Find all positive integers $ k$ such that for any positive numbers $ a, b$ and $ c$ satisfying the inequality

\[ k(ab \plus{} bc \plus{} ca) > 5(a^2 \plus{} b^2 \plus{} c^2),\]

there must exist a triangle with $ a, b$ and $ c$ as the length of its three sides respectively.

[hide]If $ k=5$, we have $ ab+bc+bc>a^2+b^2+c^2$, which is equivalent to $ (a-b)^2+(b-c)^2+(c-a)^2<0$, which is wrong.

If $ k=7$, we can take $ a=1,b=1,c=2$. The inequality is true, but $ (1,1,2)$ cannot be the lengths of a triangle.

Now assume $ 6(ab+bc+ca)>5(a^2+b^2+c^2)$. Let $ a\le b\le c$. It is sufficient to prove that $ a+b>c$. The inequality is equivalent to $ 5c^2-6(a+b)c+5a^2+5b^2-6ab<0$. Hence
\begin{align*}
c&<\frac{6(a+b)+\sqrt{(6(a+b))^2-4\cdot 5(5a^2+5b^2-6ab)}}{10}
\\&=\frac{6(a+b)+\sqrt{64ab-64(a-b)^2}}{10}
\\&\le\frac{6(a+b)+\sqrt{64ab}}{10}
\\&\le\frac{6(a+b)+\sqrt{64\left(\frac{a+b}2\right)^2}}{10}
\\&=\frac{6(a+b)+4(a+b)}{10}
\\&=a+b,\end{align*}
as desired. Hence the only possible value for $ k$ is $ 6$.[/hide]

Alternatively, completing squares, we get $ 5a^2 \plus{} 5b^2 \plus{} 5c^2 < 6ab \plus{} 6ac \plus{} 6bc \Longleftrightarrow (5a \minus{} 3b \minus{} 3c)^2 \plus{} 16(b \minus{} c)^2 < 16bc$. Now suppose $ a \ge b \plus{} c$. We have $ 16bc > (5b \plus{} 5c \minus{} 3b \minus{} 3c)^2 \plus{} 16(b \minus{} c)^2 \equal{} 4(5b^2 \plus{} 5c^2 \minus{} 6bc)$, which gives $ 10bc > 5b^2 \plus{} 5c^2$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=339662&hilit=Albanian+inequality]Albanian BMO TST 2010:[/url]
 inequality $ a^3\plus{}b^3\plus{}c^3<k(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$ where $ a,b,c$ are the sides of a triangle and $ k$ a real number.
Find the smallest value of $ k$ such that the inequality holds for all triangles.

[quote=fighter]k = 10 to find it prove that, 2(ab + bc + ca) > a^2 + b^2 + c^2[/quote]

err... I think jgnr's right. Triangle inequality should be invoked somehow.

The condition that $a$,$b$,$c$ are sides of triangle is equivalent to the condition $a=x+y$ , $b=y+z$ and $c=z+x$ for some positive real numbers  $x$, $y$ and $z$.

Therefore, we just need to find all $k$ such that ,

$$ k\sum_{cyc} (x+y)(y+z) \geq \sum_{cyc} 5(x+y)^2$$

$$\Leftrightarrow k\sum_{cyc} x^2 + 3k\sum_{cyc} xy \geq 10 \sum_{cyc} x^2 + 10 \sum_{cyc} xy $$

For $ 3 \geq k$, inequality becomes

$$ 0 > (10-k)\sum_{cyc} x^2 + (10-3k)\sum_{cyc} xy$$

which is obviously false for all positive $x,y,z$.

For $10 > k \geq 4$, inequality becomes

$$ (3k-10)\sum_{cyc} xy \geq (10-k)\sum_{cyc} x^2$$

which is false for $x \to \infty$, $y=0$ and $z = 0$ as LHS $=0$ and RHS $\to \infty$

For $ k \geq 10$, the inequality ecomes

$$ (k-10)\sum_{cyc} x^2 + (3k-10)\sum_{cyc} xy \geq 0$$

which is true for all positive real numbers $x$,$y$ and $z$.

Therefore, for all postive integers $k \geq 10$ the inequality holds.

it is better to use the fact that \[ (ab + bc + ca) > =(a^2 + b^2 + c^2),\]
\[ 5(ab + bc + ca) > 5(a^2 + b^2 + c^2),\]
for the elimination of equality
k>5  minimum value of k such that it is an integer and is greater than 5 is 6
k>6 such that k is an integer and \[ k(ab + bc + ca) > 5(a^2 + b^2 + c^2),\]


geometry
inequalities
trigonometry
geometric transformation
reflection

I actually solved this one! (I think and hope so):

"Let Q be a quadrilateral whose side lengths are a,b,c,d, in that order.  Show that the area of Q does not exceed (ac+bd)/2."

Enjoy!

-interesting_move

I think that this was problem number 4 on the Austrian-Polish Olympiad from 2001...it looks very familiar.

Anyway, the solution I came up with involves <spoiler>kites and maximizing the area</spoiler>. I don't remember it now cuz I solved it a couple of months ago, but I might post a solution later. It's pretty cool once you figure out what's going on. I think it's also found in the book "Problem Solving Strategies" along with another very similar problem.

Something to think about (in other words, an easy addition to the problem): when does equality hold?

The equality holds when the quadrilateral is square...need I prove it?  8-)

-interesting_move

The ac+bd reminded me of Ptolemy's Inequality though. Perhaps somehow proving that half the product of the diagonals >= the area?

Yes, but we must also prove that the greatest possible area can only occur when Q is cyclic.  The way I proved it was using the RMS-AM inequality, and then after working through some algebra, voila (of course, it took up an entire page with small print) :wink: .

-interesting_move

If I remember correctly, this problem can be solved by considering (ab+cd)/2 first and then comparing that to (ac+bd)/2.

That is probably a good way.  I used the RMS-AM inequality for some odd reason but still managed to prove it.  (Go Complexzeta!!! 1000+ posts!)

-interesting_move

interesting_move wrote:The way I proved it was using the RMS-AM inequality

You may be interested by another proof :[hide]

Let the quadrilaral be (A,B,C,D) with AB=a, BC=b, ...

. Let first prove that A <= (ab + cd)/2

A = area(ABC) + area(CDA)

area(ABC) = 1/2ab sin(ABC) so area(ABC) <= ab/2 with equality iff the triangle has right angle.

. Let consider the quadrilateral (A,B,C,D') AB=a, BC=b, CD'=d, D'A=c (i.e. obtained by switching c and d). This quadrilateral has same area as (A,B,C,D) and by previous result area(A,B,C,D') <= (ac + bd)/2 [/hide]

What is your proof ?

Hm...well, my proof is this:

[hide]

(The picture to this proof goes something like this: Draw quadrilateral ABCD where a=AB, b=BC, c=CD, and d=AD.  Then draw BD such that triangle ABD is region I and triangle BDC is region II.  Finally, angle A is alpha and angle B is beta.)

(note: In the following proof, angles a and b are actually supposed to be alpha and beta, respectively.  Also, the terms may be a little bit confusing, but a term like a^2d^2sin(a)^2 means (a^2)(d^2)(sina)^2.)

Proof:

We will show that the maximum area of Q is (ac+bd)/2.

[Q]=(1/2)(adsina+bcsinb).

By the RMS (Root mean square)-AM inequality, we have:

[(a^2d^2sin(a)^2+b^2c^2sin(b)^2)/2]1/2≥(adsina+bcsinb)/2

Because we want to maximize the right hand side, we will set ≥ to = and have

(a^2d^2sin(a)^2+b^2c^2sin(b)^2)/2=(a^2d^2sin(a)^2+2abcd(sina)(sinb)+b^2c^2sin(b)^2)/4

2a^2d^2sin(a)^2+2b^2c^2sin(b)^2=a^2d^2sin(a)^2+2abcd(sina)(sinb)+b^2c^2sin(b)^2

a^2d^2sin(a)^2+ b^2c^2sin(b)^2=2abcd(sina)(sinb)

a^2d^2sin(a)^2-2abcd(sina)(sinb)+ b^2c^2sin(b)^2=0

(adsina-bcsinb)^2=0

adsina-bcsinb=0

adsina=bcsinb

[I]=(adsina)/2=(bcsinb)/2=[II]

So we conclude that the two triangles areas are the same.

I will now prove that the two said triangles are congruent isosceles right triangles. (Keep in mind that Im trying to maximize the areas.)

Since 3 points determine a circle, the maximum area of triangles ABD and BCD are when points A and C bisect the arcs passing through triangles DAB and BCD, respectively, and thus a=d and b=c.  So [I]=[II]=(1/2)adsina=(1/2)bcsinb where 0<a<pi and 0<b<pi.

Analyzing the sine graph, we see that the greatest value for sina=sinb is 1 and thus angle a=b=pi/2.

So the maximum area would be when the two triangles are congruent isosceles triangles.  We see finally that these two triangles are combined to form a square, which has area (ac+bd)/2.

In the case that ABCD is not convex, where, say, A is shifted to A such that A is a reflection of A about BD:

It is apparent that the quadrilateral ABCD has even less area than region II, which means that no non-convex quadrilateral could have the maximized area of Q.

[/hide]

I hope it makes sense,

-interesting_move

Ptolemy's inequality works !
This has nothing to do with the fact whether Q's area is maximized if Q is cyclic or not.

Let |AB| = a, |BC| = b, |CD| = d and |DA| = d. Then we have (ac + bd)/2 >= |AC||BD|/2 so it is indeed enough to show that the [Q] does not exceed half the product of its diagonals. This is just easy using the formula [XYZ] = |XY||XZ|sin(YZX)/2.

Yes, that works too...much shorter than mine.  :wink:

-interesting_move


I enrolled for intermediate trig and complex numbers course, but it doesn't show up in the classes section of the forum. What's going on?

You can see it now.

After you sign up for a class, we have to manually refresh the class list before students can see the class forum, which means sometimes it takes some time.


search
complex numbers
calculus
calculus computations

bring together:
[code]sina+sin2a+sin3a+....+sin(na)[/code]
I succeeded to bting together a similar sum with cosins.
I made it only with some trigonometric identities and without "high" math.
But in this sum i stuck....i think that perhaps "complex numbers" will help...
thanks for the comming assitnce.

You mite find the following links useful:
http://joigroup.freezoka.com/knowledge/kb_show_artikel.php?id=15
Also, more on Complex numbers could be found here:
http://joigroup.freezoka.com/knowledge/kb_show_artikel.php?id=16


[quote="N.T.TUAN"][quote="chien than"]For $a;b;c$ are positive real numbers and $S=a+b+c$.Prove that:
$(a+\frac{1}{a})(a+\frac{1}{b})(a+\frac{1}{c}) \geq (1+\frac{3}{S})^{3}$[/quote]
It must is

$(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) \geq (1+\frac{3}{S})^{3}$

I think so  :wink:[/quote]
I  also think so.
$(1+1/a)(1+1/b)(1+1/c)\geq (1+1/\sqrt[3]{abc})^{3}\geq (1+3/S)^{3}$
easy by AM-GM


induction
algebra proposed
algebra
combinatorics

Given a circle and $2006$ points lying on this circle. Albatross colors these $2006$ points in $17$ colors. After that, Frankinfueter joins some of the points by chords such that the endpoints of each chord have the same color and two different chords have no common points (not even a common endpoint). Hereby, Frankinfueter intends to draw as many chords as possible, while Albatross is trying to hinder him as much as he can. What is the maximal number of chords Frankinfueter will always be able to draw?

I think this is combinatorics, and hard one...

I hope I am right...my answer is 111
First of all notice that $111\cdot 18 \leq 2006$. Therefore we can form 111 groups, aech of 18 points. In each group we have 2 points of the same color. Therefore, we get 111 segments with monocolor endpoints.
Now, if Albatross colors the circle alternatively 1,2,3,4,5,...,17 and so on we notice that each segment with monocolor endpoints has length at least 18. Because $18\cdot 112\geq 2006$ we get that 111 is the desired result.

Well, I don't think you're right. Look at the example you gave. It's true that each segment has length at least $18$. But segments can "cover" each other, say, a segment of length $20$ may "cover" a segment of length $18$, thus the segment of length $20$ actually uses only two more points.

And then Frankinfueter can join the points like this: First join two consecutive $1$s. Then look at the $17$ next to one of the $1$s we have used. Then next to the other $1$ we have used, there will be numbers $2,3,...,17$. Join these two $17$s. So this second segment uses just $17$ more points, not $18$. We can continue like this, and we will get that all segments except the first one uses $17$ more points. The number of segments we can draw will thus be $[\frac{2006-18}{17}]+1=117$. Also, it's not hard to prove that $117$ is really the maximum Frankinfueter can draw in this particular case.

I guess that $117$ is actually the answer. So we need to prove that Frankinfueter can always draw at least $117$ segments in every case...

After a lot of thinking time, I haven't come up with the solution...

Anyway, why is this thread not moved to combinatorics, so that people can try it more?

Hmmmm... this is interesting problem. does anyone have the idea?

If there are c colors and nc points, we can always draw at most n-1 lines. To find a case where we can only draw n-1 lines, consider the points numbered 1, 2, ..., c, 1, 2, ..., c... clockwise. By induction on n (we add another 1,2,...,c to the circle in the inductive step), we can draw only n-1 lines.

We can also prove that we can always draw n. Since there are c colors, there are at least n of some color, 2 of which have at most n-1 points in between. Connect those, and remove them along with the points in between them. We have at least (n-1)c points left, so we're done by induction.

Since 2006 = 118(17), we can be guaranteed to draw at most 117 lines.

Here is, hopefully, full solution.
As in #7, we conclude that it is sufficient to prove that we can always draw $117$ chords.
We prove by strong induction on $n$ that, for any $c,$ given $nc$ points and $c$ colors we can always draw $n-1$ chords.
Base $n=2$ is trivial.
Now for step. It is trivial that there is a color (say, blue) that appears at least $n$ times. We distinguish 2 cases:
1. case: Color blue appears precisely $n$ times.
1.1 case: There are $2$ blue points such that there are $\leq c-2$ points between them. This is clear, just throw them all out and we are left with at least $(n-1)c$ points and $c$ colors, and by induction we are done.
1.2 case: For every $2$ adjacent blue points there are precisely $c-1$ points between them. Enumerate clockwise blue points from numbers $1$ to $n$ and draw chords $1->2, 3->4...$ and so on. We have, for now $\left \lfloor{\frac{n}{2}}\right \rfloor $ chords and by induction we can draw $\left \lceil{\frac{n}{2}}\right \rceil -1$ more (consider all the points that are not between two blue ends of some chord, there are $\left \lceil{\frac{n}{2}}\right \rceil  (c-1)$ of them). So this case is over too.
2. case Color blue appears at least $n+1$ times. Hence, surely, there are $2$ blue points with at most $c-2$ points between them and now we proceed as in 1.1.

we wish good luck to malay mahajan for his physics practicals!?

[b]S[/b]olved with [b]mxlcv[/b]

[hide = solution]
We show that the answer is $117$. For the construction, consider the regular $2006$-gon and color the point $P_i$ with $i \bmod 17$ for all $1 \leq i \leq 2006$. Suppose we connect $P_1$ and $P_{17k+1}$, without loss of generality. Inducting down on the two regions formed after drawing this chord gives the result. Now let us show that for any coloring, we can find $117$ such chords. Again we induct. By the piegon hole principle, there are two points with the same color at most $17$ apart in indices. If they are $16$ apart, we can induct down normally, otherwise observe that the coloring is identical to the one highlighted earlier for the construction. Thus we proceed in the same way as there and finish.
[/hide]


Let $a,b,c\geq 0$ and $a+b+c=1$
find the maximum of $P_{n}(a,b,c)= \frac{(ab)^{n}}{1-ab}+\frac{(bc)^{n}}{1-bc}+\frac{(ca)^{n}}{1-ca}$ with $n \in N$ and $n\geq 2$

It's in Old and new Inequalities, proposed by harazi (if I'm not wrong).

May be ! But I think it very nice .Thankyou very much! :)


Let $x$ and $y$ are integers ; $0 \leq x <165$ , $0 \leq y <165$ . If   $y^{2}$ $\equiv$ $x^{3}+x$ $mod (165)$  then find number of $(x,y)$ pairs .

There are 51 solutions. For all $5|y$. If y=0 we have 9 solutions for x.

What if you 3 post the solution, not only the answer?

libra_gold,  can you send your solution ?


This is a very personal question, I know, so if you don't want to answer thats OK.

Wow a topic almost word for word from [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=171615&sid=783193d0e473f1e563087a33d259c646]this[/url]. Locked.


[url]http://www.deviantart.com/view/14080073/[/url]

This is COOL.

Come ON!!! This is a cool art program!! P.S. It works best on firefox.

umm what is with your source? can it make logos?

Well, if you expirement, you cna make the star in the Isreali flag.

graaaaaaargh
I can't believe I played with that thing for 10 minutes :ninja:

I did not make this. My mom found it.

P.S. Try pushing the buttons. :D

WHEEE!!!!!

I played with the last buttons with 16 thingys.


Find $ x \plus{} \sqrt{x \plus{} 1 \plus{} \sqrt{x \plus{} 2 \plus{} \sqrt{x \plus{} 3 \plus{} \sqrt{x \plus{} 4 \plus{} \ldots}}}}$ in terms of $ x$.

Let this expression be $ f(x)$.  Then $ (f(x)\minus{}x)^2 \equal{} f(x\plus{}1)$

stuck here

I would suggest induction of some sort, but we need to find some values of $ f(x)$ first...

hmmm... I suspect that limits are needed... this problem should probably be moved to the calculus form.

Taking another crack:
Let $ g(x) \equal{} f(x)\minus{}x$.  Then we have $ g(x) \equal{} \sqrt{g(x\plus{}1) \plus{} x \plus{} 1}$... still stuck.

Like I said: Find something similar to $ f(1)$.

See http://mathworld.wolfram.com/NestedRadicalConstant.html.

If the nested radical constant is $ C$, then $ f(1) \equal{} C^2$.

There is NO CLOSED FORM EXPRESSION for $ f(x)$.


topology
real analysis
real analysis theorems

Reading about Radon measures i came across the following questions.

What is the general motivation of Radon measures ? What they are useful for ? What was the motivation for inventing such a notion.

I thought that the ideas was to look at the properties of the measures derived from the Riesz representation theorem about functionals on $ C(X)$ with $ X$ locally compact, and somehow state these  properties as a definition of Radon measure and that was one of the motivations. But this is my understanding and i would be glad to hear the opinion of more experienced people.

Thanks.

Not that I know anything about topology, but have you consulted http://en.wikipedia.org/wiki/Radon_measure#Motivation ?


Can anyone tell me what the thing on the back of the GT t-shirt is? I should probably know this :oops:.

Yes, it is a fractal. Prof. Wang decided to place it there. I had prefer some other design, but I lost 3 votes to 1.

[quote="solafidefarms"]I think, as I can't look at the back right now as I'm wearing it, that it is a fractal of sorts.[/quote]

 :|  :rotfl:  :rotfl:


Sandy is thinking of a two-digit perfect square, $ x$. The sum of the digits of $ x$ is also a perfect square. What is the sum of the possible values of $ x$?


for positive reals $ a,b$
prove that
$ (1 \plus{} \frac {a}{b})^m \plus{} (1 \plus{} \frac {b}{a})^m \ge\ 2^m*2$


Given any set of 5 integers, show that you may pick 3 of them whose sum is divisible by 3.  Please help, I've spent 4 hours thinking about this and still can't prove it.

Notice first that if any three of the five leave the same remainder when divided by 3, their sum will be divisible by 3. So our only case now is if no three of the five leave the same remainder when divided by 3, which easily implies that there must be a number, when divided by 3, with no remainder, with a remainder 1, and with a remainder 2. The sum of these three numbers are divisible by 3.

(Edit: I didn't see warut_suk's post before submitting.)

[hide="Hint"]Do you know about modular arithmetic (mods)?  Every integer is equal to 0, 1, or 2 mod 3.  (Essentially that means you can assume your integers are all 0, 1, or 2.)  You can then divide the problem into two cases.  (Can you show that these two cases cover all possibilities?)

Case 1: Three of the integers are 0, 1, and 2 mod 3, respectively.  (In other words, the three are all different mod 3.)

Case 2: Three of the integers are the same mod 3.
[/hide]
There is a more general result known:  Let n be a positive integer.  Given 2n - 1 integers, we can choose n of them whose sum is divisible by n.  That result is due to Erdos, Ginzburg, and Ziv, and is pretty tough to prove.  I gave a sketch of a proof in a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10996]topic on subsets[/url].

the numbers all have to have a remainder of 0,1,and 2

so all you have to do is add the remainders and if it is 0 or 3 then it works

however if it is a set of 4, there could be 2 numbers witha remainder of 0 and 2 with remainder of 1 and it wouldn't work

I'm not to good at explaining but I hope this helps ;)

Every number leaves a remainder of 0,1, or 2 when divided by three.  In order to get the sum of three numbers to be leave 0 remainder when divided by 3, you have to three numbers with remainder of:
1,1,1
0,0,0
0,2,1
or
2,2,2
(the order doesn't matter).  If there are only 2 different remainders among the set (for instance every number leaves either 0 or 1 when divided by 3), then you will have a set of three that have the same remainder, and thus the sum of these three will be divisible by 3 as above.

Otherwise at least one number from the set will leave remainder of each 0,1,2.  In this instance the sum of these three numbers will be divisible by 3.

Thus there will always be three numbers whose sum is divisble by 3.


It's a combinatorial game of removing chips from a pile of chips. Here are the rules of the game :

1. There are two players.
2. There is a pile of , say 21, chips on a table.
3. A move consists of removing one, two or three(these are the only possibilities) chips from the pile. At least one chip must be removed. But the most important fact is that different numbers of chips must be removed in consecutive moves. For instance, if player A removes 2 chips in his/her move, player B must remove 1 chip or 3 chips in his/her next move.
4. Players alternate moves(A,B,A,B,...).
5. The player thet removes the last chip wins(If you can't move in any case, you simply lose!).

The question is : Can one of the players win this game by using a good strategy? What is that good strategy?

I'll be waiting for your valuable comments.

Just remove chips so that the remaining chips is a multiple of n+1 where n is the maximum you can take away in one move.

so for your example, just remove chips so that it makes a multiple of 4 each time, 20, then 16, 12, 8, 4, then you win

Say you remove some number of chips so it's a multiple of 4 and then they take away 2.  If you re-read the question, you'll note that you aren't allowed to take away 2 at this point, so you can't "restore order" the way you normally can in such games.  The strategy will have to be more subtle.

You're right, but fortunately only a minor variant in strategy is needed.  As long as the number of starting chips is not a multiple of 4, the first player can win.

[hide="Strategy"]If we can remove chips to leave a multiple of 4, then do so.  Otherwise, remove any legal number of chips.[/hide]

By following this strategy, we can always reduce to either the next lower multiple of 4, or the multiple of 4 below that.

I used a spreadsheet to keep track of the small winning positions, and then figured out the pattern.

But then how do you know if you can get to the "next set of 4"?

Ex. bold= you're move according to your strategy
21-[b]20[/b]-19-[b]16[/b]-15-[b]12[/b]-11-[b]8[/b]-7-[b]4[/b]-[i]2[/i]-[b]you are screwed[/b]

So you this strategy won't work here.  :roll:

sorry i misread the rules. :P

well for your scenario, if other guy takes away 2, then it doesnt matter if you take away 1 or 3, they other guy cannot take away what you took away thus never making 2+3+3=8 or 2+1+1=4. Assuming you start first and take away 1, you'll always win.

No, you aren't screwed at that point: you move to 1, and your opponent loses.

yea wthell, you move to one, then what? stalemate? :?

Rule 5 : If you can't move in any case, you simply lose. Suppose that you have removed a chip from two and one chip is left. Your opponent can't remove the last piece by rule 3, and lose the game by rule 5. This problem is much harder than it seems to be and requires some kind of an ingenuity.

If you take 1 or 3 on your first move, note that on each of your opponent's moves he is forced to take 2 chips. This makes finding a winning strategy quite easy.

If the starting pile is not a multiple of three, you win easily by going first and taking 1 each time, since your opponent is forced to take 2 on each of his moves. The position is always the same mod 3.

If the starting pile is not a multiple of five, you win easily by going first and taking 3 each time (except at the end, when you might take 1). Since your opponent is always forced to take 2 on each of his moves, the total # of chips will stay the same mod 5, and since you always have the option to take only 1 chip you will never be screwed, unless the initial position was 0 mod 5 to begin with.

If the starting position is a multiple of 15, assuming it's not 0 (in which case you're screwed before the game even begins!) you can remove 1 chip from it, and after your opponent removes 2 chips it is no longer a multiple of 5 and you can use the strategy outlined in the third paragraph above.


the base of a triangle is 15 cm. Two line segments are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into 3 equal areas. Find the lengths of the parallel segments.

any hints?
this is my partial solution:
[hide]
let the height of the triangle be $ h$ and $ a$ be the height of the topmost triangle (when divided into 3 equal parts), $ b$ be the height of the 1st trapezoid and $ h\minus{}(a\plus{}b)$ be the height of the trapezoid with base 15. Also, let $ x$ be the base of the first triangle and $ y$ be the base of the trapezoid.

So, \[ ax\equal{}5h\], \[ b(x\plus{}y)\equal{}5h\], and \[ (y\plus{}15)(h\minus{}(a\plus{}b))\equal{}5h\]...im stuck
[/hide]

Not sure if this makes sense

[hide]Let the area of the triangle be A.

Let the length of the 2 other line segments be $ a$ and $ b$, with $ a > b$.

The area of the 2nd largest triangle would be $ A\left(\frac {a}{15}\right)^2$

The area of the smallest triangle would be $ A\left(\frac {b}{15}\right)^2$

We can use the fact that

$ 2A\left(\frac {b}{15}\right)^2 \equal{} A\left(\frac {a}{15}\right)^2$

So,

$ a \equal{} b\sqrt {2}$

Using the same reasoning, we see that,

$ 15 \equal{} b\sqrt {3}$

$ b \equal{} 5\sqrt{3}$ and $ a \equal{} 5\sqrt{6}$[/hide]

[hide="Hmmm, so..."]
The top region is similar to the whole triangle, so if its area is $ \frac 13$ of the big area, then its height (and consequently its base) is $ \frac 1{\sqrt 3} \equal{} \frac {\sqrt 3}3$ of the big triangle, so segment $ a$ is $ 5\sqrt 3$ inches long.

Then, since the second-largest triangle with base $ b$ is $ \frac 23$ the area of the largest triangle, its base must be $ \frac {\sqrt 6}3 \cdot 15 \equal{} 5\sqrt 6$ inches long.[/hide]

thanks to you both! i understood it thanks

[quote="Jakey"] [color=darkred]Let $ ABC$ be a triangle. Consider the points $ \{M,P\}\subset (AB)$ , $ \{N,Q\}\subset (AC)$  so that $ M\in (AP)$ ,

$ MN\parallel PQ\parallel BC$ and $ [AMN] \equal{} [MNQP] \equal{} [PQCB]$ . Find the $ MN$ , $ PQ$ in function of $ a \equal{} BC$ .[/color] [/quote]

[u][b]Proof.[/u][/b] Denote $ MN \equal{} x$ , $ PQ \equal{} y$ .Observe that $ \left\|\begin{array}{c} \frac {[AMN]}{x^2} \equal{} \frac {[APQ]}{y^2} \equal{} \frac {[ABC]}{a^2} \\
 \\
\frac {[AMN]}{1} \equal{} \frac {[APQ]}{2} \equal{} \frac {[ABC]}{3}\end{array}\right\|$ $ \implies$ $ \frac {x^2}{1} \equal{} \frac {y^2}{2} \equal{} \frac {a^2}{3}$ $ \implies$ $ \boxed {\ \frac x1 \equal{} \frac {y}{\sqrt 2} \equal{} \frac {a}{\sqrt 3}\ }$ .

[b]Remark.[/b] $ x^2 \plus{} y^2 \equal{} a^2$ , i.e. $ x$ , $ y$ , $ a$ are the lengths of the sides of a right triangle $ T(x,y,a)\ \sim\ \Delta\ (1,\sqrt 2,\sqrt 3)$ !


limit
integration
trigonometry
calculus
calculus computations

given $y(0)=2$, $\lim_{t\rightarrow\infty}y(t)=0,\lim_{t\rightarrow\infty}y'(t)=0$

find a solution of $y''+3y^{2}-4y=0$

Mathematica Error

Solve::tdep: The equations appear to involve the variables to be solved for in an essentially non-algebraic way.

The equation is too much difficult, and to complicate more, you put initial conditions!? uhauhuahu

[url=http://eqworld.ipmnet.ru/]EqWorld[/url] describes a lot of DE techniques. Your equation is solved in [url=http://eqworld.ipmnet.ru/en/solutions/ode/ode0301.pdf]this file[/url]

could one use this formula to find a closed form for the motion of a real pendulum?

You can try, but I don't think that $\int \frac{dy}{\sqrt{\cos y}}$ has a closed form, let alone its inverse.


inequalities
quadratics
algebra
inequalities proposed

Prove that if $ a,b,c\geq0$ then
\[ \sqrt{\frac{a^2\plus{}b^2\plus{}c^2}3}\leq\frac{a\plus{}b\plus{}c\plus{}(\sqrt3\minus{}1)\sqrt{a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca}}3\leq \max(a,b,c).\]

putting $ x\equal{}a\plus{}b\plus{}c, y\equal{}ab\plus{}bc\plus{}ca$, simplifying(pure computation) kills it easilly..for the left side.

and for the right side, again computation gives equivalent inequality:

$ \sqrt 3 (b\minus{}c)^2 \leq (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2$, and using $ x^2\plus{}y^2 \leq (x\plus{}y)^2$ for $ a\minus{}b, c\minus{}a$,

gives the result.

Yes, that is right. WLOG we may suppose that $ a \plus{} b \plus{} c\equal{}1$ and let $ ab \plus{} bc \plus{} ca \equal{} q$, then we transform the desired inequality into a quadratic equation of $ t$.


algebra
function
domain
integration
calculus
calculus computations

I know that the Riemann-zeta function's domain is $x\geq{1}$ because of the divergence of values less than that. But how do you sum the powers less than one when al alternating factor makes them convergent?

For example

$ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n}}{\sqrt{n}}$

What I tried doesn't work considering the Riemann-zeta function does not extend to these values. Can someone show me how to find this and other series of the form

$ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^n}{n^{\frac{1}{p}}}\quad\text{where }0<p\leq{1}$

The Riemann zeta function's domain is actually quite a bit larger than that thanks to the magic of analytic continuation.  What you want is the [url=http://mathworld.wolfram.com/DirichletEtaFunction.html]Dirichlet eta function[/url].

Thank you very much for your reply. I was unaware that they extended that far. What gave me pause for thought in the usage of $ zeta\left(\frac{1}{2}\right)$ was when I applied the Eta Function (which I was aware of) I saw that

$ \sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^{n\plus{}1}}{\sqrt{n}}\equal{}\left(1\minus{}2^{1\minus{}\frac{1}{2}}\right)\zeta\left(\frac{1}{2}\right)$

And I did not know a closed form for $ \zeta\left(\frac{1}{2}\right)$ so I resorted to that

$ \zeta\left(\frac{1}{2}\right)\equal{}\frac{1}{\Gamma\left(\frac{1}{2}\right)}\int_0^{\infty}\frac{u^{\frac{1}{2}\minus{}1}}{e^u\minus{}1}du$

$ \equal{}\frac{1}{\sqrt{\pi}}\int_0^{\infty}\frac{du}{\sqrt{u}\left(e^u\minus{}1\right)}$

And I saw that there was an infinite discontinuity that is not solved by making the lower bound a limit. Did I do something wrong? is there a closed form value for $ \zeta\left(\frac{1}{2}\right)$

Thank you very much for your replies.

The integral representation of $ \zeta$ you are using is only valid for real $ x>1$.

Yes...that is what led me to believe that the Riemann-zeta function is only well defined on the numbers greater than one, I had not ever seen a restriction on its usage. Ok, so is there a closed form value for $ \zeta\left(\frac{1}{2}\right)$?


geometry
rectangle
circumcircle
parallelogram
cyclic quadrilateral
geometry unsolved

Let $C$ be a circle with radius $R$ and centre $O$, and $S$ a fixed point in the interior of $C$.  Let $AA'$ and $BB'$ be perpendicular chords through $S$. Consider the rectangles $SAMB$, $SBN'A'$, $SA'M'B'$, and $SB'NA$.  Find the set of all points $M$, $N'$, $M'$, and $N$ when $A$ moves around the whole circle.

[b]1st solution[/b]: The quadrilateral ABA'B' is cyclic with perpendicular diagonals, hence, its anticenter (concurency point of its maltitudes) is identical with its diagonal intersection S. Let P, Q, P', Q' be the midpoints of the sides AB, BA', A'B', B'A. The centroid G of the cyclic quadrilateral ABA'B' (the intersection of lines PP', QQ' connecting the midpoints of the opposite sides) is the midpoint between its circumcenter O and anticenter S. Since PQP'Q' is a parallelogram, its diagonals PP', QQ' cut each other in half at G. Thus the diagonals OS, PP' of the quadrilateral OPSP' also cut each other in half and this quadrilateral is also a parallelogram. Using the parallelogram rule,

$PP'^2 + OS^2 = 2(OP^2 + SP^2)$

$GP^2 + GS^2 = \frac{OP^2 + SP^2}{2}$

The segment $SP$ is the median of the right angle triangle $\triangle ABS$, hence, $AP = BP = SP$. Since P is the midpoint of the chord AB of the circle (O), $OP \perp AB$ and

$OP^2 + SP^2 = OP^2 + AP^2 = OA^2 = R^2$

where R is radius of the circle (O). Consequently, the distance GP is given by

$GP^2 = \frac{r^2}{2} - GS^2$

and similarly,

$GP'^2 = GQ^2 = GQ'^2 = \frac{r^2}{2} - GS^2$

Thus the midpoints P, P', Q', Q' all lie on the circle (P) with center G and radius independently on the direction of the perpendicular lines $AA' \perp BB$ meeting at S. The distance MM' between the vertices M, M' of the rectangles SABM, SA'B'M' is twice the distance PP' between their diagonal intersections P, P' and since the midpoint of PP' is G, the midpoint of MM' is on the line GS twice the distance GS from the common vertex S of the 2 rectangles, i.e., the midpoint of MM' is O, the center of the circle O. As a result, the locus of M, M' (and similarly,the locus of N, N') is a circle comcentric with the circle (O) and with radius

$r = 2 GP = 2 \sqrt{\frac{R^2}{2} - GS^2} = \sqrt{2R^2 - OS^2}$

[b]2nd solution[/b] (similar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61023]All the positions compatible with our problem[/url]):

$(\vec {OS} + \vec{SA})^2 = R^2,\ (\vec {OS} + \vec{SB})^2 = R^2,\ \vec {SA} \cdot \vec{SB} = 0$

$OM^2 = (\vec {OS} + \vec{SA} + \vec{SB})^2 = (\vec {OS} + \vec{SA})^2 + 2(\vec {OS} + \vec{SA}) \cdot \vec{SB} + SB^2 =$

$= R^2 - OS^2 + OS^2 + 2\vec {OS} \cdot \vec{SB} + SB^2 = R^2 - OS^2 + (\vec {OS} + \vec{SB})^2 =$

$= 2R^2 - OS^2$

which is constant and independent on the position of the point A on the circle (O).

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1995Problem4

Vo Duc Dien


Solve the undernoted system of differential equation:

dy/dx + (y^2) = x^6 - 11*(x^2)

It's a Riccati ODE.
Maple gives the particular solution y0 of the ODE s.t. y0(0)=0:
y0(x)=x^3(-2*x^4+11)/(2*x^4-3). In fact the solution y0 is defined on each of the 3 intervals separated by the roots of 2*x^4-3.
Now take y(x)=y0+1/z(x). The ODE in z is linear.....
(Remark:It is easy to obtain an antiderative of y0).


Function $f$ is differentiable on $(a,b)$ and $f(a)=f(b) =0$
let $x_0$ betwen $(a,b)$. prove there exists $c$ such that : $f(x_0)=(x_0 -a)(x_0 -b) \frac{f''(c)}{2}$ .

Sorry , moderators ! Me precedent post is a mistake . Id like delete it !
 :cool:

[quote="alekk"]I think that $f$ must be $C^2$ ?
Anyway, if $f \in C^n$ and $x_1<x_2<..<x_n$ are arbitrary real, $P_n$ the Lagrange interpolation of $f$ that goes through the points $(x_i,f(x_i)$, then for all $x \in [x_1; x_n]$ there exists $c_x \in [x_1; x_n]$ such that :
$f(x)-P_n(x)=\frac{(x-x_1)(x-x_2)..(x-x_n)}{n!}f^{(n)}(c_x)$[/quote]
As for the proof of the fact that Alekk stated, denote $w(x)=(x-x_1)(x-x_2)...(x-x_n)$ and choose $\lambda$ in order to the function $\phi = f - P_n - \lambda w$ have a zero at $x$, i.e. take $\lambda=\frac{f(x)-P_n(x)}{w(x)}$. Then by Rolle's theorem the function $\phi(x)$ having zeros at $(n+1)$ zeros would satisfy the following property: there exists $c_x \in [x_1,x_n]$ such that $\phi^{(n)}(c_x)=0$, but $\phi^{(n)}(x)=f^{(n)}(x)-\lambda n!$ from where the conclusion follows.

What about is going conversation? Can some give explanation for tchabichev function?
Abdurashid


Source: 1973 AHSME #31

[hide]...[/hide]YE

x ME

[hide].[/hide]TTT

(YE*ME=TTT)

Find E + M + T + Y.

[edited] Aie.  I feel stupid now.  That should be three Ts, not two.

Fierytycoon wrote:And are they non-zero?
[hide]Y or M has to be 0 since the product is less than or equal to 99, so[/hide]
MysticTerminator wrote:are all letters distinct digits?

[hide]No. You get ME=T and EE=T if you suppose Y=0 and YE=T and EE=T if you suppose M=0, so you either get M=T or Y=T[/hide]

Therefore, [hide]Y=0, M=1, E=1, and T=1; Y=0, M=2, E=2, and T=4; Y=0, M=3, E=3, and T=9; Y=1, M=0, E=1, and T=1; Y=2, M=0, E=2, and T=4; or Y=3, M=0, E=3, and T=9

so the answer should be 3, 8, or 15[/hide]

I'm probably wrong, oh well.

What if E*E>9?

This could lead to a carry over.

Think of it this way. The product is of the form 11*T, where T is a digit from 0 to 9. This means one of the factors must be a multiple of 11. However all 2-digit multiples of 11 have the same digit in the first and second places, and in this sort of problem, usually different letters stand for different digits. So...?

Who knows.  The problem statement and the answer choices are the only things I know, and I think you can do without the latter.  But if it helps, here they are (spoiler):

[hide](A) 19

(B) 20

(C) 21

(D) 22

(E) 24[/hide]

okay now, [hide]note that the product is a multiple of 111. Since 111=3*37, one of the factors must be a multiple of 37, i.e. either 37 or 74. Okay, if one of them is 37, we know the last digit of the other one must also be 7, and thus we know the last digit of the product is 9. Since all digits in the product are the same, the product is 999. 999/37=27. So E+M+T+Y=21.

Okay, we just check the other case. If one of the factors if 74, we proceed similarly to find that the product if 666. 666/74=9, which is not two digits, so this case does not work.

The only possible answer is therefore 21.[/hide]

[quote="mathfanatic"][quote="MysticTerminator"][quote="mathfanatic"] That should be three Ts, not two.[/quote]
okay that helps (in other words, now the problem isn't absolutely impossible ;))[/quote]

Ya, ya, don't rub it in...  :wink:

And how do you know he's right, DQ?  Although you are right in saying that he is right...[/quote]
lol, lolt, let's build a ziggurat!

Mmm, we'll need to hope Syntax Error doesn't see it or he'll get mad again, although I don't understand why.
Also this might be considered spam unless we keep on posting relevant questions and such.

Hmm... Ye is my last name... and me refers to me... oh well.

I'd be mad if someone were spaming up my website, BUT IT NOT MINE SO SPAAAAAAAAAAAAAAAAAAAAAAM it!! :twisted:  :twisted:  :twisted:  :twisted:  :twisted:


search
linear algebra
linear algebra unsolved

Show that $ \det(xI-AB)=\det(xI-BA)$ for any $ x$.

Here is a weird proof.
Since both sides are continuous in $ A$, it's enough to prove their equality on a dense subset, such as invertible matrices. If $ A$ is invertible, then $ \det(\lambda I-AB)=\det(A(\lambda A^{-1}-B))=\det((\lambda A^{-1}-B)A)=\det(\lambda I-BA)$.

It's been posted before. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64324]Link.[/url]

mlok's post is closely related to what fredbel6 put on the other topic.


i hope someone comes and guides us thru INO's :maybe:

[hide]i think par asked 4 this only.. :rotfl: [/hide]

[quote="rituraj007"]pardesi, our moderator :P[/quote]
where u frm?

[quote="rituraj007"]par is from Bhubaneshwar :P
and me,frm Delhi :oops:

Have u been directed here by Skand by any chance??[/quote]

No.           i frm bihar . So u 4 bio or what

i ahven't directed him

i know him but he won't be knowing me

he is the best student this year 4m bihar and a real genius(as can be thought from the fact that he gave 2 NSE and got selected in both) :)

Feels good to see that Aops is now getting swarmed with geniuses(class  XI & XII) from all over India. :roll:

that's quite rite :D

But not a new thing  :P

To SKAND

how'd u know me ? but i am not a genius. in that case ritu is a greater genius (topping in nstse nd all)

@tuhin

check your PM,probably you'ld recognise me

and more importantly,do you think any1 in FIITJEE don't know you  :rotfl:

and this is a place where genii are present and ritu is no exception to this rule

[quote="skand"]@tuhin
and this is a place where genii are present and ritu is no exception to this rule[/quote]
VERY true  :rotfl:
I don't think there'll be any other place in all of India with such a great gang of brainos. And its also a wonderful opportunity for all the forthcoming leaders in the scientific sphere to have known each other beforehand.   :gathering:
Lets hope together we can do something big in the future.  :chief:

Edit: I was wondering if we could
a) introduce a shoutbox on the forums
b) create an IRC channel for AoPS

cud anybody be kind enough to post a few samples , links to papers etc etc...
please :)
 :help:  :help:

INPhO 2008  [url]http://olympiads.hbcse.tifr.res.in/subjects/physics/papers[/url]

well i think we shud ourselves post the ans and solutions :)

[quote="rituraj007"]Well this is an INTERNATIONAL site and the admins aren't ur frnds :D

moreover,our mod is not too active :P[/quote]
What about the IRC option then?

i think its already there in AOPS version (perhaps!!!) :rotfl:
but not too interactive :oops:

Now here r my answers to INPhO2008  questions...plz do check them :(

[hide="1"]
$ r\equal{}\frac{n^2 B}{2 \pi C}$
$ E \equal{} \minus{}\frac{AC^2}{2 n^2}$
$ R \equal{} \frac{A}{8B^3}$
[/hide]

[hide="2"]
(a)$ x\equal{}\frac{u}{c}(1\minus{}e^{\minus{}ct})$
 $ y\equal{}H\minus{}\frac{g}{c}[t \plus{} \frac{1\minus{}e^{ct}}{c}]$

(b)How 2 take out "t"??

(c)shorter range..ending parallel to y-axis

(d)$ \approx{10 s}$
[/hide]

[hide="3"]
heavy latexing..first lets discuss these 2 :rotfl:  :P
[/hide]

[hide="4"]
$ q_r\equal{}56.19$
$ q_{cv}\equal{}44.53$
$ q_{cd}\equal{}13.8$(best insulation)
[/hide][/hide]

Has anyone received letters from HBCSE??When are they expected to get?

[quote="tuhin_fire"]How much does 1 have 2 score 2 qualify INPhO, INChO,  By the way 1 st and second answers are correct. What do u wanna discus about the 3rd?[/quote]
no idea of marks yet :)

see (b) of 2nd

and do post answers of 3rd 'coz i m not pretty sure :oops:

$ y\equal{}H\minus{}\frac gc(t\plus{}\frac xu)$.

Plug in this $ t$ in the x- equation. Thats what i did anyway... :P

well, the pt is $ ct$ in Y and $ \minus{}ct$ in X
though i m 90% sure its wrong
but still they asked to take up dirn as plus.........

@ rituraj

ur xpression for 'y' is wrong

on differentiating ur xpression ..... and takin limit t tending to infinty u get v tends to infinity

whereas v wud be approachin a constnt value [terminal velocity ]

Can anyone tell me where i can get past INAO papers???I hav only 2008 paper..

hey do post other anwers though... :lol:

Have Any information other than ANSWER/Solution   :?:

Post at
http://www.mathlinks.ro/Forum/viewtopic.php?t=252391

Hey someone told me last year cutoff was 60/80.

Crosscheck & tell the authenticity of this information.

answer to 6th ques

a)0.14 and 0.17
b)0.17 and 0.14

is it rite?? (dont care abt units)

please tell the solutions of 3rd ques--c and d part

And is the ans to 7d is 0.99??and wat about 7e(i) :?:


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geometry
inradius
circumcircle
inequalities
3D geometry
tetrahedron
sphere

[b]1st homework problem set of the IMO training 2004[/b]

This is not a TST and not a regional olympiad but just the homework which has been sent to us in order to prepare us for the second of the five IMO training seminars in Bad Homburg. The problems below could be hardly from the ISL - some of them are just well-known -, so I post them here. Enjoy them :-)

BTW, I have solved G31-G35 and G38. Thanks in advance if you find a solution of the two remaining ones (G36 and G37... I am really bad in stereometry, sorry, and I didn't have much time).

I have slightly modified the notations.

[b]G31.[/b] Let A, B, C be the angles of a triangle, measured in radians, and r and R the inradius and circumradius of the triangle, respectively.

Show that

  3A / pi * 3B / pi * 3C / pi >= 2r / R.

[Or, simply, 27ABC / pi3 >= 2r / R.]

When does equality occur?

[b]G32.[/b] Find the maximal m and the minimal n (real constants) such that for any triangle with the angles A, B, C, the inequality

  m <= sin2 A + sin2 B + sin2 C <= n

holds.

[b]G33.[/b] Let ABCDEF be a convex hexagon having a circumcircle whose radius is equal to the sidelengths AB, CD and EF.

Show that the midpoints of the sides BC, DE and FA form an equilateral triangle.

[Please allow me a comment - this is the so-called [i]Hungarian theorem[/i] or [i]Asymmetric propeller theorem[/i].]

[b]G34.[/b] For which convex quadrilaterals ABCD with sidelengths a, b, c, d, the formula

  S = sqrt((s-a)(s-b)(s-c)(s-d))

holds, where s = (a+b+c+d)/2 is the semiperimeter and S is the area of the quadrilateral ABCD ?

[b]G35.[/b] The circumcircle of a triangle ABC is denoted by k, and its radius by R. The angle bisectors of the angles of triangle ABC meet k at the points A', B', C' (apart from A, B, C). Let P and Q be the areas of triangles ABC and A'B'C', respectively. Prove the inequality

  16 * Q3 >= 27 * R4 * P.

[b]G36.[/b] Let A1A2A3A4 be a tetrahedron, and for any of the six edges AiAk (i, k = 1, 2, 3, 4, i not equal k), take an internal point Tik on the edge.

Show that the six points Tik are the vertices of a convex polyhedron, and that the volume of this polyhedron is always >= 4 * the minimum of the volumes of the four tetrahedrons AiTikTilTim (i, k, l, m = 1, 2, 3, 4, no two of i, k, l, m being equal).

[b]G37.[/b] Investigate the interrelations (i. e. the equivalences and implications) between the following properties of a tetrahedron:

(a) The centers of the circumsphere and of the insphere coincide.
(b) The tetrahedron has three axes of symmetry.
(c) The faces of the tetrahedron have equal areas.
(d) Any two opposite edges have equal length.
(e) Any two opposite edges have equal edge-angles. (The [i]edge-angle[/i] of an edge is the angle between the two faces intersecting at this edge.)

[b]G38.[/b] Given six points A, B, C, D, E, F on a sphere such that AB || DE and BC || EF. Do these conditions allow us to conclude CD || FA ? Under which circumstances can we draw this conclusion?

Well, I know much of this will look trivial to you :-) At least, I hope G31 won't be [i]as[/i] trivial! (At least, it wasn't to me.)

  Darij

G32 was done somewhere in one of the forums
G34 is Brahmagupta's formula for the area of a cyclic quadrilateral (can be proved by Ptolemy's Theorem)
Pretty interesting problems, I'm having a geometry revisiting  :D

Yes, G32 is well-known. The answer is

0 <= sin2 A + sin2 B + sin2 C <= 9/4,

where the first <= can be actually strengthened to a <.

  Darij

i guess (i'm just looking at it,havent tried the whole thing) that G-31 follows by applying JENSEN to the convex function log[(A/2)/sin(A/2)]
duznt it

Yes, this was my first approach. But try to prove the convexity of log((A/2)/sin(A/2)) by differentiating... oh oh... However, there is a way around: Prove that log((A/2)/sin(A/2)) is convex by showing Jensen for *two* values, and then you get the inequality for *three* values. This was exactly my solution.

  dg

it's not such a monstrous expression,dg.it comes down to proving x>sinx,duznt it :D  :D .anyway how close are u to getting selected for the IMO?

[quote="galois"]it's not such a monstrous expression,dg.it comes down to proving x>sinx,duznt it[/quote]

Well - it does, more or less. I used Maple to derive and got a really ugly expression, so I didn't even try to do it by hand, but I see it wouldn't be as bad at all...

[quote="galois"]anyway how close are u to getting selected for the IMO?[/quote]

Not very close - I have just written the 3rd exam, solving 1 of the 3 proposed problems (curses - one of the 3 problems was exactly the one [url=http://www.mathlinks.ro/viewtopic.php?p=14438]I've posted here[/url] and I haven't succeeded to solve it completely!). Others were definitively better than me.

PS. I have been told that the problem G35 was a problem from the IMO longlist or shortlist 1989.

  dg

For $G_{33}$, see post #4 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=180206][u][b]here[/b][/u][/url].

Best regards,
sunken rock


Prove the following: $tanx+cotx+tan3x+cot3x=\frac{8cos^22x}{sin6x}$
Is there a nice way to prove this?

yetti, i  :omighty:  you are just too good solving trig problems   :thumbup:


AwesomeMath
summer program
AMC
AIME
calculus
conics
ellipse

In April 2007, over 600 middle and high school teams competed in the

[color=indigo][size=200][b]Purple Comet[i]![/i] Math Meet[/b][/size][/color]

the international on-line mathematics competition.  The 2008 meet will take place over a four day period from April 14 through April 18.

Have your school or math circle form as many teams as it wants (up to six members on each team).  Middle School teams solve 15 problems in 60 minutes.  High School teams solve 25 problems in 90 minutes.  With the help of grant funding, we are increasing our advertising efforts and are hoping for 6,000 participating teams this year.  Don't miss out.

Have your supervisor start registering your teams now at [url]http://purplecomet.org[/url].

Is anyone actually taking this thing? if so, what do you think about it. it looks like a cool contest but way easier (at least for Middle School) than the AMC's.

I took it last year and it was quite fun. I think the high school contest is definitely harder than the AMC.

I'm taking this with a team of 3 this year :)

I'll let you know how it goes. I anticipate that there will be a (rather high) guaranteed minimum level of quality, given that Titu Andreescu is involved

Titu Andreescu? Who's that, and what does he have to do with Purple Comet?

Titu Andreescu is the former US IMO team leader (a very big position) and current director of AwesomeMath I think. Also one of the people who runs this contest.

It's pretty challenging. I did the test yesterday by myself and got ten right. If I had 5 others with me, i'm sure we would have scored high.

I took this contest yesterday with 5 other ppl but we only solved like half of it.
btw,is it harder than AIME? :o

:) Our school team got 24/25  (we missed #10 for some reason -_-?)

Did anybody get a solution of #25 without calculus?

Yes. Scale the whole diagram down horizontally by a factor of $ \sqrt {2}$ and the ellipse becomes a circle.

Our team got 20/25...did better last year with a two person team. I'll probably try to go back to the two person team next year <_<.

EDIT: here are the diagrams:

Before:
[asy]size(100);
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
pair X=(0.5,1);
draw(ellipse(X,sqrt(2)/2,0.5));
draw((0.5-sqrt(2)/2,1)--(0.5+sqrt(2)/2,1));[/asy]

After:
[asy]size(100);
draw((0,0)--(sqrt(2)/2,0)--(sqrt(2)/2,1)--(0,1)--cycle);
pair X=(sqrt(2)/4,1);
draw(ellipse(X,0.5,0.5));
draw((sqrt(2)/4-0.5,1)--(sqrt(2)/4+0.5,1));[/asy]

And this transformation linearly scales all areas, so you can recover the answer from the shrunk diagram.

[quote="sillygirl1028"]btw,is it harder than AIME? :o[/quote]
This isn't the easiest comparison to make; you get calculators, computers, and a team on this one, but it's also more problems in less time. In general, I would say it's far easier to get a higher percentage of questions right on this test than the AIME.

I don't know what our team's answers were, so I don't know if we had dumb mistakes on what we got. We ran out of time before we could do 23 and 25, both of which I got within about 15 minutes afterwards. :(

I was doing #23 in the last 5 minutes or so, and got it, but forgot to account for one of my two off-by-one errors, so we ended up with 44 instead of 45 :(

[quote="MellowMelon"]I don't know what our team's answers were, so I don't know if we had dumb mistakes on what we got. We ran out of time before we could do 23 and 25, both of which I got within about 15 minutes afterwards. :([/quote]

Unfortunately, I don't know most of our answers, either, although we got everything that I remember (except 23 and 25, of course).

Overall, it was fun.  There were some nice problems, as usual.

Yeah, I realized #25 was a simple scaling

I just did it with calculus just to be safe.

We indeed got 24/25 on this, for we got #10 incorrect by forgetting to divide by 4!

Okay you made us all nervous when in the last minute you said "WAIT I think I evaluated the wrong integral!"

Well, here's a solution to #10?

First you chose the pairs of men and women, 11*10*9.  Take the perspective of one of the remaining women and chose her partner in 7 ways.  Similarly, take another one of the remaining, chose in 5 ways, then etc.  So, 11*10*9*7*5*3 = 103950.

I liked #22 and 23.  Is there a solution to #24 that doesn't involve a teammate randomly pulling out the answer by awesome guessing skills?

I did it with a team that had 1 pro guy, me and someone else who had similar skill, then 3 people you'll never hear about. Basically, I had to do every single combo/counting problem since the pro guy didn't want to do them and the other people were brute forcing by testing out random numbers...which didn't work out. I have to say, I wish I had a TI89 or programming skills... Let's just say that there was so much brute force that the poor TI89 had to calculate at least 30 numbers above 100!.

Then the only guy on our team who knew calculus sucked, so he brute forced all the calc brute forcable ones and got none of them. Hmmm...need better team next year. I still don't get how I totally beasted it, with my crap skill.

On number 24 one guy was like "Oh wow I think I got it. I THINK I GOT IT!!! I THINK I GOT IT!!! Crap 1+1 doesn't =3." I did tons of those problems in 5th grade. I remember that they would take a very very long time for something that complex...basically you just had to notice patterns, apply rules, and then eliminate all possibilities for a few letters down to just 2 or 3 numbers. Then it would get harder...that's the only way I think there is to solve those.

The results are up now: http://purplecomet.org/winners/2081winners.htm

Yay Howard Area Homeschoolers/Chesapeake! :D (that's the team I'm on, in case it's not obvious) We knew we had a maximum of 23, since we guessed on #23 and #25 and later figured out the correct answers, so a score of 21 is pretty good :).

Wow, tied for second overall. That's better than I expected having not answered all of the questions.

(darn you TJ!)

what prizes are there? any prizes for honorable mention teams?

Wow, my team got a honorable mention.  And it's made up of people from my school.

[quote="kyyuanmathcount"][quote="E^(pi*i)=-1"][quote="kyyuanmathcount"]Is there a solution to #24 that doesn't involve a teammate randomly pulling out the answer by awesome guessing skills?[/quote]

Yes.  There is a solution involving a teammate writing a Java program to brute force it.   :rotfl:[/quote]

We tried that solution :)  our guy took a little too long, I think the program was running for like 30 minutes and didn't get too far.

Uh...I don't know why our team isn't posted on the website -_- either we misentered stuff or something went wrong in the process of the computers grading?  Our school's B-team is there though, 16/25.[/quote]
[size=75][color=darkred]Hahahahaha on the #24.  I have pwnage skillz and got it like in 20 min.  ^.^  I'm actually surprised we got a honorable mention on the high school test (since we aren't the brightest) and 1st on the middle.[/color][/size]

I brute forced #24 with a program.

For anyone who's interested, here's the code (ruby)

[code]
list = [0,1,2,3,4,5,6,7,8,9]
for p in list
if p != 0
list2 = list.clone
list2.delete(p)
for u in list2
list3 = list2.clone
list3.delete(u)
for r in list3
list4 = list3.clone
list4.delete(r)
for l in list4
list5 = list4.clone
list5.delete(l)
for e in list5
list6 = list5.clone
list6.delete(e)
for c in list6
if c != 0
list7 = list6.clone
list7.delete(c)
for o in list7
list8 = list7.clone
list8.delete(o)
for m in list8
if m != 0
list9 = list8.clone
list9.delete(m)
for t in list9
listA = list9.clone
listA.delete(t)
for z in listA
if p*100000+u*10000+r*1000+p*100+l*10+e+c*10000+o*1000+m*100+e*10+t+m*1000+e*100+e*10+t == z*111111
print p,u,r,l,e,c,o,m,t,z
print "\n"
end
end
end
end
end
end
end
end
end
end
end
end
end
end
[/code]

[quote="kyyuanmathcount"][quote="E^(pi*i)=-1"][quote="kyyuanmathcount"]Is there a solution to #24 that doesn't involve a teammate randomly pulling out the answer by awesome guessing skills?[/quote]

Yes.  There is a solution involving a teammate writing a Java program to brute force it.   :rotfl:[/quote]

We tried that solution :)  our guy took a little too long, I think the program was running for like 30 minutes and didn't get too far.

Uh...I don't know why our team isn't posted on the website -_- either we misentered stuff or something went wrong in the process of the computers grading?  Our school's B-team is there though, 16/25.[/quote]
Faster computer? :) You could've overclocked it, since CPUs can take 25% easily, and it might've been just in time.
Wow my team got an honorable mention...that seems a little too easy...considering that we probably got something like 14. (yay that means I got >40% of the questions :D)

I hacked together a script in the last five minutes for problem 24 and ran it. Realizing that is would take ~$ 10^{13}$ operations to complete no matter what, I killed it.

...and it stopped on 9 2 2 1, the correct answer.

Hooray for brute force + pure luck!

My script from above running:

$ time ruby foo.rb
3560278914
3580276914
3760258914
3780256914

real    0m26.357s
user    0m24.358s
sys     0m1.524s

As you can see, it takes quite a bit less than 10 minutes (perhaps only having to do ~10! operations instead of 10^13? :P)

Hmmm...I remembered that calculating 500 million digits of pi on my computer (e6400) took a very short amount of time with pifast, so I'm sorta surprised that brute force programs people wrote didn't work it out in time...not that I know programming so whatever, but I'd think that someone who will try that knows their program has to be efficient enough to use.

[quote="E^(pi*i)=-1"]I think our program was finished in about 10 minutes (I'm not sure because I wasn't involved with it).

In any case, beating all teams not from TJ isn't so bad.  :)[/quote]

:) Our team got posted 24/25.

I'm not sure...we had a pretty good computer person I think, maybe it just didn't finish by the time our other guy guessed it.

Wow 24/25...lucky...(lucky that you have good people)
Okay I just remembered that on our team, they couldn't find the actual answer, but they found 2 numbers it could've been, and we guessed between the 2...

CAN someone tells me how to solve no.24 without using computer program or anything...!
in case if u forgot the question:

PURPLE+COMET+MEET=zzzzzz

find the number represented by the word MEET.

They all represent a different letter

THX!

My team is slowly improving. However, our high school team is rrrrreeeeeeeeeaaaallllllllyyyyyy good at Purple Comet.


All sides of the convex pentagon $ ABCDE$ are of equal length, and $ \angle A \equal{} \angle B \equal{} 90^{\circ}$.  What is the degree measure of $ \angle E$?

$ \textbf{(A)}\ 90 \qquad \textbf{(B)}\ 108 \qquad \textbf{(C)}\ 120 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 150$

[hide]The figure is like
[u]/\
| |[/u]
With the bottom 2 corners being A,B, the next ones up being E,C, and the top one being D.
The quadrilateral ABCE is a square, so $\angle AEC=90$. The triangle EDC is equilateral, so $\angle DEC=60.$ Then
$\angle E=\angle DEC+\angle AEC=90+60=150$
[/hide]


function
trigonometry
calculus
calculus computations

[quote="xxxxtt"]Let $ f: [a,b]\to R$ a function which is continuous on (a,b).Does this imply that $ f$ has limits in $ a$(at right) and $ b$(at left)??[/quote]
$ f(x)=\sin\frac{1}{x},x\neq 0,f(0)=0$ has no limit in 0.


geometry
trapezoid
circumcircle
cyclic quadrilateral
geometry unsolved

Two parallel chords $A_{1}B_{1}$ and $A_{2}B_{2}$ are drawn in two circles that intersect at points $A$ and $B$. The lines $AA_{1}$ and $BB_{2}$ intersect at a point $X$, while the lines $AA_{2}$ and $BB_{1}$ intersect at a point $Y$. Prove that $XY\parallel A_{1}B_{1}$.

Denote $\mathcal C_{1},\ \mathcal C_{2}$ the circles intersected by $A_{1}B_{1},\ A_{2}B_{2}.$ Let $AA_{1},\ BB_{1}$ meet at U and $AA_{2},\ BB_{2}$ at V. $ABB_{1}A_{1}$ is cyclic, $A_{1}B_{1}$ is antiparallel to AB with respect to $\angle AUB.$ Let $AA_{1},\ BB_{1}$ meet $\mathcal C_{2}$ again at $A_{3},\ B_{3}.$ Since $ABB_{3}A_{3}$ is cyclic, $A_{3}B_{3}$ is also antiparallel to AB with respect to $\angle AUB,$ hence $A_{3}B_{3}\parallel A_{1}B_{1}\parallel A_{2}B_{2}.$ Thus $A_{2}B_{2}B_{3}A_{3}$ is a cyclic trapezoid, isosceles with $A_{2}A_{3}= B_{2}B_{3}$ and $\angle YAX \equiv \angle A_{2}AA_{3}= \angle B_{3}BB_{2}\equiv \angle YBX.$ It follows that $ABXY$ is cyclic, XY is antiparallel to AB with respect to $\angle AVB.$ But $ABB_{2}A_{2}$ is cyclic as well, so that $A_{2}B_{2}$ is also antiparallel to AB with respect to $\angle AVB,$ which means that $XY \parallel A_{2}B_{2}\parallel A_{1}B_{1}.$

Based on the same idea (Yetti is a faster man), another approach is as follows:

We denote as $X$ the intersection point of   $AA_{1},\ BB_{2}.$ and we draw the circumcircle of the triangle $ABX$, which intersects
the segment line $BB_{1}$ at a point so be it, $Y.$

We denote as $P,$ the intersection point of $AA_{1},\ BB_{1}.$ So, we have that $XY \parallel BB_{1},$ as antiparallel lines to the segment line $AB,$ with respect to the lines $AA_{1}, BB_{1}, ( \angle APB ).$  Hence, $A_{1}B_{1}\parallel XY \parallel A_{2}B_{2}$  and it is enough to prove that the points $A, A_{2}, Y,$ are collinear.

From cyclic quadrilateral $ABXY \Longrightarrow \angle BAY+\angle BXY = 180^{o}, (1)$ and from $XY \parallel A_{2}B_{2}\Longrightarrow \angle BXY+\angle BB_{2}A_{2}= 180^{o}, (2).$

From $(1), (2), \Longrightarrow \angle BAY = \angle BB_{2}A_{2}$ and the proof is completed.

Kostas Vittas.


I am looking for a [color=red]non inductive solution[/color] for this problem.

There are $ n>1$ balls in heap. On every operation, an heap can be chosen and it can be broken to create two heaps of non zero size say $ k_1$ and $ k_2$ respectively. We compute the product $ p_1\equal{}k_1k_2$ after every operation. We continue till we every heap has exactly one ball. Prove that, irrespective of the way the heap is broken every time, $ p_1\plus{}p_2\plus{}\dots \plus{}p_{n\minus{}1}$ equals to $ n\choose 2$.

Consider a tournament with $ n$ players which is set up as follows. to begin, the $ n$ players are arbitrarily divided into two groups, and each player plays each of the contestants in the opposite group exactly once. each round, as long as there is at least one group of more than 1 player, one of those groups is divided into two groups and players from one of the newly formed groups face off with the players from the other newly formed group. if the sizes of these groups are $ a$ and $ b$, it's clear that $ ab$ games are played this round. it's also clear that in this set-up every player faces off against every other player exactly once. thus a total of $ \binom{n}{2}$ games are played, and this proves the result.


inequalities
function
inequalities proposed

Let $ x_1, x_2,\ldots , x_n$ be positive real numbers such that $ x_1x_2\cdots x_n \equal{} 1$. Prove that
\[\sum_{i \equal{} 1}^n \frac {1}{n \minus{} 1 \plus{} x_i}\le 1.\]

[quote="mr.danh"]Let $ x_1, x_2, ... , x_n$ be positive real numbers such that $ x_1x_2 ... x_n \equal{} 1$. Prove that
$ \sum_{i \equal{} 1}^n \frac {1}{n \minus{} 1 \plus{} x_i}\le 1$[/quote]
uhm,here is also the ineq in Romania TST 1998,and AMGM works here

this is also from Romania 1999 I think:

with contradition method,assume that:

$ \sum_{i = 1}^n\frac 1{n - 1 + x_i} > 1$

hence:

\begin{eqnarray*}\frac 1{n - 1 + x_1} & > & 1 - \left(\sum_{i = 2}^n\frac 1{n - 1 + x_i}\right) \\
& = & \sum_{i = 2}^n\left( \frac 1{n - 1} - \frac 1{n - 1 + x_2}\right) \\
& = & \sum_{i = 2}^n\frac {x_i}{(n - 1)(n - 1 + x_i)} \\
& \geq & (n - 1)\sqrt [n - 1]{\frac {\prod_{i = 2}^n x_i}{(n - 1)^{n - 1}\prod_{i = 2}^n (n - 1 + x_i)}} \\
& = & \sqrt [n - 1]{\frac {\prod_{i = 2}^n x_i}{\prod_{i = 2}^n (n - 1 + x_i)}}\end{eqnarray*}

(the inequality part is deduced by AM-GM)

hence:

$ \frac 1{n - 1 + x_1} > \sqrt [n - 1]{\frac {\prod_{i = 2}^n x_i}{\prod_{i = 2}^n (n - 1 + x_i)}}$

now write similar inequalities for other $ x_i$s and multiple them,we get that:

$ x_1x_2\ldots x_n < 1$

which is a contradiction.

[quote="mr.danh"]Let $ x_1, x_2, ... , x_n$ be positive real numbers such that $ x_1x_2 ... x_n \equal{} 1$. Prove that
$ \sum_{i \equal{} 1}^n \frac {1}{n \minus{} 1 \plus{} x_i}\le 1$[/quote]
And the Cauchy Schwarz helps here, too.
The inequality equalivents with
\[ \sum {\frac{a}{{n \minus{} 1 \plus{} a}}}  \ge 1
\]
By the Cauchy Schwarz we have
\[ \sum {\frac{a}{{n \minus{} 1 \plus{} a}}}  \ge \frac{{\left( {\sum a } \right)^2 }}{{n\left( {n \minus{} 1} \right) \plus{} \sum a }}
\]
And $ \left( {\sum a } \right)^2  \ge n\left( {n \minus{} 1} \right) \plus{} \sum a  \Leftrightarrow \left( {\sum a  \minus{} n} \right)\left( {\sum a  \plus{} n \minus{} 1} \right)$, it's done, :).

[quote="Honey_S"]
By the Cauchy Schwarz we have
\[ \sum {\frac {a}{{n \minus{} 1 \plus{} a}}} \ge \frac {{\left( {\sum a } \right)^2 }}{{n\left( {n \minus{} 1} \right) \plus{} \sum a }}
\]
[/quote]
I don't think so,by cauchy schwarz we get that:

$ \sum\frac a{n\minus{}1\plus{}a}\geq\frac{\left(\sum a\right)^2}{\sum a^2\plus{}(n\minus{}1)\sum a}$

which doesn't help us at all...

Honey_8
your  solution  is not  right，because you  forget   their  products is 1。

[quote="jack2627"]Can we use Jensen's inequality for this?
Because the function is concave, so $\sum_{i = 1}^{n}\frac{1}{n-1+x_{i}} \le n $ $\times$ $\frac{1}{n-1+\frac{x_1+x_2+\dots+x_n}{n}}$ $\le$ $\frac{n}{n-1+1} = 1$
I'm not sure if I'm right though, please correct me if I'm wrong, thanks![/quote]
It's wrong because $f(x)=\frac{1}{n-1+x}$ is a convex function.

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=238609

[b]Singapore Mathematical Olympiad 2004 , Special  Round:[/b]
 Let $ x_1, x_2,\ldots , x_n\in(0,1]$ and $n\ge 1$. Prove that
\[\sum_{i \equal{} 1}^n \frac { x_i}{1+(n \minus{} 1 ) x_i}\le 1.\]

Sqing's problem can be solved by BaBak Ghalebi's approach. :)


Hunter kept all his notebooks from every class he ever took in middle school. He had a special shelf in his room for nothing but his notebooks. If he had 3 math, 3 science, 2 history, and 1 writing notebooks, how many ways were there for him to arrange all the books on the shelf?

[hide=""Big"Hint"]The number is 6 digits!  :what?:   :whistling:   :dry:[/hide]

[hide](3+3+2+1)! = 9! which I don't feel like calculating. :P[/hide]


$ A,B\in M_{n}(\mathbb{R})$,and $ A^{T}\equal{}A,B^{T}\equal{}B,AB\equal{}BA$.Show that there exists $ T\in M_{n}(\mathbb{R})$,s.t.$ T^{T}T\equal{}E$,moreover $ T^{\minus{}1}AT,T^{\minus{}1}BT$ are both diagonal. :blush:

Several related posts already, for instance http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1602887184&t=175612


ratio
arithmetic sequence
algebra unsolved
algebra

$ {a_n}$      is a real number sequence such that     $ \minus{}2005<a_n < 2005$ and $ a_{n\plus{}3}\equal{}a_{n\plus{}2}\plus{}a_{n\plus{}1}\minus{}a_{n}$ for  $ \forall n\in N$ . prove that $ {a_{n}}$ is periodic.

We have a_(n+3) - a_(n+2) = a_(n+1) - a_n. So this sequence is like an arithmetic progression with alternating ratio. Let b_n = a_(2n-1). So b_n is an arithmetic progression. If the ratio is not zero, clearly the value of |b_n|, which equals to a term in a_n, will exceed 2005. So b_n is constant, and hence a_n is periodic with period of 2.


Any chance of any MC Countdown round math jams this year, between state and nationals? There aren't any scheduled. Or are they not happening this year since MCrawford left?

If the problem is that someone doesn't have the time to actually host it, I might be willing to host the math jam (If I was given problems, because most of the problems I have have been seen by many users on these forums)

Then go steal the Mathcounts Nationals ones. :P

Well, how about you use the 2006 States, or 2007 Chapters?


Sum up two equations, we have:
$\sqrt{x}+\sqrt{32-x}+\sqrt[4]{x}+\sqrt[4]{32-x}=y^{2}-6y+21$ (1)
Follow Bunhiacopski's inequality then:
$\left(\sqrt{x}+\sqrt{32-2}\right)^{2}\leq 2(x+32-x)=64\Rightarrow \sqrt{x}+\sqrt{32-x}\leq 8$
$\left(\sqrt[4]{x}+\sqrt[4]{32-2}\right)^{4}\leq (2(\sqrt{x}+\sqrt{32-x})^{2}\leq 4.2(x+32-x)= 256\Rightarrow \sqrt[4]{x}+\sqrt[4]{32-x}\leq 4$
Hence: Left hand of (1) $\leq 12$ while the right hand is $y^{2}-6y+21=(y-3)^{2}+12\geq 12$
So (1) ocucur iff $y=3,x=16$
It is easy to check it is a solution of the system.

What's Bunhiacopski's?

EDIT: Just an odd spelling for Buniakowsky ($\iff$ Schwarz-Buniakowsky $\iff$ Cauchy-Schwarz).

It is also called Cauchy - schwatz's inequality


[quote="rhynoceros"]What are the values of x, y and z in the following equation:

 xx
 yy
 zz
____ +

xyz

x, y and z stand for digits (so x, y and z aren't numbers who exist of more then 1 digit!)[/quote]

so easy
1,8,9

[quote="rhynoceros"][quote]so easy[/quote]

It is as soon as you see that x cannot be higher then 1, wich automaticly means that y=9. Using algebra gives z=8[/quote]

i can find another way too :))))
thank u for understanding it to other
u did that what i mest be did:))


Find all the numbers n for whom the set 1,2,...,n can be partitionated in three subsets wich have the same sum of elements

consider n mod 6

n=0 mod 6 --> of course ok:  1-6, 5-2, 4-3

n=5 --> same, just pair them 0-5, 4-1, 3-2

n=1 mod 6 or 4 mod 6 --> 3 doesn't divide sum so impossible

n=2 mod 6 --> possible, same as 0 but now switch 1 and 2, and add them to the correct columns  (not possible for n=2 of course :D)

n=3 mod 6 --> same story again: number them 9-4, 8-5, 7-6, switch lowest 2 numbers, and add 1,2,3 on the correct columns

so all numbers that are 0 mod 3 or 2 mod 3.


linear algebra
matrix
algebra
system of equations

I need help. I have to solve a system of equations with eigenvalues and eigenvectors and I don't know how. I send you a link to see the system. The part of interest is at the end of the page. Please if you have some solutions help me.

[url]http://www.infogoaround.org/JBook/LSQ_Plane.html[/url]

I posted this message in the Middle school section too but I guess it's better here for high school.

Someone should move this post to:

College Playground >> Matrix - Linear Algebra >> Matrix Algebra - Computations and Tutorials

if linear algebra is really necessary.


induction
inequalities
number theory unsolved
number theory

Find all pairs $ (m,n)\in N.,m.n>0$ such that
$ m^n\minus{}n^m\equal{}7$

It is easy to chek, that $ m$ is even, $ n$ is odd.
1) If $ n\equal{}1$ then $ m\equal{}8$.
2) If $ m\equal{}2$, then $ 2^n\equal{}7\plus{}n^2\to n\equal{}5$.
For another cases $ m>3,n\ge 3$.
Because $ m^{1/m}>n^{1/n}$ we get $ 4\le m<n$.
Then $ m^n\minus{}n^m\ge m^{m\plus{}1}\minus{}(m\plus{}1)^m\ge 4^5\minus{}5^4>7.$
Therefore all solutions $ (m,n)\equal{}(8,1),(2,5).$

Nice Problem and nice solution provided by [b]Rust[/b] but I would like to ask a question. I think I have read something like this but is it true that if $ a\ge b\ge e$ then $ b^{a}\ge a^{b}$ and vice versa?

Yes for proving that you have to make induction and observe that   $ (\frac {1}{k} \plus{}k)^k <e)$  and that $ \frac {1}{k} \plus{}k> \frac {1}{t} \plus{}t$ if $ k>t$

I was doing something similar yesterday :D

Daniel

And a nice other problem
Find all pairs $ x,y\in N*$ such that

$ x^y\plus{}y\equal{}y^x\plus{}x$
(here is a problem in MYM,But it is quite old)

[quote="Allnames"]And a nice other problem
Find all pairs $ x,y\in N*$ such that

$ x^y \plus{} y \equal{} y^x \plus{} x$
(here is a problem in MYM,But it is quite old)[/quote]
I think it quite easy
We have
Case $ x\equal{}y \rightarrow$suitable
Case $ x \ne\ y$. Assum$ x < y$
.$ x\equal{}1 \rightarrow  y\equal{}2$
.$ x\equal{}2 \rightarrow y\equal{}3$
.$ x \ge\ 3$
We 'll prove $ x^y > y^x$ with  $ 3 \ge\ x < y$
$ x^y > y^x \leftrightarrow x^{\frac{1}{x}}>y^{\frac{1}{y}}$
$ \leftrightarrow \frac{1}{x}lnx>\frac{1}{y}lny$
To examine $ f(t)\equal{}\frac{1}{x}lnx$ we easy prove$ f '(t) <0$ ($ t\ge3$)$ \rightarrow f(x) > f(y)$(with $ x<y$)
$ \rightarrow x^y > y^x \rightarrow x^y \plus{} y > y^x \plus{} x$
We have done! :wink:

Assum x \le\ y

[quote="Rust"]
For another cases $ m > 3,n\ge 3$.
Because $ m^{1/m} > n^{1/n}$ we get $ 4\le m < n$.
Then $ m^n \minus{} n^m\ge m^{m \plus{} 1} \minus{} (m \plus{} 1)^m$
[/quote]
Why? Please explain inequality to help me. Thanks.


limit
real analysis
real analysis unsolved

Let $f:[0, \infty) ->R$ and $f(x)=e^{m+\sqrt{x}}+e^{m-\sqrt{x}}$ with $m \in N$. If $L_{n}= \displaystyle \lim_{x\to 0, x>0} e^{-m}f^{(n)}_{(x)}$ with $n \in N$. Compute L_2002. ($f^{(n)}$=$n^{th}$ derivate)

$ f(x)\equal{}e^{m}\left(e^{\sqrt{x}}\plus{}e^{\minus{}\sqrt{x}}\right)$.

we know that : $ e^t\equal{}\sum_{n\equal{}0}^{\infty}{\frac{t^n}{n!}}$. And we can write : $ e^t \plus{} e^{\minus{}t}\equal{}2\sum_{n\equal{}0}^{\infty}{\frac{t^{2n}}{(2n)!}}$. Now put here $ t\equal{}\sqrt{x}$, we get :

$ e^{\sqrt{x}}\plus{}e^{\minus{}\sqrt{x}}\equal{}2\sum_{n\equal{}0}^{\infty}{\frac{x^{n}}{(2n)!}}$

Now easy to calculate our limit : $ \lim_{x\to 0, x>0}{e^{\minus{}m}f^{(n)}_{(x)}}\equal{}2\cdot \frac{2000!}{4000!}$


We start from any dot (see picture). At each step we move right, left, up or down to the adjacent dot. Must visit each dot exactly once, except for the last step (the 144th), which is to the starting point.
Prove: the path can't be made of 36 steps in each direction.


Θα ηθελα τα σχολια σας για το παρακατω επιχειρημα.

[b]Προβλημα[/b]

Να αποδειξετε οτι για θετικους $ a,b,c$ ισχυει

$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}>\equal{}3$

[b]Αποδειξη[/b]

Προφανως

$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}\frac{b}{a}\plus{}\frac{a}{c}\plus{}\frac{c}{b}>\equal{}2\plus{}2\plus{}2\equal{}6$

Θετουμε

$ A\equal{}\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$

$ B\equal{}\frac{b}{a}\plus{}\frac{a}{c}\plus{}\frac{c}{b}$

Τοτε $ max(A,B)>\equal{}3$

Και εδω λεω οτι εχουμε τελειωσει γιατι θελαμε να δειξουμε οτι $ A>\equal{}3$.
Ομως οι αριθμοι $ A,B$ ειναι της ιδιας φυσεως και αφου οι $ a,b,c$ τυχαιοι δειξαμε οτι μια τετοια "φυση" ειναι μεγαλυτερη η ιση του 3.

Εχω χρησιμοποιησει σε πολλες και δυσκολες ανισοτητες αυτη την ιδεα (εδω δειχνω την πιο απλη) η οποια λογικα δεν δουλευει(τσαμπα κοπος!)
Μηπως ομως δουλευει και αν οχι που ακριβως ειναι το προβλημα?

1) $ \sum_{cyc}\frac {a}{b} > \equal{} 3\sqrt {\frac {a}{b}\cdot\frac {b}{c}\cdot\frac {c}{a}} \equal{} 3$ simply from AM-GM
2) tora ok...afou theleis na xtupiseis miga me bazoukas mazi sou (ki emena mou aresei na kano genikeuseis kai na vrisko methodous)! alla uparxei ena provlhma to "φύση" pou xrhsimopoieis-to katalavaineis kai monos sou.den nomizo oti boreis na tekmirioseis kati tetoio me austhra mathimatika, opote kata auth thn ennoia sosto den einai. alla par ola auta pisteuo oti aggizei kapoio komati theorias...dld emena mou fainetai oti h rearrangement inequality prospathei na pei kati paromoio me auto pou les esu..peripou..giati i rearrangement den borei na efarmostei amesa edo.

*** aa kati allo...auto pou les oti exoun tin idia fusi, isos einai kai logo tis kuklikis summetrias pou exoun oi duo parastaseis-opote na ki allo ena theoritiko upvathro pou borei na pei kaneis oti einai konta.

its very late...forgive me!

Λοιπόν, απλοποιώ το παράδειγμα του AlexandrosG για να δείξω σε τί προβλήματα ανοιγόμαστε με μία τέτοια συλλογιστική...

Έστω $ a$ θετικός πραγματικός. Τότε, προφανώς: $ a \plus{} \frac {1}{a}\geq 2$ (αν θέλετε, η αρχική ανισότητα για $ b \equal{} c \equal{} 1$ :P).

Έστω λοιπόν $ A \equal{} a$, $ B \equal{} \frac {1}{a}$, οπότε $ \max(A,B)\geq 1$. Μέχρι εδώ όλα καλά.

Όμως οι αριθμοί Α,Β είναι της ίδιας "φύσεως" και αφού ο $ a$ τυχαίος, δείξαμε ότι μια τέτοια "φύση" είναι μεγαλύτερη ή ίση του 1.

Άρα $ a\geq 1$ :roll:

Φυσικά αυτό δεν ισχύει ;) Το μόνο συμπέρασμα το οποίο μπορεί να εξαχθεί από την παραπάνω συλλογιστική είναι ότι είτε $ a\geq 1$ είτε $ a\leq 1$ για κάθε $ a$ θετικό πραγματικό.

Το ίδιο και για τα $ a,b,c$, είτε $ A(a,b,c)\geq 3$ είτε $ B(a,b,c)\geq 3$. Δεν μπορείς να πεις ότι οκ, αν δεν έχουμε $ A(a,b,c)\geq 3$ θα έχουμε $ B(a,b,c)\geq 3$ και άρα θέτω στο $ B(a,b,c)$ όπου $ a\to\frac {1}{a}$ κλπ, γιατί για τους καινούριους αριθμούς μπορεί να ίσχυει το $ A(1/a,1/b,1/c)\geq 3$.

Για να μπορέσεις να συνεχίσεις, πρέπει να δείξεις είτε ότι "$ A(a,b,c)\geq 3$ για όλα τα $ a,b,c$", είτε ότι "$ B(a,b,c)\geq 3$ για όλα τα $ a,b,c$". Η πρόταση "είτε $ A(a,b,c)\geq 3$ είτε $ B(a,b,c)\geq 3$ για όλα τα $ a,b,c$" δεν είναι ούτε ισοδύναμη ούτε αρκετή.

Hope this makes some sense γιατί είμαι βιαστικός και δεν προλαβαίνω να το γράψω καλύτερα :(

Cheerio,

Durandal 1707

Καλλημέρα σε όλους μετά από πολύ καιρό! (Χρονιά πανελληνίων, τι να κάνουμε...)

Κατά τη γνώμη μου μπορούμε να μιλήσουμε για 'φύση' παραστάσεων υπό μία έννοια.

Μπορούμε να ορίσουμε ότι: δύο παραστάσεις έχουν ίδια φύση, αν η μια προκύπτει από αναδιάταξη των μεταβλητών
με μια συγκεκριμένη ιδιότητα της άλλης (στη συγκεκριμένη περίπτωση η ιδιότητα είναι να βρίσκονται στους αριθμητές).
(Δεν το διατύπωσα ακριβώς όπως ήθελα, αλλά νομίζω ότι καταλαβαίνετε.)
ΠΡΟΣΟΧΗ: Αν δύο παραστάσεις έχουν την ίδια φύση, δε σημαίνει πως έχουν τις ίδιες ιδιότητες,
αλλά ότι αν η μια από τις δύο έχει κάποια ιδιότητα την έχει η άλλη (χωρίς το αντίστροφο).

Δεν ξέρω αν παραστάσεις με ίδια 'φύση' έχουν κάποιες συγκεκριμένες ιδιότητες,
εκτός από αυτές που ικανοποιούν την ανισότητα της αναδιάταξης.

Στη συγκεκριμένη λοιπόν περίπτωση, μπορούμε να παρατηρήσουμε ότι $ A >\equal{} B$ λόγω της ανισότητας της αναδιάταξης,
άρα $ 2A>\equal{}6 \Leftrightarrow A>\equal{}3$.

[quote="Nick Rapanos"]dld emena mou fainetai oti h rearrangement inequality prospathei na pei kati paromoio me auto pou les esu..peripou..giati i rearrangement den borei na efarmostei amesa edo.
[/quote]

auto pou leo kai parapano einai oti o Alexandros rotaei kati to opoio einai apoluta fusilogiko na to skeftei kaneis ka na provlhmatistei...prospathei na milisei gia "fusi" duo parastaseon-pragma pou ton odigei se lathi. anti gi auto tha prepei na milame me austhra mathimatika ergaleia, opos einai i rearrangement inequality i o,ti allo anefera parapano pou more or less kanoun ti douleia pou thelei o alexandros-dld sugrinoun "omoies" parastaseis.opote mono me auta epitrepetai n a paizoume...

alla opos grafo kai parapno i rearrangement den efarmozetai amesa edo, gi auto polu sosta soug grafei kai o silouan oti kaneis lanthasmeni xrhsh tis rearrangemnt.genika me ti rearrangement prepei na prosexei kaneis kai na mhn enthousiazetai prin na einai sigouros oti tin exei efarmoseis sosta!

[quote="silouan"][quote="maniopas"]
Στη συγκεκριμένη λοιπόν περίπτωση, μπορούμε να παρατηρήσουμε ότι $ A > \equal{} B$ λόγω της ανισότητας της αναδιάταξης,
[/quote]

Το ότι $ A\geq B$ δεν ισχύει απαραίτητα (δεν κάνεις σωστή χρήση της ανισότητας της αναδιάταξης) .
Πράγματι δείτε ότι $ A \minus{} B \equal{} \frac {(a \minus{} b)(a \minus{} c)(c \minus{} b)}{abc}$  Οπότε αν έχεις ότι $ a\geq c\geq b$ έχουμε ότι $ A\leq B$[/quote]

Exactly!  :clap2:  Ακριβώς όπως αν $ a\leq 1$ έχεις $ \frac{1}{a}\geq a$ και σου χαλάει η "απόδειξη" ότι $ a\geq 1$ για κάθε $ a>0$.

Cheerio,

Durandal 1707


Source: MAO 2001 Nationals - Proofs #12

Prove that for every integer m there is at least one integer n such that m + n + 1 is a perfect square, and mn + 1 is a perfect cube.

The first one is kinda easy I think.

Let S = {k in IN | k > n + 1}.
Then for each a in S we can take m = a - n - 1.

The second one ? Take m = n + 3n + 3.
Then indeed mn + 1 = n + 3n + 3n + 1 = (n + 1).

[color=cyan]I think it means that they are supposed to be the same number, no?  That is, n should be the same for the square and the cube.[/color]

[color=cyan]Although, curiously, your answer to the second problem
(n^2 + 3n + 3) has this property and is the desired number.  Go figure.  I wonder if there are more . . .[/color]

Oh I didn't realize that !
I'll have to re - think my "solution".

So our question is equivalent to :
For which a, n in IN is the number an - n - n + 1 a perfect cube for every n ?
Indeed a = n + 3n + 3 is a solution :D

I dunno if this is a full-fledged "solution", but we want m to be ((n+c):^3:-1)/n, in order to make mn+1 a cube, right? The only c in which (n+c):^3:-1 is a multiple of n is c=1, and if (n+c):^3:-1 is not a multiple of n, we will have a 1/n term in the expansion of m, which will make m not neccessarily a natural number. So, wec has to be 1, giving m=n:^2:+3n+3. This just happens to work in the first equation as well.

oh good point thx Simon. at least it's been narrowed down. I'll finish after dinner. bye.


I am having some problems answering some questions can anyone help?

1.   Carbolfuchsin can be used as a simple sain and as a negative stain.  As a simple stain the PH is?

2. Suppose you Gram stained a sample froma pure culture of bacteria and observed a field of red and purple cocci.  Adjacent cells were not always the same color.  What do you conclude?

Well , for your ques2:

My analysis is that :
 THere are more than one kind of bacteria in your petri dish.( it is impure  )

the ph has to be below 4 for any carboxyl group, therefore I would say it's 3.


ratio
calculus
integration
number theory

[b]First problem[/b]
[hide]
If $ 21(a^2 \plus{} b^2 \plus{} 4c^2) \equal{} (a \plus{} 2b \plus{} 8c)^2$ and $ abc \neq 0$, find the ratio $ a: b: c$. (HK Inter-school Mathematics Contest 2008)
[/hide]

[hide="hmm"]
Can anyone explain what MathLinks is for? Isn't it just a mirror image of AoPS and some additional info (e.g. MathLinks contest)? Isn't Mr Valentin Vornicu one of the staff at AoPS; why was another site created? Sheer curosity, eh?
[/hide]

Also, what about the next problem? I'm bad at number theory but would really like to learn more about it...next time, I guess?

[b]$ n^4$[/b]
[hide]
Prove that $ n^4 \plus{} 4$ is [b]never[/b] a prime number for all integral values of $ n$.
[/hide]

Another problem:

[list] How many natural numbers have no prime factors greater than 10 and have exactly 2008 factors? [/list]
and a generalised form:
[list] How many natural numbers have no prime factors greater than $ x$ and have exactly $ n$ factors? (Express your answer in terms of $ n$; assume that $ n$ is greater than $ x$)
[/list]

$ n\equal{}1$ is an obvious counterexample to the $ n^4 \plus{} 4$ problem

[quote="chickendude"]$ n \equal{} 1$ is an obvious counterexample to the $ n^4 \plus{} 4$ problem[/quote]

It is true for all integers $ n > 1$.

[hide="Solution."]
-----
Factoring using the [url=http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity]Sophie-Germain[/url] identity, $ n^4 \plus{} 4 \equal{} (n^2 \plus{} 2n \plus{} 2)(n^2 \minus{} 2n \plus{} 2)$. The first factor is greater than or equal to $ 2\ \forall n \in \mathbb{Z}^ \plus{}$. (Note that we can assume positive $ n$, since negative $ n$ would simply reverse the order of the factors.) Also, $ n^2 \minus{} 2n \plus{} 2 > 1\ \forall n \in \mathbb{Z}^ \plus{} , n > 1$, so for $ n > 1$, both factors are positive integers greater than $ 1$, and $ n^4 \plus{} 4$ is composite and not prime.
-----
[/hide]

[quote][hide="hmm"]
Can anyone explain what MathLinks is for? Isn't it just a mirror image of AoPS and some additional info (e.g. MathLinks contest)? Isn't Mr Valentin Vornicu one of the staff at AoPS; why was another site created? Sheer curosity, eh?
[/hide][/quote]

Have you seen this? [url]http://en.wikipedia.org/wiki/Art_of_Problem_Solving[/url]

If you have, then you know that MathLinks and AoPS merged in 2004 (before that, they existed as separate entities).

Erm..no one attempting the first problem?

[quote="jvenezuela716"][quote="chickendude"]$ n \equal{} 1$ is an obvious counterexample to the $ n^4 \plus{} 4$ problem[/quote]

It is true for all integers $ n > 1$.[/quote]It's true for all integers $ n^2\ne1$. And your last step should be $ (n\minus{}1)^2>0\Rightarrow n^2\minus{}2n\plus{}2>1$.


[quote="slowlearner"]I have to solve the linear inequalities and graph and shade where the points meet. i understand the graph but not how to solve the equations. any and all help is greatly appreciated, thank you.

2x+y>-2
6x+3y<6[/u][/quote]

Replace the inequalities with equal signs:

Well the points don't meet.

$6x+3y < 6$

$\frac{6x}{3}+\frac{3y}{3}< \frac63$

$2x+y < 2$

which definitely doesn't intersect with $2x+y >-2$.

[quote="jhredsox"]hah, i got a 60 on a math quiz cuz i spent 40/42 min doing elimination for an inequality

-jorian[/quote]

 :rotfl:


[hide]
Let us consider the sum of the digits in writing the numbers from 0-9999.  There are clearly 10000 numbers.  Of those, each digit is equally likely to appear in the units place, so 1000 digits are 1s, 1000 are 2s, and so on.  The units digit sum is thus $ 1000(1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\plus{}8\plus{}9)\equal{}45000$.  Similar arguments hold for the other digits, but you have to remember carrying.  I believe I get $ \boxed{49995000}$ :huh: [/hide]

[quote="sunnykim"]Find the sum of all digits used in writting the numbers between 1 and 9999. I'm looking for a better solution because I found the sum( 180 000 ) but i think i don't have the fastest method :P[/quote]
[hide="Solution (Edited)"]The sum of the $ 10$ digits is $ 0 \plus{} 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 7 \plus{} 8 \plus{} 9 \equal{} 45$, and each digit appears in each spot $ 1000$ times.

$ S \equal{} 1000\cdot45\cdot4 \equal{} \boxed{180000}$[/hide]

The correct answer is $ 180 000$. I am very sure about that(I counted it very carefully--but not so quickly-- and it's the answer in the book too)

I don't know where you got 49995000... I multiplied 45000 by four (since there are four "digits") = 180000.

yeah, I realized that a little earlier.

but it's still good practice for addition. :lol:

hey dude use arithmetic progression
solution :
s=5000(2+9999)
s=10001*5000
ans............

I have another way i found, ill use a smaller example (1+2+3+4+5)=? so there are 5 number obviously all you need to do is take the last and first numbers: like this

1+2+3+4+5
5+4+3+2+1 and add vertically
6+6+6+6+6 then divide it by two since you used two pairs. 30 divided by 2 = 15  :lol:

nice alton but i thought i would use the conventional method

NO!!! im just saying i thoguht it would get flamed or sumthing but u liked it and that nice :)

nice attitutde alton but i knew that my procedure is wrong and will correct it as soon as possible
by the way wich class r u in?

Umm well im going into 6th grade. And thanks. :)

nice to see anyone online with me
so alton ur particulaly intersted in mathematics

Favorite subject, I also entered a local math competition and won the regionals 1st place. I'm going to MIT for the nationals.

 :D

I view myslef skilled but tons of people are smarter than me at math. 6th graders in highschool!!!  :o

lol, you guys are like talking to each other.  and i do like altons solution too

:P

hey can u send me the sat scores required to get admission to get through caltech


superior algebra
superior algebra unsolved

There's something missing in that statement. Is this what you mean?

Characterize all $f : \mathbb{R}^n\mapsto\mathbb{R}^n$ such that

$||x-y||<||x-z||$ implies that $||f(x)-f(y)||<||f(x)-f(z)||.$

sorry,characterize all f:R^n------>R^n such that
 ||x-y||<||x-z|| => ||f(x)-f(y)||<||f(x)-f(z)||


inequalities
logarithms
inequalities unsolved

[quote="dina013"]If $a, b, c$ $\in$ $(0,1)$ or $(1,+\infty)$. Show that:
   $log_{a^{2}b}a+log_{b^{2}c}b+log_{c^{2}a}c \le 1$.[/quote]
I think you mean either $a, b, c$ $\in$ $(0,1)$ or $a, b, c\in(1,+\infty)$. Otherwise we can have $a^{2}b=1$ :( ...
Put $x=\log a$ etc; Then the ineq just becomes $\sum\frac x{2x+y}\le1$ for $x,y,z>0$ or $x,y,z<0$. Equivalently, $\sum\frac y{2x+y}\ge1$. Wlog $x,y,z>0$. Why then the $log$ notation? maybe you don't like fractions? :)

 $LHS= \sum\frac{y^{2}}{2xy+y^{2}}\ge\frac{(x+y+z)^{2}}{(x+y+z)^{2}}=1$  :rotfl: