When a particle is located a distance x in meters from the origin, a force cos(piex/3) newtons acts on it. How much work is done in moving the particle from x=1 to x=2? Interpet your answer by considering x=1 to x=1.5 and from x=1.5 to x=2.

May you show me how you did this and i need an answer. I need help on this one.

i know how to do the problem but what do i do with the cos(piex/3)

This post is entirely meta, i.e. it has nothing to do with the content of your question.

1) The Greek letter $\pi$ has an English name, which is two letters long and does not contain the letter "e."

2) It would be greatly appreciated (by me, and maybe others) if you took the time to format your questions, especially by learning and using LaTeX.

3) Do your own homework.

4) You say you know how to do the problem.  What does that mean exactly?  Presumably you can reduce the question to calculating an integral.  What do you think is a likely approach for this integral?  Do you know any integrals that look like it?

i'm very sorry but how can i do the problem if i can't start it off

That comment doesn't make any sense in light of the fact that you claim to know how to do the question -- either you know how to do it or you don't, and you should pick one for the purposes of this thread.  Do you know what "work" is?  Do you know how, as a general rule, to calculate it?

yes i do, i'm just stuck with the cos (\pi x/3)


Find all functions $ f: \mathbb{N} \rightarrow \mathbb{N}$ such that $ f(xf(y)\plus{}y)\equal{}yf(x)\plus{}f(y)$ for all $ x,y \in \mathbb{N}.$

$ f(x)\equal{}x$ for all $ x\in\mathbb{R}$
$ f(x)\equal{}0$ for all $ x\in\mathbb{R}$


Define a seed to be the intersection of two congruent circles.  If the distance between the two "pointy" parts is R and the overlap of the diameters of the two congruent circles is r, compute the area of a seed.

Hide answers.  Do not post processes.

I found the solution:
[hide="Solution"]
Define x to be (R^2 - 3r^2) / 2r
Define s to be 2(arcsin (R / (2r + x)))
Define a to be (360 / s) * pi * 4r^2+4rx+x^2
Define b to be Rr + Rx
The area of the seed is a - 2b
Please try to do the tedious multiplication and simplification.
[/hide][/hide]

Sorry - I posted the question incorrectly - distance between the points is 2R and overlap of diameters is 2r.


Solve in positive integers the equation:2^N-7=Y^2. Sorry for my latex guys. :)

2^n-7 is always going to be odd. So set y=2k+1. Simplifying we have 2^(n-2)=kk+k+2. Unless k=0, which gives (3,1) as a solution, put k=2m or k+1=2m. In the first case one gets 2^(n-3)-1=m(2m-1) In the second 2^(n-3)-1=m(2m+1). For m=1 one has the solutions (4,3) and (5,5) respectively. Clearly for the other solutions 2^(n-3)-1 must be composite but this is not sufficient.


what will  be the last two digits of 9^9^9 ???
     [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1516009#1516009]also,can you please find yhe number of digits in 2^2^22 ???[/url]
[size=75]
[color=darkred]Moderator note: Discuss the second problem in the linked topic.-solafidefarms[/color][/size]

#1: First, $ 9\times9\equal{}81$ and $ 81\times9\equal{}...29$ so the answer is 29.

I think he means 9^9^9 not $ 9*9*9$ as you are answering.

Woops. Then we would look at the last two digits in the powers of 9:

0 - 01
1 - 09
2 - 81
3 - 29
4 - 61
5 - 49
6 - 41
7 - 69
8 - 21
9 - 89
10 - 01
11 - 09

We see that the pattern repeats every 10 powers. We know that $ (9^9)^9\equal{}9^{81}$ so we have to find what the 81st term in the pattern would be. We know that every multiple of 10 is 01, so 80 is 01. Therefore 81 would be $ \boxed{09}$.

by the first question ,9^(9^9) is meant, not 9^81 or (9^9)^9.
 the second question can also be corrected as 2^(2^22).
i think that will make it easier to understand.
sorry for the unclear questions.

The powers of 9 cycle every 10, so we need to find the units digit of $ 9^9$, which is 9 from Powerofpi's table. Thus we need to find the last 2 digits of $ 9^9\equal{}89$

[quote="Mewto55555"]The powers of 9 cycle every 10, so we need to find the units digit of $ 9^9$, which is 9 from Powerofpi's table. Thus we need to find the last 2 digits of $ 9^9 \equal{} 89$[/quote]
Mewto55555 explained it a little here.

[hide="Table"]0 - 01
1 - 09
2 - 81
3 - 29
4 - 61
5 - 49
6 - 41
7 - 69
8 - 21
9 - 89
10 - 01
11 - 09 [/hide]

From that table, we see that the last two digits of the exponent of $ 9$ ($ 9^9$) is 89, and since the powers of nine cycle through every ten times, we only care about the units digit of the power of nine, which is $ 9$. So now we want to figure out the last two digits of $ 9^9$, which is $ 89$ according to the table.

[quote="PowerOfPi"]I think that this is getting too advanced for Middle School (logarithms?)? I am not saying I don't understand it.[/quote]
Yes. The previous logarithm problem was a hard AIME-level problem, so it's chucked up to high school. Keep it further down ;).

please do the problem with  the method of congruency along with using the binary and decimal reppresentation of integers.i think it will be much more easier. my last reply can be helpfull .please try to complete it.
p.s- note that we are asked to find the last two digit of 9 to the power 9^9.
not 9*9^9.

I didn't mean multiplication. I meant "..., we see that the last two digits of the exponent of $ 9$ ([i]which is [/i]$ 9^9$) is 89, and since the powers of nine cycle through every ten times,...".

[quote="earth"]i think i have  got an idea ,hope it"ll help
     here, we have to find  the last two digits of  9^(9^9).
so we have to get the remainder when 9^(9^9) is divided  by 100
     9^9=9(mod 10)
hence,9^9=9^10k+9
   now we also know , 9^(9^9)=89(mod 100)
    now can you please help me to proceed further  ????[/quote]

Doesn't this mean that the last two digits are just 89?

Yeah it does. I don't really know what earth wants to do.

I would recommend getting Introduction to Number Theory, they explain these types of problems perfectly :wink:

Don't hafta post it twice. Got a little too impatient there, Panther  :)


email
linear algebra
matrix
inequalities

I know $ \det\left(A \plus{} B\right)\neq 0$ and $ \det\left(A \minus{} B\right)\neq 0$ if and only if $ \det\left(\begin{array}{cc}A & B \\
B & A \end{array}\right)\neq 0$ . Can you give a characterization of  $ \det\left(\begin{array}{cc}A & \minus{} B \\
B & A \end{array}\right)\neq 0$?
If you have a solution, please email me: miwalin@163.com

[color=blue][b]Theorem.[/b] Let $ R$ be a commutative ring with unity. Let $ q\in R$, and let $ A\in R^{n\times n}$ and $ B\in R^{n\times n}$, where $ n$ is a positive integer. Then, the determinant of the block matrix $ \left(\begin{array} [c]{cc} A & q^{2}B \\
B & A \end{array} \right)$ equals $ \det\left( A \minus{} qB\right) \det\left( A \plus{} qB\right)$.[/color]

[i]Proof of the Theorem.[/i] First,

$ \left( \begin{array} [c]{cc} I_{n} & \minus{} qI_{n} \\
0 & I_{n} \end{array} \right) \cdot\left( \begin{array} [c]{cc} I_{n} & qI_{n} \\
0 & I_{n} \end{array} \right) \equal{} \left( \begin{array} [c]{cc} I_{n}\cdot I_{n} \plus{} \left( \minus{} qI_{n}\right) \cdot0 & I_{n}\cdot qI_{n} \plus{} \left( \minus{} qI_{n}\right) \cdot I_{n} \\
0\cdot I_{n} \plus{} I_{n}\cdot0 & 0\cdot qI_{n} \plus{} I_{n}\cdot I_{n} \end{array} \right)$
$ \equal{} \left( \begin{array} [c]{cc} I_{n} & 0 \\
0 & I_{n} \end{array} \right) \equal{} I_{2n}$.

Thus, the matrix $ \left( \begin{array} [c]{cc} I_{n} & qI_{n} \\
0 & I_{n} \end{array} \right)$ is invertible, and $ \left( \begin{array} [c]{cc} I_{n} & qI_{n} \\
0 & I_{n} \end{array} \right) ^{ \minus{} 1} \equal{} \left( \begin{array} [c]{cc} I_{n} & \minus{} qI_{n} \\
0 & I_{n} \end{array} \right)$.

Now,

$ \left( \begin{array} [c]{cc} I_{n} & qI_{n} \\
0 & I_{n} \end{array} \right) \cdot\left( \begin{array} [c]{cc} A \minus{} qB & 0 \\
B & A \plus{} qB \end{array} \right) \equal{} \left( \begin{array} [c]{cc} I_{n}\cdot\left( A \minus{} qB\right) \plus{} qI_{n}\cdot B & I_{n}\cdot0 \plus{} qI_{n}\cdot\left( A \plus{} qB\right) \\
0\cdot\left( A \minus{} qB\right) \plus{} I_{n}\cdot B & 0\cdot0 \plus{} I_{n}\cdot\left( A \plus{} qB\right) \end{array} \right)$
$ \equal{} \left( \begin{array} [c]{cc} \left( A \minus{} qB\right) \plus{} qB & q\left( A \plus{} qB\right) \\
B & A \plus{} qB \end{array} \right) \equal{} \left( \begin{array} [c]{cc} A & qA \plus{} q^{2}B \\
B & A \plus{} qB \end{array} \right)$,

so that

$ \left( \begin{array} [c]{cc} I_{n} & qI_{n} \\
0 & I_{n} \end{array} \right) \cdot\left( \begin{array} [c]{cc} A \minus{} qB & 0 \\
B & A \plus{} qB \end{array} \right) \cdot\left( \begin{array} [c]{cc} I_{n} & qI_{n} \\
0 & I_{n} \end{array} \right) ^{ \minus{} 1}$
$ \equal{} \left( \begin{array} [c]{cc} A & qA \plus{} q^{2}B \\
B & A \plus{} qB \end{array} \right) \cdot\left( \begin{array} [c]{cc} I_{n} & qI_{n} \\
0 & I_{n} \end{array} \right) ^{ \minus{} 1}$
$ \equal{} \left( \begin{array} [c]{cc} A & qA \plus{} q^{2}B \\
B & A \plus{} qB \end{array} \right) \cdot\left( \begin{array} [c]{cc} I_{n} & \minus{} qI_{n} \\
0 & I_{n} \end{array} \right)$
$ \equal{} \left( \begin{array} [c]{cc} A\cdot I_{n} \plus{} \left( qA \plus{} q^{2}B\right) \cdot0 & A\cdot\left( \minus{} qI_{n}\right) \plus{} \left( qA \plus{} q^{2}B\right) \cdot I_{n} \\
B\cdot I_{n} \plus{} \left( A \plus{} qB\right) \cdot0 & B\cdot\left( \minus{} qI_{n}\right) \plus{} \left( A \plus{} qB\right) \cdot I_{n} \end{array} \right)$
$ \equal{} \left( \begin{array} [c]{cc} A & \minus{} qA \plus{} \left( qA \plus{} q^{2}B\right) \\
B & \minus{} qB \plus{} \left( A \plus{} qB\right) \end{array} \right) \equal{} \left( \begin{array} [c]{cc} A & q^{2}B \\
B & A \end{array} \right)$.

Hence, the block matrices $ \left( \begin{array} [c]{cc} A & q^{2}B \\
B & A \end{array} \right)$ and $ \left( \begin{array} [c]{cc} A \minus{} qB & 0 \\
B & A \plus{} qB \end{array} \right)$ are similar. Thus,

$ \det\left( \begin{array} [c]{cc} A & q^{2}B \\
B & A \end{array} \right) \equal{} \det\left( \begin{array} [c]{cc} A \minus{} qB & 0 \\
B & A \plus{} qB \end{array} \right)$.

But $ \left( \begin{array} [c]{cc} A \minus{} qB & 0 \\
B & A \plus{} qB \end{array} \right)$ is a block-triangular matrix; thus,

$ \det\left( \begin{array} [c]{cc} A \minus{} qB & 0 \\
B & A \plus{} qB \end{array} \right) \equal{} \det\left( A \minus{} qB\right) \det\left( A \plus{} qB\right)$.

Hence,

$ \det\left( \begin{array} [c]{cc} A & q^{2}B \\
B & A \end{array} \right) \equal{} \det\left( \begin{array} [c]{cc} A \minus{} qB & 0 \\
B & A \plus{} qB \end{array} \right)$
$ \equal{} \det\left( A \minus{} qB\right) \det\left( A \plus{} qB\right)$,

qed.

You can apply the Theorem to $ R \equal{} \mathbb{C}$ and $ q \equal{} i$ to obtain

$ \det\left( \begin{array} [c]{cc} A & \minus{} B \\
B & A \end{array} \right) \equal{} \det\left( A \minus{} iB\right) \det\left( A \plus{} iB\right)$.

If the matrices $ A$ and $ B$ commute, this simplifies to $ \det\left(A^2 \plus{} B^2\right)$ (what could also be obtained using the Schur complement theory), but in the general case this won't be the same.

  darij


geometry
trapezoid
rectangle
ratio
Pythagorean Theorem

[hide="Answer - My pathetic way"]

\[\text{I forgot the formula for the area of a trapezoid, so I did this: }\]

\[\text{First, I found the length of the side \"?\", shown in my diagram.}\]

\[a^2+b^2=c^2\]
\[b^2=c^2 - a^2\]
\[b^2=5^2 - 4^2\]
\[b^2=25 - 16\]
\[b^2=9\]
\[b=3\]

\[\text{Using this data, I was able to:}\]

\[\text{Find the base length of the triangles. }\]

\[\text{My way of solving: }\]

\[\text{First find the area of both triangles: }\]

\[A_{triangleOne}=\frac{base \times height}{2}\]
\[=\frac{4\times3}{2}\]
\[=\frac{12}{2}\]
\[=6\]

\[A_{bothTriangles}=A_{triangleOne}\times2\]
\[=6\times2\]
\[=12\]

\[\text{Now the area of the rectangle:}\]

\[A_{rectangle}=bh\]
\[=4\times4\]
\[=16\]

\[\text{Trapezoid area:}\]

\[A_{trapezoidTOTAL}=A_{rectangle}+A_{Triangles}\]
\[=16+12\]
\[=28\]

\[\boxed{\text{The Area of the Trapezoid is: }28 \text{ Square Units}}\]

[/hide]

Correct answer, but there's a much easier way to do this. Memorize the trapezoid formula!

[quote="236factorial"]What's the area of an isosceles trapezoid with a shorter base of 4, a height of 4, and a slant length of 5?

hint: title[/quote]
[hide]
the trapezoid has a small base of 4, and a large base of 10 (4+3+3(3 from the pythagoream triplet). so the area is (10+4)x4/2=28
[/hide]

[quote="236factorial"]What's the area of an isosceles trapezoid with a shorter base of 4, a height of 4, and a slant length of 5?

hint: title[/quote]
[hide]area of a trapizoid=$(b1+b2)/2*h$
If the height is 4 and the slant length 5 the difference between the shorter bas and the larger base is 3+3=6 units.
so the larger base is 10
$(10+4)/2*4=7*4=28$
28 sq.un. is the area of the trapezoid.[/hide]

I have something to add to the trapezoid formula.

If you haven't noticed, $\frac{1}{2} (b_1+b_2)$ is also the midsegment. Therefore, the area of a trapezoid can also be the $midsegment*h$. This can come in handy sometimes.   :)

[quote]What's the area of an isosceles trapezoid with a shorter base of 4, a height of 4, and a slant length of 5?

hint: title[/quote]

[hide]
Using the Pythagorean Theorem: $5^2 - 4^2 = 9$, $\sqrt{9} = 3$ so the base is 3 + 3(there are 2 slanted sides) + 4, the smaller base, which equals 10, to get the area of an iscoceles trapezoid you multiply the average base lenght by the height. So: 14\s * 4 = 7 * 4=

28 [/hide]

[quote]Here's a question:

An isosceles trapezoid ABCD has legs AB and DC, E is the midpoint of AB, F is the midpoint of DC. EF is 20 inches. The ratio of the height to the area is....? [/quote]

[hide]
Using the Pythagorean theorem you get that the height is $\sqrt{40} - x = h$. Let y be the shorter base and you get: $\frac{\sqrt{40} - x}{\frac{2x +2y}{2}*(\sqrt{40} - x)}$ Which simplifies to $\frac{\sqrt{40} - x}{(x + y)*(\sqrt{40} - x)}$ By dividing by $\sqrt{40} - x$ on each side, you get: $\frac{1}{x + y}$ [/hide]
Hmm... I don't think that is correct.


This is about a calculator again. I have borrowed a ti-84 silver edition from my teacher to import the applications to my ti 83 plus. I think you can do this. I have the cables and stuff. But how do I import it???

Um.

Okay, on receiving calculator, 2nd -> LINK -> Receive.

Sending calculator:
2nd -> LINK -> Scroll down to whatever category you want to send and select whatever.

Omg. Thanks for bearing with a newbie. Or noob. thanks so much.


$k=\overline{0,2003}$ and $\frac{2003-k}{5}\in Z$ and $\frac {k}{7}\in Z$
find $k$ so that the conditions are staisfied

[quote="shobber"][quote="spx2"]$k=\overline{0,2003}$ [/quote]
what does it mean?[/quote]

 i.e $k=0,1,2,...,2003$

So I think $k= m28$ or $k=n63$ where $m, n \in N$.

can i please get a classic non s.f. ,non advanced stuff solution ?

this is what i did for it $\frac{2003-k}{5} \in Z \Longleftrightarrow \frac{3-k}{5} \in Z$ also meaning that $5\vert (3-k)$ that means that $k=3\pm5t,t\in N$
otherwise could be formulated like $k=3+5t,t\in Z$
and the other condition $\frac{k}{7}\in Z \Longleftrightarrow k=7h,h\in Z$
so i have that $k=7\cdot(3+5t)=21+35t,t\in Z$

[quote="Saul"]$k \equiv 28 (mod 35)$ by the Chinese Remainder theorem, unless I have misinterpreted the problem.[/quote]
Don't know what it is...

[quote="spx2"]why ?
did you follow what i did ?
where is it wrong ?[/quote]
I mean I knows nothing about mod, therefore I don't know what your answer really mean...
Is my answer ok??

Fiind $k \in \left\{ {0,1,2, \ldots ,2002,2003} \right\}$
such that $\frac{{2003 - k}}{5} \in \mathbb{Z}$ and $\frac{k}{7} \in \mathbb{Z}$

[b][u]Solution:[/u][/b]
$A = \left\{ {0,1,2, \ldots ,2002,2003} \right\}$
then:
[color=red]1.)[/color]
$k \in A \wedge \frac{{2003 - k}}{5} \in \mathbb{Z} \Leftrightarrow k \in A \wedge \exists p \in \mathbb{Z}:3 - k = 5p \Leftrightarrow $
$ \Leftrightarrow k \in A \wedge k = 3 - 5p,p \in \mathbb{Z}$

[color=red]2.)[/color]
$k \in A \wedge \frac{k}{7} \in \mathbb{Z} \Leftrightarrow k \in A \wedge \exists q \in \mathbb{Z}:k = 7q$

[color=red]3.)[/color]
$\exists p,q \in \mathbb{Z}:3 - 5p = 7q \Leftrightarrow p = 2 - 7r \wedge q =  - 1 + 5r,r \in \mathbb{Z} \cap \left\{ {0,1, \ldots ,57} \right\}$

then
$k =  - 7 + 35r,r \in \left\{ {0,1, \ldots ,57} \right\}$


Let $I$ be the intersection of the segment $PQ$ with the circle, and let $E,F$ be the reflections of $D,C$ respectively in the line $PQ$.

$CE,DF$ must meet in the pole of $AB$, which is $P$. $I$ is the midpoint of the arcs $DE,CF$, so $CI,DI$ are angle bisectors of $\angle PCD,\angle PDC$ respectively. This proves that the incenter of all triangles $PCD$ is $I$ (and the triangle $PAB$ is just a special case of triangle $PCD$, obtained when we take $CD=AB$).

a good problem.

it is clear that $AB$ is the polar of $P$ to the circle.

suppose $PQ$, intersects the circle at $M,N$ respectively.

we know that $M$ is the incenter of triangle $PAB$.

the we should have: $(PQMN)=-1$

and because the circle is drawn by diameter $MN$,

for each point ,$X$, on this circle we have:

the circle, is apolonius circle of triangle $XPQ$.

and it means that $XM$, is the bisector of$\angle PXQ$

and it is done.

because as i said, we should have :$CM,DM$ are the bisectors

angles $PCQ,PDQ$. and meet each other at the incenter of triangle

$PAB$.

dear shobber u posted this problem at geometry and unsolved problems.

but really u couldnt solve it?

ofcourse u solved it. ;)

then u should post it in proposed and own problem...

[quote="shobber"]$P$ is a point on the exterior of a circle centered at $O$.  The tangents to the circle from $P$ touch the circle at $A$ and $B$.  Let $Q$ be the point of intersection of $PO$ and $AB$.  Let $CD$ be any chord of the circle passing through $Q$.  Prove that $\triangle PAB$ and $\triangle PCD$ have the same incentre.[/quote]

Indeed, if the segment PO intersects the circle at a point S, then S is the common incenter of triangles PAB and PCD.

To prove this, I first note that the point Q is the inverse of the point P in the circle. In fact, the way the point Q was constructed is the usual compass-and-ruler construction of the inverse with respect to a circle of a point outside the circle. Then I will use the following lemma:

[b]Lemma 1.[/b] If a point Q is the inverse of a point P with respect to a circle, and A and B are two points on the periphery of the circle, then PA : QA = PB : QB.

[i]Proof.[/i] If O is the center and r is the radius of the circle, then the point Q lies on the line PO, and we have $OP\cdot OQ = r^2$. This is all because the point Q is the inverse of P with respect to the circle. Now, r = OA, and thus $OP\cdot OQ = r^2$ becomes $OP\cdot OQ = OA^2$. In other words, OP : OA = OA : OQ. Together with < POA = < AOQ, this shows that the triangles POA and AOQ are similar, so that PA : AQ = OA : OQ. Since OA = r, this becomes PA : QA = r : OQ. Similarly, PB : QB = r : OQ. Thus, PA : QA = PB : QB, and Lemma 1 is proven.

Now, in our problem, Lemma 1 (actually, in a limiting case!) yields PS : QS = PD : QD. Thus, the line DS is the angle bisector of the angle PDQ. In other words, the line DS is the angle bisector of the angle PDC. Similarly, the line CS is the angle bisector of the angle PCD. Hence, the point S, common to these two lines, is the incenter of triangle PCD. Now we have shown this for each position of the chord CD, implied that it passes through the point Q; one particular such position is if C = A and D = B. Hence, we obtain that the point S is the incenter of triangle PAB. And the solution is complete.

  Darij

$\angle OAP=\angle OBP=90$ so $OAPB$ is cyclic. Now $QC\cdot QD=QA\cdot QB=QO\cdot QP$ so $DOCP$ is cyclic. Now let circle $k(O,OA)$ meet $OP$ at $S$. since $OS=OC=OD$ and $OS=OA=OB$ and since $PO$ is the bisector of $\angle DPC$ and $\angle APB$ then $S$ is the incenter of both $PAB$ and $PCD$


[quote="Gibbenergy"][quote="parinaz.sa"]if $a,b,c$are positive numbers prove:
$\frac 1{a^3+b^3+abc}+\frac 1{b^3+c^3+abc}+\frac 1{c^3+a^3+abc}\leq \frac 1{abc}$[/quote]

  Use $a^3+b^3 \geq ab(a+b)$  :)[/quote]
Good job, that's really nice. totally kills the ineq :lol:


This is something I seem to have stumbled upon.
If anyone wants to use it or, PLEASE prove it.

TO find a Pythagorean triple when given the smallest number, $ x$.

If $ x$ is odd, take $ \frac{x^2}{2}$ minus 1/2 to get the smaller, and plus 1/2 to get the bigger numbers in the Pythagorean triple.

If $ x$ is a multiple of 4, take $ \frac{x^2}{4}$ minus 1 to get the smaller, and plus 1 to get the bigger numbers in the Pythagorean triple.

If $ x$ is even and not a multiple of 4, it is only in other triples of ratios earlier determined. Divide by 2, then plug into the first part.

i'm not sure if i understand you correctly, but from what i understand, 4 doesn't work for the second one

$ \frac {4^2}{2} \minus{} 1$

$ \frac {16}{2} \minus{} 1$

$ 8 \minus{} 1$

$ 7$

other number is $ 9$

4-7-9 is not a pythagorean triple....

edit: oh!
thanks BOGTRO for telling me
nvm then

[quote="vallon22"]i'm not sure if i understand you correctly, but from what i understand, 4 doesn't work for the second one

$ \frac {4^2}{2} \minus{} 1$
$ \frac {16}{2} \minus{} 1$
$ 8 \minus{} 1$
$ 7$

other number is $ 9$

4-7-9 is not a pythagorean triple....[/quote]

For the second one its x^2/4.

For the first one its for odd x.

Proofs:

Let $ x\equal{}4n$. Thus, $ 4n^2\minus{}1$, $ 4n$, $ 4n^2\plus{}1$. This obviously works.

Let $ x\equal{}2n\minus{}1$. Thus $ \frac{4n^2\minus{}4n\plus{}1}{2} \pm \frac{1}{2}\equal{}2n^2\minus{}2n, 2n^2\minus{}2n\plus{}1$.

$ 4n^2(n\minus{}1)^2\plus{}4n^2\minus{}4n\plus{}1\equal{}4n^2(n\minus{}1)^2\plus{}4n^2\minus{}4n\plus{}1$. True.

Of course, one has to set some sort of lower bound such that n>2 (clearly 2,0,-2 doesn't work)!

This is quite a well-known theorem of deriving Pythagorean Triples

I think this should go somewhere else, like in the Topics Help Forums...
mods plz move this?


This forum died, so I will try to start it up again.

Silly Sally Sue is at the origin of the Cartesian Coordinate plane. She wants to get some ice cream, and the ice cream store is at the point $ (x,y)$, where $ x,y>0$. In how many ways can she do this?

[quote="1=2"]This forum died, so I will try to start it up again.

Silly Sally Sue is at the origin of the Cartesian Coordinate plane. She wants to get some ice cream, and the ice cream store is at the point $ (x,y)$, where $ x,y > 0$. In how many ways can she do this?[/quote]

Answer is simply: $ \boxed {\infty}$

For example if x=1 and y=1, she can go to (0;52215) then (1;52215) and then (1;1)...
You can change number 52215 into another number, also statement doesn't say that x and y are integers...

[quote="1=2"]If she can only increase her x or y coordinate(but not both) by one each step?[/quote]

and x and y are real numbers?

Are they integers? Because if they aren't (which happens 100% of the time if they are selected randomly) you cannot reach it by increasing by 1!

*sigh*, you guys all know what he means.... :|

$ \binom{|x| \plus{} |y|}{|x|}$

Well, $ x,y > 0$

so I guess you can just write $ \binom{x\plus{}y}{x}$

[quote="Eulers_Apprentice"]Well, $ x,y > 0$

so I guess you can just write $ \binom{x \plus{} y}{x}$[/quote]

$ \binom{x \plus{} y}{x}$ means $ \frac {(x\plus{}y)!}{x! \cdot y!}$ ?

...  This is the cartesian coordinate plane. That means it includes the real numbers, the whole of the real numbers, and nothing but the real numbers. The origin is at (0,0). You increase either the x coordinate or y coordinate by 1 each step. Since 0 and 1 are integers, and you are adding them, you must get another integer. You must get integers the whole way. Come on, you are exceptionally bright people.

On another note, does anyone have a solution for this?

I'm not sure what you're looking for.
The answer has already been stated multiple times: $ \binom{x\plus{}y}{x}$. This is because no matter what path you choose there will be a total of $ x\plus{}y$ steps. Then, you have to choose $ x$ of them to move right, and the remaining steps will move up.

For example, if the store is in $ (1,1)$, you can reach it in $ \binom{2}{1} \equal{} 2$ ways.
If it's in $ (2,3)$, you can reach it in $ \binom{5}{2} \equal{} 10$ ways, etc...


The numbers from 1 to 1996 are written down ------ 12345678910111213.... How many zeros are written?

A. 489
B. 699
C. 796
D. 996
E. None of these

if i have not made a mistake or miss understood the problem  i think e) none of these

here's why

[1996/10]+[1996/100]+[1996/1000]=219

(because [1996/10] has at least one zero,[1996/100] has a least two but 1 of them has already been counted...., similar for [1996/1000])

You forgot that  a multiple of ten larger than 1000 but less then 1100, like 1040, actually has two zeros instead of 1.
[hide]
If you want to, you can just list all the numbers with zeros in them:
10
20
30
40
50
60
70
80
90
100
110...
1000
1010
1020
1030
1040
1050
1060
1070
1080
1090
1100...
1990

We see that there are 11 zeros per set of hundred, up to 810-900, and then there are 12 for the 910-1000, 20 in the 1010-1100, and back to 11 for the rest of the numbers. We have $ (9*11) \plus{} 12 \plus{} 20 \plus{} (9*11) \minus{} 1$ (for the case of no 2000). It simplifies to [u]229[/u], close to your answer, but different;however, I also got [b]E[/b], none of these.[/hide]

We count the number of $ 0$s in the units digits place, then the tens digit, then the hundreds digit.
[list]
[*]If $ 0$ is the units digit, then we just want the number of multiples of $ 10$, and since $ 1990$ is the largest multiple of $ 10$ in the given domain, this can be found by $ 1990 \equal{} 10 \plus{} (n \minus{} 1) \times 10 \Longrightarrow n \equal{} 199$.
[*]If $ 0$ is the tens digit, there are $ 10$ numbers with zeros in the tens digit for every hundred numbers, starting with $ 100\minus{}109$ to $ 1900\minus{}1909$. Thus we have $ 19 \times 10 \equal{} 190$ here.
[*]If $ 0$ is the hundreds digit, there are $ 100$ numbers with zeros in the hundreds digit from $ 1000$ to $ 1099$. This gives us $ 100$.[/list]

These sum up to $ 199 \plus{} 190 \plus{} 100 \equal{} 489 \Rightarrow \mathrm{(A)}$.

Remember that the question is asking for the number of zeros written, not the number of numbers that have zeros in them.

[quote="azjps"]We count the number of $ 0$s in the units digits place, then the tens digit, then the hundreds digit.
[list]
[*]If $ 0$ is the units digit, then we just want the number of multiples of $ 10$, and since $ 1990$ is the largest multiple of $ 10$ in the given domain, this can be found by $ 1990 \equal{} 10 \plus{} (n \minus{} 1) \times 10 \Longrightarrow n \equal{} 199$.
[*]If $ 0$ is the tens digit, there are $ 10$ numbers with zeros in the tens digit for every hundred numbers, starting from $ 100$ to $ 1900$. Thus we have $ 19 \times 10 \equal{} 190$ here.
[*]If $ 0$ is the hundreds digit, there are $ 100$ numbers with zeros in the hundreds digit from $ 1000$ to $ 1099$. This gives us $ 100$.[/list]

These sum up to $ 199 \plus{} 190 \plus{} 100 \equal{} 489 \Rightarrow \mathrm{(A)}$.

Remember that the question is asking for the number of zeros written, not the number of numbers that have zeros in them.[/quote]
Your second bullet point is wrong. You say there are 10 numbers with a 0 in the tens place per hundred numbers, but take 10-100:
10
20
30
40
50
60
70
80
90
100.
Only 100 has a zero in the tens place

[quote="MrRubix"]
Your second bullet point is wrong. You say there are 10 numbers with a 0 in the tens place per hundred numbers, but take 10-100:
10
20
30
40
50
60
70
80
90
100.
Only 100 has a zero in the tens place[/quote]

His bullet point is still correct, as he started counting from 100, not 0.  :D


geometry
3D geometry
sphere
geometry unsolved

Carpenters with a small workshop use the following to draw a circular arc with a big radius:
They fix together two long boards at an angle and let them glide along two distant nails. The top describes an arc of a circle.
The 3D analogon could be the following:
Let an equal sided trihedron glide along a circular hole so that the edges always touch the circle. Is the locus of the top a sphere cap?
For a rectangular trihedron I guess the radius is equal to (3*R)/(2*2^1/2) with R=radius of the circular hole.
Is this so? If so does it need to be an equal sided trihedron? Are there any internet-references for a proof (or disproof) or do you have one yourself?
kweenie

[quote="kweenie"]Is the locus of the top a sphere cap?
[/quote]
No, I just found that for a rectangular trihedron it could be a flat ellipsoid with the hole diameter as long axis and R/2^1/2 as heigt. I'm not sure if it is always 1 single surface.
But I'm still interested if there exists some 3D analogon for the 2D "inscribed angle theorem"
kweenie


If $n\in N*$ solve the equation $19\cdot [\frac{12}{n^2+3n}(1+\sqrt{2!}+...+\sqrt[n] {n!} ]=n$ where $[...]$ is the integer part of the number.

$\sqrt[k]{k!}<\frac{1+2+...+k}{k}=\frac{k+1}{2}, \forall k=2,3,...,n$
The sum will be then $< \sum{\frac{k+1}{2}}=\frac{n(n+3)}{4}$.
It follows that the integer part in the equation is < 3. It can be only 1 or 2, since if it is 0 $\Rightarrow$ n=0, which is excluded.
Then it is [b]necessary[/b] that n should be 19 or 38.
But is it also [b]sufficient[/b]?

Nice solution .With the integer part we symbolise the bigger integer which is $\leq x$.

So $n=38$.

Ok (we use the same symbolization), but how do you prove that the number inside the integer part is in in the interval $[2, 3)$ for n=38? It could be in $[1, 2)$ as well.
Also, to be sure that n=38 is the only solution, we have to prove that if  n=19 the number inside is in $[2,3)$, not in $[1,2)$.


Find the lattice point in the first octant (where $x,y,z>0$) such that $7x+6y+2z=272$ and $x-y+z=16$ that maximizes $z$.

[hide="Hint"]$z = 33$[/hide]

[hide="Solution"]Add a linear combination to eliminate $y$...
$13x+8z = 368$
Mod 8, we get that $13x \equiv 0$, so $x$ is a multiple of 8.

Eliminating $x$
$13y-5z = 160$
Mod 5, we get that $13y \equiv 0$, so $y$ is a multiple of 5.

Eliminating $z$
$5x+8y = 240$

Let $x = 8k$ and $y = 5m$, then
$40m+40k = 240$
$m+k = 6$

Go to the first equation with this substitution
$56k+30m+2z = 272$
$26k+180+2z = 272$
$13k+z = 46$

Since $x > 0$, $k > 0$. If we let $k = 1$, then $z = 33$. This is a maximum if it works.

$k = 1 \Rightarrow x = 8$
Substitute into the second equation: $8-y+33 = 16$, $y = 25$
In the first equation, $56+150+66 = 272$, so it checks.

The lattice point is $(8,25,33)$.[/hide]

[hide]
The base point that satisfies both equations is $(0, 30, 46)$. Now note that everytime you increase x by 2, y must decrease by 2, and z decrease by 1 meaning it adds 3 in $x-y+z$. However, you can increase y by 1 and decrease z by 3 meaning you subtract 4 from $x-y+z$. Since the LCM of 3 and 4 is 12, we must increase x by 4*2, y must decrease by 4*2, z must decrease by 4*1, and then we must increase y by 3*1, and decrease z by 3*3 giving $(8,25,33)$.[/hide]


Each of the figures is a square formed by connecting midpoints of the opposite sides of a larger square. What fraction of the largest square is shaded?
[asy]size(150); defaultpen(linewidth(0.8));
void drawSquare(real sidel, pair blcorner){
 draw(blcorner--(blcorner.x + sidel,blcorner.y)--(blcorner.x + sidel, blcorner.y + sidel)--(blcorner.x,blcorner.y + sidel)--cycle);
}

void fillQSquare(real sidel, pair position, pair posCorner){
 if(position == SW){
  fill(posCorner--(posCorner.x +0.5*sidel,posCorner.y)--(posCorner.x + 0.5*sidel, posCorner.y + 0.5*sidel)--(posCorner.x, posCorner.y + 0.5*sidel)--cycle);
 }
 if(position == SE){
  fill(posCorner--(posCorner.x - 0.5*sidel,posCorner.y)--(posCorner.x - 0.5*sidel, posCorner.y + 0.5*sidel)--(posCorner.x, posCorner.y + 0.5*sidel)--cycle);
 }
 if(position == NW){
  fill(posCorner--(posCorner.x +0.5*sidel,posCorner.y)--(posCorner.x + 0.5*sidel, posCorner.y - 0.5*sidel)--(posCorner.x, posCorner.y - 0.5*sidel)--cycle);
 }
 if(position == NE){
  fill(posCorner--(posCorner.x - 0.5*sidel,posCorner.y)--(posCorner.x - 0.5*sidel, posCorner.y - 0.5*sidel)--(posCorner.x, posCorner.y - 0.5*sidel)--cycle);
 }
}
drawSquare(1,origin); drawSquare(2,origin); drawSquare(4,origin); drawSquare(8,(-4,-4));
fillQSquare(1,NE,(1,1)); fillQSquare(2,NE,(2,2)); fillQSquare(4,NE,(4,4)); fillQSquare(8,SW,(-4,-4));[/asy]


If the Republican party nominee was Sarah Palin, the Democratic party nominee was Joe Biden, and the Independent party nominee was Matt Gonzalez, who would you vote for?

I would vote either Sarah Biden or Joe Palin.


group theory
abstract algebra
algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

1) Let $ K / F$ be Galois with $ G\equal{}Gal(K / F)$, and let $ L$ be an intermediate field. Let $ N \subseteq K$ be the normal closure of $ L / F$. If $ H\equal{}Gal(K/L)$, show that $ Gal(K/N)\equal{} \bigcap_{\sigma \in G} \sigma H {\sigma} ^{\minus{}1}$.

2) Give me an example of a finite extension $ K / F$ which is not normal such that there are no proper intermediate extension of $ K / F$, and $ [K: F]$ is not prime.

1) I can see some similarities to [url=http://www.mathlinks.ro/viewtopic.php?t=217995]this[/url].

2) actually, the main problem is to find a counterexample for this: Let $ K$ be a finite normal extension of $ F$ such that there are no porper intermediate extensions of $ K / F$. Show that $ [K : F]$ is prime. (it can be solved by purely inseparable closure or sth like that.) Give a counterexample if $ K$ is not normal over $ F$.

1. By Galois correspondence, $ Gal(K/N)$ should be the largest $ G$-normal subgroup in $ Gal(K/L)$. Then it is trivial that that expression is what we look for.

I don't think that we can use the Galois correspondence so offhandedly if $ [K: N]$ is infinite.  We need to point out that $ H$ is a closed subgroup, hence $ \bigcap_{\sigma\in G} \sigma H \sigma^{\minus{}1}$ is a closed subgroup.

For the second question $ Q(\sqrt[4]{2})$ should work.
It is of degree 4 over $ Q$ and is not normal since its minimal polynomial is $ x^{4}\minus{}2$ which facors to $ (x\minus{}\sqrt[4]{2})(x\plus{}\sqrt[4]{2})(x\minus{}i\sqrt[4]{2})(x\plus{}i\sqrt[4]{2})$ and $ i$ isnt in the field.

For the first question it is clear that $ H'\equal{} \bigcap_{\sigma\in G}\sigma H'\sigma^{\minus{}1}  \subseteq  \bigcap_{\sigma\in G}\sigma H\sigma^{\minus{}1}$ since $ H'$ is a normal subgroup of  $ G$ and is contained in $ H$, but I have yet to figure out the other way around.

No, $ \mathbb Q[\sqrt [4]2]$ doesn't work.

In fact, the only possible cases are the separable algebraic finite extensions:
If the extensions is not algebraic, every transcendental $ x$ gives an intermediate field $ F(x^2)$, so we are reduced to the algebraic case. If the extension is neither purely inseparable nor separable, one gets the separable closure as intermediate field.
The purely inseparable case is handled by the fact that if $ x\in K \backslash F$, then for some $ s$ we have $ x^{p^s} \in F$. Take the smallest such $ s$ and consider $ F[x^{p^{s \minus{} 1}}]$, it is strictly between $ F,K$.
The infinite separable case is done by just adjoining any element from $ K \backslash F$.

Now after knowing this, we see that we are looking for a finite separable extension. After working in a normal closure, we are looking for a Galois group $ G$ with maximal subgroup $ H$ of nonprime index. I recommend $ A_5 \subset A_6$ or $ S_5 \subset S_6$, which can be constructed from elementary symmetric functions.

[quote="Yotam"]For the second question $ Q(\sqrt [4]{2})$ should work.[/quote]

no, $ \mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt[4]{2})$.

@ZetaX: thanks, it's really helpful.


Let S(N) denote the sum of the digits of the natural number N.

Prove that there are infinitely many natural numbers N such that

S(N^2)=S(N)

if $ N\equal{}10^k$ then $ S(N)\equal{}S(N^2)\equal{}1$

Obviosly $ N\equal{}45*10^n$ work.
Let $ N\equal{}\frac 12 10^n\minus{}y, n>1\plus{}[2lg y]$, then $ S(N)\equal{}9n\minus{}S(y)$ and $ S(N^2)\equal{}9n\minus{}11\minus{}S(y)\minus{}S(y^2)$.
Therefore $ N$ is solution if $ S(y^2)\equal{}7$. It give solution $ N\equal{}49...950...0$.

[quote="Rust"]Obviosly $ N \equal{} 45*10^n$ work.
Let $ N \equal{} \frac 12 10^n \minus{} y, n > 1 \plus{} [2lg y]$, then $ S(N) \equal{} 9n \minus{} S(y)$ and $ S(N^2) \equal{} 9n \minus{} 11 \minus{} S(y) \minus{} S(y^2)$.
Therefore $ N$ is solution if $ S(y^2) \equal{} 7$. It give solution $ N \equal{} 49...950...0$.[/quote]

$ n>1\plus{}[log_{10}(y^2)]\equal{}number\ of\ gidits\ of\ y^2$

i think that $ S(\frac{1}{2}10^n\minus{}y)\equal{}S(5.10^{n\minus{}1}\minus{}y)\equal{}5\plus{}9h\minus{}S(y)$ where $ h\equal{}E(log_{10}(y))\plus{}1$
so $ S(N) \neq 9n \minus{} S(y)$.

and $ N^2\equal{}25.10^{2n\minus{}1}\minus{}y10^{n}\plus{}y^2$ gives $ S(N^2)\equal{}2\plus{}5\plus{}9h\minus{}S(y)\plus{}S(y^2)\equal{}S(N)\plus{}2\plus{}S(y^2)>S(N)$

[quote="aviateurpilot"][quote="Rust"]Obviosly $ N \equal{} 45*10^n$ work.
Let $ N \equal{} \frac 12 10^n \minus{} y, n > 1 \plus{} [2lg y]$, then $ S(N) \equal{} 9n \minus{} S(y)$ and $ S(N^2) \equal{} 9n \minus{} 11 \minus{} S(y) \minus{} S(y^2)$.
Therefore $ N$ is solution if $ S(y^2) \equal{} 7$. It give solution $ N \equal{} 49...950...0$.[/quote]

$ n > 1 \plus{} [log_{10}(y^2)] \equal{} number\ of\ gidits\ of\ y^2$

i think that $ S(\frac {1}{2}10^n \minus{} y) \equal{} S(5.10^{n \minus{} 1} \minus{} y) \equal{} 5 \plus{} 9h \minus{} S(y)$ where $ h \equal{} E(log_{10}(y)) \plus{} 1$
so $ S(N) \neq 9n \minus{} S(y)$.

and $ N^2 \equal{} 25.10^{2n \minus{} 1} \minus{} y10^{n} \plus{} y^2$ gives $ S(N^2) \equal{} 2 \plus{} 5 \plus{} 9h \minus{} S(y) \plus{} S(y^2) \equal{} S(N) \plus{} 2 \plus{} S(y^2) > S(N)$[/quote]
Let $ N\equal{}a*10^n\minus{}b$, were $ 10^n>max(b^2,2ab$.
Then $ S(N)\equal{}S((a\minus{}1)*10^n\plus{}1\plus{}(10^n\minus{}1\minus{}b))\equal{}S(a\minus{}1)\plus{}1\plus{}9n\minus{}S(b)$ (my calculate for $ a\equal{}5,b\equal{}y$ is true).
$ S(N^2)\equal{}S(a^2*10^{2n}\minus{}2ab*10^n\plus{}b^2)\equal{}S(a^2\minus{}1)\plus{}1\plus{}9n\minus{}S(2ab)\plus{}S(b^2)$ (my calculate is true).
We can solve $ S(a^2\minus{}1)\minus{}S(2ab)\plus{}S(b^2)\equal{}S(a\minus{}1)\minus{}S(b)$ and find other solutions.


z1,z2,z3 are complex numbers

 |z1|=|z2|=|z3|

  z1*z2*z3=z1+z2+z3=1

Find the value of z1,z2,z3.

This cant be done. There is not enough information. Since they all have the same magnitude, and multiply to 1, we can say that
$|z_{1}|=|z_{2}|=|z_{3}|=1$

Let them be the roots of a cubic polynomial.

The polynomial must be of the form

$x^{3}-x^{2}+Bx-1=0$

For any B, the roots of the polynomial satisfy the requirements, so there is no set value for the roots

(If we let B=1, then $i$, $-i$, $1$ works, but we get entirely different values if we let B be any other number)


Does there exist $A\subset N$ such that:
(1) For any infinite $S\subset N$ one can find $k$ distinct elements of $S$
the sum of which is from $A$.
(2) There is no infinite $S\subset N$ and $n\in N$ such that sum of any $n$ distinct numbers from
$S$ belongs to $A$?

(N is set of natural numbers as usual)

[quote="sbos"](1) For any infinite $S\subset N$ one can find $k$ distinct elements of $S$
the sum of which is from $A$.
[/quote]
Who is $k$?.


function
logarithms
Euler
induction
advanced fields
advanced fields unsolved

Let $ d$ be a natural number, and $ x\ge 1$,

prove that            $ \sum_{m\le x\atop (m,d)\equal{}1}\frac{\mu(m)^{2}}{\phi(m)}\ge\frac{\phi(d)}{d}\log([x]\plus{}1)$,

where $ \mu$ is the Mobius function, and $ \phi$ is the Euler function.

Thanks!

denote $ h(x,d) \equal{} \sum_{m\le x\atop (m,d) \equal{} 1}\frac {\mu(m)^{2}}{\phi(m)}$
 by induction, $ h(x,d) \equal{} \sum_{(p,d) \equal{} 1,\atop p is prime, p\geq x} \frac {1}{p \minus{} 1}\frac {\phi(pd)}{pd}log(\frac {x}{p}) \equal{} \sum\frac {1}{p}(logx \minus{} log p)\frac {\phi(d)}{d}$$ \geq \sum_{p\leq \sqrt {x}, (p,d) \equal{} 1}\frac{1}{p}\frac {\phi(d)}{d}\frac {logx}{2}$ since for sufficiently large $ x$, $ \frac {1}{2}\sum_{p\leq \sqrt {x}\atop (p,d) \equal{} 1}\frac {1}{p}$ will utimately larger than 1, thus complete half of the proof.

since change the factor of $ d$ for smaller one will always increase the right part and decrease the left part, it suffices to assume $ d$ is the product of the first $ k$ primes.

Well, still a part remains.


linear algebra
matrix
induction
complex numbers
superior algebra
superior algebra solved

A \in M_2(R), detA  \geq 1, a(n),b(n),c(n),d(n) s.t.
A^n=
(a(n) b(n))
(c(n) d(n))
Prove that if the four sequence converge then A=I_2

Lets denote d=det(A). Then d^n=a_n*d_n-b_n*c_n for every n>=1.
so we get that (d^n)n>=1 converges and because d>=1 we get that d=1. From Cayley's theorem we get that: A^2-tr(A)*A+I_2=0_2. SO we get that A^(n+2)-tr(A)A^(n+1)+A^n=0_2 for any n>=1 so a_n,b_n,c_n and d_n verifies the recurence: x_(n+2)-tr(A)x_(n+1)+x_n=0 for any n>=1.
It is obious that at least one of the four sequences has the limit l <>0 ; otherwise we would get d=0, contradiction. so for the sequence we get that l[2-tr(A)]=0 so tr(A)=2 which means A^2=I_2+2(A-I_2). because A=I_2+(A-I_2) using induction we get that A^n=nA+(1-n)I_2 for any n >=1.
so we get that a_n=n(a_1-1)+1; b_n=nb_1; c_n=n*c_1 and d_n=n(d_1-1)+1. because a_n, b_n , c_n and d_n are convergent we get that a_1=1; b_1=0; c_1=0 and d_1=1 from where we get that A=I_2.

cheers! :D :D

Great  problem, and beautiful solution Lagrangia.

I always believe that a mathematics problems has several different solutions.

Here another one ( found at ~ 01h25 am )
Consider A as a matrix in M_2(C), then A is similar to a triangular superior
there exist P \in M_2(C) invertible such that
A=P^(-1).T.P  where
T=
(k q)
(0 k*),  k* is the complex conjugate of k, A is real coefficients
Since a(n),b(n),c(n),d(n) are convergente sequence, A^n converge
A^n=
P^(-1)T^nP
T^n is also convergent
T^(n)=
(k^n z(n))
(0 (k*)^n) where z(n) depend of k,q
1 \leq (detA)^n=|k|^(2n) , (detA)^n convergent => |k|=1 modulus of k is 1
so k=exp(i.w)
A^n=
P((exp(inw) z(n)).P^(-1)
...(0  (exp(inw))
The convergence of A^n => exp(inw) is also convergent to L=/=0
but exp(i(n+1)w)=exp(iw).exp(inw) take the limit L=exp(iw)L
so exp(iw)=1 => w=2k \pi => k=1
A=
P(1 q)P^(-1)
..(0 1)
By induction
A^n=
P(1 nq)P^(-1)
..(0 1)
=
(a(n) b(n))
(c(n) d(n))
since the four sequences converge=> the sequence nq converge if and
only if q=0
So
A=
P(1 0)P^(-1)
.. (0 1)
=I_2


If I solve on an inequality.. and my answer is 5 < 31 but it originally had variables in it, is it all numbers because its true? The original problem is: 3 - 3(b-2) < 13 -3(b-6) by the way.
Thanks :lol:

$ 3 \minus{} 3(b \minus{} 2) < 13 \minus{} 3(b \minus{} 6)$

$ \implies \minus{} 3b \plus{} 6 < 10 \minus{} 3b \plus{} 18$

$ \implies6 < 28$, which is always true.

Our steps are reversible from the inequality $ 6 < 28$, so we can say

$ \boxed{b\in\mathbb R,\minus{}\infty\leq b\leq\plus{}\infty,b\in(\minus{}\infty,\plus{}\infty)}$, etc.

Note also that if we get a result like 1>5, this means that our inequality works for no values of the variable.

To sum it up, if you have no varibles left over, and the equation or inequality is true, then $ x \in (\minus{}\infty,\infty)$, as math154 said.  If the equation or inequality is not true, then no answer exists.


Find all values of $a$ such that there are two real solutions to the equation

\[x^{3}+(a-1)x^{2}+(9-a)x-9=0\]

[hide="source"]
I'm not even sure how I got there in the first place, but lets just say that wikipedia is not up to par when it comes to math!
Scroll down to the bottom of the page for this problem...
http://en.wikibooks.org/wiki/Talk:HSE_Complex_number
[/hide]
[hide="my solution"]
Clearly, we must have: \[(x-p)^{2}(x-q)=x^{3}+(a-1)x^{2}+(9-a)x-9\]
Since this is an identity in $x$, it is true for any particular $x$. For instance when $x=1$, we notice that the left hand side is identically zero, \[(p-1)^{2}(q-1)=0\] so one of $p,q$ is unity. At $x=0$, we have: \[p^{2}q=9 \]From these two equations, $p,q$ can be solved from.

Indeed, if $p=1$ from the first then $q=9$ from the second. If $q=1$ from the first, then $p=\pm 3$ from the second. It follows that the set of solutions is $(p,q)=\{(1,9),(3,1),(-3,1)\}$.

When $x=-1$ in the original cubic, we have \[2a-20=(p+1)^{2}(-1-q)\]
Thus all possible values of $a$ can be determined from all possible values of $(p+1)^{2}(-1-q)$. Checking, this equals $-40,-32,-8$ which yields $\boxed{a\in\{-10,-6,6\}}$ as the solutions.
[/hide]

[hide="Alternately"] $x^{3}-x^{2}+9x-9+ax^{2}-ax = x^{2}(x-1)+9(x-1)+ax(x-1) = (x^{2}+ax+9)(x-1)$

[b]Case:[/b]  $x = 1$ is a root of the quadratic.  This gives $a = \boxed{-10}$.

[b]Case:[/b]  The quadratic has one real root.  This gives $a = \boxed{ \pm 6}$. [/hide]

Yes, that's implied.
I was going to write that in but thought it was clear enough and decided to keep it word for word...
t0rajuir0u solution was exactly the same as theirs, except after the initial factorization they got stuck...


polynomial
quadratics
algebra
number theory proposed
number theory

Prove that are no a,b,c rational  and c>0 with
 2^1/3=a+b*sqrt(c)

$2^{1/3}$ is a root of $x^3-2$ which is irreducible (for example by eisenstein)
But $a+b\sqrt c$ is a root of a quadratic polynomial (Just take the other root to be $a-b\sqrt c$)
So it is obvious now!

nice solution Nima Ahmadi Pour ,it is one of the few i understand


[quote="harazi"]Condition can be written $\frac{1}{a_2+a_3}>\frac{a_1}{S}$. Add this.[/quote]

I can understand $\frac{1}{a_2+a_3}>\frac{1}{a_1+a_2+a_3}$, but can't your claim.

Or am I missing something?

Yes, you didn't carefully read the hypothesis! :)

If your claim is true, we have $S-a_1(a_2+a_3)=(a_1+a_2+a_3)-a_1(a_2+a_3)=a_1+(1-a_1)(a_2+a_3)>0$.

Please explain this, harazi.

Kunny! We have from the hypothesis $ a_1^2>a_1S-S$ from where $ S>a_1(S-a_1)=a_1(a_2+a_3)$!

Thank you for your time, harazi.

I intended to see the hypothesis, but couldn't notice how to use this.

Here is my idea.

$\frac{a_1^2}{a_1-1}=a_1+1+\frac{1}{a_1-1}>a_1+a_2+a_3 \Longleftrightarrow \frac{a_1}{a_1-1}>a_2+a_3$

Since $a_1>1$, we have $\frac{1}{a_2+a_3}>\frac{a_1-1}{a_1}=1-\frac{1}{a_1}$ etc.

We have $\frac{1}{a_1+a_2}+\frac{1}{a_2+a_3}+\frac{1}{a_3+a_1}>3-\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)$. Is this wrong basically?

kunny

$x,y,z > 1$
$\frac{x^2}{x-1} > x+y+z  \Rightarrow x+y+z > x(y+z) \Rightarrow \frac{1}{y+z} > \frac{x}{x+y+z}$

summing leads desired

$\frac{a_i^2}{a_i-1}>S \longrightarrow \frac{a_i}{a_i-1}>S-a_i\longrightarrow a_i>\frac{S-a_i}{S-a_i-1}=1+\frac{1}{S-a_i-1}\longrightarrow a_i<\frac 32\Longrightarrow S<\frac 92$

$\frac{1}{a_1+a_2}+\frac{1}{a_1+a_3}+\frac{1}{a_2+a_3}\ge \frac{9}{2(a_1+a_2+a_3)}>1$


Let $P,Q$ be two monic polynomials with complex coefficients such that $P(P(x))=Q(Q(x))$ for all $x$. Prove that $P=Q$.

[i]Marius Cavachi[/i]

they must both be of the same degree, say degree $n$.

$P(x)=x^n+a_1x^{n-1} + \ldots + a_n$

$Q(x)=x^n+b_1x^{n-1}+ \ldots + b_n$

equating the $x^{n(n-1)}$ terms in $P(P(x))$ and $Q(Q(x))$, we have $a_1^{n}x^{n(n-1)}=b_1^{n}x^{n(n-1)} \rightarrow a_1=b_1$.

we can continue in this manner, equating the $x^{n(n-i)}$ terms for $i \in {1, 2, \ldots n}$ to show that $a_i=b_i$ for those $i$, which means $P=Q$.

Also posted here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=237348

Sorry for the revival, but the above solution does not seem to be quite correct. The fact that $a_1^n = b_1^n$ does not imply $a_1 = b_1$, especially when working with complex numbers.


modular arithmetic
inequalities
calculus
integration
number theory
number theory solved
Hi

Find all positive integer solutions to $2^x - 5 = 11^y$.

[hide="Comment (some ideas)"]Obviously I suspect the only solution is $(x,y) = (4,1)$. It's easy to prove that $y \equiv 1 \pmod 4$ and $x\equiv 0 \pmod 4$, but other than that, and trying out modulo 121, I didn't get much far on this one ... I might be missing something trivial though :|[/hide]

Make a substitution $x \to 4x$ (as Valentin said it is easy to prove - look modulo $5$). So our equation will take form $(4^x-\sqrt{5})(4^x+\sqrt{5})=11^y$. Suppose that there is a prime $p$ which divides both $4^x-\sqrt{5}$ and $4^x+\sqrt{5}$. Then $p|2 \cdot 4^x$ but also $p|11^y$ - contradiction. Hence $(4^x-\sqrt{5},4^x+\sqrt{5})=1$ and both are $y$-th powers. Because $11^y=(4+\sqrt{5})^y(4-\sqrt{5})^y$ we must have $4^x+\sqrt{5}=(4+\sqrt{5})^y$. Now using binomial expansion we get (looking at $\sqrt{5}$) $1=\sum \binom{y}{2k+1}5^k4^{y-(2k+1)}$ and from it it is easy to see that we must have $y=x=1$ (because else - right side would be larger).

Hence the only solution is $(4,1)$.

Does it make any sense ?  :)

[quote="Valentin Vornicu"][quote="Megus"]Does it make any sense ?  :)[/quote]No it doesn't. Let me tell you why: because $\mathbb{Z}[\sqrt 5]$ is not an UFD. For example \[ 3\cdot 3 = 9 = (2-\sqrt 5 )(2+\sqrt 5) .  \][/quote]

Ooops :blush:

i  remember  seeing these type of diophantine equations being solved using algebraic number theory ,  and if congruences dont work then inequalities is my  biggest  bet.  but thats just my  opinion . :lol:

Could we try this:
Appearently, $(4,1)$ is a solution
Assume that there's another solution $(x_1, y_1)$ other than this one, and of  course bigger than this one.(I mean, $x_1>4, y_1>1$)

$\Rightarrow 2^{x_1}-2^4 = 11^{y_1}-11$
$\Rightarrow (2^{x_1-4}-1)(2^4) = 11(11^{y_1-1}-1)$
$\Rightarrow 11|2^{x_1-4}-1$ Simply because 11 is a prime number
$\Rightarrow \text{No Solution other than} (4,1) !!!$ :P

Unfortunately $11|2^{x_1-4}-1$ has numerous solutions, just let $x_1-4=9k+1$ ($k$ integer of course).

Disclaimer: I am tired, and I might have done something wrong. But at least the last time I checked, $11|1024-1$ etc.

Of course you cannot solve this modulo 11, since there is a solution! You have to try at least modulo 121 or 32. I tried both these, and modulo 64 also. Nothing :)

I guess the Unique factorization can be used, but we must use $Z[\sqrt{11}]$

As you yourself have shown, $x = 4m$ and $y = 4k+1$.

Let $2^{2m} = a$ and $11^{2k} = b$. Then we have that

$a^2 - 11b^2 = 5$
i.e

$(a-b\sqrt{11})(a+b\sqrt{11}) = 5$.

Now we can apply the fact that $Z[\sqrt{11}]$ is a UFD. Which is true, I think.

[quote="Valentin Vornicu"][quote="Valentin Vornicu"][quote="Megus"]Does it make any sense ?  :)[/quote]No it doesn't. Let me tell you why: because $\mathbb{Z}[\sqrt 5]$ is not an UFD. For example \[ 3\cdot 3 = 9 = (2-\sqrt 5 )(2+\sqrt 5) .  \][/quote]Now why didn't anyone notice the [b]HUGE[/b] abberation I have wrote above? :)

I ment $\mathbb{Z}[\sqrt{ - 5} ] = \mathbb{Z}[i\sqrt 5]$ is not an UFD (indeed $9\neq (2-\sqrt 5 ) (2 +\sqrt 5 ) = 4-5 = -1$ :D).

Actually both $\mathbb{Z}[\sqrt 5]$ and $\mathbb{Z}[\sqrt{11}]$ are Euclidean rings, thus UFDs also, so Megus' solution above is correct :)  :blush:[/quote]

Oh, good to hear  that :)  - but that taught me one thing - next time I'll check whether the ring I use is UFD  :D

I checked again and found those interesting stuffs
1. the endings of $11^y +5$ are always "$26$","$46$", "$66$", "$86$".(dunno how to prove this, maybe there's some exceptions)
2. the endings of $16^{x/4}$ are always "$16$","$36$", "$56$", "$96$".
(Appearently, x is a multiple of 4)

[quote="Kalimdor"]1. the endings of $11^y +5$ are always "$26$","$46$", "$66$", "$86$".(dunno how to prove this, maybe there's some exceptions)[/quote]That cannot be correct, as $11^y + 5 \equiv 16 \pmod {100}$ when $y \equiv 1 \pmod {\varphi (100) = 40}$.

you can also solve this modulo 640(you can check it out) :wink:  :D but i'm too lazzy to write it down. :blush:

Actually, you can use mods 64 and 17.
[hide]
To show no solution for $x\geq 6$, use mod 64 to get $11^{y}\equiv-5\pmod{64}$ which implies $y\equiv-3\pmod{16}$. (It's not hard to show that $11^{8}-1$ is divisible by 32 but not 64, and it just so happens that $-5\cdot 11^{3}\equiv 1\pmod{64}$.

Now mod 17 works because $11^{16}\equiv 1\pmod{17}$ by Fermat's Little Theorem, and $11^{-3}+5\equiv 12$. However, the powers of 2 mod 17 are just $\pm 1,\pm 2,\pm 4,\pm 8$, none of which are 12.
[/hide]

[quote="ZetaX"][quote="Valentin Vornicu"]Actually both $ \mathbb{Z}[\sqrt 5]$ and $ \mathbb{Z}[\sqrt{11}]$ are Euclidean rings, thus UFDs also, so Megus' solution above is correct :)  :blush:[/quote]
Just a small correction: $ \mathbb{Z}[\sqrt 5]$ is in fact not an UFD (e.g. $ 2 \cdot 2 = (\sqrt 5-1)(\sqrt 5+1)$). But $ \mathbb{Z}[\frac{1+\sqrt 5}2]$ is, implying the same way of solution.[/quote]

is there like a list of all $ \mathbb{Z}[\sqrt{d}]$ which is UFD?

could some one provide a link

thanks

Instead of $ \mathbb Z [\sqrt d ]$, it's more convenient to consider the so-called integral closure. For $ d \not\equiv 1 \mod 4$ and squarefree, this is the same. But if $ d \equiv 1 \mod 4$, you consider $ \mathbb Z [ \frac{1+\sqrt d}2]$. And if $ d$ is not squarefree, you first divide $ d$ through the largest square dividing $ d$.
If you don't want the integral closure, a bit more work is required (you consider the "conductor" [correct english word¿]).

It's conjectured that there are infinitely many such $ d$. A complete list is known for $ d<0$.
But instead of looking for a list of UFD's, you may look for a list of class numbers (class number $ 1$ $ \iff$ UFD). See http://mathworld.wolfram.com/ClassNumber.html for example.

[quote="praneeth"]2^4*(2^(x-4)-1)=11*(11^(y-1)-1)
11^(y-1) always ends in 1. So, 11^(y-1)-1 is divisible by 10.
But LHS is not divisible by 10 except for x=4.
So, x=4, y=1 is the only possible solution.[/quote]
you are wrong,the LHS is divisible by $ 10$ for all $ x\equal{}4k$

Though this is really old, but I just feel like posting my solution.

Replace 5 by 11-6

You get,
2 ( 2^(x-1) + 3 ) = 11(11^(y-1) + 1 )

Clearly, both factors on both sides are coprime.

Thus,
2 = 11^(y-1) + 1

Which gives y=1
Similarly x=4
Thus, (4,1) is the solution

[quote="ZetaX"]$6 \cdot 35 = 10 \cdot 21$, but your conclusion is wrong.[/quote]

I think the counterexample you put forward, isn't proving me wrong. My equation satisfies two conditions:

1. Both factors on either sides are coprime. (Satisfied here too).
2. 2 doesn't divide 11, But we notice that 6 divides 10 AND 6 divides 21.

Thus, you can't apply that here.

Am I right?

[quote="ZetaX"]I have some doubts about $6$ dividing $10$ or $21$ ;)
And you could use the example $2 \cdot 3 = 1 \cdot 6$, and still cannot conclude that $6=2$.[/quote]

Oops. Sorry. My bad. I was excited about the fact that I had found the solution in a simpler way. :)

[quote="ZetaX"][quote="Valentin Vornicu"]Actually both $ \mathbb{Z}[\sqrt 5]$ and $ \mathbb{Z}[\sqrt{11}]$ are Euclidean rings, thus UFDs also, so Megus' solution above is correct :)  :blush:[/quote]
Just a small correction: $ \mathbb{Z}[\sqrt 5]$ is in fact not an UFD (e.g. $ 2 \cdot 2 = (\sqrt 5-1)(\sqrt 5+1)$). But $ \mathbb{Z}[\frac{1+\sqrt 5}2]$ is, implying the same way of solution.[/quote]

In $ \mathbb{Z}[\frac{1+\sqrt 5}2]$ there exist infinitely many of units (invertible elements of this UFD). So, we need to consider the case $4^x+\sqrt{5}=\varepsilon(4+\sqrt{5})^y$ where $\varepsilon$ is arbitrary unit. Megus' solution is incomplete. Ok?

@nnosipov: It is hard to read what you wrote!
[color=#FF0000][mod edit: nnosipov's post has been fixed][/color]

We will claim that the only solution is $(x, y) = (4, 1)$, which is clearly true as $16 - 5 = 11$.

Clearly $x \le 3$ does not work.

We can prove that it is impossible for $x \geq 5$.

Taking$\pmod{5}$ gets us $2^x \equiv 1\pmod{5}$, so $4|x$.

Taking$\pmod{32}$, and $x \geq 5$ gets us $27 \equiv 11^y\pmod{32}$.

So, $y \equiv 5\pmod{8}$ for $x \geq 5$.

Letting $x = 4c$, and then taking$\pmod{17}:$

$(-1)^{c} - 5 \equiv 11^y\pmod{17}$. The RHS is equal to $11$ or $13\pmod{17}$.
The LHS is equal to $10$ or $15\pmod{17}$, so by contradiction, $x$ cannot $\geq 5.$

Hence, the only solution is $(x, y) = (4, 1)$.

The only pair that works is $(x,y) = \boxed{(4,1)}$. This can be easily checked to work so we will show it is the only one.

To begin, manually check that there are no other solutions for $x \le 4$, so assume $x>4$ for the remainder of the solution. Modulo $5$ gives $4 \mid x \implies x =4z$ for $z>1$. Thus, modulo $64$ gives

\[11^y \equiv 59 \pmod{64} \implies y \equiv 13 \pmod{16}.\]

Finally, use mod $17$ to give

\[16^z - 5 \equiv 11^y \equiv 11^{13} \equiv 7 \pmod{17}\]

which clearly has no solutions.

Take modulo $16$ first to see that we need, $$11^{y-1} \equiv 1 \pmod{16}$$ for $x \geq 4$. Then we must have $y \equiv 1 \pmod{4}$. Similarly modulo $5$ gives $4 \mid x$. Substituting this becomes, $$(2^4)^{x'} - 5 = 11 \cdot (11^4)^{y'}$$ Then taking modulo $17$ we find, $$(-1)^{x'} - 5 \equiv 11\cdot 4^{y'} \pmod{17}$$ Then this rearranges to $$ 11 \cdot 4^{y'} - (-1)^{x'} \equiv 12 \pmod{17}$$ Also taking modulo $64$ we have $$11^y \equiv 59 \pmod{64}$$ which requires $y \equiv 13 \pmod{16}$. Now it is easy to see that there is no why satisfying our modulo $17$ equation so we are done. For $x \leq 4$ our only solutions is $\boxed{(4, 1)}$.

I claim that $(4,1)$ is the only solution. Clearly it works. Plugging in all $0<x<4$ also gives that no $x<4$ works. \\\\
Now, taking both sides mod $5$, note that the RHS is always $1$ mod $5$ no matter what $y$ is, meaning that the LHS must also be $1$ mod $5$. Using our knowledge of orders, we deduce that $2^x$ is $1$ mod $5$ if and only if $4\mid x$. So let $x=4a$ for some integer $a$. We now have that
\[16^a-5=11^y.\]
Now for the sake of contradiction, assume that there's a solution $(x,y)$ with $x>4$. Taking the LHS mod $32$, since $a$ is a positive integer and $a>1$, we have that $32\mid 16^a$. Therefore the LHS is $27$ mod $32$, meaning that the RHS must also be $27$ mod $32$. \\\\
Listing the powers of $11$ mod $32$, we get
\[11, 25, 19, 17, 27, 9, 3, 1, \dots,\]
which gives us that $y$ must be $5$ mod $8$. Now we take the equation mod $17$. Since $16$ is $-1$ mod $17$, we have that the LHS mod $17$ must be either $13$ or $11$. However, listing the powers of $11$ mod $17$, we get
\[11, 2, 5, 4, 10, 8, 3, 16, 6, 15, 12, 13, 7, 9, 14, 1, \dots,\]
which gives us that whenever we have $y$ as $5$ mod $8$, the RHS must be $7$ or $10$ mod $17$, a contradiction to the $11$ or $13$. Therefore there are no solutions where $x>4$, meaning that $(4,1)$ is the only solution, finishing the problem.

The only solution where $1 \leq x \leq 5$ is $\boxed{(x,y)=(4,1)}$. Otherwise, we subtract 11 from both sides to get
\[2^x-16 = 11^y-11 \implies 16(2^{x-4}-1) = 11(11^{y-1}-1).\]

Taking modulo 5 then modulo 64, we find $x \equiv 0 \pmod 4$ and $y \equiv 13 \pmod{16}$. However, substituting these values back in, we find the equation has no solutions modulo 17 when $x \ge 6$. $\blacksquare$


This is a question I got from book so no need for you to solve but just help me to understand:

Prove that n < 2^n for all positive integers n.

1. For n = 1, the formula is true, because
1 < 2^1.
2. Assuming that
k < 2^k
k + 1 < 2^(k+1) for n = k
2^(k+1) = 2 (2^k) > 2 (k) = 2k
What's that mean????

Help me understand and I'll keep going:

2^(k+1) > 2k > k+1
k+1 < 2^(k+1)

Help!

Thanks ;)

[quote="Silverfalcon"]2^(k+1) = 2 (2^k) > 2 (k) = 2k
What's that mean???? [/quote]

First, they rewrote 2^(k + 1) as 2 * (2^k). Since we can assume that k < 2^k, we know that 2k < 2 * (2^k). So 2^(k + 1) > 2k.

Can you finish the proof from here?

they are using a method of proof called mathematical induction.

Read about it here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13443&postdays=0&postorder=asc&&start=0

I already knew it was mathematical induction because it's one of section in my trig book but thanks.

Yet, my book had 2^(k+1) = 2k, not > 2k and that's part I'm stuck on.

Ah, I see what you mean.
2^(k+1) = 2 (2^k) > 2 (k) = 2k

From this you get 2^(k+1) > 2k. If you say something like a=b>c=d>e=f, then this is short hand for a group of equations:
a=b
b>c
c=d
d>e
e=f
From which you work out that a>f. Not that a=f.

Yeah, we were told not to do that... never write a = b = c, it's nasty! :)

No, a = b = c is fine and useful.  What is not fine is to write something like 10 + 20 + 40 as 10 + 20 = 30 + 40 = 70, which I have seen too many students do.


Define an operation * on Z by

[color=green]a*b = ab - a - b +2 for a, b in Z. We know that * is commutative and associative and that * has an identity element, namely 2. Determine which elements in Z have inverses relative to *. Justify your answer. [/color]

Thanks very much   ;)

Let $a*b=(a-1)(b-1)+1$ is evidently commutative and also associative because $a*b*c=(a-1)(b-1)(c-1)+1$ and so on. The inverse g of an alement f relatively to * is such that $f*g=e$, where e is the neuter element of *. In this case if $a*e=a$ $\forall a\in \mathbb{Z}$; if we suppose to already know addition and multiplication , then $(a-1)=\frac{a-1}{e-1}$ so $e-1=1\rightarrow e=2$. If $(a-1)(b-1)+1=2\rightarrow a-1=(b-1)^{-1}$ an d both $a-1$, $b-1$ $\in$ the field $\mathbb{Z}$ then a=1 and a=2 are the only elements


algebra
polynomial
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inequalities proposed

I don't know exactly where I have found the following one, but I know is terribly difficult:
Let $f$ and $g$ be polynomials such that $f(Q)=g(Q)$ for all rationals $Q$. Prove that there exist reals $a$ and $b$ such that $f(X)=g(aX+b)$, for all real numbers $X$.

This is a result, if the difference of two continous functions vanishes on a dense set, then the functions are equal.

I think the hypothesis is { f(x), x $\in$ Q } = { g(y), y $\in$ Q } not f(x) = g(x) for x $\in$ Q (this last case being indeed obvious).

that's exactly what harazi posted :D f([b]Q[/b])=g([b]Q[/b]) is an equality of sets. and as mindspa pointed it is a well-known theorem, to which I don't have a proof (right now at least) :D

Valentin told me to indicate the source of the problems. Well, this one was given at a Miklos Schweitzer Competition.

I'm going to copy a proof verbatim (not mine) for the case of rational coefficients.

By a polynomial we shall always mean a polynomial with rational coefficients, and by the range of a polynomial we shall mean the set of its values assumed at rational numbers.

We use two classifications of polynomials. We write $f \sim g$ if the ranges of these polynomials coincide, and $f \approx g$ if $f(x) = g(ax+b)$ for all $x$ with suitable rational numbers $a \neq 0, b$. These are clearly equivalence relations, and they are compatible with multiplications by a constant, that is, if $f \sim g$or $f \approx g$ then $cf \sim cg$ or $cf \approx cg$, respectively ($c \neq 0$ rational). Our aim is to show that these relations are identical. One implication is clear: if $f \approx g$ then $f \sim g$ holds obviously.

The converse will be proved in several steps.

It is sufficient to prove the statement for polynomials with integral coefficients. Indeed assume $f \sim g$ and let $c$ be an integer such that the coefficients of $cf$ and $cg$ are all integers. We have $cf \sim cg$ thus $cf \approx cg$, assuming the integer case, which yields $f \approx g$ as wanted.

If $f \sim g$ are polynomials with integral coefficients we show that they must be of equal degree. Indeed, assume that

\[f(X) = ax^n + \ldots, \quad g(x) = bx^k + \ldots\]

where $a \neq 0, b \neq 0$ and $n \leq k$. Choose a prime $p \nmid ab$. We have $f(1/p) = c/p^n$ where $p \nmid c$. By assumption, $g$ also takes on this value at some number $u/v$ with $(u,v) = 1$. Now if $p \nmid v$, then $p$ cannot divide the denominator of $g(u/v)$, thus the denominator (in the reduced form) of $g(u/v)$ contains $p$ with an exponent $\geq k$. This yields $n \geq k$, hence $n = k$ as asserted.

It is sufficient to prove the statement for polynomials with integral coefficients where at least one of the leading coefficients is 1. Indeed let $f \sim g$, $f(x) = ax^n + \ldots$. Then we have

\[a^{n-1}f(\frac{x}{a}) \sim a^{n-1}f(x) \sim a^{n-1}g(x),\]

and here $a^{n-1}f(x/a)$ still has integral coefficients while its leading coefficient became 1. By assumption, we conclude $a^{n-1}f(x/a) \approx a^{n-1}g(X)$, which yields $f \approx g$.

Let $f \sim g$ be polynomials with integral coefficients, $f(x) = x^n + \ldots, g(x) = bx^n + \ldots$. We claim that every prime divisor of b has an exponent $\geq n$. To show this, take $p \mid b$ and consider the polynomials

\[f_1(x) = p^{n(n-1)}f(\frac{x}{p^{n-1}}) \sim p^{n(n-1)}f(\frac{x}{p^n}) \sim g_1(x)\]
\[g_1(x) = p^{n(n-1)}g(\frac{x}{p^n}).\]

We know that all coefficients of $f_1$ as well as all but the leading coefficient of $g_1$ are integral, and the leading coefficient of $f_1$ is 1. Hence every value of $f_1$ has the follwoing property: if (in the reduced form) its denominator is divisible by $p$ it is divisible by $p^n$. By $f_1 \sim g_1$ the same must hold for $g_1$. Now $g_1(1) = b/p^n + c$ with an integer $c$, and by $p \mid b$ the exponent of $p$ in the denominator is strictly less than $n$, thus it must be 0, that is $p^n \mid b$.

It is sufficient to prove the statement for polynomials with integral coefficients where both leading coefficients are 1. Let $f \sim g$ be polynomials with integral coefficients, $f(x) = x^n + \ldots, g(x) = bx^n + \ldots$. Let $c_n$ be the largets $n$-th power dividing $b$. Consider

\[f_1(x) = c^{n(n-1)}f(\frac{x}{c^{n-1}}) = x^n + \ldots,\]
\[g_1(x) = c^{n(n-1)}g(\frac{x}{c^{n-1}}) = \frac{b}{c^n}x^n + \ldots.\]

We have $f_1 \sim g_1$, by the last step the exponent of any prime in the leading coefficient of $g_1$ must be either 0 or $\geq n$, and by the definition of $c$ this coefficient must be 1 or -1. We can exclude the case -1 for even $n$, since then one domain would be bounded from below and the other from above. Finally if $n$ is odd, then replacing $c$ by $-c$ we can change a -1 into 1. So if the statement is true under this restriction then we obtain $f_1 \approx g_1$, which yields $f \approx g$.

We can also assume that the coefficient of $x^{n-1}$ is 0 in both polynomials. Indeed if

\[f(x) = x^n + ax^{n-1} + \ldots \sim g(x) = x^n + bx^{n-1} + \ldots,\]

then the polynomials

\[f_1(x) = n^nf(\frac{x}{n} - a) \sim g_1(x) = n^ng(\frac{x}{n} - a)\]

have this property, and $f_1 \approx g_1$ again implies $f \approx g$.

Now finally, if $f \sim g$ are polynomials of the form

\[f(x) = x^n + ax^{n-2} + \ldots, \quad g(x) = x^n + bx^{n-2} + \ldots\]

with integral coefficients then $f \approx g$. Observe that since the leading coefficient is 1, the only rational numbers where $f$and $g$assume integral values are the integers. Since $f(x+1) - g(x) = nx^{n-1} + \ldots,$ we have $g(x) < f(x+1)$ for large $x$, and similarly we obtain $g(x) > f(x-1)$. Thus for a large positive integer $x$, the only positive rational $y$ that can satisfy $f(y) = g(x)$ is $y = x$. If $n$ is even then there can also be a negative $y$, and similarly we find that its only possible value is $y = -x$. This means that one of the equations $f(x) = g(x)$ and $f(-x) = g(x)$ has infinitely many solutions; hence it must be an identity.

Great work, Mindspa!

I think I know how to complete proof for all variety of polynomials.

Suppose WLOG $n=\deg(f)\geq\deg(g)$. Consider set of equalities
\[
f(1)=g(q_1),\ f(2)=g(q_2),\ \dots, f(n+1)=g(q_{n+1}),\ q_i\in\mathbb{Q}.
\]
This linear system means that coefficients of $f$ are rational linear combinations of coefficients of $g$.
Consider maximal subset $\{c_i\}_{i=1}^{k}$ of linearly independent over $\mathbb{Q}$ coefficients of $g$. Then
$g(x)=c_1\cdot g_1(x)+c_2\cdot g_2(x)+\dots+c_k\cdot g_k(x)$, where $g_i(x)\in \mathbb{Q}[x]$
and
$f(x)=c_1\cdot f_1(x)+c_2\cdot f_2(x)+\dots+c_k\cdot f_k(x)$, where $f_i(x)\in \mathbb{Q}[x]$.

It is clear that $f_i\sim g_i$, $i=1..k$, so $f_i(x)=g(a_ix+b_i)$. We have $f(x)\sim g(x)\sim g(a_1x+b_1)$ $\Rightarrow$ we may assume $f_1(x)=g_1(x)$, and moreover multiplying $f$ and $g$ by appropriate integer number we may assume $g_1\in \mathbb{Z}[x]$. If $g_1(x)=a_nx^n+a_{n-1}x^{n-1}+...$ then consider $(na_n)^n\cdot g\left(x-\frac{a_{n-1}}{na_n}\right)$ and $(na_n)^n\cdot f\left(x-\frac{a_{n-1}}{na_n}\right)$ instead of $g(x)$ and $f(x)$. Then we obtain $g_1(x)=a_n'x^n+a_{n-2}'x^{n-2}+...$.

We can write $g(x)=c_1\cdot g_1(x)+\hat{g}(x)$ and $f(x)=c_1\cdot g_1(x)+\hat{f}(x)$. Due to preceding argumentation we have $\hat{f}\sim\hat{g}$. Consider arbitrary large integer $m$, then $g_1(x)$ takes value $g_1(m)$ at most twice (maybe $g_1(m)=g_1(-m)$ if $g_1$ has even degree -- see Mindspa's solution). It implies (!) that $\hat{f}(m)=\hat{g}(m)$ or $\hat{f}(-m)=\hat{g}(m)$. If $\hat{f}(m)=\hat{g}(m)$ for infinitely many $m$ then $\hat{f}\equiv\hat{g}$ $\Rightarrow$ $f=g$; if $\hat{f}(-m)=\hat{g}(m)$ (in this case we supposed $g_1(m)=g_1(-m)$) for infinitely many $m$ then $g_1$ is even function and $\hat{f}(-x)=\hat{g}(x)$, hence we conclude $f(x)=g(-x)$.

I just wanted to make sure everyone realizes that this is not my proof. ;)


Is any polynomial a continuous function?


geometry
perimeter
incenter
geometry unsolved

A triangle $ABC$ ia divided into two parts of the same area and perimeter by a line [i][b]d[/b][/i].Can you erect [i][b]d[/b][/i] and prove that your erection is true?

I think $d$ must pass incenter. But now i can't prove this.

Yes, I remember that I have done it when I was a middle school's student...
Assume that $d$ intersects the segments $[AB],[AC]$ and the bisector of angle $\hat{BAC}$ at $M,N,S$, respectively. Let $x,y$ are the distance between $S$ and $AB,BC$, respectively. We have $\left\{ \begin{array}{l}x(AM+AN)x = x(BM+CN)+yBC \\ AM+AN = MB+BC+CN \end{array}\right. \Leftrightarrow \left\{ \begin{array}{l}x(AM+AN-MB-CN)x = yBC \\ AM+AN-MB-CN =BC \end{array}\right.$
So $x=y$, which means that $S$ is the incenter of $\Delta ABC$


The answer says

f(m)=m for all m

I put like f(x)=x for all x greater or equal to 0.

Are these two the same for the question at hand?  I dont care about switching x and m (since that doesnt matter),but Im worried that the graders might think that when I said all x greater or equal 0, that it is not for all x.  But since the question is only defined on nonnegatives, wouldnt the two answers be the same?

I guarantee you that it will not be an issue (on no authority ;) ).  What you said means exactly the same thing (in context).  If someone takes off for that I'll eat my hat.

Blargh, I didn't even mention domain in my answer. I don't think that would be a problem, as the domain is stated clearly in the problem.


calculus
integration
limit
trigonometry
real analysis
real analysis unsolved

Compute the following limit: $ \lim_{n\to\infty}\int_0^{\infty}\frac {n \sin\frac {x}{n}}{x(1 \plus{} x^2)}dx$.

Thanks!

Funny, I just typed up the solution to this problem.

Note that for all $ \theta\neq 0$, $ \frac{\sin\theta}{\theta}\leq 1$ and $ \lim_{\theta\rightarrow 0}\frac{\sin\theta}{\theta}\equal{}1$.  Thus $ \frac{n\sin\left(\frac{x}{n}\right)}{x(1\plus{}x^{2})}\leq\frac{1}{1\plus{}x^{2}}\in L^{1}([0,\infty))$ for all $ n$.  So by the Dominated Convergence Theorem, we have \[ \lim_{n\rightarrow\infty}\int_{0}^{\infty}\frac{n\sin\left(\frac{x}{n}\right)}{x(1\plus{}x^{2})}\,dx\equal{}\int_{0}^{\infty}\frac{dx}{1\plus{}x^{2}}\equal{}\tan^{\minus{}1}{x}\bigg|_{0}^{\infty}\equal{}\frac{\pi}{2}.\]

Use substitution $ x \equal{} n \, z$ and since the integral converges absolutely switch the order of the integral and the limit, calculate the limit and the result is straightforward (if you know a thing or two).

[hide="Click"]
$ x \equal{} n\, z$

$ dx \equal{} n \, dz$

$ \lim _{n \to \infty}\int^\infty _ 0 \left(\frac {\sin z}{z} \frac {n}{1 \plus{} n^2 z^2}\right) \,dz \equal{} \int^\infty _ 0 \left( \frac {\sin z}{z}\lim _{n \to \infty} \frac {n}{1 \plus{} n^2 z^2}\right)\,dz \equal{}$

$ \equal{} \int ^\infty _ 0 \frac {\sin z}{z}\,dz \equal{} \frac {\pi}{2}$

[/hide]

[EDIT: Ignore my post.. It's total nonsense..]


Prove that there exist infinitely many positive integers n such that $2^n$ ends with n, i.e., $2^n=...n$

                                                                                MMO 1978

@bomb: sorry but i don't know my self which Olympiad is that ... but i guess it must be (Moscow Math Olympiad)

Davron


geometry
parallelogram
combinatorics solved
combinatorics
Enumeration

In triangle ABC, every side is divided into n equal parts and from each division point, two lines are drawn parallel to other sides. Count the number of parallelograms that can be seen inside ABC.

My try is : (n+2)(n+1)n(n-1)/8 (which happens to be 3*C(n+2,4) but I can't see why ????)

Tricky to explain without pictures.

There are 3 kinds of //grammes :

1- _\ ones :

...._\
..._\_\
.._\_\_\
._\_\_\_\
_\_\_\_\_\

2- /\ ones

..../\
.../\/\
../\/\/\
./\/\/\/\
/\/\/\/\/\

3- /_ ones :

..../_
.../_/_
../_/_/_
./_/_/_/_
/_/_/_/_/_

There are the same numbers for each kind, so just compute the number in the first case and multiply by 3.

A //gramme with 'a' vertical bars and 'b' vertical bars exists iff  a+b<=n and 1 <= a,b <= n-1,
and there are n+1-(a+b) + ... + 1 such //grammes (easy to see on a picture  :)).

So number is

 \sum _a  \sum _b (n+1-(a+b) + ... + 1) = .... = (n+2)(n+1)n(n-1)/24

And total number is (n+2)(n+1)n(n-1)/8

I checked this formula for small values of n (to be fair only for n =3  :D), seems to be correct.

To answer belenos' question of why it's 3 * C(n+2, 4):

Let the ith subdivision of ABC be the configuration where AB, BC and AC are divided into i equal parts each.

Extend the sides AB and AC of ABC to AB' and AC' respectively such that AB/AB' = AC/AC' = n/(n+1).  So the nth subdivision of ABC is part of the (n+1)th subdivision of AB'C'.  There are n+2 subdivision points on B'C' (B' and C' included) in the (n+1)th subdivision of AB'C'.

As follows from belenos' proof, we only count parallelograms in one direction, in this case, with one tip facing towards B'C'.  Extend the sides of any such parallelogram to meet B'C' at 4 of the subdivision points of B'C'.  There exists a bijection between each parallelogram with one tip facing towards A and each set of 4 points on B'C'.

Therefore total no. of parallelograms = 3 * C(n+2, 4)

Isn't this a problem from a Canadian olympiad ?

just a belenos said, we need only to compute one shape of the parallelogram. we compute the following shape:
 ________
/_______/

     /.\  line 1
    /__\  line 2
   /___\  line 3
  /____\  line 4
   ....
 /______\ line n

the upper edge and the bottom edge can be chosen from n parallel lines. if the upper edge is on the i-th line whose width is i, so there are (n-i) ways to chose the bottom edge and there are C(i+1,2) ways to chose the left and right edges of the parallelogram. hence, the total number is
\sum (n-i)*C(i,2) = \sum C(n-i, 1)*C(i+1,2).

A well-known combinatorial equation is that
\sum C(a,k)*C(b,l) = C(a+b+1,k+l+1).

From this, we know that \sum C(n-i, 1)*C(i+1,2) = C(n+2,4).
Multiplying it by 3, we get the final answer.

This problem could also be found in the national olympiad of Iran in 1995.


geometry
complex numbers
geometry solved

Please ponder this problem (I don't know where it came from, so excuse me if it was already posted)

Let ABC be a triangle, H the orthocenter and D the feet of the altitude from B. We are given that BH/HD=3. If K is the midpoint of BD, prove that angle AKC is right.

I'm looking for an euclidean solution. However, I find this problem in an article about complex numbers, so if you can provide me such a solution it'll be gratefully received :)

Thanks!

[img]http://img.photobucket.com/albums/v218/hudrea/vecrtor02.jpg[/img]
$\overline {CK}  \cdot \overline {AK} \mathop  = \limits_{\mathop {\scriptstyle \overline {AH}  = \frac{1}{2}\left( {\overline {AK}  + \overline {AD} } \right) \hfill \atop
  \scriptstyle \overline {CK}  = \frac{1}{2}\left( {\overline {CB}  + \overline {CD} } \right) \hfill}\limits^ \uparrow  } \frac{1}{2}\left( {\overline {CB}  + \overline {CD} } \right)\left( {2 \cdot \overline {AH}  - \overline {AD} } \right) \Rightarrow $
$ \Rightarrow 2\overline {CK}  \cdot \overline {AK}  = 2 \cdot \overline {CB}  \cdot \overline {AH}  + 2 \cdot \overline {CD}  \cdot \overline {AH}  - \overline {CB}  \cdot \overline {AD}  - \overline {CD}  \cdot \overline {AD} \mathop  \Rightarrow \limits_{\mathop {CB \bot AH}\limits^ \uparrow  } $
$ \Rightarrow 2\overline {CK}  \cdot \overline {AK}  = 2 \cdot \overline {CD}  \cdot \left( {\overline {AD}  + \overline {DH} } \right) - \overline {CB}  \cdot \overline {AD}  - \overline {CD}  \cdot \overline {AD}  \Rightarrow $
$ \Rightarrow 2\overline {CK}  \cdot \overline {AK}  = \overline {CD}  \cdot \overline {AD}  + 2\overline {CD}  \cdot \overline {DH}  - \overline {CB}  \cdot \overline {AD}  \Rightarrow $
$ \Rightarrow 2\overline {CK}  \cdot \overline {AK}  = \overline {CD}  \cdot \overline {AD}  + 2\overline {CD}  \cdot \frac{1}{4} \cdot \overline {DB}  - \overline {CB}  \cdot \overline {AD}  \Rightarrow $
$ \Rightarrow 2\overline {CK}  \cdot \overline {AK}  = \left( {\overline {CD}  - \overline {CB} } \right) \cdot \overline {AD}  + \frac{1}{2} \cdot \overline {CD}  \cdot \overline {DB}  \Rightarrow $
$ \Rightarrow 2\overline {CK}  \cdot \overline {AK}  = \overline {BD}  \cdot \overline {AD}  + \frac{1}{2} \cdot \overline {CD}  \cdot \overline {DB}  = \overline {BD}  \cdot \left( {\overline {AD}  - \frac{1}{2} \cdot \overline {DC} } \right) = 0$

[b][u]Conclusion:[/u][/b]
$CK \bot AK$

[quote="Jutaro"]Let ABC be a triangle, H the orthocenter and D the feet of the altitude from B. We are given that BH/HD=3. If K is the midpoint of BD, prove that angle AKC is right.[/quote]

This problem was one of the homework problems to the current IMO training seminar here in Oberwolfach, and Christopher Wulff, better known to many of you as "Werwulff", gave the following elegant solution:

The triangles HDC and ADB are similar, since their corresponding sides are perpendicular (in fact, $HD\perp AD$, $DC\perp DB$ and $CH\perp BA$, since the point H is the orthocenter of triangle ABC and the point D is the foot of the altitude from the vertex B). Thus, HD : DC = AD : DB. In other words, $HD\cdot DB=AD\cdot DC$.

Since BH : HD = 3, we have $HD=\frac14\cdot BD$. On the other hand, since the point K is the midpoint of the segment BD, we have $KD=\frac12\cdot BD$. Hence, $KD=2\cdot HD$. On the other hand, $KD=\frac12\cdot DB$, since the point K is the midpoint of the segment BD. Hence,

$KD^2=KD\cdot KD=\left(2\cdot HD\right)\cdot\left(\frac12\cdot DB\right)=HD\cdot DB=AD\cdot DC$.

Hence, KD : AD = DC : KD. Also, we know that $\measuredangle KDA=90^{\circ}$ and $\measuredangle CDK=90^{\circ}$, so that < KDA = < CDK. Thus, the triangles KDA and CDK are similar. Hence, < CKD = < KAD. On the other hand, since the triangle KDA is right-angled (with $\measuredangle KDA=90^{\circ}$), we have $\measuredangle KAD+\measuredangle AKD=90^{\circ}$. Thus, we obtain $\measuredangle CKD+\measuredangle AKD=90^{\circ}$; in other words, $\measuredangle AKC=90^{\circ}$. This solves the problem.

  darij

Let $S$ be midpoint of segment $CD$

Then KS is middle line in triangle $BCD$, so $KS||BC$ and this means $AH\bot KS$.
So $H$ is orthocenter in triangle $AKS$.
This means $HS\bot AK$.

Since segment $HS$ is middle line in triangle $CDK$ we get $HS||KC$ wich imply $\angle AKC = 90^{\circ}$

H  is the Lemoine point of AKC.
     KD is symmedian and altitude from vertex K.
     <K = 90.

    Woulgar, Esq.

well novice like me will be able to give u much simpler solution. :D

Anyhow, u can solve it very easily by co-ordinate geomtry :D

u may try it guys. this is very simple.


Prove or disprove :
Let $a,b,c,d,e$ be positive numbers than :
$\sum a(a-b)(a-c)(a-d)(a-e) \geq 0$

Wrong: take a = 1, b = 2, c = 3, d = 2.95, e = 0 (or e very close to 0).

  Darij


How many positive integers $ b$ have the property that $ \log_b729$ is a positive integer?

$ \textbf{(A)}\  0 \qquad \textbf{(B)}\  1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\  3 \qquad \textbf{(E)}\ 4$

[hide]4 such integers: 3,9,27,729[/hide]

I got this one wrong on a practice  :roll:  So mad at that.. :furious:

[hide]What numbers can be raised to integer powers and be equal to 729? Well, obviously $729^1$ is an option. Similarly, $27^2$, $9^3$, and $3^6$ work. That's 4 values of b. Knowing your perfect squares and cubes help. [/hide]

[hide="Simplest"]$729=3^6$. 6 has 4 factors, 6, 3, 2, and 1. Therefore there are E) 4 solutions.
Edit: 236factorial, you count $729$ twice (as $3^6$ and $729$)[/hide]

[quote="Hokkage"]Edit: 236factorial, you count $729$ twice (as $3^6$ and $729$)[/quote]And as $9^3$.  ;)


I found this just today, also.

There's 2 methods to find the triples really easily.

The methods are [u][b]Method of Pythagoreans[/b][/u] and [u][b]Method of Plato[/b].[/u]

[u]Method of Pythagorean:[/u]
let $ m$ be any odd number  greater than 1.

$ m, \ \frac {m^2 \plus{} 1}{2}, \ \frac {m^2 \minus{} 1}{2}$

[u]Method of Plato: [/u]
let $ n$ be any integer greater than 1.

$ 2n, \ n^2 \plus{} 1, \ n^2 \minus{} 1$

$ \star$ I don't know how those work out, but this is pretty cool.
If you know how it works, can you guys explain how those work?

This could also come in handy if you need to find the sides or area of a right triangle and are only given one side (of course, if it is a 30-60-90 or a 45-45-90 you can already do that)

DK:  please repost this in the trick and tips section.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174500]the other thread[/url] for the generalization.


The even numbers problems on the target we did last week seemed to be the hard ones (especially #4 and #6).  So here they are, waiting for solutions...

2. Divide the product of the first five positive composite integers by the
product of the next five composite integers. Express your answer
as a common fraction.

*4. How many rectangles are in the array below? (forgive my crude figure)
[quote]┌────┬────┬────┐
├────┼────┼────┤
├────┼────┼────┤
└────┴────┴────┘[/quote]

*6. If A, B and C are positive integers such that $ \dfrac{A\sqrt B}{C} \equal{} \dfrac{9}{2\sqrt3}$
what is the value of A + B + C given that A and C have no common
prime factors, and B has no perfect-square factors other than 1?

8.  Amanda has recorded how long it takes her to do her morning
activities. She does all of the listed activities each day. She starts
the first activity at 7:30 a.m. and has a two-minute break between
each pair of activities. At what time does she finish all of her
activities each morning?

[u]Activity....................Time to Complete[/u]
Shower & Dress.............15 minutes
Prepare & Eat Breakfast..20 minutes
Chores.........................35 minutes
Practice Piano...............45 minutes

[hide=" Solution for 2."]You guys should have got this question right, since we went over the primes.
Composite numbers are the ones that are not primes.
So the first 5 would be $ \ 4,6,8,9,10$.
And the next 5 would be $ \ 12,14,15,16,18$.
Then we can establish the equation $ \dfrac{4\cdot6\cdot8\cdot9\cdot10}{12\cdot14\cdot15\cdot16\cdot18}$ which would simplify to $ \boxed{\frac1{42}}$[/hide]

[hide="Solution for 4."]I actually counted this one, but I later learned that there's a formula for it.
On m by n grid the number of rectangles that can be formed is $ \ _{m\plus{}1}C_2\cdot _{n\plus{}1}C_2$
(It's Combination.You should know it if you paid attention to the lesson)
[hide="*How it works?"]It works because there are always going to be 1 more segment that is parallel to the m side and same thing on n side. And the rectangle can be formed by choosing 2 sides from m and 2 from n. So the formula would end up looking like this: $ \ _{m\plus{}1}C_2\cdot _{n\plus{}1}C_2$.[/hide]

In this case the grid is 3 by 3, so you do $ \ _{3\plus{}1}C_2\cdot _{3\plus{}1}C_2\equal{} 6\cdot6\equal{} \boxed{36}$ rectangles.[/hide]

[hide="Solution for 6."]I missed this one.. :(
You have to know the 3 rules of radicals(you will eventually learn it in Algebra) and one of them is that you can not have radical as a denominator.
So in this case, you do $ \dfrac9{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$.to get rid of radical in denominator.
Then you get $ \dfrac{9\sqrt{3}}{2\cdot{3}}$ which would simplify into $ \dfrac{3\sqrt{3}}{2}.$
Now look at the problem and plug it in and add them to get $ \boxed{8}$[/hide]

[hide="Solution for 8."]This problem is just adding.
You have to know how to add times.
If you do it well enough, you get $ \boxed{9: 31\text{a.m}}$[/hide]

i know which ones i got wrong
and one of them is #8
i thought i was so smart and didnt even glance at how many activities were there on each row
so i just added the first two and the second two and the extra 2 minutes.... :wallbash_red:

answer to #2
[hide]
"Divide the product of the first five positive composite integers by the
product of the next five composite integers. Express your answer
as a common fraction."
composite numbers are numbers that are not primes
so the first 5 compsites are 4,6,8,9,10
and next five is 12,14,15,16, and 18
4x6x8x9x10/12x14x15x16=1/42

if u think because 1 isnt prime, the first 5 numbers should be 1,4,6,8, and 9
1 is not a prime, nor a composite
composite numbers need to have more than 2 factors
1 has only one. 1x1.
[/hide]

i dont know if i got #6 right..

answer to #8
[hide]
add up the time it takes for the activities....
add the extra 6 minutes....
add the number up to 7:30 am
and u should get 9:31 am
[/hide]

[quote][hide]=Solution for 4.]I actually counted this one, but I later learned that there's a formula for it.
On m by n grid the number of rectangles that can be formed is $ \ _{m \plus{} 1}C_2\cdot _{n \plus{} 1}C_2$
(It's Combination.You should know it if you paid attention to the lesson)
[hide="*How it works?"]It works because there are always going to be 1 more segment that is parallel to the m side and same thing on n side. And the rectangle can be formed by choosing 2 sides from m and 2 from n. So the formula would end up looking like this: $ \ _{m \plus{} 1}C_2\cdot _{n \plus{} 1}C_2$.[/hide][/hide][/quote]

I really didn't get the [b][size=100]C[/size][size=59]2[/b] [/size]part........
Can you please explain it in our way??
(It's cool how you can do Latex :))

$ _{n}C_{r}$ is the same as C(n,r), that is combinations.


Hi guys !!! Does anyone know problems of IZho2010?
If you know some of them, please share with us?

Problem 1. "Find all prime numbers p, q, such that p^3-q^7=p-q"

Official site of the International Zhautykov Olympiad
http://www.izho.kz

Let's go/ :D


linear algebra
matrix
search
algebra
polynomial

let $ A,B$ be $ n\times n$ matrix, and $ det(I_n\plus{}AB)\neq0$
show that $ det(I_n\plus{}BA)\neq0$

Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=493277596&t=116836]here.[/url]

$ \det(I\plus{}AB)$ and $ \det(I\plus{}AB)$ are simply the characteristic polynomials of $ AB$ and $ BA$ evaluated at the same place; since these characteristic polynomials are the same, the result is the same.


geometry
3D geometry
sphere
analytic geometry
inequalities

Consider two solid spherical balls, one centered at $(0, 0, \frac{21}{2} )$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\frac 92$ . How many points $(x, y, z)$ with only integer coordinates (lattice points) are there in the intersection of the
balls?

$\text{(A)}\ 7 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15$

[hide="Solution"]First, we see that $\frac{9}{2}\leq z\leq \frac{11}{2}$. Thus, $z=5$. We can now plug this value of $z$ into the equations of the two spheres to get two equations for $x$ and $y$, which will help us find the number of lattice points.

Our two equations are $x^2+y^2+\left(z-\frac{21}{2}\right)^2\leq 6^2$ and $x^2+y^2+(z-1)^2=\left( \frac 92\right)^2$.
Simplifying, $x^2+y^2\leq \frac{23}{4}$ and $x^2+y^2\leq \frac{17}{4}$.

Notice that all $(x,y)$ satsfying the second inequality also satisfy the first one, so we need only find all the lattice points that satisfy the second inequality. There are $13$.

$(-2, 0, 5)$, $(2, 0, 5)$, $(0,-2, 5)$, $(0, 2, 5)$, $(-1,-1, 5)$, $(1,-1, 5)$, $(-1, 1, 5)$, $(1, 1, 5)$, $(-1, 0, 5)$, $(1, 0, 5)$, $(0,-1, 5)$, $(0, 1, 5)$, and $(0, 0, 5)$.
[/hide]

Let $S_1$ be the sphere with radius 6 and $S_2$ be the other sphere.  $S_1$ is farther along the $z$-axis than $S_2$.  So we know that if a point is going to lie in the intersection of the two spheres it must be above the lower limit of $S_1$ (i.e. $21/2-6=9/2$)  and below the upper limit of $S_2$ (i.e. $9/2+1=11/2$).

[quote="amcavoy"]How did you get that first part?

$\frac{9}{2}\leq z\leq\frac{11}{2}$[/quote]I'll use the example of two tangent circles instead of spheres, so that it's easier to explain. Let's have two circles $A$ and $B$ with centers at $(0,0)$ and $(0,10)$. The radius of the circle $A$ will be 6, and the radius of circle $B$ will be 7.

The largest that y can ever get on circle $A$ is $(0,0+6)$, because the radius is $6$. The smallest y can get for circle $B$ is $(0, 10-7)$, because the radius is $7$.

Same thing goes for spheres.

Hope this helps. :)

[quote="joml88"]Uh ok...someone asked about how 4everwise arrived at the bound for $z$...that's what my post above was explaining...but then that person must've deleted their post.[/quote]I was just about to explain it as well. Oh well.

Sorry, I deleted it right after I posted it because I saw how it was arrived at (I made a careless error subtracting and came up with a different inequality).

[quote="4everwise"]Notice that all $ (x,y)$ satsfying the second inequality also satisfy the first one, so we need only find all the lattice points that satisfy the second inequality. There are $ 13$.

$ ( \minus{} 2, 0, 5)$, $ (2, 0, 5)$, $ (0, \minus{} 2, 5)$, $ (0, 2, 5)$, $ ( \minus{} 1, \minus{} 1, 5)$, $ (1, \minus{} 1, 5)$, $ ( \minus{} 1, 1, 5)$, $ (1, 1, 5)$, $ ( \minus{} 1, 0, 5)$, $ (1, 0, 5)$, $ (0, \minus{} 1, 5)$, $ (0, 1, 5)$, and $ (0, 0, 5)$.
[/quote]

Could someone please explain this part to me?  Thanks!

I'm still waiting for someone to help me.  Thanks in advance!

[quote="4everwise"][hide="Solution"]First, we see that $ \frac {9}{2}\leq z\leq \frac {11}{2}$. Thus, $ z \equal{} 5$. We can now plug this value of $ z$ into the equations of the two spheres to get two equations for $ x$ and $ y$, which will help us find the number of lattice points.

Our two equations are $ x^2 \plus{} y^2 \plus{} \left(z \minus{} \frac {21}{2}\right)^2\leq 6^2$ and $ x^2 \plus{} y^2 \plus{} (z \minus{} 1)^2 \equal{} \left( \frac 92\right)^2$.
Simplifying, $ x^2 \plus{} y^2\leq \frac {23}{4}$ and $ x^2 \plus{} y^2\leq \frac {17}{4}$.

Notice that all $ (x,y)$ satsfying the second inequality also satisfy the first one, so we need only find all the lattice points that satisfy the second inequality. There are $ 13$.

$ ( \minus{} 2, 0, 5)$, $ (2, 0, 5)$, $ (0, \minus{} 2, 5)$, $ (0, 2, 5)$, $ ( \minus{} 1, \minus{} 1, 5)$, $ (1, \minus{} 1, 5)$, $ ( \minus{} 1, 1, 5)$, $ (1, 1, 5)$, $ ( \minus{} 1, 0, 5)$, $ (1, 0, 5)$, $ (0, \minus{} 1, 5)$, $ (0, 1, 5)$, and $ (0, 0, 5)$.
[/hide][/quote]
Is there a better way to find the number of lattice points to the last inequality rather than enumerating all the possibilities?

[quote="mathwizarddude"] Is there a better way to find the number of lattice points to the last inequality rather than enumerating all the possibilities?[/quote]

it's not really as bad as it looks. We have $ x^2 \plus{} y^2 \le 4.25$
Suppose $ x\ge y$, we have $ (0,0), (1,0), (1,1), (2,0)$ that's $ 4$ ways

Now accounting for permutations and (for lack of a better term) the "$ \pm$" factor, we multiple by $ 2$ where necessary to find;
$ 1 \plus{} 1\cdot 2\cdot 2 \plus{} 1\cdot 2\cdot 2 \plus{} 1\cdot 2\cdot 2 \equal{} 1 \plus{} 4 \plus{} 4 \plus{} 4 \equal{} 13$ ways


Billy Bob is the local king. He got 9 golden statues from the neighboring king, the Burger King. The Burger King also tells Billy Bob that one statue weighs a little less than all the others. Billy Bob has only a weighing scale (that kind where there are two sides, and you put stuff on either side, and the scale tilts and stuff) to tell the difference. An old man tells Billy Bob that he can find that statue in two weighings of the scale. How did he do it? Explain thouroughly.

[quote="Totally Zealous"][hide="just one weighing"]Actually he can do it in one weighing. Just test them and u'll end up with 4 on either side and they're balanced. The left over is the one that's a little less.[/hide][/quote]
Actually, the question is asking how to [i]guarantee[/i] that you can find the one with the lowest weight for [i]any[/i] configuration, not just a specific one.

[quote="SplashD"][hide]Put three on each side, then there's two case:

Case 1:
They weigh the same, that means the lighter one is in the other group of three. Weigh two of the other group and you can easily find out.

Case 2:
One group weighs less. That group has the lighter statue. Weigh two statues from that group and you can find the lighter statue.[/hide][/quote]
Yes, that's correct. With each weighing you can be guaranteed to eliminate 2/3 of the total number of objects.

[quote="Totally Zealous"][hide="just one weighing"]Actually he can do it in one weighing. Just test them and u'll end up with 4 on either side and they're balanced. The left over is the one that's a little less.[/hide][/quote]

This only works if he is lucky.  If the scale does not balance, then there are 4 which could be the lighter gold statue, thus it does not gaurantee a result.

[quote="mysmartmouth"][quote="nebula42"]
Yes, that's correct. With each weighing you can be guaranteed to eliminate 2/3 of the total number of objects.[/quote]
I assume we would take the floor of $\frac{2}{3}x$ and eliminate it.

Is that true for any one of these problems?  For example, pose the same question with 90 statues.  Would this mean:

Weighing 1: 30 statues
Weighing 2: 10 statues
Weighing 3: 4 statues
Weighing 4: 2 statues
Weighing 5: 1 statues

So it would take 5 trials for 90 items?  Does this work out?[/quote]

Yes, I think. 5 weighings is the maximum required to find the lightest one. With 10 left, you could weigh 3 and 3, with 4 left. This could potentially narrow it down to 3, but also possibly 4. You could also weigh 4 and 4, with 2 left.

[hide]Weight 3 against 3. If one group is lighter, it contains the statue. If they are the same, the third group of 3 contains the statue. Repeat with 1 against 1 from that group, and use the same argument.[/hide]


An isosceles trapezoid has legs of length 30 cm each, two diagonals of length 40 cm each and the longer base is 50 cm. What is the trapezoid's area in sq cm?

[quote=pascal_1623]I got the height was 24 because there was a right triangle in 30-40-50, so the area was 600 and with a base of 50, the height is 24.
Then if the leg is 30 and the height is 24, thats another 3-4-5 triangle scaled by 6 so the other side is 18. 50-36=14

(14+50)/2 * 24 = 768[/quote]

Yes, thats the correct answer. The Alcumus answer states:
We can pick a diagonal and a leg of the trapezoid such that, along with the longer base, these lines form a triangle with sides of length 30, 40, and 50. This is a Pythagorean triple, so the triangle is a right triangle. It follows that the altitude to the longer base of the trapezoid is $30\cdot 40/50 = 24$. This altitude is the same length as the height of the trapezoid.

We now look at the right triangle formed by this altitude, the adjacent leg of the trapezoid, and part of the longer base. These three sides form a right triangle, with hypotenuse of 30 and one leg (the altitude) of length 24. It follows that the other leg has length 18.

Because this is an isosceles trapezoid, we can now calculate the shorter base to have length $50 - 2\cdot 18 = 14$. Therefore, the area of the trapezoid is $\dfrac{(50 + 14)(24)}{2} = \boxed{768}$.


Just a quirky little idea I came up with while fiddling with my smartphone. Basically, the idea goes that there is a small (byte-wise) webpage (http://www.mathlinks.ro/mobile, for instance) that has the problem of the day archive on it. It should just be the current problem of the day and a list of previous ones. It's best without the frills and images, simply HTML and some formatting.

What is the feasibility of this, and is it a good idea at all? I think it would be fun during times of boredom. Naively, I would say it would be somewhat easy to implement. Thoughts?

I had such an attempt at some point in the past, but the way technology is evolving this would be a waste of my time. I can now use my phone for example to browse the full version of the page (and mine is 2 years old -- today's popular phones have huge screens that support almost all features normal browsers do.)

Still it might give some respite to poeple like me whose phone browser does nt support much !