn't work..   :rotfl:
and one more thing..
can u use TI-89 at MATHCOUNTS  :?:
because my mathcounts coach said she wouldn't allow anything more than a TI-83 and i wuz like wut?? i can't bring my TI-84 (that's my sister's)??? becuz that TI-84 is the same thing as the 83, just way more memory..
soooooo could someone get me out of this calculator confusion??  :help:   :stretcher:

I'm pretty sure they let you bring any calculator without a QWERTY keypad, so most TIs (at least, that I know of)
should be fine.

You can use TI-89 in MathCounts, but only on the Team Round and Target Round, obviously. 84's are okay too.

I brought and 86 last year...(there pretty old, but more functions than 83/84)

I like this program for these types of problems (only for TI-89, I'm not sure if someone has written one for other calculators):

[url]http://www.ticalc.org/pub/89/basic/programs/arithm.zip[/url]

Note: If you wanted the last $ n$ digits of $ a^{b}$, you need to enter [code]fi/pmod(a,b,10^n)[/code]

It automatically spits out the last $ n$ digits (very quickly). I am not sure about the limits of this thing, but it can surely handle this particular problem.

For MathCounts, your calculator isn't that useful because you can't use it in the sprint round, but these are good for the AMCs (if they let us use our calculators next year).

wow..
thanks..
that's a good program!!
:)

i have a variation. this was from this years state target round but i cant remeber which number. it was:

[hide]what is the product of the tens and ones digit when (2^1)(2^2)...(2^99)(2^100) is expanded[/hide]

[quote="abacadaea"]i have a variation. this was from this years state target round but i cant remeber which number. it was:

what is the product of the tens and ones digit when (2^1)(2^2)...(2^99)(2^100) is expanded[/quote]
[hide="Solution"]$ 2^{1}\cdot2^{2}\cdots2^{99}\cdot2^{100}=2^{1+2+\ldots+99+100}=2^{5050}$

$ 2^{5050\bmod40}\equiv2^{10}$

The last two digits are 24, and their product is $ \boxed{8}$.[/hide]

It's soooo easy:

5x5x5x5....they all have a 25 at the end, (e.g. 25, 125, 625) well, besides 5 ($ 5^{1}$).

Add the 2 and 5 (2+5), which equals 7.

For #2, anything times 6 ends in 6 (e.g. 6, 36, 216), so #2 has to be 6.

So the answer are:

1. 7

2. 6

:starwars:

yes.. i said that they were super easy.. :lol:

uhm, you better hope you get top 5, because your school doesn't look like they are going to help you get to state...

Not exactly top 5, unless you're in a REALLY hard state. At my chapter (in Michigan), someone who didn't make countdown advanced as an individual.

Oh, and as everyones already pointed out, the answers are 1.7 and 2. 6

2 things,

1. xpmath you said in michigan but in your info under your avatar it says your from ohio.

2. sometimes its top 6 because everyone on our team made to the cdr at chapter and we placed 2nd, 3rd, 4th, and 6th(our chapter stunk that year except for 2 teams). plus the 1st place guys team was the top 2nd place team in the state so they were a wild card team at states so really, the only person who made it soley as an individual was the guy who came in 5th.

I just moved two weeks ago (ask Temperal), but the competition was in Febuary :wink:


Suppose I wanted to use some theorem or rule that isn't commonly known but that was proved on, say, a mathworld page. Like, say I wanted to find the "point-plane distance" for some reason. Could I just say:

... the distance between the point and the plane is such-and-such (a proof of this can be found on http://mathworld.wolfram.com/Point-PlaneDistance.html) ...

Note that I picked out that mathworld article completely randomly and am not making any reference to any of the Round 2 problems.

Anyway, would that be okay? The proof is kinda messy and it was already proven by someone else, so is it okay to just use their result?

Actually, the formula for distance between a point and a plane we would view as commonly known.

Using a formula from a reference is fine. We graders, though, like to know exactly what you are using. To be easy on us, please quote the formula before using it. If it has a name, then give the name too.

Erin Schram
USAMTS grader


We denote an arbitrary permutation of the integers $1$, $2$, $\ldots$, $n$ by $a_1$, $a_2$, $\ldots$, $a_n$.  Let $f(n)$ denote the number of these permutations such that:

(1) $a_1 = 1$;

(2):$|a_i - a_{i+1}| \leq 2$, $i = 1, \ldots, n - 1$.

Determine whether $f(1996)$ is divisible by 3.

No,It is not divisble by $3$ because$f(1996)\equiv f(4)\equiv 1 (mod 3)$ :)

[quote=vietnamesegauss89]We have:$f(n+3)=f(n)+f(n+2)+1 \forall n>1$
$\Rightarrow f(8k-1)\equiv f(8k+3) \equiv 2(mod 3)$
$f(8k)\equiv f(8k+5)\equiv f(8k+6)\equiv0 (mod 3)$
$f(8k+1)\equiv f(8k+2)\equiv f(8k+4)\equiv 1 (mod 3)$
$1996=249.8+4\Rightarrow f(1996)\equiv 1(mod 3)$[/quote]

Can anyone help me with this solution as I can barely understand how he got what he got.


limit calculus integration derivative ratio function

Show $ \lim_{n \to \infty}\frac {\sum_{k \equal{} 0}^{n}e^{\sqrt {k}}}{2\sqrt {n} e^{\sqrt {n}}} \equal{} 1$

It's easy if you know about something like Cezaro Stolt's theorem.
we need to prove that $ \sum e^{...}\approx 2\sqrt{n}e^{...}$
by this theorem you only need to chek that :
$ \lim_{n \minus{} > \infty}\frac {e^{\sqrt {n \plus{} 1}} \minus{} e^{\sqrt {n}}}{2\sqrt {n \plus{} 1}e^{\sqrt {n \plus{} 1}} \minus{} 2\sqrt {n}e^{\sqrt {n}}} \equal{} 1$
try to prove it  yourself.
I can give you some hint I think this is $ \approx \frac {e^{\frac {1}{2\sqrt {n}}}}{2(1 \plus{} \frac {1}{2\sqrt {n}})(\sqrt {n} \plus{} \frac {1}{2\sqrt {n}}) \minus{} \sqrt {n}}\approx... yourself... \equal{} 1$ :wink:

$ \sum_{k \equal{} 0}^{n}e^{\sqrt {k}}$ with comparaison serie integral it is

equivalent to $ \int_{0}^{n}e^{\sqrt{x}}dx$ with a change varaible $ u\equal{}\sqrt{x}$

you can compute with an integration by parts and it is equivalent $ 2\sqrt{n}.e^{\sqrt{n}}$

the limit is 1

[quote="Gib Z"]but the difference between the series and the integral diverges :([/quote]
It does, indeed, but you need not the difference but the ratio ;) In other words, you just need to show that the difference grows much slower than the integral itself, which is true for any reasonably regular increasing function $ f$ of subexponential growth (basicly, all you need for that is that $ f(n)\minus{}f(n\minus{}1)$ be much less than $ f(n)$ for large $ n$ and $ f(x)\equal{}e^{\sqrt x}$ clearly satisfies this property).

Is the Euler-Maclaurin formula of ANY help to me? I could swear it looks like exactly what I need but I dont know how to use it...

It can be employed here to get an asymptotic expansion of the sum (the first term will be the integral $ \int_0^x e^{\sqrt t}\,dt$). The problem is that you'll still have to show by hand that the remainder in the Euler-Maclaurin is much less than the last term in the asymptotic expression that you want to keep, which is approximately as easy/hard as the question you started with.

[quote="Gib Z"]Extremal - Thank you, that was what I was thinking of as well, but I was not looking forward to showing that to my friend who asked that question :( We are both in high school, and whilst I'm sure I can't prove it, he probably hasnt even heard of it lol. I need another line of thought =] EDIT: I just read the wikipedia article about the theorem and it reminded of the interpretation that the theorem is like a L'hopitals rule for sequences. So im thinking to prove it, go through the normal proof for L'hopitals rule with Taylor series, but replacing derivatives with finite differences. That sounds like it should work yea?

Moubinool - I also tried that, but the difference between the series and the integral diverges :([/quote]
Now I have no idea about any other way method (but maybe there exist something nonstandart ways)
When mathematicians  thought about calculation sums like this  or find asymptotical expression  they found method's by Integrals and Cezaro, and fedja had already posted full explanation and Essence of these ideas.


[quote="Ragingg"]Is there a difference between an international school and a regular American private school?  What kind of students usually go?  Is it more/less rigorous than the country's "real" schools?[/quote]

I hope you realize your question is ambiguous. Is there a difference? Well, first, they are in different countries, that is a big difference to start. On the other hand, probably they speak another language, they probably teach different curricula (specially in the non-science subjects). Now, what do you mean with [b]country's "real" schools[/b], are there in america imaginary schools? (well, besides those that say that you can get a Ph.D. in two years after you got your HighSchool Diploma and without going to classes  :D ).

[quote="Ragingg"]Actually, I was under the impression that international schools spoke english... Which is their main difference.  I'm pretty sure there are International Schools in Taiwan which are conducted in English, but I wanted to know what the cirriculum was like etc.  BTW, International Schools=high schools

http://www.hkis.edu.hk/ -- for example.[/quote]

Well, then, it depends on your deffinition of International Schools. If I understand well, you are asking for private schools who have campus on several countries rather than just private schools out of the United States. In such a case, you will get suprised that most of them actually do not speak in English. Most of these schools (at least those working in Costa Rica) use to teach english as a second (or pretty often a third) language, often with high standards, but they don't lecture in English. Most of them either lecture in the native language of the country where the campus is located or in the native language of the original school (or both). In Costa Rica, I remember, the Humbolt, that lecture mainly in German, a little in Spanish. The Brittish School, that teaches completely in brittish english, the Franco, that teaches mainly in Spanish and some in French, there is other who teaches in Japanese and other that teaches in Italian, and a few, originally Spaniard, that teaches, obviusly, in Spanish, and I have heard that latelly (I have a few years out of my country) there is a new school teaching in Jiddish  :? .

Any way, if this is the case, the answer is very sparse. I mean, in my experience, after the HS where I was in my country (kind of magnet school), the two best are the Humbolt and the Brittish (with respect to the curriculum). But others are pretty bad.

Don't most of these schools have the IB programs?  Are those as bad as you say they are?

Ragingg, I think I understand your question. ;)

Most of these schools that you refer to began as embassy schools for the UK, Canada, U.S., Australia, and New Zealand. This is why they all speak English. The other expatriate schools refer to themselves as the Japanese School, the German School, the Korean School, etc.

Today, these schools have grown as the economic opportunities in the host country have also grown. Many international schools have 50+ nationalities and 20+ languages represented. This results in an extremely rich but tolerant culture. And the culture of each international school is slightly different.

The quality of international schools varies greatly, and for an average (and even for some exceptional students), a good international school is an excellent environment. (But for individuals who are extremely talented in one area, esp. non-academic areas, international schools are often a poor choice. However, this can be overcome if you are willing to assimilate into the local culture and seek opportunities within your host country.)

To answer one of your questions, yes, the instruction is in English, and yes, most international schools offer the IB program. The primary reason for this is the transient nature of the student body. Many schools have students who move every 2-3 years, and as a service to their students, the international schools make an attempt to make the students' education as continuous as possible, despite their many moves. However, the way IB is approached still varies differently, and the style of IB advising also varies greatly.

Do you have a specific school in mind?

[quote="JBL"][quote="djimenez"]I have heard that latelly (I have a few years out of my country) there is a new school teaching in Jiddish  :? .[/quote]

Perhaps you mean Yiddish?[/quote]

 :blush: Yep. Sorry for the typo.


linear algebra
matrix
Functional Analysis
advanced fields
advanced fields unsolved

Let A in $C^{nxn}$. Show that for any unitary matrices P,Q in $C^{nxn}$ there holds
||PAQ||=||A||

How to do this? Unitray means $U^{H} = U^{-1}$...

That's the 2-norm, as are all that follow here. Using the fact that $||Pu||=||Qu||=||u||$ for any $u$, we have that $\frac{||PAQv||}{||v||}=\frac{||PAQv||}{||Qv||}=\frac{||AQv||}{||Qv||}$. Since $Q$ is invertible, $||PAQ||=\sup_{v\neq0}\frac{||PAQv||}{||v||}= \sup_{Qv\neq0}\frac{||AQv||}{||Qv||}=||A||$.

This all works in any Hilbert space; composition with a unitary operator (an invertible isometry) on either side leaves the norm unchanged.


If $ a$ , $ b$ , $ c$ , $ x$ , $ y$ and $ z$ are positive reals such that $ a\plus{}b\plus{}c\equal{}x\plus{}y\plus{}z\equal{}1$ prove or disprove that $ \frac{a}{x}\plus{}\frac{b}{y}\plus{}\frac {c}{z} \geq 3$

[quote="canada"]If $ a$ , $ b$ , $ c$ , $ x$ , $ y$ and $ z$ are positive reals such that $ a \plus{} b \plus{} c \equal{} x \plus{} y \plus{} z \equal{} 1$ prove or disprove that $ \frac {a}{x} \plus{} \frac {b}{y} \plus{} \frac {c}{z} \geq 3$[/quote]
Try $ a \equal{} \frac {3}{5}$, $ b \equal{} c \equal{} \frac {1}{5}$, $ x \equal{} \frac {1}{2}$, $ y \equal{} z \equal{} \frac {1}{4}.$ :wink:


I have come across the fact that all perfect squares are congruent to $1 \pmod{4}$ or  $0 \pmod{4}$

Regarding this, I have 3 questions:
1) What is the proof for this? (aside from squaring evens (2n), giving you  $0 \pmod{4}$ and odds (2n+1), giving $1\pmod{4}$, is there any other proof using modular congruences?)

2) Is there any similar statement for numbers other than 4?

3) Are there any similar statements for higher powers like cubes and such?

I greatly appreciate your help. Thanks a lot ;)

For question 1 , you can divide cases . Let number be $(4k+r)^2$ where $0\leq r\leq 3$ is the remainder . Then for $r=0,1,2,3$ you will get $0,1,0,1 \mod 4$ respectively . So there are only two possibility .

Question 2 : Well mod 3 also gives you $0,1$ . Mod 5 gives you $-1,0,1$ . Basically I think that you can list out your self ( not hard taking mod for small number since if you take  (mod n) , then there are $n$ number to deal with) .

3) similarly I think this can be found by listing out the cases . ;)

Thank you for your prompt reply, shyong and dondigo. This really helps ! :thumbup:


geometry
geometric transformation
reflection
\/closed

Okay so I used to have a rating (like an hour ago) but now on my profile it says "not yet rated"!!! But i KNOW i was rated before!

Hmm...the other day izzy also said something like that happened.  i checked mine, and it was still there.  she said that i had "escaped the bug" or something like that...but i didnt do anything abnormal.

this should be moved to support.

[quote="isabella2296"]Cough. I was testing the post ratings the day AoPS went down, and every time I rated a post, the rating changed to "Not Rated Yet". The next time someone rates a post of yours, it'll be like I never changed your post rating by accident. :)[/quote]
It's okay Izzy. It's fine now =]

ya, my rating right now is "Not Rated Yet"
it's just a bug

Do the post ratings really reflect anything? Everyone says when you see spam, rate it 1. But with a maximum post rating per hour, you can only effectively rate a few posts, and unless I see a great solution, I never give any post a rating.

[quote="mewto55555"]Do the post ratings really reflect anything?[/quote]  Yes.  For example, everyone who has posted before me on this thread has a rating below 3, and this reflects the fact that all of you (with the exception of Isabella) have a strong tendency to make posts with close to 0 interest for everyone else, e.g., most of the posts on this thread.  For a recent discussion of this, I suggest http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227261


Such include circle, triangles, word problems, inscribeds, others, etc. Like a specific thing.

Like I am bad at drawing pictures from descriptions, as well as not good at some triangle lengths.

[quote="Treething"]You plain can't make such broad topics.[/quote]''

I just wanted to know what people have trouble with. simply

There is really no one subject that trips up everybody - for everybody it's different. Work on your weaknesses and don't worry about what others are bad at.

[quote="now a ranger"]oh okay. That is a good idea. I still can't understand how to draw good diagrams of word problems =( I trip up a lot in the back half of problems really often. What should I do to improve my interpretations of words ?[/quote]
Dude, stop posting pointless topics. Just work on the problems and you will get better.

Seriously, don't make pointless posts please.

Your question is "what can I do to improve my reading?"

Read more. Practice more. Simple as that, what do you think? Everybody else has some magic trick that if they share with you, you will suddenly get better?

I don't think this is quite as pointless as mosts of you make out: sure, perhaps its in the wrong forum or something (don't know where else it would go though) but it's a genuine question that he asked to help improve his geometry...  :maybe:

I find questions that say ".... Find the locus of all points X such that..." hard. I don't know why, they just seem much harder than other geometry questions. For these type of questions you usually need to guess the answer and then prove that a general point X lies on this locus and you're done.

I used to hate ones where you had to prove a relation to do with lengths as well (like ... Prove $\frac{AB\cdot CD}{BC}=\frac{AD\cdot BC}{BD}.$) but I found that these can be done well with trig or other algebraic-type methods.

Circles and triangles are nicest. Quadrilaterals are ok, particularly if they have a special property. Other polygons are ugly, unless inscribed in circle.

I found [url=http://www.mathcomp.leeds.ac.uk/Publications.htm]this book[/url] (scroll down) and [url=http://www.amazon.com/Episodes-Nineteenth-Twentieth-Euclidean-Mathematical/dp/0883856395]this book[/url] helped me improve my geometry greatly.


If the area of a 5-cm-by-5-cm square is decreased by 9 square cm to form a smaller, what is the perimeter in centimeters of the new square?

The area of a $ 5$ by $ 5$ square is $ 25$. $ 25\minus{}9\equal{}16$. So the dimensions of the new square are $ 4$ by $ 4$, so the perimeter is $ 4\times{4}\equal{}\boxed{16}$.

it's obviously decreased to a 4x4 square(just use intuition, check $ 5^2\minus{}4^2$ if you're that kind of person)
perimeter of square is $ 4\cdot\mathfrak{side length}$ so it's $ 4\cdot 4$
a generic way to get 4x4 is using $ 5^2\minus{}s^2\equal{}9$ $ s^2\equal{}16$ $ s\equal{}4$
the answer is
16


Find all functions $f: \mathbb R \to \mathbb R$ that satisfy
\[(x+y)(f(x)-f(y))=(x-y)(f(x+y))\]
for all reals $x$ and $y$.

[quote="Erken"]Find all functions $ f: R\rightarrow R$,that satisfy to the following condition:
For all $ x,y\in R$: $ (x+y)(f(x)-f(y))=(x-y)(f(x)+f(y))$.[/quote]

Developping the products, we have $ xf(y)=yf(x)$, or also $ \frac{f(x)}{x}=\frac{f(y)}{y}=c$ $ \forall x,y\neq 0$

And so $ f(x)=cx$ (even for $ x=0$ since $ f(0)=0$
and putting back in the original equation, any value $ c$ fit.

So the general solution is $ f(x)=cx$.

[quote="Erken"]Sorry for my mistake;
Here is the right version:
Find all $ f: R\rightarrow R$,that satisfy the following condition:
For all $ x,y\in R$ we have that $ (x+y)(f(x)-f(y))=(x-y)(f(x+y))$[/quote]

With $ y=2$, we have $ (i)$ : $ f(x+2)=\frac{x+2}{x-2}(f(x)-f(2))$

With $ y=1$, we have $ f(x+1)=\frac{x+1}{x-1}(f(x)-f(1))$

So $ f(x+2)=\frac{x+2}{x}(f(x+1)-f(1))$

So $ f(x+2)=\frac{x+2}{x}(\frac{x+1}{x-1}(f(x)-f(1))-f(1))$

So $ (ii)$ : $ f(x+2)=\frac{x+2}{x}(\frac{x+1}{x-1}f(x)-2f(1)\frac{x}{x-1})$

Comparing $ (i)$ and $ (ii)$, we get :

$ \frac{x+2}{x-2}(f(x)-f(2))=\frac{x+2}{x}(\frac{x+1}{x-1}f(x)-2f(1)\frac{x}{x-1})$

$ \frac{1}{x-2}(f(x)-f(2))=\frac{x+1}{x(x-1)}f(x)-2f(1)\frac{1}{x-1}$

$ 2f(1)\frac{1}{x-1}-\frac{1}{x-2}f(2)=(\frac{x+1}{x(x-1)}-\frac{1}{x-2})f(x)$

$ f(x)=\frac{f(2)}{2}x(x-1)-f(1)x(x-2)$

$ f(x)=ax^{2}+bx$

Putting back this expression in original equation, we can verify that this solution works for any values of $ (a,b)$

Nice solution,i've solved it the same way :lol:.


function trigonometry calculus integration derivative algebra polynomial

The function $f$ is defined by the power series

\[f(x) = \sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n}}{(2n+1)!}= 1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}{7!}+\cdots+\frac{(-1)^{n}x^{2n}}{(2n+1)!}+\cdots\]

for all real numbers $x$.

(b) Show that $1-\frac{1}{3!}$ approximates $f(1)$ with error less than $\frac{1}{100}$.

The solution I found in the internet used something called alternating series. While I liked this one and simple, I was wondering if anyone can tell me by using Lagrange Error Bound. That's one concept of series/sequences that I still do not understand.  Thanks a lot.

(c) Show that $y = f(x)$ is a solution to the differential equation $xy'+y = \cos x$

The solution took integral respect to $x$. Okay this may sound really dumb how do you do that? I know how to take integral respect to $t$ and $\theta$ but with $x$? Never really done it before. Can anyone help me?

For my solution, this was my final line.

$ \sum^{\infty}_{n=1}\frac{(-1)^{n-1}x^{2n-2}}{(2n-2)!}+\sum^{\infty}_{n=1}\frac{(-1)^{n-1}2n x^{2n}}{(2n+1)!}$

which got me nowhere.. help? I understood solution in internet (slick and simple) but I want to know if I can finish this way too. Thanks!

For (b), the alternating series test is the easiest and simplest way to estimate the remainder, and is undoubtedly what the test writers intended. The alternating series test requires two things: that the terms alternate in sign (obvious here) and decrease in size. The latter is easy in this case as long as $|x|\le 1.$ To use the Lagrange form of the Taylor series would require us to know the identity of the function. Now it turns out that we do know that, since $f(x)=\frac{\sin x}x,$ but finding and estimating the fourth derivative of that function would not be a nice thing.

For part (c), I usually suggest to the less experienced that they not try to use sigma notation, but instead use more paper to write things out. We have:

$y=1-\frac{x^{2}}6+\frac{x^{4}}{120}-\frac{x^{6}}{7!}+\cdots$

$y'=-\frac{x}3+\frac{x^{3}}{30}-\frac{6x^{5}}{7!}+\frac{8x^{7}}{9!}+\cdots$

$xy'=-\frac{x^{2}}3+\frac{x^{4}}{30}-\frac{6x^{6}}{7!}+\frac{8x^{8}}{9!}+\cdots$

$xy'+y=1-\frac{x^{2}}2+\frac{(4+1)x^{4}}{120}-\frac{(6+1)x^{6}}{7!}+\frac{(8+1)x^{8}}{9!}+\cdots$

$xy'+y=1-\frac{x^{2}}2+\frac{x^{4}}{24}-\frac{x^{6}}{6!}+\frac{x^{8}}{8!}+\cdots$

And we recognize that as the series for $\cos x.$ We should go back and do the generic term to make sure the pattern holds, but we've probably seen enough.

In this particular case, the differential equation is easy enough to solve directly, so it isn't that much of an advertisement for the uses of power series methods.

Part (b) using Lagrange form of the remainder: (note that it doesn't solve under the condition they gave and you pretty much have to use a calculator)
[hide]There exists a value $z$ between the center of our series, $0$, and the number we are trying to approximate, $1$.  Also it is a second degree polynomial  so $n=2$:

$R_{n}(x)=\frac{f^{(n+1)}(z)}{(n+1)!}\cdot (x-c)^{n+1}$
$R_{2}(1)=\frac{f^{(2+1)}(z)}{(2+1)!}\cdot (1-0)^{2+1}$; Plug in all values we know.
$R_{2}(1)=\frac{f^{(3)}(z)}{6}$

Now we gotta fnid what $f^{(3)}(z)$ is:

$f(x)=\frac{\sin x}{x}$
$f'(x)=\frac{x \cdot \cos x-\sin x}{x^{2}}$
$f''(x)=-\frac{2x \cdot \cos x+(x^{2}-2) \cdot \sin x}{x^{3}}$
$f^{(3)}(x)=\frac{3(x^{2}-2) \sin x-x(x^{2}-6) \cos x}{x^{4}}$

Thus:

$R_{2}(1)=\frac{\frac{3(z^{2}-2) \sin z-z(z^{2}-6) \cos z}{z^{4}}}{6}$
$R_{2}(1)=\frac{3(z^{2}-2) \sin z-z(z^{2}-6) \cos z}{6z^{4}}$

Note that the maximum error will occur when $z=1$ because at $0$ it is technically going to be approaching $0$ due to the $z^{4}$ dominating at most $z^{3}$ times $\cos z$.  You can take a derivative of the third derivative, draw a derivative line for $0<z<1$ to prove that it is strictly monotonically increasing anyways.  This is where you'll need a calculator to evaluate the maximum error by letting $z=1$.  Because this is # $6$, you couldn't use a calc and the other test would be appropriate, but I'm just showing you how to do it this way.  Okay.. Thus:

$R_{2}(1)=\frac{3(z^{2}-2) \sin z-z(z^{2}-6) \cos z}{6z^{4}}$
$R_{2}(1)|_{z=1}<\frac{5 \cos 1-3 \sin 1}{6}$
$R_{2}(1)|_{z=1}<\frac{5 \cos 1-3 \sin 1}{6}$
$\boxed{R_{2}(1)|_{z=1}<.029516}$

Which doesn't prove anything (the condition that it is $<\frac{1}{100}$).  Its definitely true but it doesn't show what they wanted and you'd have to use a calculator.  So use alternating series remainder theorem.[/hide]

Part (b) using alternating series remainder:
[hide]I'm sure you saw their solution but its simple.  The remainder is less than the next term.  So it is $R<\frac{1}{120}<\frac{1}{100}$ which is what I'm sure they said.  If not that's it in a nutshell.[/hide]

Part (c) a different way than Kent's:
[hide]We know $f(x)=\frac{\sin x}{x}\iff y=\frac{\sin x}{x}\iff x \cdot y = \sin x$.  Differentiate implicitly:

$y \cdot \frac{dx}{dx}+x \frac{dy}{dx}= \cos (x) \cdot \frac{dx}{dx}$
$\boxed{x \cdot y'+y = \cos x}$[/hide]

Also if you want a good practice problem for Lagrange before your AP test (assuming you're taking it on Tuesday), check out 2004 Form B #2 on the last part.  I liked that.

Thanks guys! I especially liked how it's easier to do series without writing sigmas  :blush: . My habit of using sigmas isn't very useful in AP I suppose (worked awesome during AMC/AIME though hehe).


Okay.. I have little problem here. I'm doing good with all my schools and sports but my videogame is pretty ancient. I have PS1 but man, its graphic is way below PS2,Xbox or Gamecube. Plus, you can't really find PS1 games in regular stores like Walmart anymore. You have to go to EB Games or GameStop for those. So, I realized that I needed a new videogame machine and my decision is to buy PS2. I know couple of good games in there (not much fan of Halo.. I heard it was excellent though) but my problem is this: How can I get it..? I used most of my money for AoPS classes and books and my birthday is way way long time and far far away. I wonder if there is any tip on buying PS2 machine easily without paying so much money (144 bucks for new one seems still too much for me.. :-/). I wonder if I can still buy the old machine (not the slim one but the fat one; I think their graphic is same) because I think their price is bit cheaper (?).

Anyway, give me a good tip.. I'll appreciate that.

Well... it's like this.  I believe you can still find the big PS2 in some places (especially if you look online), but you won't save all that much money on it; it's only about 20 dollars cheaper, if I remember correctly.  Furthermore, the big PS2 is a bit less reliable than the slimline version (the disc drive is prone to acting up).  You could always buy used but... the PlayStation line isn't exactly renowned for being reliable.  I don't know if I'd trust a used PS2.

Rock and a hard place.

Then there's the fact that after you buy the system itself, you're not done buying yet.  You still need to by a new Memory Card (that's another 30 bucks right there-you can't save PS2 games on your PS1 cards), and you still haven't gotten a game yet.

If $\$$144/$\$$124 (plus the other stuff) seems too expensive to you, the question you need to ask yourself is

[quote="Silverfalcon's Mind"]How badly do I want a PS2, and how soon?[/quote]

This summer (spring, if, unlike me, you believe Sony), the PS3 will be released, and the PS2 will almost undoubtedly drop in price significantly.  You might want to wait until then.

If you don't want to wait until then, and are still unwilling to pay the price, you might consider a different system, like the less-expensive GameCube or GameBoy Advance.  If you decide to take this path, I can recommend some good games.

Good luck on your quest for a new system!

Wait for the PS3, PS2 is just not worth it

I did Google search and saw that PS3's price is still undetermined. Any idea on what will be the price? (guess..?)

Also, how much do you mean "significant" drop in price?

[quote="Silverfalcon"]I did Google search and saw that PS3's price is still undetermined. Any idea on what will be the price? (guess..?)[/quote]

Well, let's see.  XBox 360 is 300-400 dollars, depending opn if you get the hard drive.  PS3 will be [i]at least[/i] as expensive as the basic 360, seeing how it will be more powerful, so... probably at least 350, 400 dollars.

[quote="Silverfalcon"]Also, how much do you mean "significant" drop in price?[/quote]

Beats me.  I'd guess at least 20 dollars, though.

with the xbox 360 already released and the ps3 coming out soon (or at least by fall), you can probably find someone who will sell you their old system pretty cheap. if you want to get a new one asap, then set up a lemonaide stand, bake sale, yard sale, or something else to make some money.

or just kill some one and steal their ps2 :ninja:

I think the ps3 will be around 500, if they have all these goodies sony claim ps3 will have.

[quote="sponge008"]OK, stop right there. You are obviously posting here, so save yourself a bundle and just game on the PC. There are plenty of good, cheap (or "free") games that can run on most any computer (Starcraft, Tribes, Quake 3 (MOST computers)), just to name a few. Unless your computer is ancient, you should be fine. Tribes and Quake 3 are in fact now de facto free. Starcraft can be found for ubercheap, and I'm sure you can find other PC games that you like. How good is your computer, anyways?[/quote]

LOL. I played Starcraft for long time. Probably more than some of people here played but whatever.

Computer games are indeed fun but they're not good for your eyes. That's why I'm looking for video games. So, please go sideways about talking computer games. That's clearly irrelevent to what I'm asking.

Hmm.. 500 bucks for PS3? I just hope PS2 drops below or really close to 100 dollars. I won't be able to do anything before summer anyway.