OK, i have quite big problems with this chapter - posibility. You can expect me to stop by several times in the near future since this surely isn't easy for me.

First two different tasks:

1) A person has 10 pair of shoes in a bag. She picks 3 shoes. What is the posibility that there will be at least 1 pair of shoes?

2) In a country every 3 person has a car. What is the posibility that in a gathering of 12 people are:
a. max. 4 people has a car.
b. exactly 4 people has a car.
c. at least 4 people has a car.

Plz set up the equation - i don't have any problems putting the numbers into the calculator. Thx in advance for any help.

(for my own help to remember what tasks we are talking about in this topic - 1024+1025)

[hide="1"]The first two shoes she picks can be any two shoes. The third shoe must be of the same pair as one of the first two shoes. There are 18 shoes left in the bag, and we can pick between 2 of them to have a complete pair of shoes. The probability is $\displaystyle\frac{1}{9}$.[/hide]

First of that isn't entirely correct as the second shoe could make a pair with the first one, so the possibility has to be less than 1/9. Secondly i use these kind of calculation methods and would like the result answered in a simular way if you could do that plz.

(K(13,7)*K(39,6))/K(52,13)

(The chance of getting 6 spades out of a hand of 13 cards)

Sorry, I think this should account for it: [hide]If she gets a pair on the 2nd pick, then the 3rd pick doesn't matter, so the probability would be $\frac{1}{19}$. If she gets a pair on the 3rd pick, then she cannot get a matching shoe on the 2nd pick, so the probabilty is $\frac{18}{19}\cdot\frac{2}{18}=\frac{2}{19}$. Adding these together gets a probabilty of $\frac{3}{19}$.[/hide]
I'm not sure how your method works... What does the K stand for? :?

Have you never used this method before?

K(n,r)=n!/(r!*(n-r)!)
K(7,4)=7!/(4!*3!)
[u]K(7,4)=35[/u]

Or you could do it easier on of the newer calculator models:
7 nCr 4
[u]35[/u]

nCr is a shortcut on my calculator.

I believe your result is right, but since this chapter is about plotting numbers into these equations and some more sophisticated ones i was hoping for learning something about how to write it this way and calculate this results from there.

Thx for your help so far though ;)

Oh, yes, I know what you mean now. I'm just more used to seeing it this way: $\binom{7}{4}$. I'm not sure how to use it for this problem though... sorry :(

so is the answer to the 1st one $\frac{3}{19}$?

I believe that could be true, but i'm really looking forward to see someone showing me how to do the two tasks using K(n,r) and ! (maybe in a more sophisticated method).


I read the "Introduction to Mathematical Induction".  What comes next in learning more about induction?

I'm not sure if this is Getting Started stuff, but in the last problem on the ARML Math Jam (transcript at [url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=30[/url] ), we didn't do what you read in the intro to induction.  We figured that it would be impossible (or at least very very messy) to induct on our actual number, n I think it was, so we inducted on the number of digits n had.


Let n be a even number in N and x,y in R so:
x^n + y^(n+1) > n^n
y^n+x^(n+1)>n^n

 Prove that x+y>1.   :)

If both x and y are positive . It's too obvious to see that x>1 or y>1 .
   If x < 0 < y , we denote : a = -x.
   We have : a^n + y^(n+1) > n^n
                       y^n > a^(n+1) + n^n
        If x + y  \leq  1 then y  \leq  1 - x = 1 + a .
   So    (1+a)^n > a^(n+1) + n^n ( * )
        <=>  (1/a + 1 )^n > a + (n/a)^n (a)
   But from (*) we see that : a  \geq  n
         => (1/a + 1 )^n  \leq  (1/n + 1 )^n  \leq  3 (1)
   And : a + (n/a)^n  \geq  3 ( because n is even ) (2)
    (1) , (2)  => (1/a + 1)^n  \leq  a + (n/a)^n (b)
    (a) and (b) give contradiction .

[quote="Tung Lam"]But from (*) we see that : a  \geq  n[/quote]
I don't see it. I see that $a> n-1$. Anyway, (1) holds for $n\geq 4$.

[quote="Tung Lam"]And : a + (n/a)^n \geq 3 ( because n is even ) (2)[/quote]
And again, we can state that only for $n\geq 4$.

So we should consider $n=2$ separately.

I done this problem long time ago.
Thank you for showing my mistake.  :oops:

I like to remind people their old mistakes :)
It is not first partitially wrong proof I found last days.


How can you, and I am in MATHCOUNTS as are a lot of people so maybe you could make a section for MATHCOUNTS problems?

I hope the admins will allow me to field this one. Unfortunately, MATHCOUNTS copyright laws do not allow for us to post a compilation of their problems here. Sorry.


Find all real solutions to the equation:

$ (2\plus{}\sqrt{3})^x\plus{}1\equal{}\left( 2\sqrt{2\plus{}\sqrt{3}} \right)^x$.

[hide]Let $ a \equal{} \sqrt {2 \plus{} \sqrt {3}}$. Note that $ 1 < a < 2$.

We have: $ a^{2x} \plus{} 1 \equal{} (2a)^x \Longleftrightarrow a^x \plus{} a^{ \minus{} x} \equal{} 2^x$.

Since $ 2^x \equal{} a^x \plus{} a^{ \minus{} x} \ge 2$, we have $ x \ge 1$.

Let $ f(x) \equal{} a^x \plus{} a^{ \minus{} x}$ and $ g(x) \equal{} 2^x$.

Note that $ f'(x) \equal{} \ln(a)(a^{x} \minus{} a^{ \minus{} x})$ and $ g'(x) \equal{} \ln(2)2^x$.

Since $ 0 < \ln(a) < \ln(2)$, and $ 0 < a^{x} \minus{} a^{ \minus{} x} < a^x < 2^x$ for all $ x \ge 0$, we have $ f'(x) < g'(x)$ for all $ x \ge 0$.

Since $ 2 \equal{} f(0) > g(0) \equal{} 1$, and $ f'(x) < g'(x)$ for all $ x \ge 0$, there must be exactly one solution to this equation.

By inspection $ x \equal{} 2$ yields: $ f(2) \equal{} (2 \plus{} \sqrt {3}) \plus{} (2 \minus{} \sqrt {3}) \equal{} 4$, and $ g(2) \equal{} 2^2 \equal{} 4$.

Thus, the only solution is $ x \equal{} 2$.[/hide]


[quote="Phelpedo"]Let $x,y,z \geq 0$ with $xyz=1$. Prove that
\[x+y+z \leq x^{2}+y^{2}+z^{2}\]
Once again, I don't really see the point of the $xyz=1$, as this can be proved without that condition. Does it somehow make the question easier?[/quote]

Actually, you need the $xyz=1$ this time. I would like to see you prove the inequality without it, especially for $x,y,z <1$  :P

btw, this is #2 from mildorf's article.  :D

Oops. I thought the question was something else...

[quote="Phelpedo"]Let $x,y,z \geq 0$ with $xyz=1$. Prove that
\[x+y+z \leq x^{2}+y^{2}+z^{2}\]
Once again, I don't really see the point of the $xyz=1$, as this can be proved without that condition. Does it somehow make the question easier?[/quote][hide]From Chebyshev, \[(x+y+z)(x+y+z)\le 3(x^{2}+y^{2}+z^{2}),\] \[\left(\frac{x+y+z}{3}\right)(x+y+z)\le x^{2}+y^{2}+z^{2}.\]AM-GM gives \[\sqrt[3]{xyz}(x+y+z)\le x^{2}+y^{2}+z^{2},\] and since $xyz=1$, then \[x+y+z\le x^{2}+y^{2}+z^{2}.\][/hide]

The first line also follows from Cauchy.

As $\left(2,0,0\right) \succ \left(\frac{4}{3},\frac{1}{3},\frac{1}{3}\right)$, we know by Muirhead that \[x^{2}+y^{2}+z^{2}= \frac{1}{2}\sum_{\textrm{symm}}x^{2}\geq \frac{1}{2}\sum_{\textrm{symm}}x\sqrt[3]{xyz}= \frac{1}{2}\sum_{\textrm{symm}}x = x+y+z\]


I know that this forum is very very inactive, and this topic is almost 2 months old, does someone have the results and/or questions from the 2008 Manhattan Mathematical Olympiad?

http://www.math.ksu.edu/~soibel/Olympiad/

I think they updated for this year also.


conics
parabola
analytic geometry
graphing lines
slope
geometry
symmetry

Prove that two parabolas with the same focus but opposite axes, intersect each other at a right angle.

By opposite axes, do you mean that the axes are perpendicular, or that the axis is the same and the direction of opening is opposite?

[hide="Potentially useful fact"]Consider a parabola with focus point F, directrix line L, and some point on the parabola, X. Let P be the projection of X onto directrix L. Then, if M is the midpoint of F and P, then the line MX is the tangent to the parabola at point X.[/hide]
[hide="Hint"]Using this, it may help to coordinatize with common focus (0,0), and use the fact that slopes of perpendicular lines multiply to -1, along with the fact above, to quickly elucidate the answer.[/hide]

I remembered this problem a while ago, and decided it deserves a solution.

I am assuming the axis is the same line for both parabolas, but that they open in different directions.

[hide="Chalk"]
(may take a few seconds to show up)
[img]http://img.photobucket.com/albums/v194/mukmasterxps/chalkparabola.gif[/img]
[/hide]
[hide="Solution by Geometry using Chalk"]
In the chalk, $ F$ is the common focus, the lime-green vertical line is the common axis of symmetry, and $ I$ is a point of intersection.

$ \overline{IN_1}$ is normal to the downward parabola at $ I$, and $ \overline{IN_2}$ is normal to the upward parabola at $ I$.

We know that an incident ray parallel to the axis of a parabola, and coming from the "inside" will reflect towards the focus.

Let $ R_1$, $ R_2$, and $ I$ be collinear, and let the line formed thereof be parallel to the axis.

Then, $ \angle R_1 I N_1 \cong \angle N_1 I F$, and $ \angle R_2 I N_2 \cong \angle N_2 I F$

Let the measures of those pairs of angles be $ \theta$ and $ \phi$, respectively.

Then, $ 2 \theta \plus{} 2 \phi \equal{} \text{m} \angle R_1 I R_2 \equal{} 180 ^{\circ} \implies \theta \plus{} \phi \equal{} 90 ^{\circ} \implies \text{m} \angle N_1 I N_2 \equal{} 90^{\circ}$

Thus, the normals to the two parabolas at the intersection point are perpendicular. Because the normal is perpendicular to the tangent, we note that $ N_2$ is a tangent segment to the downward parabola at $ I$ and $ N_1$ is a tangent segment to the upward parabola at $ I$, and thus the tangent lines to the two parabolas at $ I$ are perpendicular as well.
[/hide]
[hide="Solution by Calculus"]
Let the common focus be the origin, and let the axis of symmetry be the y axis. With positive numbers $ a$ and $ b$, let the directrix of one parabola be $ y \equal{} \minus{} 2a$ and the directrix of the other parabola be $ y \equal{} 2b$.

Then, by the distance formula, the equations of the parabolas are:
$ x^2 \plus{} y^2 \equal{} (y \plus{} 2a)^2 \iff y \equal{} \frac {1}{4a} x^2 \minus{} a$
$ x^2 \plus{} y^2 \equal{} (y \minus{} 2b)^2 \iff y \equal{} b \minus{} \frac {x^2}{4b}$

Solving for the point of intersection, we have $ \frac {1}{4} \left( \frac {1}{a} \plus{} \frac {1}{b} \right) x^2 \equal{} a \plus{} b \iff x^2 \equal{} 4ab$

Now, the derivatives of the two parabolas in terms of $ x$ are $ \frac {x}{2a}$ and $ \minus{} \frac {x}{2b}$, respectively.

Their product is $ \minus{} \frac {x^2}{4ab}$. At a point of intersection, this is equal to $ \minus{} \frac {4ab}{4ab} \equal{} \minus{} 1$, so the tangent lines at a point of intersection are perpendicular.
[/hide]

[b]Note:[/b] You may as well disregard my previous post.


geometry
angle bisector
geometry unsolved

draw 3 degree angle only by ruler and compa :lol:

That's not too bad.
[hide] Just construct a bit of a pentagon, giving a 108 degree angle, then construct an equilateral triangle sharing a side and vertex outside the pentagon.  The total angle will be 168 degrees, so the complementary is 12 degrees.  Finally, take the angle bisector of this twice, for a 3 degree angle.[/hide]


A $ 13$ by $ 13$ by $ 13$ cube consists of $ {13}^3$ unit cubes. What is the maximum number of cubes that can be seen from one viewpoint.

I think 3 sides, so $ 3\times 13^2\minus{}13\times 3\minus{}1$??

Your calculations give 467. The actual answer is 469.

Anyway, could you explain your method?

The answer is simply $ 13^3\minus{}12^3\equal{}\boxed{469}$. You can see three faces, and the cube that they cover up is hidden, prompting my calculations.

I'm sorry, the last minus should be replaced with plus. I worked some kind of Inclusion-Exclusion


People have complained that these questions belong here. I have moved them to Basic High School.

Some of these aren't mathcounts lvl as far as i know...

Usually recursion problems do not appear in Mathcounts. I don't think it belongs here.


Any ideas to this one?
You are to find all pairs of integers such that their sum is equal to the given integer number N and the second number results from the first one by striking out one of its digits. The first integer always has at least two digits and starts with a non-zero digit. The second integer always has one digit less than the first integer and may start with a zero digit.
Input

The first line of the input file is the integer number t ( 1 ≤ t ≤ 20 ), the number of test cases. Then t lines follow, each test case in one line; the line consists of a single integer N (10 ≤ N ≤ 10^9).
Output

For each test case:

On the first line write the total number of different pairs of integers that satisfy the problem statement. On the following lines write all those pairs. Write one pair on a line in ascending order of the first integer in the pair. Each pair must be written in the following format

X + Y = N

Here X, Y, and N, must be replaced with the corresponding integer numbers. There should be exactly one space on both sides of '+' and '=' characters.
Example

Input:
2
302
11

Output:
5
251 + 51 = 302
275 + 27 = 302
276 + 26 = 302
281 + 21 = 302
301 + 01 = 302
1
10 + 1 = 11

If you know the last (least significant) digit of X, and you also know what digit you are going to strike out and at which position it is, then you can (using the number N) find all digits of number X. (There may be two possibilities for this but never more than two.) This means that you can find all pairs X, Y in about O(10^4) time.


Given 3 f(x) + 2f(1-x) = x^2

I just can't figure this out.

Can somebody help me?

From GPML

Note that this function is cyclic so you can plug in 1-x.

thanks a lot!

But how do you know it is a cyclic function or what is a cyclic function?

It's because if you plug in $x=1-y$ (different variable to clarify), you have $3f(1-y)+2f(1-(1-y))=3f(1-y)+2f(y)$, so they're the same variable (just with different coefficients). From there, you solve the system of equations. So, cyclic in this case indicates that there are two different function values, but if you plug in a certain number, they'll just switch.

For example, there was a problem I saw recently (maybe AMC12 2006A?), which had a problem like this. It had $f(x)=f\left(\frac{1}{x}\right)$. The only values for which this is the case are $x=\pm 1$, so that was the solution set. It's not quite the same, but it's similar, where you're putting in a value to make one equal to the other. You have to learn to look for which value to put in to make it work.

That was indeed AMC 12 2006A, Problem 18.


Let triangle ABC with AB+BC=3AC the incircle touches th sides AB and BC respectively . Let K and L be symmetric points of D and E with respect to I then show that the quadrilateral ACKL is cyclic

Denote J, Q the intersection of AL and BC, CK and AB.
We have $ BQ\equal{}AD, BJ\equal{}EC$
$ \Rightarrow BE\equal{}JC, BD\equal{}AQ$
But $ BE\equal{}BD\equal{}\frac{AB\plus{}BC\minus{}AC}{2}\equal{}AC$ then $ AQ\equal{}CJ\equal{}AC$
$ \Rightarrow \angle QCA\equal{}90^o\minus{}\frac{A}{2}$
$ \angle JLK\equal{}\angle JLE\plus{}\angle ELK\equal{}90^o\minus{}\angle AJE\plus{}90^o\minus{}1/2\angle LIK$
$ \equal{}90^o\minus{}(90^o\minus{}C/2)\plus{}B/2\equal{}90^o\minus{}A/2\equal{}\angle QCA$
$ \Rightarrow$ QED


AMC USA(J)MO USAMO Putnam LaTeX email geometry

I've been thinking of some USAMOish problems recently, and I thought maybe ya'll'd like to take a look at them. Admittedly I'm not a great proof-grader, so if someone else wants to help grade the proofs, I'd be grateful :). Also, if the same person wants to look over my problems and make sure they are solvable and reasonable, more thanks would be extended :).

I was thinking July 10 I'd post the test, and would post the best solutions and winners July 31st. (I'd let ya'll find 2 days to devote 4.5 hours of to the test.)

Anyone interested?

I'll try it, but I probably won't be able to get any of the problems. :D

Spending 9 hours over 2 days staring at a piece of paper with problems on it sounds fun, so I'm in.

My special way of suicide. Yay.
My brain is going to be fried.
BUt who cares? It will be fun. :D

Hmmm I'll try to do it, but I'm making no promises.

In my life, I think I have solved maybe 2 or 3 USAMO problems. And those were the easiest of the easiest.

solafidefarms, please include at least one question that is really easy, so even stupid people like me can solve them :P

I might be able to [i]try[/i] it. but i havent really done much usamo in my life.

Pakman, is that you expressing interest in taking the test or helping with the test?

seems interesting...i would like to check over problems and maybe propose some, i am not sure that i would have that much time to devote to the test though, 9 hours is just wow...i have a job during the summer and the weekends are usually not that free

it would take a while to make a problem though, orignial problems take a while to find something that works and is difficult enough, yet you can still solve it

I'm in, but I can't write proofs very well:(

I'll try...but if I don't send in answers it doesnt mean I couldnt find time :dry:

I'll try to take the test even if I'll be away  :)

Should we attatch a .pdf file to the pm or is it ok to just have the solution as text and latex in the pm?

wait, do you guys give us the ID numbers? Can mine be 666?

No, you can't; I am afraid. Your ID will be less than 50, and randomly assigned.

Many people will LaTeX their solutions.  Perhaps we could modify the USAMTS LaTeX for use with the mock USAMO.  I'm working on it now, and if I get anywhere, I'll post the template here, along with the instructions.  What do you think?

Billy, you can decide if you want to use this or not.  The source is rather messy, but the output looks great.

Instructions:
[b]Download both files[/b], and [b]put them in the same folder[/b].  Then, [b]open the file called "solutions."[/b]  You do not need to open the other file.  Once you open "solutions" all you have to do is [b]insert your Mock USAMO ID #[/b] where it tells you to.  Then, p[b]ut your solutions where it tells you to[/b].  DON'T EDIT ANYTHING THAT IS SURROUNDED BY "DON'T EDIT" SIGNS.  If you need help, just post questions or PM me.

[b]Next, download [url=http://www.artofproblemsolving.com/LaTeX/Images/fancyhdr.sty]this[/url] by right-clicking on it.  Put it in the same folder as the two files above.[/b]

NOTE: If you are trying this out, it will to output to PDF unless you have some typing where your solutions ought to go.  Just type in anything there to test it out.

Man, this feels like USAMTS all over again... :rotfl:

[quote="MysticTerminator"]ok, how about -666[/quote]

I'm quite sure he means positive integers less than 50... (wait, maybe non-negative...)

I say we pm him solutions without complicated, impractical, and pretty (hooray) formatting. Let's just stick with that and not try to inflict anything more complicated or time-consuming on ourselves.

Do we submit one solution at a time or all solutions at once? :maybe:

It is preferred that you submit all of a set's solutions at one time; I don't mind people submitting all at once, just don't submit, say, 1,4,5 and then 2, 3,6 later.

Thanks for the clarification.
I will now start the contest. :)

Is it too late to join this contest? I'd like to give it a try if it isn't...

[quote="sapphyre571"]Is it okay to take the test over two nonconsecutive days?
I know on the real USAMO you can't
but if we don't look at the problems, it is all right if someone takes set 1 on a Tuesday and the next set on a Friday?

concerned with availability...[/quote]

Of couse it's fine to take it on 2 non-consecutive days, as long as you don't look at the other problems.


Prove that of two dinstinct 2 digit numbers, we can select two with a two digit difference of the form aa.

can you reword that a little better?
I'm not sure what numbers we're taking the difference of. Are they the ones we select or the two distinct ones?

I think you mean of $ 12$ distinct two-digit numbers.


How many positive three-digit integers are divisible by neither 2 nor 3?

There are $450$ numbers divisible by 2.
There are $300$ numbers divisible by 3.
There are $150$ numbers divisible by 6.

Add the ones divisible by 2 and 3, and subtract one divisible by 6 since we overcounted. So there are $600$ numbers divisible by 2 or 3.

Therefore, there are $300$ numbers.

techically isn't the number of 3 digit integers divisible by 2 be $499$?

[quote="perfect628"]No, there are $5*10*9=450$[/quote]

I don't get it?

Extend your post a little more, perfect

There are 9 ways for the first digit. (You can't use 0)
There are 10 ways to choose the middle digit.
Since it has to be even, there are only 5 ways to choose the last digit.

$9 \times 10 \times 5 = 450$.

The way I did it was:

The first 3 digit number is 100, and the last is 999, so there are 900 3 digit numbers. Dividing by 2 we get 450, dividing by 3 we get 300, and dividing by 6 we get 150.

If the number wasn't as perfect as 900 I would find the first and last numbers that work, subtract, divide by 2, 3, or 6, and add one.

[quote="perfect628"]Sorry. For the units digit, there are 5 possibilities, as only 5 are even. For the tens digit, there are 10. as its a 3 digit #, 0 can't be the hundreds digit, so there are 9 choices. Is that clearer?[/quote]

Yes it is much clearer...thanks!


would anyone be interested in starting an AIME problem solving thread?

even though this isn't the AMC/AIME/USAMO board, it seems like more problems of the sort are solved in this board than in the other one (just checking it out, the AMC/AIME/USAMO board is more of a discussion of the tests themselves instead of the actual content)

im mainly looking to do a lot of problem solving, discuss general methods for approaching these kinds of problems, and just to look at all the differnet, hopefully full, solutions of the problems, instead of kalva's one-liners  :P

another problem solving thread would be nice, since the material through AMC and AIME are about intermediate lvl, imo. until that happens, work w/ the intermediate daily problems.

well i set up a poll just to see how many people would be interested/uninterested

I don't see why not. I'll start, unless there  are objections.

There is an AIME marathon thread in the AMC forum, but starting another one wouldn't hurt.


Let $ a, b, c$ positive real numbers such that $ ab \plus{} bc \plus{} ca \equal{} abc, a, b, c > 1$. Prove or disprove that

$ \sum(\frac{a^4 \plus{} 4a \minus{} 4}{(a \minus{} 1)^2})\ge 54$.

_________
 :?:

Again, my answer is yes, your inequality is tremendously true.
Actually, the inequality can be solve beautifully by algebraic substitution. Indeed, we set
$ \frac{1}{a}\equal{} x$, $ \frac{1}{b}\equal{} y$, $ \frac{1}{c} \equal{} z$
Hence, we get $ x\plus{}y\plus{}z\equal{}1$
Therefore, we get the following expression:
$ \frac{a^4\plus{}4a\minus{}4}{(a\minus{}1)^2}\equal{}\frac{1\plus{}4x^3(y\plus{}z)}{(y\plus{}z)^2}$
Now, it is completely familiar to us.
Anyway, applying Cauchy for 3 positive real numbers
$ \sum_{cyc}\frac{1}{(y\plus{}x)^2} \ge 3\sqrt[3]{\frac{1}{[(x\plus{}y)(y\plus{}z)(z\plus{}x)]^2}}$
and
$ \sum_{cyc}\frac{x^3}{y\plus{}z}\ge \frac{x^2\plus{}y^2\plus{}z^2}{2}$


probability
geometry
number theory
\/closed

[quote="rrusczyk"]We are going to experiment with offering classes during the school day, primarily with homeschool students in mind.  We are offering our Introduction to Counting & Probability class on Mondays this fall, 1 PM - 2:30 PM ET (10 AM - 11:30 AM PT), starting October 5.  Please share this with homeschoolers who might be interested, and help us make this experiment a success![/quote]
But won't this only work for homeschool people? Are you still offering 7:30-9:00 PM ET?

We are offering both times.  Students who have already enrolled in the evening class can switch to the afternoon class by emailing us at classes@artofproblemsolving.com.

Daytime classes would be much more convenient for those of us who have sports practice in the evenings.

TJ

I have two homeschooled sons and am very excited about enrolling them for the Alg. 3 class starting in March. The time for that class works for us; however, we have conflicts with many of the evening class times, and as I hope my sons will also eventually take Number Theory, Counting & Probability, Precalculus, etc., having some midday times will be a wonderful option! I will definitely spread the word in our homeschool community! Thank you.

~Laura

I just saw this.  I am so sorry I missed it earlier as my homeschooled son was just complaining that he wishes he could take the classes instead of just using the textbooks.  Unfortunately, he has gymnastics practice every evening and cannot make any of those class times.

I am off to see what is being offered in the spring!  Please keep offering these timeslots.  It may take time for it to become known that they are available.

Our expectation is to wait until next fall to re-start this experiment with classes in the day.  Currently, we expect to offer Algebra 1 and Intro Geometry at homeschooler-friendly times in fall, 2010 (as well as at times in the evening), and then offer Algebra 2 and Intro Number Theory in spring, 2011.

What about the classes at 9:00PT/6:00 PT?

We'll likely continue doing occasional problem series at those times.


From point $A$ outside circle we draw tangents, which intersect circle in points $B$ and $C$. THen we draw line through A which intersect our circle in $Q$ and $P$.  Lines $BC$ and $QP$ intersect in point D. prove that :
\[ \frac{AP}{PD}=\frac{QA}{QD} \]

It is really well-known: $BC$ is a polar of $A$ - I guess it has been posted many times before.


Let $ c( a, b, c ) \equal{} \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}$. Compute $ c(2, 12, 9)$.

hello, we have $ c(2,12,9)\equal{}\frac{1}{6}\plus{}\frac{4}{3}\plus{}\frac{9}{2}\equal{}6$.
Sonnhard.


Let $ ABC$ is triangle with $ AB \equal{} AC$. Let $ H$ is midpoint of $ BC$, let point $ D$ is the project of point $ H$ on line $ AC$. Let point $ M$ is midpoint of $ HD$. Prove that $ AM \perp BD$.

check this :

http://www.mathlinks.ro/Forum/viewtopic.php?t=198301

Hi! Who can slove my problem by other way?

[asy]size(250);
    import olympiad;
    import math;
    pair A,B,C,H,D,M,N,E,F,G,J;
    H=(0,0);
    A=(0,15);
    B=(-8,0);
    C=(8,0);
    draw(H--C--A--B--H--A,blue);
    D=foot(H,A,C);
    M=midpoint(H--D);
    draw(H--D,blue);
    draw(A--M,blue);
    draw(B--D,blue);
    N=intersectionpoint(A--M,B--D);
    dot(A--B--H--C--D,red);
    dot(M--N,red);
    label("$A$",A,N);label("$B$",B,S);label("$H$",H,S);label("$C$",C,S);label("$D$",D,NE);label("$M$",M,SE);label("$N$",N,NE);
    draw(rightanglemark(A,H,B,20));draw(rightanglemark(H,D,A,20));
    add(pathticks(B--H,2,0.5,0,25,red));add(pathticks(H--C,2,0.5,0,25,red));
    add(pathticks(M--H,2,0.5,7,25,red));add(pathticks(M--D,2,0.5,7,25,red));
    draw(circumcircle(A,N,H),orange);draw(circumcircle(A,N,D),orange);
draw(H--N,dashed+blue);[/asy]

The method in the link is called proof by contradiction.
So if you want a  direct proof just go backwards step by step.
First draw circumcircles to $ \triangle ANH$ and $ \triangle AND$
Clearly,$ HD$ is tangent to circumcircle of $ \triangle AND$ and the common chord $ AN$ bisects $ HD$.Hence $ HD$ is also tangent to circumcircle of $ \triangle ANH$.
Now,$ \angle HBD \equal{} \angle CHD \minus{} \angle NDH$
$ \Rightarrow \angle HBD \equal{} \angle HAD \minus{} \angle MAD \equal{} \angle HAM$
$ \Rightarrow ABHM$ is a cyclic quadilateral.
$ \Rightarrow \angle BNA \equal{} 90^\circ$


geometry
AMC
AIME
USA(J)MO
USAMO
ARML
summer program

I'm sorry because I'm sure this has been discussed plenty of times before.

Could someone give me a comprehensive list of all the different high school math contests in the NYC metropolitan area, and explain what they are (or provide links)?  Thank you sooooooooo much.

I go to Benjamin N. Cardozo High School, Bayside.
We are participating in:

[b]NYCIML[/b] - new york city interscholastic math league, ask coaches like JBL, Penny77114, or Jack Po more about this
[b]NYML[/b] - http://www.mathleague.com
[b]Atlantic Pacific[/b] - it is too easy for you maybe. I have scored perfect score for three years in a row.

[b]AMC[/b]
[b]AIME[/b]
[b]USAMO[/b]
http://www.unl.edu/amc

as a New York City Math Team,
[b]NYSML[/b]
[b]ARML[/b]
you can get more information at nycmathteam.com

[b]the Queens College Math Competition[/b] - next friday!

In addition, I participated in:
[b]AHSIMC[/b] (with mathcamp people)
[b]USAMTS[/b] - http://www.usamts.org
[b]LG/KSEA Mathematics Olympiad[/b] (for korean)

I'll post more if i remember more :)

THank you.  I want to build a math club when I get to high school, so that was very very helpful.

is it ok for me to assume that you're living in NYC?
which high school are you going to attend?

if you go to stuy, bronx sci, or brklyn tech
even cardozo, bayside, francis lewis or john bowne,

they are probably participating some math contests.

are there even any schools in NYC that don't have some kind of organized math team?
i know that midwood doens't, but there seems to be many schools that do...

I believe that Midwood is one of the only ones...I know at least one person in every close to every other school on Brooklyn and Manhattan area that have a math team or math club, so starting one up most likely will not be necessary...

I think you are probably greatly overestimating the number of schools in the city with math teams, given the very large number of high schools in New York.  (Just for the record, I think bubala is from the suburbs, not the city itself.)  I would bet that only a fairly small number have a math team with any kind of central organization.

All I know is that I have many junior highschool friends that email me to talk about the math discussions they have in their schools...this is not math team, and this is not class activities...it's when their teachers take the time off to show them things that are not offered at school...

No, I live in Scarsdale, in Westchester County.  It's very near NYC.  We are a very exellent school, but we don't really have a math culture...yet.  See my blog for more info (if u actually care--it's called Think Globally Act Locally)

from Scarsdale, I know teddy singer and steve margorlin from mathcamp,
and wendy whos going to princeton.

They were good math students and also they were at ARML.

omg yes I know them 2!

Are you going back to Mathcamp?

probably not.
I was trying to work at CCNY, well, I'm still trying.

but mathcamp is awesome.

So is Hampshire!!! If you want, you can still apply to Hampshire and do the IT to get an idea of the type of mathematics they do...You can apply at HCSSIM.org...

I knew someone was gonna say that  :lol:

YAY Hsampshire!!!

i also gotta admit, though, the class offerings at Math Camp look great  :|


calculus
derivative
limit
analytic geometry
graphing lines
slope
LaTeX

This is calculus - a mod should move it.

Just take the derivative of $x^2$: $\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}$

(Basically, that's taking the slope $\left( \frac{\Delta y}{\Delta x}\right)$ of a very small part of the curve (from $x$ to $x+h$) as $h$ gets very very small.)

i think that you are asking for the derivitive...this is misplaced, but anyway, here is how you should do it

if you have a line tangent to the curve, then when you try to find the intersections, you will only have one solution,
(consider a point (x,x^2) is on the curve, so using point slope form)

$y-x_0^2=m(x-x_0)$
$y=x^2$

solving these...

$x^2-x_0^2=mx-mx_0$
$x^2-(m)x-(x_0^2-mx_0)=0$ using the discriminant
$m^2-4(1)(-1)(x_0^2-mx_0)=0$
$m^2-4mx_0+4x_0^2=0$
$(m-2x_0)^2=0$
$m=2x_0$

umm please can u tell me a bit more on how u got $2x$? I fail to understand after the discriminant part. And why cant u look at it this way???
$m=x$?
Because applying only formulas it should come.

where are you getting that m=x? you cannot just assume that the slope of a line to a certain graph is equal to x at that point

[quote="riddler"]Find gradient of a line,
$y=x^2$

[hide="See this is my approach...."]
$y=mx+c$
$m=x$????
But the answer is $2x$
[/hide]
Please help[/quote]

riddler, have you taken calculus yet?

im not sure how to write limits in latex so ill just say:

use the power rule so bring the ^2 infront of the x and subtract one from the exponet to get 2x

So,
$y=x^n$
Generalised formula would be....?
2x^n?

The generalized formula, I believe it's called the power rule, for $f(x) = x^n$ is $f'(x) = nx^{n-1}$

Can somebody prove it USING finite maths? no calculus please Im just starting.


limit
inequalities
real analysis
real analysis unsolved

Suppose  that $ a_{n}$ and $ b_{n}$  are two sequences  of nonnegative numbers  such that
for some real number $ N_{0}\geq 1$, the  following recursion inquality holds:
$ a_{n\plus{}1}\le{a_{n}\plus{}b_{n}}$, for any $ n\ge N_{0}$.
Prove that  if $ \sum b_{n}<\infty$  , then $ a_{n}$is converges .

[quote="bookworm_vn"]Put $ a_{0}\equal{} 0$, then $ a_{n}\equal{}\sum_{n \equal{} 1}^{n}{\left({a_{n}\minus{}a_{n\minus{}1}}\right)}$. Since, $ a_{n}\minus{}a_{n\minus{}1}\leqslant b_{n\minus{}1}$ and $ \sum{b_{n}}<\plus{}\infty$ then $ a_{n}$ is converges[/quote]
It is quite easy but you use this for $ n>N_{0}$ don't use for all n.

[quote="TTsphn"][quote="bookworm_vn"]Put $ a_{0}\equal{} 0$, then $ a_{n}\equal{}\sum_{n \equal{} 1}^{n}{\left({a_{n}\minus{}a_{n\minus{}1}}\right)}$. Since, $ a_{n}\minus{}a_{n\minus{}1}\leqslant b_{n\minus{}1}$ and $ \sum{b_{n}}<\plus{}\infty$ then $ a_{n}$ is converges[/quote]
It is quite easy but you use this for $ n > N_{0}$ don't use for all n.[/quote]

this $ n > N_{0}$ doesn't matter much.
bookworm_vn's proof is incomplete because some of $ a_{n}\minus{}a_{n\minus{}1}$ may be negative.

I am sorry, here is my correct proof. Since
\[ \begin{gathered} a_{n \plus{} 1} \leqslant a_n \plus{} b_n \hfill \\
a_{n \plus{} 2} \leqslant a_{n \plus{} 1} \plus{} b_{n \plus{} 1} \leqslant a_n \plus{} \left( {b_n \plus{} b_{n \plus{} 1} } \right) \hfill \\
... \hfill \\
\end{gathered}
\]
then
\[ a_m \leqslant a_n \plus{} \sum\limits_{k \equal{} n}^{m \minus{} 1} {b_k } \leqslant a_n \plus{} \sum\limits_{k \equal{} n}^{ \plus{} \infty } {b_k }
\]
provided $ m \geq n \plus{} 1$. Now taking the $ \limsup$ with respect to $ m$ we have
\[ \mathop {\lim \sup }\limits_{m \to \plus{} \infty } a_m \minus{} \sum\limits_{k \equal{} n}^{ \plus{} \infty } {b_k } \leqslant a_n .
\]
Now consider the $ \liminf$ with respect to $ n$ we deduce
\[ \mathop {\lim \sup }\limits_{m \to \plus{} \infty } a_m \leqslant \mathop {\lim \sup }\limits_{m \to \plus{} \infty } a_m \minus{} \mathop {\lim \inf }\limits_{n \to \plus{} \infty } \sum\limits_{k \equal{} n}^{ \plus{} \infty } {b_k } \leqslant \mathop {\lim \inf }\limits_{n \to \plus{} \infty } a_n
\]
since $ \mathop {\lim }\limits_{n \to \plus{} \infty } \sum\limits_{k \equal{} n}^{\plus{} \infty } {b_k } \equal{} 0$. This and the fact that $ \mathop {\lim \sup }\limits_{m \to \plus{} \infty } a_m \geqslant \mathop {\lim \inf }\limits_{n \to \plus{} \infty } a_n$ yields the conclusion.

thanks for all. I think last solution is correct

\documentclass{report}
\begin{document}
 I have another solution.
if  $ 0\le a_{n \plus{} 1}\le a_n \plus{} y_n$ and $ \sum y_n$ is convergent then
we will show that  $ a_n$ is convergent.

 For this let we consider following sequence  $ a_n*: \equal{} sup\{a_k: k\ge n\}$
 From following result we obtain that
$ a_n\le a_{n \minus{} 1} \plus{} y_{n \minus{} 1}\le\dots\le a_1 \plus{} y_1 \plus{} y_2 \plus{} \dots y_{n \minus{} 1}$ our
sequence $ \{a_n\}$ is bounded . Now we compare the sequences
$ \{a_n\}$ and  $ \{a_n*\}$ We can check that sequence  $ \{a_n*\}$
is monotone decreasing and bounded from below . That is why this
sequence is convergent. Let we use sandwich theorem. Following
statement obvously

                   $ a_n\le a_n*$
 From another hand following inequality holds.
$ a_n*\le a_n \plus{} \sum_{k \equal{} n}^{\infty}y_n$. Consequencely we obtain
\[ a_n\le a_n*\le a_n \plus{} \sum_{k \equal{} n}^{\infty}y_n
\]
from this we get  $ lim_{n\to \plus{} \infty}|a_n \minus{} a_n*| \equal{} 0$ relation. We
know that sequence  $ \{a_n*\}$ is convergent therefore sequence
$ \{a_n\}$ is convergent.
\end{document}

apply Cauchy convergence to [b]bookworm_vn[/b]'s proof, i.e.

$ \forall\epsilon >0$ choose $ N$ such that $ \sum_{v\geq N}b_v<\epsilon$, then $ \forall n>k>\max\{N_0,N\}: |a_n\minus{}a_k|\leq\sum_{v\equal{}k}^{n\minus{}1}b_v<\epsilon$

hello, I am sorry for my interesting to your solution , because in this instance it is difficult to apply Cauchy convergence and you can meet some confusion, please write you solution clearly as soon as possible.  (hjbrasch)
 with best regards Akmal


Hallo,

Wie kan mij helpen met het (zeg maar) breken van volgende stelling? Ik zelf krijg er momenteel kop noch staart aan.

[img]http://img83.imageshack.us/img83/1622/ana0nu.jpg[/img]

Groeten. Dank bij voorbaat.

Hmm... en wat is precies de vraag? Hoe "breek" je een stelling?

Ik vind dat toch een beetje vaag
Kan je niet zeggen wat je precies dwars zit : alles?
een bepaalde regel in het bewijs?
wat de opgave van de stelling is?
wat voor nut deze stelling eigenlijk heeft

ik zou het wel verwoorden met lijnintegrale, gradienten en scalaire velden, heb het nooit gehad voor 'totale differentiaal' toestanden

Mja, wat hebben jullie gezien als definitie van totale differentiaal enzo? En is dit middelbaar of universiteit?

1ste jaar universteit.

Euh de bedoeling is dat we proberen te bewijzen dat één voorwaarde waar moet zijn omdat de differentiaal vergelijking een totale differentiaal is van een functie.

Dit staat in mijn boek inleiding tot differentiaal vergelijkingen. Maar heb de uitleg daar niet over gehad omdat ik ziek was.

Waar ik dwars zit tja kan er niet aan volgen

Groeten.

tja in elk geval, maar ik weet dus of het dat is wat je wil he!, is dit wat ze bedoelen met totale differentiaal :

$p(x,y) \ dx+q(x,y) \ dy$ is een totale differentiaal  als en slechts als er een functie $f(x,y)$ bestaat zodat

$p(x,y)=\frac{\partial{f}}{\partial{x}}$

$q(x,y)=\frac{\partial{f}}{\partial{y}}$

Men bewijst dus de 2 implicaties $a\Rightarrow b, b\Rightarrow a$. Tot voor "omgekeerd" is $a\Rightarrow b$, de rest van het bewijs is $b\Rightarrow a$, en dat laatste bewijs je hier in het speciale geval dat $D$ het inwendige van een rechthoek is.

Dan zoek je dus die functie $z=g(x,y)$. Door de eerste vergelijking heeft ze zeker die speciale vorm integraalvorm $z=\int... + \phi(y)$, en dat vul je in in de tweede vergelijking, zodat je uiteindelijk uitkomt dat $\phi'(y)=q(x_0,y)$, waardoor $\phi(y)$ de integraal daarvan is.

Vul nu die $\phi(y)$ in in $z=\int... + \phi(y)$, en je krijgt dat je $z$ (of $g(x,y)$, da's 'tzelfde) van de gedaante (9.3) is. Op volledig dezelfde manier, maar nu beginnend met de tweede vergelijking $q=\frac{\delta z}{\delta y}$ [kan er even niet meer opkomen hoe je die krullekes schrijft in latex]

Aangezien de gelijkheid moet opgaan voor alle $x,y$ en dus ook voor $x=x_0,y=y_0$, moet $c=g(x_0,y_0)$. En alzo heb je dus de gezochte functie geconstrueerd.

Schnappi?

[quote]Schnappi?[/quote]

begint idd door te dringen bedankt voor de hulp. Groeten.

Dank Ihnen in Nizza Themen kann man das lösen meine Probleme hier, ich kann nicht gefunden werden ein CD-Key auf meiner Software dazu führen, dass, wenn ich es zu kaufen gibt es nichts info und wie wird verwendet, wenn ich die Software installieren es braucht eine Software, aber meine CD-Software hatte keine Software.

_____________________
[url=http://www.bestegutscheine.de]Gute Ratgeber[/url]


Find an integer $ x$ such that $ \frac{2}{3} < \frac{x}{5} < \frac{6}{7}$.

Set a common denominator of 105, so we get that $ \frac{70}{105}<\frac{21x}{105}<\frac{90}{105}$.  We see that $ x\equal{}4$ works.

You could also use cross multiplying and plug in numbers between 2-6 and do guess/check.

You could also just know 2/3=.666... So x is greater than 3 but less that 5 so it's 4

[quote="mathman523"]How could you use cross multiplication?[/quote]

From $\frac{2}{3}<\frac{x}{5}$, we have $10<3x$ or $3 \frac{1}{3}<x$ and from $\frac{x}{5}<\frac{6}{7}$ we have $7x<30$ or $x<4 \frac{2}{7}$. Hence $x=4$.


let $a,b,c>0$
Prove that:
$\frac{b^2+ac}{b^2+bc}+\frac{c^2+ba}{c^2+ca}+\frac{a^2+cb}{a^2+ab} \geq 3$

Inequality above is equivalent to the following:

$3 - \sum _{cyc} \frac{(b-a)c}{b(b+c)} \geq 3$

$\sum _{cyc} \frac{c(a-b)}{b(b+c)} \geq 0 $

$\sum _{cyc} ac^2(a-b)(a+b)(a+c) \geq 0$

$\sum _{cyc} (a^3c^2 - ab^2c^2)(a+c) \geq 0$

$\sum _{cyc} a^4c^2 - \sum _{cyc} c^3b^2a \geq 0$

and the last inequality is true due to rearrangement (or am I wrong ?  :? )

can you clarify that last ineq?

edIt:
nevermind, its obvious $\sum a^3(ac^2 - bc^2) \geq 0$ from rearrangement.


Let $ f$ be a function from the non-negative integers to the non-negative integers such that $ f(nm) \equal{} n f(m) \plus{} m f(n), f(10) \equal{} 19, f(12) \equal{} 52,$ and $ f(15) \equal{} 26.$ What is $ f(8)$?

A. 12
B. 24
C. 36
D. 48
E. 60

Such a function shares properties with the [url=http://web.mit.edu/lwest/www/intmain.pdf]number derivative[/url].

[hide]We can factor the different numbers to find a solution.

$ f(10)\equal{}5f(2)\plus{}2f(5)\equal{}19$
$ f(12)\equal{}3f(4)\plus{}4f(3)\equal{}52$
$ f(15)\equal{}3f(5)\plus{}5f(3)\equal{}26$

Also,
$ f(4)\equal{}2f(2)\plus{}2f(2)\equal{}4f(2)$

so

$ f(10)\equal{}5f(2)\plus{}2f(5)\equal{}19$
$ f(12)\equal{}12f(2)\plus{}4f(3)\equal{}52$
$ f(15)\equal{}3f(5)\plus{}5f(3)\equal{}26$

Solve this system as you wish (using matrices or whatever); I found that $ f(2)\equal{}1$ or $ f(2)\equal{}3$ by looking at the first equation.  Because the value of each function is negative and $ f(x)$ is always an integer, only these two values are possible for $ f(2)$.  It turns out that the solution is $ (f(2),f(3),f(5))\equal{}(3,4,2)$

Now,
$ f(8)\equal{}4f(2)\plus{}2f(4)\equal{}4f(2)\plus{}8f(2)\equal{}12f(2)\equal{}(12)(3)\equal{}36\equal{}>C$[/hide]

[hide="Two Ways"]

I'll first introduce extremely slick solution given by actual creators.

$ f(120) \equal{} 12 f(10) \plus{} 10 f(12) \equal{} 15 f(8) \plus{} 8 f(15)$

So, $ f(8) \equal{} \frac{12 \cdot 19 \plus{} 10 \cdot 52 \minus{} 8 \cdot 26}{15} \equal{} 36$. Really simple.

Here is how I solved it, not elegant but pretty straightforward.

$ 2 f(5) \plus{} 5 f(2) \equal{} 19$
$ 3 f(5) \plus{} 5 f(3) \equal{} 26$
Subtracting yields $ 5f(2) \equal{} 5 f(3) \plus{} f(5) \minus{} 7$

$ 3f(4) \plus{} 4 f(3) \equal{} 52$
$ 3 f(5) \plus{} 5 f(3) \equal{} 26$
Subtracting yields $ 3f(4) \equal{} 3f(5) \plus{} f(3) \plus{} 26$

Note that

$ f(8) \equal{} f(2 \cdot 4) \equal{} 4 f(2) \plus{} 2 f(4)$

Which can be also written as (substitution)

$ \frac{4}{5} (5 f(3) \plus{} f(5) \minus{} 7) \plus{} \frac{2}{3} (3 f(5) \plus{} f(3) \plus{} 26$

Simplifying this gives

$ \frac{14}{3} f(3) \plus{} \frac{14}{5} f(5) \plus{} \frac{176}{15}$

Note that $ \frac{14}{15} (5 f(3) \plus{} 3 f(5)) \plus{} \frac{176}{15}$ is same as above and we know the inside quantity to be 26. So, putting that in gives the answer of 36. [/hide]

[hide="Discussion"] Letting $ g(n) \equal{} \frac{f(n)}{n}$, we have $ g(nm) \equal{} g(n) \plus{} g(m)$.  Letting $ h(n) \equal{} e^{g(n)}$, we have $ h(mn) \equal{} h(m) h(n)$; in other words, $ h(n)$ is a completely multiplicative arithmetic function.  Such a function is uniquely determined by its values at the primes.  The given information allows us to determine $ f(2)$, which is all the information we need. [/hide]


This year I just barely made AIME on the AMC 10b (121.5)  :D

but I might have made a scantron mistake (I didn't fill in a wrong bubble, but I might have had some random lines on the scantron that I forgot to erase that the machine might  misread)

So what happens if I didn't erase the lines? Can I do something to fix the score?

I think there is an additional fee you can pay to have your test rescored by hand, but it may be too late for that now. Check into it in the Teacher's Manual on the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2008-9pub/TM-AMC_1012A,2009.pdf]AMC 10 Archives[/url]. It may take a while to load.

By the way, I freak out about this kind of stuff all the time, but it usually turns out okay.

Talk about scantron mistakes. At the VERY end of the test, I erased an answer i wasn't sure of and it ripped my scantron.... FML.

oh yeah... and the answer I tried to erase was correct.

In my years of experience here, with over 2 million forms scanned, the only errors that we have ever found have always been traced to incomplete erasures, circles not completely filled in, double marks in a row, wrinkled papers, using a pen to mark the form and other failures to follow the directions on the forms.

If your contest manager received the email with the initial score reports, then at the end of the report is a CSV file with the marks as the scanner read and reported them for your school.  Ask your Contest Manager for the information before proceeding further.

[quote]Can I do something to fix the score?[/quote]

There is a rescoring request form on page 12 of the AMC 10/AMC 12 Teacher's Manual.

Note that in the contest directions in the Teachers' Manual, Section II, Item 12, page 5, it says "the official answers will be the ones blackened on the answer form."

Steve Dunbar
MAA Director, American Mathematics Competitions

Thanks!  :)

especially AMCDirector... You've been a great help!


(1) Consider on the sides of a polygon orthogonal vectors of lengths proportional to the lengths of the sides, pointing outwards. Show that the sum of these vectors is zero.

(2) Orthogonal to each of a polyhedron consider a vector of length numerically equal to the area of that face, pointing outwards. Prove that the sum of these vectors is zero

 :D  :D

Here's a brief solution.

(1) This is almost self-evident. cuz the sum of vectors is as same as teh sum of sides of the polygon

with their direction $90\deg$ turned.

(2) We can use $S_1 = S_2 cos \theta$ when the angle between $S_1$ and $S_2$ is $\theta$.

considering some cases, we get exactly two opposite vectors. ...

I think my solution for (1) is the shortest one. But my solution for (2) is a little bit complicated.

I hope someone already find fantastic solution for (2). Can someone post it?  :D


Hi, what is wrong with the following "proof":

1 = squareroot((-1)(-1)) = squareroot(-1)*squareroot(-1) = i*i = -1

hence:

1 = -1

I suppose the sqrt((-1)(-1)) --- >  sqrt(-1)*sqrt(-1) is illegal, since my calc sais so.. but I dont understand why.

please help me on this one

I think this should be posted in the other topics forum.
Crus-
$1=\sqrt{(-1)^2}$
But not neccesarily is $1=-1$,
as,
$a^2=b^2$
$a=b$?
no
$a=+-b$
and $x^2=4$
$x=+-2$
So, that doesnt mean,
$-2=2$
Did that help?

oh, sorry, in which forum should it be posted?

hmm.. what you say is obviously true. but what is wrong with my statements?

You're not allowed to simplify roots like that.  It's a specific rule while defining imaginary numbers.

hmm, what are the rules for simplifying roots then?

a nice explination for that statement and this sites get a big PLUS (+)

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53709]my post[/url] in another thread.

Also, please do not double post.

I believe the rule is for taking negatives out of square roots.  I never learned the rule formally, but it was basically that for the square root of positive term will always be positive, and that you can't take out negative factors into their own square roots.


algebra
polynomial
function
functional equation
algebra proposed

Let the functional equation
$f(x)f(x+1)=f(x^{2}+x+1)$ (1)

a)find all polynomial solutions of (1)
b)find all continuous functions that satisfay (1)
c)find all defferentiable functions that satisfay (1)

umm... sure b and c don't need to be switched? Differentiable functions are a subspace of continuous functions. ;)

Here my solution
$P(x) = 1$ or $P(x) = 1$ for all $x$, satisfy that condition.
Assume that $P(x_{0}) = 1$ then $P(x_{0}^{2}+x_{0}+1) = 0$, implies that $degP = 2m$ where $m$ is an positive integer.
We set $P(x) = a(x^{2}+1)^{n}+Q(x)$.
From the hypothesis, we easily to show that $Q(x) = 0$ for all $x$and $a = 1$, then $P(x) = (x^{2}+1)^{n}$.

I'm sorry about my poor English!!


AMC USA(J)MO USAMO geometry probability MATHCOUNTS quadratics

It depends on your level really. I think PSS is good. If you're more of a begginer to average, I'd get First steps for math olympians.

//First steps for math olympians//

Huh? That book is like just AMC 12 stuff.

That's why I said "If you're more of a begginer to average", but yeah I shouldn't have posted that here. Sorry.

[quote="FantasyLover"]Hi,

I wanted to know what books are mostly used in preparing USAMO.

I know of ACoPS and Engel, but I wanted to know how lots of people think.

What books are used in preparing olympiad algebra, combinatorics, geometry, etc...

Thanks. :)[/quote]

Hmmm...both of those books are pretty complete books(i.e. they cover most of the subjects required), but both of them will have problems that arent really that good. I own both, but i dont really use them that much. Also, Engel is harder that ACoPS(if you were wondering). For individual subjects, this is what i use:
NT: 104 NT problems
Geo: I use some chinese book....so i cant really help here.
Algebra: I dont have any specific book...darn, i am not helpful again
Combo: i still use Intermediate Counting and Probability, although Engel combo is pretty good. I also heard "A path to combinatorics for undergraduates" is good.

Meh, these are just my preferences. However, you should always pick a book that challenges you. If you pick one thats around your level, you will find it boring very fast.(like i did with ACoPS).

[quote="abacadaea"] I also heard "A path to combinatorics for undergraduates" is good.[/quote]

It has some really nice problems and solutions, but it's way, way too notationally heavy; good perhaps for undergraduates who will have to learn to get used to that sort of thing, but terrible for Olympiad training. There are some problems with extremely ingenious and short solutions in there which, when written out, take up a full page of obfuscating symbols. It's still worth a read, though, particularly the bijections and Fubini's principle (though the latter chapter is particularly guilty of overemphasizing notation).

[quote="Ubemaya"]I also heard (and know from personal experience) that the MathCounts tests from 1995+ are excellent sources for practicing speed and not making mistakes on olympiads.[/quote]

What? Tests from before 1995 are actually even better in doing this.

I'd say that AoPS Volume 2 is a must have.

Here are my resources

Combo: erm... I used Inter Combo... Now just doing more problems... ask fwolth, he is beast at combo

Algebra: erm... I used Inter Alg... Now I just do problems...

Geo: Geometry Revisited is good

Number Theory: Number Theory as a reference in case I need anything... Most NT problems just need Order, PR, Quadratic Reciprocity, Wilson's, Fermat's, and Euler's... sometimes you need algebraic NT, rings, fields, etc. on harder problems...

C&P-Intro counting, Inter. Counting (I actually picked up most of these stuff myself by doing problems, my favorite topic  :lol: )

Alg- Inter. Algebra

Geo- Geometry Revisited (it's my weakest subject, so I'm gonna get Challenging problems in Geometry)

NT: Number Theory: Structures, Examples, and Problems

I haven't actually started the last two books, so I can't really account for them.

Also, you can find problems listed by topic on the wiki at various levels.

Hmm...this is what a certain anonymous black mopper suggests:

Algebra: do mop problems(you can find like 5/6 years of hw problems via google)
NT: do PEN problems
Combo: do Russia problems
he "fails" at geo, so he has no suggestions for that.

[quote]Ubemaya
I also heard (and know from personal experience) that the MathCounts tests from 1995+ are excellent sources for practicing speed and not making mistakes on olympiads.[/quote]

Not much funnier the second time :wink:

As for USAMO, there arent much books i know but Canadian olympiad is good for beginners i heard.

Ubemaya has posted "mathcounts to increase speed" on like every Olympiad, AIME, thread I've visited. It's much more than the 2nd time by now.

This is how I prep for USAMO:

AoPS 2
interm. Alg
geo revisited
ACoPS
PSS

and hopefully the result will be: WIN!!!!!

Well he (batteredbutnotdefeated) needs to start somewhere, he can't just jump into Olympiads and expect to do well.

Yeah... Actually I started with AoPS volume 1, then I went on to intro to C&P, and then intro to geo, and now AoPS volume 2 & interm alg... I'm looking forward to reading PSS, ACoPS, and GR(geometry revisited)

It helps if you really like the topic. I find challenging geometry very interesting, which makes thinks a whole lot easier. Unfortunately, I find NT very tedious... I also really enjoy learning, whenever I learn something new(any subject), it makes me feel like I've grown. Previous losses are wonderful motivators btw. failing at MC regionals really made me rethink my life.

gl FanFan.

What is PSS?

EDIT: And the author is Arthur Engel? Is that the same as the book Engel?

a book. It stands for Problem Solving Strategies.

Alright thanks everyone! :D

[quote="batteredbutnotdefeated"]gl FanFan.[/quote]
Huh I didn't know somebody was still calling me fanfan... :wink:

[quote="Zhero"][quote="abacadaea"]Hmmm...both of those books are pretty complete books(i.e. they cover most of the subjects required), but both of them will have problems that arent really that good. [/quote]
At that, Engel also has some [i]exceptionally good[/i] problems. By that, I mean problems that, if done, will literally trivialize problems on Olympiads a few years from now. By that, I mean a USA TST 2008 problem, and USAMO 2008 #1, and a German Olympiad problem, and few others that aren't on the top of my head.

Also, AOPS 2 mostly covers late AMC to easy AIME level math. It's nothing close to enough for Olympiads. I would say that PSS is probably the best book out there.[/quote]

It has some exceptionally good problems, but there are also some very "non-legit" problems in it as well (for example, direct applications of Muirhead). I wouldn't say it's the best source of problems.

I'm surprised that the IMO Compendium hasn't been mentioned yet. It's... expensive, but it's very useful.

IMO Compendium is simply the collection of all shortlists, some longlists, and IMO problems. Many foundable at the resources.

Engel's Problem Solving Strategies. Just... get it. Do ALL the problems. You will suddenly find that you are [i]pretty good at Olympiads[/i], especially because there are always shortlist problems almost every year that are VERY SIMILAR to Engel problems.

Geo specifically: Geometry Revisited.

I don't know enough of the rest ;) but just do problems. Contest section, PEN for NT, Olympiad/pre-Olympiad forums. Books are slightly overrated, but they are good for doing focused sets of problems and for actually explaining some stuff.

well, problems are easy to find... so i don't think we should mention this stuff

I'm surprised someone hasn't mentioned Problems from the Book... It has a lot of basic stuff in it, as well as challenging problems and solutions. It is pretty much the only book that i use. Then for geo, i just do problems

[quote="dnkywin"][quote="iYOA"]about how long does it take most people to [b]properly[/b] finish PSS?[/quote]

I don't know... it depends on what you mean by "proper". If you mean just understanding the concepts within the book, then usually most people take at most 2-3 months. On the other hand, if it means solving every problem in the book, I don't think anyone has done that yet... so $ \infty$  :)[/quote]

People have done it, but all of the people I have heard of who have done it are black moppers and have taken at least 2 months.

So it's definitely possible, just not that easy.

[quote="bpgbcg"]Problems from the Book is awesome.
Additionally, I just find random problems and do them.  (I don't have a very organized way of studying; I just find interesting problems and work on them.)  Also, WOOT is a really good source of problems, at least for me.[/quote]

In my opinion, doing random problems that look interesting is the best way of training for olympiads... but maybe not for geometry i guess?


Since it is going to be my first mathcounts year, I was wondering what are the average scores that you need to get in the state round to move on. (Target + Sprint)

And what is the average time per question in countdown.

All I know is that they give you 45 seconds to solve in countdown, but usually people get it within 10 seconds for the first 2 or 3, depending on which state you're in.

[quote="Ihatepie"]alanchou, are u from connecticut..[/quote]

No, not really  :P

[quote="Phelpedo"]High 39's probably to make top 4 in state.[/quote]

thanks phelpedo

@alancho

I could tell  :P

Let's see... two years ago when I got first it was 37/35/34/33.

35 is probably good enough.

[quote="jrshoch"]I'm reviving super old topics that no one will look at for a very long time because I can.

in 2004 the top four scores were 41/40/40/39
in 2005 the top four scores were 42/36/36/33

phelpedo yell at me if i'm wrong[/quote]Those sound right.

Huh...now who was the noob who got 33?  :wink:

I don't know but he can't be smart at all...

haha he almost didn't make it because that girl also got a 33  :P

Oh yeah. That was mad awkward, especially because he guessed #28 right  :ninja:

[quote="jhredsox93"]ha who was getting those 40's?

there hasn't been a 40+ for a long time now[/quote]jrshoch was a MC beast for a while. He won every year I went that he was there.


[quote="mr. math"]Im not sure how 2 do this problem:

Rick has 6 different pairs of socks. What's the probability that two randomly selected socks will be from a mathcing pair? Express your answer as a common fraction.

 :spider:  :starwars:[/quote]
[hide]There are 12 socks, $\binom{12}{2}$ ways to choose two socks. There are 6 ways to choose matching socks so $\frac{6}{\binom{12}{2}}=\frac{1}{11}$[/hide]

[hide]there's 6 pairs and 12C2 or 66 possibilities, so it's $\frac{6}{66}=\frac{1}{11}$[/hide]

[hide]If we think about it, the first sock we draw doesn't affect whether or not it will be a pair.  The second one will decide if it is a pair or not.  Since it is 1 sock that is needed and 11 total the answer is

$\frac{1}{11}$[/hide]

[hide]6 pairs of socks makes 12 socks to choose from to pick the first sock.  It does not matter what sock is chosen.  There is only one sock in the remaining 11 socks that will be the match, so the probability is $\frac{1}{11}$.

This is easier than doing number that work over total number, because there is less computation.[/hide]

Yes, but when more fractions are needed, then the methods that are not like mine and b-flats should be used.


doubts:
1) Glucose reacts with phenyl hydrazine(osazone formation) but it doesnt give 2,4 DNP test? why?
2) It reacts with hydroxylamine and hydrogen cyanide, but doesnt give sodium bisulphate test?

how do we know which addition reactions it gives?

Can someone explain structure of glucose penta-acetate? Straight chain or hemiacetal form?

Glucose doesnt show 2,4-DNP because glucose exisits in the pyranose (cyclic) structure and so it doesnt have any free aldehydic group...

the questions ur asking abt the possibility ...Some reactions are possible because the enrgy is enough to converted the cyclic pyranose structure to the linear struture...
some reactions are not possible due to more energy requirements...

I think that this is the only possible explantion.

The cyclic forms of carbohydrates are always in equilibrium with the open chain, aldehydic (or ketonic) form, and that form can react with a suitable nucleophilic reagent.

so basically we need to know which nuclephiles actually react with glucose and there is no way of predicting

The pyranose structure is what I have given. This does not have a free -OH at the anomeric C1 carbon and hence does not reduce tollen's and fehling's solution and does not show mutarotation.


[quote="srulikbd"]does $n!$ can be a square? (not 1 of course)[/quote]
No. You can easily prove that you will have a prime in the prime factorisation of $n!$, so that that prime occurs only once.
Proof: Assume that there will not be a prime in the factorisation so it will occur only once. Take the largest prime in prime factorisation of $n!$, let that prime be $p$. If that prime occurs more than once, then $2p\le n$. But by Bertrand's postulate(see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98535[/url] post from shalevbd), there will be a prime $q$ so that $p<q<2p$. Bu this contradicts the statement that $p$ is the largest prime in prime factorisation.
QED

Nice  :yup:

What about:

If $n$ is even, then there's a prime number $p$ such that $\frac{n}{2}< p< n$. (Bertrand  :agent:) This implies that $n< 2p$, so it appears just once in the prime decomposition of $n!$.
If $n$ is odd, then there's a prime number $p$ such that $\frac{n-1}{2}< p< n-1< n$. This implies that $n-1 < 2p$. So the only possibility for $p$ to appear twice in the prime decomposition of $n!$ is when $n=2p$. But $n$ was odd, so we're done
It's kind of the same of course  :D

can you prove it without bertrand? (in a simple way?)

[quote="JBL"][quote="rem"]bertrand is simple, I think the simplest way...[/quote]

Yes, just like it's simplest to prove $x^{4}+y^{4}= z^{4}$ has no positive integer solutions using Fermat's Last Theorem.  That doesn't mean you shouldn't try to do it other ways.[/quote]
The solution is like three lines, there are probably other ways, but longer. I will try to find some.
EDIT: it's hard to find other ways :( anyone found another way? :maybe:

ok use the fact that factorial canno't be square, prove the following:
1# $2\cdot4\cdot....\cdot2n$ canno't be a square.
2# $1\cdot3\cdot...\cdot(2n+1)$ cannot be a square.
3# $1\cdot2\cdot5\cdot6\cdot9\cdot10...$ cannot be a square.
maybe i'm wrong...put more please. thanks.

Well, for number 1, we can write:
$2*4*...*2n=(2^{n})n!$, and we can use the fact that we can find a prime that is present only once in prime factorisation of $n$, and the prime will not divide $2n$, so $2*4*...*2n$ cannot be a perfect square. The only case when the prime doe divide $2^{n}$ is when $n=1,2$ but it's easy to verify that in those cases $2*4*...*2n$ is not a square, QED

[quote="SBHD"]Here's something that's a bit harder (Austrian Olympiad)

Are there integers n, for which $n!+19^{93}$ is a perfect square?[/quote]

There doesn't exist any integer $n$ satisfying this condition.
The main thing to notice is that $19^{93}\equiv 3\pmod 4$. Proof is easy: $19\equiv-1\pmod 4$, so $19^{93}\equiv-1\equiv 3\pmod 4$. Now we see that if $4\mid n!$, then $n!+19^{93}$ can't be a perfect square. We have only to check if $19^{93}+1,2,6$ are squares. For sure $19^{93}+1$ is not a square, because $19^{93}+1=x^{2}\iff (x-1)(x+1)=19^{93}$ and it means that $19^{93}$ can be factorised into $2$ factors whose difference is $2$-contradiction, because there doesn't exist a pair of powers of $19$ whose difference is $2$. We will prove that also $19^{93}+2$ and $19^{93}+6$ aren't squares.
[b]1.[/b] $19^{93}+2$
We will consider it $\bmod{8}$. $19^{93}\equiv 3^{93}=9^{46}\cdot 3\equiv 3\pmod 8$, so $19^{93}+2\equiv 5\pmod 8$ and $5$ isn't quadratic residue $\bmod{8}$ which ends the proof. $\blacksquare$
[b]2.[/b] $19^{93}+6$
We will consider it $\bmod 7$. $19^{93}\equiv \left(-2\right)^{93}=-2^{93}=-\left(2^{6}\right)^{15}\cdot 8\equiv-1^{15}\cdot 1=-1\pmod 7$, so $19^{93}+6\equiv 5\pmod 7$, but $5$ isn't qudratic residue $\bmod 7$ which ends the proof. $\blacksquare$

Do you know the official solution?

The official solution is nearly the same as yours.


geometry
modular arithmetic
number theory

how many pairs (x,y) of positive integers satisfy $ 2x\plus{}7y\equal{}1000$?

Well first you find a solution, such as $ x \equal{} 150$ and $ y \equal{} 100$.
Then we find that possible solutions are in the form of $ (150 \minus{} 7d,100 \plus{} 2d)$

Since each pair of $ (x,y)$ needs to be positive, we look for the minimum positive value of $ 100 \plus{} 2d$, which is $ 2$ when $ d \equal{} \minus{} 49$. Then the minimum positive value of $ 150 \minus{} 7d$ which is $ 3$ when $ d \equal{} 21$. So the total amount of 71 pairs.

how did you figure out that they are in the form $ (150\minus{}7d,100\plus{}2d)$ ?

Well it's actually something I learned from a number theory book, I'm not sure how to explain it.

But if you plug in those values for $ x$ and $ y$ this is what you get:

$ 2x \plus{} 7y \equal{} 1000$
$ 2(150 \minus{} 7d) \plus{} 7(100 \plus{} 2d) \equal{} 1000$
$ 300 \minus{} 14d \plus{} 700 \plus{} 14d \equal{} 1000$
$ 1000 \equal{} 1000$

So it does work, and here is a link to a site that contains a similar problem and a lengthy solution with the method that I used to get that [url=http://mathforum.org/library/drmath/view/51595.html]link[/url]... One specific area that I used was this part:

If $ x \equal{} x_0$ and $ y \equal{} y_0$ are the initial solutions, then the family of
solutions is given by:

     $ x \equal{} x_0 \plus{} \left(\dfrac{b}{d}\right)\cdot t$

     $ y \equal{} y_0 \minus{} \left(\dfrac{a}{d}\right)\cdot t$

where $ d \equal{} gcd(a,b)$ and t is an arbitrary integer.

The solution to your problem is basically the solving of a linear congruence.  So, to make things easier for myself  :lol: , I'm going to assume you know basic modular arithmetic (although I don't see how there's an intermediate or advanced level of modular arithmetic knowledge).  If you don't know modular arithmetic, then my solution probably won't make sense to you, and then I'll post again with an explanation of modular arithmetic and a (hopefully) easy-to-understand solution.

Solution:
We start by viewing both sides of the given equation $ \bmod{7}$ (we do this because it eliminates one of the variables, namely $ y$).  This yields $ 2x\equiv6\pmod{7}$.  We can "divide" both sides by $ 2$ since $ gcd(2,7)\equal{}1$ (you should understand why this division is allowed if you know enough modular arithmetic).  This yields $ x\equiv3\pmod{7}$.  No further simplification is possible, so we can transform this congruence back to an equation by writing $ x\equal{}7d\plus{}3$ for any integer $ d$.  Substituting this expression for $ x$ in the given equation, we can then solve for $ y$ in terms of $ d$: $ 2x\plus{}7y\equal{}1000\Rightarrow14d\plus{}7y\plus{}6\equal{}1000\Rightarrow y\equal{}142\minus{}2d$.
So we can finally express our answer completely in terms of $ d$ such that $ d$ can be any integer (so there is an infinite number of solutions): $ \boxed{(x,y)\equal{}(7d\plus{}3,142\minus{}2d)}$.

Now, note that my boxed answer is actually just another way of writing tonypr's solution (so we are both correct).  Basically, we're saying that $ x$ can be any multiple of $ 7$ added to $ 3$ (tonypr's $ x$ value does indeed allow this), and that $ y$ is any even integer (although it depends on $ x$ because it depends on the choice of $ d$).

Sorry if I'm being unclear; it's late at night and I don't have much time to post right now.

Sorry for your long post, however, I have no knowledge of modular arithmetic whatsoever, and I don't know where i can go to learn about it.  :mad:

First observe that $ 2x,\ 1000$ are even, so $ y$ should be also even, let $ y \equal{} 2m$ for some positive integer $ m$,

we have $ 2x \plus{} 7\cdot 2m \equal{} 1000\Longleftrightarrow x \plus{} 7m \equal{} 500$.


integration
calculus
logarithms
algebra
partial fractions
calculus computations

This is driving me crazy now.

I am supposed to find a primitive to $\frac{1}{x\sqrt{x^2 + 1}}$.
We set $y = x + \sqrt{x^2 + 1}$. From this follows that $(y - x)^2 = x^2 + 1$ from which we get that $x = \frac{y^2 - 1}{2y}$. From this it follows that $\sqrt{x^2 + 1} = y - x = \frac{y^2 + 1}{2y}$ and that $dx = \frac{y^2 + 1}{2y^2}dy$.

From all this it follows that
\[ \int\frac{dx}{x\sqrt{x^2 + 1}} = \int\frac{2y}{y^2 - 1}\cdot\frac{2y}{y^2 + 1}\cdot\frac{y^2+1}{2y^2}dy = \int\frac{2}{y^2 - 1}dy  \]

Partial fractions gives that $\int\frac{2}{y^2 - 1}dy = \ln|y-1| - \ln|y+1| + C = ...$

Now, this is not the right answer. I've been looking for my mistake for like half an hour, and this is bugging me like mad. Please help! :(

edit: I just figured out that this is actually right. What a waste of time. The book answers $\ln|x| - \ln(1 + \sqrt{x^2 + 1}) + C$. Maple answers $-g(\frac{1}{x^2+1})$, where g is "arctanh". How are these my answers at all?

Here's how I evaluated the integral:

Make the substitution: $x=\sinh t \Rightarrow dx = \cosh t \ dt$, then we have
\[ \int \frac{1}{x\sqrt{x^2+1}} \ dx = \int \frac{\cosh t}{\sinh t \sqrt{\sinh^2 t +1}} \ dt = \int \frac{1}{\sinh t} \ dt = \ln(\tanh\frac{t}{2})  \]

Using the identites $\tanh \frac{\alpha}{2} = \frac{\tanh \alpha \sinh\alpha}{\tanh \alpha + \sinh \alpha}$ and $\tanh(\sinh^{-1} \alpha ) = \frac{\alpha}{\sqrt{\alpha^2+1}}$, we get
\[ \ln(\tanh\frac{\sinh^{-1} x}{2}) = \ln(\frac{\frac{x}{\sqrt{x^2+1}}\cdot x}{\frac{x}{\sqrt{x^2+1}}+x}) = \ln(\frac{x}{\sqrt{x^2+1}+1})  \]

and thus the integral is

\[{ \ln|x|-\ln(\sqrt{x^2+1}+1}) + C  \]

Also, your answer is the same as mine (and the book):

$\frac{x+\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}+1} =\frac{x}{\sqrt{x^2+1}+1}$

Here's the proof:

Let $x = \sinh t$, then on the left, we have:
\[ \frac{\sinh t+\cosh t -1}{\sinh t+ \cosh t +1} = \frac{e^t-1}{e^t+1} = \tanh \frac{t}{2} = \tanh(\frac{\sinh^{-1} x}{2}) = \frac{x}{\sqrt{x^2+1}+1}  \]


Given the equation $ x^3\minus{}2x^2\plus{}x\minus{}3\equal{}0$, write the equation whose roots are each 2 less than the roots of the given equation.

Two forces, one of 120 lbs and the other of 200 lbs, act on the same object at angles 30 and -30, respectively, with the positive x-axis. Find the direction of the resultant of these two forces.

The first is obviously $ (x\plus{}2)^3\minus{}2(x\plus{}2)^2\plus{}(x\plus{}2)\minus{}3\equal{}0$

[quote="dgreenb801"]The first is obviously $ (x \plus{} 2)^3 \minus{} 2(x \plus{} 2)^2 \plus{} (x \plus{} 2) \minus{} 3 \equal{} 0$[/quote]

Which does in fact simplify: $ x^3\minus{}3x\minus{}1$ was what I got.

For the first one you need a horizontal translation of 2 to the left, so you must change all x to x+2. Doing this gives you $ (x \plus{} 2)^3 \minus{} 2(x \plus{} 2)^2 \plus{} (x \plus{} 2) \minus{} 3 \equal{} 0$, which simplifies into $ x^3\minus{}3x\minus{}1\equal{}0$ if I did things correctly.


calculus
integration
trigonometry
limit
geometry
function
3D geometry

Let $f(x)=\pi x^{2}\sin \pi x^{2}.$ Prove that the volume $V$ formed by the revolution of the figure surrounded by the part $0\leq x\leq 1$ of the graph of $y=f(x)$ and $x$ axis about $y$ axis is can be given as $2\pi \int_{0}^{1}xf(x)\ dx$ then find the value of $V.$

Consider a cylindrical shell of width $\delta{x}x$ height f(x) and radius x

$\delta{V}\V=2\pix{xf(x)}\delta{x}x$
$V=\lim_{\delta{x}x\to 0}\sum_{x=0}^{1}\ 2\pi{xf(x)}\delta{x}x =2\pi\int_{0}^{1}xf(x)dx =2\pi^{2}\int_{0}^{1}{x^{3}}\sin\pi{x^{2}}{dx}$

letting $u={x^{2}}$ reduces the integral into

$\pi^{2}\int_{0}^{1}usin\pi{u}{dx}=\pi^{2}\frac{1}{\pi}[-ucos\pi{u}]_{0}^{1}+\pi^{2}\int_{0}^{1}cos\pi{udu}=\pi$

$2\pi \int_{0}^{1}xf(x) \ dx = 2\pi\int_{0}^{1}\pi x^{3}\sin\left(\pi x^{2}\right) \ dx$
Let $u=\pi x^{2}$, then we have
$V= \int_{0}^\pi u\sin u \ du = \pi$

That's O.K. but How about the former question, Jimmy.

Was there anything wrong with yip's answer to the first question?

We can't confirm the explanation of yip's one.

Regrettably maybe this explain will get 8 points at best of full marks 20 points in case of Tokyo University.

Each cylindrical shell has a surface area of $A= 2\pi xf(x)$, then using the formula $V=\int_{a}^{b}A(x) \ dx$, we have \[V=2\pi\int_{0}^{1}xf(x) \ dx\] To get the formula for volume, take a function $f(x)$, $a\le x \le b$  and divide the interval into $n$ segments, and then $\Delta x = \frac{b-a}{n}$, then choose a point from each subinterval $x_{i}^{*}$ and then $V_{i}= A(x_{i}^{*})\Delta x$, where $A(x_{i}^{*})$ denotes the area of the face of each disk, and finally $V = \lim_{n\to\infty}\sum_{i=1}^{n}A(x_{i}^{*}) \Delta x = \int_{a}^{b}A(x) dx$

Maybe you get full marks. In my opinion, the volume of the tube with the hight $f(x),$ the width $\delta x,$ inner radius $x$ is $\pi f(x)\{2x\delta x-(\delta x)^{2}\}$, thus....

You can find the volume in using basic method with integral by parts.

Someone will post the solution. :lol:

Im not sure about Japan, but in Australia high schools, they generally use that reasoning for the proof of volume by cylindrical shells, where u take one shell, "roll it out", and find the volume as a rectangular prism, then sum it over the required limits


linear algebra
matrix
linear algebra unsolved

I have a problem. I don´t know, how to calculate this. pls help me.
a + 3b +5c - 4d + 0 = 1
a + 3b + 2c -2d + e = -1
a -  2b + c - d - e = 3
a -  4b + c +d - e = 3
a + 2b + c - d +e = -1
After I must do this.
| 1  3  5 -4  0 | 1 |
| 1  3  2 -2  1 | -1 |
| 1  -2  1 -1  -1 | 3 |
| 1  -4  1 1  -1 | 3 |
| 1  2  1 -1  1 | -1 |
And now I don´t know how is next.


Find all solutions (x,y,z) between 0 and 1 not inclusive ie. (0,1), satisfying x^2 + y^2 +z^2 + 2xyz = 1

[hide="Just by observation"] .5, .5, .5 :D [/hide]

Identity, HI!
Say $a+b+c = S$ and $x = \cos a$, $y = \cos b$, and $z = \cos c$.
Then
$\cos^2 a + \cos^2 b + \cos^2 c = 1 - 2\cos a \cos b \cos c$
only holds if what about $S$ is true. Hope this helps.


Q1The sales price of a car is $12,590, which is 20% off the original price. What is the original price? Q2 if Leah is 6 years older than her sister, Sue, and John is 5 years older than Leah, and the total of their ages is 41. Then how old is Sue? Q3Alfred wants to invest
$4,000 at 6% simple interest rate for 5 years. How much interest will he receive?

[quote="ekta"]Q1The sales price of a car is 12,590, which is 20% off the original price. What is the original price? Q2 if Leah is 6 years older than her sister, Sue, and John is 5 years older than Leah, and the total of their ages is 41. Then how old is Sue? Q3Alfred wants to invest 4,000 at 6% simple interest rate for 5 years. How much interest will he receive?[/quote]
Is this your homework or something? This is really easy to do. You should probably get a pre-algebra book.


trigonometry
geometry
circumcircle
analytic geometry
number theory
relatively prime
trig identities

In triangle $ABC$, $AB=13$, $BC=15$, and $CA=14$. Point $D$ is on $\overline{BC}$ with $CD=6.$  Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$

[quote="4everwise"]Point $D$ is on $BC$ with side length $6$.[/quote]

Huh? :huh:

does anybody care to clarify, because i am sure that the wording was more clear on the actual test, and if you say that either DB or DC =6, then you get 2 different values for BE

I fixed the problem statement.  You can now solve it :D

[hide="hint"]before i saw the math jam solution, i could not solve it without a terrible mess of equations, but the solution lies in using the 2 congruent angles in an elegant way, (i.e. not using the law of cosines), think parallel postulate[/hide]

By using the cosin theorem on the triangle ABC,ACD and ABE we have:cosC,AD,cos CAD,cos BAE,BE,q=827

Applying Stewart's Thm thrice also seems to work. ;)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=192501]Here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=193339]here[/url] are a few places this has been discussed.

Urgh, that's hard...I only noticed that one side is 13 and tryed with Pythagorean triples but didn't do more :blush:

By law of cosines,we can figure out COS DAC=13/ 205,then  cos BAE=13/ 205   sin BAE=6/ 205     cos ABE=33/65  sin ABE=56/65  sin AEB=sin( EAB+ ABE)=926/65/ 205    by extended law of sines   a/sinA=b/sinB   BE/AB=sin BAE /sin AEB      SO,BE=2535/463     SO, q=463

Extend $AD$ to meet the circumcircle of the triangle $\Delta{ABC}$ at $F$.

By power of a point, we have $DF = \frac{BD\cdot DC}{AD}$, So $AF=AD+DF= \frac{AD^2+BD\cdot DC}{AD}$

Since $\Delta{CFD} \sim \Delta{ABD}$, we have $CF = \frac{AB\cdot CD}{AD}$.

Since $\Delta{ABE} \sim \Delta{AFC}$,  we have $EB = \frac{AB\cdot CF}{AF} = \frac{AB^2\cdot CD}{AD^2+BD\cdot DC}$

By Stewart's theorem: $BC\cdot (AD^2+BD\cdot DC) = AC^2\cdot BD + AB^2\cdot CD$

Therefore, $EB = \frac{AB^2\cdot CD\cdot BC}{AC^2\cdot BD + AB^2\cdot CD} = \frac{13^2\cdot6\cdot15}{14^2\cdot9+13^2\cdot6}=\frac{13^2\cdot15}{463}$

Before we start, let it be noted that the infamous 13-14-15 triangle is just a 5-12-13 right triangle and a 9-12-15 right triangle being put together with the sides of 12 put together as the height and the sides of 5 and 9 running together to make the side of 14. Using this fact we can easily find the sines and cosines of the angles of the 13-14-15 triangle.

Say $\angle DAC = \theta$
First we must find $\tan\theta$. Reason will be obvious later.
Given $\Delta ADC$, we drop the altitude from $D$ to a point $R$, so $\tan\theta = \frac{DR}{AR} = \frac{DR}{AC-RC}$. We know $\sin(C) = \frac{4}{5}$ and $\cos(C) = \frac{3}{5}$. These give us two facts: $\sin(C) = \frac{4}{5} = \frac{DR}{6} \Rightarrow DR = \frac{24}{5}$ and $\cos(C) = \frac{3}{5} = \frac{RC}{6} \Rightarrow RC = \frac{18}{5}$ Using these, we get $\tan\theta = \frac{\frac{24}{5}}{14-\frac{18}{5}} = \frac{6}{13}$.
Now to set up a coordinate system.
Points:
$A(0,0),B(5,12),$ and $C(14,0)$ give us
Lines:
$\displaystyle AD \Rightarrow y=\tan(\theta)x = \frac{6}{13}x \\ AB \Rightarrow y=\tan(A)x=\frac{12}{5}x \\ AE \Rightarrow y=\tan(A-\theta)x \\ BC \Rightarrow \frac{y-0}{x-14}=\frac{12-0}{5-14} \Rightarrow y = -\frac{4}{3}(x-14)$
Also, $\displaystyle\tan(A-\theta)=\frac{\tan(A)-\tan(\theta)}{1+\tan(A)\tan(\theta)}=\frac{\frac{12}{5}-\frac{6}{13}}{1+\frac{12}{5}\frac{6}{13}}=\frac{12(13)-6(5)}{13(5)+12(6)}=\frac{126}{137}$
We find point E based on the intersection of lines BC and AE, but we only need find the x-coordinate of E.
\begin{eqnarray*} y=\frac{126}{137}x&=&-\frac{4}{3}(x-14)\\ \frac{189}{274}x&=&(14-x)\\ \frac{463}{274}x&=&14\\ x&=&14\left(\frac{274}{463}\right)=\frac{3836}{463}\\ \end{eqnarray*}
If we drop the altitude from $B$ to a point $P$, and if we drop the perpendicular from $E$ to $AC$ we get a point $Q$. Since $\angle BPC \cong \angle EQC \cong 90$ and $\angle BCP \cong \angle ECQ$, $\Delta EQC \sim \Delta BPC$. Also, $CQ = AC - AQ = 14 - x$ where $x$ is the x-coordinate of E. By similir triangles,
\begin{eqnarray*} \frac{EC}{BC}&=&\frac{QC}{PC}\\ EC&=&BC\frac{14-x}{9}\\ BC-BE&=&BC\frac{14-x}{9}\\ BE&=&BC\left(1-\frac{14-x}{9}\right)\\ BE&=&5\left(\frac{x-5}{3}\right)\\ BE&=&5\left(\frac{\frac{3836}{463}-5}{3}\right)\\ BE&=&5\left(\frac{1521}{3(463)}\right)\\ BE&=&5\left(\frac{507}{463}\right)\\ \end{eqnarray*}

Ha...wow I remember spending so much time on this problem. Looking back, it is totally easy.

Lemma: $ \frac{BE}{EC}\cdot \frac{BD}{DC}\equal{}\left(\frac{AB}{AC}\right)^2$. Proof: Apply the sine area formula with angles at $ A$. Then

$ \frac{BE}{EC}\equal{}\frac{6}{9}*\frac{13^2}{14^2}\implies \frac{BE}{BE\plus{}EC}\equal{}\frac{BE}{15}\equal{}\frac{2*13^2}{2*13^2\plus{}3*14^2}$

Sorry for the revive, but can somebody please prove Altheman's lemma?

That's [b]Steiner[/b] relation!

Best regards,
sunken rock

Yeh, if you know steiner then you'll solve it in no time, but other wise you might get stuck with an ugly, very not general, solution.
there are many proofs of steiner I'll write one now and try to show the others later .[let $\angle CAD=\angle EAB=\alpha$,$\angle DAE =\beta$]
applying law os sines:
$\triangle ACE:\frac {CE}{sin(\alpha +\beta)}=\frac {AE}{sin C}.....(1)\\\triangle ABD :\frac {sin(\alpha +\beta)}{BD}=\frac {sinB}{AD}.....(2)\\\triangle AEB: \frac {AE}{sinB}=\frac {EB}{sin\alpha}\Rightarrow AE=\frac{sin B}{sin\alpha}BE.....(3)\\\triangle ADC:\frac {sinC}{AD}=\frac{sin\alpha}{CD}\Rightarrow \frac {1}{AD}=\frac{sin\alpha}{sinC}\frac{1}{CD}.....(4)$
multiplying (1),(2) and plugging (3),(4):
$\frac{CE}{BD}=\frac{EB}{CD}.\frac{sin^2 B}{sin^2 C}\\\Rightarrow \frac{CE}{BD}.\frac{CD}{BE}=\frac{CE}{BE}.\frac{CD}{BD}=(\frac{sin B}{sinC})^2$
but from the law of sines on triangle ABC:
$\frac{sinB}{sinC}=\frac{AC}{AB}$
plugging this gives us the relation we want :)

there is anther way which is nicer, but harder to see :
we draw a line parallel to AC and passes through B, Let the intersection between this line and the line AE be the point E', similarly we define D'.
again let the angles be denoted by alpha and beta ( in the same way as the pervious comment)
now notice that :
$in\;\triangle ACD':\\\angle D'CA=B+C\\\angle CAD'=\alpha\\in\;\triangle AE'B:\\\angle E'AB= \alpha \\\angle ABE'=B+C \\\so\;\triangle ACD' \;is\;similar\;to\; \triangle ABE'\\\Rightarrow \frac {AC}{AB}=\frac{CD'}{BE'}.....(1)\\ AB||CD'\Rightarrow \triangle CD'D \;is\;similar\;to\; \triangle BAD \\\Rightarrow \frac{CD'}{AB}=\frac{CD}{BD}\\\Rightarrow CD'=\frac{CD}{BD} . AB...(2)\%Error. "similarly" is a bad command.
\;:\\E'B=\frac{EB}{EC} AC\\\Rightarrow \frac{1}{E'B}=\frac{EC}{EB.AC}.....(3)$
multiplying (2),(3):
$\frac{CD'}{E'B}=\frac{CD.AB.EC}{BD.EB.AC}$
plugging in (1):
$\frac{AC}{AB}=\frac{EC}{EB}.\frac{CD}{BD}.\frac{AB}{AC}$
multiplying be $\frac{AB}{AC}$ we're done
(sorry for skipping a lot  :roll: )
there is anther way (maybe you should try to do it by your self) which the construction in it is a parallel to AC that basses through E (and the same with D,AB)

Note that $ \frac{13sin \angle BAE}{14sin \angle CAE}= \frac{BE}{15-BE}$ and $ \frac{13sin \angle BAD}{14sin \angle CAD}=\frac{9}{6}.$ Then use $ \angle BAE\cong\angle CAD$ and $ \angle BAD\cong\angle CAE$ and solve for $BE.$

This is killed by barycentrics. Note that $D = (0:2:3)$. Observe that since $AD$ and $AE$ are isogonal conjugates, we have $E = \left(0: \frac{b^2}{2} : \frac{c^2}{3} \right ) = \left (0 : \frac{14^2}{2} : \frac{13^2}{3} \right)$. Thus, letting $BE = x$, we have
$$\frac{x}{15} = \frac{\frac{13^2}{3}}{\frac{13^2}{3} + \frac{14^2}{2}} \implies x = \frac{15 \cdot 13^2}{463}$$

Let $\angle{DAC}=\theta$, $\angle{DAE=\varphi}$, and $DE=x$. From two applications of Ratio Lemma (both on $\triangle{ABC}$, with cevians $AE$ and $AD$ respectively), $$\frac{13}{14}=\frac{3\sin{(\theta+\varphi)}}{2\sin{\theta}} \text{  and  } \frac{13}{14}=\frac{(9-x)\sin{\theta}}{(6+x)\sin{(\theta+\varphi})}.$$ Hence,
\begin{align*}
\left(\frac{13}{14}\right)^2 &= \frac{3(9-x)}{2(6+x)} \\ \implies 9-x & = \frac{2535}{463}.
\end{align*}
So $q=\boxed{463}$.

This problem becomes extremely trivial given one observation:

[hide]
Observe that because $\angle BAE \cong \angle CAD$, it follows that $AE$ and $AD$ are isogonal; therefore,  $$\frac{BE}{EC}\cdot \frac{BD}{DC}\equal{}\left(\frac{AB}{AC}\right)^2$$ Letting $BE=x$ and $EC=15-x$ and solving for $x$, we see that $x=\frac{2535}{463}$, so $q=\boxed{463}$.

[/hide]

@Gibby (even though its four years later) the general result is a corollary of the isogonal results presented in EGMO.

Applying Stewart's Theorem on $\Delta$ $ABC$, we have that $$(9)(6)(15) + DA^2(15) = 14^2 \cdot 9 + 13^2 \cdot 6 \implies DA = \sqrt{\frac{656}{5}}.$$
Using the Law of Cosines on $\Delta ADC$, we have that $$196 + \frac{656}{5}-28\sqrt{\frac{656}{5}}\cos(DAC) = 36 \implies \cos(DAC) = \frac{13}{\sqrt{205}}, \sin(DAC) = \frac{6}{\sqrt{205}}.$$ We also know that $$[ABC] = 84 = \frac{1}{2}cb\sin(BAC) \implies \sin(BAC) = \frac{12}{13}, \cos(BAC) = \frac{5}{13}.$$ Hence, $$\sin(EAC) = \sin(BAC - BAE) = \sin(BAC - DAC)= \sin(BAC)\cos(DAC)-\sin(DAC)\cos(BAC) = \frac{126}{13\sqrt{205}}.$$ Additionally, $$\frac{[BAE]}{[EAC]} = \frac{7AE\sin(EAC)}{\frac{13AE\sin(BAE)}{2}}=\frac{294}{169}.$$ Since $\frac{[BAE]}{[EAC]} = \frac{BE}{EC} = \frac{BE}{15-BE}$, we know that $\frac{BE}{15-BE} = \frac{294}{169}$, so $BE = \frac{2535}{463}$. $\boxed{463}$.

Side-Note: There's some messy computation, and I left that out on the presumption that you can do pre-algebra level arithmetic. I didn't really want to clutter my post with things people can already do. Plus, time is precious!

[hide=Solution (why is this in HSO?)]

Let $\angle BAE = \angle CAD = \theta ,$ and let $h_1, h_2$ be the (perpendicular) distances from $D$ to $\overline{AC}$ and $E$ to $\overline{AB}$, respectively.
Using the [hide=well known angles of a 13-14-15,]It is just a 9-12-15 right triangle joined with a 5-12-13 right triangle at the leg of length 12; from this we can derive $\angle BCA = \cos^{-1} \tfrac{3}{5}, \angle BAC = \cos^{-1} \tfrac{5}{13}$, and lastly $\angle ABC = \cos^{-1} \tfrac{33}{65}$ from angle sum formulas.[/hide], we have $ \tfrac{h_1}{\tan \theta} + 6 \cos \angle BCA = \tfrac{h_1}{\tan \theta} + \tfrac{18}{5} = 14.$ In addition, $h_1 = 6\sin \angle BCA = \tfrac{24}{5}$.

So, $\tfrac{h_1}{\tan \theta} = \tfrac{52}{5}, \tfrac{1}{\tan \theta} = \tfrac{13}{6} .$

We can derive similar equations with $h_2$ and $BE$: $h_2 = BE \sin \angle ABC = \tfrac{56}{65} BE ,$ and $\tfrac{h_2}{\tan \theta} + BE \cos \angle ABC = 13.$ Since we know what $\tfrac{1}{\tan \theta}$ is we can solve for BE:
\begin{align*} BE \cdot \tfrac{56}{65} \cdot \tfrac{13}{6} + BE \cdot \tfrac{33}{65} = 13 \\ BE \left( \tfrac{56 \cdot 13 + 6\cdot 33}{6 \cdot 65} \right) = 13 \\ BE \cdot \tfrac{463}{3\cdot 65} = 13 \\ BE = \tfrac{3\cdot 13 \cdot 65}{463} \end{align*}

A quick test shows that $\boxed{463}$ does not divide $3, 5, $ or $13$, so it is the answer.

[/hide]

[quote=franzliszt]Notice that $AD$ and $AE$ are isogonal.

Theorem: Let $D$ and $E$ be points on $\overline{BC}$ so that $\overline{AD}$ and $\overline{AE}$ are isogonal. Then$$\frac{BD}{DC}\cdot\frac{BE}{EC}=\left(\frac{AB}{AC}\right)^2.$$Proof. We use directed angles and the Law of Sines and the fact that $\sin \theta = \sin (180^\circ-\theta)$.

Notice $\frac{\sin \measuredangle BAD}{BD}=\frac{\sin \measuredangle ADB}{AB}$ and $\frac{\sin \measuredangle CDA}{AC}=\frac{\sin \measuredangle DAC}{DC}$. Multiplying, dividing, and rearranging gives$$\frac{AB}{AC}=\frac{BD}{DC}\cdot \frac{\sin \measuredangle DAC}{\sin \measuredangle BAD}.$$Notice $\frac{\sin \measuredangle EAC}{EC}=\frac{\sin \measuredangle CEA}{AC}$ and $\frac{\sin \measuredangle BAE}{BE}=\frac{\sin \measuredangle AEB}{AB}$. Multiplying, dividing, and rearranging gives$$\frac{AB}{AC}=\frac{BE}{EC}\cdot \frac{\sin \measuredangle EAC}{\sin \measuredangle BAE}.$$Multiply the two equations to get the desired$$\frac{BD}{DC}\cdot\frac{BE}{EC}=\left(\frac{AB}{AC}\right)^2.$$$\square$

So now this is trivial. We have $\frac96\cdot\frac{BE}{15-BE}=\left(\frac{13}{14}\right)^2 \Rightarrow BE=\frac{2535}{463}$.[/quote]

nice solution! how didnt i think of that!

Thanks! The theorem I used was Theorem 4.22 in EGMO (Isogonal Ratios). So EGMO does help with AIME. :)

kinda quick but requires heavy computation

Let $\angle BAE = \angle CAD = \theta$. We know that the area of $\triangle CAD = 6 \cdot \frac{56}{5}\cdot \frac{1}{2},$ since the height of the triangle is $\frac{56}{5}$ with base $CD=6$ (this is because the area of the triangle is $84$, so the altitude is $\frac{84\cdot 2}{15}$.) By Pythag, we can find that $AD=\frac{4\sqrt{205}}{5}$. Then using the Sin area formula, we get $$3\cdot \frac{56}{5}=\frac{4\sqrt{205}}{5} \cdot 14 \cdot \frac{1}{2} \cdot \sin\theta\implies \sin\theta = \frac{6}{\sqrt{205}}.$$ Now let $BE=x$. We again know the area of $\triangle ABE=\frac{1}{2}\cdot x \cdot \frac{56}{5}.$ Then by Pythag, we find that $AE = \sqrt{x^2-\frac{66x}{5}+169}.$ We can use the Sin area formula to get $$\frac{28x}{5}=\frac{1}{2}\cdot 13 \cdot \sqrt{x^2-\frac{66x}{5}+169}\cdot \frac{6}{\sqrt{205}}.$$ Simplifying, we get $\frac{28\sqrt{205}x}{15\cdot 13} = \sqrt{x^2-\frac{66x}{5}+169}.$ Squaring both sides, we get $\frac{784 \cdot 205 x^2}{15^2 \cdot 13^2} = x^2-\frac{66x}{5}+169.$ Multiplying both sides by 5, we get $$\frac{784\cdot 41 x^2}{169 \cdot 9} = 5x^2-66x+845.$$ We can simplify this further to get $\frac{24539}{1521}x^2+66x-845=0.$ Now we can apply the quadratic formula to get $$x=\frac{-33\pm \sqrt{1089+845\cdot\frac{24539} {1521}}}{\frac{24539}{1521}}.$$ This simplifies further to $$x=\frac{1521(-33 \pm \sqrt{1089 + \frac{5\cdot 24539}{9}})}{24539}.$$ We can remove the square root and the $\pm$ sign because we know $x$ is positive, so we get $$x=\frac{1521(-33+\frac{364}{3})}{24539} = \frac{507(364-99)}{24539} = \frac{507(265)}{24539}=\frac{2535}{\boxed{463}}.$$

first of all why BC=15 AC=14 and not AC=15 BC=14

also how is this even an aime 14

Let $BE=x$. Note that $AE$  and $AD$ are isogonal, so we have $$\frac{9x}{6(15-x)}=(13/14)^2,$$ from which $x=2535/463$ so the answer is 463

By isogonal lines, BE*BD/(CD*EC)=AB^2/BC^2, which implies 15/BE-1=(BC-BE)/BE=CE/BE=294/169. Solving yields the denominator of 463.


AMC
AIME
ratio
geometric series
number theory
relatively prime

From this year's Alternate AIME:

An infinite geometric series has sum $2005$.  A new series, obtained by squaring each term of the original series, has sum $10$ times the sum of the original series.  The common ratio of the original series is $\frac{m}{n}$, where $m$ and $n$ are relatively prime integers.  Find $m+n$.

This is what I did, but I came up with an incorrect answer...
$a+ar+ar^2+ar^3+ar^4+...=2005$

$a^2+a^2r^2+a^2r^4+a^2r^6+...=2005\times10$.

Then $\frac{a}{1-r}=2005$, and $\frac{a^2}{1-r^2}=2005\times10$.  Dividing these equations we have $\frac{a}{1+r}=10$.

This means $a-ar+ar^2-ar^3+ar^4-ar^5+ar^6...=10$.  Add this to $a+ar+ar^2+ar^3+ar^4+ar^5+ar^6...=2005$ yields $2a+2ar^2+2ar^4+2ar^6...=2015$.  This means $a+ar^2+ar^4+ar^6+...=\frac{2015}{2}$.

Then multiply by $a$:
$a^2+a^2r^2+a^2r^4+a^2r^6...=\frac{2015a}{2}$.

Then $\frac{2015a}{2}=2005\times10$.  But solving for $a$ gives me an incorrect answer... :?

Rargh.   :blush:.  For some reason I thought $a$ was the common ratio...woopsies.

[quote="eryaman"]
Then $\frac{a}{1-r}=2005$, and $\frac{a^2}{1-r^2}=2005\times10$.  Dividing these equations we have $\frac{a}{1+r}=10$. [/quote]
FWIW - once you get that, that is $\frac {1+r}{1-r}= \frac {2005}{10}=\frac {401}{2} $ you get
 $r=\frac {401-2} {401+2} =\frac {399} {403}$ right away ..

[quote="riddler"][quote="towersfreak2006"]Are you sure? This gives me the correct answer.
$2005=\frac{a}{1-r}=\frac{\frac{2005*10*2}{2015}}{1-r}$
$1-r=\frac{2*2}{403}$
$r=\frac{399}{403}$
$m+n=802$[/quote]
Hey how dyou just add them???
$\frac{m}{n}=\frac{399}{403}$
so how can you just add them to get $802$?
you are assuming that $m=399$ and $n=403$.
Can you explain please.[/quote]

because it is given that $m$ and $n$ are relatively prime to each other

sorry I didnt see that. :blush:  :blush:  :blush:  :blush:


linear algebra
matrix
induction
algebra
polynomial

Firstly I hope this is in the right section, and hello everyone! I have recently learnt about the Fibonacci numbers (Im sure you are aware of them), and that the equation of the nth number in this sequence is given by the rather random looking equation:

u_n = (1/sqrt5)(a^n - b^n),

where a = (1 + sqrt5)/2 and b = (1 - sqrt5)/2

Whilst proving this I was shown an interesting piece of maths behind this, as a and b are roots of the equation:

x^2 = x + 1

I can't figure out why this is and would like to learn a bit more about this, so could anyone explain the link and where this equation comes from? Any ideas how it was first found? I'm not so sure it was just trial and error!

Thanks for any feedback!

Mike

It seems I write up this method every month or so :?

See the bottom post here

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=344268&highlight=#344268

for a brief introduction (the method I have outlined there is insufficient for dealing with repeated roots in the characteristic equation; see if you can figure out what to do there).

you might just notice that

$\left[ \begin{array}{ccc} F_{n+1}\\ F_n \\ \end{array} \right]= \left[ \begin{array}{ccc} 1 & 1\\ 0 & 1 \\ \end{array} \right] \left[ \begin{array}{ccc} F_n\\ F_{n-1} \\ \end{array} \right]$
So you can calculate $F_n$ by calculating powers of this matrix.

Indeed, if you know how to diagonalize a matrix, that leads to another proof of the formula, since

$\left( \begin{array}{ccc} F_{n+1} & F_n\\ F_n & F_{n-1} \\ \end{array} \right)=\left( \begin{array}{ccc} 1 & 1\\ 1 & 0 \\ \end{array} \right)^n$

I'm afraid this matrix can't be diagonalized, but its n'th power may be calculated using some other method.

Or might just do this kind of trial and error:

$f(n) = A * a^n+B * b^n$
solve $f(1)=1$,...,$f(4)=3$ ... for $A,a,B$ and $b$, I think that's not very difficult but I haven't tried yet.
Now prove by induction that the formula holds for all n. This will be really easy.

That matrix definitely can be diagonalized.  It has characteristic polynomial $p(x)=(x-1)x-1=x^2-x-1$, which is what we want.

Sorry, I meant that the matrix I gave can't but I didn't notice the other one is different.
I worked it out here
$\left( \begin{array}{cc} 1 & 1\\ 1 & 0 \end{array} \right)^n=E \cdot \left( \begin{array}{cc} \frac{1+\sqrt{5}}{2} & 0 \\ 0 & \frac{1-\sqrt{5}}{2} \end{array}\right)^n \cdot E^{-1}$

 where $E=\left( \begin{array}{cc} \frac{1+\sqrt{5}}{2} & 1 \\ \frac{1-\sqrt{5}}{2}& 1 \end{array}\right)$.


A castle and a bishop are placed at random on different squares of a chessboard. Find the probability that at least one piece threatens the other. Express your answer as a common fraction reduced to lowest terms.

so putrting bishop down we got
$ P(\text{castle threatens bishop}) \equal{} \frac {7}{63} \plus{} \frac {7}{63} \equal{} \frac {14}{63} \equal{} \frac {2}{9}$.
Putting castle down:
I am stuck on the $ P(\text{bishop threatens castle})$.

Because of their unique threat spaces, they will never threaten each other.
Say we put down the castle. It threatens everything in the same row and column as itself, which are 14 squares out of the 63 not occupied by it. (I agree with your first calculation).

Now, say we put down a bishop. By similar logic, we need to check the squares it threatens.

This is different depending on where we place it.
Check the probabilities of four independent events: (assume it's a black bishop, the color doesn't matter due to symmetry

It's in the middle four squares (placed in here, it will always threaten [b]13[/b]) (P = 13/63)
It's in the surrounding squares (placed in here, it will always threaten [b]11[/b]) (P = 11/63)
It surrounds the previous layer (placed in here, it will always threaten [b]9[/b]) (P = 9/63)
It encompasses the border of the board. (placed in here, it will always threaten [b]7[/b]) (P = 7/63)

The probability of getting the middle four squares is 4/64.
The probability of getting the first ring is 12/64.
The probability of getting the second ring is 20/64.
The probability of getting the border is 38/64.

So: 4/64 of the time we have a 13/63 chance of threatening..
12/64 of the time we have a 11/63 chance of threatening..
i.e. we multiply the probabilites by their chances of happening:

P(Bishop Threatens Castle) = $ \frac{4}{64} \cdot \frac {13}{63} \plus{} \frac{12}{64} \cdot \frac {11}{63} \plus{} \frac{20}{64} \cdot \frac {9}{63} \plus{} \frac{38}{64} \cdot \frac {7}{63}$ You can multiply and add if you want.

Then add that to the first probability. (and subtract the case where they both attack each other, formally, but this has zero probability.)

How did you come up with how the bishop placed in any squares in a ring will threatens the same number of spaces?

The board has four way symmetry, and I noticed that bishops take up different amounts of spaces depending on where they are on the board. I realized there had to be some sort of zones where they behaved the same, and I went out and found them.

Yes, they are. In the problem statement however, the OP used the term Castle, so I stuck with that for the answer  :lol:


induction
number theory
greatest common divisor
number theory unsolved

let $n,a,b$ be natural numbers and $(a,b)=1$ .Prove that:

$\frac 1{a}+\frac 1{a+b}+\frac1{a+2b}+...+\frac 1{a+nb}$

isn't a natural number

Hard!!!
I worked on it for 3 hours and I couldn't solve it. :blush:
Can someone send the answer?

I have found the sulotion of problem .
Sulotion is for "Hesam Mahdavifar". he has silver Medall of IMO 1999

isn't this the same as the proof for the harmonic sum is never an integer:
here: $a,b=1$ but you can use the same idea, i guess...
if $h_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
Let $k$ be the largest power or $2$ less than or equal to $n$

Now if sum is $h_n=p$, multiply both side by $\frac{k}{2}*$(product of all odd numbers up to $n$ ) then except for the term corresponding to $\frac{1}{k}$ every other term would be an integer so: contradiction.
maybe now you can use induction to prove it for every $a,b\in\mathbb{N}$

No. The case a=0 is too simple because there is only one term in a+bi that is divisible by 2^k.
but for any p>2 there are more terms. So we can't solve the problem that easily if [b]b is divisible by 2 and 3.[/b]

Simple. I first encountered this problem 4 years ago... :)
Take p to be the smallest prime that does not divide b, but divides some term a+bi.
Now let e be the highest exponent of p in a, a+b, ... , a+nb
also let i be the smallest index such that a+bi=m p ^e
we then see that if a+bj=k. e^p we should have that j=i+k.p^e
so that the terms divisible by p^e are: mp^e,(m+b)p^e, (m+2b)p^e,...,(m+kb)p^e where k=[n-i/p^e]and also k< p-1
Now note that gcd(m+ib,m+sb)= gcd(s-i,m+ib). but s-i < p and hence gcd =1. otherwise a+bi has a DIVISOR LESS THAN P.
This means that all the terms m,m+b,...,m+kb are relatively prime.
Let us consider the sum A= (1/m + 1/m+b +... + 1/m+kb)*s
where s= lcd(a,a+b,...,a+nb)/p^e
clearly  A needs to be divisible by p.
but since s=m(m+b)...(m+kb) t
we have that: A=t(k+1)  and so k+1 should be divisible by p, which is impossible.
-Ali

altough some typo in previous solution, someone can explain me why A=t(k+1) ?? i don t understand how it comes out!


Hi, I am having some trouble working out the whole "encoding matrix" thing.

If I had to use A = 2  5  as an encoding matrix, how would i code the word MATH
                          1  3
and what "word" would be sent using this code? Also, if you coded the message MATHS IS GREAT what message would be recieved?

Sorry, it's just me so very confused. Any help would be greatly appreciated.

What you're talking about is completely mysterious to me.  Perhaps you can provide some background and context?