One stamp is randomly selected from a 10-by-10 sheet of 100 stamps. What is the probability that the stamp selected is not along an outer edge? Express your answer as a common fraction.

the number of stamps that are not on the edges is $ (10\minus{}2)^2\equal{}64$, and the whole number of stamps is 100.

so the probability that the stamp selected is not along an outer edge is $ \frac{64}{100}$, which is $ \frac{16}{25}.$

Answer : $ \frac{16}{25}$

[quote="forthegreatergood"]I don't think this is right.  I think it's $9^2 = 81$ stamps, thus $\frac{81}{100}$[/quote]

Nope, it's $8^2 = 64$ stamps. Consider any row of stamps (not on the edge). All but the two stamps on the sides are not on the edges, hence there are 8 in that row. We do the same with the column, and we get 8 in each column.

Oh right, because of the other edge :blush:


The spinner below is designed so that landing in each of the
five sectors is equally likely. What is the probability that the
sum of the numbers determined in two spins is 10? Express
your answer as a common fraction.

[asy]draw(Circle((0,0),20));
draw((0,0)--(0,20));
draw((16,12)--(0,0));
draw((-16,12)--(0,0));
draw((-12,-16)--(0,0));
draw((12,-16)--(0,0));
label("5",(8,12),NW);
label("4",(6,-2),NE);
label("8",(0,-6),S);
label("3",(-6,-2),NW);
label("7",(-8,12),NE);[/asy]

there's total 25 (5*5) outcomes...

and the number of ways that the sum of the spins is 10 is 3.

(5,5 - 3,7 - 7,3)

so answer is 3/25.

answer : 3/25

[b]ya but wouldn't (3, 7) and (7, 3) be the same.......[/b]

[quote]Don't compare apples to oranges.[/quote]


If $ \alpha, \beta$ are the roots of
$ (x \minus{} 2)(x \minus{} \sqrt3) \equal{} \sqrt7$,
find the roots of
$ (x \minus{} \alpha)(x \minus{} \beta) \plus{} \sqrt7 \equal{} 0$

We have $ x^{2}\minus{}4\sqrt{3}x\plus{}2\sqrt{3}\minus{}\sqrt{7}\equal{}0$ so
$ (x\minus{}\alpha)(x\minus{}\beta)\plus{}\sqrt{7}\equal{}x^{2}\minus{}4\sqrt{3}x\plus{}2\sqrt{3}\minus{}\sqrt{7}\plus{}\sqrt{7}\equal{}x^{2}\minus{}4\sqrt{3}x\plus{}2\sqrt{3}\equal{}(x\minus{}2)(x\minus{}\sqrt{3})$.
So the two roots are just $ 2$ and $ \sqrt{3}$.

[quote="JRav"]We have $ x^{2} \minus{} 4\sqrt {3}x \plus{} 2\sqrt {3} \minus{} \sqrt {7} \equal{} 0$.[/quote]

Shouldn't that be $ x^{2} \minus{} (2 \plus{} \sqrt {3})x \plus{} 2\sqrt {3} \minus{} \sqrt {7} \equal{} 0$?

By bad.  Yes it should.  But that doesn't affect the final answer.

Its easier to see why when it is written this way:

$ (x\minus{}2) (x\minus{}\sqrt 3) \equal{} \sqrt 7 \Rightarrow (x\minus{}2)(x\minus{} \sqrt 3) \minus{} \sqrt 7 \equal{} (x \minus{} \alpha)(x\minus{}\beta)$

$ \Rightarrow (x \minus{} \alpha)(x\minus{}\beta) \plus{} \sqrt 7 \equal{} (x\minus{}2)(x\minus{} \sqrt 3)$.

 Hence we easily conclude that the solution set for $ (x \minus{} \alpha)(x\minus{}\beta) \plus{} \sqrt 7 \equal{} 0$ are $ {2, \sqrt 3}$


geometry
inequalities
inequalities unsolved

give a, b, c is the sides of a triangle. prove that

\[\frac{\sqrt{3}}{2} \sum a + \frac{1}{4} \sum \mid a-b \mid \geq \frac{3}{2\sqrt{2}} \sqrt{a^2 + b^2 + c^2 + 4\sqrt{3}S} \]
where $S$ is the area of ABC
 :D

I have some ideas about this one:
\[( \sqrt 3/2 \sum a+1/4 \sum|a-b|) ^2 \geq 3/4\sum a^2 +3/2 \sum ab +1/8 \sum a^2-1/8 \sum ab+ \sqrt 3/4a(a-b)+ \sqrt 3/4b(b-c)+ \sqrt 3/4c(c-a) \]
\[=9/8 \sum a^2+ (\sqrt 3-1)/4 \sum a^2+(11-2 \sqrt 3)/8 \sum ab \geq 9/8  \sum a^2+9/8 \sum ab\]
But \[ \sum ab \geq 4 \sqrt 3S\]

So it is done. :)

congratulation !!
in fact we have this inequality

$\frac{\sqrt{3}}{2} \sum a + \frac{1}{4} \sum \mid a-b \mid \inq \sum m_a \gneq \frac{3}{2\sqrt{2}} \sqrt{a^2 + b^2 + c^2 +4\sqrt{3} S}$

the first, i think, is an interesting inequality
so try it  :D

no one care about this inequality?

$\frac {\sqrt{3}}{2} \sum a + \frac 1{4} \sum \mid a-b \mid \geq \sum m_a \geq \frac{3}{2\sqrt{2}} \sqrt{a^2+b^2+c^2+4\sqrt{3}S}$

 :(

Well,this one is easy if we are well-known at inequalities about Geometry.
For the first one:
 [tex] (\frac {\sqrt3}{2} \suma +\frac 14\ \sum \mid a-b \mid \ ) ^2 \geq (m_a+m_b+m_c)^2 [/tex]
Expand it and use this inequality:
 [tex] 2(a^2+b^2+c^2)+ab+ac+bc \geq 4(m_am_b+m_bm_c+m_cm_a) [/tex]
For the second one:
 [tex] (m_a+m_b+m_c)^2 \geq \frac98 (a^2+b^2+c^2+4\sqrt3 S) [/tex]
In fact,if x,y,z are three sides of a triangle then:
  [tex] 2(xy+xz+yz) \geq 4 \sqrt3 S_{xyz} + x^2+y^2+z^2 [/tex]  where $ S_{xyz} $ is the area of triangle.
And $ S_m $ which is area of the triangles with three side $ m_a,m_b,m_c $ is [tex] \frac34\ S [/tex]
So ,expand it again and we have done. :)


Where do i go in the forum for geometry? Everything in this high school "basics" part is above geometry.


inequalities
vector
inequalities proposed

Let x,y,z>0 satisfying xy+yz+zx=1. Prove that
     $ \sqrt{2*(x^2+y^2+z^2+1)}\geq x*\sqrt(1+y^2)+y*\sqrt(1+z^2)+z*\sqrt(1+x^2) $

Well,nice title but not a nice hint :) .
We have:
[tex] x\sqrt{1+y^2}+y\sqrt{1+z^2}+z\sqrt{1+x^2} [/tex]
[tex]= \sqrt{(x^2+xy)(x^2+xz)}+\sqrt{(y^2+yz)(y^2+xy)}+\sqrt{(z^2+xz)(z^2+yz)} [/tex]
[tex]\leq \sqrt{2(x^2+y^2+z^2+xy+xz+yz)}=\sqrt{2(x^2+y^2+z^2+1)}[/tex]

maybe i'm misunderstanding something, but i think the second line must be

$\sqrt{(x^2+xy)(xy+xz)}+\sqrt{(y^2+yz)(yz+yx)}+\sqrt{(z^2+zx)(zx+zy)}$

and i don't understand anyway how you get the third line from the second(it sure is meant to be AM-QM for the three roots, but I don't get your terms).

Peter

Oh,I am so sorry about that crazy mistake.I don't understand how I can wrote something like that.In my mind is exactly what you have written,but when I wrote it,it was not.
In fact,what I wrote is clearly wrong because we must have the opposite one.
Hence,it leads me up to another one,and may be harder,who knows :) :
Let $ x,y,z \geq0 $ satisfying $ xy+xz+yz=1 $.Prove that:
[tex] x\sqrt{1+x^2}+y\sqrt{1+y^2}+z\sqrt{1+z^2} \geq \sqrt{2(1+x^2+y^2+z^2)} [/tex]
So try it again.

 :D

Oh,again,my problem is thrown into a bin :( So boring?
Maybe I must open a new topic for it.

Have you a simple or a very hard solution , Treegoner ?

Here is my solution :
     First I want to tell you an inequality
   Let ABC be an acute triangle. Then for any positive m,n,p we obtain
  $\sum mcosA \geq \sum (n^2+p^2-m^2)sin^2A$
    You can see its proof in  http://mathlinks.ro/viewtopic.php?t=6523
  Back to our problem, take $cosA =  \frac {x}{\sqrt{(x+y)(x+z)}} $ and $m=x+y$.
we finally obtian my inequality. :P

With your lemma I don't obtain your ineq, Treegoner?

Can you please detail it better ?

Thank you very much.

I'm sorry. Let $m=\sqrt{x+y}$
 Thank you , Manlio.

Hmm. We can use Cauchy on treegoner's inequality:

Using Cauchy on the vectors [tex]\displaystyle\{\sqrt{x^2+xy}\}[/tex] and [tex]\displaystyle\{\sqrt{xy+xz}\}[/tex] gives

[tex]\displaystyle\\ \sqrt{2(x^2+y^2+z^2+1)}\\
= \sqrt{((x^2+xy)+(y^2+yz)+(z^2+zx))((xy+xz)+(yz+yx)+(zx+zy))}\\
\ge \sqrt{(x^2+xy)(xy+xz)}+\sqrt{(y^2+yz)(yz+yx)}+\sqrt{(z^2+zx)(zx+zy)}\\
=x\sqrt{y^2+1}+y\sqrt{z^2+1}+z\sqrt{x^2+1}[/tex]


[quote="sam-n"]prove that if $n$ is arbitrary integer then $\sum_{r=0}^{n}(-1)^{r}{n \choose r}{2n-2r \choose n-1}=0$.[/quote]
are these things usuful or the are only practice for calculating.can you help me?because i really don't like these equations. :(

[quote="jensen"]are these things usuful or the are only practice for calculating[/quote]

actually Calculating these kinds of problem is not nice,finding a combinatorics proof for it is nice.

yes,i am agree with you.you are right.combinatoric is nice but combinatoric calculating is not so nice.so you are right and we should find a combinatoric proof for thes equations and then we will enjoy our selvese. :D

I think this'll work for a combinatorial proof

We have a $2 \times n$ strip divided into $n$ $2 \times 1$ bricks.  We color the strip such that the following two conditions are satisfied:

1. If a brick is colored at all, its two squares are colored some combination of red and blue (not all bricks have to be colored)
2. Exactly $n-1$ squares are colored blue.

The number of ways of doing this such that $r$ bricks remain uncolored is (up to sign) the summand, so it suffices to show there are an equal number of colorings with an even number of uncolored bricks and an odd number of uncolored bricks.  To do so, we note the following bijection between them:

Examine the bricks in order from left to right until we encounter a red brick or an uncolored one, and either uncolor it (if it is red) or color it red (if it is uncolored).  This is well defined since if all $n$ bricks are colored one of them can't have any blue squares, and can easily be seen to be its own inverse.


[quote="nguyenvanhuyen"]Solve an equation
$ log_2x \plus{} log_3x \plus{} log_4x \plus{} ... \plus{} log_{2009}x \equal{} log_{2010}x$[/quote]

The solutions are in $ (0,\plus{}\infty)$.

Let $ f(x)\equal{}\log_2x\plus{}\log_3x\plus{}\log_4x\plus{}\cdots \log_{2009}x \minus{} \log_{2010}x$

$ f'(x)\equal{}\frac1x \left(\frac1{\ln 2}\plus{}\frac1{\ln 3}\plus{}\frac1{\ln 4}\plus{}\cdots \plus{}\frac1{\ln 2009}\minus{}\frac1{\ln 2010} \right) > 0, \qquad \forall x>0$

So $ f(x)$ is strictly increasing and therefore $ f(x)\equal{}0$ has just one solution: $ \boxed{x\equal{}1}$


[quote="maggot"]Let $A,B,C,D,P$ be points on a circle $O$ such that $AB=CD$. Let $X,Y$ be the incenter of triangles $PAB,PCD$. $PX,PY$ cut the circle $O$ again at $E,F$. Show that $XE=YF$.[/quote]
Just prove that $XE=EA=EB$,$YF=FC=FD$ and $\triangle{EAB}=\triangle{FCD}$ :wink:


There's a sheperd that has a wolf, a sheep, and some grass. He also has a boat. The sheperd wants to take all of the objects and animals to the other side of a river with his boat, but he can only take two objects at a time. The sheperd also doesn't want to get one of the objects or animals eaten.

[hide]The wolf would eat the sheep when he goes to the other side of the river for the grass, and the sheep would eat the grass when he gets the wolf, so he would take the wolf with the grass to the other side, then fetch the sheep.  :rotfl: [/hide]

Here's a slightly harder version:

Same thing, but he can only take one item at a time.

He takes the wolf, then the sheep, then the grass. Nice one! By the way, how do you always get to my blogs? Is there a certain way to get to a certain person's blogs?  :wink:

But the sheep will eat the grass, if you take the wolf first.

Wow! This time it isn't wrong: the sheep first, then the wolf, then the grass.  :D

Still a little problem: You can't leave the sheep and wolf together while you get the grass.

i wonder if there is a true answer... do you know the right one, i_like_pie? Also, have you checked out my "An Easy Riddle?"  :lol:

There is a solution. It isn't obvious, but can be found with a little thought.

The Steps:
1. Take Sheep and go back
2. Take Wolf and bring sheep back
3. Take Grass and go back
4. Take sheep :D


Let x,y,z be integers. Solve the equation: x^5+y^5+z^5=2009.

Suppose that $ x$ is not divisible by $ 11$, then by fermat's little theorem:
$ x^{10}\equiv 1\bmod{11}$
$ (x^5\minus{}1)(x^5\plus{}1)\equiv 0\bmod{11}$
So $ x^5\equiv 1\bmod{11}$ or $ x^5\equiv \minus{}1\bmod{11}$
Then looking at the values of $ x^5 mod(11)$, which are $ 1,0, \minus{}1$ we conclude that
$ x^5\plus{}y^5\plus{}z^5 \not\equiv 2009\bmod{11}$ so there aren't any solutions

Daniel


1. Joe and I begin jogging around a track, starting at the finish line and going the same direction. Joe completes a lap every $ 78$ seconds, while I take $ 90$ seconds to complete each lap. At the end of our workout, we cross the finish line together in a whole number of minutes for the first time. How many more laps did Joe run more than me?

2. Ken gets his hair cut every $ 20$ days. Larry gets his cut every $ 26$ days. Ken and Larry get their hair cut on the same Tuesday. What day of the week is it the next time they get their hair cut on the same day?

3. Add $ {1/37} \plus{} {1/9} \plus{} {1/3}$. Express the result as a decimal using bar notation.

4. Express the following as a decimal: (NO CALCULATOR)

a.) $ {41/99} \plus{} {55/990}$

b.) $ 2/9999$

c.) $ 2/9900$

d.) $ 2/33$

e.) $ 5/7$

5. What is the smallest 4-digit number that is divisible by 2,3,4,5,6,7,8,9,and 10?

6. Using only $ 1's$ and $ 2's$, what is the smallest integer you can create which is divisible by both $ 3$ and $ 8$?

7. Simplify: (NO CALCULATOR).

a.) $ \sqrt {25600}$

b.) $ \sqrt {48/15}$ (NOTE THAT $ \sqrt {\dfrac{a}{b}}$ = $ \sqrt {a}/\sqrt {b}$).

c.) $ \dfrac{5}{2 \plus{} \sqrt5}$

d.) $ 16^{ \minus{} 3/4}$

e.) $ 256^{3/2}/ 16^{3}$

f.) $ 3125^{ \minus{} 3/5}$

1. 4 laps[hide="Solution"] you can simplify this problem to an equation of 3 variables: Time, Fast, Slow, Time finishes a cycle every 60 seconds, fast every 78 seconds, slow every 90 seconds. reduce each cycle by a factor of 6, it then becomes a least common multiple equation of 10, 13, and 15, the lcm of which is 390. this is equivalant to 39 t cycles, 30 f cycles and 26 s cycles. the rest is self-explanatory[/hide]
2.Thursday[hide="Solution"] another lcm problem, lcm is 520/7= 74 remainder 2, tuesday+2 days= Thursday[/hide]
3.0.471471471... rest will be done later g2g

@ garfield: Look at astuteverbosity's solution for #2. Express #3 into a decimal using bar notation. On #4 d, you probably made a computation error. Also, check your work on b and c on #7 (B can be simplified further).

Besides that, good job.  :lol:

What about me? :read:
[quote="mr.k_74"]@ garfield:...Besides that, good job.  :lol:[/quote]
jk,jk :jump:  :harhar:


Problem 5. Suppose we have triangle ABC and its circumcircle.  Consider any two diametrically opposite points (not A, B, or C) on the circumcircle and draw their Simson lines.  Prove: These lines are perpendicular and that their intersection lies on the nine-point circle of ABC.

Well-written proofs would be appreciated.

this well-known and classic problem is for TST? :o

I'm also unsure of why this problem was included in TST. As I am aware, no member of the Taiwanese team did not receive full score on it.

However, although this problem may be classic and well-known in the West, I have the impression that that is not the case here.  I think different topics are emphasized in Taiwan!


. solve for real x,
   sin3x+cos3x=2sinx * cosx

I'm not sure...would expanding help? I don't know what to do...
[hide]\[ 4\sin x \cos^2 x\minus{} \sin x \plus{} \cos x \minus{} 4 \sin^2 x \cos x \equal{}2 \sin x \cos x\][/hide]

[hide="may be a hint, or just wasting your time"]double angle formula? BUt that would cause a 2x to appear, which doesn't really help.[/hide]

Man, I can't do trig.

hmm i dont think expanding will help much

If we square both sides and apply double angle formulas we get $ sin6x\plus{}1\equal{} (sin2x)^2$. Then if we use a triple angle formula on $ sin6x$ we get $ \minus{}4(sin2x)^3\minus{}(sin2x)^2\plus{}3sin(2x)\plus{}1\equal{}0$. However this polynomial-ish expression doesn't seem to have nice roots which can inverse sine( unless I messed something up).

I don't know if this will be helpful

sin3x + cos3x = (sin3x*cos45 + cos3x*sin45)*sqrt(2) = sqrt(2)*sin(3x + 45)


[quote="James Potter"]Let be given a natural number $n\ge 2$.Find all real numbers $\alpha$ such that for all sets of positive real numbers $a_1\ge a_2\ge a_3\ge ...\ge a_n$ we have:
$\frac{a_1}{a_n}+\frac{a_2}{a_{n-1}}+....+\frac{a_{n-1}}{a_2}+\frac{a_n}{a_1}\le n+\frac{(a_1-a_n)^2+(a_2-a_n)^2+...+(a_{n-1}-a_n)^2+(a_n-a_n)^2}{a_n^\alpha}$[/quote]
The inequality is true for $\alpha=2$ ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=440638#p440638]proof is here[/url]).
For all other $\alpha$ we can multiply all $a_i$ by $c>0$ and for $c\rightarrow 0$ or $c\rightarrow \infty$ (depending on $\alpha$) we get $RHS\rightarrow n+0$ while the LHS is constant. Obviously we don't have $LHS\leq n$ for all such sets, so the inequality is wrong for $\alpha\not=2$.


conics
ellipse
search
geometry proposed
geometry

[quote][color=blue]Construct the triangles $ABN$, $ACM$ out of the triangle $ABC$. Define the point $I\in MB\cap NC$. Prove that if two from the following statements are truly, then and the third statement is truly:
$1^{\circ}.\blacktriangleright\ \triangle AMC\sim\triangle ABN$ (with the mentioned order of letters) .
$2^{\circ}.\blacktriangleright$ The quadrilateral $MNBC$ is cyclically .
$3^{\circ}.\blacktriangleright$ The lines $AI$, $BN$ and $MC$ are concurrently .[/color][/quote][b]Demonstration.[/b]

$\boxed {1^{\circ}\ \wedge\ 2^{\circ}\Longrightarrow 3^{\circ}}\ : \ldots$
$\boxed {1^{\circ}\ \wedge\ 3^{\circ}\Longrightarrow 2^{\circ}}\ : \ldots$
$\boxed {2^{\circ}\ \wedge\ 3^{\circ}\Longrightarrow 1^{\circ}}\ : \ldots$

Hello Virgil,

  I'm truly delighted that you still get inspiration from some of the fine ideas in

  the my solutions(?).

  But solutions to the problems of 'this kind' (specifically prove 1 when 2 and 3 are true)

   I usually leave for Yetti to deliver,   for he is a master of presentation:

   a whole lot of the finest  detail in multiple color ink,

  wonderful  graphics, and application of very special theorems, known to a very narrow

  circle (is it an ellipse?) of mathematicians. Later we all can search for the bright

  ideas of his in those solutions, and enjoy.

   T.Y.

   M.T.

You are permanently for me a "[u]inspiration source[/u]" ! I read all you write. Like you this last problem inspirated from the your unique of its kind [u]great[/u] ideas ? [quote][color=red][b]The classical geometry is the oldest section of the Math. The unique ideas in this geometry are very rarely ! Still all come from this  Geometry and return back cyclically...[/b][/color] [color=blue]
(Dan Barbilian - a great Romanian geometer).[/color][/quote]


do you need to send it every year because I thought if you participated in previous years, you only needed to verify your grade or something.

If you do need to send one in every year and I didn't send in my entry form for year 18 but already sent my solutions, can I still send it in and be eligible?

I'm pretty sure it said on the top "submit one per year."


geometry
3D geometry
sphere
number theory

Why is my approach and answer to the following incorrect?

[b]8. (Geometry Post-Test)
Three spheres are tangent to a plane at the vertices of a triangle and are tangent to each other. Find the radii of these spheres if the sides of the triangle are 6, 8, and 10.[/b]
[hide="My Incorrect Approach"]See my attached diagram.
x+y=6
y+z=8
x+z=10
Adding the above equations together, we get 2x+2y+2z=24 so x+y+z=12.
Using this equation, we find integer answers. However, the correct answers are 12/5,15/4, and 20/3.
[/hide]

[b]8. (Number Theory Post-Test)How many natural numbers are divisors of exactly two of the three integers 240, 600, and 750?[/b]

How would you attack and solve the above number theory problem?

are you sure those are the right answers? i got the same solutions as you did.
for the number theory one, it would help to prime factorize.

[hide="number theory"]
We factor to get it as
$ 2^4*5*3$
$ 2^3*5^2*3$
$ 2*5^33$.
If we have a multiple of $ 2^2$, it will clearly divide only 2. We can pair it with 5,3, or 15.
Same for $ 2^3$,$ 5^2$, so 3*4=12.[/hide]

If I understood the problem correctly, I'm pretty sure your solution is right..  :huh:

Thanks guys.
I understand the number theory problem.
Thanks for pointing that out, Igd0612.

the answers from the 3 equations are the distance between each tangent points of the three spheres on the plane (see diagram, Ac and Bc)       not the radii           [/img]

sorry here is the diagram   that was my first post, i dont really know how to insert images on my post       here is just an example of 2 spheres touching the plane,  the distance between their tangent points(A,B) on the plane is not the sum of their radii


Let $A_{1}A_{2}\ldots A_{n}$ be a regular polygon. Find all points $P$ in the polygon's plane with the property: the squares of distances from $P$ to the polygon's vertices are consecutive terms of an arithmetic sequence.

I found, if I'm not wrong, that the answer is trivial for $n\ge 5$. I would like to see a second opinion.

See case $n=4$. Maybe this will make the problem interesting...


floor function
combinatorics unsolved
combinatorics

Consider a rectangular grid of $ 10 \times 10$ unit squares. We call a [i]ship[/i] a figure made up of unit squares connected by common edges. We call a [i]fleet[/i] a set of ships where no two ships contain squares that share a common vertex (i.e. all ships are vertex-disjoint). Find the least number of squares in a fleet to which no new ship can be added.

For a n X n  board  the minimum number of ships in the fleet so that no new ship can be added is
\[ \big(\frac{n}{3}\big)^2\quad if\quad3|n\]
\[ \big(\lfloor\frac{n}{3}\rfloor\plus{}1\big)^2\quad if \quad3\nmid n\]

We start with observing that a 3 X 3 board requires exactly one ship in the fleet at the centre. Hence the case when $ 3|n$ is straightforward. When $ n\equal{}3k\plus{}1$ add an extra column to $ 3k$ board.Keeping the ships of the $ 3k$ board intact, we need to add exactly $ 3k\plus{}1$ ships. This suffices for $ 3k\plus{}2$ board too. If we rearrange the ships in the $ 3k$ board, we still end up requiring at least $ 2k\plus{}1$ ships.

Consider a 3(m+1)+1 X 3(m+1)+1 board,rearrangement of the ships in the 3m X 3m (top left corner) would not yield to a configuration where one needs to add lesser ships ( than $ 2(m\plus{}1)\plus{}1$ ships) as the bound is tight in 3m X 3m board(this comes inductively by the bound in 3 X 3 board). We are left with a intersecting vertical and horizontal strips that can be broken down into  $ m$ 4 X 3 (or 3 X 4) boards and one 4 x 4 board. It can seen that both 4 X 3 and 4 X 4 boards are tight bounds of 2 and 4 respectively, the 7 X 4 board has the tight bound of 7 and the result follows.


9. Find the value of x that satisfies the equation
$ 25^{\minus{}2} \equal{} \frac{5^{48/x}}{5^{26/x} * 25^{17/x}}$.

$ (A) 2 \qquad (B) 3 \qquad (C) 5 \qquad (D) 6 \qquad (E) 9$.

$ 5^{ \minus{} 4} \equal{} \frac {5^{48/x}}{5^{26/x} \cdot 5^{34/x}} \Longrightarrow 5^{ \minus{} 4 \plus{} 26/x \plus{} 34/x} \equal{} 5^{48/x} \Longrightarrow \frac {12}{x} \equal{} 4 \Longrightarrow x \equal{} 3\ \mathrm{(B)}$

[quote="JRav"]Honestly, on such a question, since you get a calculator, just use it.  That's how I would solve it if I was taking it.[/quote]

Calculators aren't allowed anymore. I think solving by exponents is faster anyways.  $ 25^{ \minus{} 2} \equal{} 5^{ \minus{} 4}$ for example

Really, they banned calculators?  Is this just on the 10 or the 12 also?

\[ \displaystyle\frac{1}{25^2}\equal{}\displaystyle\frac{1}{\sqrt[x]{5^{12}}}\] from where x=3


Solve the equation :
 $ x^4 + 4ax^3 + 6b^2x^2 + 4c^3x + 1 = 0 $ , where $ a,b,c $ are possitive real numbers , $ a\leq 1 $ , knowing that it has four real roots

Since $a,b,c > 0$ the roots, say $x_1,x_2,x_3,x_4$, are clearly negative.
Let $y_i = -x_i>0.$
Thus $a = \frac 1 4(y_1+y_2+y_3+y_4)$ and $y_1y_2y_3y_4 = 1.$
Using AM/GM, it leads to $1 \geq a \geq 1$. Thus $a=1$ and we are in the equality case of AM/GM, which means that $y_1 =y_2=y_3=y_4 = 1.$
Therefore $a=b=c=1$ and the equation reduces to $(x+1)^4 = 0$, so that $-1$ is the only solution.

Pierre.


Consideremos el conjunto del 1 al 2001. Demostrar que si tomamos un subconjunto cualquiera de 1336 numeros distintos, entonces en dicho subconjunto existen 4 numeros $a<b<c<d$ tales que uno es la suma de los otros tres.

[color=blue]¿Podría la longitud del subconjunto ser de $1335$ en lugar de $1336$ números, como solución al problema? Mis cálculos me inducen a ello, aunque bien puede aún faltar un poco más de trabajo.[/color]

Pues RGAM, el problema te lo esta pidiendo resolver con 1336, pero si lopuedes hacer con 1335 pues significa que as mejorado el resultado, y elproblema se sigue cumpliendo. (Aunque a decir verdad dudo que se pueda con 1335) Si terminas hay posteas tu solucion!

[color=blue]Estoy analizando la situación aplicando combinatoria, pero aún tengo algunas complicaciones.

Hallé el valor $1335$ de este modo: Sabemos que $2001=3\times667$. Pronto deducimos que $2001=666+667+668$. Si establecemos un conjunto $A=[667,2001]$, comprobamos que la suma de tres números cualesquiera es un valor mayor que $2001$, pero si agregamos cualquier valor del conjunto $[1,666]$ a $A$, siempre habrá cuatro números en $A$ que cumpla la condición requerida.
Entre $667$ y $2001$ hay $2001-667+1=1335$ términos (¡Ups, me equivoqué, no son de $1335$ términos!), y adicionando cualquier otro valor, el conjunto $A$ tendría $1336$ elementos.

Estuve asumiendo que ése es el mejor caso. ¿Quién podría darme una mano para demostrar que no hay otro mejor?

[b]P.D.:[/b] Me da la impresión de que la solución del problema $1$ del tópico [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108283[/url] nos podría ser de ayuda.[/color]


What is the coefficient of $ x^{3}$ in the expansion of

                                    \[ (x\plus{}2\sqrt3)^{7}? \]

Using the binomial theorem.

$ \binom{7}{4} \cdot x^3 \cdot (2\sqrt{3})^4 \equal{} 35 \cdot x^3 \cdot 144 \equal{} 5040x^3$

So the coefficient is $ \boxed{5040}$.


Mrs. Joespehm is giving a pop quiz today. It is a quiz with 10 multiple choice questions. The choices for each question are A or B, and there is exactly one correct answer for each question. She hands out the answer sheets to all the students in her class. When she realized that the Xerox machine messed up on making copies of the question sheets, she blamed it all on ProtestanT. She made ProtestanT do the pop quiz without the question sheets. Assuming that ProtestanT picks an answer for every question, what are the chances that he gets exactly 3 out of the 10 questions right? Express your answer as a common fraction.

[hide="Hint (Shortcut)"]There is a binomial distribution present: constant number of attempts, independent events, probability does not change... You could just do this completely oblivious of it being called a binomial distribution, of course - but I like the standard methods.[/hide]
[hide="My Solution"]The number of questions that the unlucky ProtestanT gets right is a binomial distribution with parameters $ n = 10$, $ p = 0.5$. A binomial distribution gives the probability of $ x$ successes across $ n$ attempts with a probability $ p$ of success by the formula:

$ \binom{n}{x} \cdot p^{x} \cdot (1 - p)^{n - x}$

Therefore, the chance of him getting $ 3$ out of $ 10$ ($ x = 3$) right will be:

\begin{align*} \binom{10}{3} \cdot \left(\frac {1}{2}\right)^{3} \cdot \left(\frac {1}{2}\right)^{7} & = \frac {10!}{(10 - 3)! \cdot 3!} \cdot \frac {1}{8} \cdot \frac {1}{128} \\
& = \frac {8 \times 9 \times 10}{6} \cdot \frac {1}{8} \cdot \frac {1}{128} \\
& = \frac {720}{6} \cdot \frac {1}{8} \cdot \frac {1}{128} \\
& = 120 \cdot \frac {1}{8} \cdot \frac {1}{128} \\
& = \frac {120}{8} \cdot \frac {1}{128} \\
& = 15 \cdot \frac {1}{128} \\
& = \boxed{\frac {15}{128}} \end{align*}

At least, that's what I think the answer is.

(Isn't the align environment exciting?)[/hide]


I'll add one.

(5) $\frac{1}{1\cdot 2\cdot 3},\ \frac{1}{2\cdot 3\cdot 4},\ \frac{1}{3\cdot 4\cdot 5},\cdots$

[quote="lingomaniac88"][hide="3"]The $n^\text{th}$ term is $\frac{(2n)^{2}}{(2n-1)(2n+1)}$, or $\frac{4n^{2}}{4n^{2}-1}$.  Clearly, all the terms are greater than $1$.  So the answer is $\infty$, I believe.[/hide][/quote]

I think you're suppose to answer in terms of $n$

[hide="1"]this is a common one

$\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$
prove by induction[/hide]

[hide="4"]rationalize the denominators

$\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+\sqrt4}+\cdots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-1$[/hide]

Oh... I guess that makes sense...

It should probably have been stated that it should be in terms of $n$.

[quote="Kurt Gödel"][quote="SplashD"]
[hide="1"]this is a common one

$\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$
prove by induction[/hide]

[hide="4"]rationalize the denominators

$\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+\sqrt4}+\cdots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-1$[/hide][/quote]

Funny: you solve the second one by telescoping, while the first one is a classic in telescoping  :P [hide="1bis"] \[\sum_{i=1}^{n}\frac{1}{i(i+1)}= \sum_{i=1}^{n}\left(\frac{1}{i}-\frac{1}{i+1}\right) = 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{n}-\frac{1}{n+1}= 1-\frac{1}{n+1}.\] [/hide][/quote]

I don't get what's so funny, what's bis?

[quote="Kurt Gödel"][quote="SplashD"]
[hide="1"]this is a common one

$\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$
prove by induction[/hide]

[hide="4"]rationalize the denominators

$\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+\sqrt4}+\cdots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-1$[/hide][/quote]

Funny: you solve the second one by telescoping, while the first one is a classic in telescoping  :P [hide="1bis"]

What is bis? \[\sum_{i=1}^{n}\frac{1}{i(i+1)}= \sum_{i=1}^{n}\left(\frac{1}{i}-\frac{1}{i+1}\right) = 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{n}-\frac{1}{n+1}= 1-\frac{1}{n+1}.\] [/hide][/quote]

Ugh. "Bis" in this case means "second time", "second version". It's Latin for "twice".
Oh yeah, there isn't any connotation of "better" or something like that...if that's what you thought I meant. :lol:

i'm sure he has solved the sequence by telescoping, it's just that it is a very memorable series and is memorized by most people.

[hide="#5"]Clearly, \[\frac{2}{(k-1)k(k+1)}= \frac{1}{k-1}-\frac{2}{k}+\frac{1}{k+1}.\] If you want to know how to get this result...[hide]Let \[\frac{2}{(k-1)k(k+1)}= \frac{a}{k-1}+\frac{b}{k}+\frac{c}{k+1}= \frac{ak(k+1)+b(k-1)(k+1)+c(k-1)k}{(k-1)k(k+1)}= \frac{(a+b+c)k^{2}+(a-c)k-b}{(k-1)k(k+1)},\] so $a+b+c=0$, $a-c=0$ and $-b=2$, i.e. $b=-2$, $a=c$ and $0=a+b+c=2a-2$, thus $(a,b,c)=(1,-2,1)$, from which we get the desired result.[/hide]Or...[hide]Just read the nice identity Virgil Nicula just posted  :lol: [/hide]
Therefore, \[\begin{aligned}\sum_{k=2}^{n}\frac{1}{(k-1)k(k+1)}& = \frac{1}{2}\sum_{k=2}^{n}\left(\frac{1}{k-1}-\frac{2}{k}+\frac{1}{k+1}\right) \\ & = \frac{1}{2}\left(\frac{1}{1}-\frac{2}{2}+\frac{1}{2}+\frac{1}{n}-\frac{2}{n}+\frac{1}{n+1}\right) \\ & = \frac{1}{2}\left(\frac{1}{2}-\frac{1}{n}+\frac{1}{n+1}\right).\end{aligned}\] (Result by telescoping: for $3\leq k\leq n-2$, all the fractions clear out.)
[/hide]

That's a really nice identity, Virgil Nicula  :)

Problem 2:
Since we can get that $a_{n}= \frac{1}{(3n-1)(3n+2)}= \frac{1}{3}\left( \frac{1}{3n-1}-\frac{1}{3n+2}\right)$, thus the sum is $\frac{n}{2(3n+2)}$.


Suppose that $a,b,c,d,e$ are positive real numbers. Prove the inequality
\[\left(\sum_{cyclic}a^{2}\right)\left(\sum_{cyclic}a^{3}\right)\geq \left(\sum_{cyclic}a^{5}\right)\Bigg(8-\frac{3(a^{5}+b^{5}+c^{5}+d^{5}+e^{5})}{5abcde}\Bigg). \]


Talk about your coaches, give props and such...

my coach doesn't do anything except workouts and warmups, which are already on teh mathcounts test

my coach lead some national temas, but i dunno wut ranks they got...any1 know where i can find out?

[quote="mwpl11"]@indianamath: does ur school get coached by Mr. Fischer?[/quote]

no! Mr Fischer is arguably the best coach in the US. he even made up his own practice sprints (about 6-7 for every year) to help his students improve. He has turned his decent HC team into the State Champion Team. Truly remarkable.

[quote="kstan013"]my coach led a team to 2nd at state, but never 1st. I'm only the 3rd person in the history of my school to make nationals[/quote]

Wow I was the 2nd ever to make it to state and the first to place in top 100 at state and our chapter sends 24 people out of 90 to state from my school.

indianamath, I don't think he makes 6-7 for every year, but he has made them for like 1980 years (or whatever his age - 20 is :P) [ok 1991 is the oldest one I've seen so far].

Mr. Fisher, I know *somebody* will relay this to you so you'll...

Yes, I'm proud to say Mr. Fisher has led a team to 1st, with several teams to 2nd to boot.

Wow I can't believe I accidentally spelled "Fisher" "Fischer" the first time around...

At State my coach led my team to 1st this year and last year.  He was the Wisconsin coach last year and I think WI got 19th?

this is the first year my school has had mathcounts and my team didn't make state.

Geez Ignite what state are you? Ure post doesn't make much sense and that's not really much of an accomplishment either.

[quote="mwpl11"]my coaches have never made it to nats. we come second in states quite often, but never first. my coaches are really nice ppl, but they like don't have the fighting spirit that it takes to make it to nats. i think that they think it's like one of those "walk-in-the-park" contests where "as long as u have fun, that's all that matters", but it's way more serious than that. i think that spirit is what Mr. Fisher (Fischer?) has that my coaches so conveniently lack...plus Mr. Fisher like sleeps, eats, and thinks math 24/7, doesn't he? i wish my coaches were that dedicated, then who knows, maybe our team would come first once in a while :([/quote]

well this year our team was 3rd at state, so that happens quite often as well

[quote="mathgeniuse^ln(x)"][quote="ProtestanT"][quote="mathgeniuse^ln(x)"]Our coach, has some huge records.

I would know that in

1986:11th,
1993: Don't know
1994: Don't know
2004:7th
2005:5th
2006:13th

So 5th[/quote]
where do u find those records? got any idea how i can do the same kind of record-finding on my coach?[/quote]

Our school holds the plaques on the wall, so I knew the years, and then she told us that stuff happened.[/quote]
ooo...
My coach put all the plaques in the shelf for display all the way up to 2005, wen it ran outta room. Then she says she jus gave most of them to the Technology/woodworking guy to use as extra wood.
But we aren't the only school my coach coaches. There's 2 middle schools in my district, and she coaches both of them. So the national ones aren't all in our school.

My coach:
1) Has never went to nats
2) had 2 individuals go to state (one was in 2003, other is me... tomorrow :D )

The reason for all of this is because we live in California.

my coach is amazing.... nobody knows of him, but he is amazing

his name is Mr.(john) Lindle.  He is retired but still teaches advanced math classes at our school.

20 years ago he coached 4 students, 2 of which took 1 and 2 at state.  The team won and he coached the national team to 2nd place.  My math teacher for next year is actually one of those nats students and went to college on a MC schollarship.  If you ask me- he is the best coach in the nation.  he eats and breaths math---- he has a collection of over 30 4inch binders full of MC study sheets he made.  Basically- he knows every formula ever made.

Best math coach ever.  No argument.

Hi Richard.

Mr Fischer's team did have a break in his streak when the Benjamin's team beat his...and then they moved to his school so he would be their coach!!!

I believe that Leona Penner of Nebraska has the current and all-time streak.  She has been there every year I have and I want to say as many as 20 in a row.

I also think that Takoma Park Middle School in MD has won every year of MATHCOUNTS.  Their current coach, Sara Manchester, was on a winning team as a kid and now coaches her Alma Mater.  I am pretty sure she's been there every year I have but one, and that was only because she couldn't go for family reasons.

So I have a ways to go, but thanks for the encouragement.

What I really want to know is am I the first to coach two different states to Nationals?  I know Bob Fischer has coached different schools to state wins in IN, but anyone else is different states?

Anyone know?

This is the first time my coach will be coaching NY. Her team got first place a few years back, but sommething will always hold her back.

Our team had the biggest blowout yet in NY, and it sent two team mambers to nationals this year.

[quote="mwpl11"]

is ur coach Mrs. Gupta?[/quote]

No, she is the coach of our rival school.

[quote="13375P34K43V312"]wait so like Ravi B isn't going?

man, I wanted to meet him

he seems cool[/quote]
Thanks for the kind words!  We'll probably meet at some high-school competition some year.  I wasn't a coach this year.  You should meet this year's NY coach, Mary O'Keeffe.  Her daughter was an IMO gold medalist about 3 years ago.

The person that is actually supposed to be our coach
hired a coach/friend, who is a senior at M.I.T.
brilliant person for a 22 year old.
Learned Calculus in 7th grade
We didn't have a team back then,
he made one they won counties, and went on to states
and got on the team. :lol:
Not all of them, just one.

[quote="frost13"]Hi Richard.

Mr Fischer's team did have a break in his streak when the Benjamin's team beat his...and then they moved to his school so he would be their coach!!!

I believe that Leona Penner of Nebraska has the current and all-time streak.  She has been there every year I have and I want to say as many as 20 in a row.

[/quote]

Which school does Leona Penner teach?  Is it Lux Middle School?

Anyway, this was the first year my coach coached the MATHCOUNTS team.  We took first at Chapter, but nobody got anything at state (not even me).

Our coach has only been doing Mathcounts for two years, but we've gotten 1st in state both times (this year and last) and 2nd at Nats last year.

It was our school's first year (I had to beg to counsler)

and ...... Nothing for team.

Just happy that I'm 5th in state...... :blush:

6th grade got first in our state  :bomb:

i think that the poll has nothing to do with the question given.

my teacher has only led teams to even states just a few times, but i feel that she is better than the coach of a nearby school who has led her team to nationals before.  My coach understands that all of us are doing this because we like math, and she doesn't grill us for 3+ hours a day.  Plus she is really super nice, determined, and just a really good coach.

Just saying.

[quote="Karth"]aargh...my coach doesn't care as much as others, but i still have a good chance of making nats.[/quote]
Aren't you a junior?


Find $tanA$ in a triangle $ABC$. If:

$sinA=n(sinBsinC)$
$cosA=n(cosBcosC)$

Well clearly $\tan A = \tan B \tan C$
If you want to go further... perhaps use $\tan A \tan B \tan C = \tan A + \tan B + \tan C$.  I couldn't get anywhere with it though.

Here's my solution:

Obviously:$tgA=tgB.tgC$

Then:$cosA=-cos(B+C)=-cosBcosC+sinBsinC=n(cosBcosC)$

Dividing by $cosBcosC$:$-1+tgBtgC=n \rightarrow tgBtgC=tgA=n+1$

almost similar idea with Mario

we substract the first equation from second

$\cos A-\sin A=n(\cos B\cos C-\sin B\sin C)=n\cos (B+C)=-n\cos A$

$\Longleftrightarrow \sin A=(n+1)\cos A$

$\Longleftrightarrow \tan A=n+1$


From any letter in the diagram, a move can only be made to a letter diagonally adjacent and below. In how many ways can a path that spells MATHCOUNTS be taken?

[hide]
It should be easy to see that there are $2^4=16$ ways to get to the $C$s.
Now you have to think about how many ways to get from the Cs to the S.
Think backwards.
There are 16 ways to get to the Cs from the S, same as before.
Thus it is $16\cdot16=256$
[/hide]

That's not what I'm getting, but I might be rwong.

[hide] There is exactly one way to get to each of the "A"'s. You can then write down "1" next to the A's. Then, to figure out how many ways you can get to the "T"s, you add the numbers from the contiguous (touching) "A"s, so the left T is 1, the middle T is 1+1 = 2, and the third T is 1.  You can keep doing this going down. For the H's, you can get 1-3-3-1, for the C's you can get 1-4-6-4-1, for the O's you can get 5-10-10-5, for the U's you can get 15-20-15, for the N's you can get 35-35, for the T you can get 70, and for the S you'll get 70. That's my final answer [/hide]

I'm not seeing how it's 4x4 = 16 ways to get to the C :blush: [/hide]

Youre probably right.
I got the 16s by knowing that you have 2 choices at each intersection.

Eh, sorry for reviving an old thread...but there's a nice identity that goes along with this.  See if you can figure out why this method works to get the answer:

[hide]
\[ \binom{4}{0}^2+\binom{4}{1}^2+\cdots+\binom{4}{4}^2=\binom{8}{4}=70 \]

More generally:

\[ \binom{n}{0}^2+\cdots+\binom{n}{n}^2 = \binom{2n}{n} \][/hide]

What does $n$ represent?

This is not too short!

[hide]It's the same as asking how many ways to get to T. Block walking approach: we must move 8 times, and 4 must go to our right. ${8\choose 4}=\boxed{70}$[/hide]


search
symmetry
percent
geometry
3D geometry
probability
trapezoid

Can anyone guess topics that might come up on the AMC 10? or tell topics that have come up in during the past 4 years?

Do a search on this site and on the AMC site and you'll find the information you need.

Apparently the website is heavily out-of-date, but:

From the FAQ:

An item-by-item analysis of the content in the 25 questions on the 2001 AMC 10 Contest is as follows:

   1. Understanding of median and mean
   2. Understanding of number line, “reciprocal? and “inverse?
   3. Elementary algebra
   4. Intersection of circle and triangle
   5. Symmetry of irregular figures
   6. Base ten representation of a number
   7. Decimal notation and number representation
   8. Reasoning about integers from a word problem, understanding Least Common Multiple
   9. Word problem about percents
  10. Solving three equations in three unknowns, (can be done by educated guessing)
  11. Pattern recognition in figures and numbers
  12. Products, primes, and divisors
  13. Reasoning about integers
  14. Word problem about two variables
  15. Geometry reasoning, similar triangles
  16. Mean and median of set of integers
  17. Geometry of a cone
  18. Area of a square and proportions
  19. Combinations
  20. Definition of an Octagon, and Pythagorean Theorem
  21. Geometry of a cylinder and a cone
  22. Solving a system of equations, using reasoning
  23. Elementary probability
  24. Pythagorean theorem applied to a simple trapezoid
  25. Reasoning about integers


I am suppose to "co-lecture" about modular arithmetic for an upcoming math club at school.  Does anyone know introductory-level modular arithmetic problems or where there is a nice selection of them online?  I have a few but want a bigger variety.  :)

Thanks.

I dug up a post from the (private) So Cal ARML forum, and smuggled them out... lots of them aren't what'd you'd call introductory, but the first few may help...
[hide="long post"]
[quote="JAM"]1)(a) Find the last digit of the sum $ 109489143588 + 9810754958543$
(b) Find the last digit of the product $ 109489143588\times 9810754958543$

2) Find the remainder when $ 6^{2007}$ is divided by $ 7$

3) Find the remainder when $ 1^2 + 2^2 + 3^2 + \cdots + 2007^2$ is divided by $ 5$.

4) Prove that a number is equivalent to the sum of it's digits modulo $ 9$ (and therefore, a number is divisible by $ 9$ if and only if the sum of it's digits is divisible by $ 9$). Find similar rules for $ 3$ and $ 11$. Using these ideas, can you find rules for divisibility by other integers?

5) (AMC 2007) For each positive integer $ n$, let $ S(n)$ denote the sum of the digits of n. For how many values of $ n$ is $ n + S(n) + S(S(n)) = 2007$?

6)  (ARML 2007) Let $ a_1 = 1$ and $ a_n$ be the number formed by putting the digits of $ n$ at the end of the digits of $ a_{n - 1}$.  Thus, $ a_9 = 123456789$ and $ a_{10} = 12345678910$.  For $ k = 1$ to $ 2007$, compute the number of $ a_k$ that are divisible by $ 3$.

7)  (AMC 2003) Let $ n$ be a $ 5$-digit number, and let $ q$ and $ r$ be the quotient and remainder, respectively, when $ n$ is divided by $ 100$. For how many values of $ n$ is $ q + r$ divisible by $ 11$?

8)  (USAMTS 2003-2004) For four integer values of $ n$ greater than six, there exist right triangles whose side lengths are integers equivalent to $ 4$, $ 5$, and $ 6$ modulo $ n$, in some order. Find those values. Prove that at most four such values exist.

9) (a) Prove that $ n^2 \equiv 0 \text{ or } 1 \pmod{4}$ for all integers $ n$.
b) Prove that $ n^2 \equiv 0 \text{ or } 1 \pmod{3}$ for all integers $ n$.
(c)Prove that $ n^2 \equiv 0,1,4,5,6 \text{ or } 9 \pmod{10}$ for all integers $ n$.

10) (a) Find the remainder when $ 2^{123456789}$ is divided by $ 11$.
(b) For each of $ n = 1,2,3,4,5,6$ find the smallest positive integer $ k$ such that $ n^k\equiv 1\pmod{7}$
(c) For each of $ n = 1,3,7,9$ find the smallest positive integer $ k$ such that $ n^k\equiv 1\pmod{10}$

11) (a)  Calculate $ \phi(2),\phi(3),\phi(4),\cdots, \phi(12)$.
(b) Prove that $ \phi(p^e) = (p - 1)p^{e - 1}$ for all primes $ p$. In general, if the prime factorization of $ n$ is $ n = p_1^{e_1} p_2^{e_2}\cdots p_m^{e_m}$ then
\[ \phi(n) = (p_1 - 1)p_1^{e_1 - 1}(p_2 - 1)p_2^{e_2 - 1}\cdots(p_m - 1)p_m^{e_m - 1}.
\]
12) Prove that $ 2^{70} + 3^{70}$ is divisible by $ 13$.

13) Find the last two digits of $ 2007^{2006^{2005^{2004}}}$.

14)   (IMO '75 #4) Let $ A$ be the sum of the decimal digits of $ 4444^{4444}$, and B the sum of the decimal digits of $ A$. Find the sum of the decimal digits of $ B$.}

15)   Let $ n$ be an integer, and let $ a$ be an integer \emph{relatively prime to $ n$}. Prove that there is some integer $ b$ such that $ ab \equiv 1 \pmod{n}$ (Hint: Pigeonhole principle). We say the $ b$ is the inverse of $ a$ modulo $ b$, and write $ b = a^{ - 1}$. Sometimes we write $ \displaystyle b\equiv \frac {1}{a} \pmod{n}$.

16) Find all integers $ x$ such that $ 3x + 4 \equiv 0 \pmod{8}$.

17) (a) Let $ n$ be a positive integer. Let $ x_1,x_2,\ldots,x_{\phi(n)}$ be the positive integers less then $ n$ which are relatively prime to $ n$. Let $ a$ be any integer relatively prime to $ n$.  Prove that $ \{ax_1,ax_2,\ldots,ax_{\phi(n)}\} \equiv \{x_1,x_2,\ldots,x_{\phi(n)}\} \pmod{n}$, that is, prove that each of the remainders $ x_1,x_2,\ldots,x_{\phi(n)}$ occurs exactly once among the numbers $ ax_1,ax_2,\ldots,ax_{\phi(n)}$.
(b) Prove Fermat's and Euler's theorems.

18) [i]Wilson's Theorem:[/i] Prove that $ (p - 1)! \equiv - 1\pmod{p}$ for all primes $ p$.

19)Let $ N$ be the positive integer $ \underline{A_{n}}\underline\,\underline{A_{n - 1}}\cdots\underline{A_{2}}\,\underline{A_{1}}\,\underline{A_{0}}$ where the $ A_{i}$ are digits. Thus $ N = A_{N}\cdot10^{n} + A_{n - 1}\cdot10^{n - 1} + \cdots + A_{2}\cdot10^{2} + A_{1}\cdot10 + A_{0},$ where $ 0\le A_{i}\le9$ and $ A_{n}\ne0.$ For any integer $ r$ we define the function
\[ F(N,r) = A_{0}r^{n} + A_{1}r^{n - 1} + \cdots + A_{n - 1}r + A_{n}
\]
Prove: If $ N$ is an integer and $ p$ is any prime number greater than $ 5,$ there exists at least one positive integer $ k$ such that if $ p\mid N,$ then $ p\mid F(N,k).$

20) [i]Chinese Remainder Theorem:[/i] Let $ m_1,m_2,\ldots,m_k$ be $ k$ [i]pairwise relatively prime[/i] (that is, any two of them are relatively prime) positive integers. Let $ a_1,a_2,\ldots,a_k$ be arbitrary integers. Prove that there is \emph{exactly} one $ N$ between $ 1$ and $ m_1m_2\cdots m_k$ satisfying the equations

$ N \& \equiv a_1 \pmod{m_1}$
$ N \& \equiv a_2 \pmod{m_2}$
$ \cdots$
$ N \& \equiv a_k \pmod{m_k}$

(Hint: First consider the case where $ a_i = 1$ and all the other $ a_j$'s are $ 0$.)

21) If $ N \equiv 1 \pmod{3}$, $ N\equiv 2 \pmod{5}$ and $ N \equiv 3 \pmod{7}$ find the remainder when $ N$ is divided by $ 105$.

22) (USAMTS 2006-2007) For how many integers $ n$ between $ 1$ and $ 10$, inclusive,  are the last $ 2007$ digits of $ n$ and $ n^3$ the same? (If $ n$or $ n^3$ has fewer than $ 2007$ digits,  treat it as if it had zeros on the left to compare the last $ 2007$ digits.)

23) (AIME 1998) Find the smallest positive integer whose cube ends in $ 888$.

24) Find the remainder when $ 1 + 2007 + 2007^2 + \cdots + 2007^{2007}$ is divided by $ 1000$.

25) Prove that if the prime factorization of $ n$ is $ n = p_1^{e_1} p_2^{e_2}\cdots p_m^{e_m}$ then $ \phi(n) = (p_1 - 1)p_1^{e_1 - 1}(p_2 - 2)p_2^{e_2 - 1}\cdots(p_m - 1)p_m^{e_m - 1}.$ (Hint: Prove that if $ a$ and $ b$ are relatively prime then $ \phi(ab) = \phi(a)\phi(b)$.)

26) For polynomials $ p(x)$ and $ q(x)$, we say that $ p(x)|q(x)$ if there is a polynomial $ k(x)$ such that $ q(x) = p(x)k(x)$. We say that $ q(x) \equiv r(x) \pmod{p(x)}$ if $ p(x)|q(x) - r(x)$. If $ a(x) \equiv c(x) \pmod{p(x)}$ and $ b(x) \equiv d(x) \pmod{p(x)}$ then prove that:

$ a(x) + b(x) \& \equiv c(x) + d(x) \pmod{p(x)}$
$ a(x)b(x) \& \equiv c(x)d(x) \pmod{p(x)}$

This theorem can be summed up by saying that modular arithmetic "works" for polynomials, just like it does for integers. This is a useful (and not widely known) technique for solving polynomial problems. (Side note: can you find polynomial analogues for the existance of inverses and the Chineese remainder theorem?)}

27) (MA$ \Theta$) Find the remainder when $ x^{203} - 1$ is divided by $ x^4 - 1$.

28)  (MDATB 05) Find the remainder when $ x^{135} + x^{125} - x^{115} + x^5 + 1$ is divided by $ x^3 - x$.}

29) Show that $ x^{9} + x^{8} + \cdots + + x + 1|x^{9999} + x^{8888} + \cdots + x^{1111} + 1$.

30) For any integer $ n$ and any integer $ a$ relatively prime to $ n$, let $ \text{ord}_n(a)$ denote the smallest positive integer $ k$ such that $ a^k\equiv 1 \pmod{n}$. Prove that $ \text{ord}_n(a)|\phi(n)$.

31) Let $ a$ be an integer greater than $ 1$. Prove that for every integer $ n$ there is a prime $ p_n$ such that $ \text{ord}_{p_n}(a) = 2^n$. Conclude that for all integers $ n$ there are infinitely many primes of the form $ 2^nk + 1$.

32) Prove that $ \text{ord}_{5^k}(2) = 4\cdot 5^{k - 1}$ for all positive integers $ k$.

[b]Challenge Problems:[/b]

33) (USAMO 2003 #1) Prove that for every positive integer $ n$ there exists an $ n$-digit number divisible by $ 5^n$ all of whose digits are odd.

34) (IMO 2005 #4) Determine all positive integers relatively prime to all the terms of the infinite sequence
\[ a_n = 2^n + 3^n + 6^n - 1,\ n\geq 1.
\]
35) (USAMO 2006 #1) Let $ p$ be a prime number and let $ s$ be an integer with $ 0 < s < p$. Prove that there exist integers $ m$ and $ n$ with $ 0 < m < n < p$ and
\[ \left \{\frac {sm}{p} \right\} < \left \{\frac {sn}{p} \right \} < \frac {s}{p}
\]
if and only if $ s$ is not a divisor of $ p - 1$.

36) (German TST 2006, Day 3, #3) Let $ a,b$ be positive integers such that for any positive integer $ n$ we have $ a^n + n\mid b^n + n$. Prove that $ a = b$.

37) (USA TST 2003 #3) Find all ordered triples of primes $ (p, q, r)$ such that
\[ p \mid q^r + 1, \quad q \mid r^p + 1, \quad r \mid p^q + 1.
\]
[/quote][/hide]


Let $ p(x) \equal{} x^2 \plus{} x \plus{} 1$. Show that for every positive integer $ n$, the numbers $ n, p(n), p(p(n)),...$ are relatively prime.


Why is number 2 the only even number that is also a prime number?

Definition of prime: A number > 1 whose factors are only one and itself.

2 fits this definition, and is therefore a prime.

All other even numbers are divisible by 2 and are not prime.


trigonometry
inequalities unsolved
inequalities

:omighty:  [u][b]HELP ME PLEASE![/b][/u] :omighty:

[quote="James Potter"]$\forall a_1,a_2,a_3,...,a_n \in R$[/quote]
$R$???

Does everybody understand this problem?You can ask!!

[quote="tmbtw"]This is a easy sum.
The last Equality what zhaobin posted is old![/quote]
Yes,this is old.

[quote="James Potter"][quote]The last Equality what zhaobin posted is old![/quote]
Can you give me the link!
[quote]for $f(a_1,a_2,a_3,...,a_n)$
two steps.
first: try to prove  get its maximum when
 $a_1=\frac{1}{a_1}+a_2=\frac{1}{a_2}+a_3=\cdots =\frac{1}{a_{n-1}}+a_n=\frac{1}{a_n}$

second:
slove that equality:$a_1=\frac{1}{a_1}+a_2=\frac{1}{a_2}+a_3=\cdots =\frac{1}{a_{n-1}}+a_n=\frac{1}{a_n}$[/quote]
What about $g(a_1,a_2,a_3,...,a_n)$[/quote]
see [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x_1%3Dx_2%2B%5C%5Cfrac%7B1%7D%7Bx_1%7D&t=34278]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x_1%3Dx_2%2B%5C%5Cfrac%7B1%7D%7Bx_1%7D&t=34278[/url]
I think $g(a_1,a_2,a_3,...,a_n)$ is  the same....


rotation LaTeX induction function trigonometry vector integration

So the previous problem I posted caused a rather short (or not? I [i] am [/i] new here) discourse but nevertheless helped me solve it  :D This is why I've decided to post a series of problems under this topic. The problems I'll propose will be from nearly all fields of high school physics. Although they won't necessarily be hard, the solution will probably not be obvious ;) I hope more people will join in as time passes and that the moderator(s) will not complain about my posts for whatever reason there can be...

Enough of the small talk, let's do some physics  :P

[color=green] [size=150] New problem [/size][/color]  [color=indigo][size=100][b] (SOLVED) [/b][/size][/color]

Two disks are set on a long, thin stick (whose torsion constant is $C$) so that the stick passes through their centers of mass and is perpendicular to their planes. The moments of innertia of these disks are $I_1$ and $I_2$. Determine the period of small torsion oscillations of this system. The axis around which the system is twisted is, clearly, the axis of the stick.

 [i] Brief reminder: A stick (string, rope or whatever can be elastically deformed --twisted around its axis) with a torsion constant $C$ will exert an 'elastic' torque of ${\mathcal T} = - C\phi$ if rotated (twisted) by angle $\phi$ from its equilibrium state. This is clearly the analogy of the elastic force given by $F = - k x$ where $x$ is the elongation and $k$ the elasticity.[/i]

I hope that the problem is clear enough... Happy problem solving everybody!

this problem is the rotational analog of a problem posted earlier on the forum. Arguing by analogy, the answer is

$T=2\pi \sqrt{\frac{1}{C(\frac{1}{I_1} + \frac{1}{I_2})}}$

[color=blue]Do you mean your answer is
$\large T=2\pi\sqrt{\frac{1}{C(\frac{1}{I_1}+\frac{1}{I_2})}}$[/color]

[quote="psychicalchemist"]yeah, i meant that...

$\text{wow, this is the first time ive ever written in LateX. This is so cool.}$[/quote]

[color=blue]$\large T=2\pi\sqrt{\frac{1}{C(\frac{1}{I_1}+\frac{1}{I_2})}}$
How can you get that answer?
My solution is
The energy of the system
$\large E=\frac{1}{2}I_1\omega^2+\frac{1}{2}I_2\omega^2+\frac{1}{2}C\phi^2=\text{const}$

$\large\Longrightarrow \frac{dE}{dt}=I_1\ddot{\phi}\dot{\phi}+I_2\ddot{\phi}\dot{\phi}+C\dot{\phi}\phi=0$

$\large\Longrightarrow \ddot{\phi}+\frac{C}{I_1+I_2}\phi=0$

$\large\Longrightarrow T=2\pi\sqrt{\frac{I_1+I_2}{C}}$

Where am I wrong?[/color]

You got the answer right, it's $T=2\pi \sqrt{\frac{I_1I_2}{C(I_1+I_2)}}$. The problem, of course, is to determine why is this so :) I am familiar with the 'translational' analogue of the problem, but I have no idea how to obtain this answer.

When I first tried to solve this problem I reasoned like this: the moment of innertia is additive, so the total moment of inertia would be $I = I_1 + I_2$. Twisting the stick by $\phi$ would result in a torsion torque of $\mathcal{T} = - C\phi = I \ddot{\phi}$ and now it was supposed to be straightforward. I was wrong. Anyone know the reason for this?

@ Psychicalchemist:

Don't over$\LaTeX$ your posts and keep an eye on those backslashes :P

Is the axis, or the stick, fixed or can it rotate??

i basically did the same thing arnav did w/ center of mass, making the displacement angle theta for $I_2$ equal to -$\frac{I_1}{I_2}\theta$.

Then I used $\gamma = -C\theta$

$I_1 \alpha = -C(\theta_1 - \theta_2)$

i subtracted $\theta_2$ since $\theta_2$ is always negative and thus would add to the restoring torque as it grew in magnitude.

thus, you get $\alpha = -C(\frac{\theta}{I_1} + \frac{\theta}{I_2})$
and the rest is obvious...

woo, check out my $\LaTeX$

for phunc87 and Djole: the object isnt rigid, so you cant add the rotational inertias to get I for the entire object or assume that the angular velocity of both of them are equal.

Look, the [i]axis[/i] itself [i]is[/i] fixed. But anytime you 'wind' the disks on the stick (like you do to that little wheel on the side of an analogue wrist watch, for examle), it will 'unwind' itself. Like [b]psychicalchemist[/b] said, it's the same situation with a spring: if you pull it, it will exert an elastic force and try to return to its innitial length. In our problem the stick will exert an elastic torque and try to return to its innitial angle, but pulling the disks with it.

Now, when you've placed the disks on the stick, then they can't slide -- when you've rotated one disk by an angle, the other disk will be rotated by the same angle.

I hope I cleared things up a bit. Please tell me if you don't understand it yet...

BTW, @ [b]psychicalchemist[/b]: can you post a link to the topic where the problem with springs you mentioned is discussed? There are several different problems which have the solution like the one you mentioned, and I don't know exactly which one you're referring to...

And @ [b]phucnv87[/b]: I like your approach to this problem ;) But did you find out where is your (our) mistake?

Ummm, sorry I'm so thick-headed... But I don't get it!  :blush: Well, the stuff after your first statement is obvious, but [i]the[/i] first statement leads me to the conclusion that I've completely misunderstood the problem! Could you clarify it a bit more, pls?

Oh yes, I'll put aside my own stupidity and take it for granted that the rest of you who are viewing this discussion have unterstood the solution. Thus, I'll soon post the next sweet problem :P Soon, as in within minutes!

@ [b]psychicalchemist[/b]:
Nice $\LaTeX$... Keep up the good work!

Well, here it is: the next problem. Electromagnetism this time, it's a classical Russian problem which I try and try to solve and never get right:

[color=green] [size=150] New problem [/size][/color] [color=indigo][size=100][b] (SOLVED) [/b][/size][/color]

A flat spiral consisting of $N$ tightly wrapped coils is placed in a homogenous magnetic field perpendicular to the plane of the spiral. The outer radius of the spiral is $R$. The magn. induction of the field is a function of time given by $B=B_0 \sin \omega t$, where $\omega$ and $B_0$ are constants. Find the amplitude of the induced EMF in the spiral, assuming that $N$ is very large.

I've thought about posting a "hidden" answer, but that wouldn't be fair -- who am I to bring you into such a moral dilemma ("should I reveal it or not?"). So I'll let you work for a while... But I consider this one a fairly easy problem... But still I can't get it right  :mad:

[quote]Ummm, sorry I'm so thick-headed... But I don't get it!  Well, the stuff after your first statement is obvious, but the first statement leads me to the conclusion that I've completely misunderstood the problem! Could you clarify it a bit more, pls? [/quote]

arnav's solution is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70112&start=20]here[/url]

I think you made the error when you tried to consider the rotational inertia of the object as $I_1 + I_2$. I don't think many people realize this, but the $\gamma = I\alpha$ equation only works on rigid objects. The two disks by themselves are rigid, but once you put them together, the resulting system is not. This is why you have to consider the disks separately, but realize that since there are no external torques acting on the objects that the center of rotational inertia stays the same. I hope this helps.

Heh, thanks for the compliment on my $\LaTeX$. I still cant get over how cool it is.[/quote]

[color=blue]I still can't imagine what a [b] flat spiral[/b] is.
Can you draw a picture?[/color]

@ [b]psychicalchemist[/b]: Ok, I think I get your idea :| I'll probably comprehend it eventually... Don't mind me, carry on with the problems ;) Speaking of which,

@ [b]phucnv87[/b] It means a regural spiral ("flat" signifies that it's entirely in one plane). You can consider the spiral to be a large number of rings with increasing diameters coined together (or, more physically said, a large number of serially connected conducting circles whose radii increase from 0 to $R$).

BTW, this is such an easy task (though I never had the patience to complete it), that I will now disclose the final solution. I hope that a full solution will be posted by someone during the next couple of hours.

[hide] {\mathcal E}_0 = {a^2\omega\pi NB_0 \over 3} [/hide]

[color=blue]Oh, now I see.
Thanks [b]Djole[/b] so much.
Here's the solution to your problem.
 $\large |e_i(t)|=|-\frac{d\phi_i}{dt}|=\pi r_i^2B_0\omega\cos\omega t$
The total induced EMF in the spiral
$\large E(t)=\sum e_i(t)=B_0\omega\cos\omega t \sum_{i=1}^{N}\pi r_i^2$

But $\large\sum_{i=1}^{N}\pi r_i^2=\frac{1}{3}\pi R^2N$

$\large \Longrightarrow E(t)=\frac{1}{3}\pi R^2NB_0\omega\cos\omega t$

$\large \Longrightarrow \text{The amplitude of the EMF is}E=\frac{1}{3}\pi R^2NB_0\omega$
[/color]

Great job, [b]phucnv87[/b]! Now, people usually like problems from mechanics... Enjoy!

[color=green][size=150] New problem [/size][/color] [color=red] [size=100][b] (NOT SOLVED) [/b][/size][/color]

Consider an equilatteral triangle $\triangle ABC$. At $t=0$, a small ball of mass $m$ is shot from point $A$ towards point $C$ of the triangle with the innitial speed of $v_0$ -- call this ball $A$. At the same time, another small ball of mass $2m$ and the same radius as the first one is shot from $B$ towards $C$ with the innitial speed of $v_0$; call this ball $B$. The collision of the balls is elastical. Find the numerical value of the angle between their speeds $\vec v_A$ and $\vec v_B$ after the collision.

sigh, i told myself i wasnt going to do physics today... but i went on AoPS anyway

ok, so if momentum and energy are conserved, then the collisions in the horizontal and vertical directions are elastic as well.

[b]Vertical[/b]

vertical component of Mass A velocity: $-v_o\sin 60^\circ$

$-mv_o\sin 60^\circ=mv_a + 2mv_b$
$-v_o\sin 60^\circ = -v_a + v_b$

solving, we get $v_a = \frac{\sqrt{3}}{6}v_o$ and $v_b=-\frac{\sqrt{3}}{3}v_o$

[b]Horizontal[/b]

horizontal component of Mass A velocity: $v_o\cos 60^\circ$ or $\frac{1}{2}v_o$
horizontal component of Mass B velocity: $v_o$

$\frac{5}{2}mv_0 = mv_a + 2mv_b$
$\frac{1}{2}v_o = v_a - v_b$

solving, we get $v_a = \frac{7}{6}v_o$ and $v_b = \frac{2}{3}v_o$

Now, we get the angles both velocities make with the horizontal and add them

Angle of Mass A = $\arctan(\frac{\sqrt{3}}{7})$
Angles of Mass B = $-\arctan(\frac{\sqrt{3}}{2})$

And the rest is history...

Final answer: $\arctan(\frac{9\sqrt{3}}{11})$

This solution seems really long, but i tried doing some kind of eloquent thing with center of mass or something but it didnt work out. I'm just waiting for someone to post a really cool solution which has about 2 steps.

@ [b]psychicalchemist[/b]: your answer evaluates to about $54^o$, which is not the answer I have :( I don't see if there is a mistake in your reasoning...

This is -- I think -- the toughest problem I've posted. Conservation laws are often tricky to get right as sometimes you need to solve systems of equations with more unknown variables than equations. The catch is usually to make some variable to cancel out, but I haven't been lucky with this one. Using the most radical approach, I got two equations (conservation of energy and conservation of momentum -- calculated as a sum of vectors), but I always end up with three variables ($v_A,\;v_B$ and $\varphi$ -- the angle between the speeds). I just can't solve it :mad:

I wonder if the fact that the radius of the balls is same plays a role in the solution  :huh:

I assumed it would be a head-on collision in my equations, although this might not be the case. It would make it much more complicated if it wasnt.

I think that the solution that has been posted is the solution for 2 points' collision, while we have 2 circles. I believe there is a difference.

I'm not happy with the lack of a solution, people :mad:

Nah, just joking. But I've been looking at [b]psychicalchemist[/b]'s solution and I really, really can't make out how he did what he did. I tried to reproduce his approach but ended with more unknown variables than equations, as I had already posted. Could you please explain your solution, step by step? I [i]do[/i] think that the only way to solve this one is to use a special reference frame or something. Any suggestions?

If there are no solutions by tomorrow (within 24h from... NOW) I'll postpone the answering of this problem and post a new one... I hope that will not need to happen ;)

I just split the velocities into components and reasoned that, if the total collision is elastic, then I could pretend that the horizontal and vertical components of the collisions were elastic. This is wrong because the "rate of approach equals rate of separation" equation i used only works in a head on collison (the line through the velocity vectors intersects the center of mass of both objects). In this case, this is clearly not true, indicating that we must try something with vector algebra or something similar.

I have a gut feeling that the angle between the velocities of two objects in a non-headon collision is constant from the fact that it is always $90^\circ$ in an oblique collision between a mass at rest and an equal mass moving at constant velocity. This does not depend on radius, and I believe that conservation of momentum never depends on the shape of the objects involved.

$m\vec v_a + 2m\vec v_b = m\vec v_c + 2m\vec v_d$

$\frac{1}{2}m(|\vec v_a|)^2 + m(|\vec v_b|)^2 = \frac{1}{2}m(|\vec v_c|)^2 + m(|\vec v_d|)^2$

finally,

$\vec v_c \cdot \vec v_d = |\vec v_d ||\vec v_c | \cos \theta$

this just might give us the angle theta between them. If this is the method we must pursue, then I believe it is the hardest mechanics problem ive ever seen.

Well, the time is almost up and there are no solutions... I'll modify the post where I proposed this problem accordingly and I'll post a new one (hopefully, a solvable. I'm really not happy with the fact that we couldn't solve this one :( I even though that there's maybe a mistake in the solution I have ($49^\circ$), but it's awful that we still don't know how to solve the problem...

BTW, I believe that this problem also comes from Russia, so I address this to anybody out there familiar with it: your help would be appreciated :)

[size=150][color=green] New problem[/color][/size] [size=117][color=indigo][b] SOLVED [/b][/color][/size]

Consider a liquid flowing through a cylindrical pipe of length $L$ and radius $R$. The flow is stationary and laminary (there is no turbulence and the speed of the liquid is constant along a line parallel to the axis of the pipe). Prove that the speed of the liquid changes with its distance to the axis of the pipe $r$ as $v(r) = v_0 (1 - \frac{r^2}{R^2})$.

I was wondering, djole, where do you get these problems? I mean, at what specific level of competitions do these problems come from?

Well, they are mostly downloaded from the internet (like the one with the collisions). The one with the spiral comes from a Russian problem book I have, and the thermodynamics problem is from some preparation material for the national physics olympiad in my country -- these olympiads are composed of three to five problems which seem to be harder than those that appear in the open response problems in the final and semi-final round of your USAPhO :)

These problems could as well appear on an IPhO but only as a part of one problem -- you know, when you have three or four independent sub-problems...

BTW, I could post the last year's national olympiad here so that you can enjoy yourselves... ;) I'll think about it...

[quote="Djole"] I could post the last year's national olympiad here so that you can enjoy yourselves... ;) [/quote]

Please do so!!! I would love to see what type of problems you've got.

[quote="Djole"][size=150][color=green] New problem[/color][/size]

Consider a liquid flowing through a cylindrical pipe of length $L$ and radius $R$. The flow is stationary and laminary (there is no turbulence and the speed of the liquid is constant along a line parallel to the axis of the pipe). Proove that the speed of the liquid changes with its distance to the axis of the pipe $r$ as $v(r) = v_0 (1 - \frac{r^2}{R^2})$.[/quote]

[size=150][color=green] Solution [/color][/size]

Mmm... let's start. I don't know how to include images, yet. Thus
you'll either have to figure it out yourself, or somebody will
post a drawing.

I believe this is the basic relation for a laminary flow:

$\pi{r}^2dp=2\pi{r}{dl}\tau\frac{dv}{dr}$

Where $r$ is the radius of the cylinder that I'm analyzing; $dl$
is its length; $dp$ - pressure difference at its either sides;
$\tau$ - some coefficient :)  and $\frac{dv}{dr}$ is obviously the
speed gradient.

Transforming we get:

$\int(\frac{dp}{dl})\frac{1}{\tau}rdr=\int{dv}$

Integrating and using the conditions that $v(0)=v_0$ and $v(R)=0$
we get rid of all the unknowns to get
$v(r)=v_{0}(1-\frac{r^2}{R^2})$

That's a nice solution... The problems will be posted shortly...

That's a nice problem you proposed :) Let's see if there's somebody with an answer... (I'm familiar with it, so I won't compete immediately)...

@ [b]AndersoN[/b]: I think that it says where I'm from in my profile ;)

Now, I'll post the last years' national olympiad in my country. I'm not too thrilled with the problems -- they used to be so much better...

[color=green][size=150]New problems[/size][/color]

[color=green][size=150]1.[/size][/color] [color=indigo][size=117][b](SOLVED)[/b][/size][/color] The Chancellor is planning a secret meeting of the University Board in order to sanction the Facculty of Physics which has broken through its budget once too often. The Chancellor's Assistant is booking luxury suites in an intergalactic hotel on an attractive planet 10 light years away from Earth.

a) The Chancellor stated that if he takes a sufficiently fast spaceship, the journey can last only one day according to his wrist watch.
b) The Secretary in charge of appointment then claimed that no matter what the Chancellor's watch shows, the time the Chancellor spends in the spaceship (measured from the spaceship) will be much longer than one day.
c) The Assistant (who has to attend a conference on Earth prior to his departure) said that he can start his voyage 12 hours after the Chancellor's space ship takes off. Furthermore, he asserted that if he uses the ultra-fast Gamma Express, he will reach the hotel at the same time as the Chancellor.
d) The clerk at the Social and Pensions Fund warned that no matter how long the Chancellor and the Assistant find their journey, at their arrival to the hotel they will be 10 years older than when they took off from Earth.
e) The accused Dean of the Facculty of Physics (who somehow found out about this meeting of the University Board) said that the report from the Board meeting can't get to the Earth within less than 20 years (and, thank god, by then he'll be retired).

Determine whether the statements a) - e) are true. (20 points)

[color=green][size=150]2.[/size][/color] [color=indigo][size=117][b](SOLVED)[/b][/size][/color] Consider a thin convex lens of focal length $f$. Its central part of width $a$ is removed, and the two remaining parts are pressed tightly together (the center of the removed part is on the optical axis passing through the center of the lens). A small light source is positioned on the optical axis, in front of the obtained bi-lens. A cover is positioned behind the lens and interference lines have appeared. The distance between two adjacent interference lines on the cover is $\Delta x$ and doesn't depend on the distance between the bi-lens and the cover. The wavelength of the light is $\lambda$. Find $a$ in terms of $f$, $\Delta x$ and $\lambda$. (25 points)

[color=green][size=150]3.[/size][/color]  [color=indigo][size=117][b](SOLVED)[/b][/size][/color]  A battery of EMF $\mathcal{E}$ is serially connected to a capacitor of capacitance $C$, a solenoid of inductance $L$, an ideal diode and an open switch. At $t = 0$ the charge on the capacitor is zero. After closing the switch the capacitor starts charging. Find the electric current in the circuit as a function of time and the final voltage on the capacitor. Neglect all resistances in the circuit. (25 points)

[color=green][size=150]4.[/size][/color] [color=red][size=117][b](NOT SOLVED)[/b][/size][/color] A fisherman (whose mass is $m$) is standing on one end of a boat of mass $M$ and length $L$. The boat is at rest on a lake. The fisherman then walks across the boat and stops at its other end. Find the total displacement of the boat in the following two cases:

a) the viscosity of the water is negligible (there is no friction);
b) the force of friction acting on the boat is proportional to its speed.

How much time will pass until the boat stops in both cases? The fisherman requires a time of $\tau$ to walk across the boat. (30 points)

To what grade do these problems belong? 12th?

I'll try to post the problems from my country's national.

You're expected to know some stuff from the 12th grade, yes. But there was never an accent on relativity, nuclear or quantum physics; the hardest problems are always from electromagnetism (well, they used to be).

[quote="Djole"]
[color=green][size=150]1.[/size][/color] The Chancellor is planning a secret meeting of the University Board in order to sanction the Facculty of Physics which has broken through its budget once too often. The Chancellor's Assistant is booking luxury suites in an intergalactic hotel on an attractive planet 10 light years away from Earth.

a) The Chancellor stated that if he takes a sufficiently fast spaceship, the journey can last only one day according to his wrist watch.
b) The Secretary in charge of appointment then claimed that no matter what the Chancellor's watch shows, the time the Chancellor spends in the spaceship (measured from the spaceship) will be much longer than one day.
c) The Assistant (who has to attend a conference on Earth prior to his departure) said that he can start his voyage 12 hours after the Chancellor's space ship takes off. Furthermore, he asserted that if he uses the ultra-fast Gamma Express, he will reach the hotel at the same time as the Chancellor.
d) The clerk at the Social and Pensions Fund warned that no matter how long the Chancellor and the Assistant find their journey, at their arrival to the hotel they will be 10 years older than when they took off from Earth.
e) The accused Dean of the Facculty of Physics (who somehow found out about this meeting of the University Board) said that the report from the Board meeting can't get to the Earth within less than 20 years (and, thank god, by then he'll be retired).

Determine whether the statements a) - e) are true. (20 points)
[/quote]

Supposing that the human body can resist ANY accelerations, then:
a) is true
b) is false
c) is true
d) is false
e) it's not clear enough who is where, but i think it's true.

So, Djole, am I right?

[quote="AndersoN"]
Supposing that the human body can resist ANY accelerations, then:
a) is true
b) is false
c) is true
d) is false
e) it's not clear enough who is where, but i think it's true.
[/quote]

About e) - it's supposed that the Dean is staying on Earth while the Chancellor et al. are moving with near-light speed. However, since the distance to the hotel is 10 light years, it's obvious that -- from the Dean's point of view -- they will nead more than 10 years for a one-way journey. So the Dean is right (he is a physicist :))

All of your responses are ok. You can post the solutions if you like :) Now, that was the easy one. What about the other three?

[quote="Djole"]
[color=green][size=150]3.[/size][/color] A battery of EMF $\mathcal{E}$ is serially connected to a capacitor of capacitance $C$, a solenoid of inductance $L$, an ideal diode and an open switch. At $t = 0$ the charge on the capacitor is zero. After closing the switch the capacitor starts charging. Find the electric current in the circuit as a function of time and the final voltage on the capacitor. Neglect all resistances in the circuit. (25 points)
[/quote]

Well, again no drawing :(
Btw, I'll use $\varepsilon$ for $\mathcal{E}$.

$\varepsilon-L\frac{dI}{dt}-\frac{q}{C}=0$

Using $I=\frac{dq}{dt}$ we get a differential equation in respect
to $q$ with the following solution:

$\frac{q(t)}{LC}-\frac{\varepsilon}{L}=A\sin(\omega{t})+B\cos(\omega{t})$

where $\omega=\frac{1}{\sqrt{LC}}$

Including the initial conditions of $q(t)=0$ and $I(t)=0$ we get:

$q(t)=\varepsilon{C}(1-\cos({\omega{t}})$ and
$I(t)=\varepsilon{C}\omega\sin({\omega{t}})$

Drawing the graph, analyzing it and taking into consideration the
diode, we can understand that for $t>\frac{\pi}{\omega}$ the
charge $q(t)=2\varepsilon{C}$ and the current $I(t)=0$ because it
can not flow in the opposite direction.

So the final answers:

$V_f=2\varepsilon$ and

$I(t)=\varepsilon{C}\omega\sin({\omega{t}})$ for $t\leq\frac{\pi}{\omega}$

$I(t)=0$ for $t>\frac{\pi}{\omega}$

Right again!

I see you're solving the problems according to their difficulty;) The next is, of course, task number 4 (unless you're one of those people who [i]worship[/i] optics (unlike myself) :)

Well, I am most displeased... Several days have passed and nobody seems to care for these problems... Let me post the solution of one of the two problems I've posted (leaving the other for enthusiasts who want to type $\text{\TeX}$ the entire day  :P). Hopefully we will now get into some more problem-solving. I'll also post a new problem soon enough.

2) Let $p$ denote the distance between the light source and the bi-lens; we will presume that $p<f$ (taking $p>f$ would lead to a contradiction). Each part of the bi-lens will form an image of the source according to the equation for convex lenses \[ {1\over f} = {1\over p} + {1\over l},  \]  where $l$ is the distance between the image(s) and the bi-lens. We then have $l = pf/(p-f) < 0$ which means that the images will be virtual. These images will be positioned symmetrically in respect to the optical axis, and the distance between them will be $d = ap/(f-p)$.

Now, let $r$ denote the distance between the bi-lens and the cover. The two virtual images are coherent light sources and interference will occur. We can now use the standard formula for the distance between two adjacent interference stripes: \[ \Delta x = \lambda\frac{r+l}{d}=\lambda\left(\frac{r(f-p)}{ap}+\frac{f}{a}\right).  \] Because $\Delta x$ does not depend on $r$ then the condition $p\rightarrow f$ must be satisfied. Using this conclusion, the above formula leads to \[ \boxed{a = \frac{\lambda f}{\Delta x}}  \].

[color=green][size=150][b]New problem[/b][/size][/color]

A thin stick of uniformly distributed mass $M$ and length $2\ell$ is lying on a frictionless table. A small body of mass $m$ approaches the stick with speed $v_0$. The trajectory of this body is perpendicular to the stick and passes through one of its ends. As a result of their collision, the stick begins to move both translationally and rotationally (at the same time). After rotating by $180^\circ$, the stick hits the small body with its other end (the end which wasn't hit by the body). Determine the ratio $\mu = M/m$ which must be satisfied in order for the described collision to be possible. Also determine the speeds of the stick and the small body immediately after the second collision. All collisions are to be considered absolutely elastical.

[quote="AndersoN"]With Djoles permission I will post a new problem. It's different from those we've seen yet, but it's still worth a try.

[size=150][color=green] New problem[/color][/size]

Estimate the difference of distances, from the center of the Earth to the sea level on the pole $R_p$ and at the equator $R_e$, that appears due to Earth's rotation.

Click here to REVEAL the HIDDEN answer:
[hide] $R_e-R_p\approx{10km}$ [/hide][/quote]

I just know im approaching this wrong, but I was hoping someone could tell me where im making my mistake.

let us consider an arbitrary mass $m$ on the equator.

since we are merely approximating the change in the radius, we can write the sum of the forces on this mass $m$ as

$mg - F_n = \frac{mv^2}{r}$

if $mg < \text{the centripetal force}$, then the radius would have to expand until F_n equals $0$. The period would stay constant, however, since $m$ is still rotating with the Earth's rotation. Thus, we can designate a $v = \frac{2\pi r}{T}$. This is where i start running into problems. The fact that the $r$ is in the numerator indicates that the centripetal force increases with the radius of the orbit. If this were the case, then $mg$ would never equal it unless the radius of the planet contracted at the equator. Since it is known that the radius of the planet Earth at the equator is larger than its radius at the poles, my method is obviously wrong. I was wondering if someone could explain the error in my reasoning.

I was sitting around doing comp sci today and started thinking about physics (cuz binary trees are just about the most boring thing I have ever encountered). I thought about the fact that a pendulum will actually rotate a clockwise path in the Earth's rotational field even though it seems that it just goes back and forth. soon, I had a physics problem on my hands and... here it is. I don't think its quite as challenging as the one's which are in this thread, but I want to see if my reasoning is correct.

A rod of length $L$ is suspended from a stationary hinge off the ground. Given that the planet's radius is R $R$ m and the period of its rotation is $T$ seconds, find the amplitude of the swings of the rod $\theta$ given that $\theta$ is small and the acceleration due to gravity on the planet is $g$. Note: the rod is at rest initially with the Universe but not with respect to the planet's rotation

[hide]

I got the answer $\theta = \frac{R \sqrt{6g}}{T L^\frac{3}{2}}$[/hide]

@[b]psychicalchemist[/b]: Ummm, sorry, could you present your problem a bit more clearly? Did you just add centrifugal force into the equations of the pendulum's motion? I just don't get it...

PS: Don't underestimate the importance of binary trees ;)


Real numbers $ a_i,b_i (1\leq i\leq n)$ are such that $ \sum_{i\equal{}1}^na_i^2\equal{}\sum_{i\equal{}1}^nb_i^2\equal{}1$. and $ \sum_{i\equal{}1}^na_ib_i\equal{}0$. Prove that

$ (\sum_{i\equal{}1}^na_i)^2\plus{}(\sum_{i\equal{}1}^nb_i)^2\leq n$

[quote="massnet"]Real numbers $ a_i,b_i (1\leq i\leq n)$ are such that $ \sum_{i \equal{} 1}^na_i^2 \equal{} \sum_{i \equal{} 1}^nb_i^2 \equal{} 1$. and $ \sum_{i \equal{} 1}^na_ib_i \equal{} 0$. Prove that

$ (\sum_{i \equal{} 1}^na_i)^2 \plus{} (\sum_{i \equal{} 1}^nb_i)^2\leq n$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152850


Ring Theory
number theory
greatest common divisor
algorithm
superior algebra
superior algebra unsolved

Consider the ring $ \mathbb{Z}[i\sqrt {5}]: \equal{} \{a \plus{} bi\sqrt {5}|a,b\in\mathbb{Z}\}$. Determine whether there exists a Euclidean division on $ \mathbb{Z}[i\sqrt {5}]$ (i.e. $ \forall s,t\in\mathbb{Z}[i\sqrt {5}],\exists q,r\in\mathbb{Z}[i\sqrt {5}]$ such that $ s \equal{} qt \plus{} r$ and $ 0\leq N(r) < N(t)$ where $ N(r) \equal{} N(a \plus{} bi\sqrt {5}) \equal{} (a \plus{} bi\sqrt {5})(a \minus{} bi\sqrt {5}) \equal{} a^2 \plus{} 5b^2$).

For an euclidean division, you normally don't fix the corresponding norm, so you have to show it for all norms, not just the standard norm. To do this, note that the ring is not factorial.

And this is not really an Open problem.

It is known that $ \mathbb{Z}[i]$ and $ \mathbb{Z}[i\sqrt{2}]$ are Euclidean ring. But the construction of the Euclidean division for them can not be applied to construct one for $ \mathbb{Z}[i\sqrt{5}]$. So could you please give a proof or disproof for this, Mr.[b]ZetaX[/b]?

He already did: Euclidean $ \implies$ UFD, but $ \mathbb{Z}[\sqrt{\minus{}5}]$ is not a UFD. The first inference may be proved in the same way as it is for $ \mathbb{Z}$, using the Euclidean GCD algorithm and consequent Bezout identity. Non-UFD may be proved by noticing that $ 2$ is irreducible but not prime, e.g. see [1].

[1] http://google.com/group/sci.math/msg/db4c79a018d9f733


algebra
polynomial
functional equation
algebra unsolved

Find all polynomials $ p(x)$ such that:

$ (x\minus{}1)p(x\plus{}1)\minus{}(x\plus{}2)p(x)$ $ \equiv$ $ 0$

Clearly, $ 1$ is a zero of $ p$, which implies $ 0,\pm 1$ are zeroes of $ p$. Therefore we make the substitution $ p(x)\equal{}x(x\plus{}1)(x\minus{}1)q(x)$ into the functional equation and can verify that $ q(x)$ can be arbitrarily chosen, i.e. the solutions are all polynomials of the form $ p(x)\equal{}x(x\plus{}1)(x\minus{}1)q(x)$.

[quote="hjbrasch"] into the functional equation and can verify that $ q(x)$ can be arbitrarily chosen, i[/quote]
Are you sure about that?
When substituting in the equation i get:
$ (x \minus{} 1)x(x \plus{} 1)(x \plus{} 2) ( q(x \plus{} 1) \minus{} q(x) ) \equal{} 0$. Hence $ q(x)$ must be the constant polynomial. So the only solutions are $ p(x) \equal{} c \cdot (x \minus{} 1)x(x \plus{} 1)$ for all $ c \in \mathbb{C}$.

[b]Mathias_DK[/b], you are absolutely right. That happens, when you solve without paper and pen. Sorry, for the mix-up. :blush:

[quote="hjbrasch"]Sorry, for the mix-up. :blush:[/quote]
It wouldn't be interesting to read solutions if one knew they were without mistakes, would it ? :P


if $n$is a postive integer  $\{\sqrt[k]{n}\}>0$ prove that
$\{\sqrt[k]{n}\}\>\frac{1}{k\sqrt[k]{n^{k-1}}}$
where $\{x\}$is the fractional part of the nmber $x$

[quote="Hawk Tiger"]if $n$is a postive integer  ${\{\sqrt[k]{n}}>0$ prove that
$\{\ \sqrt[k]{n}\ }\>\frac{1}{k\sqrt[k]{n^{k-1}}$
where ${x}$is the fractional part of the nmber $x$[/quote]

I think it is
$\left\{\sqrt[k]{n}\right\}>0$ prove that
$\left\{\sqrt[k]{n}\right\}> \frac{1}{k\sqrt[k]{n^{k-1}}}$
where $\left\{x\right\}$is the fractional part of the nmber $x$


Let $M,N,P$ be the midpoints of the sides $BC,CA,AB$ respectively, of a triangle $ABC$. Let $G$ be the centroid of the triangle $ABC$. Prove that if $4BN^2 = 3AB^2$, and the quadrilateral $BMGP$ is cyclic, then $ABC$ is an equilateral triangle.

this one is easy:
4*BN^2 = AB^2 means, after some compuatations, 2*a^2 = b^2 + c^2 (#)
since BMPG is cyclic, let PM meet BG meet at X. PX*MX=BM*MG, so
 BN^2 / 12 = b^2 / 16, and thus 4*BN^2 = 3*b^2, so then b=c and from (#) a=b=c we are done.
oh and whats junior balkan? is it lyk a junior IMO?


Bill has 61 sticks, of lengths 1, 2, … ,59, 60, and 2007. He then makes another stick, of positive integral length, but not 60 or 2007. How many different lengths can the stick that he take, if the resulting 62 sticks must be able to form a convex 62-gon when put together?

[hide]$1\le x\le2007+60\triangle-1$

But $x$ can't be $2007$ or $60$, so can $x$ take on $3836-2=3834$ different values.[/hide]

[quote="i_like_pie"][quote="dragonx"]the answer is 3834 becuase 3836-2=3834[/quote]
Some kind of explanation please? Just posting an answer without a solution is utterly pointless, and helps nobody.

@ ZzZzZzZzZzZz - How did you get that?[/quote]

The largest length you can have is 2007 plus the rest of the sticks minus 1=61*30+2007-1=3836

The least length you can have is 2007 minus the rest of the sticks plus one=2007-60*61+1=178

3836-178+1=3659  Subtract one because you can't include 2007 to get [b]3658[/b]

Oh... I think you're right. I forgot about the lower bound. :oops:

[quote="magixter"]But, the last stick can't be less than 2007-60*61/2 can it?
shouldn't the answer be
[hide]
177>x>3837
So the answer would be 3658.[/hide][/quote]
I think that's a typo :huh:

[quote="erdogankerem123"][hide]its 3659, because of the same inequality right?[/hide][/quote]
You have to subtract one because the last side can't be 2007.

well i worked it out and got that in order to have the values be 2008 or above the value has to be at least 178 and the other spectrum at most 3836 so not counting the 2007 the answer is 3836-178 so i got 3658 as my final answer.

All right, so we agree that the answer is [size=125][b]3658[/b][/size]


Prove that $ (\mathbf{a}\times\mathbf{b})\times\mathbf c\equal{}(\mathbf a\cdot\mathbf c)\mathbf b\minus{}(\mathbf b\cdot\mathbf c)\mathbf a$.
[hide="My proof (not done)"]If $ \{\mathbf a, \mathbf b\}$ are linealy dependent, then it is clear. Suppose that $ \{\mathbf a, \mathbf b\}$ are linearly independent, then so is $ S\equal{}\{\mathbf a, \mathbf b, \mathbf a\times\mathbf b\}$. Thus $ S$ is basis for $ \mathbb R^3$. Hence there exist $ x, y, z\in\mathbb R$ such that

$ (\mathbf a\times\mathbf b)\times\mathbf c\equal{}x\mathbf a\plus{}y\mathbf b\plus{}z(\mathbf a\times\mathbf b).$

But $ (\mathbf{a}\times\mathbf{b})\cdot[(\mathbf{a}\times\mathbf{b})\times\mathbf c]\equal{}(\mathbf{a}\times\mathbf{b})\cdot [x\mathbf a\plus{}y\mathbf b\plus{}z(\mathbf a\times\mathbf b)],$

i.e. $ z\|\mathbf a\times\mathbf b\|^2\equal{}0$. Since $ S$ is linearly independent, $ \|\mathbf a\times\mathbf b\|\neq0$. Therefore $ z\equal{}0$. Besides, we have

$ \mathbf c\cdot[(\mathbf a\times\mathbf b)\times\mathbf c]\equal{}\mathbf c\cdot(x\mathbf a\plus{}y\mathbf b),$

i.e. $ x\mathbf a\cdot\mathbf c\plus{}y\mathbf b\cdot\mathbf c\equal{}0$. Thus $ y\equal{}\minus{}\frac{\mathbf a\cdot\mathbf c}{\mathbf b\cdot\mathbf c}x$ ...
======================[/hide]

Moreover, how to prove the following without using the above formula?
1. $ (\mathbf a\times\mathbf b)\times \mathbf c\plus{}(\mathbf b\times\mathbf c)\times\mathbf a\plus{}(\mathbf c\times\mathbf a)\times\mathbf b\equal{}\mathbf 0.$
2. $ [(\mathbf a\times\mathbf b)\times\mathbf a]\times\mathbf b\equal{}\mathbf 0$, where $ \mathbf a$ and $ \mathbf b$ are perpendicular.

Two $ k$-linear expressions are equal if they agree on all possible combinations of a basis. Testing these on the 27 possible combinations of the standard basis vectors suffices.


linear algebra
matrix
function
linear algebra unsolved

Let A be a real square matrix of order n, diagonalizable.
Give an explicit formula for the trace of $A^k.$

In function of what?

If you don't state that, I simply answer: sum of the eigenvalues of A^k :roll:

Well,  :P
it is not explicitly written, but you must prove your claim...
Be careful: the text does not say that A is diagonal, but it says that A is diagonalizable, i.e. similar to a diagonal matrix.

$A$ is similar to $D=diag(a_1,...,a_n)$

$A^k$ is similar to $D^k=diag(a_1^k,...,a_n^k)$ then

$tr(A^k)=\sum a_j^k$


conics
ellipse
geometry
calculus
trigonometry
area of a triangle
Heron\u0027s formula

? anybody find one...i was looking at it and i was like wtf, i hate cooridnates...

i have been looking into a few geometric loci for ellipses and i have nothing...the whole distance from C=2 is sort of an awkward condition to use in a geometric sense...

I wasn't able to find one.  On the other hand, in general geometry problems don't have "geometric" solutions, and one has to turn to algebra and calculus (differential geometry, anyone?) to solve very difficult problems.  Generally math contest problems are carefully designed so that that's not necessary, but in this case there may just have been no more elegant solution than coordinates.

Well, it wasn't exactly geometric, but my solution didn't use analytical geometry. I just said A[ABC]+A[AXC]=A[ABX]+A[BXC]. I then just used Heron's formula on all of these letting AX=x, and BX=8-x. I then just solved for x and used Heron's formula again once I had the lengths.

I also did what asquared did.  It took tons of algebra and trig to find $x$ and $8-x$ though and then to plug them into Heron's formula... and then you get a nested radical which I assumed you had to simplify (I wasn't taking any chances with graders  :lol: ).  You can see the solutions thread for mine.


I know this isn't really that relevant to math but good luck everybody the AMC's in two days.

Have a great time with the tests, everyone.

I'm having AMC 12 tomorrow. Gee, a little bit nervous  :oops:
Thanks a lot for the g'luck  :)


Evaluate $\lim_{n \rightarrow \infty}F(n,n)$, where $F(m,1) = m$ and $F(m,k+1) = (m-k)^{2}+\frac{1}{2}F(m,k)$ for $k, m, n \in \mathbb{N}$.


calculus
integration
function
trigonometry
calculus computations

I understand that this is child's play compared to some of the problems you all normally work, but there is no other place to post them, I think.  Thanks for all the help in advance.

How would one take the integration of something in the form $\sqrt{a^2 + u^2}$, where u is the function?

Also, how would one integrate something in the form $\sqrt{a^2 - u^2}$ or $\sqrt{u^2 - a^2}$

I think that this forum is for the very such as your questions. Don't you think so Kent Merryfield?
Anyway, for $\sqrt{u^2+a^2},$ you can set $u=a\tan \theta,$ for $\sqrt{a^2-u^2},$ you can set $u=a\sin \theta.$
Actually you can see the formula for these, which is Japanese student aren't allowed to use directly in university entrance exam.

kunny


geometry
3D geometry
sphere
combinatorics proposed
combinatorics

In space are given $ 1997$ points. Let $ M$ and $ m$ be the largest and smallest distance between two of the given points. Prove that $ \frac{M}{m}>9$.

Without loss of generality, let the smallest distance be $ 1$. We claim that the greatest distance must be at least $ 9$. To show this, assume the greatest distance was $ <9$. This means that the entire set of points would fit into a sphere of radius $ 4.5$. Note however that each pair of points is at least $ 1$ apart; thus there are no points within a sphere of radius $ 1$ of a given point - if we can show that we cannot fit all of these inside a sphere of radius $ 5.5$, we are done. Each sphere intersects at most 12 other spheres, so if we remove $ \frac{11}{12}$ of them, we can attain a set of at least $ 167$ disjoint spheres. But the sum of the volumes of these is greater than the volume of a sphere with radius $ 5.5$, so we are done.


trigonometry
modular arithmetic
geometry unsolved
geometry

Let  $ x \equal{} {20^ \circ }$ prove that:
                   $ {(\sin x \plus{} \sin 2x \plus{} \sin 4x)^5} \minus{} {(\sin x \plus{} \sin 2x \minus{} \sin 4x)^5} \minus{} {(\sin x \minus{} \sin 2x \plus{} \sin 4x)^5} \minus{} {( \minus{} \sin x \plus{} \sin 2x \plus{} \sin 4x)^5} \equal{} 15\sqrt 3$

In general, $ (a+b+c)^5+(-a-b+c)^5+(-a+b-c)^5+(a-b-c)^5=$
$ \sum_{i+j+k=5} (1+(-1)^{i+j}+(-1)^{i+k}+(-1)^{j+k}) a{}^ib{}^jc{}^k \binom{5}{i,j,k}$

Notice that $ 1+(-1)^{i+j}+(-1)^{i+k}+(-1)^{j+k}=(1+(-1)^{i+j})(1+(-1)^{i+k})$. So this expression gets only the values $ 0$ and $ 4$. We need to evaluate the previous sum only in the case when our expression equals 4. It happens when $ i+k \equiv i+j \equiv 0 \pmod 2$, or: $ i \equiv j \equiv k \pmod 2$. The possible solutions are $ (1,1,3)$ and their permutations.
The sum becomes:
$ 4\sum_{i,j,k=1,1,3} a{}^ib{}^jc{}^k \binom{5}{1,1,3}=80\sum_{i,j,k=1,1,3} a{}^ib{}^jc{}^k=80abc \sum_{i,j,k=0,0,2} a{}^ib{}^jc{}^k=80abc\sum a^2$

In our case, $ abc=\sin 20^\circ\cdot\sin 40^\circ\cdot\sin 80^\circ=\frac{\sqrt{3}}{8}$, and it is left to calculate $ a^2+b^2+c^2$, which is $ 1.5$:
$ \sin^2 20^\circ+ \sin^2 40^\circ ^2+ \sin^2 80^\circ=1.5-\frac{\cos 40^\circ+\cos 80^\circ +\cor 160^\circ}{2}=1.5$

[quote="bambaman"]In general, $ (a + b + c)^5 + ( - a - b + c)^5 + ( - a + b - c)^5 + (a - b - c)^5 =$
$ \sum_{i + j + k = 5} (1 + ( - 1)^{i + j} + ( - 1)^{i + k} + ( - 1)^{j + k}) a{}^ib{}^jc{}^k \binom{5}{i,j,k}$

Notice that $ 1 + ( - 1)^{i + j} + ( - 1)^{i + k} + ( - 1)^{j + k} = (1 + ( - 1)^{i + j})(1 + ( - 1)^{i + k})$. So this expression gets only the values $ 0$ and $ 4$. We need to evaluate the previous sum only in the case when our expression equals 4. It happens when $ i + k \equiv i + j \equiv 0 \pmod 2$, or: $ i \equiv j \equiv k \pmod 2$. The possible solutions are $ (1,1,3)$ and their permutations.
The sum becomes:
$ 4\sum_{i,j,k = 1,1,3} a{}^ib{}^jc{}^k \binom{5}{1,1,3} = 80\sum_{i,j,k = 1,1,3} a{}^ib{}^jc{}^k = 80abc \sum_{i,j,k = 0,0,2} a{}^ib{}^jc{}^k = 80abc\sum a^2$

In our case, $ abc = \sin 20^\circ\cdot\sin 40^\circ\cdot\sin 80^\circ = \frac {\sqrt {3}}{8}$, and it is left to calculate $ a^2 + b^2 + c^2$, which is $ 1.5$:
$ \sin^2 20^\circ + \sin^2 40^\circ ^2 + \sin^2 80^\circ = 1.5 - \frac {\cos 40^\circ + \cos 80^\circ + \cor 160^\circ}{2} = 1.5$[/quote]

thankyou bambaman,i have a bad proof ,use $ cosx + cos2x = cos4x ,x=20^ \circ$ :D

$ (a\plus{}b\plus{}c)^5 \minus{} (a\plus{}b\minus{}c)^5 \plus{} (a\minus{}b\plus{}c)^5 \minus{} (\minus{}a\plus{}b\plus{}c)^5 \equal{} 32((a\plus{}b)^5\minus{}a^5\minus{}b^5) \equal{} 160ab(a\plus{}b)(a^2\plus{}ab\plus{}b^2) \equal{} 80abc(a^2\plus{}b^2\plus{}c^2)$ :lol:


Patio blocks that are hexagons $1$ unit on a side are used to outline a garden by placing the blocks edge to edge with $n$ on each side. The diagram indicates the path of blocks around the garden when $n=5.$

[asy]
size(250);int i,j;
real r=sqrt(3);
for(i=0; i<6; i=i+1) {
for(j=0; j<4; j=j+1) {
draw(shift(((j*r)*dir(60*i+150)).x, ((j*r)*dir(60*i+150)).y)*shift((4r*dir(60i+30)).x,(4r*dir(60i+30)).y)*polygon(6));
}}[/asy]
If $n=202,$ then the area of the garden enclosed by the path, not including the path itself, is $m(\sqrt{3}/2)$ square units, where $m$ is a positive integer. Find the remainder when $m$ is divided by $1000.$

[hide="solution"]
I hope I remembered this correctly...

It becomes obvious as soon as you fill in the picture (may take a few moments to show):
[img]http://img.photobucket.com/albums/v194/mukmasterxps/diagram.gif[/img]

Number of hexagons is $1+6\frac{(n-2)(n-1)}{2}$, in this case, $1+6\frac{(200)(201)}{2}=120601$

Since these are unit hexagons, each has area $6\frac{\sqrt{(3)(1)(1)(1)}}{4}=\frac{3\sqrt{3}}{2}$

The total area is therefore $(120601\frac{3\sqrt{3}}{2})=(120601)(3)\frac{\sqrt{3}}{2}$

$m=(120601)(3)$, and in $(mod 1000)$, this is $601*3\underline{=}\boxed{803}$
[/hide]

Don't ask where I got this.
EDIT: I also have the thing as a png, I can pm that to you.

[hide]The inside space can be tesselated by hexagons, each of area $3\sqrt{3}/2$. There are $201+\cdots+400+401+400+\cdots+201=120601$( :P ) of these hexagons. $m=3\times120601\equiv\boxed{803}\pmod{1000}$.[/hide]

[quote="pianoforte"][hide]The inside space can be tesselated by hexagons, each of area $ 3\sqrt {3}/2$. There are $ 201 \plus{} \cdots \plus{} 400 \plus{} 401 \plus{} 400 \plus{} \cdots \plus{} 201 \equal{} 120601$( :P ) of these hexagons. $ m \equal{} 3\times120601\equiv\boxed{803}\pmod{1000}$.[/hide][/quote]

sorry to revive an old topic.

but why'd u start with 201 in ur sum instead of 202?

Try it yourself with the case when $ n\equal{}5$.  You'll notice that you have to start with $ n\minus{}1$.

oh,  :oops:

i included the ones on the boundaries too..

thx a bunch

Morning problem. If we fill in the original picture, we notice that if we count the amount of hexagons in rows, we see that it goes $4, 5, 6, 7, 6, 5, 4$. This pattern will clearly continue, so in the $202$ case, we get $201+202+\ldots+401+400+399+\ldots+201$ total hexagons. We can split this up into $201+202+\ldots+400 = 60100$ and $400+399+\ldots+401 = 60501$, so the sum is $120601$. The value of $m$ is just $3 \cdot 120601 = 361803$, giving an answer of $\boxed{803}$.