AMC AIME ratio geometry algebra polynomial parameterization

So, the competition is over and it's discussion time! The test can still be found [url=http://wangsblog.com/jeffrey/2007mockAIME6.pdf]here[/url]. First, the top scorers.

[hide="Highest Score"]beta - 12
SplashD - 12[/hide]

[hide="2nd Highest Score"]K81o7 - 11[/hide]

Now, the answers. Note that for #4 I accepted two answers even though only the first is technically right.

[hide="Answers"]1.  751
2.  450
3.  374
4.  572 (286)
5.  063
6.  210
7.  528
8.  648
9.  084
10. 439
11. 111
12. 006
13. 968
14. 301
15. 134[/hide]

And for the average scores on each problem and the whole test.

[hide="Averages"]1. 0.73
2. 0.33
3. 0.6
4. 0.47
5. 0.73
6. 0.6
7. 0.27
8. 0.93
9. 0.53
10. 0.4
11. 0.2
12. 0.13
13. 0.33
14. 0.27
15. 0.13
Total: 6.67[/hide]

Pretty high for an AIME, but we have smart people here. Anyway, thanks for taking the test!

I have a question about #13.

I solved it correctly but I kinda burte forced it.

I managed to do this.

Call the point where the radius of the small circle intersects PB "X".  So we have the similarity ratio-(PX)/r=(20sqr2+PX)/11.  Furthermore, we have PX^2+r^2=9  where r is the radius of the smaller circle.

Also, from the sr area formula I found the area of the triangle was (20sqr2+PX)r.  However, with my two top equations I could not find a easy way to solve the two equations without having to go into 4th degree polynomials, so I brute forced it and I got lucky to find that PX=2sqr3 and r=1.  This satisfied both equations.  So I found the area to be (20sqr2+2sqr2)(1), meaning the area was 22sqr2.  So the area squared was 968, which was the correct answer.  However, is there an easier way to do this problem or simplify my two equations without having to brute force the problem?

hmm.. i didnt participate in this but i was looking over the test and i seem to have missed #2(by percentages, it was actually quite hard..) so can someoen please explain to me how it is 180+180+90=450

yes, I thought the wording on 4 was a little ambiguous because 1) the points were not numbered and 2) the number of triplets and the number of angles was not the same.

and would you mind posting the actual answers? i.e. if the answer asked for the sum of the numerator and denominator of a fraction, can you give us the actual fraction? that would be nice. so would the basic solution to #14, [hide="because I dont know where I went wrong"] to terminate in one of these bases, the only possible denominators are $a^{k}$ or $2^{m}c^{n}$ where $a\in\left\{2,3,5,7\right\}$ and $c\in\left\{3,5,7\right\}$ and $k,m,n$ are integers, right?[/hide]

@usaha, if two diagonals intersect, we need to count both the angle and its supplement. there are 3 distinct angles at which two diagonals can meet, so you would think 180+180+180, but one of these angles is 90, whose supplement is itself, so the answer is 180+180+90=450.

I enjoyed this, even though I totally failed... I misread the first questions as "positive" instead of "nonnegative" and in question 7 i thought that the roots of the polynomial $1+x+x^{2}+...+x^{n}$ were the nth roots of unity (except for one) and they're really the (n+1)th roots...

Atleast I got #10 right... since I spent about an hour on it trying to figure out which direction to rotate.

[quote="usaha"]uhh..3 distinct.. hmm i counted 4 distinct, with one of them being 90.. so i got 630..hmm ill redraw my picture..[/quote]

If you got $630$, you probably counted $\frac{45}{2}$ and $\frac{315}{2}$, which are not formed by the intersections of diagonals in the interior of the octagon.

For #13, try proving that $AB$ is equal to the distance between the points of tangency on the two circles (i.e. the length of the external tangent segment).

For #14, note that each number's prime factorization must be completely contained within one of the $S_{k}$. For example $\frac{1}{15}$ is not in any of them.

As for real answers:

[hide]
1. $3751$
6. $160-48\sqrt{2}$
11. $\frac{47}{64}$
12. $2006$
14. $\frac{241}{60}$[/hide]

[quote]For #13, try proving that AB is equal to the distance between the points of tangency on the two circles (i.e. the length of the external tangent segment). [/quote]

Wait, thats what I did..thats how I got that similar triangle ratio and that pythagorean theorm.. but I just dont know how to simplify them-is there an easier way to do that?

[quote="white_horse_king88"][quote="paladin8"][quote="white_horse_king88"]Could someone please post their solution to number 12? Thanks.[/quote]

[hide="hint"] Note that the equation is equivalent to

$x+2007^{-k}= \frac{1}{\sqrt[k]{2007-x}}$.

If we let $a = 2007$ be a parameter, what can we say about the two sides of the equation as functions of $a$?[/hide][/quote]

Ah... nice. [hide]The general solution for $x_{k}$ is just $2007-\frac{1}{2007^{k}}$, right? [/hide][/quote]

Yup.

[quote="SuperThread"]I'd appreciate it if somebody would post a good solution for 15, too. (And some elaboration on 12 would be nice. :))[/quote]

The solution to #15 isn't that "elegant" I guess, but it's a counting problem so what do you expect. I'll give you three hints.

[hide="hint 1"] Think recurrences on the number of superpositive sequences that have terms $\le n$. [/hide]

[hide="hint 2"] If $A$ is a superpositive sequence, then $f(A)$ is a superpositive sequence. [/hide]

[hide="hint 3"] Consider the number of superpositive sequences whose terms sum to $\le n$. [/hide]

And for #12, if you read what was above the next step is

[hide] If $f(a)$ is the LHS, then the equation is $f(a) = f^{-1}(a)$ so $f(f(a)) = a$. [/hide]

Any nice way to do #13? I still can't work with paladin's hint.


geometry
inequalities unsolved
inequalities

[quote="thanhnam2902"][color=darkblue]Let $ ABC$ is triangle. Let $ m, n, p \in R$ and $ \begin {cases} m \plus{} n > 0 \\
n \plus{} p > 0 \\
p \plus{} m > 0 \\
mn \plus{} np \plus{} pm > 0\end {cases}$.

Prove that: $ \boxed {m.a^2 \plus{} n.b^2 \plus{} p.c^2 \ge 4S\sqrt {mn \plus{} np \plus{} pm}}$.

[i]Note:[/i] $ AB \equal{} c,BC \equal{} a,CA \equal{} b$, $ S$ is area of $ \triangle ABC$[/color].[/quote]
Easy to see that $ ma^2 \plus{} nb^2 \plus{} pc^2>0.$ After squaring we need to prove that
$ (m\plus{}n)(m\plus{}p)a^4\plus{}(n\plus{}m)(n\plus{}p)b^4\plus{}(p\plus{}m)(p\plus{}n)c^4\minus{}$
$ \minus{}2p(m\plus{}n)a^2b^2\minus{}2n(m\plus{}p)a^2c^2\minus{}2m(p\plus{}n)b^2c^2\geq0.$
Thus, it remains to prove that
$ \left(p(m\plus{}n)b^2\plus{}n(m\plus{}p)c^2\right)^2\minus{}$
$ \minus{}(m\plus{}n)(m\plus{}p)\left((n\plus{}m)(n\plus{}p)b^4\plus{}(p\plus{}m)(p\plus{}n)c^4\minus{}2m(p\plus{}n)b^2c^2\right)\leq0,$
which is equivalent to $ \minus{}(mn\plus{}mp\plus{}np)\left((m\plus{}n)b^2\minus{}(m\plus{}p)c^2\right)^2\leq0.$


group theory
abstract algebra
superior algebra
superior algebra unsolved

$ G$ is a group. Prove that the following are equivalent:
1. All subgroups of $ G$ are normal.
2. For all $ a,b\in G$ there is an integer $ m$ such that $ (ab)^m\equal{}ba$.

$ \Rightarrow$ Consider $ \langle ab \rangle$

Since  $ \langle ab \rangle$ is normal, $ b(ab)b^{ \minus{} 1} \in \langle ab \rangle \Rightarrow ba \equal{} (ab)^m$.

$ \Leftarrow$

There exists $ m$ such that $ ((kg^{ \minus{} 1})g)^m \equal{} gkg^{ \minus{} 1}$

Suppose $ k \in H$, therefore $ gkg^{ \minus{} 1} \equal{} k^m \in H$ and hence $ H$ is normal.

[b]Wow![/b] That was a great job, [b]Simon[/b].  :clap2:

Keep up the good work!


In LaLaland, the only currence is the 7-zorkmid and 11-zorkmid coins. Some prices, like 42,43 or 44 zorkmids can be paid with exact change, while others, like 5 or 20 zorkmids cannot. How many prices cannot be paid with exact change?

Okay, number 1, your answer is wrong because 7 can be made as well as 11, so I suggest you don't make stupid posts like that...no offense or anything...

uh... the question was how many prices CANNOT be paid with exact change.

Added:  $1 can be paid by giving 9 11-dollar bills and receiving 14 7-dollar bills.
$2 can be paid by giving 4-11 dollar bills and receiving 6 7-dollar bills.

 $3 can be paid by giving 6 11-dollar bills and receiving 9 7-dollar bills....etc..

I think my answer is correct or am i misunderstanding something here?

that's not exact change, exact change as in you pay the exact amount and the other dude doesn't owe you anything.

[hide="Hint"]draw a table of possible values, then use modular arithmetic to find what doesn't work.[/hide]

[hide="Other hint"] The Chicken McNugget Theorem tells us that the highest unattainable amount is 7*11-7-11=59.[/hide]


geometry
search
Support
radical axis
geometry proposed

Do you mean $ (AA_1A_2),(BB_1B_2),(CC_1C_2)$? Please read my problem carefully!

[quote]Given triangle $ ABC$ and its incircle $ (I)$. Let $ A_1,B_1,C_1$ be the intersections of $ AI$ and $ BC,BI$ and $ CA,CI$ and $ AB$, respectively; $ A_2$ be the intersection of a line through $ I$ and perpendicular to $ AI$ and $ BC$, similarly define $ B_2, C_2$. Show that $ (AA_1A_2), (BB_1B_2), (CC_1C_2)$ concur at two points.[/quote]

Look like we have to invert something around here! :P. Anyway, nice problem, [b]Linh[/b]! Here is my solution.

Denote $ D,$ $ E,$ and $ F$ respectively by the tangency points of $ (I)$ with $ BC,$ $ CA,$ $ AB$. Consider the inversion though pole $ I$, power $ k \equal{} r^2$, where $ r$ is the radii of $ (I)$. We have $ \mathcal {I}(I,k)$ maps $ A$ into the midpoint $ M_a$ of $ EF$, $ A_1$ is mapped into $ A'_1$, which is the projection of $ D$ onto $ IM_a$ and $ A_2$ is mapped into $ A'_2$, which is the projection of $ I$ onto the $ D \minus{}$ altitude of $ \triangle DEF$. Thus, under $ \mathcal {I}(I,k)$, $ (AA_1A_2)$ $ \mapsto$ $ (M_aA'_1A'_2)$, let denote this circle by $ (O_a)$ with $ O_a$ is its circumcenter. The same argument and definition holds for $ (O_b)$ and $ (O_c)$.

Now, just taking our concern into $ \triangle DEF$. Since $ \angle DA'_2I$ $ \equal{}$ $ \angle DA'_1I$ $ \equal{} 90^{\circ}$, thus both $ A'_1$ and $ A'_2$ lie on the circle with diameter $ DO$. Let $ A'$ be intersection of the $ 9 \minus{}$ point circle $ (\mathcal {E})$ wrt $ \triangle DEF$ and the $ D \minus{}$ altitude of $ \triangle DEF$ ($ A'$ $ \notin$ $ EF$) $ \Longrightarrow$ $ A'$ is the orthocenter of $ \triangle DM_bM_c$. Then it is well-known that $ A'$ and $ A'_2$ are reflected each other across to $ M_bM_c$. Therefore, so as to $ A'_1$ and $ M_a$. As the consequence, $ A',$ $ A'_2,$ $ M_a$ and $ A'_1$ are concyclic, which implies that $ A'$ also $ \in$ $ (O_a)$. Hence $ O_a$ is the intersection of the perpendicular of $ A'M_a$ and $ A'_1M_a$. Or in other words, $ O_a$ is the intersection of $ M_bM_c$ and a line which passes through $ \mathcal {E}$ and perpendicular to $ M_a\mathcal {E}$, let call this line $ d_a$. The same argument holds for $ O_b$ and $ O_c$.

To sum up, in triangle $ \triangle M_aM_bM_c$, we have $ M_a\mathcal {E}$, $ M_b\mathcal {E}$, $ M_c\mathcal {E}$ are concurrent at $ \mathcal {E}$, and $ O_a$ is the intersection of $ M_bM_c$ and $ d_a$, so as $ O_b$, $ O_c$. Then by [i][1][/i], we conclude that $ O_a,$ $ O_b,$ $ O_c$ are collinear. Also note that $ (O_a)$, $ (O_b)$, $ (O_c)$ have $ \mathcal {E}$ as their radical center. Therefore, $ (O_a)$, $ (O_b)$ and $ (O_c)$ together create a system of [i]Coaxial circles[/i]. As the result, $ (AA_1A_2),$ $ (BB_1B_2),$ $ (CC_1C_2)$ concur at two points. Our proof is completed then. $ \square$
_____________________
Reference
[1] http://www.mathlinks.ro/viewtopic.php?search_id=1737274958&t=313257

Nice solution dear [b]Minh[/b]! But why not a synthetical proof?  :lol:

if M is midpoint of arc BC and N is meetpoint of (O) and line MA_2 easy to pove that MN*NA_2 = MA_1*MA = MI*MI angle MIA_2 = 90 and MN*MA_2 = MI*MI so angle INM = 90 if line NI intersect (O) at point P than PM goes throw center O of (O) and AM throw I so esy to prove (Pascal theorem) that AN BN' and CN" intersects on line OI (N' N" have same construction as N but on other arc) and trivial that if they intersects at Q than Q is radical center of (O) (AA'A_2) (BB'B_2) (CC'C_2).
deg(M) wrt point I = deg(M) wrt (AA'A_2). if point V is on CB and angle VIA_1 = IA_2A_1 then easy to prove that deg(V) wrt point I =  deg(V) wrt  (AA'A_2). if line MV intersect (O) at R then easy to prove that angle MRI = VIA_1 (MV*MR = MI*MI) = A_1AH were AH is perpendicular from A on BC. if AH intersect (O) at point H' then trivial that A_1 is on RH' and linne AM goes throw I and AH' goes throw orthocenter so lines MV M'V' M"R" intersects on line throw I and orthocenter in point Q' and trivial that Q' is radical center of point I and (AA'A_2) (BB'B_2) (CC'C_2) so Q' and Q is 2 radii centers so problem done.

This problem is equivalently with the following : $\left\|\begin{array}{ccc}
D\in BC\cap AI\ ,\ X\in BC & \mathrm{so\ that} &  XI=XD\\\\
 E\in CA\cap BI\ ,\ Y\in CA & \mathrm{so\ that} &  YI=YE\\\\
F\in AB\cap CI\ ,\ Z\in AB & \mathrm{so\ that} &  ZI=ZF\end{array}\right\|\ \implies\ Z\in XY$ .

Upper I denoted $BC$ - the sideline (support line) of the side $[BC]$ a.s.o. Prove easily that $\frac {XD}{\frac {a}{\left|b^2-c^2\right|}}=\frac {YE}{\frac {b}{\left|c^2-a^2\right|}}=\frac {ZF}{\frac {c}{\left|a^2-b^2\right|}}=2Rr$ .

Hello,
[quote="Virgil Nicula"]$\implies\ Z\in XY$ [/quote]
Barycentrics are $0:b \left( {b}^{2}+ab-{c}^{2} \right) :-c \left( ac-{b}^{2}+{c}^{2}\right)$ etc. This leads to your relation $\frac {XD}{\frac {a}{\left|b^2-c^2\right|}}=2Rr$, etc.  But \[ \det(X,Y,Z)=-abc \left( b-c \right)  \left( a-c \right)  \left( a-b \right) \left( b+a+c \right) ^{3} \]
Best regards, Pierre.

The same circles appear in this problem: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=199419[/url]


modular arithmetic
geometry
geometric transformation
reflection
number theory
prime numbers
divisibility tests

Here's a purely original problem (I hope) using prime numbers ($\mathbb{P}$). :)

[i]Prove that if $p>5\in\mathbb{P}$ does not divide $k(1...1)$, with $1...1$ having $p-1$ digits, for $k\in\mathbb{Z}$, then $1...1\in\mathbb{P}$.[/i]

Masoud Zargar

Could you better define what $1...1$ is?

Maybe provide a couple examples of numbers that you are talking about of this form.

I'm a bit lost

A numebr of the form $1...1$. 1111 is an example. 11 is an example. These are called repunits. :)

I don't understand:  if k = p, then all p divide the k(1...1)

Well, I said if $p$ does not divide $k(1...1)$, which implies that p=k is excluded. :)

Masoud Zargar

I don't understand the use of $k$ here.  Is it multiplication?  Concatenation?  All $k$ or a given value?   :huh:

$k$ is just any integer mutiplied by $1...1$.  It is not substantial in solving the problem, other than thr fact that at the end, it generalizes the problem a little more.

11 divides any of these numbers, doesnt it?

If we choose a prime greater than $5$, then $p-1$ is even.

If we have a number that has all the same digits, and has an even number of digits, then it is divisible by $11$, and is obviously not prime.

Maybe put some stipulations on possible $k's$

Let's say we are to find a repunit with $n$ digits such that it divides a prime $p$.

A repunit with $n$ digits is $\frac{10^n-1}{9}$.

Divisibility by $9$ does not affect congruences, thus  we have
\[ \\10^n-1\equiv\pmod p\\10^n\equiv 1\pmod p \]

By Fermat's Little Theorem, $n=p-1$, so $p|1...1$, where $1...1$ has $p-1$ digits.
If $p|1...1$, where $1...1$ has $p-1$ digits, it also follows that it divides $k(1...1)$ for $k\in\mathbb{Z}$.

Since all repunits end in $1$, $5$ does not divide $1111$.  By a similar reasoning, $3$ does not divide $11$, and $2$ doe snto divide $1$.  This is of course if $k\neq p$.

Since prime numbers do not have any non-trivial divisors, $p$ does not divide $k(1...1)$, where $1...1\in\mathbb{P}$ has $p-1$ digits.

The converse is true, implying that if $p>5\in\mathbb{P}$ does not divide $k(1...1)$, where $1...1$ has $p-1$ digits, then $1...1\in\mathbb{P}$.

Masoud Zargar

[quote="boxedexe"]\[ 10^n\equiv 1\pmod p  \]

By Fermat's Little Theorem, $n=p-1$.[/quote]

$p-1$ is not the only possible value of $n$, not even necessarily the smallest. Take $2^n\equiv 1 \pmod {7}$ for example. By fermat, $n=6$. But $n=3$ also works and is smaller. In fact $k$ satisfies the equation iff $3|k$.

I don't quite understand what you're trying to say.  Could you please explain.  So far, as my understand tells me, if $n=3$, then we're saying that $p=4\notin\mathbb{P}$.

The implication that $n=p-1$ is wrong. There are other values of $n$ that work as well. For example if $p=11$ we have $10^n\equiv 1 \pmod {11}$. So we know that $10^{10}\equiv 1 \pmod {11}$ by FLT. But $10^2\equiv 1 \pmod {11}$ also... see what I mean? You can't say that $10^n\equiv 1 \pmod {p}\implies n=p-1$.

just to clarify,

$\frac{10^n-1}{9}$ is an integer, but it is more rigirious to look at it
$10^n-1 \mod 9*p$, and we can use CRT to break this up (note that since $p>5$, and it is prime, we have that $(p,9)=1$, so look mod $p$ and $9$, it is 0 in both mods, so it is $0 \mod9p$, and that implies $0\mod p$

[quote="4everwise"][quote="Petoria"]11 divides any of these numbers, doesnt it?[/quote]Only in the case where there are an even number of 1's. :)[/quote]

There are $p-1$ 1's, and $p \in \mathbb{P} > 5$ ...   :huh:

[quote="4everwise"][quote="Petoria"]11 divides any of these numbers, doesnt it?[/quote]Only in the case where there are an even number of 1's. :)[/quote]
There are always an even number of ones.

[quote="JBL"]Let me re-phrase the question, for clarity:

Let $p$ be a prime larger than 5, and let $n$ be the number whose base-10 representation is $p - 1$ digits long, each digit being equal to 1 (the repunit of length $p - 1$).  Prove that if $p \not | n$ then $n$ is prime.

The $k$ was meant to give generality to the question -- in fact, though, it doesn't really add anything.

Since, as some have noted, for $p > 3$ $n$ will [i]never[/i] be prime, the approach must be to prove that $p | n$.  This is what boxedexe did, although in a rather convoluted way.  Neither chess64 nor Altheman's comments reflect a flaw in the mathematics, although perhaps they reflect flaws in the presentation.[/quote]

Well, alright then,

[hide="hopefully straightforward solution"]
$1..1 = \frac{10^{p-1} - 1}{9}$

By Euler's, $10^{p-1} \equiv 1 \bmod p$.  Hence $p | n$.  Q.E.D.
[/hide]

that was a nice clear solution, good job.

Cool so that is the easy part. Ummm ... Good luck guys because now I have no clue how to move on! HAVE FUN!

-Sid

Indeed the k is really something unnecessary since p not divide ab iff p not divide a and p not divide b.

To prove primality of 1...1, it only left to prove that for repunit length p-1, this repunit is not divisible by 2,3,5. But case for 2 and 5 are obvious. For 3, any repunit divisible by 3 necessary have a prime factor bigger than 5, therefore the result follows.

In fact, if we want some generality, maybe we can drop the condition that the repunit is p-1 length. Nevertheless, the condition that 1...1 is not divisible by p>5 already set the question to be quite simple.


Let P(x)=x^4+ax^3+bx^2+cx+d be a polynomial where a,b,c,d are real numbers. Suppose P(0)=2006,P(1)=2007, P(2)=2008.
What is the smallest possible value of P(3)?

[hide]
$ P(0)\equal{}d\equal{}2006\\
P(1)\equal{}1\plus{}a\plus{}b\plus{}c\plus{}2006\equal{}2007\Leftrightarrow a\plus{}b\plus{}c\equal{}0\\
P(2)\equal{}16\plus{}8a\plus{}4b\plus{}2c\plus{}2006\equal{}2008\Leftrightarrow 8a\plus{}4b\plus{}2c\equal{}\minus{}14\Leftrightarrow 4a\plus{}2b\plus{}c\equal{}\minus{}7\\
3a\plus{}b\equal{}\minus{}7\\
P(3)\equal{}81\plus{}27a\plus{}9b\plus{}3c\plus{}2006\equal{}9(3a\plus{}b)\plus{}c\plus{}2087\equal{}c\plus{}2087\minus{}63\equal{}c\plus{}2024$
$ c$ can be made arbitrarily small, so the smallest possible value doesn't exist... :huh:
[/hide]

[hide]
Let $ f(x) \equal{} P(x)\minus{}x\minus{}2006 \equal{} x^6\plus{}ax^3\plus{}bx^2\plus{}(c\minus{}1)x\plus{}(d\minus{}2006)$.

Then $ f(0)\equal{}f(1)\equal{}f(2)\equal{}0$ so $ f(x) \equal{} x(x\minus{}1)(x\minus{}2)(x\minus{}r)$, where $ r$ is the last root of $ f$.

Setting $ x\equal{}3$, we have $ f(3) \equal{} P(3) \minus{} 3 \minus{} 2006 \equal{} 3(2)(1)(3\minus{}r)$ so $ P(3) \equal{} 6(3\minus{}r)\plus{}2009$. We can choose any number to be $ r$, and as $ r$ gets arbitrarily large, $ P(3)$ approaches negative infinity.

So I guess there's no bound.[/hide]


calculus
derivative
function
calculus computations

Show that the map $ f: \mathbb{R}^{n}\setminus{\{0}\} \to \mathbb{R}^{n}: f(x) \equal{} \frac{x}{\parallel{}x\parallel{}^2}$ is differentiable at each point $ x \in \mathbb{R}^{n}\setminus{\{0}\}$.

Is finding each partial derivative and showing continuity of each one of them the straightforward way of approaching such problems?

The quotient of differentiable nonzero functions is differentiable and both $ x$ and $ \parallel{}x\parallel{}^2$ are polynomial.

Thank you t0rajir0u. But that would make the problem utterly trivial for an exam (unless the student provides the proof of this fact).

So I guess the interesting part will be to ask for the Yah -ko bee - un [Jacobian].


induction
inequalities
algebra
square of a binomial

The principle of mathematical induction is an important and powerful tool of mathematics. It is especially useful in proving statements involving the natural numbers; often when it is known to be true for the first few, and it is suspected or [i]conjectured[/i] to be true for all natural numbers. The method of proof involves the following steps:

1. First prove the statement is true for n = 1 (or some other integer).
2. Assume that the statement is true for n = k, where k is [i]any[/i] integer greater than or equal to 1.
3. From the assumption in 2 prove that the statement must be true for n = k + 1. This is the part of the proof establishing the induction. This is the step that usually involves the most work.

Since the statement is true for n = 1 (from step 1) it must be true for n = 1 + 1 = 2 (from step 3), and from this for n = 2 + 1 = 3 (step 3 again), etc. etc. -- and so must be true for all positive integers!

Proving something by mathematical induction is a lot like tipping over a set of dominoes. First you push over the first dominoe (step 1). Then if the dominoes are aligned correctly, each dominoe that falls will knock over the subsequent dominoe (step 3). So it follows that all of the dominoes will fall.

Here are some examples to try:

1. Prove that 1 + 3 + 5 + 7 + ... + (2n - 1) = n2.

2. Prove that [tex] \displaystyle {\frac {1}{2} + \frac {1}{4} + \frac {1}{8} + ... + \frac {1}{2^n} = 1 - \frac {1}{2^n}} [/tex].

3. Prove that (1 + x)n :ge: 1 + nx for n = 1, 2, 3,... if x > -1.

In case anyone wants an example, I will prove a different one, namely, that:

1 + 2 + .... + n-1 + n = [n(n+1)]/2

1. 1 = [1(1+1)/2] = 2/2 = 1, so it is true for n=1

2.  Assume 1 + 2 + .... + n-1 + n = [n(n+1)]/2 holds for n=k (for some induction problems, it is necessary to assume n holds for n :le: k

3.  We need to show that (2) holds when n = k+1

From (2), we have that 1 + 2 + .... + k-1 + k = [k(k+1)]/2

Therefore, we can add k+1 to both sides to obtain:

1 + 2 + .... + k + k+1= [k(k+1)]/2 + k+1

RHS = (k+1)(k/2+1) = (k+1)(k+2)/2 = [(k+1)((k+1)+1)]/2

Therefore 1 + 2 + .... + k + k+1= [(k+1)((k+1)+1)]/2, and so our statement holds for n=k+1.

It is then true for all natural numbers by the Principle of Mathematical Induction.

joml88, if you still need help, here is the hint:

[hide]Given (1+x)^n  :ge: 1+nx, multiply both sides by 1+x and simplify.[/hide]

[quote="JBL"]Just a suggestion:
when you are trying to prove a statement that looks like "A = B," it is helpful not to start with the statement "A = B" and turn it into a statement "C = C."  It's far clearer if you start with the ending statement which demonstrates a known truth, and work backwards to the thing that you are trying to prove.  This doesn't mean that while solving one shouldn't do what you've done, but it means when you come back and write it up, turn your chain of logic upside-down.[/quote]

I'm not sure that I understand you totally.  If you could just tell me what I should end up with (that is what would be the A=B in my proof) I would proobably understand what you mean.

Thanks

He's saying that when you write a proof, don't write it backwards. It is easier to follow the logic of a proof if it flows forwards from known facts, rather than working backwards. Here is an example of a backwards proof and a forwards proof. (notice that the forward proof is alot easier to understand)

Question: Prove [tex]\displaystyle\frac{a+b}{2} \ge \sqrt{ab}[/tex] for positive reals a and b.

Bad Backwards Proof:

We want to prove [tex]\displaystyle\frac{a+b}{2} \ge \sqrt{ab}[/tex]. Since fractions are not very easy to deal with, we multiply the inequality by 2, so that the given inequality is equivalent to [tex]a+b\ge2\sqrt{ab}.[/tex] Now, we move everything to one side, so we want to prove [tex]a-2\sqrt{ab}+b\ge0.[/tex] Recognizing the square of a binomial, we see that this is equivalent to [tex](\sqrt{a}-\sqrt{b})^2\ge0,[/tex] which is true since all squares are [tex]\ge0.[/tex] So we're done. [tex]\blacksquare[/tex]

While this proof is correct and shows how we solved the problem, it is not as easy to follow as the next 'forward' proof:

Since all squares are greater than or equal to 0, we have

[tex](\sqrt{a}-\sqrt{b})^2\ge0.[/tex]

Now, we use some simple algebra to prove the desired inequality:

[tex]\sqrt{a}^2-2\sqrt{ab}+\sqrt{b}^2\ge0[/tex]

[tex]a+b\ge 2\sqrt{ab}[/tex]

[tex]\displaystyle \frac{a+b}{2}\ge\sqrt{ab},[/tex] as desired. [tex]\blacksquare[/tex]

Ahh... ok.  I understand now.  That is a much better approach.  I'll start trying to do it that way.

Since 3 is unproved, I will complete it.

Moderator's note: I put the solution in spoiler text, which is the preferred practice for solutions to challenge problems, so that others can take up the challenge. By the way, how old is everyone who is replying in this thread?

[hide]

(1+x)^1 = 1+(1)x for k=1

Assume (1+x)^n  :ge: 1+nx for n  :le: k

Then

(1+x)^k  :ge: 1+kx implies

(1+x)^(k+1)  :ge: (1+kx)(1+x)

(1+x)^(k+1)  :ge: 1+x+kx+kx^2

(1+x)^(k+1)  :ge: 1+(k+1)x+kx^2

But kx^2  :ge: 0, and so:

(1+x)^(k+1)  :ge: 1+(k+1)x as desired, and we are done.[/hide]

Reply to moderator's question:  I am 15 and a sophomore.(not sure if that was a rhetorical question or not so just replied anyway)


$\sqrt{\frac{1}{9} + \frac{1}{16}} =$

$\textbf{(A)}\ \frac15 \qquad
\textbf{(B)}\ \frac14 \qquad
\textbf{(C)}\ \frac27 \qquad
\textbf{(D)}\ \frac{5}{12} \qquad
\textbf{(E)}\ \frac{7}{12}$

(click smiley)[hide="  :rotfl: "]
$\sqrt{\frac{1}{9}+\frac{1}{12}}=\sqrt{\frac{25}{144}}=\boxed{\frac{5}{12}}$ so the answer is $D$[/hide]

if this was on the AMC 12 I should have no problem with it


function
topology
real analysis
real analysis unsolved

[quote="math2319"]I don't really understand continuity in topological spaces. Our prof gave us the definition but i'm still really confused. So I guess I have two questions. 1. could anyone try to INTUITIVELY explain it to me? And also I have to prove a couple things about it which I would appreciate help with (although I don't want the complete answer just help leading me in the right direction.)

1. if f:X-->Y is continuous prove that if {x_n} is a convergent sequence in X then {f(x_n)} is a convergent sequence in Y.

and

2. if f:X-->Y is continuous and S is contained in X, prove that cl(f(S)) = f(cl(S))[/quote]

Write down the definition of continuity in a general topological space in your next post, and answer this question: Does this look like anything you've seen before in a more concrete setting?

Here's the paradigm: recall the delta epsilon definition of continuity. It basically says that if have a function $ f: \mathbb{R}\rightarrow \mathbb{R}$, and you take an open ball in the codomain, taking every point in its preimage has an open ball contained fully within the preimage. So basically, we call sets such that each point has an open ball in the set "open". A function is continuous then if the preimage of an open set is open. That being said, in topology, we basically think like this. We just pick out distinguished sets as being open and then we define functions to be continuous if preimages of open sets are open (motivated by the real line case). Alright. Now, lets take a look at your problems.

Suppose $ \{x_n\}$ converges to $ p$ in $ X$ but $ \{f(x_n)\}$ doesn't converge in $ Y$. Take any point that has $ p$ in its preimage. Take an open set around that point. Pull it back via preimage to an open set around $ p$. Does it intersect $ \{x_n\}$?

Second, closures are obtained by adjoining all the limit points of a set. Then proceed in a similar manner.

I don't blame you for being confused, the first time exposed to this stuff it seems really contrived.

thanks that actually helps a lot! i don't completely understand it yet but I'm letting it settle in my brain and musing it over. My prof says it'll take us a good week of thinking to fully understand what this is and exactly what we're doing and why.


probability
geometric series
number theory theorems
number theory

I'm looking for a proof that the probability of the relation $\gcd (n,m)=1$ is $\frac{6}{\pi ^2}$

Also can anyone have any good papers about random integers in the given interval?

The probability that n is divisible by 1/p and likewise for m.  The probability that they both are is 1/p^2, and therefore, the probability that neither is is 1-1/p^2.  Take the product over all the primes and you get 1/(zeta(2))

And how do you pass from the product to the sum?

By geometric series and the fundamental theorem of arithmetic, see http://www.mathlinks.ro/Forum/viewtopic.php?highlight=probability&t=31028


5 people stand in a line where people are labeled as person A,B,C,D, and E.
how many combinations of these people can stand in a line where person A and B cannot stand next to each other?

[hide]
First, solve without restrictions: $ 5! \equal{} 120$.

Next, we subtract the possibilities where $ A$ and $ B$ are next to each other. We can look at this problem like treating $ (AB)$ and $ (BA)$ as one unit. For each, there are $ 4! \equal{} 24$ possibilities, for a total of $ 48$ invalid possibilities.

[b]Final answer:[/b] $ 120 \minus{} 48 \equal{} \boxed {72}$[/hide]

Thank I had 48 as my answer originally I just forgot to subtract it from 120.

[hide]Here's one approach for which you don't need any special idea:

1. A is on the edge of the row (WLOG first) place. Now you have 3 (no A and no B) possibilities for place two, 3 (no A and no person on the place two)possibilities for place three, 2 for place four, 1 for place five. Multiplying gives you 18.
2. A is not on the edge (WLOG the second place). Now 3 possibilities for place 1, 2 for 3, 2 for 4, 1 for 5. Multiply and get 12.

Now: case 1 yields A on 1 and 5, and case 2 is there for A on 2,3,4. hence the number of combinations should be

18x2+12x3=72.

Other approach. Make a block out of A and B and call it F. Now, you have 4 persons: C, D, E, F. Numer of combinations is 24. Make it double (for AB and BA). It's 48. Now, Take the total number of combinations for ABCDE. It's 120. Edit: Yongyi gave this one already.

48 combinations don't work. So, 120-48=72.[/hide]


number theory
prime numbers
prime factorization

for primes $ p, q$
find the number of solutions p ,q satisfying $ (p \plus{} q) \equal{} (p \minus{} q)^3$

[hide]all prime numbers are of the form 6k-1 or 6l+1 exept 2 and 3
as p,q are syymetric
CASE 1
p=6k-1 , q=6l-1
putting it in the form of the question
RHS is a multiple of 6 whereas LHS is not. so no solution of thios form
CASE 2
p=6k-1 , q=6l+1
setting it in the form of the question
LHS is a multiple of 6 whereas RHS is not. so no solution of this form
CASE 3
p=6k+1 , q=6l-1
according to the question
RHS is a multiple of 6 whereas LHS is not. so no solution of this form

only case remains is p=3 , q=2 or vice versa which is also not a solution
there are no such primes p,q[/hide]

er.. i dont know abt the solution but the equation is pretty much true for $ 5$ and $ 3$
so i am trying to find a solution or a mistake in your solution rahul1993

ya there is a mistake in my solution  :blush: actually i have proved that either of p or q = 2 or 3. Actually I have not considered all cases. considering primes as 6k-1 or 6k+1 excludes 2 and 3.

[hide=" srinath.r"]dont spam here .[/hide]

consider the primes $ p,q$
obviously for the equation to hold good $ p\ge\q$
$ p + q = (p - q)^2(p - q)$
$ \frac {p + q}{p - q} = (p - q)^2$
so we can have $ p + q = (p - q)^2 k$   and $ p - q = k$       where $ k$ is a natural number
adding the above we get
$ p = \frac {k}{2}((p - q)^2 + 1)$
which implies the factors of p are $ \frac {k}{2}$ and $ (p - q)^2 + 1$
and we cannot have the factors as $ k$ and $ \frac {(p - q)^2 + 1}{2}$
reason- $ (p - q)^2 + 1$cannot be 2 because then we have p=k which is not possible
also $ (p - q)^2 + 1$ cannot be other  even numbers then $ p$ will have a factor which is not equal $ p$ or one ($ p$ is a prime number)  this also implies that for $ p$ to be natural $ k$ is even which means $ p$ and $ q$ are odd  primes  :wink:
so the factors are $ \frac {k}{2}$ and $ (p - q)^2 + 1$
obviously $ \frac {k}{2}$ can either be $ p$ or 1
$ \frac {k}{2}$ cannot be $ p$  so $ \frac {k}{2} = 1$
implies $ k = 2$
so now substituting the value of k in $ p - q = 2$ and $ p + q = 8$
solving we get the one and only solutions are $ p = 5$ and $ q = 3$

ps:sorry for the spam earlier

set $ p \minus{} q \equal{} t$
so we have to  solve for,
$ t \plus{} 2q \equal{} t^3$
or $ 2q \equal{} t^3 \minus{} t$
or $ 2q \equal{} (t \minus{} 1)t(t \plus{} 1)$
note that the $ R.H.S$ is divisible by six as there are three consecutive terms
therefore there should be a no. which has $ 3$ on its prime factorization in $ L.H.S.$
but since $ q$ is a prime and the only no. other than 2, $ q \equal{} 3$
now by sight it is obvious that the consecutive numbers are $ 1,2,3$
therefore $ t \equal{} 2$,
or $ p \equal{} 5$
therefore $ q \equal{} 3 ,p \equal{} 5$
hence solved

p.s. sry for the spam earlier!!
it is not my solution but i wish it was!!

Any prime but 2 and 3 mod 12 is either 1,5,7,or 11, we find no two of these works. So we try out 2 and 3, we find only (3,5) works.

[quote="I"][hide]all prime numbers are of the form 6k-1 or 6l+1 exept 2 and 3
as p,q are syymetric
CASE 1
p=6k-1 , q=6l-1
putting it in the form of the question
RHS is a multiple of 6 whereas LHS is not. so no solution of thios form
CASE 2
p=6k-1 , q=6l+1
setting it in the form of the question
LHS is a multiple of 6 whereas RHS is not. so no solution of this form
CASE 3
p=6k+1 , q=6l-1
according to the question
RHS is a multiple of 6 whereas LHS is not. so no solution of this form

checking for p=3 , q=2 or vice versa which is also not a solution
[/hide][/quote]
continuing
now we know p or q = 2 or 3
looking at the RHS we have q=2 or 3
CASE 4
setting q=2 we get values of p are roots of equation $ p^3 \minus{} 6p^2 \plus{} 11p \minus{} 10 \equal{} 0$
from rational roots theorem, possible prime values for p are 2 and 5. putting each of them in the cubic equation we see LHS $ \ne 0$
CASE 5
setting q=3 we get values of p are roots of the equation $ p^3 \minus{} 9p^2 \plus{} 26p \minus{} 30 \equal{} 0$
again from rational roots theorem, possible prime values of p are 2,3,5. putting each of them in the equation, we see it is satisfied only for p=5

so (p,q)=(5,3)


A solid sphere of radius R has a uniform charge density d and total charge Q. Derive an expression for its total electric potential energy.

[hide="hint"]
actually there are a number of ways of doing this problem.First just use the basic definition of potential energy of a system  that is it is the minimum work an external agent has to do to assemble the system.
Here our system is a sphere and we already have been provided with all kinda point charges with us .What one has to do is bring them from infinite(Assumed) seperation and make a sphere of Radius $ R$.
now say at any time during this process u have already a sphere of radius $ x$ then to increase it's radii to $ dx$ u need $ 4 \pi x^{2}dx d$ amount of charge which has to be bought from infinity and kept on the sphere.the potential on the sphere is known to u so u can find out the work done
(I am leaving this so that u do solve it by urself)
now integrate that from $ x \equal{} 0$ to $ x \equal{} R$
[/hide]

[hide="answer"]
$ \frac {3Q^{2}}{20 \pi \epsilon_{0} R}$
[/hide]


I was wondering which chapters are important for the USPhO open exam?

I'm talking about the textbook "Fundamentals of Physics", if there is any chapters in this book will help me prepare for USPhO?

btw, i mean the first round of USPhO the F=ma contest.

The $ F\equal{}ma$ test is purely mechanics, fluids, and the like.  That's usually the first half of most physics textbooks.


Let $n$ be a positive integer. Let $S_n$ be the set of all nondecreasing lists $ a=\left(a_1,a_2,...,a_n\right)$ of $n$ integers such that $1\leq a_j\leq 2j$ for all $1\leq j\leq n$. Find the number of elements in $S_n$.

Let $x_n$ be the desired number.
Am I wrong or $x_2 = 7$?
($a_1 = 1$ leaves $a_2=1,2,3,4$ and $a_1=2$ leaves $a_2 = 2,3,4$)
So Kuba I think your answer is wrong....

Pierre.

If I am not wrong, the first terms are 2, 7, 30, 143, 728.
Looking on Sloane's sequences encyclopedia, it might be  A006013 (Formerly M1782)

so that the answer would be $\frac 1 {3n+2} C_{3n+2}^{n+1} $.
Does anyone want to prove it?

Pierre.

oups... seems that I missunderstood the word "nondecreasing" :( ...

the number of such sequencences of length n ending with k (denoted here as $S_{n,k}$) can be defined as:

$S_{n,k} = \sum \limits^k_{i=1} S_{n-1,i} $ for $n>1$ and $k\leq 2n$

and $S_{n,k} = 0$ for $n>1$ and $k \geq 2n+1$.

(how do I make sigma appear as a symbol?)

Beyond this I have nothing, especially why the sum seems to magically match "3n+2 choose n+1" divided by 3n+2.

Well, there is something wrong about this problem. Notice that the only strictly decreasing list is $(2,1)$ for $n=2$. $a_1\in {1,2}$ so the only decreasing lists are $(1,1,1...1)$, and $(2,2..2,1,1.....1)$ and their number is $1+(n-1)$. So now we only have to subtract these from the total number of $\prod{2j}=2^nn!$.
Am I wrong????

The converse of 'decreasing' is not 'non-decreasing'...
English words are sometimes surpsrising :
'Increasing' means 'strictly increasing', so that 'non-decreasing' means 'increasing but not necessary strictly increasing' (I know, it sounds a bit strange). The same is true for 'decreasing' and 'non-increasing'.

Pierre.

wow
in Romanian it is just increasing for "non-decreasing".
thx. Now I can think on the problem


[quote="lamefif"]Hi
I need the knowledge of algebra at the moment, and I thought why not try this forum.
[img]http://uk.geocities.com/ones8zeros/index_picture1.gif[/img]

It will help me a lot if anyone can show the work done to the above.[/quote]
Sure thing :D First, I'll give you a hint and see if you can work it out. If you can't, then look at the solution.

For the hint, remember the distributive property? What does this remind you about the distributive property?

[hide="Solution"]You'll see this quite common so be ready to recognize it. The denominator of 2 really doesn't contribute anything to this problem at all so I'll just talk about the numerators.

You have $m(m+1)+2(m+1)$. Let's set $x=m+1$. Now, we have $mx+2x$. This reminds you of distributive right? We can distribute and get $x(m+2)$. Then, substitute $m+1$ for $x$, and we get

$m(m+1)+2(m+1)=(m+1)(m+2)$[/hide]

[quote="lamefif"]Hi
I need the knowledge of algebra at the moment, and I thought why not try this forum.
[img]http://uk.geocities.com/ones8zeros/index_picture1.gif[/img]

It will help me a lot if anyone can show the work done to the above.[/quote]

Would this method be usuable as well.

m(m + 1) + 2(m + 1) = (m + 2)(m + 1)

m(m + 1) <=> m^2 + m
2(m + 1)  <=> 2m + 2

 m^2 + m + 2m + 2 = m^2 + 3m + 2

m^2 + 3m + 2 <=> (m + 1)(m + 2)

since

(m + 1)(m + 2) <=> m^2 + 2m + m + 2 = [b]m^2 + 3m + 2[/b]


If $ a,b,c \ge 0$  then prove

$ \frac {a}{b\plus{}2c}\plus{}\frac {b}{c\plus{}2a}\plus{}\frac {c}{a\plus{}2b} \ge 1$

\[ \frac {a}{b \plus{} 2c} \plus{} \frac {b}{c \plus{} 2a} \plus{} \frac {c}{a \plus{} 2b} \equal{} \sum_{cyc}\frac {a^2}{ab \plus{} 2ac} \]\[ \geq \frac {(a \plus{} b \plus{} c)^2}{3(ab \plus{} bc \plus{} ca)} \geq 1
\]
By Cauchy (Engel Form) and Newton's Sums inequality

[quote="soruz"]how prove Newton Inequality?[/quote]
$ (a\plus{}b\plus{}c)^2 \ge 3(ab\plus{}bc\plus{}ca) \Leftrightarrow (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2 \ge 0$!!!
Done!


why do you even post this here??? there is no point in the problem. ;)   you have your numbers, and the formula..

all you have to do is plug in the numbers... ;)  ;)

Please avoid posting Homework problems here!

[quote="toshiba1"]A sample of gas absorbs exactly 577 J of heat from its surroundings. At the same time, the gas expands from 4.37 L to 12.93 L against a constant external pressure of 0.985 atm. What is the internal energy change for the gas?

now the formula for this is
Delta U= DELTA h - P DELTAV
v= 12.93-4.37
P= 0.985

(1 atm L = 101.325 J)[/quote]

dU=dq+dw (pertains to the work of expanstion)


rotation
combinatorics unsolved
combinatorics

Players $ A$ and $ B$ alternately move a knight on a $ 1994 \times 1994$ chessboard. Player $ A$ makes only horizontal moves, i.e. such that the knight is moved to a neighboring cell to the right or to the left, while $ B$ makes only vertical moves, similarly. Initially player $ A$ places the knight to an arbitrary square and makes the first move. The knight cannot be moved to a square that was already visited during the game. A player who cannot make a move, loses. Prove that player $ A$ has a winning strategy.

If A moves horizontal, A moves from column to column, not to another row. And why call it a knight when it moves like a rook?

Anyway, after each move of A the number of occupied cells in each row is even, so A always has a move when it is his/her turn on a board with an even number of columns. Boring game. A cannot lose.

[quote="MrOoijer"] so A always has a move when it is his/her turn on a board with an even number of columns. [/quote]

What do you mean? Why?

$ A$ can certainly lose if he does something stupid, so I'm also not sure I understand the argument given by MrOoijer. Anyway, the argument I do see is similar: Player $ A$ imagines the board tiled by $ 1 \times 2$ dominos, all horizontal. On his first move, he moves to the other cell in the tile. By always taking the other cell on the current tile, $ A$ forces $ B$ to always move into a new tile on his move, and $ A$ can then take the other cell on that tile. Thus $ B$ will eventually lose.

Example of $ A$ losing on a $ 4 \times 4$ board: Let the cells be

[code]
ABCD
EFGH
IJKL
MNOP
[/code]

and suppose that $ A$ starts at B, moves to C, $ B$ moves to G, $ A$ moves to F, $ B$ moves to J, $ A$ moves to K, $ B$ moves to O and $ A$ stupidly moves to P. Then $ B$ moves to L and $ A$ loses.

If $ A$ followed our strategy and started at B, the game would go BAEFJIMN and $ B$ is stuck.

Does this work?

Suppose Player $ B$ always has a winning strategy. Player $ A$ [imagines that he] places the knight on the square right below the top left square (on randomgraph's diagram, E). He then has only one option for his first move, to the square to the right of it. If player $ B$ then moves the knight a square up, player $ A$ can move the knight into the top left corner, and then player $ B$ has no moves, contradiction. So player $ B$ must move the knight down, and there is no way for the knight to return to the top left corner. By assumption, $ B$ now has a way to win.

Player $ A$ instead rotates the board $ 90^{\circ}$ and plays the knight on the new location of the top-left square (A'), and moves it to the square that corresponds to the original location of the previous imaginary knight (E'). He then copies player $ B$'s strategy from the previous game, so player $ B$ cannot always have a winning strategy.

That's a nice strategy-stealing argument. I think my approach has two advantages:

1. It works no matter what the starting position is (so it shows $ A$ can win even if $ B$ gets to choose the starting position. In fact, it shows that $ A$ wins even if $ B$ is allowed to jump to an arbitrary empty cell at each move!)

2. It's explicit. Strategy-stealing arguments don't tell you what the strategy actually is (see the sorry state of Chomp).


[quote="zaizai-hoang"][color=green][b]Problem of zaizai: [/b][/color] (Nguyễn Thúc Vũ Hoàng)
Let a,b,c are positive numbers such that $abc=1$. Prove that:
$3 \sqrt[3]{(a^{3}+1)(b^{3}+1)(c^{3}+1)}+1\ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2 \sqrt{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$[/quote]
This follows from the following inequalities
$3 \sqrt[3]{(a^{3}+1)(b^{3}+1)(c^{3}+1)}\ge a+b+c+ab+bc+ca$
and
$1+a+b+c\ge 2\sqrt{1+ab+bc+ca}$
The first is quite easy and the second might (or not) be found in pvthuan's book  :).

A similar problem: for any positive reals such that $abc=1$,
$\sqrt[3]{(1+a)(1+b)(1+c)}\ge \sqrt[4]{4(1+a+b+c)}$

i want see your solution
 :D can you post ?

Now, Everybody let try prove that a inequality stronger than above inequality.
Let $a,b,c\ge 0$ such that $abc=1$. Prove that:
$\sqrt[3]{(a^{3}+1)(b^{3}+1)(c^{3}+1)}+1\ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
My solution used mixing variable method, but I think it is ugly  :( I hope anyone will post in here a nice solution by another method  :wink:


Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}$.

[hide="Answer"]$10^{10}=2^{10}\cdot 5^{10}$, $15^7=3^7\cdot 5^7$, $18^{11}=2^{11}\cdot 3^{22}$
$121$ factors of the first, with $11$ of those being factors of $15^7$ and $11$ being factors of $18^{11}$; $64-8=56$ new factors of the second, with $7$ of those being factors of $18^{11}$; $276-11-7=258$ factors of the third, so we have $121+56+258=\boxed{435}$ distinct postive integers.[/hide]

[b]EDIT:[/b] That was an odd mistake. :huh:

[hide="Solution"]First, we write the prime factorizations of all of our numbers to help us see what's going on. $10^{10}=2^{10}5^{10}$, $15^7=3^75^7$, and $18^{11}=2^{11}3^{22}$.

Our total number of divisors are: $121$, $64$, and $276$. However, we realize that we have overcounted. To compensate for this, we apply PIE, also known as the Principle of Inclusion-Exclousion.

Basically, this is just finding the number of elements in a group of overlapping sets by subtracting elements common to two, adding back for three, subtracting for four, adding back for five, and etc.

Therefore the answer is $(121+64+276) - (8+11+8) + 1= 435$.
[/hide]

[hide]$10^{10}= 2^{10} \cdot 5^{10}$, so it has $121$ factors. $8$ of them are also factors of $15^7$, and $11$ of them are also factors of $18^{11}$.

$15^{7}= 3^{7} \cdot 5^{7}$, so it has $64$ factors. $8$ of them are also factors of $10^{10}$, and $8$ of them are also factors of $18^{11}$.

$18^{11}= 2^{11} \cdot 3^{22}$, so it has $276$ factors. $11$ of them are also factors of $10^{10}$, and $8$ of them are also factors of $15^{7}$.

By the Principle of Inclusion-Exclusion, our answer should be \[ 121+64+276-8-11-8+1=\boxed {435}.  \][/hide]

I knew i had overcounded with 461 but how do you use POIE to find how many you have overcounted? I just did alot of casework and got 437  :wink:

[hide="easy one"]
\begin{eqnarray*} & a = 2^{10}\cdot 5^{10} & \\
& b = 3^7\cdot 5^7 & \\
& c = 2^{11}\cdot 3^{22} & \\
\end{eqnarray*}
Where $ d(x)$ denotes the number of divisors of some positive integer $ x$,
\[ D = d(a) + d(b) + d(c) = 11^2 + 8^2 + 12\cdot 23 = 461
\]
Now with $ (a,b) = 5^7$, $ (a,c) = 2^{10}$, $ (b,c) = 3^7$ we have obviously overcounted. Precisely, $ 8 + 11 + 8 - 1$ divisors too many because we must count [b]one[/b] at least once... So the overcount $ o = 26$. Thus, $ D - o = 461 - 26 = \boxed{435}$.[/hide]
Mathemonster, it's clear when looking at the prime factorizations of the three numbers that the only common divisors between any given two of $ a,b,c$ must be of the form $ x^n: x\in\mathbb{P}$. Hence, it is the sum of such possible divisors minus one that gives the overcount.

edit: Perhaps someone else can explain it better. These types of things aren't exactly my strong point.

[quote="mathemonster"]so we overcounted 2^10 , 5^7, 3^7, because there are numbers with those bases raised to higher powers. i get why we subtract but why add a one on the end?[/quote]

Consider $ 2_0\equal{}5^0\equal{}3^0\equal{}1$. We counted this $ 1$ in the 461 three times. But if we actually use the same theorem to calculate $ o$ we count $ 1$ three times again but we only want to count it twice. To compensate, we add one to the number 434.

[url=http://www.mathlinks.ro/Wiki/index.php/PIE]This[/url] may interest you.

Theres another method I thought of (besides PIE) but it doesn't get the right answer but sorta make sense. Sorry if it may seem really weird, but just wanted to clear up logic misunderstanding.

So to get a number that doesn't have too many primes so that its bigger than all three and would be multiple, not divisor would be maximum 10 fives, 22 thres and 11 twos. So you get number of factors add to $ 11x23x12 $. But if there are at least one of each prime (2,3,5) then it can be not a divisor to any because none of them have all. So subtract $ 10x22x11 $. But then you get 616.

I want to know why this is wrong seems like even though I got right answer with Pie that this sorts of method seems valid but its not so if you could explain why that would really help understanding counting types better Thanks.


Let $\mathcal{M}(A)$ denote the $\sigma$-algebra generated by $A$, that is, the intersection of all $\sigma$-algebras containing $A$.

Prove that

$\mathcal{M}(\mathcal{E})=\bigcup_{\mathcal{F}\subseteq \mathcal{E}\atop{\mathcal{F} \text{ is countable}}} \mathcal{M}(\mathcal{F})$

In particular, I can prove that the object on the RHS is a $\sigma$-algebra, but not much else.

It seems to me that you've done most of the work.
Since each $\{A\} \subseteq \mathcal E$ is countable (singleton), it follows that RHS is some sigma algebra containing $\mathcal E$. So LHS $\subseteq$ RHS, by definition of LHS.
However, RHS $\subseteq$ LHS since each $\mathcal F \subseteq \mathcal E \Rightarrow \mathcal M(\mathcal F) \subseteq \mathcal M(\mathcal E)$, so $\bigcup_{\mathcal F} \mathcal M(\mathcal F) \subseteq \mathcal M(\mathcal E)$.

I hope that there are no silly mistakes.

It still is not clear to me why $\mathcal{E}$ is contained in the RHS.  Similarly, I do not see why $\mathcal{F}\subset\mathcal{E}\implies \mathcal{M}(\mathcal{E})\subset\mathcal{M}(\mathcal{F})$.

First we prove that $\mathcal E \subseteq$ RHS.
Let $A \in \mathcal E$.
Then $\{A\}$ is a countable subset of $\mathcal E$. Since $\{A\} \subseteq \mathcal M( \{A\})$, it follows that $\{A\} \subseteq \bigcup_{\mathcal F} \mathcal M(\mathcal F)$, for each $A\in\mathcal E$. So it follows that $\mathcal E \subseteq$ RHS.

Now for the claim that $A \subseteq B \Rightarrow \mathcal M (A) \subseteq \mathcal M (B)$:

By definition of $\mathcal M$, $\mathcal M(A) = \bigcap_\sigma \sigma$ over all $\sigma$ such that $A \subseteq \sigma$. Same for $\mathcal M(B)$. And simply note that if $A \subseteq B$ then every sigma algebra containing $B$ also contains $A$. So $\mathcal M(A)$ is the intersection of at least all the sigma algebras containing $B$, possibly more.
That's why $\mathcal M(A) \subseteq \mathcal M(B)$.

I hope its correct.

Thank you so much for your help.  I completely misunderstood something, but now it's clear.


invariant
linear algebra
matrix
superior algebra
superior algebra unsolved

Hi

I am taking my first shy steps in representation theory
However I will rephrase the problem a bit :

Let A be the set of 3*3 symmetric matrices with trace (sum diagonal elements) zero over the reals
(that is a five dimensional space)

let G be the group of orthogonal 3*3 matrices

Let the action of a g in G on a in A be

$g^{-1}Ag$

Show that A cannot be write as the direct sum of two subspaces, both invariant under action of G

(This would make it an irreducible representation)

My apologies if I said something silly, or if this is the wrong place for this kind of stuff.
I am willing to listen to more general ideas, hints or whatever...

just an idea:
Suppose that $A=E+F$, where the sum is direct, $E,F$ invariant under the action induced by $G$.
$W$ the 2-dimensional space of diagonal matrices with zero trace.

Take $M \in E$. $M \neq 0$ is a real symetric matrix, its spectrum is real.

Suppose that its eigenvalues are $(a,b,c)$, $a+b+c=0$.
At least two eigenvalues are different. For example $a \neq b$, then the matrices $D_1=diag(a,b,c)$
and $D_2=diag(b,a,c)$ are in $E$. ($E$ is invariant and permutaion matrices are orthogonal).
But these two matrices form a basis of $W$. So $W$ is included in $E$.
But every matrix in $E$ can be written under the form $UwU^*$, where $U$ is orhtogonal
and $w \in W$. So $E=A$

So there is no invariant non trivial subspace. (maybe computation mistakes)

Looks good

I only see a small subtle problem if c=0 then your two diagonal matrices are linear dependent
But in that case you can use (a,-a,0) and (a,0,-a)

Anyway, thanks for the good ideas

Do you know any generalisations of this kind?


geometry
3D geometry
sphere
ratio
integration

[color=blue] A charge $q$ is distributed equally in a sphere with radius $R$.
Consider the constant of insulator is unit.
Find the ratio $\frac{W_1}{W_2}$ of the internal and external potential energy of the static electricity.  [/color]

I have a few questions:

Are you talking about electric potential (as in V), or the potential energy of the sphere? Also, does internal potential energy mean taking the center of the sphere as the reference point and external mean taking infinity as a reference point?

I think it's obvious that he's talking about the energy, that's what he asked: "...potential energy...".

[color=blue]The ratio is $\large\frac{1}{5}$
[quote="psychicalchemist"] does internal potential energy mean taking the center of the sphere as the reference point and external mean taking infinity as a reference point?[/quote]
That's right[/color]

ok, in retrospect the first part of my question may have been superfluous. But if my assumption in the 2nd part of my question is correct, then shouldnt the ratio be negative (the potential energy of a sphere with respect to its center is negative, while with respect to infinity it is positive).

come to think about it.... I've solved it relative to infinity

[color=blue]Here's the solution
The internal potential energy of the static electricity is

$\large W_1=\frac{\epsilon_0}{2}\int_0^R(\frac{\rho}{3\epsilon_0}r)^2 4\pi r^2dr$

$\large\Longrightarrow W_1=\frac{q^2}{8\pi \epsilon_0 R^6}\int_0^R r^4dr$

$\large\Longrightarrow W_1=\frac{q^2}{40\pi \epsilon_0 R}$
The external potential energy... is
$\large W_2=\frac{\epsilon_0}{2}\int_R^{\infty}\frac{q^2}{16\pi^2\epsilon_0^2 r^4}4\pi r^2dr$

$\large\Longrightarrow\large W_2=\frac{q^2}{8\pi\epsilon_0}\int_r^{\infty}\frac{dr}{r}$

$\large\Longrightarrow W_2=\frac{q^2}{8\pi\epsilon_0 R}$
where $\large\frac{\rho}{3\epsilon_0}r$ is the electric field of sphere's entrail and $\large\frac{q^2}{16\pi^2\epsilon_0^2 r^4}$ is the electric field outside the sphere
Then the ratio is $\large\frac{1}{5}$[/color]


Let $ O$ be the point of intersection of the diagonals $ AC$ and $ BD$ of the quadrilateral $ ABCD$ with $ AB \equal{} BC$ and $ CD \equal{} DA$. Let $ N$ and $ K$ be the feet of perpendiculars from $ D$ and $ B$ to $ AB$ and $ BC$, respectively. Prove that the points $ N$, $ O$, and $ K$ are colinear.

[color=darkred][b]April[/b], I think that is such (I corrected - boxed) :[/color]
[quote="April"][b]Let $ O$ be the point of intersection of the diagonals $ AC$ and $ BD$ of the quadrilateral $ ABCD$ with $ AB \equal{} BC$ and $ CD \equal{} DA$. Let $ N$ and $ K$ be the feet of perpendiculars from $ D$ and $ B$ to $ \boxed {AB}$ and $ \boxed {CD}$ respectively. Prove that the points $ N$, $ O$, and $ K$ are colinear.[/b][/quote]
[color=darkblue][b]Proof.[/b] Define the point $ M\in BD$ for which $ MA\perp AB$. Thus,

$ MC\perp AC$ and $ \left\{\begin{array}{c}
\widehat {MAD}\equiv\widehat {MCD}\\\
\widehat {NDA}\equiv\widehat {MAD}\\\
\widehat {KBC}\equiv\widehat {MCD}\end{array}\right\|$ $ \implies$ $ \widehat {NDA}\equiv\widehat {KBC}$.

The quadrilaterals $ OAND$ and $ OCKB$ are cyclically. Therefore,

$ \left\{\begin{array}{c}
\widehat {NDA}\equiv\widehat {NOA}\\\
\widehat {KBC}\equiv\widehat {KOC}\end{array}\right\|$ $ \implies$ $ \widehat {NOA}\equiv\widehat {KOC}$ $ \implies$ $ O\in NK$.

[b]Otherwise[/b], since the quadrilaterals $ OAND$ and $ OCKB$ are cyclically obtain

$ \left\{\begin{array}{c}
m(\widehat {NOA})\equal{}m(\widehat {NDA})\equal{}90^{\circ}\minus{}\left[m(\widehat {ABD})\plus{}m(\widehat {ADB})\right]\\\\
m(\widehat {KOC})\equal{}m(\widehat {KBC})\equal{}90^{\circ}\minus{}\left[m(\widehat {CBD})\plus{}m(\widehat {CDB})\right]\end{array}\right\|$ $ \implies$

$ \widehat {NOA}\equiv\widehat {KOC}$ $ \implies$ $ O\in NK$.[/color]


geometry
3D geometry
search
number theory
prime numbers
AMC

Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.


topology
abstract algebra
advanced fields
advanced fields unsolved

Compute the homology groups of the projective plane $ \mathbb{R}P^2$ using Mayer-Vietoris.

(I only know about singular homology groups as Vick defines them in his book.)

Anyone? I have some good ideas but I don't get the details right...

cut out a $ D^2$, so $ A \cong S^1, B \equal{} D^2, A \cap B \equal{} S^1$. Then you have to calculate the mapping degrees.

That was my idea, too, but how to find these mapping degrees (in a rigorous way)?

Hi Arne. As myan said, you can cover $ {\mathbb R}P^2$ by two open sets, one, say $ B$, being an open disc, the other, say $ A$, its complement thickened a bit such that it is open. Then $ B$ is contractible and $ A$ is homotopy equivalent to $ {\mathbb S}^1$ by projection onto the boundary $ {\mathbb R}P^1$ and noting that $ {\mathbb R}P^1\cong {\mathbb S}^1$. Next, $ A\cap B$ is thickened $ {\mathbb S}^1$ and thus homotopy equivalent to $ {\mathbb S}^1$. Letting $ \iota: A\cap B\to A$ the inclusion, we have to determine the induced map $ \iota_{\ast}: H_1 (A\cap B)\to H_1(A)$ in homology. Since the fundamental groups of both spaces are $ {\mathbb Z}$ and thus in particular abelian, we can alternatively try to describe $ \iota_{\ast}: \pi_1(A\cap B)\to \pi_1 (B)$. To determine the image of the generator $ {\mathbb S}^1\to A\cap B$ in $ \pi_1(A)$, we use the homotopy equivalence $ A\simeq {\mathbb S}^1$ which is given by the projection onto the boundary $ {\mathbb R}P^1$ followed by the homeomorphism $ {\mathbb R}P^1\cong {\mathbb S}^1$. Since the composition $ {\mathbb S}^1\to {\mathbb R}P^1\cong {\mathbb S}^1$ is the map $ z\mapsto z^2$, a generator of $ \pi_1(A\cap B)$ is mapped to two times a generator in $ \pi_1(A)$. This should be enough to compute $ H_{\ast}({\mathbb R}P^2)$ via Mayer Vietoris.

Hi Hanno, thanks for your answser. A few questions: $ A \cap B$ is thickened $ \mathbb{S}^1$: why exactly? I'm confused because we're messing around with points of $ A$, identifying antipodal points. Same question for $ A$. Also, how to do this without using the link between $ H_1$ and $ \pi_1$?

Hi Arne. I'm talking about $ {\mathbb R}P^2$ as the space obtained by identifying antipodal points of the boundary of $ {\mathbb D}^2$. It's not difficult to see that this is the same as the more common definition as the quotient space of $ {\mathbb S}^2$ by the antipodal operation of $ {\mathbb Z}/2{\mathbb Z}$. Ok, so looking at the set of equivalence classes of the boundary points, we get an $ {\mathbb R}P^1$, which is simply a $ {\mathbb S}^1$. Thickening this set of equivalence classes, that is, taking the set of equivalence classes of a thickened boundary, we get the set which I called $ A$. Now, $ B$ can be defined as the set of equivalence classes of points that lie inside $ {\mathbb D}^2$, which is this interior itself (since there are no identifications). Is it clear now? If you don't want to use the connection between $ H_1$ and $ \pi_1$ you can also look at some arbitrary generator of $ H_1(A\cap B) \cong H_1({\mathbb S}^1) \cong {\mathbb Z}$ (whatever you like) and compute it's image in $ H_1(A)$ explicitly - this shouldn't be much more difficult than the argumentation via $ \pi_1$.


[quote="Robert0201"]Have any of you taken AP Computer Science?  Is there any math in it?[/quote]
I've not taken it, but computer science is technical lingo for programming, which generally does not involve copious amounts of math, and the math that it does involve is generally not difficult. However, programming does require strong logic skills, and the lower-level languages require far more logic than the higher-level languages.

[quote="Spoon"]you will be required to understand different base systems... there was a problem this year that asked you to calculate 100_16 - 10_16 and give your answer in base 10.... other than that, not really.[/quote]
What do you need, Base 16, maybe base 8 (octal, no longer really used), base 2, and base 10?

I don't remember running across any problems dealing with other bases in AP Compsci.

I have found good uses for binary and hexadecimal in programming.

Binary: boolean arrays.

Booleans take up one byte, although they really only need one bit, but a byte is the smallest addressable unit of memory, so they take up a byte, which is kind of wasteful (this matters on USACO.)

Hexadecimal: Using infinity.

I don't care to memorize the largest number, in base 10, that can fit into an int.

But its easy to figure out (and memorize) in hexadecimal.

Its 0x7FFFFFFF. Good to know.

[quote="Spoon"]I'm just letting you know the question was on the test this year...[/quote]

Aww, man, they're definately going to cancel your score now, and send you to prison.  :-)

[quote="Nukular"]I don't care to memorize the largest number, in base 10, that can fit into an int.

But its easy to figure out (and memorize) in hexadecimal.

Its 0x7FFFFFFF. Good to know.[/quote]
In C++, you can write numeric_limits<int>::max() for the largest integer, provided you include the header <limits>.  In C (or C++), you can also use the macro INT_MAX, provided you include the header <limits.h> (or <climits>).  Both solutions are more portable than assuming that an int has 4 bytes.

[quote]Why take AP Computer Science when you could teach yourself?[/quote]

Well this goes back to the question of why take any AP class when you could learn it on your own.  Well, for one you would get an AP credit which always looks good on college apps.  You could possibly learn more in a class where you have a teacher to ask questions even though I know that a lot of you are good at self teaching.  Some college will give you credit for AP.  Finally it can just fill a blank spot in your schedule.

[quote="Duck-Billed Platypus"]Why take AP Computer Science when you could teach yourself?[/quote]

Why go to school at all when you could teach yourself?

I agree with gauss. Although I have managed to self-study a lot of things, self-study just isn't the same as studying with a teacher. I admit that you do learn things faster and more efficiently, but, when you take a class, you are forced to review old concepts and focus on important ideas. When you self-study, you miss out on this, and-- if you are like me-- you forget things very quickly. Also, taking a class makes it easier to ask questions and get help.

Off the top of my head, I can't think of any math that was tested on in APCS this year. Just your ability to reason.

I think that Computer Science is probably the worst taught subject in school. The official curriculum is weak, and teachers are math teachers. Our other programming teacher was a math teacher/track coach. I did not learn anything from my teacher this year. Most of the things he wrote on the board had to be corrected. All he did was take things from the book. He did not prepare me for the AP test. He did not teach me anything useful, that I did not know, or that could not be learned much quicker from the textbook...

If you have a good teacher for computer science then I'm sure you will enjoy the class. However, under the circumnstaces I had to take it in I would rather have taken it self-study. I think this goes for most people.

I'm 13, and I don't care too much about college apps, and I don't plan to.

Why go to school? Because it's required by the law where I live. If it wasn't required by the law, I'd probably not go, unless my parents made me...

What exactly is taught in Computer Science?

Since it's about computers, I'm sure that if you go the right places on the Internet then you would find some helpful people there.

There are people [url=http://forums.gentoo.org/]here[/url] that know a lot of programming (Java included). You could ask there.

Why Java? If it's because it's OOP, then why not Python? I hear that it's easier. The code is also cleaner looking.

I'm teaching myself to program in C, and I'm having a hard time, but figuring things out is part of the fun, isn't it?

Why is school so important? Is it because everyone thinks that it's important? Just because it seems important in modern society doesn't make it important.

For most people, it's important since there aren't that many people who are sufficiently motivated to learn the material on their own. But those who are often find school completely useless. There is a boy who went to my high school who was really motivated and studied a lot on his own (at least math and sciences; I doubt he ever would have opened a history book on his own). He thought school was a waste of time, and his parents agreed with him, so they would excuse any absence of his, and therefore he only showed up two or three times a week. Now he's a freshman at Berkeley, and he says all the classes are trivial. But I think people like him are quite rare.

Indeed, most (80+%) time in school is not worthwhile.  The main reason is that the teachers don't want to challenge the students too much.  This is OK, but it has gone a bit too far.  How can you expect them to be interested if you give them coloring books to color?  They could at least give puzzles out so that the children would actually think.  Now this is in elementary school.  Then, in middle school and high school, many kids don't really want to learn or think about things anymore.

[quote="ComplexZeta"]Some teachers give students coloring books in high school. Some of my homework assignments (the ones I didn't do, naturally) were to color maps and similarly ridiculous things.[/quote]

That is not the same as not caring about going to college, as Duck mentioned... From what he described, I would consider it seriously underplaying the importance of high school.

Today was my last day btw, I'm free :)

my history teacher this year was one of the best teachers I have ever had.  She actually challenged me and my fellow classmates.  Though she had some stupid ideas for projects...Who wants or sees purpose in classifying 20 Texas Wildflowers?

[quote="probability1.01"][quote="Spoon"]umm... If you actually want to do something with your life, school is the most important thing... anyway

Comp Sci is basically a beginning Java Programming class, it doesn't really deal with "computers", Comp Sci is the fancy name for programming.[/quote]

OR

if (school == mine) {
comp_sci = learn_to_use_microsoft_word
}[/quote]

Thats sad learning to use Micrsoft Word ???  I have been using it since 2nd grade!

[quote="probability1.01"][quote="Spoon"]umm... If you actually want to do something with your life, school is the most important thing... anyway

Comp Sci is basically a beginning Java Programming class, it doesn't really deal with "computers", Comp Sci is the fancy name for programming.[/quote]

OR

if (school == mine) {
comp_sci = learn_to_use_microsoft_word
}[/quote]

At my city's high school, you get to learn Excel.  But then, I looked at the college's course descriptions and found word processing...

How does learning either of those prepare you for the AP exam... I was referring to Comp Sci AP in my description... not just a normal computer class.

[quote="Duck-Billed Platypus"]
Why go to school? Because it's required by the law where I live. If it wasn't required by the law, I'd probably not go, unless my parents made me...
[/quote]

homeschool then... if you feel you're not learning anything in school, you'll probably be better off teaching yourself throughout highschool. and you'll feel better about learning.

unless you feel like that'd be too sheltering... in that case, you have a reason to go to school.

whoops, almost forgot why I was posting... can anyone recommend a method of teaching oneself Java for AP compsci? i haven't had much java before, so I'm using "Sams teach yourself in 24 hours blah blah". would that be sufficient?

Here's the course description.
http://www.collegeboard.com/prod_downloads/ap/students/compsci/ap03_compsci.pdf

There are several books specifically geared for CS AP exam, here's one:
http://www.eimacs.com/LM/LMPlain.asp?F=beprepared&N=Welcome

And I believe, you can subscribe for an internet course at eImacs site

There's also a specific "case study" in the course; you can find info on it (and other info) at apcentral (it's actually for teachers, but I'd say, if you are teaching yourself, you'll be the teacher)
http://www.skylit.com/cblinks.html

Oh, and yes, homeschooling is perfectly legal in California. As well as in any other state.

My parents won't let me homeschool, largely due to my irresponsibility...

Learning how to use Microsoft Word is crazy.
1.  There are word-processors other than MS Word.
2.  You not only have to pay for it, but it's expensive. Some of you may say that free software sucks becuase it's free, but I say otherwise.

Generally, I'm against all things Microsoft. Mainly because I disagree with their business practices (or whatever you call it), but also partly because I believe that there are better things out there than the software that MS produces. But I do admit, they make nice mice...

For programming, if they have one, Oreilly books are generally nice, though very often a bit expensive...

They should teach how to write object code in hex in Computer Science... Not that I know how to, but it'd be useful.

Back to the oringinal question -

There is some math on it, but only the AB test.  That''s just using BigO notation, which is really easy to learn.


Prove that there exist infinitely many positive integers $ n$ such that $ 2^n$ ends with $ n$ , in other words, $ 2^n\equal{}...n$ .

Obviosly $ 4|n$, therefore $ 2^n\equal{}1\mod 5\to n\equal{}1\mod 5\to n\equal{}16\mod 20$.
If $ n\equal{}16\mod 20$, then $ 2^n\equal{}...36$, therefore $ n_2\equal{}36$ first solution.
If we had solution $ n_k$ - had k digits, then consider $ n_{k\plus{}1}\equal{}2^{n_k}\mod 10^k,k\ge 2$. Easy to chek, that $ 2^{n_{k\plus{}1}}\equal{}2^{n_k}\mod 10^{k\plus{}1}$, therefore $ 10^{k\plus{}1}|2^{n_{k\plus{}1}}\minus{}n_{k\plus{}1}$ or $ n_{k\plus{}1}$ is next solution. For example $ 2^{36}\equal{}736\mod 1000\to n_3\equal{}736.$


geometry
3D geometry
sphere
Columbia
HCSSiM

hola from the MAs

6th place team!! 8 on team =D

Me: 112th place, fun, fun. 15 on sprint. 12 OF LAST 15 WRONG.
everyone cheer now.

hey jorge silvis came in 8th.

I got 24th place :D !!! Made stupid mistake on 1 target which would have gotten me 14th probably.
And 15th and 16th were both girls I think.
And MA got 6th!! With 3 7th graders!!

Note: Jorge Silvis is actually George Silvis from MA.

56th! Someone from CT was the last person in the top half!

37th with 31 (19/6) :| (dropped from 30th due to tiebreakers)
I still think they screwed up grading; I remember specifically 2 sprints they marked wrong Im sure I would have gotten. That would have put me up at least 10 places, but O well.

Sprint : 22, not bad....

Target : [hide="bad is an understatement"]only 3.... (6+8=10, right.)[/hide]

Total : 28, 70th place,.

MICHIGAN : 13th place team.

16/7 which is really weird considering I've gotten a 2 on a practice target before but 20+ on practice sprints. Usually best in CT on sprints and worst on targets and this time, I was 2nd on Sprint score, and best on targets.

Michigan isn't that overrated.

It was only filletwho, who told everyone that they should vote for Michigan, we had nothing to do with it.  Though I did think that Michigan could go all the way.

As for me, I am also not that overrated.  I thought I was going to make countdown, but I just missed it. :(

My score was 23 on sprint, 12 on target, 35 total.  Place was 14th.

Team was 13th

State Michigan

I won the tiebreakers.

I should've gotten the target problem correct (the one about the test questions, find the order...)

Now, I hope to move on.

[quote="Ignite168"][quote="biffanddoc"][quote="pgpatel"]I got a 21/6 giving me a 33 and 27th place. Funniest thing happened though.
Biffanddoc: What did you get?
Me: A 33
Biffanddoc: Hah, loser. I got a 36.
Next day I look at the rankings and I was ranked higher then biff :D.[/quote]
I still think I should have gotten 36. (at least 33)[/quote]
Maybe if you concentrated less on stealing CA pins and more on preserving your arm, you would've done better, or was that after the competition?[/quote]
Preserving my arm??
There was no such thing as CA pins, it was CT, and I wasnt trying to steal them.

Biffanddoc tried to steal a michigan from kyyuanmathcounts (I think). When kyyuan said no, biff borrowed a dollar from someone and tried to buy it, but the Michigan coach said no. The whole thing was hilarious.

i was kinda desparate for a michigan pin. I eventually got one though

People from Michigan just asked to trade with me :P .  My teammates didn't like Michigan, but I don't really care...

edit: I am from Ohio, neighbor of Michigan

[quote="mathgeniuse^ln(x)"]Michigan isn't that overrated.

It was only filletwho, who told everyone that they should vote for Michigan, we had nothing to do with it.  Though I did think that Michigan could go all the way.

[/quote]

Yeah, I doubt anyone who voted MI actually thought we'd win. But winning the polls is good enough for me.

[quote="xxm0rg07hxx"]
18th. team-12th. I am amazed we beat Michigan, New York, New Jersey, and some others.[/quote]
namely
Alabama
Alaska
Arizona
Colorado
Connecticut
Delaware
Department of Defense
District of Columbia
Georgia
Guam
Hawaii
Idaho
Iowa
Kansas
Kentucky
Louisiana
Maine
Minnesota
Mississippi
Missouri
Montana
Nebraska
Nevada
New Hampshire
New Mexico
Northern Mariana Islands
North Dakota
Ohio
Oklahoma
Oregon
Puerto Rico
Rhode Island
South Carolina
South Dakota
State Department
Tennessee
Utah
Vermont
Virgin Islands
West Virginia
Wisconsin
Wyoming

Three years later...

We get the point,

Inother words Arkansas beat all the teams except for the teams that they lost to. :?:

[quote="mathcaunt"]Wow. Thirteen people voted for top twelve...   :?:[/quote]

Daesun = Treething = Iron Fist

He might have voted twice or else someone's lying. Besides, someone said that 7 were AoPSers....not 12.

Does anyone know the AOPSers in the top 12?

[quote="Ignite168"][quote="pgpatel"]1st- Treething/Iron Fist
2nd- Hexahedron
S- Binouniquineist(sp?)
Q-RC-7th
Q-Samath
Q-(I_Would_Say)
P-Hamster1800[/quote]

2 of those got a bye to where they are

And your teammate also stole pins off the ground, and after that ran from people who asked for them.[/quote]
He also stole fl pins from a teammate of mine.

Ok, let's please drop the pin-stealing grudges.

Anyhow, we know that AoPS polls don't exactly have a good reputation for being too accurate...

ummm just a thought...
how is it that there are 14 people said they came in 1-12 in the poll?

[quote="Treething"]XD no, I suggested to Kevin and Daniel that we make 13 votes in the top 12. I'm pretty sure they got other people to vote.[/quote]
that answer your question?

[quote="Phelpedo"][quote="Treething"]btw Phelpedo, who are you?[/quote]

[i]Someone you know...[/i]

from camp

last summer

(hint- the coolest guy there.)[/quote]

seriously, who?

I know you're from Connecticut and you're probably in the yearbook but I'm too lazy to check.

EDIT: Tony Feng?


In 1990, the average age of Tad and his older sister was 6.  In 2002, the average age of Tad, his older sister, and their twin brothers was 13.  In what year were the twin brothers born?

A.1993	B.1994	C.1998	D.1999	E.2000

In 1990, the sum of Tad's and his sister's age is $ 6$ x $ 2$ = $ 12$ years
In 2002, their sum becomes $ 12 \plus{} 24 \equal{} 36 years$
If we say that their brothers were $ x$ years old in 2002, then the following equation is satisfied:
$ 36 \plus{} 2x \equal{} 52$
$ 2x \equal{} 16$
$ x \equal{} 8$
Therefore, the twin brothers were born in the year $ 2002 \minus{} 8$ = $ 1994$
B

[hide=" click to reveal hidden content"]

I think this one is $ 1994$.

Let the ages of Ted, his sister and the twins be $ t, s, w$.

Then $ t\plus{}s\equal{}2 \times 6\equal{}12$

and

$ t\plus{}12 \plus{} s\plus{}12 \plus{} 2(w\plus{}12) \equal{}4 \times 13 \equal{} 52$

Therefore $ w\equal{}\minus{}4$, so the twins were born in $ 1994$.

[/hide]


I don't know how to type the character "¬", the reason I need it is because I am making an Excel sheet to study for my spanish test, I got an app to handle spanish characters, but The only way to run the macro is to begin your character with the ¬ thing, the problem is I don't know how to type it on the keyboard. (If you haven't figured it out, I was copying that character when I'm writing this)

Hold alt and type 0172 with the number pad on the right (with num lock on, of course).

¬

Thanks, but those alt commands only work if you're using a keyboard

¬ Hooked up USB Keyboard now!

Wait, how do you type [i]without[/i] a keyboard?

well by keyboard I meant the ones that desktops use (I'm using a laptop right now)

Oh, I see. Well you have to use numpad to type the numbers but you can do it on a laptop keyboard as well. If it has a numpad. For example my laptop keyboard has the numpad mapped to Fn + 789 UIO JKL M keys so it's a little hard to do it because you have to be holding alt and fn at the same time, but it can be done ;).


what difference betwen
 [color=red]IMO
IMO Shortlist
IMO Longlists [/color] :?:

[quote="altusi"]i think you speak about problems,
imo: the 6 problems choosen
imo shortlist: a list of 20 problems (i am not sure about the number)
imo longlist:a list of problems certainly bigger than the precedent!

    :D[/quote]

Well, each country is allowed to send up to 6 problems to the IMO Problem Selection Committee (several months before IMO starts). These problems together (100 problems or something like that) form the IMO Longlist. The Problem Selection Committee (PSC) then judges these problems, and selects between 20 and 30 "good" and suitable problems. These problems then form the Shortlist. Finally, in jury meetings, 6 problems are chosen from the Shortlist. These 6 problems are the IMO problems :)

Doesn't that mean that if a country's problem gets chosen then the participants from that country will all be able to do it?

Well couldn't whoever thinks up the problems make sure that the participants from their country understand it? Or do they rely on people's honesty for not doing that?

IMO Proposed Problems should be kept secret. Very few cases of "cheating" are known. :) So that works quite well.

Hmm, I went the first year after 2001... let us just say cheating does happen unfortunately, simply by letting a team leader send you the answers after he has seen them at selection from the shortlist.  A certain team in 2001 all scored perfectly on a question with exactly the same solution as proposed, none of them managed to solve another problem completely... :(   but we will never know

Cyprus, problem 2. I heard that story...

Also, lots of people think the Bulgarian team cheated in 2003...

Arne I did not really wanna say it as it was not proven, but you did it anyway  :P
oh well ....

Bulgaria was fourth when I went, their score in 2003 is very good, what are the grounds to claim they cheated?

oh yes i was in imo 2003 ,bulgarian scors was ...all the members of the team got more than 36 ,very strong , i heard somthing about cheating when i participate in imo2004


function
inequalities
induction
inequalities unsolved

Maybe I have reinvented the wheel, but at least it’s a nice one.
$x_1,\ldots,x_n\geq0$. For natural $k\leq n$ and real $s>0$, define the “symmetric mean” $M_{k,s}$ by
$M_{k,s}=\left( \frac{\sum_{sym}(x_1\cdots x_k)^s}{{{n}\choose{k}}} \right)^{\frac{1}{s}}$.
(Note that the sum has ${{n}\choose{k}}$ monomials as its terms.) For real $r$ such that $kr\geq s$, prove $M^k_{1,kr}\geq M_{k,s}$, i.e.
$\left({\frac{\sum_{i=1}^n x_i^{kr}}{n}} \right)^{\frac{1}{r}} \geq \left( \frac{\sum_{sym}(x_1\cdots x_k)^s}{{{n}\choose{k}}} \right)^{\frac{1}{s}}$.
Looks harder to prove than it is. Take a closer look at it and enjoy :)

Nobody? Here we go.
Take the LHS as a function of $r$. As it’s just the power means of the $x_i^k$’s, it is well known that this decreases when $r$ decreases, (By the way, similarly the RHS increases with $s$.) So it’s sufficient to prove the inequality for $kr=s$, i.e.
$\left( \frac{\sum_{i=1}^n x_i^{s}} n \right)^{\frac ks} \geq \left( \frac{\sum_{sym}(x_1\cdots x_k)^s} {{n \choose k}} \right)^{\frac1s}$.
But now we can assume WLOG $s=1$ (replace $x_i^s$ by $x_i$), leaving us with
$\left( \frac{\sum_{i=1}^n x_i} n \right)^k \geq \frac{\sum_{sym}x_1\cdots x_k} {{n \choose k}}$.
We haven’t done anything serious yet, and yet the thing looks much simpler now.  :) We could prove it by induction on $k$ (nice exercise!) but it is even easier with Muirhead:
The LHS, if developped, will become a sum of "Muirhead partition functions"
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61207:
$LHS=\sum\lambda_j[\alpha_j]$, where the $\alpha_j$ are partitions of $k$ and the (rather tiny) positive coefficients $\lambda_j$ sum up to $\frac1{n !}$, as the RHS is equal to $\frac1{n !}[\underbrace{1,\ldots,1 }_{k}]$ and the denominators guarantee the correct weighting. As all the $\alpha_j$ dominate $[\underbrace{1,\ldots,1 }_{k}]$, the result follows from Muirhead.

nice inequality.but can you explain the last inequality more??
because I can't understand the link compeletely.
thanks

[quote="spanferkel"]I suppose by "last inequality" you mean the one that results when $kr=s=1$?
It's RHS is the sum over all ${n\choose k}$ possible products of exactly $k$ of the $x_i$'s (i.e. the elementary symmetric function).[/quote]
Thanks.I also find it is easy when I consider the Mac Laurian ineq post by Vasc :)
thanks again

[quote="spanferkel"]Now let's generalise the problem slightly. [hide]Already gotten rid of $s$ right away  :)[/hide]
 For integers $k>\ell$, prove $M_{\ell,1}^{k}\ge M_{k,1}^\ell$, i.e.

$\sqrt[\ell]{{{n \choose{\ell}}}^{-1}{\sum_{sym}x_{1}\cdots x_{\ell}}}\ge{\sqrt[k]{{n \choose k}^{-1}{\sum_{sym}x_{1}\cdots x_{k}}}}$  :D  :wink:[/quote]

I think this is Maclaurin's one. see http://www.mathlinks.ro/Forum/viewtopic.php?t=73460 .


Timea, who is always in a hurry, went up the escalator by making one step every second. This way it took her 20 steps to get upstairs. Next days she made 2 steps every second on her way up, and this time it took her 32 steps to get upstairs. How many steps would it take her to go upstairs if the escalator did not work?

Would a system of equations work?

let $x = \text {the speed of the elevator in steps per second.}$
let $y = \text {the number of steps between downstairs and upstairs}$
$20x + 20 = y$

$16x + 32 = y$

transitive property gives:

$20x + 20 = 16x + 32$

$4x = 12$

$x = 3$

so the elevator is moving at 3 steps per second. This gives $y$ to be: $20(3) + 20 = \boxed{80\text{ steps}}$
:jump:

I came really close to that equation, except that little off part made the whole thing screwed up, so I was sort of cluelesss  :blush:


[img]http://www.imagedropbox.com/images/mediya/hatmen.gif[/img]

Four men are lined up on some steps. They are all facing in the same direction. A wall seperates the fourth man from the other three.
So to summarise :-

Man 1 can see men 2 and 3.
Man 2 can see man 3.
Man 3 can see none of the others.
Man 4 can see none of the others.
The men are wearing hats. They are told that there are two white hats and two black hats. The men initally don't know what colour hat they are wearing. They are told to shout out the colour of the hat that they are wearing as soon as they know for certain what colour it is.
They are not allowed to turn round.

They are not allowed to talk to each other.

So the question is -
Who is the first person to shout out and why?

I think Man 2 would be the first to shout.

Man 1 thinks:
"Man 2 is wearing a white and Man 3, a black hat. Man 4 could be wearing either a black or white hat, since there are two black hats and two white hats. Therefore I don't know what the colour of my hat is."

Man 2 thinks:
"Man 3 is wearing a black hat. I could be wearing a black or white hat. If I'm wearing black, Man 1 would immediately know that his hat was white. Since Man 1 didn't know what colour his hat was, my hat is white."

The first who can logically shout it out depends on all the possible combinations of the two white and the two black hats among the four men.

There are 6 possible combinations.

Man 1	Man 2	Man 3		Man 4
-------       White       White                       --------
-------       Black        Black                         --------

In either of the above two cases Man 1 would be the one cry out correctly the colour of the hat he is wearing.

If Man 1 remains silent, then it is obvious that Man 2 and Man 3 are wearing hats of different colours. Therefore, in the remaining of the 4 cases, it would be the Man 2 who would shout out the colour of his hat for ceratin which would be just the opposite of what the Man 3 is wearing and he is able to see it.

Therefore, in 33.33% of the cases it would be the Man 1 and in the remaining 66.67% of the cases Man 2 would be the person to shout the colour out. Unfortunately, Man 3 and Man 4 are just passive watchers.

Man 1-----	Man 2-----	Man 3-----------------Man 4
B-----------	B-----------	W----------------------W       ==> Man 2 would shout that he is wearing the Black hat.
B-----------	W-----------B----------------------W       ==> Man 2 would shout that he is wearing the White hat.
B-----------	W-----------W---------------------B        ==> Man 1 would shout that he is wearing the Black hat.
W-----------B-----------B----------------------W       ==> Man 1 would shout that he is wearing the White hat.
W-----------B-----------W---------------------B        ==> Man 2 would shout that he is wearing the Black hat.
W-----------W----------B----------------------B        ==> Man 2 would shout that he is wearing the White hat.

well HATS OFF to both Aiden and Shamik Banerjee. Tx Shamik .. well explained.

Hmm Shamik Banerjee is more correct... I assumed that the arrangement of the hats on the men's heads would be as shown in the picture.


probability
geometry
analytic geometry
trapezoid
integration
conditional probability

Choose a point at random in the square with sides
0 <-   x   <-  1 and    0 <-   y  <-  1. This means that the probability that the point falls in any region within the square is the area of that region. Let X be the x coordinate and Y the y coordinate of the point chosen. Find the conditional probability P(Y < 1/2 | Y > X)

First, I think you should make it clearer that you mean the square $(0, 0), (1, 0), (0, 1), (1, 1)$.

Second, a nudge in the right direction:

[quote="8parks11"]This means that the [b]probability[/b] that the point falls in any region within the square is the [b]area[/b] of that region. Let X be the x coordinate and Y the y coordinate of the point chosen. Find the conditional probability P([b]Y < 1/2 | Y > X[/b])[/quote]

;)

Does that notation mean that we want to find the probability that $y < 0.5$ AND $y > x$, or is it $y < 0.5$ OR $y > x$?
[hide="Solution"]
For now, I'll assume you mean AND.

Take the unit square, and sketch it.
Draw the line $y = x$
Draw the line $y = 0.5$

You want to shade the region such that $y > x$, so we shade in the top left half of the square.

We also want to shade in the region such that $y < 0.5$, so we shade in the bottom half of the square.

We now have a square with two shaded regions. The conditional probability is the intersection of the regions divided by the total area of the square.

Some quick geometry on the shaded trapezoid reveals it has area $\frac{3}{8}$, so the conditional probablity that the point selected meets the demands of the problem is

$\boxed{\frac{3}{8}}$

This is, of course, interpreting the notation as meaning AND.
I'm a little unclear on what to do if it says OR.
[/hide]

The notation is y < 0.5 given-that y > x

tthe answer is

1/8 / 1/2 = 1/4

simple than i thought
didnt know the given

If you want to do it [i]very [/i]formally:

[hide]
The x-coordinate and y-coordinate are uniformly distributed over the interval $(0,1)$.

$P(y>x, y<5) = \int^{\frac{1}{2}}_{0}\int^{\frac{1}{2}}_{x}dy dx$

$\int^{\frac{1}{2}}_{0}\frac{1}{2}-x dx$

$\frac{1}{4}-\frac{1}{8}= \frac{1}{8}$.

$P(y>x) = \frac{1}{2}$.

$P(y<5 | y>x) = \frac{P(y<5, y>x)}{P{y>x}}= \frac{\frac{1}{8}}{\frac{1}{2}}= \frac{1}{4}$.
[/hide]

You could also do it graphically (which is probably easier).

@seesaw:  try to keep the solutions at HSB level.  Thanks.

Just draw the square & shade the regions.


A rather long but interesting exercise:

Find the last four digits of $2005^{2006}$ and call it $a$.  Then find solutions x and y such that $x^{2}+y^{2}=a$.

[hide="Hint"]
There's three.

[hide="Another hint"]
For the first part, use the totient function (and CRT, as t0r pointed out).

[hide="Another hint"]
For the second part, look at it mod 3, then mod 9.

I'm just joking; there's more no hints.  :rotfl:

[/hide]
[/hide]
[/hide]

$x^{2}+y^{2}=75^{2}$

blah..both numbers are 0 mod 3, so we need solutions to

$x^{2}+y^{2}=5^{4}$

$x+yi=(2+i)^{k}*(2-i)^{2-k}$
taking absolute values, $(x,y)=(7,24),(15,20),(25,0)$
correspond to $k=4,3,2$

now we just switch them, and switch the signs to get all solutions, then multiply by 3 to get the solutions for the original equation


Develop a method for finding n natural numbers such that for any two of them, their sum divides their product. Does your method finds the minimal numbers?

P.S. The original problem was for n=6, and you wasn't asked about the minimality.

I suppose you want $n$ distinct numbers, right? Otherwise we could simply take them all to be $2$ :).

In order to find $n$ such numebrs, we could take them to be of the form $a,2a,4a,\ldots,2^{n-1}a$, for example. Since we want $(2^i+2^j)a|2^{i+j}a^2\iff 2^i+2^j|2^{i+j}a$, if $i<j$, all we need is $2^{j-i}+1|a$, i.e. $2^t+1|a,\ \forall t\in\overline{1,n-1}$. We can have this by choosing an appropriate $a$.

However, I don't know about the minimality of this system of numbers, nor do I know what minimality means when we talk about a system of $n$ numbers :?. Maybe you could clarify that? Their sum should be minimal? Or if we order them $a_1<\ldots<a_n$, each $a_i$ is minimal in its position? Something else?


1 . Να βρεθεί η γενική λύση της μερικής διαφορικής εξίσωσης
     $ (x_1 +2x_2)u{}_x{}_1+(x_2-x_1)u{}_x{}_2=x_2$

2.  Να βρεθεί το γενικό ολοκλήρωμα που επιλύει την μερική διαφορική εξίσωση
     $ x{}_2u{}_x{}_1+u{}_x{}_2=x_1$
     αφού πρώτα βρεθεί η κανονική της μορφή

Για την 2.
θεώρησε το σύστημα $ \frac{dx}{y}\equal{}\frac{dy}{1}\equal{}\frac{dz}{x}$
Η 1η ισότητα δίνει $ d(x\minus{}\frac {y^2}{2})\equal{}0 \Rightarrow x\minus{}\frac {y^2}{2}\equal{}c_1$
Η 2η δίνει $ d(c_1y\plus{}\frac {y^3}{6})\equal{}dz \Rightarrow \minus{}z\plus{}c_1y\plus{}\frac {y^3}{6}\equal{}c_2 \Rightarrow\minus{}z\plus{}xy\minus{}\frac {y^3}{6}\equal{}c_2$
Λύση είναι η  $ f(c_1,c_2)\equal{}0$ όπου f αυθαίρετη συνάρτηση

Για την 1. Είναι  παρόμοια με λίγο πιο δύσκολες ολοκληρώσεις , πιθανόν να χρειαστείς την αντικατάσταση $ y\equal{}ux$

Από συνήθεια έβαλα $ x_1\equal{}x , x_2\equal{}y , x_3\equal{}z$


geometry
ratio
AMC
AIME
summer program
MathPath

Consider a triangle with an area  of 1. There are 3 lines passing each through A,B and C and dividing the opposite side in a ratio of 2:1. Show that the area of the triangle in the middle is 1/7.

$[ABC]=1;\ M\in (BC),\ \frac{MB}{MC}=m$$;\ N\in (CA),\ \frac{NC}{NA}=n;\ P\in (AB),\ \frac{PA}{PB}=p.$

$X\in BN\cap CP,\ Y\in CP\cap AM,\ Z\in AM\cap BN$ such that $X\in (YC)\cap (ZN).$

[u]Indication.[/u]

$[XYZ]=[ABC]-[ABM]-[BCN]-[CAP]+[AYP]+[BZM]+[CXN]$.
The Menelaus' theorem applied to the transversal $\overline {CYP}$ and the triangle $BCP\Longrightarrow$

$\frac{AP}{AB}\cdot \frac{MB}{MC}\cdot \frac{YC}{YP}=1\Longrightarrow \frac{YC}{p+1}=\frac{YP}{mp}=\frac{CP}{mp+p+1}\Longrightarrow$
$[AYP]=\frac{[AYP]}{[ACP]}\cdot \frac{[APC]}{[ABC]}=\frac{PY}{PC}\cdot \frac{AP}{AB}=\frac{mp^2}{mp+p+1}$ a.s.o. $\Longrightarrow$ (final results)

$\blacksquare\ 1^{\circ}.\ [XYZ]=\frac{(mnp-1)^2}{(mn+m+1)(np+n+1)(pm+p+1)}.$
$\blacksquare\ 2^{\circ}.\ \frac{YZ}{AM}=\frac{(m+1)(1-mnp)}{(mn+m+1)(pm+p+1)}$ a.s.o. [u]Particular case[/u] $(mnp\ne 1)\ :$
$\frac{YZ}{AM}=\frac{ZX}{BN}=\frac{XY}{CP}\Longleftrightarrow m(n+1)=n(p+1)=p(m+1).$
[u]Remark.[/u] If $m=n=p=\frac 12\Longrightarrow [XYZ]=\frac 17.$

I knew that the formula of the area for the triangle $XYZ$ is clasically.
But I wished to offer a more clear solution of this.

Whoa pretty crazy! Essentially the same question in my mock AIME:

problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65275
solution
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67284

Hey this was also on the MathPath qualifying quiz 2005...

I think the problem can be solved in two main passes:

- For a equilateral triangle, with side = 1
- And then, transform it to any triangle, using transformation that mantains the ratio inter areas.

First, let ABC a equilateral triangle.
Let D on AC, such 3 AD = AC
Then triangle BAD has an area that is 1/3 of the total area.
Let the line BD. It can be proved that the length of BD is
Let E on CB such 3 CE = CB. Let P the point of intersection of BD and AE. P is one of the vertices of the inner triangle. Then, triangle APD is similar to BAD (two angles are equals). The side BD has length sqr(7)/3. The corresponding side on APE, is AD with length 1/3. Then, since area BAD = 1/3 of the original, so area APE = 1/7 BAD.

It's clear that the inner triangle is the sum of the three triangles similar to APE. Then its area is 1/7 of the original.

The second part of the proof (generize from equilateral triangle to any triangle), can be obtained, noting/proving that "streching" a triagle with some transformations, alter the areas, but not its ratio.

Angel "Java" Lopez

Well, I strongly disagree with using this problem in any contest. This is too old. I myself have seen this problem more than 5 times in the High School forum after I joined the Mathlinks....

I am in accord with the your opinion. The "moderator" must move to other section same problems.

I saw a lovely proof of the general case in terms of the three ratios that gave a very elegant proof of Ceva's as a special case... I believe the proof was done with mass points, but I did not understand very clearly.  Can anyone provide a general proof in terms of the three side ratios $1 : \alpha, 1 : \beta, 1 : \gamma$?  That would be more interesting.

[quote="t0rajir0u"]I saw a lovely proof of the general case in terms of the three ratios that gave a very elegant proof of Ceva's as a special case... [/quote]
Click my link at post 3.


A state requires that all boat licenses have the letter A or M followed by any five digits. What is the number of combinations of letters and numbers available for boat licenses?

2 letters, and 10^5 number combos.

so 200,000


calculus
integration
calculus computations

Find the integral of (3/2x^3)= ???+C
Also, can you please show your work?
Thanks a lot.

$ \int\frac{3}{2x^3}dx\equal{}\frac{3}{2}\int x^{\minus{}3} dx\equal{}\frac{3}{2}\frac{x^{\minus{}2}}{\minus{}2}\plus{}C\equal{}\minus{}\frac{3}{4x^2}\plus{}C$

If you meant $ \frac{3}{2}\int x^3dx$, then we have $ \frac{3x^4}{8}\plus{}C$.


logarithms
real analysis
real analysis unsolved

Is this a test of convergence for the series $\sum{a_n}$: for each $\epsilon>0$ there is a sequence $b_n$ with $|1-b_n|<\epsilon$ for all $n$ and $\sum{a_nb_n}$ converges?

Take $a_{2n}=\frac 1{\ln n}$, and $a_{2n+1}=\left(\frac 1n-1\right)\frac 1{\ln n}$. For each $\varepsilon>0$ take, for sufficiently large $n$ (greater than some $n(\varepsilon)$) $b_{2n}$ to be $1-\frac 1n$ and $b_{2n+1}$ to be $1$.

I hope it works.