Sequence [url=http://www.research.att.com/~njas/sequences/A112956]A112956[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] counts the number of ways the set of the first $ n$ integers can be partitioned so that all pieces of the partition have the same sum.  The sequence in the OEIS contains terms for $ n \leq 18$ -- can you improve on this?  (In particular, running over all set partitions seems unrealistic for $ n$ this large.)

Let a(s,n) be the number of ways to reach sum s using the first n numbers.
a(s,n) = a(s-n, n-1) + a(s,n-1)
The result is a(S,N), where S = n(n+1)/2 and N is the given integer.

I'm not sure what problem you're solving, but I don't think it's the one I've asked :).  Set partition means we write the set of integers from 1 to $ n$ as a disjoint union of some smaller sets, so each integer between 1 and $ n$ belongs to exactly one of the smaller sets.  In this case, we require that each of these smaller sets have the same sum.  So, for example, for $ n \equal{} 7$ we can take the six partitions $ 1234567$, $ 167 \minus{} 2345$, $ 1346 \minus{} 257$, $ 1256 \minus{} 347$, $ 1247 \minus{} 356$ and $ 16 \minus{} 25 \minus{} 34 \minus{} 7$ and no others (where, for example, $ 1247 \minus{} 356$ means the partition with parts $ \{1,2,4,7\}$ and $ \{3,5,6\}$, each having sum 14).

Well, I opted for a reasonably brute force memoization; ramping up the java memory to 1024m *just* squeaked in the value for 31 before it entirely crashed my computer :)

[code]1 0
2 0
3 1
4 1
5 1
6 1
7 5
8 11
9 10
10 1
11 79
12 165
13 1
14 664
15 2917
16 3308
17 9295
18 23729
19 31874
20 301029
21 422896
22 1
23 13716866
24 71504979
25 100664384
26 54148590
27 880696661
28 498017758
29 27450476786
30 111911522818
31 179459955553
[/code]

[hide][code]import java.util.*;

public class A112956 {
	public static void main(String[] args) {
		for (int i=1;;i++) {
			solve(i);
		}
	}
	static Hashtable<Sequence,Long> table;
	static int eachgroup;
	static void solve(int N) {
		int tot = N*(N+1)/2;
		long answer = 0;
		for (int groups = (N+1)/2; groups>=2; groups--) {
			if (tot%groups!=0) continue;
			table = new Hashtable<Sequence,Long>();
			table.put(new Sequence(new int[groups],0),1L);
			eachgroup = tot/groups;
			int[] start = new int[groups];
			for (int i=0; i<groups; i++) {
				start[i]=eachgroup;
			}
			Sequence init = new Sequence(start,0);
			long here = solve(init,N);
			answer+=here;
		}
		System.out.println(N+" "+answer);
	}
	static long solve(Sequence s, int upto) {
		Long hash = table.get(s);
		if (hash!=null) return hash;
		hash=0L;
		for (int i=0; i<s.seq.length; i++) {
			if (s.seq[i]<upto) continue;
			s.seq[i]-=upto;
			Sequence next = new Sequence(s.seq,i);
			s.seq[i]+=upto;
			hash += solve(next,upto-1);
			if (s.seq[i]==eachgroup) break;
		}
		table.put(s,hash);
		return hash;
	}
} class Sequence {
	int[] seq;
	public Sequence(int[] s, int changed) {
		seq = (int[])s.clone();
		while (changed>0 && seq[changed]<seq[changed-1]) {
			int tmp = seq[changed];
			seq[changed]=seq[changed-1];
			seq[changed-1]=tmp;
			changed--;
		}
	}
	public boolean equals(Object o) {
		Sequence s = (Sequence)o;
		for (int i=0; i<seq.length; i++) if (seq[i]!=s.seq[i]) return false;
		return true;
	}
	public int hashCode() {
		// not particularly clever, just enough for small cases
		int ret = 0;
		for (int i=seq.length-1; i>=0; i--) ret=1000*ret+seq[i];
		return ret;
	}
}[/code][/hide]

[quote="JBL"]I'm not sure what problem you're solving, but I don't think it's the one I've asked :).[/quote]

This was actually a solution to split the original set into two sets.


It was on MOEM Maraton, and I couldn't solve it.

In a group of 5 friends, what is the probability that any one of them shares a birthday with another?
Note: Ignore flip years to make it less complicated.

Would it be $ \frac1{365\cdot365}$?

I need help..

P:S I saw this problem somewhere,but can't recall where..

This kind of problem is a good example of the principle of [i]complementary counting.[/i]  In this case, rather than calculating the probability that any of them share a birthday, calculate the probability that [b]none[/b] of them share a birthday.  :)

eh......
in a way 1/365x365 makes sense
but im thinking that there's gotta be more than that....
but if i was on pressure to solve that real fast,
id go with 1/365x365

im gonna think on this a bit :huh:

[quote="t0rajir0u"]You haven't used the fact that there are $ 5$ of them.

DonkeyKong:  Offhand, I'm not sure why you posted this question in a local forum.  Most people don't frequent the local forums, and you're more likely to get a response posting in the MOEMS board.

Try solving the problem for $ 2$ friends first.  It might help you gain some insight.  That, or you can follow the first hint :)[/quote]
Like I typed in the previous one, it's already on the MOEM section.
It's on MOEM Marathon, and that question hasn't been answered for quiet a bit..(like 5 days or so..)
If you know the answer and solutions, can you post it there?

The probability that two people don't have the same birthday is $ \frac{364}{365}$.  This is because the first person's birthday doesn't matter; it can be on any day, but the second person has $ 364$ days that his birthday can be on that are different and $ 1$ that are the same.

In a similar vein, the probability that five people don't have the same birthday is $ \frac{364 \cdot 363 \cdot 362 \cdot 361}{365^4}$.  Again, the first person's birthday doesn't matter; there are $ 364$ different days for the second person; $ 363$ for the third; and so forth.  The answer is

$ 1 \minus{} \frac{364 \cdot 363 \cdot 362 \cdot 361}{365^4}$

[quote]You multiply by every person's respective probabilities.
[/quote]

could you expound on that a bit more?

im still confused

[quote="spiders_hate_chestnut"]i kinda get it

i get why the probability of the two people not having the same birthday is 364/365,
but i dont get the part 364x363x362x361/365^4
why is the numberater 364x363x362x361??[/quote]
Because they have 5 people.

Person1(Control) can be birthdays in any date, so the probabillity is $ \ 1$
Person2, we are trying to avoid from this person having the same birthday with person3, and it can be any date except that particular date.So the probabillity is $ \dfrac{364}{365}$
Person3, is trying to avoid from having the same birthday with person1 and 2.
So the probabillity is $ \dfrac{363}{365}$

Person4, the probabillity would be $ \dfrac{362}{365}$

Person5, the probabillity would be $ \dfrac{361}{365}$

Now those combined($ \dfrac{364\cdot363\cdot362\cdot361}{365\cdot365\cdot365\cdot365(365^4)}$ are the probabillity that they will NOT have the same birthday.
But the question asks you what is the probabillity that they will share the same birthday, so $ 1\minus{}\dfrac{364\cdot363\cdot362\cdot361}{365\cdot365\cdot365\cdot365(365^4)}$.

Get it?

When you typed $ \frac{364\cdot363\cdot362\cdot361}{365\cdot365\cdot365\cdot365(365^4)}$ some peopl e will interpret it has $ \frac{364\cdot363\cdot362\cdot361}{365^{8}}$. So chose one or the other.

[quote="hunter34"]When you typed $ \frac {364\cdot363\cdot362\cdot361}{365\cdot365\cdot365\cdot365(365^4)}$ some peopl e will interpret it has $ \frac {364\cdot363\cdot362\cdot361}{365^{8}}$. So chose one or the other.[/quote]

Oh.. Yeah.. if that's the case, I should probably edit it, but there was already a solution there and Spider_hate_Chestnut got confused..So I tried to explain it.. :)


Let a>1 be a real number.  If S is the set of real numbers x that are solutions to the equation $a^{2log_{2}x}=5+4x^{log_{2}a}$, then...

S is the empty set
S contains infinitely many real numbers
S contains exactly two real numbers
S contains more than two, but finitely many, real numbers
S contains exactly one real number.

Just a hint: $a^{\log_2{x^2}}=x^{2\frac{ln a}{ln 2}}=(x^{log_2{a}})^2$. Now if you substitute $u=x^{log_2{a}}$ you get the quadratic equation

$u^2-4u-5=0$.

I think you know how it goes on :lol:

Misha

So is the answer only 1 in the set?

Btw...
How did you get that conversion? I'm not seeing it!

Thanks soooo much

$a^{2\log_2{x}}=a^{2\log_a{x}\log_2{a}}=a^{\log_a{x}}^{2\log_2{a}}=x^{2\log_2{a}}=(x^{\log_2{a}})^2$.

These are all the steps of the convertion: the first is the base transformation rule and the others are just rules for exponents.

Sorry for bad explanation in the first post :blush: .

Misha


analytic geometry
algebra
polynomial
geometry
angle bisector
Galois Theory
geometry unsolved

On a given circle C construct point P so that tangent at P is an angle

   bisector of APB,  where A and B are two given points outside C.

   Thank you.

    M.T.

[quote="armpist"]On a given circle C construct point P so that tangent at P is an angle

   bisector of APB,  where A and B are two given points outside C.[/quote]

This cannot be solved by compass and ruler.

As an example, I consider the case when the circle C has equation $x^{2}+y^{2}=9$ in Cartesian coordinates, and the points A and B have coordinates A (-1, 0) and B (0, 2). These two points A and B lie inside the circle C, but this doesn't really matter as the problem will be the same for interior and exterior points.

The x-coordinate of the required point P is then a root of the polynomial $16X^{4}+144X^{3}+261X^{2}-648X-405$. This polynomial has Galois group $S_{4}$, and its roots are therefore not constructible using compass and ruler.

  darij


In 2 triangles $ ABC$ and $ A_1B_1C_1$ $ AB\equal{}A_1B_1$ the angle $ A\equal{}A_1$ and $ BC\plus{}AC\equal{}A_1C_\plus{}B_1C_1$. Show that the 2 triangles are concurente with each other.

We can use the cosin theorem to prove this problem.

[quote="Elio (n)"]Show that the 2 triangles are [b]concurente[/b] with each other.[/quote]In proper English, it is "Show that the 2 triangles are [b]congruent[/b] with each other."

@quangchuc: "Geometry first, trigonometry second" ;)

[hide="Partial Solution (so other AoPSers can try it themselves)"]Extend side $ AC$ through point $ C$ so that $ BC \equal{} B'C$. Similarly, extend side $ A_1C_1$ through point $ C_1$ so that $ B_1C_1 \equal{} B_1'C_1$.
[u][i]Remark[/i][/u]: Why is it possible to do such extended sides for all triangles? <--For some people to ponder :)
Then, $ BC \plus{} AC \equal{} B'C \plus{} AC \equal{} AB'$ and $ B_1C_1 \plus{} A_1C_1 \equal{} B_1'C_1 \plus{} A_1C_1 \equal{} A_1B_1'$

Step-exercise#1: Prove that $ \triangle AB'C\cong\triangle A_1B_1'C_1$
Step-exercise#2: Prove that $ \angle ACB \equal{} \angle A_1C_1B_1$
Step-exercise#3: Finally show that the two triangles are congruent.
[/hide]

My English is not very good.
However, I will try to post the solution:
 Draw angle $ \large\widehat{xOy}\equal{}\large\widehat{C}\equal{}\large\widehat{C'}$.
Let $ A, A_1\in Ox$  and  $ B, B'\in Oy$ satisfy the condition. Let AC=b, BC=a, $ A_1B_1\equal{}a_1, B_1C_1\equal{}b_1$.
Because c=c', $ a^2\plus{}b^2\minus{}2abcosC\equal{}a'^2\plus{}b'^2\minus{}2a'b'cosC'$. Hence ab=a'b' (because a+b=a'+b' and C=C'), and because a+b=a'+b'. Hence {a;b}={a';b'}. So we had proved the problem.


$I$ is the incentre of $\triangle ABC$. $3$ points $A_{1}, B_{1}, C_{1}$ are chosen on $AI, BI, CI$ respectively. The perpendicular bisectors of $AA_{1}, BB_{2}, CC_{2}$ intersect to form $\triangle A_{2}B_{2}C_{2}$. Show that if the incentre of $\triangle ABC$ and circumcentre of $\triangle A_{2}B_{2}C_{2}$ coincide then this point will also be the orthocentre of $\triangle A_{1}B_{1}C_{1}$.

[quote="Andreas"]I think there are some typos on the text. Try to read it again!  :)[/quote]

I don't think so. Where are you referring to?

[quote="lotrgreengrapes7926"][quote="Andreas"][quote="Franz Joseph"] The perpendicular bisectors of $AA_{1}, BB_{2}, CC_{2}$ intersect to form $\triangle A_{2}B_{2}C_{2}$.[/quote]

How are defined $A_{2}, B_{2}, C_{2}$?[/quote]

It's obviously "The perpendicular bisectors of $AA_{1}, BB_{1}, CC_{1}$ intersect to form $\triangle A_{2}B_{2}C_{2}$."[/quote]

It doesn't really matter where about $A_{2}, B_{2}, C_{2}$ are on the bigger triangle.

If I haven't misunderstood the problem statement I think that it's true only if $A_{1}B_{1}C_{1}$ is homethetic w.r.t. $ABC$.  :maybe:


solve for x:

2^x + 4x - 2 = 0

how do you do this?

You can't get an exact answer unless there's something simple (for example, if it's an integer). I don't see it being anything simple.

But the only answer he'll be able to get is a numerical approximation -- there ain't no nice form for solutions to that kind of problem, unless it's conveniently tangent or something, or has integral solutions as Simon said.  (You might show for fun that there can't be any non-integral rational solutions to that equation.  Not very hard at all.)

Oh.. ok. It was a question of my mom's friend taking a log class... so does it have anything to do with log???

but then... at first she told my mom that 2^x + 4^x - 2 = 0 which is easy to solve... thanks for your input

Well, you can re-write it as a log equation, too.  But that just gets you to a different-looking unsolvable equation.

So why does that equation have no solutions?

-interesting_move

not no solutions (although it might have no solutions) -- just that there isn't any way to find what the solutions are except numerical approximation.  There's no way to solve the question he gave.


inequalities
algebra
polynomial
real analysis
real analysis theorems

My understanding of Archimedean property is that
if x > 0 and if y is any arbitrary real number, there exists a positive integer n such that nx > y.

But then, I was confused with wikipedia's definition that involves order field from abstract algebra.

Here are the questions -
It is probably my lack of understanding of Group/Ring/Field, but then why is abstract algebra needed to explain this property?
It also mentions completeness, what is completeness? How is archimedean property relates to completeness?
I recalled reading somewhere that "real number is only conditionally complete". Unclear what is the condition. Anyone knows the condition?

I can follows from the above definition, that in real number system, there is no nontrivial infinitesimals. However, I do not understand why an infinitesimal is non-archimedean?

I believe I must have some conceptual issues here and I do want to get this all clear up before moving on. Please help me to piece this together.

Thanks,

Taking on some of your points:

- Why an ordered field?
Because we need that much structure for the definition to make sense at all. We have arithmetic, which leads us to a field, and we have inequalities, which imply an order on the set. An ordered ring makes sense as well, but it doesn't really add anything interesting.

- Why does completeness come up?
Because the axioms of an ordered field don't imply the Archimedean property. Those axioms together with completeness do. I've posted proofs of that a few times, most recently [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1588986#1588986]here[/url].

- Infinitesimals non-archimedean?
There are non-archimedean ordered fields. Every ordered field contains a copy of the rationals. In the non-archimedean case, there are positive numbers which are less than every rational; we call those infinitesimal. It's just a name and a definition.

Thank you jmerry.

Your answers are helpful.
I've been digging a few hours on this "order field" stuffs and I've read your suggested thread, but some of discussions on formal proof confused me. But I do understand your least upper bound property

Specific questions - is completeness property the same as the least upper bound property?

Does it simply mean that all non-empty set which is bounded from above has a supremum?

But, going back to my original post on real number, neither x and y are bounded, and there is no supremum involved, how is this property affect the Archimedean property? That is, how does it affect finding n so that nx>y?

Thanks again.

[quote="MathDaoist"] but some of discussions on formal proof confused me.[/quote]

That's because that was one of τρικλινο's threads - you have to ignore him. Pay attention just to what jmerry wrote there.

[quote]Specific questions - is completeness property the same as the least upper bound property?

Does it simply mean that all non-empty set which is bounded from above has a supremum?[/quote]

Yes. The simplest way to state completeness is to state that every nonempty set which is bounded above has a least upper bound. The word "supremum" is synonymous with "least upper bound," and this completeness property is often informally referred to as the least upper bound property.

[quote]But, going back to my original post on real number, neither x and y are bounded, and there is no supremum involved, how is this property affect the Archimedean property? That is, how does it affect finding n so that nx>y?[/quote]

Let $ M\equal{}\{n\in\mathbb{N}: nx\le y\}.$ (We're assuming here that $ x$ and $ y$ are positive real numbers.)

$ M$ is bounded above, because it is bounded by the real number $ \frac yx.$ $ M$ might be empty, but if it's empty, we have what we want: a natural number $ n$ (which can now be any natural number at all) such that $ nx>y.$ Otherwise, it's nonempty and hence has a least upper bound. Let $ w$ be the least upper bound of $ M.$ This means that $ w\minus{}1,$ being less than $ w,$ is not an upper bound of $ M.$ Hence there exists $ n_0\in M$ such that $ n_0>w\minus{}1.$ Now let $ n\equal{}n_0\plus{}1.$ Since $ n>w$ and $ w$ is an upper bound of $ M,$ it follows that $ n\ne M,$ meaning that $ nx>y.$

=====

The simplest construction I know of a nonarchimedean ordered field looks like this. Take the set of all polynomials with real coefficients. That's a ring. (Think, by analogy, of the ring of integers.) Form the set of quotients of polynomials, with the understanding that $ \frac{p(x)}{q(x)}$ will be considered "the same" as $ \frac{r(x)}{s(x)}$ provided $ q(x)$ and $ s(x)$ are not the zero polynomial and $ p(x)s(x)$ is the same polynomial as $ q(x)r(x).$ (More formally, I'm defining an equivalence relation, and I'm constructing the field of quotients. The construction is completely analogous to constructing the rational numbers from the integers.)

We can show that this is a field - the field of rational functions. (Note that the polynomials are included, just as the integers are included in the rational numbers.) Now we define an order: if $ q(x)$ is not the zero polynomial, we can always choose to write $ \frac{p(x)}{q(x)}$ using a $ q(x)$ whose highest order term has a positive coefficient. We then define $ \frac{p(x)}{q(x)}$ to be positive if and only if $ p(x)$ has a highest order term with a positive coefficient.

There's some work to do, but we can show that this creates an ordered field - all of those axioms are satisfied. Some tidbits from the order: $ x^2\minus{}100>100x\plus{}1.$ $ x^2\plus{}2x>x^2\plus{}10.$ $ x^2\plus{}2x>x^2\plus{}x.$ $ 1>\frac1x.$ And so on.

Now, consider the polynomials $ 1$ and $ x.$ Is there an $ n\in\mathbb{N}$ such that $ n\cdot 1>x?$ No, there isn't, no matter how big $ n$ is. Hence the archimedean property does not hold for this ordered field; it's non-archimedean. And it's not complete, either. The set $ \{1,2,3,\dots\}$ is bounded above, by $ x,$ but has no least upper bound.

wow! with detail examples and explanation. Thank you Kent! and jmerry for taking the time.

A bit of digesting; well, I believe I now have a much better understanding of this. A bit fuzzy; I suspect that I am just not comfortable with this yet. All can be cure with time, more reading and work problem, I am sure :-)

Thank you all


inequalities
geometry
analytic geometry
algebra
binomial theorem
AMC

The expression
\[ (x \plus{} y \plus{} z)^{2006} \plus{} (x \minus{} y \minus{} z)^{2006}
\]is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

$ \textbf{(A) } 6018 \qquad \textbf{(B) } 671,676 \qquad \textbf{(C) } 1,007,514 \qquad \textbf{(D) } 1,008,016 \qquad \textbf{(E) } 2,015,028$

[hide]Using the binomial theorem, we get

\begin{eqnarray*} (x+(y+z))^{2006}+(x-(y+z))^{2006} &=& \sum_{k=0}^{2006}\binom{2006}{k}x^k(y+z)^{2006-k}+\sum_{k=0}^{2006}(-1)^k\binom{2006}{k}x^k(y+z)^{2006-k}\\ &=& 2\left(\binom{2006}{0}(y+z)^{2006}+\binom{2006}{2}x^2(y+z)^{2004}+\cdots+\binom{2006}{2006}x^{2006}\right) \end{eqnarray*}

So we now have the sum of a bunch of binomial expansions.  However, we notice that each is multiplied by a different power of $x$.  Therefore it suffices to find how many terms are in $(y+z)^{2006}, (y+z)^{2004},\ldots, (y+z)^0$ and then add them up.  So consider $(y+z)^{2006}$.  Each term's total degree is 2006 and the individual degree of $y$ is between 0 and 2006.  Each of these 2007 values corresponds to exactly one term in the expansion.  So there are 2007 terms in $(y+z)^{2006}$.  Likewise, for $(y+z)^i$ there are $i+1$ terms in its expansion.  So the answer is \[ 2007+2005+\cdots+1 = 1004^2 =\boxed{1,008,016}. \][/hide]

[quote="joml88"]chess, there's a typo.  It should be $(x-y-z)^{2006}$.[/quote]

Oops, :blush: fixed.

[quote="Hikaru79"]There's a much easier way to do it. If you just do the expansion for smaller terms, you'll notice that that the number of terms (when there is an even power) is always a perfect square. The only perfect square among the choices is $D)$[/quote]

Yeah, but we here at AoPS like mathematical rigor :P  That's a good observation though.  I did a little bit of experimenting with smaller cases as well during the test but ultimately did the equivalent of my solution...probably a little bit of wasted time there..o well.

in general for any $n=2k$
there will be $(k+1)^2$ terms left...

[quote="Hikaru79"]There's a much easier way to do it. If you just do the expansion for smaller terms, you'll notice that that the number of terms (when there is an even power) is always a perfect square. The only perfect square among the choices is $D)$[/quote]

Yeah, I just noticed what amirhtlusa said, $1004^2\implies \boxed {D}$.

[quote="Hikaru79"]There's a much easier way to do it. If you just do the expansion for smaller terms, you'll notice that that the number of terms (when there is an even power) is always a perfect square. The only perfect square among the choices is $D)$[/quote]

That's not really a solution, if you know what I mean. That's more of a lucky observation, but for the purposes of the AMC it works fine I guess.

Go go TI-89! :lol:

[hide="Answer"]By using low powers, we find the pattern $1,4,4,9,9,16...$. Therefore, the answer is $1004^2=1,008,016\Rightarrow \boxed{D}$.[/hide]

[hide]You will notice if you factor it to the originally incorrectly typed but still equivalent form of (x+(y+z))^2006 + (x-(y+z))^2006 that if x is raised to an odd power then (y+z) is also raised to an odd power, and the (y+z) in the left-hand expansion will cancel with the -(y+z) in the right-hand expansion. Therefore we find all even powers of x and how many possibilities there are for y and z. Well, if x is raised to the a, then (y+z) is raised to the 2006-a, and there will be 2006-a+1 terms in the expansion of (y+z)^(2006-a). So we just find the sum from i = 0 to 2006 in steps of 2 of 2007-i, and if we work this out (by any number of methods, including by hand if you are me and don't have a TI-89), you get the correct answer.[/hide]

[quote="Hikaru79"]There's a much easier way to do it. If you just do the expansion for smaller terms, you'll notice that that the number of terms (when there is an even power) is always a perfect square. The only perfect square among the choices is $D)$[/quote]

There's certainly nothing wrong with making a highly educated guess on a problem, which is essentially what you're doing when you don't verify the validity of the pattern you found.  While joml88 pointed at that AoPSers have a certain appreciation for mathematical rigor, I wanted to point out the pragmatic side of digging in to the complete "proof" of the solution.  When you fully understand the solution to a problem at one level, you can dig into more problems at the next level.

I found another very different approach to this solution: let's expand the first trinomial.  We'll end up with terms of the form $x^a y^b z^c$, where $a + b + c = 2006$.  The second trinomial will end up with similar terms, except terms where $b + c$ is odd will be negative and cancel out terms from the first trinomial - i.e. terms where $a$ is odd will cancel out.  So, we need to find all the ways to select distinct non-negative solutions $(a,b,c)$ to $a + b + c = 2006$ with the restriction that $a$ is even.  So, we have

$N = \sum^{1003}_{k=0} \binom{2k + 1}{1} = \sum^{1003}_{k=0} 2k + 1 = 1008016$.

We got this equation by running through all even $a$ so that we're summing the individual solutions to $b + c = 2006, b+ c = 2004, ..., b + c = 0$.  In fact, this method can be generalized to any even power of the sum of trinomials by replacing $1003$ with $n/2$, so we have

$N = \sum^{n/2}_{k=0} 2k + 1 = \frac{(n+2)^2}{4}$

so, if we let $n = 2p$, as some in this thread have done, we have

$N = (p + 1)^2$

We can treat the odd form as easily by noting that in this case we must have $a$ odd, meaning we run through $b + c = n - 1, b + c = n - 3,..., b + c = 0$.  This leads to the sum

$N = \sum^{(n-1)/2}_{k=0} \binom{2k + 1}{1} = \frac{(n+1)^2}{4}$

or, with $n = 2p - 1$, we have

$N = p^2$

QED  :lol:

Ah I shouldn't have omitted this one...it's a cool problem.

Way old revive, but I think this solution is different enough to justify it! :)
[hide="Solution"]We can rewrite the second term is -x+y+z, since changing the sign doesn't matter because we are taking it to an even power. Now we have the expression $ (x\plus{}y\plus{}z)^{2006}\plus{}(\minus{}x\plus{}y\plus{}z)^{2006}$. Each combination of nonnegative integer $ x,y,z$ such that $ x\plus{}y\plus{}z\equal{}2006$ will produce a corresponding term; however, when $ x$ is odd, then the corresponding terms will cancel out. Therefore, we are looking for the number of solutions to $ x\plus{}y\plus{}z\equal{}2006$, where $ x\equal{}2t$ is even.

There are several ways to count this, but my way involves using [url=http://www.artofproblemsolving.com/Wiki/index.php/Pick%27s_theorem]Pick's Theorem[/url]:

Any choice of $ x,y$ will uniquely determine what $ z$ is; therefore, we need only consider the cases where $ x\plus{}y \leq 2006$. We also use the fact that $ x$ is even and arrive at the inequality \[ 2t \plus{} y \leq 2006,\] where $ t,y$ are nonnegative integers.
Representing this on a ty-axis, we have the area bounded by $ t \geq 0, y \geq 0, 2t\plus{}y \leq 2006$. Any point on the boundary or interior of this will count as a solution. We can easily casework to find the amount of boundary points, and by seeing that our region is a nice right triangle with lattice point coordinates , we realize that we can use Pick's to count all the interior points.

We have 1004 points (0 through 1003) on the t-axis, 2007 points on the y-axis, and 1004 points on the line $ 2t\plus{}y\equal{}2006$. ($ t\equal{}0,1,\ldots,1003$) We have an overcount of the 3 vertices, so subtracting 1 from each side, we get $ 1003\plus{}2006\plus{}1003\equal{}2\cdot2006$ boundary points.
We clearly have a triangle with base 1003 and height 2006, so it's area is $ 1003^2$.
Now, using Pick's, we have $ 1003^2\equal{}I\plus{}2006\minus{}1 \iff 1003^2\minus{}2\cdot1003\plus{}1\equal{}I\equal{}(1003\minus{}1)^2\equal{}1002^2$.

The total amount of solutions is $ I\plus{}B\equal{}1002^2\plus{}2\cdot1003\equal{}1003004\plus{}5012\equal{}\boxed{1008016}$.

Not as good as joml88's solution, in my opinion, and kinda long (although most of the length is due to explaining the procedure more rigorously), but still worth my time to type it. :P[/hide]

just like Brut3Forc3, this is a really really really old revive, but take a look!

[hide="my solution"]

Use the distribution of number of solutions by noting the sum of the exponents have to be 2006. I don't have time to explain too much now, but essentially...

Count the number of solutions to the sum of the exponents of $ (x \minus{} y \minus{} z)^{2006}$ = $ 2006$ where the exponents of y and z are both odd and are both even. We get $ \binom {1004}{2}$ + $ \binom{1005} {2}$. Now we take the number of solutions, $ \binom{2008}{2}$ - ($ \binom{1004} {2}$ + $ \binom{1005} {2}$) = 1007012. Now we take $ \binom {2008}{2}$ $ \minus{} 1007012 \equal{} 1,008,016$.
[/hide]


calculus
inequalities
limit
real analysis
algebra
binomial theorem

Hello Guys,

Here it is a nice problem that is very easy using "Super Bombs", that is, those big theorems that you will get to know after one or two years in Math Contests, but without them is solvable, even though, it could be a little more trickly.

Suppose that $a\leq b<c$ are the sides of the non degenerated $\triangle ABC$. Prove that there is some $n\in\mathbb{N}$ one has that \[a^n + b^n < c^n\]. Just in case, a triangle is said not degenerated if, with the notation of this case, $c < a+b$.

Enjoy it. Best,

Am I missing something?..

Because if  $a<c$ and $b <c$  there exist an $n$ such that  $a^n + b^n <(\frac {c^n}{100000000000})$  :D:D

[quote="blahblahblah"]Um, by big theorems do you just mean that
[hide]
$\forall \epsilon>0 \text{ and } x\ni \vert x\vert<1, \exists n\in\mathbb{N}\ni \vert x^n\vert<\epsilon$?[/hide][/quote]

Actually But this property can help to an easy solution

[hide]
\[\forall \epsilon>0, \mu>0 \exists N\in\mathbb{N}\mbox{ such that }N\epsilon>\mu\]
Known as archimidean property.
Of course, this can help too
\[n\in\mathbb{N}, n>1, x\in\mathbb{R}, x>-1, x\neq0\Rightarrow (1+x)^n > 1 + nx\]
[/hide]

Also, using calculus it is pretty easy to proof.

By the way,
\[\displaystyle{ a^n + b^n <\left(\frac {c^n}{100000000000}\right)}\]
Is also correct for some value of $n$.

As I remember, Bernoulli's inequality (the second property you posted) is the way that we usually rigorously prove that if $\vert x\vert<1$, then $\lim_{n\to\infty} x^n=0$.  I'll sketch the proof so some people who haven't had exposure to analysis type proofs can see the basic ideas instead of taking them for granted.

This method is pretty nice - instead of considering the size of $x^n$, we consider $\frac {1}{y^n}$, where $y>1$ so that we can write $\frac{1}{(1+z)^n}$ for some $z>0$.  Bernoulli's inequality yields

$\frac{1}{(1+z)^n}\leq\frac{1}{1+nz}\leq\frac{1}{nz}$.

Now, for any $z$, we can pick an $n$ so large that $\frac{1}{nz}$ is really small, and then we're done (I omitted the $\epsilon$s in the interests of readability.)

(There are other ways to do this; among the analysis books I own, in Rudin and Stoll, they use the binomial theorem in a nifty way)


calculus
integration
logarithms
calculus computations

find
$ \int_0^a\frac{\arctan x}{x^2\plus{}ax\plus{}2}\,dx$
where $ a>0$.


linear algebra
matrix
function
linear algebra unsolved

$A$ is a commutative ring and $f: A^n\to A^n$ is a $A$-linear function with matrix $M \in M_n(A)$.
1/ show that $f$ is injective iff $det(M) \in A$ is not a divisor of zero.
2/ show that $f$ is surjective iff $det(M) \in A$ is invertible.

I noticed something strange about your statement, as any element in A can be made a determinant, it would imply that being invertible is equivalent in a commutative ring to not being zero divisor, for which $\mathbb{Z}$ is a counterexample

take this case $A=\mathbb{Z}$ and $n=1$
let $f$ send $a$ to $2 a$
the determinant is two
the map is injective
however 2 is not invertible

Or am i missing something?

you are perfectly right  :blush:  I've edited ..

I think this theorem will help you now

I will do part of  b

let $Adj(M)$ denote the adjoint matrix

then $M Adj(M)= det(M) I$

if $det(M)$ is invertible with inverse $z$

then $M (Adj(M)z)=I$ and thus f is surjective

need more think time now


Let $ A\equal{}\{ 1 ,2 , \dots , n  \}$ . How many solutions $ \{ X , Y , Z \}$ with $ |Z| > |Y| > |X| \geq 2$ has the equation $ X \cup Y \cup Z \equal{} A$ .

[b]Solution[/b]

Let $ (X,Y,Z)$ be a triple such that
(i) None of them is empty.
(ii) $ |X| < |Y| < |Z|$
(iii) $ X\cup Y\cup Z \equal{} \{1,\dots,n\}$
Lets take care of the other condition of the problem later.

Consider the sets $ X,Y \minus{} X$ and $ Z \minus{} (X\cup Y)$.
(i) They are mutually disjoint.
(ii) Their union is $ \{1,\dots,n\}$.
(iii) Only $ Z \minus{} (X\cup Y)$ can be empty, $ X$ and $ Y \minus{} X$ are always non empty.

$ \Rightarrow$ We have a bijection between number of ordered triples $ (X,Y,Z)$ and the number of arrangement of { partitions of $ \{1,\dots,n\}$ into either 2 or 3 sets except when $ |X|\equal{}n\minus{}1$}. Hence we have $ 3!.S(n,3) \plus{} 2!.S(n,2)\minus{}n$ where $ S(n,k)$ is stirling's number of second kind.

Taking care of the restriction( looks artificial, does not seem to make sense in combinatorial perspective) : $ |X|\geq 2$, we have $ 3!.S(n,3) \plus{} 2!.S(n,2) \minus{}n \minus{} {n\choose 1}(1 \plus{} 2!.S(n \minus{} 1,2))$

Unfortunately, I don't think your edit necessarily fixes things.

More generally , $ Y \minus{} X$ ever doesn't "see" the parts of $ Y$ which are contained in $ X$, and $ Z \minus{} (X \cup Y)$ never sees the parts of $ Z$ contained in $ X$ and $ Y$.

So whenever $ Y$ is somewhat larger than $ X$ and somewhat smaller than $ Z$, we can take elements of $ X$ and pull them in and out of $ Y$, without ever having any effect on $ X$, $ Y \minus{} X$, or $ Z \minus{} (X \cup Y)$.

For example, if $ n \equal{} 10$, how do you distinguish $ \{\{1,2\}, \{3,4,5,6\}, \{5,6,7,8,9,10\}\}$ from $ \{1,2\}, \{1,2,3,4,5,6\}, \{1,2,3,7,8,9,10\}\}$?


advanced fields
advanced fields unsolved

Do you mean a language on the alphabet {a,b,c}  or do $a,b,c$ in the definition refer to arbitrary words?

[quote="student21"]HI everybody,
since some days a mathmatic problem causes my head trouble;) Maybe you can help me!?
I'm trying to improve my skills, but it doesn't work out as fast as I like;)

Would be great if you could help me soon, because I need it for an presentation
Thanks
Anna

So here my Problem:
Does anyone here has an idea how I can solve this question and post it!?[/quote]

For a, try to find a pushdown automaton wich recognizes that language by empying the pushdown memory
For b,try to apply the pumping lemma-it's really easy at all!

b) Assume, for the sake of a contradiction, that L is regular. Let p be the pumping length and s = c^2p b^p a^m be a string in L. Then, by the pumping lemma, s can be slip into a string xyz, where:
xy^iz is in L
|xy| <= p

Because |xy| <= p, then y must be composed entirely of "c"s. However this means that  the string xy^2z is not in L since it has more "c"s than twice the number of "b"s.
Hence L is not regular.

Hopefully i didn't make a mistake, it's been a while.


1. Let n be an integer greater than 1. Prove that $ 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} \cdots \frac {1}{n}$ is not an integer.

2. Find all pairs of positive integers a and b such that gcd(a,b) = 12 and lcm(a,b) = 360.

:wink: The first one has been posted here a couple times before.

For 2., use the relation $ \gcd(a,b)\text{lcm}(a,b)\equal{}ab$ and the fact that $ 12\le a,b\le360$.

[quote="t0rajir0u"]That's actually not enough.  You need to consider prime factorizations.[/quote]Right, that's why it is a hint...

I'm new so I think Ill try the first one.

isn't it because since all the numbers are all 1/(an integer), the answer itself will never be an Integer, no matter what what integer value of N is.

$ \frac{1}{2} \plus{} \frac{1}{3} \plus{} \frac{1}{6} \equal{} 1$.  You'll need to do better than that :)

[hide="Approach 1"] Bertrand's Postulate. [/hide]
[hide="Approach 2"] Add all the fractions.  What's the greatest power of $ 2$ dividing the numerator?  The denominator? [/hide]

i see, i did not think this one through enough, thanx

Number 2:
Suppose $ a \equal{} 12p, b \equal{} 12q$, $ \gcd(p,q) \equal{} 1$, then $ pq \equal{} \frac {\text{lcm}(a,b)}{\gcd(a,b)} \equal{} 30 \equal{} 2\cdot 3\cdot 5$. Since there are four such (unordered) pairs of $ p,q$: $ (1,30), (2,15), (3,10),(5,6)$, all such (unordered) pairs $ (a,b)$ should be:
$ (12,360), (24,180), (36,120),(60,72)$.


Today is September 15. What month was it 1.75 years ago?

one year ago it was still September.
1 year=12 months
12 months*3/4=9.
If you count back 9 months, you get December.


¿Qué les parece si vamos poniendo acá los equipos para la Olimpíada del Cono Sur?

Equipo Argentino:

Mauro Schilman - Córdoba

Nicolás Ponieman  - Ciudad de Buenos Aires

Pablo Nicolás Zimmermann - Rosario -  Santa Fe

Fernando Martín Vidal - Villa Cañas - Santa Fe

Haber, aki puedo ir adelantando a los chicos de Peru ,
a quienes conosco:

[color=blue]PER1: Daniel Soncco (chen)    (Lima)
PER2: Franco Vargas.             (Lima)
PER3: Freddy Ochoa.             (Lima)
PER4: Maria xxxxxxxx           (Huanuco)

SUP1: Jossy Alva G.               (Lima)
SUP2: XXXXXXXXX[/color]

Haber si alguien completa la lista ,

Gracias
Carlos Bravo ;)

El equipo seleccionado por la Sociedad Matemática Peruana fue:

 [b]Daniel Chen Soncco[/b]
 [b]Franco Vargas [/b]
 [b]Maria Huanuco[/b] (Su apellido es Huanuco... no es que sea de Huanuco)
 [b]Freddy Ochoa[/b]

Suplentes :
Jossy Alva
Juan Vasquez Cahua

______________________________________________________________

$Tipe$

[quote="hucht"][quote="carlosbr"]ah ok,

Gracias por la correccion ....

PD: Haber si alguien mas envia la lista
de sus seleccionados eh??

 Carlos  Bravo ;)
Lima Peru[/quote]

 :D  Yo tengo los seleccionados de Takijistan pero no tengo letritas takijistanies  :?[/quote]

Eso es un problema, ejejejeje

Desde cuando Takijistan va a la cono sur o acaso es un nuevo pais particpante ?


A jar contains 2 nickels, 3 dimes, 2 quarters, 1 half dollar and 4 pennies. If a coin is drawn randomly from the jar, find the expected value of the coin.

[hide]
From [url=http://en.wikipedia.org/wiki/Expected_value#Mathematical_definition]this[/url], the expected value is
\[ \frac{(2\cdot5)\plus{}(3\cdot10)\plus{}(2\cdot25)\plus{}50\plus{}4}{12}\equal{}\boxed{12}
\]
[/hide]


Just wanted to ask because I'm not sure if I'm the only one or not who hasn't gotten their materials.

I haven't. Also, note MOP phone calls are not out yet.


calculus
integration
real analysis
real analysis unsolved

Solve integral for $n \in \mathbf{N}$:

\[ \int \frac{dx}{1+x^{2n}} \]


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

CS-Engel works.
Notice that $ LHS \equal{} \sum \frac{a^2}{2ac\plus{}ab} \geq \frac{(a\plus{}b\plus{}c)^2}{3ab\plus{}3bc\plus{}3ca} \geq 1$

Well there are much nicer proofs,
but I made a good (not short, but beatiful proof) this tech is also used to prove nesbitt, so i used it here and it works!
Sorry if this proof has been posted before :maybe:

let: $P: \sum{\frac{a}{b+2c}}$,   $Q=\sum{\frac{b}{b+2c}}$,  and $R=\sum{\frac{2c}{b+2c}}$

Then trivially, $Q+R=3$ ------------------------------------[1]
And by simple AMGM we get $P+2Q \ge 3$ ------------[2]

                                      and  $4P+R \ge 6$------------[3]

By calculating this: $[2]+2 \circ [3]-2 \circ [1]$
$9P\ge 9$  $\leftrightarrow P \ge 1$

Therefore the Proof is completed  :) i hope it's not only me who likes this solution.

Inequality from JBMO 2003 after substitution is the same as your's.
Check this out:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=20974&sid=a267fe909e1bb8bbee248101cc0ca48a#p20974


geometry
linear algebra
combinatorics solved
combinatorics

Suppose that $C_{1}$, $C_{2}$, ..., $C_{n}$ are $n$ circles of radius $1$ in the plane such that no two of them are tangent and the subset of the plane formed by the union of these circles is connected (i.e., for any partition of $\{1, 2, ..., n\}$ into nonempty subsets $A$ and $B$, $\bigcup_{a \in A}C_{a}$ and $\bigcup_{b \in B}C_{b}$ are not disjoint). Prove that $\left|S\right| \geq n$, where $S = \bigcup_{1\leq i<j \leq n}C_{i}\cap C_{j}$ is the set of pairwise intersection points of the circles. (Each circle is viewed as the set of points on its circumference, not including its interior.)

this is really nice problem here is nice approach, assume that $A_i$ is the intersection points and $a_i$ is the number of the circles which contains $A_i$ now suppose that m8n table s.t if j th circle contains $A_i$ we put $1/a_i$ in the intersection between i th row and j th collumn otherwise put 0.then note that sum of the number in each row is 1 and in each column is atleast 1.

Really nice approach.

What is your solution, Pascual2005?

I know you are asking Pascual, but each solution I saw for this problem is similar to the one given above (even if sometimes the linear algebra tool is masked).

Pierre.


algebra
polynomial
geometry
3D geometry
ratio
quadratics
Rational Root Theorem

If I have a polynomial of degree $ 3$ in the form $ P(x)\equal{}ax^{3}\plus{}bx^{2}\plus{}cx\plus{}d$, how would I "factor" something like this without attempting to turn it into a difference/sum of perfect cubes problem? Is it all guess and check or is there a systematic approach to it?

Also, for the rational roots theorem, I'm not completely understanding it. I have two textbooks and both have enormously long formulas while I get the feeling that it's not actually that complicated. I tried wikipedia for it but I was having trouble understanding that too.

There is a very complicated formula for polynomials of degree 3 and ever more complicated for degree 4.
Otherwise its more or less guess and check (quickest way is with synthetic division), until you can factor it into polynomial of degree 2.

For ration root theorem check this link http://www.mathlinks.ro/viewtopic.php?t=195808

Yeah, as the above poster mentioned, it's pretty hairy:
http://en.wikipedia.org/wiki/Cubic_equation
You're probably looking for the material under Cardano's method.

The general case is as complicated as mentioned above. But in most cases it is not much hard by trial and error(guesswork as you said) in cubics. Try to think of those $ \frac{p}{q}$ so that $ p|d$ and $ q|a$ usually there aren't many such $ p,q$. Is this the rational root theorem? I dont know. Usually $ p|d$ and $ q|a$ because $ \alpha\beta\gamma\equal{}\minus{}\frac{d}{a}$. In cubics there is always a linear real-coefficiented factor. But in quartic there could be none, where there are two quadratic factors. It might be known, but I have developed a partially systematic approach to do it [b]when it is simple looking[/b]. It is a bit trial and error also. After all middle-term-factor is also a bit of trial and error unless one uses the direct formula.

Let $ P(x)\equal{}ax^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}e$
Try to think of some $ c'$ such that $ ay^2\plus{}c'y\plus{}e$ can be middle-term-factored. There are limited such numbers. Now check if $ (c\minus{}c')y^2\plus{}dy\plus{}e$ can be middle-term-factored. If no, think of some other $ c'$. If yes, a nice thing happens, $ ay^2\plus{}by\plus{}(c\minus{}c')$ can be middle-term-factored too.

Now write it as $ ax^4\plus{}bx^3\plus{}c'x^2\plus{}(c\minus{}c')x^2\plus{}dx\plus{}e$. factor the last three terms and take $ x^2$ common in the 2nd and 3rd terms. $ ax^4\plus{}(bx\plus{}c')x^2\plus{}(fx\plus{}g)(hx\plus{}i)$. Now it will be apparent that $ ay^2\plus{}(bx\plus{}c')y\plus{}(fx\plus{}g)(hx\plus{}i)$ can be middle-term-factored too. Thus we are done.

[hide="Example1"]
$ 3x^4\plus{}10x^3\plus{}22x^2\plus{}22x\plus{}15$
[After some trial and error we see that $ c'\equal{}14$]
$ \equal{}3x^4\plus{}10x^3\plus{}(14\plus{}8)x^2\plus{}22x\plus{}15$
$ \equal{}3x^4\plus{}(10x\plus{}14)x^2\plus{}(8x^2\plus{}22x\plus{}15)$
$ \equal{}3x^4\plus{}(10x\plus{}14)x^2\plus{}(8x^2\plus{}(12\plus{}10)x\plus{}15)$
$ \equal{}3x^4\plus{}(10x\plus{}14)x^2\plus{}(2x\plus{}3)(4x\plus{}5)$
$ \equal{}3x^4\plus{}(3(2x\plus{}3)\plus{}(4x\plus{}5))x^2\plus{}(2x\plus{}3)(4x\plus{}5)$
$ \equal{}(x^2\plus{}2x\plus{}3)(3x^2\plus{}4x\plus{}5)$[/hide]
[hide="Example2"]
$ 2x^4\minus{}3x^2\plus{}13x\minus{}21$
[$ c'\equal{}\minus{}1$]
$ \equal{}2x^4\minus{}x^2\minus{}(2x^2\minus{}13x\plus{}21)$
$ \equal{}2x^4\plus{}((2x\minus{}7)\minus{}2(x\minus{}3))x^2\minus{}(x\minus{}3)(2x\minus{}7)$
$ \equal{}(x^2\minus{}x\plus{}3)(2x^2\plus{}2x\minus{}7)$
[/hide]
[hide="Example3"]
$ x^4\minus{}x^3\plus{}11x\minus{}15$
[$ c'\equal{}2$]
$ \equal{}(x^2\plus{}x\minus{}3)(x^2\minus{}2x\plus{}5)$
[/hide]

Other problems
[b]Dont see the hints first[/b]
$ 2x^4\minus{}4x^3\plus{}9x^2\minus{}11x\plus{}5$
[hide="Hint"]$ c'\equal{}11$[/hide]
$ 3x^4\plus{}2x^3\plus{}7x^2\plus{}4x\plus{}5$
[hide="Hint"]$ c'\equal{}8$[/hide]
$ 6x^4\minus{}2x^3\minus{}25x^2\minus{}9x\minus{}1$

PS.What is synthetic division?


Where can I get problems from before year 10?

You can try the IMTS http://www.cms.math.ca/Competitions/IMTS/ :D
The problems may be the same as the USAMTS.  :maybe:

We do not have the rights to the problems before Year 10.  Sorry.  :(


What is the largest integer that is a divisor of
\[ (n\plus{}1)(n\plus{}3)(n\plus{}5)(n\plus{}7)(n\plus{}9)
\]for all positive even integers $ n$?

$ \textbf{(A)}\ 3 \qquad
\textbf{(B)}\ 5 \qquad
\textbf{(C)}\ 11 \qquad
\textbf{(D)}\ 15 \qquad
\textbf{(E)}\ 165$

[hide="Solution"]
Because we are restricted to positive even integers $ n$, we know that all the factors are odd. Because we have 5 consecutive odd integers, we know that $ 5$ must divide it because the units digits of any 5 consecutive odd integers must be 1, 3, 5, 7, and 9. We also know that 3 must divide the product because every odd multiple of $ 3$ is odd itself and at least one of the odds in any range of numbers will be a multiple of three. Thus, the largest integer that is a divisor of the product is $ 3\cdot 5 \equal{} \fbox{15\;(D)}$.
[/hide]

[hide="And"]we also know that $ 11$, and by extension $ 165\equal{}11X15$, is not a divisor when $ n\equal{}12$[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $\left( G, \cdot \right)$ be a group and consider a prime $p$. Let $f_{p}$ be the (finite) number of elements of order $p$.

Prove that $(p-1)|f_{p}$. Can $f_{p}$ be equal to $2(p-1)$?

$(p-1) | f_{p}$ is a consequence of the famous proof of cauchy's theorem (via the action of $G$ on the $p$-tuples whose product is $1$).

Objections:

1. Don't we get that $f_{p}\equiv-1 \left( \bmod \, p \right)$?
(I've just woken up, so sorry if I'm saying stupid things.)

2. Will the fact that $G$ can be infinite cause a problem?

By the way, the first part has a very easy proof.

Those objections are exactly right- divisibility by $p-1$ is a consequence of partitioning into subgroups, and the argument referred to by -oo- proves that $f_{p}\equiv-1$ if $G$ is finite and has order divisible by $p$. I've posted a more general version- a similar result holds at least for squarefree $n$ in place of $p$.

The sticking point is if $G$ is infinite.

A very ad hoc argument: Suppose $a$ and $b$ are two incommensurable elements of order $p$, and consider the subgroup generated by $a$ and $b$. We will show it has elements of order $p$ which are neither powers of $a$ nor powers of $b$.
Consider $a^{-1}ba$; this has order $p$. If $a^{-1}ba=a^{r}$ for some $r$, $b=a^{r}$, which is impossible. If $a^{-1}ba$ is neither a power of $a$ nor a power of $b$, we are done. This leaves only the possibility that $a^{-1}ba=b^{r}$ for some $r$; similarly, $b^{-1}ab=a^{s}$. WLOG, $1\le r,s\le p-1$.
Computing inductively, $a^{-k}ba^{k}=b^{r^{k}}$ for each $k$, and $b=a^{-p}ba^{p}=b^{r^{p}}=b^{r}$ (Fermat's theorem), so $r=1$. Similarly, $s=1$ and $a,b$ commute. Then $ab$ is another element of order $p$, and we are done.

This is a lot less satisfying than an extension of the $f_{p}\equiv-1$ result.


[quote="ckck"][quote="sharkboy"]This problem sounds easy but I just can't get it for some reason. Find the surface area of a cube with a diagonal of 8 units.  The diagonal of a cube is a segment whose endpoints are vertices that are not on the same face.[/quote]
[hide]That would be called the space diagonal. The space diagonal is $\sqrt{3}$ of a side length. So set up an equation. $s\sqrt{3}=8$. Solving for the side, we get that the side is $\frac{8\sqrt{3}}{3}$. The surface area of a cube is $6a^{2}$ where a is the side length of the cube. Solving for the surface area, we get that the surface area is $(\frac{8\sqrt{3}}{3})^{2}=\frac{8*8*3}{3*3}=\frac{64}{3}$. Multiplying this by 6 to get our final answer, which is $128$.
[/hide][/quote]

lol dont help him out with homework questions, especially not by giving him the work and answers.

OK, nvm.  I was mixing up the 30, 60, 90 triangles and 45, 45, 90 triangles. Thanks


Let's say I am typing inequalities and equations in conjunction in the align environment, which one is more accepted in the math community?

Cramming one line (shows my intention):

when $ x > 2$;

$ x^{2}> 2 \cdot 2 = 4$

1)
[code]
\begin{align*}

x^{2}& > 2 \cdot 2\\
         & = 4

\end{align*}
[/code]

2)

[code]
\begin{align*}

x^{2} & > 2 \cdot 2\\
          & > 4

\end{align*}
[/code]

Hope you get what I mean...thank you.  (Of course, in real situation I have 5 lines +, so I can't cram...)

edit: somehow it won't render...[/code]

I think it should be the first - otherwise you're saying 2.2 > 4

I don't know if this is just on my side, but I got rid of the line at the top and the bottom, and it worked fine:
[code]\begin{align*}
x^{2}& > 2 \cdot 2\\
         & = 4
\end{align*} [/code]

Well because I think technically when nothing's on the left side of the equation array then it means implicitly that you copy the left side down...But somewhere I see the first one being used, so I don't know...

jshen's version is the correct one and the way I expect my students to write their inequalities.
\begin{align*}x^{2}& > 2 \cdot 2\\ & = 4 \end{align*}
Although
\begin{align*}x^{2}& \ge 2 \cdot 2\\ & \ge 4 \end{align*}
would be technically correct, I still prefer to use = 4 to show that line is not an inequality. Sometimes you are asked to prove an inequality and then ask under what conditions the two sides are equal. Using = where appropriate helps to pin down what has to happen.

[quote="pkerichang"]edit: somehow it won't render...[/quote]As a workaround put \par just before \begin{align*} which is what I've done above.


Is it possible to simplify this $ \binom{2m}{m\plus{}a}\minus{}\binom{2m}{m\plus{}a\plus{}2}$ more than this $ \binom{2m}{m\plus{}a\plus{}2}*\frac{4am\plus{}6m}{(m\minus{}a\minus{}1)(m\minus{}a\minus{}2)}$ and is it possible to get $ \binom{2(m\plus{}1)}{(m\plus{}1)\plus{}a}\minus{}\binom{2(m\plus{}1)}{(m\plus{}1)\plus{}a\plus{}2}$ from $ \binom{2m}{m\plus{}a}\minus{}\binom{2m}{m\plus{}a\plus{}2}$ ?

first  one  i am  also  not able to procced further from the step where u have stopped. :)


(x+a)(x+b)+(x+c)-5=0, find a+b+c

the answer is 2.
But how did they got it?
this was a competition question.

my solution is
3x+a+b+c=-5
a+b+c=-5-3x
i let x=1
my answer is -8

help pls!

1. Did you make a typo?
2. If so, then a+b+c is expressed in terms of x. 2 is the incorrect answer.
3. If not, we get a quadratic in x. $ x^2\plus{}(a\plus{}b\plus{}1)x\plus{}ab\plus{}c\minus{}5\equal{}0$. Dunno what to do from here.

The question is weird...

I think it's $ (x\plus{}a)(x\plus{}b)(x\plus{}c)$ or $ x\plus{}a\plus{}x\plus{}b\plus{}x\plus{}c$

henyomzky's steps guide us to the $ x\plus{}a\plus{}x\plus{}b\plus{}x\plus{}c$, with a typo.

$ 3x\plus{}a\plus{}b\plus{}c\equal{}5$

$ a\plus{}b\plus{}c\equal{}5\minus{}3x$

It depends on $ x$...

my solution is incorrect..

Given that (x+a)(x+b)(x+c)-5=0
then find the value of a+b+c

Doesn't change anything - still infinitely many solutions. Cannot be all positive, since then the product is at least $ 8$. And what is the point, anyways  :mad:


geometry
parallelogram
ratio
congruent triangles
similar triangles

Given that$ GF \equal{} 24$, $ EF \equal{} 32$ and $ AC$ is the diagonal of the parallelogram, calculate $ BE$.

[img]http://i204.photobucket.com/albums/bb126/agerador/trig2.jpg[/img]
Please explain it step-by-step. If you can solve it using only similar and congruent triangles better!

Ooops! :blush:  I forgot to upload the figure.

Besides A and D, what are the other two vertices of the parallelogram?

[hide="Solution"]
Note that $ \triangle AEF \sim \triangle CEB$ and $ \triangle GEC \sim \triangle AEB$. So $ \frac {32}{BE} \equal{} \frac {AE}{EC} \equal{} \frac {BE}{8}$, so $ BE \equal{} 16$.[/hide]

There are two pairs of similar triangles in that figure. You can find them by looking at the intersection of lines FB and AC. GEC and AEB are opposite angles, so we know they are congruent. So both triangles GEC and AEB have a congruent angle. But we also know that the ratios of the sides are the same, because GC and AB are parallel. We can also show that AEF and CBE are similar the same way. Let BE equal $ x$, let EC equal $ y$, and let EA equal $ w$. We know that

$ \frac{GE}{EB} \equal{} \frac{EC}{AE}$
$ \frac{8}{x} \equal{} \frac{y}{w}$ (Equation 1)

We also know that;

$ \frac{EF}{BE} \equal{} \frac{EA}{EC}$
$ \frac{32}{x} \equal{} \frac{w}{y}$
Taking the reciprocal of both sides, we have;
$ \frac{x}{32} \equal{} \frac{y}{w}$
Now, plugging this value of $ \frac{y}{w}$ into Equation 1, we get;

$ \frac{8}{x} \equal{} \frac{x}{32}$

Cross multiplying, we get;

$ 8 \times 32 \equal{} x^2$
$ 2^3 \times 2^5 \equal{} x^2$
$ 2^{3 \plus{} 5} \equal{} x^2$
$ 2^8 \equal{} x^2$
Taking the square root of both sides, we get;
$ 2^4 \equal{} x$
$ 16 \equal{} x$.

So 16 is the answer.

TIP: The only thing you know about the figure is a few side lengths and you have the intersection of two lines, which yields similar triangles. You don't know any angles or areas, so this should tell you to look for ratios between sides.


be dalile estaghbal binazire mardom mahe ramezan 24 saat tamdid shod,azinke az khodetoon tahamol dar vakardid manoonam.

Eide saeid fetr ra niz be hameye olympiadiha tabrik migam.

[quote="amirhtlusa"]zaheran mosabegheye [color=blue]esteghlal[/color]-[color=red]Piroozi[/color]  gharar e shanbe bargozar she...

rasti bar o bach inja esteghalai hastin ya perspolisi???

[color=red]Esteghala SOOORAKH e [/color]  :P[/quote]
Khodetoon sooraakhid baa oon timeh tah jadvalitoon :rotfl:
[color=blue]Perspolis :!:  sooraakh[/color]

faghat va faghat [color=blue]esteghlal[/color] :winner_first:

akhe kesi harfi bezaneh ke tooye league dahom nabashe :rotfl:

be nobeye khodam eide fetro tasliat migam,
chon majboorim dobare dir berim khooneh...

Ashegh faghate faghat dar in ye mored bahat movafegham

Bia khoob .... [size=150]BEHET[/size]

bale [color=blue]esteghlale[/color] dige shokhi ke nist. :D

[quote="amirhtlusa"]perspolis behtar bazi kard. :P[/quote]
joob zadi :D

:D  engar hamin parsal bood ke man eide fetro tabrik goftam :D

Dobare in eid ro be hame tabrik migam

Movafagh bashid.

avalan manam eide fetro be hame tabrik migam dovoman asasan bache haye allame tabatabei tarafdare esteghlal hastan va man ham hamintor vali motmaen bashid prespolis soooooooorakhe. :first:

[quote="saman"]bebin azize man bejaye inke be ham dg eide fetro tabrik begin , ye soale tarkibiate khoshgele mamani bedin ta hase aghal ye kam hal konim...kosh shodim baba.....ino ham begam ke bachehaye allameh tabatabai hameghe esteghlalian :D[/quote]

tavanayie anjam dadande joftesho baham nadarid,ha?

are, kheily zood gozasht, manam in eido tasliat migam chon alave bar

inkeh, majboor hastim dirtar az madrese tatil shim,mizane jarayeme

mokhtalef niz dar sathe keshvar, ba payane mah ramezan,afzayesh

yafteh :!:

yek sale dige gozasht,

eide fetro dobare tabrik migam be hameye bacheha :wink:

rahmate khodavand hamrahe ma bad,chenan ke ba hameye hastist :)

man ham tarafdariye khodamo nesbat be esteghlal elam mikonam :D


Show that $ \lim_{x \to a}f(x)\equal{}\lim_{h \to 0}f(a\plus{}h)$

Let $ h \equal{} x\minus{}a \rightarrow 0$ and you're done...?

yes, the first thing i thought was that... But i think this is not true if $ f$ is not continuous at $ a$. But this is stated as in the book, pag 133 number 9.


AMC
AMC 12
probability
AIME
geometry
number theory

I am trying to pick a math book to study this summer in preparation for the AMC12 and possibly the AIME. I am kinda new to the competitive math so any information and tips would be helpful.
In my Precal class last year we covered probability and counting, however not in to depth. So my question is: would it be wise to get the Intro to Probability and Counting or dive right into Intermediate Counting and Probability? In addition I am interested in Number Theory and am aware of congruences but don't know much else so would the Intro to Number Theory be good? I don't want to get a book that's going to be too easy and seeing as the books are kinda expensive I would like to make a prudent choice. Keep in mind I want a book that will be challenging, but not too challenging that I want to go in a corner in cry. So any tips from owners of these books would be great, and thanks!

intro books first!!! Intermediate books cover hard AMC 12 and AIME problems.

Ok so the intro first, but which intro? Which subject would be more beneficial to the AMC12 and AIME: Probability and Counting or Number Theory? Moreover is the Number Theory book going to have an intermediate follow up soon? I know about permutations and combinations already as well as the fundamental counting theorem and the and/ or statements in probability also vaguely familiar with geometric probability. So given all the above information which book would you (general public) suggest for preperation as well as sheer enjoyment of the subject.
Add On- Right now I am considering Intermediate Probability and Counting only because I know a lot of the basics and the stuff that I don't know is covered in the first chapter. Also the excerpts were really insightful and interesting and not too difficult to comprehend. If no one has any word of caution or more advice I think I might go ahead with this one.

Well...first of all...intro books are more toward Easy AMC-Med. AIME level. Intermediate books are sort of more Hard AMC-AIME-Olympiad level. Well, I recommend you buy whichever book you need most help with. If you're not exactly sure what you're good at, then you should consider getting the AOPS Vol. 2, since it covers most topics, and doesn't get [i]too[/i] advanced.

Wait...by "possibly AIME", do you mean that you're not sure if you'll even make it in? Because generally, if you can't make AIME, you should more build your weaker points. Working on improving your probability and counting to AIME-Olympiad level won't really help you. Work on all subjects until they're at least easy AIME level.

The AMC 12 is not a subject test. It covers all basic fields of Mathematics in each test: Algebra, Geometry, Number Theory, Combinatorics. The Introduction books are a great review if you have covered it a long time ago, and the intermediate books are excellent in covering some advanced material. I have the whole AoPS library right next to me.

well on the bookstore section, if you select a book, it has a pre-test and a post-test next to it, so you can decide after doing the tests.

I ended up getting Intermediate Counting and Probability and so far it seems fairly easy. I'm only in chapter 2 doing sets and logic but it's not hard at all so far, the problems are straight forward and require little thinking. It's a great book, and the material is just awsome. I think i will get AOPS volume 1 and 2  next, to work on my other skills. I am just intrigued by counting and probability and find it the most enjoyable math skill i've learned so far.
Thanks for the advice people! :lol:


function
Princeton
college
algebra proposed
algebra

Does there exist some pair of functions $ p: \mathbb{Z}\to \mathbb{Z}$ and $ q: \mathbb{Z}\to\mathbb{Z}$ such that $ (\forall a,b\in \mathbb{Z})(a\neq \minus{}b)$
\[ \frac{1}{a\plus{}b}\equal{}\frac{p(a)}{q(b)}?\]

My first instinct is to say no, there isn't, but first instincts have (often) been wrong.

[quote="foxjwill"]Does there exist some pair of functions $ p: \mathbb{Z}\to \mathbb{Z}$ and $ q: \mathbb{Z}\to\mathbb{Z}$ such that $ (\forall a,b\in \mathbb{Z})(a\neq \minus{} b)$
\[ \frac {1}{a \plus{} b} \equal{} \frac {p(a)}{q(b)}?
\]
My first instinct is to say no, there isn't, but first instincts have (often) been wrong.[/quote]
Well.. You have a good instinct :D

$ b\equal{}0: \frac {1}{a} \equal{} \frac {p(a)}{q(0)} \iff p(a) \equal{} \frac{q(0)}{a}$

$ a\equal{}0: \frac {1}{b} \equal{} \frac {p(0)}{q(b)} \iff q(a) \equal{} b p(0)$

So $ \frac {1}{a \plus{} b} \equal{} \frac {q(0)}{ab \cdot p(0)}$
Which obviously can't be true!!

[quote="Mathias_DK"][quote="foxjwill"]Does there exist some pair of functions $ p: \mathbb{Z}\to \mathbb{Z}$ and $ q: \mathbb{Z}\to\mathbb{Z}$ such that $ (\forall a,b\in \mathbb{Z})(a\neq \minus{} b)$
\[ \frac {1}{a \plus{} b} \equal{} \frac {p(a)}{q(b)}?
\]
My first instinct is to say no, there isn't, but first instincts have (often) been wrong.[/quote]
Well.. You have a good instinct :D

$ b \equal{} 0: \frac {1}{a} \equal{} \frac {p(a)}{q(0)} \iff p(a) \equal{} \frac {q(0)}{a}$

$ a \equal{} 0: \frac {1}{b} \equal{} \frac {p(0)}{q(b)} \iff q(a) \equal{} b p(0)$

So $ \frac {1}{a \plus{} b} \equal{} \frac {q(0)}{ab \cdot p(0)}$
Which obviously can't be true!![/quote]

Thanks! I can't believe I didn't think of such an easy solution! I guess it takes practice not to complicate stuff, eh?

[quote="foxjwill"]Thanks! I can't believe I didn't think of such an easy solution! I guess it takes practice not to complicate stuff, eh?[/quote]
Yeah... There's a book called "How to solve it" by some author - I forgot his name.. It is pretty good, and will probably help you a lot :)

The best advise I can get on these kind of things is a "brute forcing" technique:
Take a look at the equation from all kind of positions.
Fx $ \frac {1}{a \plus{} b} \equal{} \frac {p(a)}{q(b)}$ is equivalent to $ a\plus{}b \equal{} \frac {q(b)}{p(a)}$. This "transformation", does not help nescesarrily help you at all. (I didn't garantuee nothing:P) But the equation $ p(a)(a\plus{}b) \equal{} q(b)$ could get also you a idea maybe. For example: If we fix $ b$, then for some $ a$ we have that $ (a\plus{}b, q(b))\equal{}1$, and thus if $ q(b)>1$ we have that $ p(a)$ is a fraction (contradicting the original statement).
What helped for me was this tactic:
I set $ a\equal{}0$. simple as that. I now had a expression for $ p(a)$ (and in the same way for $ q(b)$) and with substition, it was obvious that such a solution could not exist.
I could have taken a lot of other approaches to. For example $ a\equal{}b$. Then we have $ 2a \cdot p(a) \equal{} q(a)$ But substituting into the equation we get: $ a\plus{}b \equal{} \frac{q(b)}{p(a)} \equal{} \frac{2b \cdot p(b)}{p(a)}$ And thus $ p(a) \equal{} (a\plus{}b)(2b \cdot p(b))$(You notice how we got almost the same result earlier, but went a totally different way? Sometimes you only see the solution, when you look at it from that special angle - so even though you may've missed the other solution, we get it again down here :)) Which again can't be true. Proof: Fix $ a$. Then for some $ b$ $ (p(a),a\plus{}b)\equal{}1$ and $ a\plus{}b>p(a)$ and thus $ 2b \cdot p(b)$ is a fraction.(Not possible

Hope i helped you:)

[quote="Mathias_DK"][quote="foxjwill"]Thanks! I can't believe I didn't think of such an easy solution! I guess it takes practice not to complicate stuff, eh?[/quote]
Yeah... There's a book called "How to solve it" by some author - I forgot his name.. It is pretty good, and will probably help you a lot :)
[/quote]

Maybe George Polya!  :D

[quote="Mathias_DK"]
Yeah... There's a book called "How to solve it" by some author - I forgot his name.. It is pretty good, and will probably help you a lot :)
[/quote]

Was this it: http://www.amazon.com/How-Solve-Aspect-Mathematical-Method/dp/0691023565

[quote="Mathias_DK"]Hope i helped you:)[/quote]
Totally!

[quote="orl"]Maybe George Polya!  :D[/quote]
Yes :)

[quote="foxjwill"]Was this it: http://www.amazon.com/How-Solve-Aspect-Mathematical-Method/dp/0691023565[/quote]
Yes. That's the one. (I've got it on my computer. If you're interested just pm me)

Edit: A lot of people(2:P) have pm'ed me about the book. You can get it here: http://www.4shared.com/file/48711175/406fb4fa/George_Polya_-_How_to_Solve_It_A_New_Aspect_of_Mathematical_Method_Princeton__University_Press_1971_.html?s=1


How many two-digit numbers are there in which the tens digit is greater
than the ones digit?

There are 10 possible digits. If we choose any two of them that are distinct, then our two-digit number is forced. Therefore, the answer is $ \binom{10}{2} \equal{} \boxed{45}$.

Just think about it. Look every ten digit number has that many of numbers that have the same tens digit but less. Example. When the tens digit is 9 there are 9 2-digit numbers that have 9 as there tens place. 90-98
So you find the sum: 10 times 9 divided by 2

I know how to do it using 10 choose 2, but when I look at it in [hide="a different way"]Each two-digit number has digits that are either the same or different.
There are 90 two-digit numbers, and 9 of these have the same digits.
Now, using common sense, half of the remaining numbers satisfy the property of the tens digit being greater than the ones digit.
So, there are $(90-9)/2=$ [color=#f00]wait whattt.[/color].... $81/2=40.5$ I'M BAMBOOZLED! what did I do wrong??? :maybe:  :what?: [/hide], I don't even know what's right. Can someone help me?

Also, you can't have half a number... I'm so dumb all of a sudden

What did I do wrong though?

You could just write something like this...
9=0,1,2,3,4,5,6,7,8,
8=0,1,2,3,4,5,6,7
and then again again and again
then you could add up all the ways and get...
[hide= ]$\boxed{45}$[/hide]

[hide=Alternate Solution]
We have to think about it.
So we know that only half of the two-digit numbers have the tens digit greater than the ones digit. There are 90 ten digit numbers.
Hence, 90/2 = 45 two digit numbers have the tens digit greater than the ones digit.[/hide]


[hide="Solution"]
We have that if $ \frac {10^{210}}{10^{10} \plus{} 3} \equal{} k$, then $ 2 \plus{} f(k) \equal{} \frac {10^{210} \plus{} 3^{21}}{10^{10} \plus{} 3}$, where $ f$ is the floor function.  This is because $ \frac {10^{210} \plus{} 3^{21}}{10^{10} \plus{} 3}$ is an integer since $ (a \plus{} b)|(a^k \plus{} b^k)$ for odd $ k$ and $ 10^{10} < 3^{21} < 2\cdot 10^{10}$.  We now find the units digit of $ \frac {10^{210} \plus{} 3^{21}}{10^{10} \plus{} 3}$.  The numerator of the top has a units digit of $ 3$ since $ 10^{210} \plus{} 3^{21}\equiv 3^{21}\equiv 3\bmod 10$ and that of the bottom is $ 3$ since $ 10^{10} \plus{} 3\equiv 3\bmod 10$.  Now, we have that $ \frac {3}{3}\equiv 1\bmod 10$, so $ f(k) \plus{} 2\equiv 1\bmod 10$, meaning that $ f(k)\equiv 9\bmod 10$.  We conclude that the units digit of the term is $ \boxed {9}$. [/hide]


Consider the Banach Space $ l^1$.  Let S = $ \{ x \in l^1 \mid \| x \| \le 1\}$. Prove that $ S$ a compact subset of $ l^1$.

Are you missing the word "not" somewhere in your statement? The closed unit ball of any infinite dimensional normed space is not compact.

oh yes.....I'm sorry...it should read Prove [i]or disprove[/i]

It is easy to see that if $ x^{(m)}_{n} \equal{} \begin{cases}0,\ n\neq m \\
\frac{1}{2},\ n \equal{} m\end{cases}$
then $ \|x^{(k)} \minus{} x^{(l)}\| \equal{} 1$ for $ k\neq l$, and $ (x^{(m)})_{m \equal{} 1}^{\infty}\subset S$ cannot have a convergent subsequence.


A = 11^100 -1

How many zero the in the end of A ?

3

Given $ 11 \equal{} (10 \plus{} 1)$

Look at the binomial expansion of $ (10 \plus{} 1)^{10^n}$, $ n > 1$

We consider the last term

$ 1^{10^n}$, clearly it will end in one, ensuring that the $ \minus{} 1$ gives us ending $ 0$s

The next term will be

$ 10^{1}1^{10^n \minus{} 1}$

Which will also end in zero

The next terms will be $ {{10^n} \choose {x}} 10^{x}1^{10^n \minus{} x}$

Clearly, all but the binomial coefficient will be divisible by 10.

Therefore we need to find a point when the sum isn't 0.

In the case for n = 2, this is 3. Although we can generalise this for any n (coming up)

Let $ a_n\equal{}11^{10^k}\minus{}1$, by the lifting the exponent theorem,

$ 2\parallel{}11\minus{}1$ and $ 2^k\parallel{}10^k$, so $ 2^{k\plus{}2}\parallel{}A$ and

$ 5\parallel{}11\minus{}1$ and $ 5^k\parallel{}10^k$, so $ 5^{k\plus{}1}\parallel{}A$.

Hence $ 10^{k\plus{}1}\parallel{}A$.

[quote="SimonM"][quote="diophantient"][quote="Altheman"]Let $ a_n \equal{} 11^{10^k} \minus{} 1$, by the lifting the exponent theorem,

$ 2\parallel{}11 \minus{} 1$ and $ 2^k\parallel{}10^k$, so $ 2^{k \plus{} 2}\parallel{}A$ and

$ 5\parallel{}11 \minus{} 1$ and $ 5^k\parallel{}10^k$, so $ 5^{k \plus{} 1}\parallel{}A$.

Hence $ 10^{k \plus{} 1}\parallel{}A$.[/quote]
That doesn't work, since 2||(11-1) ;)[/quote]

Doesn't that mean 2 divides 10 or am I not understanding the notation?[/quote]
That means that 2 divides 10 but 4 doesn't.
That means that Altheman's method doesn't work.
For example, by his logic, $ 11^4\minus{}1$ wouldn't be a multiple of 16, but it is.


calculus
derivative
inequalities
inequalities proposed

for $ x,y,z >0$ with

$ \sum x^2\equal{}1$

prove that:

a)$ \sum \frac{x}{1\plus{}x^2} \le \frac{3\sqrt{3}}{4}$.

b)$ \sum \frac{x}{1\minus{}x^2} \ge \frac{3\sqrt{3}}{2}$

sorry if it has been posted before.
since i want a solution different of mine plz solve it without using derivatives and things like that,thx for your help :)

we have

$ \sum_{cyc}\frac {a}{1 \plus{} a^2} \equal{} \sum_{cyc}\frac {a}{\frac {2}{3} \plus{} \frac {1}{3} \plus{} a^2} \le \sum_{cyc}\frac {a}{\frac {2}{3} \plus{} \frac {2a}{\sqrt {3}}} \equal{} \sum_{cyc}\left(\frac {\sqrt {3}}{2} \minus{} \frac {3}{2\sqrt {3} \plus{} 6a}\right)$

hence it's sufficient to show that:

$ \sum_{cyc}\frac {1}{6a \plus{} 2\sqrt {3}} \ge \frac {\sqrt {3}}{4}$

By Cs

$ \sum_{cyc}\frac {1}{6a \plus{} 2\sqrt {3}} \ge \frac {9}{6(a \plus{} b \plus{} c) \plus{} 6\sqrt {3}} \ge \frac {9}{6\sqrt {3} \plus{} 6\sqrt {3}} \equal{} \frac {\sqrt {3}}{4}$

because $ \sqrt {3(a^2 \plus{} b^2 \plus{} c^2} \ge a \plus{} b \plus{} c$

we have

$ \frac {x}{1 \minus{} x^2} \ge \frac {3\sqrt {3}x^2}{2}\Leftrightarrow x^3 \plus{} \frac {2}{3\sqrt {3}} \ge x$

which is true by AM GM   since $ x^3 \plus{} \frac {1}{3\sqrt {3}} \plus{} \frac {1}{3\sqrt {3}} \ge 3.\sqrt [3]{x^3.\frac {1}{27}} \equal{} x$

hence

$ \text{LHS} \ge \frac {3\sqrt {3}}{2}(x^2 \plus{} y^2 \plus{} z^2) \equal{} \text{RHS}$

 :oops:

Hope I 've help you  :lol:

my solution for the first part was this :
we can prove that:

$ \frac{x}{1\plus{}x^2} \le \frac{3\sqrt{3}}{16}(1\plus{}x^2)$

$ \leftrightarrow 0 \le (x\minus{}\frac{1}{\sqrt{3}})^2(\frac{3\sqrt{3}}{16}x^2\plus{}\frac{3}{8}x\plus{}\frac{9\sqrt{3}}{16})$

which is true.
now by using this inequality for each of the fractions we are done.

and my solution for the second part was the same as Abdek's answer.


You omited "2" in problem 8 . The correct form :
 $\frac{1}{a\sqrt{a+b}} +\frac{1}{b\sqrt{b+c}}+\frac{1}{c\sqrt{c+a}} \geq \frac{3}{\sqrt{2abc}}$

Bai 2 co le suy ra tu bai so "buong binh" TST VN 2000!

Hi there !

Just a question.....

Can we discuss these problems on the forum or is the contest under progress??

 :)

Hey

What about problem 8??

Seems too tough (at least for me..... :( )

[quote=MrMATH][b]Vietnamese Mathematics e - Olympiad (VMEO)[/b]

Problem 5 [hide]Let be positive integers $a,b$. How many integers numbers which can be written in the form $ap+bq$. In here $p,q$ are nonnegative integers and $p+q\le{2005}$[/hide]
Problem 6 [hide]For a given convex quadrilateral $ABCD$, let $I$ be avariable point on the diagonal $AC$ such that $I$ and $A$ are on the same of the diagonal $BD$. Assume $E,F$ lie on the diagonal $BD$ such that $IE||AB$ and $IF||AD$. Show that $\widehat{BIE}=\widehat{DCF}$[/hide]

Problem 7 [hide]Find all monotonous function $f: [0,+\infty)\to{\mathbb{R}}$ satisfying the equatinon $(f(x)+f(y))^2=f(x^2-y^2)+f(2xy)$ for all $x\ge{y}\ge{0}$[/hide]
Problem 8 [hide]Let $a,b,c>0$. Prove the inequality $\frac{1}{a\sqrt{a+b}}+\frac{1}{b\sqrt{b+c}}+\frac{1}{c\sqrt{c+a}}\ge{\frac{3}{\sqrt{2abc}}}$ [/hide]

P.S: I'm very honour to present some problems in VMEO contest to you. And, I hope there isn't many solutions of them in here before VMEO comes to and end (about 2 months after my post)[/quote]

in problem 6 is missing the [b]cyclic [/b] condition for ABCD as, BlackMax noticed [url=https://artofproblemsolving.com/community/c6h2668808p23126078]here[/url]

[quote=Hong Quy][quote="bomb"]Is Q6 wrong? Bomb[/quote]
No, it's true .[/quote]
It missed the [b]cyclic[/b] condition.

the VMEO II post collection is [url=https://artofproblemsolving.com/community/c2463154_vmeo_ii]here[/url]

you may solve those problems mentioned above


[quote="arqady"]Let $a,$ $b$ and $c$ are positive numbers. Prove that:
$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq2\left(\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}\right).$[/quote]
It's equivalent to:

$\sum \left( \frac{a^{2}}{b}-2\frac{a^{2}}{b+c}\right)\geq 0$ $\Leftrightarrow$ $\sum \frac{a^{2}(c-b)}{b(b+c)}\geq 0$ $\Leftrightarrow$ $\sum a^{3}c(a+b)(c+a)(c-b)\geq 0$ $\Leftrightarrow$
$\sum (a^{4}c^{3}+a^{5}c^{2}+a^{3}bc^{3}+a^{4}bc^{2})\geq \sum (a^{4}bc^{2}+a^{5}bc+a^{3}b^{2}c^{2}+a^{4}b^{2}c)$ ???

I thought i had proved this last inequality by am-gm but i'm wrong.

EDITED

[quote="Svejk"]
$\sum (a^{4}c^{3}+a^{5}c^{2}+a^{3}bc^{3})\geq \sum (a^{5}bc+a^{3}b^{2}c^{2}+a^{4}b^{2}c)$ which is AM-GM[/quote]
Svejk, detail pleas.  :wink: Thank you.

I am sorry Svejk. The inequality $\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq2\left(\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}\right)$ is wrong. I have corrected it.

Hope I am not wrong at my calculations .
The inequality is equaivalent to
$\sum\frac{(a^{2}-b^{2})^{2}}{ab^{2}}-\sum\frac{(a-b)^{2}}{a}\geq \frac{(a-b)^{2}(a+b+c)}{(b+c)(a+c)}$

or
$\sum (a-b)^{2}(\frac{(a+b)^{2}}{ab^{2}}-\frac{1}{a}-\frac{a+b+c}{(a+c)(b+c)})\geq 0$

which is easy with SOS

[quote="arqady"]Let $a,$ $b$ and $c$ are positive numbers. Prove that:
$\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}}\geq2\left(\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}\right).$[/quote]

Another proof should be using Cauchy:

$4 \cdot (\sum{\frac{a^{2}}{b+c}})^{2}\leq 4 \cdot (\sum{\frac{a^{3}}{b^{2}}}) \cdot (\sum{\frac{ab^{2}}{(b+c)^{2}}}) \leq \sum{\frac{a^{3}}{b^{2}}\cdot \sum{\frac{ab}{c}}}$.

So it only remains to prove that $\sum{\frac{ab}{c}}\leq \sum{\frac{a^{3}}{b^{2}}}$, which is equivalent with $\sum{a^{2}b^{5}}\geq \sum{a^{3}b^{3}c}$, which is true.
($11a^{2}b^{5}+b^{2}c^{5}+7c^{2}a^{5}\geq 19a^{3}b^{3}c$ and the other cyclic ones.)


quadratics
modular arithmetic
number theory unsolved
number theory

prove that all  positive prime divisors except 2 of $a{}^2{}^n+1$ are greater than
$2^n+1^$

 the exponent on the first term  is $2^n$ not 2n and the exponent on the last  term  is  n+1 .  i  just  cant seem  to  get  them  right on the latex.   :blush:

do u mean ?

  prove that all  positive prime divisors except $2$ of $a^{2^n}+1$ are greater than
$2^{n+1}$

Let $p>2$ be a prime divisor of $a^{2^n}+1$, and let $d$ be the least positive integer such that $a^d \equiv 1 \pmod{p}$. Since $p$ divides $a^{2^n}+1$, we have $a^{2^n} \equiv -1$, so $a^{2^{n+1}} \equiv 1$. It's well-known that $d$ (which is the order of $a$) must then divide $2^{n+1}$. But since $2^{n+1}$ is a power of two, $d$ must be a power of two, as well. Let $d=2^k$. If $k \leq n$, then squaring the congruence $a^{2^k} \equiv 1$ $n-k$ times we get $a^{2^n} \equiv 1$, absurd, since $a^{2^n} \equiv -1$ and $p>2$. Then $k>n$, that is, $k \geq n+1$. But since $d \leq p-1$ (as $d$ is the order of $a$, again) we have $p-1 \geq d=2^k \geq 2^{n+1}$, which is what we wanted to prove.


Indiana is a strong math state, so we need to get things going in here! :starwars:  :jump:  :wow:

[quote]indianamath WE NEED MORE PEOPLE in THIS FORUM!!!
nobody ever uses this

--------------------------------------------------------------------------------

Indiana is a strong math state, so we need to get things going in here!    [/quote]
I concur


Find the remainder when a polynomial P(x) is divided by (X-1)(X-3).
Given:
the remainder when dividing by (X-1) is 2;
the remainder when dividing by (X-3) is 1.

[hide]So we can write $ P(x)\equal{}Q(x)(x\minus{}1)(x\minus{}3)\plus{}r$. Since $ deg(r)<deg(x\minus{}1)(x\minus{}3)$, we can say r=ax+b.
Subbing in x=1, we have a+b=2, and subbing in x=3, 3a+b=1. We find $ a\equal{}\frac{\minus{}1}{2},b\equal{}\frac{5}{2}$, so $ \frac{\minus{}x}{2}\plus{}\frac{5}{2}$.[/hide]


[quote="t0rajir0u"]Show that for an angle $0 < \theta < \frac{\pi}{2}$ we have

$\sqrt{ \sin^{2}\theta+\frac{1}{\sin^{2}\theta}}+\sqrt{ \cos^{2}\theta+\frac{1}{\cos^{2}\theta}}\ge \sqrt{10}$

And determine where equality holds.

I have a somewhat annoying multi-step proof of this, and I was wondering if someone could come up with a more clever manipulation.   :D[/quote]
Put $a=\sin^{2}\theta$, $b=\cos^{2}\theta$ with $a+b=1$ and $0<a,b<1$. Then Just use Jensen for convex function $\sqrt{t+\frac1t}$.

Heh...my math teacher actually gave us a copy of this magazine, and I submitted my solution...I did the same thing as Rzeszut...

$\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}\ge 2\sqrt{\sqrt{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}}}$

It is clear that $\frac{a}{b}+\frac{b}{a}\ge 2$.  Now all that is left is to prove that $ab+\frac{1}{ab}\ge 4+\frac{1}{4}$, but that is true because $ab \le \frac{1}{4}$ by AM-GM, and a few more steps and we're done.


probability
combinatorics theorems
combinatorics

There are 9 books on the shelves A,B,C.
We are searching for the probability of that there are 2 on A, 2 on B.

If the books are same,  what is the probability?
If the books are different, what is the probability?

If we are searching for number of books on shelves, is the similarity matter?

We must find in how many ways 9 can be separated into three parts.
x + y + z = 9        x >= 1, y >= 1, z >= 1.
a + b + c = 6        a >= 0, b >= 0, z >= 0.
C (8; 2) = 28. So, if we think it is one of these to separate 2 2 and 5, we will get that the probablity is 1/28. But in your permission, I don't have any idea 1/28 belongs to which of these variants as an answer.


Dear Valentin,

I am very interested in the solution of the very nice problem of 4th Mathlink Contest round 6

Let $n \geq 3$ be an integer. Find the minimal value of the real numbers $k_n$  such that for all positive numbers $x_1,x_2,...,x_n$ with product $1$ we have

$\frac{1}{\sqrt{1+k_nx_1}}+\frac{1}{\sqrt{1+k_nx_2}}+...+\frac{1}{\sqrt{1+k_nx_n}} \leq n-1$

Can you please tell me when it will be published?

Thank you very much.

[quote="manlio"]I am very interested in the solution of the very nice problem of 4th Mathlink Contest round 6[/quote]

Are you sure about [b]4[/b]th? In this case you have a remarkable talent of looking into the future...

SCNR.

  darij

Hi Manlio, I moved this topic, since it refers to ML Contest 3rd Edition. This problem is Harazi's proposal. I am planning to write up solutions, and I will announce them as they will get uploaded, so stay tuned.

Ohh.......sorry, Darij, for my mistake, I am losing my brain :blush: .

Thank you very much, Valentin.


THIS IS NICE:

LET a,b,c>0 then prove that:

$\frac{1}{1+2ab}+\frac{1}{1+2bc}+\frac{1}{1+2ac}+\frac{4}{9}(a+b+c)\geq7/3$

and this is my compose :D   maybe easy for you :D (not for my friend )

$b\sqrt{b}+a\sqrt{b}+c\sqrt{c}+b\sqrt{c}+a\sqrt{a}+c\sqrt{a}\geq\frac{3(a+b+c)+3}{2}$

with abc=1, a,b,c>0

For the second one,
power mean yields
\[
\begin{array}{l}
 \left( {\frac{{a^{\frac{3}{2}}  + b^{\frac{3}{2}}  + c^{\frac{3}{2}} }}{3}} \right)^{\frac{2}{3}}  \ge \frac{{a + b + c}}{3} \\
 a^{\frac{3}{2}}  + b^{\frac{3}{2}}  + c^{\frac{3}{2}}  \ge 3\left( {\frac{{a + b + c}}{3}} \right)^{\frac{3}{2}}  \\
 \end{array}
\]
amgm yields
\[
a\sqrt b  + b\sqrt c  + c\sqrt a  \ge 3
\]
Let $a + b + c = 3x^2$
it remains to prove
\[
\begin{array}{l}
 3x^3  + 3 \ge \frac{{9x^2  + 3}}{2} \\
  \Leftrightarrow 2x^3  - 3x^2  + 1 \ge 0 \\
  \Leftrightarrow (2x + 1)(x - 1)^2  \ge 0 \\
 \end{array}
\]
which is true.

for 1,
\[
\begin{array}{l}
 \frac{1}{{1 + 2ab}} + \frac{2}{9}a + \frac{2}{9}b \ge \frac{7}{9} \\
  \Leftrightarrow \frac{2}{9}a + \frac{2}{9}b + \frac{2}{9} \ge \frac{{2ab}}{{1 + 2ab}} \\
 \end{array}
\]
amgm on the denominator of r.h.s. yields
\[
\frac{{2ab}}{{1 + 2ab}} \le \frac{2}{3}\left( {ab} \right)^{\frac{1}{3}}  \le \frac{2}{9}(a + b + 1)
\]
which completes the proof.

dear siuhochung:

the second is not so hard
i have a nice solution:

the first maybe
only using $a+b+c\geq\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$

let $t=\sqrt{ab}$
and so on :D

for the second:

$lhs=\sqrt{b}(a+b)+\sqrt{c}(b+c)+\sqrt{a}(a+c)$

applying CAUCHY we have:

$lhs\geq2b\sqrt{a}+2c\sqrt{b}+2a\sqrt{c}$

so:

$2lhs\geq(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})+b\sqrt{a}+c\sqrt{b}+a\sqrt{c}$

with abc=1 then complete the proof :D  :D

for the firt:

just CAUCHY we have:

$a+b+c\geq\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$

we only prove that:

$\frac{1}{1+2ab}+4/9\sqrt{ab}\geq7/9$ :D  :D  :(  :lol:  :P

it 's true by let $t=\sqrt{ab}$           :lol:

and complete the proof :D


geometry
incenter
circumcircle
ratio
angle bisector
perpendicular bisector
geometry unsolved

Given the incenter, the foot of the altitude from the vertex $ A$ and the midpoint of the minor arc $ BC$. Find vertices of the triangle $ ABC$.

Hint : The angle bisector $ AI$ passes through the midpoint of arc $ BC$. This along with the foot of the perpendicular  uniquely determine the vertex $ A$

Ah... This hint is not sufficient for me (I was already trying to use it...).

Can you give another hint (or solve it)?

[b]befuddlers[/b], foot of the altitude is a point, not a line. How do you determine the vertex $ A$?

Hello sorry for not being clear. But according to the question the circumcircle of the triangle is already given if I am not mistaken. :wink:

So the line connecting the midpoint of arc $ BC$ and the incenter of the triangle $ I$ intersects the circumcircle of the triangle at a second unique point. This point is the vertex $ A$.

I is the incenter, M midpoint of the circumcircle minor arc BC and H foot of the A-altitude. Invert H into H' in a circle (M) with center M and radius MI. (M) is coaxal with H, H'. Perpendicular bisector of HH' cuts MI at P. Circle (P) with center P and radius PH = PH' is perpendicular to the circle (M). (P) cuts MI at A, D outside and inside of (M). D is foot of the internal bisector of angle <A. HD cuts (M) at B, C.

Yetti,

Could you explain how H' lies on the circumcircle of triangle ABC. Otherwise how is circle (M) coaxal with circle (O) .

Thanks in advance.

[quote="feliz"]Hm... Why don't you prove it works?[/quote]
Sorry, I thought that I gave sufficient reason for a proof. Anyway, if D is foot of the internal bisector MI of the angle <A, the angle <AHD is right and AD must be diameter of some circle centered on MI and passing through H. The question is what kind of a circle. Inversion in the circle (M) with center M and radius MI = MB = MC takes the line BC into the circumcircle (O) of the triangle ABC and (O) into the line BC. Therefore it takes A to D and D to A. Consequently, it takes the circle AHD into itself, so this circle must be perpendicular to the circle (M). The next question is how to construct such a circle.

[quote="befuddlers"]Could you explain how H' lies on the circumcircle of triangle ABC.[/quote]

If you invert H into H' in (M), since this inversion takes BC to (O) and H is on BC, then H' is on (O). Back to [b]feliz[/b]. By the basic property of inversion, $ \overline{MH} \cdot \overline{MH'} \equal{} MI^2.$ If K is the midpoint of HH' then power of K to (M) is
$ KM^2 \minus{} MI^2 \equal{} \frac {_1}{^4}(\overline{MH} \plus{} \overline{MH'})^2 \minus{} \overline{MH} \cdot \overline{MH'} \equal{}$ $ \frac {_1}{^4}(\overline{MH} \minus{} \overline{MH'})^2 \equal{} \frac {_1}{^4}{HH'}^2 \equal{} KH^2 \equal{} KH'^2$
This means that powers of K to the circle (M) and to points H, H' are equal, they are coaxal. All circles passing through H, H' form a pencil perpendicular to the pencil defined by H, H', (M). Any circle from first pencil is perpendiculer to any circle in the second one. This includes the circle (P) from the first pencil being perpendicular to the circle (M) in the second one. The reason is that inversion in a circle (H) with center H and radius HH' takes the first pencil into a pencil of lines concurrent at H' and the second one into a pencil of concentric circles with the common center H'. Any line from the first pencil is perpendicular to any concentric circle from the second one.

Of course there is a construction without using inversion. The A-excenter J is diametrically opposite point of the incenter I on the circle (M) with center M and radius MI. D in BC is foot of the internal bisector of the angle <A. Since the cross ratio (A, D, I, J) = -1 is harmonic and the angle <AHD is right, it follows that HD bisects the angle <IHJ internally and HA externally. The internal bisector of the angle <IHJ cuts (M) at B, C and the external bisector cuts MI at A. I know what you are going to ask next: Is there a proof without cross ratios ?


trigonometry
calculus
derivative
function
calculus computations

I feel like these questions are very easy, but I just don't know which way I should use the chain rule... hopefully you guys can help:

1. Find the first derivative:
h(x) = sec x^2

2. Find the first derivative:
g(t) = 5cos^2(pi * t)

[/list]

From the outside in. In assembling the function, what was the last thing you did? That's the first thing you take care of in integrating.

$h(x)=\sec^2(x).$ What's the last thing you did in evaluating it? You squared something. So that's the first thing you take care of. The derivative of (something$)^2$ is $2\cdot$(something).

So, $2\sec x\,\cdot$ (the derivative of what's inside).

That would be the derivative of $\sec x,$ which happens to be $\sec x\tan x.$

$h'(x)=2\sec x\sec x\tan x=2\sec^2x\tan x.$

Always from the outside in.

Sorry Kent, but I should have specified that the first problem was actually h(x) = sec (x^2); how would this change the solution?

be h(x)=secx^2
and
$\dfrac{d\sec x}{dx}=\sec x\tan x$

$\dfrac{dh}{dx}=\dfrac{d}{dx}\sec(x^2)=\sec(x^2)\tan(x^2)2x$

Be
$g(t)=5cos^2( \pi t)$

By the Power Rule combined with the Chain Rule
$\dfrac{dg(t)}{dt}=2.5cos(\pi t)\dfrac{dcos(\pi t)}{dt}$
By Chain Rule
$=-10cos(\pi t)sen(\pi t)\dfrac{d(\pi t)}{dt}$
By Procut Rule

$=-10cos(\pi t)sen(\pi t)[\pi \dfrac{dt}{dt}+t\dfrac{d \pi}{dt}]$

 we know that $\dfrac{d(C)}{dx}=0$ being C a constant

then
$\dfrac{dg}{dt}=-10 \pi cos(\pi t)sin(\pi t)$


For k=1,2,3.... consider the k-tuples $ (a_1,a_2,a_3........a_k)$ of positive integers such that
$ a_1\plus{}2.a_2\plus{}3.a_3\plus{}.......\plus{}k.a_k\equal{}1979$
Show that there are as many such k-tuples with odd k as there are with even k

This problem is equivalent to http://www.mathlinks.ro/viewtopic.php?t=312574 due to the identity:
\[ \sum_{k \equal{} 0}^{\infty} ( \minus{} 1)^k \prod_{j \equal{} 1}^k \frac {x^j}{1 \minus{} x^j} \equal{} \prod_{k \equal{} 1}^{\infty} (1 \minus{} x^k).\]


A circle with the center O is internally tangent to two circles inside it at points S and T. Suppose the two circles inside intersect at M and N with N closer to ST. Show that OM is perpendicular on MN if and only if S,N,T are collinear.

look at the following problem:

Let (W) a circle with center O. Let A,B,C,D be a convex
quadrilateral inscribed in (W). Let P be the intersection of AC end BD.
The circles circumscribed about A,B,P and D,C,P intersect at N and M.
We are going to prove that OM is perpendicular to MN.
===
AMD=AMP+NMD=ABD+ACD=2*ABD=AOD
so A,M,D,O are concyclic.
hence:
OMN=OMA+AMN=ODA+ABD=ODA+(1/2)*AOD=PI/2  because ODA isocele on O.

So i think that your problem is just the degenerate case if A=B
and C=D, isn't it ??

bye.


I am a bit confused, might be trivial.

So if a,b,c and d are the internal angles and Given, a>b and b=c and c>d

The max value of 'b' would be? What would be its minimum?

[hide="solution"]

i could think of 119 and 2, but don't have a concrete explanation.[/hide]

well do all angles have integer measures?...
if so
i think you are right
but heres why
[hide]
we have $ a > b \equal{} c > d$ and because they are internal angles of a quadrilateral, they add up to 360
maximixing b:
we know $ a > b$ so let $ a \equal{} b \plus{} x$ for a natural number value of x
then $ a \plus{} b \plus{} c \plus{} d \equal{} 360 \implies 3b \plus{} x \plus{} d \equal{} 360$
since both x and d must both be natural numbers, they add to at least 2. suppose this sum is y
so $ 3b \plus{} y \equal{} 360,y \ge 2 \implies 3b \le 358 \implies b < 119\frac13$ and since b is a natural number, b is 119
and of a quick check shows that one possible solution for $ b \equal{} 119$ is $ (a,b,c,d) \equal{} (120,119,119,2)$

minimizing b:
since $ b \equal{} c > d, b > d$ and both b and d are natural numbers, d is at least 1 and b is at least 2
we can easily check that the solution $ (a,b,c,d) \equal{} (355,2,2,1)$
noter- this is a concave quadrilateral
[/hide]

Wonderful.

I tried out this method for a pentagon to get the values of 134 and 2. Much easier instead of just playing around with numbers ,which i did the last time to get the answer :)

Thanks

but make sure a solution exists such that no angle is greater than 360 degrees
then the figure looks like a star or something

Not quite a star, but two of the sides may intersect in a place other than the vertices. (Note that, depending on the angles, it may end up looking like a normal quadrilateral after all, and the sides might not cross.)


[i]Let $k\circ m$ mean $k\geq m+2$.  Show that every positive integer $n$ has a [u]unique[/u] representation of the form $n=F_{k_{1}}+...+F_{k_{r}}$, where $F_{k_{i}}$ are Fibonacci numbers and $k_{1}\circ k_{2}\circ \hdots\circ k_{r}\circ 0$.[/i]

Masoud Zargar

strong induction?
Just build up the fibonacci sum like binary sums.


A quadrilateral $PQRS$ has an inscribed circle, the points of tangencies with sides $PQ$, $QR$, $RS$, $SP$ being $A$, $B$, $C$, $D$, respectively. Let the midpoints of $AB$, $BC$, $CD$, $DA$ be $E$, $F$, $G$, $H$, respectively. Prove that the angle between segments $PR$ and $QS$ is equal to the angle between segments $EG$ and $FH$.

[quote="k2c901_1"]A quadrilateral $PQRS$ has an inscribed circle, the points of tangencies with sides $PQ$, $QR$, $RS$, $SP$ being $A,B,C,D$, respectively. Let the midpoints of $AB$, $BC$, $CD$, $DA$ be $E,F,G,H$, respectively. Prove that the angle between segments $PR$ and $QS$ is equal to the angle between segments $EG$ and $FH$.[/quote]

We will work with directed angles modulo 180°.

Let I be the center of the incircle of the quadrilateral PQRS. Since the incircle touches the sides SP and PQ of the quadrilateral PQRS at the points D and A, these points D and A are symmetric to each other with respect to the line PI. Thus, $DA\perp PI$, and the midpoint H of the segment DA lies on the line PI. Similarly, $AB\perp QI$, $BC\perp RI$, $CD\perp SI$, and the midpoints E, F, G of the segments AB, BC, CD lie on the lines QI, RI, SI, respectively.

Since $DA\perp PI$, we have < IHD = 90°; equivalently, < IHD = - 90° (since 90° = - 90° as we are using directed angles modulo 180°). On the other hand, since the incircle of the quadrilateral PQRS, centered at I, touches the side SP at the point D, we have $ID\perp SP$ and thus < IDP = 90°. Hence, < IHD = - < IDP. Also, trivially, < HID = - < DIP. Thus, the triangles IHD and IDP are inversely similar. Therefore, $\frac{IH}{ID}=\frac{ID}{IP}$, or, equivalently, $IH\cdot IP=ID^2$. Similarly, $IF\cdot IR=IB^2$. But ID = IB, since the points D and B both lie on the incircle of the quadrilateral PQRS, which is centered at I; thus, we get $IH\cdot IP=IF\cdot IR$. Hence, by the converse of the intersecting chords theorem, it follows that the points H, P, F and R lie on one circle. Thus, < PRF = < PHF; equivalently, < (PR; RI) = < (PI; FH). Similarly, < (QS; SI) = < (QI; EG). On the other hand, since $BC\perp RI$ and $CD\perp SI$, we have < (RI; BC) = 90° and < (SI; CD) = 90°, so that

< (RI; SI) = < (RI; BC) + < (BC; CD) - < (SI; CD) = 90° + < (BC; CD) - 90°
= < (BC; CD) = < BCD.

But similarly, < (QI; PI) = < BAD, and since the points A, B, C, D lie on one circle (namely, on the incircle of the quadrilateral PQRS), we have < BCD = < BAD. Thus, < (RI; SI) = < (QI; PI). Hence,

  < (PR; QS) = < (PR; RI) + < (RI; SI) - < (QS; SI) = < (PI; FH) + < (QI; PI) - < (QI; EG)
                     = < (PI; FH) + < (EG; PI) = < (EG; FH).

And the problem is solved.

  darij


I'm having trouble reasoning a purely functional algorithm (no mutation!) to the n-queens problem. I'm programming it in scheme if any one is interested. Any help would be appreciated!

In pseudocode:

[code]main()
    initialize n*n board, count etc.
    call searchNext on the board
function searchNext(parameters)
    if array already has n queens
        display board, increment count, etc.
    else for each unchecked spot after the last placed queen
        add queen to this spot
        if board is still valid then call searchNext on this board
        delete queen on this spot
function checkValid(chessboard)
    for each row, column, and diagonal
        if queen count on said row, column, or diagonal > 1 then return FALSE
    return TRUE[/code]

I used this basic format on one of my programming challenges, and by only changing n, I could solve any similar question (4*4, 7*7, and the like)

Using java, NetBeans, it solves the 8 queens question in 1.969 seconds. 9 queens solves in 19.863 sec.

Hope this helps.


geometry
circumcircle
parallelogram
geometric transformation
reflection
Euler
vector

This is the fifth problem from OBM (Brazilian Mathematics Olympiad in 2001).

[quote="mathVNpro"]This problem is very classic. Here is my outline solution about this problem:

[b]LEMMA[/b]: Let ABCD be the cyclic quadrilateral, AC intersects BD by the point E. If M is the midpoint of AB then ME is perpendicular to CD.

[i]proof[/i]: Leave for the readers (This is very easy, just simple angle chasing then you will get the result :lol: )
Apply this lemma 4 times in the problem, we shall get the result.[/quote]

I don't think your lemma is correct...

$ \mbox{Let }\Omega \mbox{ the circumcenter of cyclic quadrilateral ABCD and } \{G\}\equal{}MO\cap NP,$
$ \mbox{then the perpendiculars from to M,  to CD, from N to DA, and from O to AB,}$
$ \mbox{and to P to BC are concurrent in the symetric of point } \Omega \mbox{ respect to G.}$

Oops :blush: . I am so sorry, I have thought that these diagnols of this ABCD are perpendicular to each other :D . Sorry for my big mistake.

[quote="hollandman"]Let $ ABCD$ be a cyclic quadrilateral, and $ M,N,O,P$ the midpoints of sides $ AB,BC,CD,DA$ respectively. Prove that the perpendiculars from $ M$ to $ CD$, from $ N$ to $ DA$, from $ O$ to $ AB$, and from $ P$ to $ BC$ are concurrent.[/quote]
Let I be the intersection of MO and PN. E be the circumcenter of ABCD, H be the symetric of E respect to I
Since $ MNOP$ is  a parallelogram, we get $ MI\equal{}OI$, moreover $ HI\equal{}EI$ then $ MHOE$ is a parallelogram
Hence $ MH \perp DC$. Similarly, $ NH \perp AD, OH\perp AB, PH\perp BC$

Let the perpendiculars from $M$ to $CD$ and $O$ to $AB$ meet at $U.$ If $K$ is the circumcenter of $ABCD,$ then $KM \perp AB$ and $KO \perp CD$ implies that $KMUO$ is a paralelogram $\Longrightarrow$ $U$ is the reflection of $K$ about the midpoint $G$ of $OM.$ But $G$ is also the midpoint of $NP.$ Hence, perpendiculars from $N$ to $DA$ and $P$ to $BC$ also pass through $U.$

[b]Remark:[/b] It's easy to see that U coincides with the Euler point of ABCD.

[quote="hollandman"]Let $ ABCD$ be a cyclic quadrilateral, and $ M,N,O,P$ the midpoints of sides $ AB,BC,CD,DA$ respectively. Prove that the perpendiculars from $ M$ to $ CD$, from $ N$ to $ DA$, from $ O$ to $ AB$, and from $ P$ to $ BC$ are concurrent.[/quote]

1) The point of concurrency, the mid points of opposite sides and the circumcentre form a parallelogram.
Let the circumcentre be $ G$ and the point of concurrency be $ F$. Since we can drop perpendiculars from $ G$ to $ AB$, $ CD$ at their midpoints $ M$ and $ O$ respectively, $ MGOF$ is a parallelogram. Similarly, $ NGPF$ is also a parallelogram.

2) Moving on to the Argand plane, with circumcentre as the origin, denote $ z_1, z_2, z_3, z_4$ as the vertices of $ ABCD$. The point of concurrency is nothing but the vector sum of the midpoints of opposite sides.

So, $ F \equal{} \frac {z_1 \plus{} z_3}{2} \plus{} \frac {z_2 \plus{} z_4}{2}$ which is the same point as $ \frac {z_1 \plus{} z_4}{2} \plus{} \frac {z_2 \plus{} z_3}{2}$. Thus, we arrive at the same point for each pair of opposite sides.

I think this proof can be generalized to a cyclic $ n$-gon, $ n$ being even, if I'm not mistaken.

Here is my solution:
Denote $ M_1$, $ N_1$, $ O_1$, $ P_1$ the projections of $ M$, $ N$, $ O$, $ P$ onto $ CD$, $ DA$, $ AB$, $ BC$, respectively. Now, It is easy to see that $ MNOP$ is parallelogram. Let $ I$ be the intersection of its diagonals, $ K$ be the circumcenter of $ ABCD$.
The symmetry through point $ I$ maps $ K$ into $ G$, maps $ M$ into $ O$, hence, the line $ KM$ is maped in to the line $ OG$. But KM is perpendicular to $ AB$, therefore, $ OG$ is also perpendicular to $ AB$.
With the same argument, we also have $ PG$, $ MG$, $ NG$, respectively, is perpendicular to $ BC$, $ CD$, $ AD$.
Then we get the result of this problem.
Our proof is completed. :lol:

Dear Mathlinkers,
can some one explain why the meetpoint is called "anticenter".
Who gives this name?
Any reference after Honsberger?
Sincerely
Jean-Louis