Just tried to upload my solutions for Round 2 through the web uploader. I selected the PDF file and hit "Upload," and the uploader doesn't seem to be responding. (See picture attached).

I checked my profile and it doesn't show the green checkmark next to round 2. However, when I try to submit again, it tells me that I've already submitted the solutions (again, picture attached).

Email usamts@usamts.org and include your user ID#.

If you can't get the uploader to work, please mail your solutions (postmarked today).

Actually, nevermind, we have received all 7 pages of your solution.

In general, if the uploader runs for a few minutes but then "hangs", and then you try to upload it again but it says "you've already sumbitted a solution", then we have almost certainly received your original submission, but for whatever reason (e.g. your connection temporarily dropped) your computer didn't receive the "submission completed" acknowledgment from our computer.


Let $ABCD$ be a rectangle. Let $P$ and $Q$ be points on $BC$ and $CD$ respectively such that the areas of $ADQ$, $CPQ$ and $ABP$ are equal to 5, 4 and 3 respectively. Find the area of triangle $APQ$.

[hide]Defining $S$ as the area of the rectangle and doing: $\frac{PC}{BC}=m$ and $\frac{CQ}{CD}=n$,we got:

$\frac{\Delta PAC}{\Delta ABC}=\frac{\Delta PAC}{\frac{S}{2}}=m \implies {\Delta PAC=m \cdot \frac{S}{2}}$

$\frac{\Delta ACQ}{\Delta ACQ}=\frac{\Delta ACQ}{\frac{S}{2}}=n \implies {\Delta ACQ=n \cdot \frac{S}{2}}$ in consequence:

$\Delta ABC=\Delta ACD \implies m \cdot \frac{S}{2} + 3=n \cdot \frac{S}{2} + 5 \implies n=\frac{m \cdot S-4}{S}$ and:

$\frac{\Delta CDP}{\Delta BCD}=\frac{\Delta CDP}{\frac{S}{2}}=m \implies {\Delta CDP=m \cdot \frac{S}{2}}$ thus:

$\frac{\Delta CPQ}{m \cdot \frac{S}{2}}=n=\frac{m \cdot S-4}{S} \implies m^2 \cdot S - 4m -8=0$, solving the quadratic equation for a [b]positive[/b] real number:

$m=\frac{4 + \sqrt{16 + 32S}}{2S}$ but: $m \cdot \frac{S}{2} + 3=\frac{S}{2} \implies \sqrt{16 + 32S}=2\cdot S -16 \implies 4\cdot S^2 -96\cdot S + 240=0$

Choosing the quadratic solution that do not leave $\Delta APQ$ to a negative number, we got:

$\Delta APQ=S - 5 - 4 - 3=2\sqrt{21}$[/hide]

put AB, BP, PC, CQ, QD, DA, as c+b, a-d, d, c, b, a.

[BPA] = 3 implies (c+b)(a-d) = 6

[PCQ] = 4 implies dc = 8

[AQD] = 5 implies ab = 10

So $ab + ac - cd - bd = 6$ so that $ac - bd = 4$.

We solve for ac.  Let x = ac.  Then $x - \frac{80}{x} = 4$, so that $x^2 - 4x - 80 = 0$.

Finally, the area of the entire rectangle is $a(b+c) = 10 + x$.  Then we subtract 12, from the triangles so we just get the center triangle.  So we want $x - 2$.

Put $y = x-2$ so that $0 = y^2 + 4y + 4 - 4(y+2) - 80 = y^2 - 84$ and clearly $y = 2\sqrt{21}$.

*Note: there may be arithmetic errors because i rushed*

Your answers are correct two guys.

Let $AB=x,AD=y$ we can set $BP=\frac{6}{x},DQ=\frac{10}{y}$,hence $\triangle CPQ=\frac{1}{2}\left(x-\frac{10}{y}\right)\left(y-\frac{6}{x}\right)=8$ solve for $xy$, we get $xy=12\pm 2\sqrt{21}$

The answer is
$\triangle {APQ}=\text{rectangle\ ABCD}-\triangle {ADQ}-\triangle {CPQ}-\triangle {ABP}$
$=12+2\sqrt{21}-(5+4+3)=\boxed{2\sqrt{21}}$


in each step ,we will change the location of two digits.
beginning from $12345$, after even number ($2n$) steps, which one cant be got?  (e.g $12345 \Rightarrow 13245 \Rightarrow 43215$)
$a. 13425$
$b. 21435$
$c. 35142$
$d. 43125$
$e. 53124$

Also is this correct topic for this type of question?

d. Write permutations as a product of transpositions and calculate their signs.

Permutations can be found in basic algebra books more probably than in combinatorics books.


can anyone tell me how to solve problems using steady state appproximation with eg.

If I is an intermediate in a given reaction mechanism, then the steady state approximation consists in considering $ \frac{d[I]}{dt} \equal{} 0$.

In summary, to use the steady state approximation in order to deduce a rate law from a mechanism, follow the following steps:

1) consider the reaction rate as equal to the rate of formation of a product;

2) eliminate the concentrations of any intermediate $ I$ that appear in the rate expression obtained in 1) by using $ \frac{d[I]}{dT} \equal{} 0$ in order to find the concentration of each such intermediate $ I$ as a function of other species/intermediates and rate constants;

3) if the previous step introduces concentrations of additional intermediates, repeat step 2) to these intermediates until you get a rate law expressed only in terms of reacting species and rate constants.

Here's a (perhaps common) example for you to play with:

For the reaction in aqueous acidic solution $ H_2O_2 \plus{} 2H^\plus{} \plus{} 2I^\minus{} \longrightarrow I_2 \plus{} 2H_2O$, one of the mechanism that occur is

$ H^\plus{} \plus{} I^\minus{} \rightleftharpoons HI$

$ HI \plus{} H_2O_2 \longrightarrow H_2O \plus{} HOI$

$ HOI \plus{} I^\minus{} \longrightarrow I_2 \plus{} HO^\minus{}$

$ HO^\minus{} \plus{} H^\plus{} \longrightarrow H_2O$

(a) Use the steady state approximation to derive the rate law for this reaction.

(b) Compare the rate law obtained in (a) with the one that can be derived by using the rate determining step approximation (knowing that the first step is a fast equilibrium, the second is slow, and the last two are fast).

[quote="VARUNARASU"]to what intermediates can i apply steady state approximation in this case??[/quote]

Well for instance see what substances are formed and disappear in this rn  :)

but how do you know that ssa is applicable over here. i mean how do you know rate of production= rate of consumption for each of the intermediates?

You do not know that, its just an assumption that is made. There are two main approximations in order to derive a rate law from a mechanism: the steady state approx. and the rate determining step approx. They are just approximations.

I came up with some crap like this:
D[hi]/dt=k1[h+][i-]( rate of production)
Applying ssa:
K1[h+][i-]=k2[hi][h2o2]=k3[hoi][i-]=k4[oh-][h+]
As d[hi]/dt=d[hoi]/dt=d[oh-]/dt=0.
Therefore k1[i-]=k4[oh-]
I know this is wrong can u tell me where I am wrong.
And carcul, can you give the solution to this. I know you don’t like giving away the solution straightaway but I’ve got the Olympiad tc selection round coming up the day after, and I have to know this. Plzzzzzzzzzzzzzz…


For each integer $n>1$, let $s_n$ be the number of permutations $(a_1,a_2,...,a_n)$ of $1,2,...,n$ which satisfy $1\leq |a_k-k|\leq 2$ for every $k=1,2,...,n$. Prove that $\frac{7}{4}s_{n-1}<s_n<2s_{n-1}$

In case anyone is interested, I just accidentally noted that this was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=3710 .

  darij


MATHCOUNTS
calculus
geometry
probability
number theory
relatively prime

I need solutions for numbers 1, 3, 5, and 10 from [url=http://yashar.biz/mathcounts/mathcounts_files/2001/2001%20Chapter%20Team.pdf]the 2001 MathCounts Chapter Team Rond[/url].  I need solutions, not answers.  I would appreciate it if you could give me solutions that don't invole calculus or some other method that won't show up in MathCounts.

Thanks,
That is all

#1
[hide]The area of the total region is 8x4 = 32, the area of the region that fits the question is 2x2=4, so the probability is 4/32 = 1/8.

The common mistake on this problem is to only count the integer values and say the answer is (3x3)/(9x5) = 1/5[/hide]

#5
[hide]Since you want there to be as few relatively prime dates, you want the month with the most distinct prime factors.  Most people jump to December(12), but June has just as many prime factors and less days.

June(6) 1,5,7,11,13,17,19,23,25,29

10 dates[/hide]

That is why I did not directly post the problem, just the link!

[hide="Number One"]
Ok this one seems pretty easy, It says that 0 <= x <= 8, and 0 <= y <= 4. So the probability that x <= 2 is 1/4 (2/8). The probability that y <= 2 is 1/2 (2/4). So the answer is 1/4 * 1/2= [b]1/8[/b][/hide]

I dont feel like doing the others now...

#3
[hide]In the given picture the dotted lines are each equal to twice the radius, so 2x2=4 each.  There are three dotted lines, so 3x4=12.  The three times the belt is up against the circles, the belt contacts one third the circles circumference, and all three are equivalent to the entire circumference.

Since the circle has circumference 4pi, the length of the belt is 4pi + 12. [/hide]

Ah, this one:
[hide="10"]So you have a square ABCD
1 square that is ABCD
4 squares with sharing edges facing out.
4 squares with thier opposite corners as the sides of ABCD
4 squares with a side being ABCDs diagonal
1+4+4+4=13
>>13<<[/hide]

[hide="3"]You can split this into 2 parts.
part I: The belt that is touching the sylindars
part II: The belt that is not touching the cylindars

I: touching=circ of one cyl.=4pi
II:not touching=distance between centers of 2 cyl.*3=4*3=12
>>4pi+12<<[/hide]

[quote="dwx314"][hide="1"]just graph it and it becomes super easy.
Area of All:4*8=32
Area of Good:2*2=4
4/32=1/8
>>8<<[/hide][/quote]

What about my way..which dosnt require graphing, but simple probability.


MATHCOUNTS trigonometry AMC AIME USA(J)MO USAMO ratio

So, in this "competition", there will be teams, one team gives a specific team a problem to solve (hsb lvl) and if they solve it they get a point, if they don't, the opposing team gets a point, after all teams have solved one problem, that will be counted as the end of a round. After a round, the team with the lowest score is kicked off! Sign-ups start now. (I'm signed up!)

I would recommend harder problems because MATHCOUNTS is almost equal to HSB, so maybe a competition in Pre-Olympiad?

[quote="suma_milli"]I'll sign up.  Oh and I request gauss1181 to be on my team.  Our team will be Team :D.

... :D[/quote]
YAY!!!

Oh darn, not again!
Too many smilies in your post: 5
You can have at most 3 smilies in this message. Please remove some smilies before submitting this message.

 :D Yay, there we go.

[quote="pythag011"]Hmm.. I assume we can't give them a problem like "Prove the Riemann Hypothesis?"[/quote]
I guess not. :wink:

join.
fkdjalfkdjalfkdjlkfjdkajfkldjfkladjfkldjafkljdakfj. --> message was too small.

I'll join. If possible, Team :D.

AMC or mathcounts level is best for a middle school tournament, in my opinion.

ok.
so someone moved this. crap. i was hoping for trig problems. can someone move this back to HSB? pretty please? I really need trig and log problems!!!!! aaaaaaah until then make the problems around HSB lvl to [hide][hide][hide]and intermidiate will rock too. [/hide][/hide][/hide][/hide]

ok yeah,
so here is the new problem:
In triangle ABC, on side AC, there is a point X such that AX=4, XC=2 and BC=5. Find the smallest possible integer value BX such that the ratio of BX to BA is 5 to 7 and BA is also an integer.
Deadline: Tomorrow 3:00 pm eastern

oops i totally spaced out and forgot about this...

hehe....

Ok, I am extending the deadline another day because some people may not have that much time.

What are the chances that this thread gets locked for using complex numbers in the Mathcounts forum :| ? Actually, not that high. Meh.

[quote="Mewto55555"]Hm so I lose? No prod pm? Don't they have to answer first? In any case,

Let $ y \equal{} x \plus{} \frac {1}{x}$

Expanding (I will let reade fill in details) we have $ x^3 \plus{} x^2 \plus{} x \plus{} 1 \plus{} x^{ \minus{} 1} \plus{} x^{ \minus{} 2} \plus{} x^{ \minus{} 3} \equal{} 0$

As $ x\neq 0,1$ we multiply both sides by $ x^3(x \minus{} 1)$

$ x^7 \minus{} 1 \equal{} 0$

Thus $ x \equal{} \text{cis}\left(\frac {2k\pi}{7}\right)$ such that k is an integer which is not a multiple of 7.

Substituing back into $ y \equal{} x \plus{} \frac1x$ and applying DeMoivre's theorem, we have $ y \equal{} 2\cos{\frac {2\pi}{7}},2\cos{\frac {4\pi}{7}},2\cos{\frac {6\pi}{7}}$

Hm, good job guys. Now ask your question?[/quote]

Thanks for essentially getting the whole motivation of this problem from me without citing your sources and basically copying [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1295344#1295344]my work from a few months ago[/url] (to name one source; ignore all the weird $ \operatorname{cis}$ stuff). :P

[quote="Mewto55555"]Hm so I lose? No prod pm? Don't they have to answer first? In any case,

Let $ y \equal{} x \plus{} \frac {1}{x}$

Expanding (I will let reade fill in details) we have $ x^3 \plus{} x^2 \plus{} x \plus{} 1 \plus{} x^{ \minus{} 1} \plus{} x^{ \minus{} 2} \plus{} x^{ \minus{} 3} \equal{} 0$

As $ x\neq 0,1$ we multiply both sides by $ x^3(x \minus{} 1)$

$ x^7 \minus{} 1 \equal{} 0$

Thus $ x \equal{} \text{cis}\left(\frac {2k\pi}{7}\right)$ such that k is an integer which is not a multiple of 7.

Substituing back into $ y \equal{} x \plus{} \frac1x$ and applying DeMoivre's theorem, we have $ y \equal{} 2\cos{\frac {2\pi}{7}},2\cos{\frac {4\pi}{7}},2\cos{\frac {6\pi}{7}}$

Hm, good job guys. Now ask your question?[/quote]
Ahem. Did I not say he wasn't leaving?

Back on topic: Does mewto still get a point.

Yeah, uh, are we allowed to use complex numbers, etc. in mc forum?

why don't people ever look at the "Click here to reveal hidden content" in the first few posts?

I FIRST PUT THIS IN HSB! ok sorry
yeah and I tried putting it in intermidiate and it got moved here a looooooooong time ago (like 4 months ago)

Hmm..this tourney supports my argument for the new HS Tournaments forum (Questions and Suggestions Forum)

I asked for that also. didn't work mind you. Well, anyway deadline for a solution is tomorrow 3:00 pm Eastern.

No solutions were sent in! Come ON people! Anyway, my team wins 3.5 points.
next match up is Team Team Pwnage-Team We will fail! ! I will send out reminder pm's.


linear algebra
matrix
superior algebra
superior algebra theorems

How can you exprim $TrA^*$ only with $TrA$ if $A \in M_3(C).$

Characteristic of $A$ is

$x^3-(trA)x^2+tr(A^{*})x-det(A)$

$A$ is trigonalisable

if $k_1,k_2,k_3$ eigenvalues of $A$ then $tr(A^{*})=k_1k_2+k_2k_3+k_3k_1$

some computation shows $tr(A^{*})=\frac{1}{2}((trA)^2-tr(A^2))$

[quote="silviu"]\[A^{-1}= \frac1{detA}A^{*},A \in M_n(C),detA \ne 0\][/quote]

As far as I know, that is not the standar matrix associated with the notation $A^{*}$, usually, as far as I know, that matrix represent the adjoint, that is, the transpose of the conjugate matrix, and not the transpose if the matrix of cofactors.

Best,


there was once a mock AIME question that read:
2006 points are evenly spaced on a circle.  Given one point, find the number of other points that are less than 1 radius distance away from that point.

I said the answer was 668, but I was wondering how could this be?

[hide="Solution"] Inscribe a hexagon into the circle.  The side length of the hexagon is the radius (because a hexagon is composed of six equilateral triangles), so the number of other points less than $1$ radius distance away is twice the the number of points on an arc of one of the triangles (but not on the vertices).

Since $2 | 2006$ but $3$ does not, starting from one of the $2006$ points and drawing the hexagon with that point as one of its vertices, only one other point (opposite the original) is also a vertex of the hexagon, so there are $\lfloor \frac{2006}{6}\rfloor = 334$ points strictly within $1$ radius of the original on either side.  This gives a total of $\boxed{ 668 }$ on both sides. [/hide]

Alright thx, I just got confused because if we tried a circle with only 6 points, the answer to this type of question would be 0-and not floor(6/6)*2


What is the sum of the first 10 positive integers?

Note that we can derive the formula for the first $ n$ integers as follows.

We want to find $ 1 \plus{} 2 \plus{} 3 \plus{} \dots \plus{} n$. Hey, this is an arithmetic sequence! The common difference is 1, so now the question is, how do we find the sum of an arithmetic sequence?

Well, let's pair up the numbers.

$ (1 \plus{} n) \plus{} (2 \plus{} n \minus{} 1) \plus{} \dots$

The sum of this is the last term plus the first term multiplied by half the total number of terms(because that's how many pairs there are.) Thus the formula is:

$ \frac{n}{2} \cdot (n \plus{} 1) \equal{} \boxed{\frac{n(n\plus{}1)}{2}}$

You could prove this using induction, like this:

First prove it works for some base case, $ k$. Say, 1. The sum of the first 1 number is just 1, right? And testing this in our formula works!

So now that we proved it works for 1, we have to prove it works for $ k \plus{} 1$. So...

$ 1 \plus{} 2 \plus{} 3 \plus{} \dots \plus{} k \plus{} k \plus{} 1 \equal{} \frac{k(k\plus{}1)}{2} \plus{} k \plus{} 1 \equal{} \frac{k(k\plus{}1)}{2} \plus{} \frac{2k \plus{} 2}{2} \equal{} \frac{k(k\plus{}1) \plus{} 2k \plus{} 2}{2} \equal{} \frac{(k\plus{}1)(k\plus{}2)}{2}$

QED

Blargh izzy or you could do it simple way:

$ S\equal{}1\plus{}2...\plus{}N$

$ S\equal{}N\plus{}N\minus{}1...\plus{}1$

Adding equations:

$ 2S\equal{}(1\plus{}N)\plus{}(1\plus{}N)....\plus{}(1\plus{}N)$ Note, there are N terms.

$ 2S\equal{}N(N\plus{}1)$

$ S\equal{}\frac{N(N\plus{}1)}{2}$


The base of compression-type spring balance is rigidly bolted to the floor inside the car.
The spring inside the blance has a spring constant $k=10^4 Nm^{-1}$.  box of mass $m=100kg$ is rigidly bolted to the weighing pan of the balance. (The mass of the box is much greater than the mass of the weighing pan, so you may ignore the weighing pan in your analysis.) When the car is parked on a horizontal road, the balance correctly registers the weight of the box, i.e. $980N$

This car is now travelling on a slightly undulating road at a constant horizontal velocity $v$. The profile of the road is shown in figure 2 with $\lambda = 100m$ and $A=1m$. Note that $\lambda \gg A$,and $\lambda$ is much longer than the body length of the car. The shape of this profile may be fitted by a sinosuidal function.

(a) Sketch a graph showing the balance reading $W$, as a function of time $t$, if $v$ is kept constant at $15ms^{-1}$ at all times.
Indicate the time scale on your graph.

(b) Explain clearly how you arrive your answer,giving any derivation if necessary.

I pressume they want you to model it as if you are just moving the spring up and down. The length between peaks is 100m and the speed is 15 ms^-1 so the time period between peaks is 6.7 seconds.

the equation describing the time period of natural oscillations of this sort of body is T = 2(pi)sqrt(m/k), k = 1000 and m = 100 so m/k = 0.1, sqrt(m/k) is 0.32, so T = 2.

So you have a system which oscillates at natural frequency 0.5 and driving frequency 0.15. as the driving frequency is a multiple of the natural frequency you will get some degree of resonance.

Have you studied how to graph things like this? Becuase I haven't..

My take is that they want you to find the normal force on the mass needed to traverse the road, taking into account the centripetal force from going on the road at that velocity, as well as the normal from the road being inclined at that point.

but fi you model the car as having two points of contact (ie. wheels) with the road, you'd get tilting and the motion of the mass would have a horizontal component which makes calculations much harder.. I think. Also the shape of the road is sinusodal so i can't see how dynamics of circuklar moption will help.

But I don't have the answer, so my critisism of your method doesnt have much weight as I dont know how to interpret the problem myself.

My take on it is that the car is mentioned only as a means by which the box-and-spring contraption would travel across the road at a uniform velocity.  Because they have not given a distance of separation for the wheels, I think it would be safe to say it doesn't matter, and that we can ignore the car entirely.

Even though the road is sinusoidal, you get centripetal acceleration every time the car turns.  Because the road is not straight, those effects would occur in this problem.  Anyway, I don't believe they mean for us to calculate the radius of curvature at every point- just estimate enough such that we can provide a rough sketch of what forces will be required.

No, forget about it, it's just too complicated. And you'd end up with the same thing as degreestudent in his first post. Much more interesting question is how to add up those two oscillations...


George is planning a dinnetr meeting for 150 people.  Each table will be set for nine people.  If there are enough tables to accommodate everyone, what is the fewest possible number of unused table settings?

[hide]since $9*17=153$ and you cant have FEWER seettings than there are people, there will be 3 leftover settings. I'm not sure what you mean by 'fewest' - is there another answer?[/hide]


The mean price of a set of tapes and CDs is $ \$8.00$. The tapes average $ \$7.00$ and the CDs average $ \$10.00$. What is the ratio of the number of CDs to the number of tapes? Express your answer as a common fraction.

We let $ t$ = the number of tapes, and $ c$ = the number of CD's. We are given that the tapes average $ \$8$ each and the CD's average $ \$10$ each. So we can set up an equation.

$ \frac{\$7t \plus{} \$10c}{t \plus{} c} \equal{} \$8 \Rightarrow 7t \plus{} 10c \equal{} 8t \plus{} 8c \Rightarrow 2c \equal{} t \Rightarrow \frac{c}{t} \equal{} \boxed{\frac{1}{2}}$


superior algebra
superior algebra solved

show Sn isnt isomorphic to anysubgroup of A(n+1),n>1.

I remember seeing this at least twice on this forum. Try searching.


Given that 1025/1024 = 1.0009765625, what is the sum of the digits of $5^{10}$?

Well, from the given we know what $\frac{1}{1024}$ is, so we can find what $\frac{10^{10}}{1024}= 5^{10}$ is.


[hide]
I cheated and used finite differences/generating functions to get

$\frac{(x)(x+1)(x^{2}+5x+6)}{24}$

I'll be back with a proper solution in a bit.
[/hide]


induction
binomial coefficients
combinatorics solved
combinatorics

Prove identity:
\[ \sum_{k \equal{} 1}^{n} 2^{2n \minus{} 2k}\binom{2n}{2k} \binom{2k}{k} \equal{} \binom{4n}{2n}
\]

Let's extend your problem by writing $n$ instead of $2n$, thus adding the case of $n$ being odd:

[color=blue][b]Theorem 1.[/b] If $n\geq 0$ is an integer, then

$\sum_{k = 0}^n 2^{n - 2k} \binom{n}{2k} \binom{2k}{k} = \binom{2n}{n}$. [/color]

This is a particular case (for $u = 0$) of the following fact:

[color=blue][b]Theorem 2.[/b] If $n\geq 0$ and $u$ are integers, then

$\sum_{k = 0}^n 2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k} = \binom{2n}{n + u}$.[/color]

[i]First proof of Theorem 2.[/i] Consider a table consisting of two rows and n columns. A [i]coloring[/i] of the table will mean a way to write in every cell of the table a 0 or a 1. Such a coloring will be called [i]fair[/i] if it satisfies the equation

(the number of 1's in the first row) = (the number of 1's in the second row) + u.

How many fair colorings exist?

On the one hand, a coloring is fair if and only if

(the number of 1's in the first row) = (the number of 1's in the second row) + u,

i. e. if and only if

(the number of 1's in the first row) = n - (the number of 0's in the second row) + u

(since we always have (the number of 1's in the second row) = n - (the number of 0's in the second row)), i. e. if and only if

(the number of 1's in the first row) + (the number of 0's in the second row) = n + u,

i. e. if and only if the 1's in the first row and the 0's in the second row, together, occupy exactly n + u cells.

Thus, in order to get a fair coloring, we have to mark n + u of the 2n cells of the table; then, the marked cells in the first row will be filled with 1's, the unmarked cells in the first row will be filled with 0's, the marked cells in the second row will be filled with 0's, and the unmarked cells in the second row will be filled with 1's. This way, we get a fair coloring, and conversely, each fair coloring can be obtained uniquely this way. Thus, the number of fair colorings equals to the number of ways to mark n + u of the 2n cells of the table, and this number is obviously $\binom{2n}{n + u}$.

On the other hand, let's construct a fair coloring in a different way: In the following, an (x,y) column will mean a column which has the number x in the first row and the number y in the second row (where x and y are numbers from {0; 1}). Then, for any coloring, every column is either a (0,0) column, or a (0,1) column, or a (1,0) column, or a (1,1) column. A coloring is uniquely determined if we know which columns are (0,0) columns, which ones are (0,1) columns, which ones are (1,0) columns, and which ones are (1,1) columns.
Obviously, for every coloring, the following two equations hold:

(the number of 1's in the first row) = (the number of (1,1) columns) + (the number of (1,0) columns);
(the number of 1's in the second row) = (the number of (1,1) columns) + (the number of (0,1) columns).

Thus, the equation

[color=green][b](1)[/b][/color] (the number of 1's in the first row) = (the number of 1's in the second row) + u,

holds if and only if

[color=green][b](2)[/b][/color] (the number of (1,0) columns) = (the number of (0,1) columns) + u.

In other words, a coloring is fair (i. e. it satisfies the equation [color=green][b](1)[/b][/color]) if and only if it satisfies the equation [color=green][b](2)[/b][/color].

This yields the following construction of a fair coloring:

- Specify an integer k such that $0\leq k\leq n$. This number k is going to be the number of (0,1) columns in our coloring, and the number of (1,0) columns in our coloring is going to be k + u (in order to satisfy [color=green][b](2)[/b][/color]).
- Now let's determine which columns will become (1,0) columns and which ones will become (0,1) columns. This is done as follows: First choose 2k + u columns of the table, then select k out of these 2k + u columns and mark these k columns as (0,1) columns, and then mark the remaining k + u of the 2k + u chosen columns as (1,0) columns. The number of possibilities to do this is $\binom{n}{2k + u}\binom{2k + u}{k}$ (in fact, there are $\binom{n}{2k + u}$ possibilities to choose 2k + u of the n columns of the table, and $\binom{2k + u}{k}$ possibilities to select k out of these 2k + u columns).
- Finally, we have to decide which of the remaining n - 2k - u columns become (0,0) columns and which become (1,1) columns. This can be decided freely, so there are $2^{n - 2k - u}$ possibilities (2 for every column).

Altogether, there are $\sum_{k = 0}^n \binom{n}{2k + u}\binom{2k + u}{k}\cdot 2^{n - 2k - u} = \sum_{k = 0}^n 2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k}$ possibilities to perform such a construction. They all yield different fair colorings, and each fair coloring can be constructed uniquely in such a way. So the number of fair colorings must be $\sum_{k = 0}^n 2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k}$. But above, we saw that the number of fair colorings is $\binom{2n}{n + u}$. Comparing these numbers, we obtain $\sum_{k = 0}^n 2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k} = \binom{2n}{n + u}$, and Theorem 2 is proven.

[i]Second proof of Theorem 2.[/i] Recall that for any integer $u$, we have

$\binom{0}{u} = \left\{\begin{array}{c} 1,\text{ if }u = 0; \\
0,\text{ if }u\neq 0\end{array} \right.$ and $\binom{u}{0} = 1$.

We will use Knuth's sum convention (see [url=http://www.artofproblemsolving.com/viewtopic.php?p=1426350#1426350]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259730 post #4[/url] for its definition). We are going to prove that

[color=green][b](a1)[/b][/color] $\sum_{k = 0}^{n}2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k} = \sum_{k}2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k}$

and that

[color=green][b](a2)[/b][/color] $\sum_{k}2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k} = \binom{2n}{n + u}$

for any integers $n\geq 0$ and $u$.

In fact, [color=green][b](a1)[/b][/color] is easy to prove:

[hide="Proof of (a1)"]

First we show that

[color=green][b](a3)[/b][/color] For every $k\in \mathbb{Z}\setminus \left\{0,1,...,n\right\}$, we have $\binom{n}{2k + u}\binom{2k + u}{k} = 0$.

[i]Proof of [color=green][b](a3)[/b][/color].[/i] Let $k\in \mathbb{Z}\setminus \left\{0,1,...,n\right\}$. For the sake of contradiction, assume that $\binom{n}{2k + u}\binom{2k + u}{k}\neq 0$. Then, $\binom{n}{2k + u}\neq 0$ and $\binom{2k + u}{k}\neq 0$.

From $\binom{n}{2k + u}\neq 0$ and $n\geq 0$, we conclude that $2k + u\geq 0$ and $2k + u\leq n$. From $\binom{2k + u}{k}\neq 0$ and $2k + u\geq 0$, we conclude that $k\geq 0$ and $k\leq 2k + u.$ Now, $k\leq 2k + u$ and $2k + u\leq n$ yield $k\leq n$.

Now, $k\geq 0$ and $k\leq n$ yield $k\in \left\{ 0,1,...,n\right\}$, contradicting $k\in\mathbb{Z}\setminus \left\{ 0,1,...,n\right\}$. This contradiction shows that our assumption $\binom{n}{2k + u}\binom{2k + u}{k}\neq 0$ was wrong. Hence, $\binom{n}{2k + u}\binom{2k + u}{k} = 0$, and [color=green][b](a3)[/b][/color] is proven.

Now, [color=green][b](a1)[/b][/color] follows from

$\sum_{k}2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k}$
$= \sum_{k = 0}^{n}2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k} + \sum_{k\in \mathbb{Z}\setminus \left\{ 0,1,...,n\right\} }\underbrace{2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k}}_{ = 0\text{ by \textbf{(a3)}}}$
$= \sum_{k = 0}^{n}2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k}$.

[/hide]

Let us now prove [color=green][b](a2)[/b][/color] by induction over $n$:

[hide="Induction base"]

[i]Induction base.[/i] For $n = 0$, we have

$\sum_{k} 2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k} = \sum_{k = 0}^{n} 2^{n - 2k - u}\binom{n}{2k + u}\binom{2k + u}{k}$ (by [color=green][b](a1)[/b][/color])
$= \sum_{k = 0}^{0}2^{0 - 2k - u}\binom{0}{2k + u}\binom{2k + u}{k}$ (since $n=0$)
$= 2^{0 - 2\cdot 0 - u}\binom{0}{2\cdot 0 + u}\binom{2\cdot 0 + u}{0} = 2^{ - u}\binom{0}{u}\underbrace{\binom{u}{0}}_{ = 1} = 2^{ - u}\binom{0}{u}$
$= 2^{ - u}\left\{\begin{array}{c}1,\text{ if }u = 0; \\
0,\text{ if }u\neq 0\end{array}\right. = \left\{\begin{array}{c}2^{ - u},\text{ if }u = 0; \\
0,\text{ if }u\neq 0\end{array}\right. = \left\{\begin{array}{c}2^{ - 0},\text{ if }u = 0; \\
0,\text{ if }u\neq 0\end{array}\right.$
$= \left\{\begin{array}{c}1,\text{ if }u = 0; \\
0,\text{ if }u\neq 0\end{array}\right. = \binom{0}{u} = \binom{2\cdot 0}{0 + u} = \binom{2n}{n + u}$ (since $0=n$).

Thus, [color=green][b](a2)[/b][/color] is proven for $n = 0$.

[/hide]

[i]Induction step.[/i] Let $N\geq 0$ be an integer. Assume that [color=green][b](a2)[/b][/color] is proven for $n = N$. We have to prove [color=green][b](a2)[/b][/color] for $n = N + 1$.

Let $U$ be an integer. We have assumed that [color=green][b](a2)[/b][/color] is proven for $n = N$; thus, we can apply it to $n = N$ and every $u\in\mathbb{Z}$.

In particular, we can apply [color=green][b](a2)[/b][/color] to $n = N$ and $u = U$ and get

[color=green][b](a4)[/b][/color] $\sum_{k}2^{N - 2k - U}\binom{N}{2k + U}\binom{2k + U}{k} = \binom{2N}{N + U}$;

also, we can apply [color=green][b](a2)[/b][/color] to $n = N$ and $u = U - 1$ and get

[color=green][b](a5)[/b][/color] $\sum_{k}2^{N - 2k - \left( U - 1\right) }\binom{N}{2k + \left( U - 1\right) }\binom{2k + \left( U - 1\right) }{k} = \binom{2N}{N + \left(U - 1\right) }$;

also, we can apply [color=green][b](a2)[/b][/color] to $n = N$ and $u = U + 1$ and get

[color=green][b](a6)[/b][/color] $\sum_{k}2^{N - 2k - \left( U + 1\right) }\binom{N}{2k + \left( U + 1\right) }\binom{2k + \left( U + 1\right) }{k} = \binom{2N}{N + \left(U + 1\right) }$.

Now,

$\sum_{k}2^{\left( N + 1\right) - 2k - U}\binom{N + 1}{2k + U}\binom{2k + U}{k}$
$= \sum_{k}2^{\left( N + 1\right) - 2k - U}\left( \binom{N}{2k + U} + \binom{N}{2k + U - 1}\right) \binom{2k + U}{k}$
(since $\binom{N + 1}{2k + U} = \binom{N}{2k + U} + \binom{N}{2k + U - 1}$ by the recurrence relation of binomial coefficients)
$= \sum_{k}\underbrace{2^{\left( N + 1\right) - 2k - U}}_{=2 \cdot 2^{N - 2k - U}}\binom{N}{2k + U}\binom{2k + U}{k} + \sum_{k}2^{\left( N + 1\right) - 2k - U}\binom{N}{2k + U - 1}\binom{2k + U}{k}$
$= 2\cdot \sum_{k}2^{N - 2k - U}\binom{N}{2k + U}\binom{2k + U}{k} + \sum_{k}2^{\left(N + 1\right) - 2k - U}\binom{N}{2k + U - 1}\binom{2k + U}{k}$
$= 2\cdot \sum_{k}2^{N - 2k - U}\binom{N}{2k + U}\binom{2k + U}{k}$
$+ \sum_{k}2^{\left( N + 1\right) - 2k - U}\binom{N}{2k + U - 1}\left( \binom{2k + U - 1}{k} + \binom{2k + U - 1}{k - 1}\right)$
(since $\binom{2k + U}{k} = \binom{2k + U - 1}{k} + \binom{2k + U - 1}{k - 1}$ by the recurrence relation of binomial coefficients)
$= 2\cdot \sum_{k}2^{N - 2k - U}\binom{N}{2k + U}\binom{2k + U}{k}$
$+ \sum_{k}2^{\left( N + 1\right) - 2k - U}\binom{N}{2k + U - 1}\binom{2k + U - 1}{k}$
$+ \sum_{k}2^{\left( N + 1\right) - 2k - U}\binom{N}{2k + U - 1}\binom{2k + U - 1}{k - 1}$
$= 2\cdot \sum_{k}2^{N - 2k - U}\binom{N}{2k + U}\binom{2k + U}{k}$
$+ \sum_{k}2^{N - 2k - \left( U - 1\right) }\binom{N}{2k + \left( U - 1\right) }\binom{2k + \left( U - 1\right) }{k}$
$+ \sum_{k}2^{N - 2\left( k - 1\right) - \left( U + 1\right) }\binom{N}{2\left( k - 1\right) + \left( U + 1\right) }\binom{2\left( k - 1\right) + \left( U + 1\right) }{k - 1}$
$= 2\cdot \sum_{k}2^{N - 2k - U}\binom{N}{2k + U}\binom{2k + U}{k}$
$+ \sum_{k}2^{N - 2k - \left( U - 1\right) }\binom{N}{2k + \left( U - 1\right) }\binom{2k + \left( U - 1\right) }{k}$
$+ \sum_{k}2^{N - 2k - \left( U + 1\right) }\binom{N}{2k + \left( U + 1\right) }\binom{2k + \left( U + 1\right) }{k}$
(here we substituted $k$ for $k - 1$ in the last sum)
$= 2\cdot \binom{2N}{N + U} + \binom{2N}{N + \left( U - 1\right) } + \binom{2N}{N + \left( U + 1\right) }$
(by [color=green][b](a4)[/b][/color], [color=green][b](a5)[/b][/color] and [color=green][b](a6)[/b][/color])
$= 2\cdot \binom{2N}{N + U} + \binom{2N}{N + U - 1} + \binom{2N}{N + U + 1}$
$= \left( \binom{2N}{N + U} + \binom{2N}{N + U - 1}\right) + \left( \binom{2N}{N + U + 1} + \binom{2N}{N + U}\right)$
$= \binom{2N + 1}{N + U} + \binom{2N + 1}{N + U + 1}$
(since $\binom{2N}{N + U} + \binom{2N}{N + U - 1} = \binom{2N + 1}{N + U}$ and $\binom{2N}{N + U + 1} + \binom{2N}{N + U} = \binom{2N + 1}{N + U + 1}$ by the recurrence relation of binomial coefficients)
$= \binom{2N + 2}{N + U + 1}$
(by the recurrence relation of binomial coefficients)
$= \binom{2\left( N + 1\right) }{\left( N + 1\right) + U}$.

If we rename $U$ into $u$, this becomes

$\sum_{k}2^{\left( N + 1\right) - 2k - u}\binom{N + 1}{2k + u}\binom{2k + u}{k} = \binom{2\left( N + 1\right) }{\left( N + 1\right) + u}$.

Thus, [color=green][b](a2)[/b][/color] holds for $n = N + 1$.

This completes the induction step. Hence, [color=green][b](a2)[/b][/color] is proven for any integers $n\geq 0$ and $u$.

Combining [color=green][b](a1)[/b][/color] and [color=green][b](a2)[/b][/color], we immediately see that Theorem 2 holds.

  Darij

[hide="Old proof of Theorem 1"]
[i]Proof of Theorem 1.[/i] Consider a table consisting of two rows and n columns. A [i]coloring[/i] of the table will mean a way to write in every cell of the table a 0 or a 1. Such a coloring will be called [i]fair[/i] if the number of 1's in the first row equals to the number of 1's in the second row. How many fair colorings are there?

On the one hand, a coloring is fair if and only if the number of 1's in the first row equals to the number of 1's in the second row, i. e. if the number of 1's in the first row equals n minus the number of 0's in the second row, i. e. if the 1's in the first row and the 0's in the second row are, together, n cells. Thus, in order to get a fair coloring, we have to mark n of the 2n cells of the table; then, the marked cells in the first row will be filled with 1's, the unmarked cells in the first row will be filled with 0's, the marked cells in the second row will be filled with 0's, and the unmarked cells in the second row will be filled with 1's. This way, we get a fair coloring, and conversely, each fair coloring can be obtained uniquely this way. Thus, the number of fair colorings equals to the number of ways to mark n of the 2n cells of the table, and this number is obviously $\binom{2n}{n}$.

On the other hand, let's construct a fair coloring in a different way: In the following, an xy column will mean a column which has the number x in the first row and the number y in the second row (where x and y are numbers from {0; 1}). Then, a coloring can have 00 columns, 11 columns, 01 columns and 10 columns. Obviously, we have:

number of 1's in the first row = number of 11 columns + number of 10 columns;
number of 1's in the second row = number of 11 columns + number of 01 columns.

Thus, the equation

number of 1's in the first row = number of 1's in the second row

holds if and only if

number of 10 columns = number of 01 columns.

In other words, a coloring is fair (i. e. the number of 1's in the first row equals to the number of 1's in the second row) if and only if the number of 10 columns equals to the number of 01 columns. This yields the following construction of a fair coloring:

- Specify an integer k such that $0\leq k\leq n$. This number k is going to be the number of 10 columns, or, equivalently, the number of 01 columns in our coloring (in fact, as we know, in a fair coloring these numbers must be equal).
- Now let's determine which columns will become 10 columns and which ones will become 01 columns. This is done as follows: First choose 2k columns of the table, then take k out of these 2k columns and mark them as 10 columns, and then mark the remaining k of the 2k columns as 01 columns. The number of possibilities to do this is $\binom{n}{2k}\binom{2k}{k}$ (in fact, there are $\binom{n}{2k}$ possibilities to choose 2k of the n columns of the table, and $\binom{2k}{k}$ possibilities to take k out of these 2k columns).
- Finally, we have to decide which of the remaining n - 2k columns become 00 columns and which become 11 columns. This can be decided freely, so there are $2^{n - 2k}$ possibilities (2 for every column).

Altogether, there are $\sum_{k = 0}^n \binom{n}{2k}\binom{2k}{k}\cdot 2^{n - 2k} = \sum_{k = 0}^n 2^{n - 2k}\binom{n}{2k}\binom{2k}{k}$ possibilities to perform such a construction. They all yield different fair colorings, and each fair coloring can be constructed in such a way. So the number of fair colorings must be $\sum_{k = 0}^n 2^{n - 2k}\binom{n}{2k}\binom{2k}{k}$. Comparing this with the number $\binom{2n}{n}$ that we obtained above, we get $\sum_{k = 0}^n 2^{n - 2k}\binom{n}{2k}\binom{2k}{k} = \binom{2n}{n}$, and Theorem 1 is proven.
[/hide]

$ \sum_{k=0}^{n}2^{n-2k}\binom{n}{2k}\binom{2k}{k}$. if you take ,k=n,you can`t choose 2n things from n,you forget some thing in the exercice :!:

If $ n$ is a nonnegative integer, then $ \binom{n}{k}$ is [b]defined[/b] as $ 0$ if $ k<0$ or $ k>n$.

  darij

let 2n = m
(LHS) = constant term of (2+x+1/x)^m
(RHS) = constant term of (root(x) + root(1/x))^2m

so, LHS = RHS.


modular arithmetic
number theory
number theory unsolved

Prove that there does not exist an integer $ n>1$ such that $ n|(3^n\minus{}2^n)$.

A very standard trick is to consider the minimum prime divisor $ p$ of $ n$ and then show that $ (n,p\minus{}1)>1$, reaching a contradiction
If you have troubles with this idea I can post the full solution, but I think it is easy with this trick :)

[u]Solution[/u]. Let $ p$ be the smallest prime divisor of $ n$.  Clearly, $ p\neq 2,3$; otherwise $ 2|3^n\minus{}2^n$ or $ 3|3^n\minus{}2^n$.  Therefore, $ 3^n\equiv 2^n\pmod p\implies (3\cdot 2^{\minus{}1})^n\equiv 1\pmod p$.  Let $ a>1$ be the order of $ 3\cdot 2^{\minus{}1}$ modulo $ p$.  Then $ a|n$.  Along with $ a|p\minus{}1$ by Fermat's Little Theorem, we have $ (n,p\minus{}1)\geq (n,a)>1$, a contradiction to the minimality of $ p$. Ergo, no such $ n>1$ exists. $ \Box$


LaTeX
greatest common divisor
analytic geometry
graphing lines
slope
ratio

I have two questions here. I guess I'm very depending on aops members since I'm asking questions every day :).

Anyway, here is the first one.

Least Squre Regression Line
The least squres regression line, y = ax + b, for the points [tex](x_1,y_1), (x_2, y_2), (x_3, y_3) .... (x_n, y_n)[/tex] is

Um.. They got this complicated sigma but I don't seem to be able to put them in latex or regular form.

I tried googling to get that sigma but nope, no luck.  Anyway, how do I use this thing anyway?

-------
Second question:
Let's say x = 12, y = 10, and z = 24. Then what's z:x:y in simplest form?

Well, I know what that meant but when I tried to do something like that, the answer just don't come out at once.  I got 12:6:5 but well, it came after many trials.  So, is there any easier way to play with this?

Help :P [/tex]

For the least squares I wouldn't bother trying to learn how to use it.  Just ues a graphing calculator.

The second part:

We want to find 24:12:10 in lowest terms.  So we need to take their GCF out of each of them.  In this case the GCF is 2.  Divide them all by 2 to get 12:6:5.

Actually, you would need to know how to derive a least squares regression line in statistics. Of course, I just finished getting a 5 on the AP Statistics test, so naturally, I remember nothing about the class.  :blush:

A little bit of googling says that for given points
[tex]\\{(x_1,y_1),(x_2,y_2),...(x_n,y_n)}
\\ \bar{x}=\frac{1}{n}\displaystyle\sum_{i=1}^{n}x_i
\\ \bar{y}=\frac{1}{n}\displaystyle\sum_{i=1}^{n}y_i[/tex]
Let [tex]s_x=\sum_{i=1}^{n}(x_i-\bar{x})^2,
s_x_y=\sum_{i=1}^{n}(x_i-\bar{x})(y_i-\bar{y})[/tex].

Then for the equation, [tex]\hat{y}=a+bx_i[/tex] where [tex]\hat{y}[/tex] is the best fit y-coordinate for the x-coordinate [tex]x_i, b=\frac{s_y}{s_x},[/tex] and [tex]  a=\bar{y}-b\bar{x}.[/tex]

Basically, in all that gibberish, the slope of the line of best fit is the ratio of the sum of the squares of the differences between the [tex]x[/tex]-values and the mean of the [tex]x[/tex]-values, over the sum of the product of the difference between the values and the mean of the [tex]x[/tex]-values with the difference between the values and the mean of the [tex]y[/tex]-values. Given the slope of the line, calculate the [tex]y[/tex]-intersect for the line of best fit by plugging in the mean values of [tex]y[/tex] and [tex]x[/tex]. Most of the time you will need this, such as on a statistic exam, this formula will be given to you.

Most graphing calculators can already do this. On the TI-83, go to STAT->Edit... and store the x-values in one list, and the y-values in another. Then go to STAT->CALC->8: LinReg(a+bx). You should see LinReg(a+bx) on the main screen, then type in the two lists you stored the values under with the x-list coming first, comma, y-list (Ex. LinReg(a+bx)[tex] L_1,L_2[/tex]) and hit enter. You should then see a as the y-intersect, b as the slope, the coefficent of correlation and its square, depending on whether or not the diagnostics are set on. The last two values help measure the accuracy of the line. Hope that helped.


Find all functions $f: \mathbb N \to \mathbb N$ for which
\[ f(f(m) + f(n)) = m + n\]
holds for all positive integers $m$ and $n$.

[hide]Fix $ m$, and let $ a,b\in\mathbb{N}$ such that $ f(a)\equal{}f(b)$. We have $ f(f(m)\plus{}f(a))\equal{}f(f(m)\plus{}f(b))$, so $ m\plus{}a\equal{}m\plus{}b$ and $ a\equal{}b$. Thus $ f$ is injective. We have $ f(f(m\plus{}1)\plus{}f(n\minus{}1))\equal{}m\plus{}1\plus{}n\minus{}1\equal{}m\plus{}n\equal{}f(f(m)\plus{}f(n))$. Therefore $ f(m\plus{}1)\plus{}f(n\minus{}1)\equal{}f(m)\plus{}f(n)$ or $ f(m\plus{}1)\minus{}f(m)\equal{}f(n)\minus{}f(n\minus{}1)$. Substitute $ m\equal{}x,n\minus{}1\equal{}y$, we get $ f(x\plus{}1)\minus{}f(x)\equal{}f(y\plus{}1)\minus{}f(y)$ for all positive integers $ x,y$. Thus there exists a constant $ k$ such that $ f(x\plus{}1)\minus{}f(x)\equal{}k$, or $ f(x\plus{}1)\equal{}f(x)\plus{}k$. Clearly $ k>0$ (otherwise $ f(x)<0$ for some $ x$ or $ f$ is constant which are both impossible).

$ f(x)\equal{}f(1)\plus{}(x\minus{}1)k$.

$ 3\equal{}f(f(2)\plus{}f(1))\equal{}f(2f(1)\plus{}k)\equal{}f(1)\plus{}(2f(1)\plus{}k\minus{}1)k\equal{}(2k\plus{}1)f(1)\plus{}k^2\minus{}k\ge 2k\plus{}1\plus{}k^2\minus{}k\equal{}k^2\plus{}k\plus{}1$

So $ k\equal{}1$.

$ m\plus{}n\equal{}f(f(m)\plus{}f(n))\equal{}f(2f(1)\plus{}m\plus{}n\minus{}2)\equal{}3f(1)\plus{}m\plus{}n\minus{}3$. Hence $ f(1)\equal{}1$. We get $ f(x)\equal{}1\plus{}(x\minus{}1)1\equal{}x$.

Answer: $ \boxed{f(x)\equal{}x,\forall x\in\mathbb{N}}$[/hide]

[quote="Johan Gunardi"]Fix $ m$, and let $ a,b\in\mathbb{N}$ such that $ f(a) \equal{} f(b)$. We have $ f(f(m) \plus{} f(a)) \equal{} f(f(m) \plus{} f(b))$, so $ m \plus{} a \equal{} m \plus{} b$ and $ a \equal{} b$. Thus $ f$ is injective. We have $ f(f(m \plus{} 1) \plus{} f(n \minus{} 1)) \equal{} m \plus{} 1 \plus{} n \minus{} 1 \equal{} m \plus{} n \equal{} f(f(m) \plus{} f(n))$. Therefore $ f(m \plus{} 1) \plus{} f(n \minus{} 1) \equal{} f(m) \plus{} f(n)$ or $ f(m \plus{} 1) \minus{} f(m) \equal{} f(n) \minus{} f(n \minus{} 1)$. Substitute $ m \equal{} x,n \minus{} 1 \equal{} y$, we get $ f(x \plus{} 1) \minus{} f(x) \equal{} f(y \plus{} 1) \minus{} f(y)$ for all positive integers $ x,y$. Thus there exists a constant $ k$ such that $ f(x \plus{} 1) \minus{} f(x) \equal{} k$, or $ f(x \plus{} 1) \equal{} f(x) \plus{} k$. Clearly $ k > 0$ (otherwise $ f(x) < 0$ for some $ x$ or $ f$ is constant which are both impossible).

$ f(x) \equal{} f(1) \plus{} (x \minus{} 1)k$.

$ 3 \equal{} f(f(2) \plus{} f(1)) \equal{} f(2f(1) \plus{} k) \equal{} f(1) \plus{} (2f(1) \plus{} k \minus{} 1)k \equal{} (2k \plus{} 1)f(1) \plus{} k^2 \minus{} k\ge 2k \plus{} 1 \plus{} k^2 \minus{} k \equal{} k^2 \plus{} k \plus{} 1$

So $ k \equal{} 1$.

$ m \plus{} n \equal{} f(f(m) \plus{} f(n)) \equal{} f(2f(1) \plus{} m \plus{} n \minus{} 2) \equal{} 3f(1) \plus{} m \plus{} n \minus{} 3$. Hence $ f(1) \equal{} 1$. We get $ f(x) \equal{} 1 \plus{} (x \minus{} 1)1 \equal{} x$.

Answer: $ \boxed{f(x) \equal{} x,\forall x\in\mathbb{N}}$[/quote]

Quite OK, but you should be a little bit quicker in the end : from $ f(x) \equal{} f(1) \plus{} (x \minus{} 1)k$, you get $ f(x) \equal{} ax \plus{} b$ and you just have to plug back in the original equation to obtain $ a^2 \equal{} 1$ and $ b \equal{} 0$ and so $ f(x) \equal{} x$


linear algebra
matrix
linear algebra unsolved

Take $AB=-BA$ and left-multiply by $A:$

$B=-ABA$

Since $A=A^{-1},$ that shows that $B$ is similar to $-B.$

Similar matrices have the same traces, so

$\text{trace}(B)=\text{trace}(-B)=-\text{trace}(B).$

Hence $\text{trace}(B)=0$.

Right-multiply by $B$ to get $A=-BAB$ and draw a similar conclusion about $A.$

Sayeh sent me a PM but it makes more sense to me to reply in the open:

[quote="Sayeh"]Dear Dr. Marryfield,

Thanks for your answer. But I have 2 problems. First you have written:
B=-ABA and since A=A^-1 then B=-B. How can we say that? Isn't it true that ABA is not equal to BAA?[/quote]
I didn't say that $B$ was equal to $-B;$ I said that $B$ was similar to $-B.$ $X$ is similar to $Y$ iff $\exists P$ such that $P^{-1}XP=Y.$ Then $\text{trace}(P^{-1}XP)= \text{trace}(XPP^{-1})=\text{trace}X.$ Similar matrices have the same trace.

[quote]The other problem is that I think this question is wrong! When A.B= -B.A, A^2 and B^2 could not be equal to I. Let me explain why:[/quote]
I didn't look all that closely at your explanation because this is very much possible.

Let $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ and $B=\begin{bmatrix}0&1\\1&0\end{bmatrix}.$

Do the computations: $A^{2}=B^{2}=I$ and $AB=-BA.$


For $ \forall$ $ n \in \mathbb{Z}$ and $ n>0$

we have: $ 0<a_{n}<a_{2n}\plus{}a_{2n\plus{}1}$

show that: $ \sum_{n\equal{}1}^{\infty}a_{n}$ is divergent

It seems to use Cauchy convergence principle can work out it..... :D

Write this as $ \sum_{k\equal{}0}^{\infty}\sum_{n\equal{}2^k}^{2^{k\plus{}1}\minus{}1}a_n.$

(We're simply putting in parentheses in an appropriate way.)

Let $ A_k\equal{}\sum_{n\equal{}2^k}^{2^{k\plus{}1}\minus{}1}a_n$ so that our whole sum is $ \sum_{k\equal{}0}^{\infty}A_k.$

But $ A_k\equal{}\sum_{n\equal{}2^k}^{2^{k\plus{}1}\minus{}1}a_n<\sum_{n\equal{}2^k}^{2^{k\plus{}1}\minus{}1}(a_{2n}\plus{}a_{2n\plus{}1})\equal{}\sum_{m\equal{}2\cdot 2^k}^{2\cdot 2^{k\plus{}1}\minus{}1}a_m\equal{}A_{k\plus{}1}.$

So $ 0<A_k<A_{k\plus{}1}$ and $ \sum_{k\equal{}0}^{\infty}A_k$ diverges because its terms don't go to zero.

[quote="Kent Merryfield"]

 $ \sum_{n \equal{} 2^k}^{2^{k \plus{} 1} \minus{} 1}(a_{2n} \plus{} a_{2n \plus{} 1}) \equal{} \sum_{m \equal{} 2\cdot 2^k}^{2\cdot 2^{k \plus{} 1} \minus{} 1}a_m \equal{} A_{k \plus{} 1}.$

[/quote]

Thank you very much !

Excellent !

$ \sum_{n \equal{} 2^k}^{2^{k \plus{} 1} \minus{} 1}(a_{2n} \plus{} a_{2n \plus{} 1}) \equal{} \sum_{m \equal{} 2\cdot 2^k}^{2\cdot 2^{k \plus{} 1} \minus{} 1}a_m \equal{} A_{k \plus{} 1}.$

Elegant and not visual.... :D


function
induction
quadratics
logarithms
algebra unsolved
algebra

Find all continuous functions $f\: (0, \infty) \to (0,\infty)$, for which
\[ f(x)f(y)=f(xy)+f(x/y) \] for any positive $x$ and $y$.

Well, yes, this is quite challenging! We have $ f^2(x)=f(x^2)+2$. Thus, $ f(x)\geq\sqrt{2}$ for all x. Thus, $ f(x)\geq\sqrt{2+\sqrt{2}}$ for all x and an easy induction and a limit process shows that $ f(x)\geq 2$ for all x. Substitute $ x$ by $x^2$, y by $y^2$ and use the above relation to find that $ f(xy)f(\frac{x}{y})+4=f^2(x)+f^2(y) $. Use the initial relation and eliminate $f(\frac{x}{y})$ between the two relations to find that $ f^2(x)+f^2(y)+f^2(xy)=4+f(x)f(y)f(xy)$. Now since $ f(x)\geq 2$ for all x we find that there exists a function g such that $ g(x)\geq 1$ for all x and $ f(x)=g(x)+\frac{1}{g(x)}$ for all x. From the above relation we also find that $ g(xy)=g(x)g(y) $ for all x and y and since g is continuous we have $g(x)=x^k$ for a certain k. Therefore $ f(x)=x^k+\frac{1}{x^k} $ for all x, and these functions verify.

You got $g(xy)=g(x)g(y)$ from the condition $g(x)\geq 1$ (I checked - it is important), but $g(x)=x^{k}$ doesn't satisfy the inequality.

Indeed, that's quite strange. Anyway, I don't think that changes a lot, because we still now that $g(x)$ must be $x^{k}$ (we work with implications and not equivalences), so $f(x)=x^{k}+1/x^{k}$, which is solution to the equation. Am I wrong again, as usual?

Actually, if $f(x)=x+1/x$ then your function $g(x)=x$ for $x\geq 1$ and $g(x)=1/x$ for $x<1$. It means, that all implications you get must be satisfied with this function. I just want to find mistake wich is present in your proof.

I found the mistake: we cannot deduce that $g(xy)=g(x)g(y)$ because it may happen that $f(xy=g(x)/g(y)+g(y)/g(x)$ when solving the quadratic equation. Sorry.

Sadly that some solutions with mistakes are unchecked for years.

Sadly that instead of coming up with new ideas we start searching the past for revenge...

You make me laughting. I am cleaning up all problems in this section without distinction.

Ok, because Myth is too occupied searching the past and because this deserves however a solution, here is a possible way to obtain a solution for this problem. As I said, the relation $f^{2}(x)+f^{2}(y)+f^{2}(xy)=4+f(x)f(y)f(xy)$ combined with $f(x)=g(x)+\frac{1}{g(x)}$ does not imply that $g(xy)=g(x)g(y)$, but rather that $g(xy)=g(x)g(y)$ or $g(xy)=\frac{\max(g(x),g(y))}{\min(g(x),g(y)}$. The initial relation shows that $f(x/y)=f(y/x)$, so $f(1/x)=f(x)$ and so $g(1/x)=g(x)$. Let $h$ be defined on $R$ by $h(x)=lng(e^{x})$. Then $h$ is nonnegative, continuous and even and also $h(x+y)=h(x)+h(y)$ or $h(x+y)=|h(x)-h(y)|$ for all $x,y$.

   Lemma 1: if $h$ does not vanish on $(0,\infty)$, then it is injective on this interval.

   Suppose otherwise that $x>y>0$ are such that $h(x)=h(y)$. Because $h(x+y)\ne 0$, we must have $h(x+y)=2h(x)$. Because $h(x-y)\ne 0$ and $h(x)=h(y)+h(x-y)$ or $h(x)=|h(y)-h(x-y)|$, it follows that $h(x-y)=2h(x)$. Also, because $h(2x)\ne 0$, we must have $h(2x)=2h(x)$. Thus $h(2x)=h(x-y)$. Applying the same argument, but this time with $2x>x-y>0$ we find that $h(x+y)=h(2x-(x-y))=2h(2x)=4h(x)$ and so $4h(x)=2h(x)$ and $h(x)=0$, contradiction.

    Lemma 2: $h$ does not vanish on $(0,\infty)$ if it is non the zero function.

    Suppose that $z>0$ is a zero of $h$ and note that $z$ is a period of $h$. Because $h$ is continuous nonzero, it cannot have two periods whose quotient is irrational. Also, because $h$ is continuous and periodical, it is bounded. We know that for all $x$ we have $h(2x)=0$ or $h(2x)=2h(x)$. By the previous arguments, this implies $h(2^{n}\sqrt{2}z)=2h(2^{n-1}\sqrt{2}z)$ and so $h(2^{n}\sqrt{2}z)=2^{n}h(z\sqrt{2})$. Because $h$ is bounded, it follows that $h(z\sqrt{2})=0$ and so $z\sqrt{2}$ is also a period of $h$, contradiction.

    Ok, by lemma 1 and 2 we know that $h$ is injective on the set $A$ of positive numbers and by continuity it is strictly monotonous. Because it is always greater than or equal to $h(0)=0$, it must be increasing. Thus $h(x+y)=h(x)+h(y)$ for $x,y\in A$ and by continuity $h$ is linear on $A$. Because $h$ is even, $h(x)=a|x|$ and we are done.

    I hope the new antivirus Myth will kindly point me out another mistake.

[quote="harazi"]I hope the new antivirus Myth will kindly point me out another mistake.[/quote] No doubts

I am tired searching numerous mistakes in various topics.

The official solution goes as follows.
1) They prove $f(x)\geq 2$ and then consider the same function $g(x)\geq 1$.
2) Substitute $g$ into $f^{2}(x)=f(x^{2})+2$. It follows that $g(x^{2})=g^{2}(x)$.
3) Using induction and the initial equation they immediately get $g(x^{n})=g^{n}(x)$. It also implies that $g(x^{a})=g^{a}(x)$ holds for all rational $a$, and consequently for all real $a$.
4) Finally, $g(x)=g(e^{\ln x})=g^{\ln x}(e)=x^{\alpha}$.

You may see that authors made a similar mistake. Actually, $g(x^{a})=g^{a}(x)$ is true for all $a\geq 0$. So $g(x)=x^{\alpha}$ for $x\geq 1$ and $g(x)=x^{\beta}$ for $x<1$. The rest is clear.

I am tired and bored by this Myth-type useless remarks...

[quote="orl"]Find all continuous functions $f\: (0, \infty) \to (0,\infty)$, for which \[f(x)f(y)=f(xy)+f(x/y)\] for any positive $x$ and $y$.[/quote]
1. x=y=1 give f(1)=2, and one solution $f(x)=2 \forall x\in R_{+}$.
2. x=y give $f(x)^{2}=f(x^{2})+2$. Therefore $f(x)=\sqrt{2+f(x^{2})}=\sqrt{2+\sqrt{2+f(x^{4})}=...\ge 2.}$
3. x=1 give $f(y)=f(\frac{1}{y})$.
Therefore exist g(x): $f(x)=g(x)+\frac{1}{g(x)}, g(\frac{1}{x})=\frac{1}{g(x)}, g(x^{2})=g(x)^{2}.$ Let $u_{n}=g(y^{n})$, from $x=y^{n}$ we have
$(u_{n}+u_{n}^{-1})(u_{1}+u_{1}^{-1})=u_{n+1}+u_{n+1}^{-1}+u_{n-1}+u_{n-1}^{-1}$. By induction $u_{n}=u_{1}^{n}$, therefore exist u, suth that $f(x)=u^{\ln(x)}+u^{-\ln(x)}\ \forall x\in Q$. Because f(x) continiosly it is true for all x. If f(x) can be write $f(x)=x^{c}+x^{-c}.$

[quote="Myth"]Are you sure $g(1/x)=1/g(x)$? And why post actually the same solution?[/quote]
$g(1/x)=g(x)$ or $g(1/x)=1/g(x)$. We can choose g(1/x)=1/g(x) (it is our chose, and it give same f(x)=g(x)+1/g(x)).
I don't read another solutions. From harazi post I undestand, that are not completely solution.

It's your problem you don't read other solutions ( what better proof of limitation?...), but why posting exactly the same thing, just for fun? By the way, you cannot really make choices as you wish by keeping the function continuous...

[quote="harazi"]you cannot really make choices as you wish by keeping the function continuous...[/quote]
Why?
I can consider only $x\ge 1$ find $g(x)=x^{c}$ and define $g(x)=x^{c}=\frac{1}{(1/x)^{c}}$ for x<1. We can chose g(x), suth that g(x) is not continiosly but $f(x)=g(x)+\frac{1}{g(x)}$ continiosly.


inequalities
algorithm
geometry
perimeter
quadratics
function
linear algebra

Hi there,

Today we were assigned a sheet of questions in Math History class, and there have been 4 that I haven't been able to manage. The questions might appear quite basic for such a high-level, but I'm stuck on them. As it is a historical math course, we are expected to use classic techniques, good old algebra/trigonometry/geometry to solve them, not brute force. Anyway, here are the 4 I haven't been able to do:

[quote]1. Suppose a rectangle has area A and perimeter P. Find formulas for the length and width of the rectangle (in terms of A and P). What restrictions are there on A and P?[/quote]

I've been struggling to get it purely in terms of A and P, I'm always left with either an x or a y.

[quote]2. If a and b are positive numbers, prove that $ \sqrt {a}\sqrt {b} \leq \frac {a \plus{} b}{2}$. What can be said about a and b when equality occurs?[/quote]

I can't get this inequality, I tried the Sandwich Rule but didn't get anywhere. I've deduced that for equality to occur, a and b must be equal (I think).

[quote]3. If $ a \equal{} \sqrt {2}$ and b=1.4142135623, which of the following is true: $ a < b, a \equal{} b, a > b$? (This is not a calculator problem)[/quote]

Well the answer is clearly a>b, but I don't know how to show this. I tried multiplying b through by 10,000,000,000 so I could write it as a surd but that didn't really help I don't think.

[quote]4. What type of curve is the graph of the equation $ x^2 \plus{} xy \plus{} y^2 \equal{} 1$? Find at least 6 points on this curve and make a sketch[/quote]

I'm pretty hopeless with graphs and such. The only points I've figured out are the trivial ones, (0,1) and (1,0).

Many thanks for any assistance!

OK I solved 3 by doing LOOOOOOOONG multiplication. I showed that $ b^2$ = 1.99999999979325598129 < 2

For the second one, it's equivalent with $ (\sqrt{a} \minus{} \sqrt{b})^2 \geq 0.$  :)

I'm really interested in the solution to number three, since I haven't been able to come up with anything interesting. You can compare $ \frac{b}{2}$ and $ \frac{1}{b}$ and show $ \frac{b}{2} < \frac{1}{b} \iff b^2 < 2 \iff b < a$. $ \frac{b}{2}$ is easy to calculate, but $ \frac{1}{b}$ is not at all.. If I were to do the problem on paper I would probably just find $ \sqrt{2}$, and then compare it to $ b$..

You can use the algorithm to calculate square roots by hand [url]http://www.wikihow.com/Calculate-a-Square-Root-by-Hand[/url].

For #1, the sides $ x$ and $ y$ add up to half the perimeter, or $ \frac{P}{2}$, and they multiply to $ A$. Consider the quadratic with roots x and y, which is $ (n\minus{}x)(n\minus{}y)\equal{}0$, or $ n^2\minus{}\frac{P}{2}n\plus{}A\equal{}0$. Solving this quadratic, we get $ x$ and $ y$ are
$ \frac {P/2 \plus{} \sqrt {\frac {P^2}{4} \minus{} 4A}}{2}$and$ \frac {P/2 \minus{} \sqrt {\frac {P^2}{4} \minus{} 4A}}{2}$. We must have $ P^2 > 16 A$.

[quote="dgreenb801"]You can use the algorithm to calculate square roots by hand [url]http://www.wikihow.com/Calculate-a-Square-Root-by-Hand[/url].[/quote]
That takes more time than calculating $ B^2$.

For two, I sort of worked backwards:

$ 0\leq (a \minus{} b)^2$
$ 0\leq a^2 \minus{} 2ab \plus{} b^2$
$ 2ab\leq a^2 \plus{} b^2$
$ 4ab\leq a^2 \plus{} 2ab \plus{} b^2$
$ 4ab\leq (a \plus{} b)^2$
$ \sqrt {ab}\leq \frac {a \plus{} b}{2}$

To show that equality holds only for $ a \equal{} b$, just do that exact proof in reverse order.

For 4), you can solve that quadratic for y as a function of x (will get two functions) and take a few x values to get an idea for the y values. Here is a picture, assuming I can upload it...


I was retaking the 07-08 School Round for practice today, and I made a few (careless) mistakes.  However, I do see one problem that I consider actually to be challenging.
As the title said, this problem that I am referring to is #9 on the Team Round.

Here's the link for the picture and the problem's text:
http://www.mathcounts.org/competition/2008%20School%20Team%20Round.pdf

Here's the actual problem's text, but without a picture:
"A wood, rectangular box, with no top, is constructed with five panels that are each one centimeter thick. The external dimensions of the base are 16 cm by 24 cm, and the external height is 10 cm. When the box is totally submersed in paint, how many square cm of painted surface will there be?"

[hide]The answer is 2216, but I keep getting 2220 as my answer.[/hide]

[quote]Did you overcount the little 1x1 squares on the corners on the top surface?[/quote]

Come to think of it, I did!  I hate problems like this...


Hi everybody,

Is there any adjective in English language which one can apply it instead of "necessary and sufficient". Mainly for the expression "a necessary and sufficient condition".

"A" being both a necessary and a sufficient condition of "B" (in logic) is the same as saying "A if and only if B," "A iff B," or "A <-> B." This is also called material equivalence.

Here is a link that explains further:

[url=http://www.sfu.ca/philosophy/swartz/conditions1.htm]Necessary and Sufficient Conditions[/url]

Source: http://thesaurus.reference.com/browse/necessary%20

Main Entry:   necessary
Part of Speech:   adjective
Synonyms:   alimental, compulsory, contingent, crucial, de rigueur, essential, fundamental, imperative, important, incumbent, indispensable, indispensableinevitable, inevitable, inexorable, inherent, integral, intrinsic, mandatory, needed, needful, obligatory, paramount, perforce, prerequisite, required, requisite, statutory, unavoidable, unpreventable, vital
Antonyms:   casual, discretional, needless, nonessential, optional, unnecessary

Source: http://thesaurus.reference.com/browse/sufficient

Main Entry:   sufficient
Part of Speech:   adjective
Definition:   Being what is needed without being in excess.
Synonyms:   adequate, comfortable, competent, decent, enough, satisfactory

there were a lot more synanyms on thesaurs.com than i though there would be

[quote="myyellowducky82"][quote="isabella2296"]How about, "essential"?[/quote]
Izzy is right.[/quote]

no, that's a synonym for "necessary", not "necessary and sufficient"

yo, RussianRocket, think a bit more before you type, kay

in all of your posts I've seen so far, your response is at best tangentially related and usually not of interest. I know you're trying to contribute -- just put more thought into it

"Necessary and sufficient" is one of those characteristically precise terms that only mathematicians use, so I doubt it.  Linguistically, it's rare to need to talk about a necessary and sufficient condition - usually we talk about causality instead ("if it rains, I will not go to the park," but I may not go to the park for other reasons as well).  In other words, most of the sets we refer to in natural speech by different qualifiers (the set of rainy situations, the set of situations where I won't go to the park) are very rarely identical.

That having been said, "the two conditions are equivalent" works fine depending on how it fits the flow of a paper.

[quote="MysticTerminator"]yo, RussianRocket, think a bit more before you type, kay

in all of your posts I've seen so far, your response is at best tangentially related and usually not of interest. I know you're trying to contribute -- just put more thought into it[/quote]

@QuantumTiger: Um, I think I'd have to side with MysticTerminator on this one... his response was clear and concise, and I think your post describes itself pretty well  :wink:  (in addition to MysticTerminator's post and this post  :)  )

And I think "adequate" only covers the "sufficient" part.

Yes, leoxnlin, "adequate" does not cover "necessary".

[quote="QuantumTiger"][quote="alton66"]Ummm yeah.... this is a MATH forum. might wanna ask a dictionary or something. :|[/quote]
 Ummm yeah.... this is the LINGUISTICS subforum...might wanna read the title before you randomly post here.

@Mystic - please refrain from making posts consisting of "no" and reprimanding others in the way you did in your posts. It's clear that you're trying to help, but please try to modify your method.

@the actual question - adequate, perhaps?[/quote]

[i]no[/i]

equivalent.

and lol, you're trying to correct a 2-year-old post by someone who knows what they're talking about (and probably has more credentials than you will for a long long long time)? oops?

[quote="alton66"]Ummm yeah.... this is a MATH forum. might wanna ask a dictionary or something. :|[/quote]
Well, this is the linguistics section, so it's actually quite relevant.

[quote=alton66]Ummm yeah.... this is a MATH forum. might wanna ask a dictionary or something. :|[/quote]

HUH?????


I need help on how to prove huron's formula.

I can't prove it.

help

You mean Heron's Formula.  Huron is the name of one of the Great Lakes.

See [url=http://en.wikipedia.org/wiki/Heron's_formula#Proof]here[/url] for both an algebraic version, and one using the Pythagorean theorem.  The algebraic one is rather tedious.


ratio
trigonometry
geometry
incenter
angle bisector
geometry proposed

($\Leftarrow$)

assume that $MN \| UV$,we know that $AM=AN$ hence we get that $AU=AV$...

now point $A$ and points $U,V$ are completely similar on $w$ respect to $BC$...so the intersection of two similar lines
$AU,AV$ with line $BC$ must give us two similar points respect to $BC$,so $X,Y$ are similar on $BC$ so we get that $AX=AY$

hence we get that $XM=YN$ and $MU=NV$ so the result follows easily...  :wink:

($\Rightarrow$)

assume that $\frac{MX}{MU}=\frac{NY}{NV}$,assume that $CN$ intersects the circle at $T$,we know that $AN=AB=AC$ so
we get that $\angle ANC=\angle ACN$ so we get that $TB=TC$ so $T$ is the midpoint of arc $BC$...

so we get that $AO,CN,BM$ are concurrent at the midpoint of arc $BC$...

but with $\sin$law in $\triangle BMX,\triangle BMU$ we get that $\frac{MX}{MU}=\frac{\sin MBX}{\sin MXB}.\frac{\sin BUM}{\sin MBU}$ and also in a similar way we get that $\frac{NY}{NV}=\frac{\sin NCY}{\sin NCY}.\frac{\sin NVC}{\sin NCV}$

so we get that $\sin CYV=\sin BXU$ but $\angle CYV,\angle BXU<90$ so we get that $\angle CYV=\angle BXU$ so we get
that the arcs $BU,CV$ are equal so $AU=AV$ and we had $AM=AN$ so we get that $MN\|UV$

and also you can combine these two solutions to give this general solution:

first we prove that $CN,BM,AO$ are concurrent at a point on circle $w$,then with $\sin$ law in $\triangle BMX,\triangle BMU$ we
get that $\frac{MX}{MU}=\frac{\sin MBX.\sin BMU}{\sin MXB.\sin MBU}$ and with $\sin$ law in $\triangle CNY,\triangle CNV$ we get that $\frac{NY}{NV}=\frac{\sin YCN.\sin CVN}{\sin CYN.\sin NCV}$

so we have:

$\frac{MX}{MU}=\frac{NY}{NV}\Leftrightarrow \sin CYV=\sin BXU \Leftrightarrow \angle CYV=\angle BXU \Leftrightarrow CV=BU \Leftrightarrow AV=AU \Leftrightarrow MN\|UV$

so the problem is solved...

[quote="BaBaK Ghalebi"]

 so we get that $TB=TC$ so $T$ is the midpoint of arc $BC$...

so we get that $AO,CN,BM$ are concurrent at the midpoint of arc $BC$...

[/quote]

 :maybe: Why? I can not see why the three line are concurrent on the circle...

we dont need that...

assume that $CN$ intersects the circle at $T$ then just note that from $\angle ANC=\angle ACN$ we get that $BT=TV$ so
$CN$ is the angle bisector of $\angle YCV$ (because of $AB=AC$) so we get that $\frac{NY}{NV}=\frac{CY}{CV}=\frac{\sin CVY}{\sin CYV}$

and also $\frac{MX}{MU}=\frac{BX}{BU}=\frac{\sin BUX}{\sin BXU}$

so we get that:

$\frac{MX}{MU}=\frac{NY}{NV}\Leftrightarrow \frac{\sin BUX}{\sin BXU}=\frac{\sin CVY}{\sin CYV}$
$\Leftrightarrow \sin BXU=\sin CYV \Leftrightarrow BU=CV \Leftrightarrow AU=AV \Leftrightarrow MN\|AU$

[quote="Virgil Nicula"][color=darkred]Let $w=C(O)$ be a circle with the diameter $[BC]$ and let $A\in w$ be a point for which $AO\perp BC$.
For two points $\{\begin{array}{c}X\in (OB)\\\ Y\in (OC)\end{array}$ denote : the second intersections $U$, $V$ of the circle $w$ with the lines $AX$, $AY$ respectively ;
the points $M\in (XU)$, $N\in (YV)$ for which $AM=AN=AB$. Prove that $\frac{MX}{MU}=\frac{NY}{NV}$ $\Longleftrightarrow$ $MN\parallel UV$. [/color][/quote]
[color=darkblue][b]Proof.[/b] The points $M$, $N$ are [b]the incenters[/b] of the triangles $BUC$, $BVC$ respectively because $AM=AN=AB=AC$.

Thus, $\frac{MX}{MU}=\frac{NY}{NV}$ $\Longleftrightarrow$ $\frac{BC}{UB+UC}=\frac{BC}{VB+VC}$ $\Longleftrightarrow$ $UB+UC=VB+VC$. Apply [b]the Ptolemeu's theorem[/b] for $ABUC$, $ABVC$ :

$\{\begin{array}{c}UB\cdot AC+UC\cdot AB=AU\cdot BC\\\ VB\cdot AC+VC\cdot AB=AV\cdot BC\\\ (\ AB=AC\ )\\\ (\ UB+UC=VB+VC\ )\end{array}\|$ $\Longrightarrow$ $AU=AV$ $\implies$ $UV\parallel BC$ $\implies$ $MN\parallel UV$. Reciprocally is evidently.[/color]


Anyone have recommendations for a good AP Computer Science AB textbook?

Supplementary materials and review book reviews are also appreciated.

Thanks.

hello, :)

i have not taken Computer Science AB test,
but my friend has.

he recommends me Princeton Review's book and Barron's.

if you finish and go over those two books, you will get 5 points out of 5.

i also read those books, and i think they are really good.

good luck!

Yea, but those are review books, not text books.

And I liked the book Java Programming: From the Beginning, but that only covers the A material.


Hi, I'm currently a freshman in college, majoring in math.

My school has a rule that I need to take two physical science courses. I've taken a semester of Physics, and I have another physical science credit from high school dual enrollment, so I am pretty much done with that. I could take Physics 2, but honestly, I don't like the subject at all, and would rather not. I am not all that great at Physics, and I have no desire in continuing, but I feel that as a math major, it might be wise. (considering I would like to pursue a non-academia career)

Any suggestions? I could possibly take a computer science course instead. Anything but physics.

Amit?

I'd take computer science, but then again I AM two years younger than you are, so my opinion probably doesn't count.

Learning a modern programming language (Java, C++, Visual Basic,...) is essential for any math career.  The equivalent of six semester hours of study in the chosen language is the minimum I'd recommend.  Additional CS courses in data structures, database design, systems analysis & design would be beneficial as well.

Physics is important for those working with engineers or physical scientists in research labs.  On the other hand, there are plenty of math jobs/careers that require no knowledge of Physics.  If you're not sure what you want to do with your math degree, and you're sure that Physics isn't for you (not just a bad experience with a poor instructor), try some other applied areas.

However, many math majors end up choosing a career in an applied area.  Applied courses are a good idea for folks who are not committed to theoretical math.

Pick something you like- math is closer to being self-contained than any other scientific discipline. Everything else uses math, but you don't need all of the other areas to do math, and therefore this requirement is merely there to broaden your education.

If you choose an applied field, you may have to go back and take courses in that application- but that can be when you make that choice.


find all the roots of the given equation :
 $ x^{3}\plus{}4x^{2}\plus{}x\plus{}6\equal{}0$

[quote="Anandiscool"]find all the roots of the given equation :
 $ x^{3} \plus{} 4x^{2} \plus{} x \plus{} 6 \equal{} 0$[/quote]

Let $ x\equal{}\frac{y\minus{}4}3$. The equation becomes $ y^3\minus{}39x\plus{}254\equal{}0$

Cardan's method gives then the unique real root of this equation $ u\equal{}\sqrt[3]{18\sqrt{43}\minus{}127}\minus{}\sqrt[3]{18\sqrt{43}\plus{}127}$ so the real root of $ x^3\plus{}4x^2\plus{}x\plus{}6\equal{}0$ is $ v\equal{}\frac{u\minus{}4}3$

Then, writing $ x^3\plus{}4x^2\plus{}x\plus{}6\equal{}(x\minus{}v)(x^2\plus{}(v\plus{}4)x\plus{}v^2\plus{}4v\plus{}1)$, we get the three roots :

$ v\equal{}\frac 13 \sqrt[3]{18\sqrt{43}\minus{}127}\minus{}\frac 13 \sqrt[3]{18\sqrt{43}\plus{}127}\minus{}\frac 43$

$ z_1\equal{}\minus{}2\minus{}\frac v2\minus{}i\frac{\sqrt{3v^2\plus{}8v\minus{}12}}2$

$ z_2\equal{}\overline{z_1}\equal{}\minus{}2\minus{}\frac v2\plus{}i\frac{\sqrt{3v^2\plus{}8v\minus{}12}}2$

Hello Anandiscool!

Since you posted in this forum, it means you have a solution to this problem.

Since your subtitle was "easy", I think you have an easier solution than mine.

Could you, please, kindly give us your solution ?

And, btw, could you kindly also answer to the very simple questions I asked you in http://www.mathlinks.ro/Forum/viewtopic.php?t=295357

I have solved this by the same cardan's method in decimals.

[quote="Anandiscool"]I asked that question bcoz I wanted to know that whether there exist any method to find A or not ?
And I don't want to be bounded by any question rather want to explore my ideas and thats why I have joined this great site bcoz I know you all are brilliant in mathematics and this is the place where I can get proper solution. And main thing is that to a single question, many geniuses give their own method to tackle a particular problem which is very hard to find from elsewhere.

And I thought that you all will enjoy this question.

If my this question hurt you then sorry.
I will never post any modified question in the future.[/quote]

Dont hurry and keep cool.

You can post what you want but try to be clear enough.
You posted in an "Algebra proposed & own problem" forum. So it means you know the answer and ask for other answers in order to compare.
You used a subtitle "easy", so we all thought - and looked for - a clever interesting solution, rather easy.

I personnaly did not find and use the terrific ugly Cardan method to find the result. So I just was interested in your own. That's all.

It seems to me rather fair to give help to users if you ask for help.
And I think rather unfair that you keep your easy solution of this problem for yourself.   :(

[quote="Dimitris X"][quote="pco"][quote="Anandiscool"]find all the roots of the given equation :
 $ x^{3} \plus{} 4x^{2} \plus{} x \plus{} 6 \equal{} 0$[/quote]

Let $ x \equal{} \frac {y \minus{} 4}3$. The equation becomes $ y^3 \minus{} 39x \plus{} 254 \equal{} 0$

[/quote]
How did you think this clever substitution [b]pco[/b]?[/quote]

I know that to extinguish the second term of a cubic equation we use the transformation $ x\equal{}y\minus{}\frac{a_{n\minus{}1}}{na_{n}}$ but Patrick here uses the $ x\equal{}\frac{y\minus{}a_{n\minus{}1}}{na_{n}}$ that also works too.

To Patrick--> I think you have a typo in the above equation. That is, $ \minus{}39x$ should be replaced by $ \minus{}39y$

@Γιώργος : Yes, you're right about the typo error $ 39x\to39y$

@Dimitris X : I first use the normal transformation $ x\equal{}z\minus{}\frac 43$ as suggested Γιώργος and i got $ z^3\minus{}\frac {13}3z\plus{}\frac{254}{27}\equal{}0$ and I then used a new transformation $ z\equal{}\frac y3$ in order to get the prettier equation $ y^3\minus{}39y\plus{}254\equal{}0$


function
induction
strong induction
combinatorics unsolved
combinatorics

A convex polygon P with n sides.P is partioned into triangles with diagonals which are not intersect pairwise.
In each move with  two triangles ABD and BCD ,we change into ABC and ACD.Let P(n) be the minimum of moves
such that begin with any partion we can get any other partion.Prove that
1) P(n) is not less than n-3 and P(n) is not more than 2n-7
2) P(n) is not more than 2n-10 (for n>12)

Really interesting problem.  I really enjoyed this one.

Okay, first notice that the number of diagnols needed to divide a $n$-gon up into disjoint triangles is $n-3$.  This can be proved by a very simple induction.  Also, for any vertex $P_{i}$, let $deg(P_{i})$ be the number of diagnols that have $P_{i}$ as one of its endpoints.

Now, let us start with the lower bound.  To show $P(n)\ge n-3$, consider the following two partitions: let the partition $\mathcal{P}=P_{1}P_{2}\cdots P_{n}$ be divided so that $P_{1}P_{k}$ exists for all $3\le k\le n-1$.  Define the other partition $\mathcal{P'}=P_{1}\cdots P_{n}$ to be divided so that $P_{n}P_{k}$ exists for all $2\le k\le n-2$.  Note that for each move, the degree for each vertex can only increase by at most $1$.  Thus because $deg(P_{n})=0$ in $\mathcal{P}$ and $deg(P_{n})=n-3$ in $\mathcal{P'}$, we know that we need at least $n-3$ moves.  This shows that for all $n\ge 3$, we have that $P(n)\ge n-3$.

To prove the upper bounds, we define the function $F(n,P_{i})$ to be the least number of moves required to orientate all the $n-3$ diagnols of a divided $n$-gon to have one of their endpoints as $P_{i}$.  I claim the following as a lemma:

[b]Lemma:[/b] For any arbitrary divided $n$-gon, $F(n,P_{i})=n-3-deg(P_{i})$.

We proceed to prove this by strong induction on $n$.  The statement is obviously true for $n=3$ (and $n=4$ for that matter).  Suppose that it is true for all integers less than $n$.  Let the divided $n$-gon $\mathcal{P}=P_{1}P_{2}\cdots P_{n}$.  W.l.o.g. let us examine $F(n,P_{1})$ only.  If $d=deg(P_{1})\ge 1$, then let $P_{m}$ be the vertex such that $P_{1}P_{m}$ is a diagnol and $m$ is minimal.  Then if we divide up $\mathcal{P}$ into the two smaller polygons $P_{1}P_{2}\cdots P_{m}$ and $P_{1}P_{m}\cdots P_{n}$, then we have by induction

$F(n, P_{1}) = F(m, P_{1}')+F(n-m+2, P_{1}'') = m-3-0+[n-m+2-3-(d-1)] = n-3-d$ as desired. (To try to avoid confusion, $P_{1}'$ is the point as it exists in the first partition, and $P_{1}''$ is the point as it exists in the second partition).

If $deg(P_{1})=0$, then $P_{2}P_{n}$ must exist, and there also must be a $m$ such that $P_{2}P_{m}P_{n}$ is a drawn triangle.  Then in one move we can erase $P_{2}P_{n}$ and draw $P_{1}P_{m}$.  Then by induction we have

$F(n, P_{1}) = 1+F(m, P_{1}')+F(n-m+2, P_{1}'') = 1+(m-3-0)+(n-m+2-3-0)=n-3$ as desired.  This proves the Lemma.

Now, because there are $n-3$ diagnols, we know that $\sum_{i}deg(P_{i})=2(n-3)$ for all $n\ge 3$.  Thus for all $n$ we know that there is a vertex $P_{M}$ such that $deg(P_{M})\ge \frac{1}{n}\sum_{i}deg(P_{i})=2-\frac{6}{n}$, and so for $n\ge 4$ we know that $deg(P_{M})\ge 1$.  Notice that from the above lemma, from any configuration, we can order the diagnols in some sequence of $F(n,P_{i})$ moves, and by reversing the order of these moves, we can also obtain any confguration of the polygon.  Thus we have from the Lemma

$P(n)\le F(n, P_{M})+F(n, P_{M}')\le (n-3-1)+(n-3)=2n-7$.

For $n>12$, let the two partitions $\mathcal{P}$ and $\mathcal{Q}$ have vertices $P_{i}$ and $Q_{i}$ respectively.  Then we know

$\sum_{i}deg(P_{i})+deg(Q_{i})=4(n-3)$

which means that there is an $M$ such that $deg(P_{M})+deg(Q_{M})\ge \frac{4(n-3)}{n}=4-\frac{12}{n}$.

But since $n>12$, we know that $deg(P_{M})+deg(Q_{M})\ge 4$.  Then from our same reasoning above $P(n)\le F(n, P_{M}')+F(n, P_{M}')\le 2n-10$ as desired.

OK this problem is interesting and your solution is also very nice.
thank you very much.
 :)


[quote="Peter"]Prove that no Fibonacci number can be factored into a product of two smaller Fibonacci numbers, each greater than 1.[/quote]

Let us make a contradicting suggestion. Let's suppose that there are three positive integer numbers $a,b,c>2$ so that $F_{a}=F_{b}F_{c}$. We know that $\gcd(F_a;F_b)=F_{\gcd(a,b)}$ then $b|a$. As $a=bm \Rightarrow F_{bm}=F_{b}F_{c}$
and also we should have that $F_c|F_{bm} \Rightarrow c|bm$ then $2c<bm+1$. On the other hand $F_{bm}=F_{b(m-1)+b}=F_{b(m-1)+1}F_{b}+F_{b(m-1)}F_{b-1}$ $\Rightarrow$ $F_{b}F_{c}=F_{bm}>F_{b(m-1)+1}F_{b}$ then $F_{c}>F_{b(m-1)+1}$ so $c>b(m-1)+1$ but we have proved that $2c<bm+1$ thus $bm>2b(m-1)+1 \Rightarrow 2b>bm+1>bm$ so $m=1$.Then $F_c=1$,contradiction!


Hi! could someone please help me with this task? thanks a million!!

There are two water tanks providing water to a station: a primary water tank, which is a sphere with a 4.5 metre radius, and a much larger backup water tank. The primary water tank has become empty and bucketfuls of water from the larger backup water tank to the primary water tank has to be carried.

The tank must be filled so that the water is exactly 6.5 metres deep at the center. If one bucket of water holds exactly 10 litres, how many bucketfuls of water are required to fill the tank to the required depth? Round up to the nearest whole number.

I`m very very thankful if someone replies!! :D

Solution:
[hide]
Consider the attached circular cross-section of the tank. The bottom part of the sphere is easy; that's half the volume, so $\frac{4}{3}\pi(4.5)^3 x \frac{1}{2}= \frac{243\pi}{4}$.

Now consider the triangular portion above the halfway mark. The chord that marks the water surface line is $2\sqrt{4.5^2+2^2}=\sqrt{97}$. The volume of that region is then the cone with $\sqrt{97}$ as the base and height $2$, so the volume is $\frac{1}{3}(\sqrt{97})(2)= \frac{2\sqrt{97}}{3}$.

The last bit, the two arc regions between the halfway mark and triangle, must be found with trig. To find the central angle that subtends the arc, find the center angle of one of the small right triangles and subtract it from 90. Thus one of those angles is $cos^-1 (2/4.5) = 63.6122$ degrees. The area of this cross-section is then $\frac{1}{2}(4.5)(4.5)sin(63.61)$. To get the volume, multiply by $2\pi$ because we're swinging the section all the way around (I think that's right, but not positive-- sorry if wrong). Then the volume of that section is $\frac{1}{2}(4.5)(4.5)sin(63.61)(2\pi)$.

Adding them all together, we get 254.4064 cubic meters, which makes 254.4064(1000) liters = $254406.4$ liters.  Then $\frac{254406.4}{10}= 254406$ buckets.

Answer: $254406$ buckets. Someone please correct me if I computer the volume of the arc part incorrectly. [/hide]


i am not in the U.S so i can not buy the aops so what other books on aops vol1 and 2 do you reccomend(live in africa)

Do they not ship to where you live or something?

they do but the Aops books is restricted to the U.S

Whoa, triple-post.

I thought AoPS could ship to other countries, but it might take a longer time...?


inequalities
LaTeX
algebra unsolved
algebra

I'm quite sure this one already appeared...
Anyway, could you use latex please?

Pierre.

[quote="bomb"]Let a_1,a_2,... be an arbitrary infinite sequernce of positive numbers. Show that the inequality 1+a_n>a_(n-1)*2^(1/n) holds for infinite;y many n.

Bomb[/quote]
in latex :
let $a_1,a_2,...$ be an arbitrary infinite sequence of positive numbers.show that the inequality $1+a_n>a_{n-1} \times (2)^{\dfrac{1}{n}}$ holds for infinite;many $n$.

I think this is a recent IMO shortlist problem, possibly 2001.

Indeed, this is the first problem of the Algebra part of the IMO 2001 Shortlist.


calculus
integration
logarithms
function
complex analysis
complex analysis unsolved

Here is the beginning:

I prefer to consider $ f(z)\equal{}\frac{1}{1\minus{}\log(1\minus{}z)}\equal{}\sum_{n\ge 0}a_n z^n$ to eliminate powers of $ \minus{}1$ and factorials. Then our task is to determine the asymptotics of $ a_n$. Write
\[ a_n\equal{}\frac 1{2\pi i}\oint_\Gamma \frac {f(z)\,dz}{z^{n\plus{}1}}\]
where $ \Gamma$ is the contour on the picture
[asy]size(100);
MC("\Gamma",D(CR((0,0),2,1,359)--reverse(CR((1,0),0.2,10,350))--cycle),SW);
D("1",8,(1,0)); D("0",8,(0,0));
MP("\Omega",(-1,0));[/asy]
with a very small radius of the small circle around $ 1$ and some fixed slightly greater than $ 1$ radius of the outer circle centered at $ 0$.

Note that the integral over the small circle tends to $ 0$ as its radius tends to $ 0$. Also, the integral over the outer circle tends to $ 0$ exponentially in $ n$. Thus, we just need to evaluate the integral over two horizontal pieces, which results in a nice real-valued integral approximation for $ a_n$ whose asymptotics is not too hard to find ;)

I wonder to know how you estimate or calculate asymptotics of an, can you show your solving way?

If you set up the integral I mentioned, you'll get
\[ a_n\equal{}\minus{}\int_0^\Delta\frac{1}{(1\minus{}\log t)^2\plus{}\pi^2}(1\plus{}t)^{\minus{}n\minus{}1}\,dt\plus{}O(e^{\minus{}\Delta n})\]
where $ 1\plus{}\Delta$ is the radius of the outer circle. Now just observe that $ \log t$ is a very slowly changing function near $ 0$ and $ (1\plus{}t)^{\minus{}n\minus{}1}$ is essentially concentrated at the values of $ t$ comparable to $ 1/n$ in the $ L^1$ sense, so we can make the rough estimate replacing the first factor with logarithm by its value at $ 1/n$, which immediately yields the desired asymptotics $ a_n\asymp \minus{}\frac 1{n\log^2 n}$.

I was intentionally a bit vague here to see if you can recover the technical details yourself ;). If you have severe troubles with some step, let me know and I'll go over it in more detail :).


geometry
geometric transformation
rotation
inequalities proposed
inequalities

$ x,y,z \geq0,r\geq1$,prove that

$ f(r) \equal{} \prod_{i \equal{} 1}^r{(x \minus{} iy)(x \minus{} iz)} \plus{} \prod_{i \equal{} 1}^r{(y \minus{} iz)(y \minus{} ix)} \plus{} \prod_{i \equal{} 1}^r{(z \minus{} ix)(z \minus{} iy)}\geq 0$
BQ

(x,y,z) into ($ \sigma_1,\sigma_2,\sigma_3$) program is(run in Maple):

eg1:

f:=(x+y)^10*(2*z-x-y)^10+(y+z)^10*(2*x-y-z)^10+(z+x)^10*(2*y-z-x)^10;
    (while x=y=z,f=0)

>read "lhdc3_4.txt":
>f:=sgm((x+y)^10*(2*z-x-y)^10);
>dc3(f,degree(f)):#(in Symmetric);

Rotation Symmetric:
lhdc3(f,degree(f))


The game of Ibish is played in rounds.  In the first round, you earn 0, 10 or 11 points; in the second round, you earn 0, 10, 11, 12 or 13 points, and so one, including the next greatest integer in the possible point values.  What is the fewest number of rounds after which your total score can have a 9 in the units digit?

[color=indigo]I don't understand if the possible scores increases by one or two... the example seems like two but the explanation seems like one. [/color]

you're right, LadyKn1ght, so [color=blue]assuming[/color] that in the third round, you can earn 0, 10, 11, 12, 13, 14, or 15 points,
[hide]In the first round, you earn 11 points, in the 2nd round, you earn 13pts, and in the third round, you earn 15 pts, giving you a total or 39 pts. [/hide]

that might not be the case though. [/hide]

[hide="i'm confused"]
i'm confused but i think it's the 3rd round[/hide]


[hide]
For this problem, we want to find an integer that is less than or equal to $-17$ but also has the least difference when it is subtracted from $-17$. Since 0 is the least difference between two numbers, and $-17$ is an integer, the answer would be $-17-0 = \boxed{-17}$.[/hide]

[quote="Klebian"]

So we would only take the floor of 4....., which would be 4.[/quote]
actually, we would take the [b]opposite[/b] of the square root of 17, (notice negative sign outside of radical but inside floor function) or [b]-[/b]4......, the floor of which would be -5.

[quote="dodobird"][quote="myc"]Find $\lfloor-{17}\rfloor$.

shouldn't $\lfloor{-17}\rfloor$ be -4, 5.
$\sqrt{17}= \pm{4....}$
so floor of $-4...=-5$
and floor of $4....=4$,
but the answer is only -5.

thanks :)[/quote]

the reason is that the floor function does round but it doesnt round like the way you are looking at it. It rounds down.

$\lfloor{sqrt{17}}\rfloor=4$
BUT
$\lfloor{-sqrt{17}}\rfloor=-5$

because look at it this way
$sqrt 17$ is alittle above 4 but once it is negative it is to round to the smaller of the 2 numbers in the negatives the larger the smaller it is so common sense would be that $-1>-2$
so $-4>-sqrt{17}>-5$
flooring goes to $-5$[/quote]

He was rounding correctly, what he was doing incorrectly was he was taking both the postive and negative root of 17.

By the way, k8reindeer is correct.

[quote="Klebian"][quote="myc"]Find $\lfloor-{17}\rfloor$.

shouldn't $\lfloor{-17}\rfloor$ be -4, 5.
$\sqrt{17}= \pm{4....}$
so floor of $-4...=-5$
and floor of $4....=4$,
but the answer is only -5.

thanks :)[/quote]

The question is actually:
Find $\lfloor-\sqrt{17}\rfloor$

What you're doing is taking both the positive and negative root of 17. However, (as stated on page 6 of AOPS book 1), "A good rule of thumb is: if the radical sign or the fractional power was in the problem to start with, then we are only looking for the positive root; if we force the problem to have a fractional exponent, then we must find all answers."

So we would only take the floor of 4....., which would be 4.[/quote]

I don't think that's a rule of thumb, I think that the radical is defined as positive square root only.


The number 2367 is not divisible by 11. Of all possible permutations of these four digits, how many result in a number divisible by 11?

You need $ 2$ digits to equal the other $ 2$ (or be apart by a factor of $ 11$, not possible here), because if $ \overline{abcd}$ is divisible by $ 11$, $ a \plus{} c \equal{} b \plus{} d$.  (Once again, this is only if you can't form the two pairs with sums that have a difference of a factor of $ 11$.)

Now, we know that $ 2$ and $ 7$ could be $ a$ and $ c$, and $ 3$ and $ 6$ could be $ b$ and $ d$, or it could be the other way around.

[i]Case 1: 2 or 7 is the first digit.[/i]

We have $ 2 \times 2 \times 1 \times 1$ or $ 4$ ways.

[i]Case 2: 3 or 6 is the first digit.[/i]

We have, once again, $ 4$ ways.

Thus, the answer is $ 4 \plus{} 4$, or $ \boxed{8}$.


If $A$ and $B$ are $n*n$ square matrices, and if $AB=I_{n}$, prove that A and B are invertible.

$\det{A}.\det{B}=\det{I_{n}}=1$, therefore $\det{A}\neq 0$ and $\det{B}\neq 0$.

hmm.. is there a way to prove it using only the definitions of a matrix? (i.e. associativity?) That's more of the solution I had in mind.

Suppose that there is a column vector $v$ such that $Bv=0$. Then, on the one hand, $ABv = A(Bv) = 0$. But, on the other hand, $ABv = (AB)v = I_{n}v = v$, so $v = 0$.
Thus, the system $Bv = 0$ has the unique solution $v=0$. [b]By the theory of linear systems, any equation $Bv = w$ has a solution and, furthermore, it is unique.[/b] It follows that there is $C$ such that $BC = I_{n}$.
We have that:
- $ABC = (AB)C = I_{n}C = C$;
- $ABC = A(BC) = A I_{n}= A$.
Therefore, $C=A$, so both $A$ and $B$ are invertible.

*The bolded part is the key stuff. You can't get away using only the ring properties of $n \times n$ matrices, since, in general, $ab=1$ does not imply $ba=1$ (as grobber kindly pointed out to me in another thread).


Those who r close (e.g. 1330) may seek help from me to get it. dont worry if u dont have 1337 u can still join i just couldnt think of a name =P  :D

edit:anyone can join!

YES!

I AM LEET 1337

IM GONNA JOIN.

HAHA MAYBE IF I GET MORE CARRIED AWAY I'LL MAKE A WEBSITE :D

yay that means will have a new website too!

ME ES JOINZES, NO?
JUST KIDDING. BUT SERIOUSLY, I WANNA JOIN.

these people hve joined so far

1. Angie
2. Mz
3. JoeJack
4. Me eddy!

president and vice prez elections auditions will be coming soon join them now! (to join pm me) also i need someone to help me make a list cause i suck at it  :blush:

edit: meant an election ballet

i have decided to host eleet meetings in the aops fmily chatroom (with angie's permission) and if people decide to watch then i will host them in a passwrd room of my making
passwords will be given to eleet memmbers yea secret society! w00t

i'll join the 1337 whatever. I like 1337

I want to join . My rating is 1309. I will run for something.

IM RUNNING FOR PREZ!

VOTE FOR ME!!!!

:D (IM SO SPASTIC TODAY)

results of a meeting earlier today ANGIE IS PREZ MZ is VICE prez JJx is chairman who works with us. erabel is board member and since i am founder i work with alll

Well, I have had a 1337 rating many times... I'll join.

i want to join

1337 is an awesome number

ok so ive decided to post a few things. first we need a goverment. I as founder assign jobs and later we may use eleet nicknames  :D ok bck to buisiness these are the eleet roles so far:

xoangieexo: (eleet nickname:angie) is president and my equal
mz94: (eleet nickname: cutie) is vice prez and works with me too!
isabella2296 (she asked me to sign her up): (eleet nickname:izzy) my advisor
jjx1: (eleet nickname: charmander) chairman of the board one of the originals (along with cutie angie and me!)
mathblitz (me!): (eleet nickname: eddy) founder.

erabel, chensi, the prodigy, and lotsof math, also JoeJack r board members and should pm me what they want their eleet nicknames to be.

remember anyone can join but we can only have one more board member everyone else r just members. BUT their r promotions!  :D

we will be playing pranks sometimes or getting revenge on those who have been annoying us. all decisions must be approved by angie and me first then its prankster time! we also make nicknames for people! go the eleet! w00t  :P

edit: meetings will be held in aops family chatroom: [url]http://www.freewebs.com/xoangieexo/chat.htm[/url]

post will be given for times (aops time)

[quote="mathblitz"]
mz94: (eleet nickname: cutie)
[/quote]

and just when i thought Mystic and ding calling me that was badd enough.......

 :dry:

and i wanted to be tyrant and dictator-for-life. poop

fine mz u choose just pm me it

anyway... making a goverment i too hard instead we will have only the members and the founders (me, mz, angie, charmander  :lol: ) everone else give me nicknames plz preferabaly by pm

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
I LOST THE THING THAT WHERE BUBBLES AGREES TO LET ME CALL HIM CHARMANDER
NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

OH WELLS I HAVE WITNESSES :D

1337! W00T!

hm can i join? (i want LEET but im too bad at math to get there)

actually i couldnt think of a different name so i gave it the eleet club. leet stands for 1337 and it is the aops fav number.

So do our FTW ratings have to be around 1337?

[quote="chenhsi"]yay... (in an unenthusiastic way)...[/quote]

how nice....

and dragon, anyone can join ;)


[u]Leaderboard[/u]
Mystic Terminator: 7 Points
Cats: 1 Point
The rest of you: Stop playing those name games and do some math

Source: Past MMPC 2 Exams
[u]Round 3[/u]
1987 1)
Let D(n) denote the number of positive factors of the integer n. For example, D(6) = 4, since the factors of 6 are 1,2,3 and 6. Note that D(n) = 2 iff n is a prime number.
a) Describe the set of all solutions to the equation D(n)=5
b) Describe the set of all solutions to the equation D(n)=6
c) Find the smallest n such that D(n) = 21

1986 2)
A bowl contains some red balls and some white balls. The following operation is repeated until only one ball remains in the bowl:

Two balls are drawn at random from the bowl. If they are different colors, then the red one is discarded and the white one is returned to the bowl. If they have the same color, then both are discarded and a red ball (from an outside supply of red balls) is added to the bowl.
a) Show that if the bowl originally contained exactly 1 red ball and 2 white balls, then the color of the ball remaining at the end (i.e. after two applications of the operation) does not depend on chance, and determine the color of this remaining ball.
b) Suppose the bowl originally contained exactly 1986 red balls and 1986 white balls. Show again that the color of the ball remaining at the end does not depend on chance and determine its color.

1986 3)
Let a, b and c be three consecutive positive integers, with a<b<c.
a) Show that ab cannot be the square of an integer
b) Show that ac cannot be the square of an integer
c) Show that abc cannot be the square of an integer

1986 5)
Let n and m be positive integers. An n*m rectangle is tiled with unit squares. Let r(n,m) denote the number of rectangles formed by the edge of these unit squares. Thus, for example, r(2,1) = 3.
a) Find r(2,3)
b) Find r(n,1)
c) Find, with justification, a formula for r(n,m)

1. [hide]alright, all of you should know the formula for d(n) by now so i won't state it. if d(n)=5, n=p^4, where p is prime. This follows directly from the formula for d(n). If d(n)=6, 6 can be either written as 6 or 2*3, i.e. (5+1) or (1+1)(2+1), so n can be either p^5, or p^2q, where p and q are primes. If d(n)=21, 21=21 or 3*7, i.e. n=p^20 or p^6*q^2. The smallest n in the first case is 2^20, while the smallest n in the second is 2^6*3^2. The latter is obviously smaller, making the answer 576.[/hide]

Mystic scores another point on the first problem

3. [hide]a=b-1, c=b+1, ab=b:^2:-b. However, this is always bounded by (b-1):^2: and b:^2: for b:ge:2, and can thus never be a square (if it's in between two consecutive squares). ac=b:^2:-1, and there are no positive squares that differ by 1. abc=b(b-1)(b+1), and if this =k:^2:, b|k:^2:, meaning that either b is a square or b:^2:|k:^2:. The second case is impossible as (b,b-1) and (b,b+1) both equal 1, meaning there are no extra factors of b in abc. THe first case gives us :sqrt:b=x:, so abc=x^6-x^2, which is bounded on both sides by (x^3-1)^2 and (x^3)^2 for positive x.[/hide]

Mystic gets the third question. Is he the only one playing?

4. [hide]r(n,m)=n(n+1)m(m+1)/4. To see why, partition these rectangles up according to their width, and assume that there are A rectangles that can be formed from a 1*m column. Well then there will be n*A rectangles of width 1, (n-1)*A rectangles of width 2, (n-2)*A rectangles of width 3 ... 1*A rectangle of width n. When we add these up, we get n(n+1)A/2. Ok, now we need to find A. We partition the rectangles that can be formed from a 1*m column similarly, except with respect to their length. There will be (in a similar fashion) m rectangles of length 1, m-1 rectangles of length 2, ..., 1 rectangle of length m, i.e. A=m(m+1)/2. Putting all this together gives the formula at the beginning. We now see that r(n,1)=n(n+1)/2 and r(2,3)=18[/hide]

Just a comment to the students about these contests - when someone posts a solution to a problem, don't let that stop you from working on it too. Give it a try before you look at the other solutions, and see if you can solve it.

And confuted, you should leave the contests open longer to give more people a chance to work on the problems. I have a feeling not everyone checks the board so frequently or has enough time to work on them.

2. [hide]if a red ball is drawn at any time along with a white ball, the cumulative effect will be the removal of a red ball. If two red balls are drawn, the same effect will take place. If two white balls are drawn however, the cumulative effect will be the addition of a red ball and the removal of two white balls. Note that this last is the only move which can remove white balls. Note also that as long as the number of each ball is above, say, three, the order of these moves is irrelevant as one does not affect the next. The reason why this may not be true when the number of a certain color ball goes under three is that there may not be enough balls to initiate one of the moves. For example, if we had just one red ball, we could not perform any of the moves, even though if we just considered the cumulative effect of each of these moves, we could perform either of the first two. Okay, for part a), we can never remove just one white ball (look at the moves above), so we can never end up with simply one white ball. Therefore, the last remaining ball is red. For part b), same logic. I'll try to generalize to r and w red and white balls later.[/hide]

Mystic gets #4. This isn't much of a competition.

Mystic also got #2. I think I'll make a new round right now with quite a few questions. Expect it... as soon as I get it done.