A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $ \$1$ to $ \$9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were $ 1, 1, 1, 1, 3, 3, 3$. Find the total number of possible guesses for all three prizes consistent with the hint.

There are $ 12$ ways to divide them, not $ 14$. The three unallowable cases are $ (5,1,1), (1,5,1), and (1,1,5)$ So the ans is $ 12(35) \equal{} 420$

[hide="solution"] First, we know that there are $ \frac{7!}{3!4!}\equal{}35$ different arrangements of four ones and three threes. If we count the number of ways to divide each arrangement into three groups, with each group having one to four numbers, we may let each group be one of A, B, and C and then find our answer.

Each arrangement will still have the same number of groups possible, so we can just count the number of groups possible of 1111333. We have four cases:

Case 1: A is just the first 1-
Then B can either be 11, 111, or 1113, giving three groups here.

Case 2: A is 11-
Then B can be 1, 11, 113, or 1133, giving four groups.

Case 3: A is 111-
Then B can be 1, 13, or 133, which are three more possible groups.

Case 4: A is 1111-
Then B is 3 or 33, just two more groups.

In all, we have 3+4+3+2=12 groups, giving a total answer of $ 35\cdot{12}\equal{}\boxed{420}$.[/hide]

[hide]
We first find the number of ways to assign a number of digits to each prize.  We seek the number of solutions to $ a\plus{}b\plus{}c\equal{}7$, where $ a$, $ b$, and $ c$ are positive integers less than or equal to $ 4$.  Balls and urns gives us $ 15$ total solutions, but $ 3$ of these come from permutations of $ (5,1,1)$, so there are only $ 12$ which meet the requirements.

Now, there are $ \binom{7}{3}\equal{}35$ ways to choose three digits to receive 3's rather than 1's.  So the answer is

$ 35(12)\equal{}\boxed{420}$.
[/hide]

[quote="xpmath"][hide="solution"] First, we know that there are $ \frac {7!}{3!4!} \equal{} 35$ different arrangements of four ones and three threes. If we count the number of ways to divide each arrangement into three groups, with each group having one to four numbers, we may let each group be one of A, B, and C and then find our answer.

Each arrangement will still have the same number of groups possible, so we can just count the number of groups possible of 1111333. We have four cases:

Case 1: A is just the first 1-
Then B can either be 11, 111, or 1113, giving three groups here.

Case 2: A is 11-
Then B can be 1, 11, 113, or 1133, giving four groups.

Case 3: A is 111-
Then B can be 1, 13, or 133, which are three more possible groups.

Case 4: A is 1111-
Then B is 3 or 33, just two more groups.

In all, we have 3+4+3+2=12 groups, giving a total answer of $ 35\cdot{12} \equal{} \boxed{420}$.[/hide][/quote]
Instead of doing casework, I partitioned it so that there were $ \binom{6}{2} \equal{} 15$ ways to group seven elements, but forgot to subtract overlaps before multiplying by 35...

At first I did that as well, but then I decided to double check with the casework solution.

[quote="username718"]This problem was very misleading, I would have got it except the problem leads you to think that the contestant must also get the right order of the 3 numbers. Consequently, I did some tedious casework where I multiplied every case where I had 3 different numbers by 6, and multiplied by 3 when there were two of the same number.[/quote]
The contestant DOES have to get the right order of the 3 numbers.

There will be one solution counted twice:
11, 11, 333

as 11 and 11 are two identical 2-digit numbers.

Instead of generating 6 different guesses, these should only generates 3 different guesses, right?

(12 - 3) * 35 = 280

did I miss something?
thx.

I don't believe they are counted twice. If you notice the way I did it, I only counted each group once.

Prize values do not have to be distinct. Also, you multiplied by $ 8$, not by $ 9$.

I liked this problem. (My first AIME too :P)
[hide]At first I added brute forceish and got 305 (xD FAIL). I gotta get more organized now :).

But yeh. For every possible arrangement of the four 1's and the three 3's and ordered pair $ (a,b,b)$, where $ a\plus{}b\plus{}c\equal{}7$, take the first $ a$ digits to be the the price of prize $ A$, the first b digits to be the price of prize B, and the first c digits to be the price of prize C. We know that A, B, and C have to be less than 4, so the possibilities to "place" the dividing lines are $ \binom{6}{2}\equal{}15$, however (1,1,5), (1,5,1), and (5,1,1) will not work. Therefore, there are 12 ways to take different number of digits A, B, and C. There are $ \binom{7}{3}\equal{}35$ ways to "arrange" the 1's and 3's, and 12 ways to make prize value numbers (a,b,c) from each of them, so the answer is $ 12\cdot35\equal{}\boxed{420}$

[/hide]

[quote="username718"]This problem was very misleading, I would have got it except the problem leads you to think that the contestant must also get the right order of the 3 numbers. Consequently, I did some tedious casework where I multiplied every case where I had 3 different numbers by 6, and multiplied by 3 when there were two of the same number.[/quote]

i did the exact same thing

I brute forced this one, it only took me about 45 minutes :D

[quote="simo14"]I got 490, so could someone tell me what i did wrong:
[hide]first i found out the number of ways to order the numbers 1,1,1,1,3,3,3. that's 7!/(4!3!) = 35
then i found out the number of solutions to x_1 + x_2 + x_3 = 7, for which none of them is 0 or greater then 4. and that's (6 choose 2) -1 =14
so 14*35=490 [/hide][/quote]

Instead of (6 choose 2) -1, it's (6 choose 2) -3, because x_1, x_2, and x_3 being 5 are 3 different cases.
So its 35 X 12 = 420.

I don't understand why you multiply $ \binom{6}{2}\minus{}3$ by 35. 12 Is how many ways you can partition A,B,C right? I don't see where multiplying by $ \binom{7}{3}$ comes in.

A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $1$ to $9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were $1, 1, 1, 1, 3, 3, 3$. Find the total number of possible guesses for all three prizes consistent with the hint.

[hide="solution"]
Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$, then the string is

\[1131331.\]

Since the strings have seven digits and three threes, there are $_7C_3$ arrangements of all such strings.

In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups.

Let's look at the example. We have to partition it into 3 groups with each group having at least 1 digit. In other words, we need to find the solution to

\[x+y+z=7, x,y,z>0.\]

This gives us

\[_6C_2\]

ways by balls and urns. But we have counted the one with 5 digit numbers; that is, $(5,1,1),(1,1,5),(1,5,1)$.

Thus, each arrangement has \[_6C_2-3=12\] ways per arrangement, and there are $12\times35=\boxed{420}$ ways.
[/hide]

I am wondering why we don't need to worry about over-counting such as having A and B being the same amount, etc

[color=red]  Mod note:  Merged [/color]

[quote="InProgress"]A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $1$ to $9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were $1, 1, 1, 1, 3, 3, 3$. Find the total number of possible guesses for all three prizes consistent with the hint.

[hide="solution"]
Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$, then the string is

\[1131331.\]

Since the strings have seven digits and three threes, there are $_7C_3$ arrangements of all such strings.

In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups.

Let's look at the example. We have to partition it into 3 groups with each group having at least 1 digit. In other words, we need to find the solution to

\[x+y+z=7, x,y,z>0.\]

This gives us

\[_6C_2\]

ways by balls and urns. But we have counted the one with 5 digit numbers; that is, $(5,1,1),(1,1,5),(1,5,1)$.

Thus, each arrangement has \[_6C_2-3=12\] ways per arrangement, and there are $12\times35=\boxed{420}$ ways.
[/hide]

I am wondering why we don't need to worry about over-counting such as having A and B being the same amount, etc[/quote]
Can you give an example of a triplet (A, B, C) that is counted twice/three times/ect by this method? If not, overcounting hasn't occurred.

For a more formalized explanation:

Each string of the 7 digits is distinct. So thus, it is impossible to get the same triplet (A, B, C) from two strings. Each partition of the string is distinct, so you cannot get the same triplet from one string.

Sorry for revive, can someone explain why we cannot do 35*6 for each case of how long the integers can be and account for the overcounts? I found there were 3 cases of the lengths of integers (4,2,1), (3,3,1), (3,2,2). Multiplying by 6 comes from the fact that once we have a possible arrangement we have 6 ways to rearrange the numbers as A, B and C.

In other words, how can you tell this is a ball and urn problem? Thanks.

[quote=v_Enhance][quote="TennisVictor"]I don't understand why you multiply $ \binom{6}{2} \minus{} 3$ by 35. 12 Is how many ways you can partition A,B,C right? I don't see where multiplying by $ \binom{7}{3}$ comes in.[/quote]

There are two steps involved: sorting the $ 1,1,1,1,3,3,3$ and then partitioning it into $ A$, $ B$, and $ C$.

$ \binom{6}{2}\minus{}3\equal{}12$ is the number of ways to partition it (balls and urns).  (e.g. $ 11: 113: 33$, $ 1: 111: 333$, etc.)

In contrast, $ \binom{7}{3}\equal{}35$ is the number of ways to rearrange  $ 1,1,1,1,3,3,3$.  (e.g. $ 1331311$, $ 1311331$, etc.)

Therefore, we multiply these two values together to get $ 35(12)\equal{}\boxed{420}$, which is our final result.[/quote]

This mostly makes sense, but why are we subtracting $3$ from $\binom{6}{2}$?

The three cases we remove are when one of the partitions has 5 digits, for example 1 | 1 | 1 3 3 3 3. There's a limit of 9999 for each prize, which we have to take into consideration.

Thanks - I understand now!

[hide=solution in my own words (this is just for myself, i'm aware its similar to v_enhance's)]First we arrange $1,1,1,1,3,3,3$ for $\frac{7!}{3!4!}=35$ ways. Next, the amount of ways to place two dividers is just $\binom{6}{2}=15$ (since there are six positions) but since we don't want to create any five digit numbers, we must eliminate 1|1|11333, 11113|3|3, and 111133|3|, for $15-3=12$ ways. Multiplying, the answer is $\boxed{420}$.[/hide]


Show that there is no function $ f: \mathbb R \to \mathbb R$ such that  for all $ x,y\in \mathbb R$,
\[ f(f(x\plus{}y))\equal{}f(x)\plus{}f(y)\minus{}|x\minus{}y|.\]

[quote="Rose_joker"]Show that there is none function from $ R$ to $ R$ such that
for all $ x,y\in R$
$ f(f(x \plus{} y)) \equal{} f(x) \plus{} f(y) \minus{} |x \minus{} y|$[/quote]

Let $ P(x,y)$ the assertion $ f(f(x\plus{}y))\equal{}f(x)\plus{}f(y)\minus{}|x\minus{}y|$

$ P(x,x)$ $ \implies$ $ f(f(2x))\equal{}2f(x)$

$ P(2,0)$ $ \implies$ $ f(f(2))\equal{}f(2)\plus{}f(0)\minus{}2$ and so  $ (E1)$ : $ 2f(1)\equal{}f(2)\plus{}f(0)\minus{}2$
$ P(\minus{}2,0)$ $ \implies$ $ f(f(\minus{}2))\equal{}f(\minus{}2)\plus{}f(0)\minus{}2$ and so  $ (E2)$ : $ 2f(\minus{}1)\equal{}f(\minus{}2)\plus{}f(0)\minus{}2$
$ P(1,\minus{}1)$ $ \implies$ $ f(f(0))\equal{}f(1)\plus{}f(\minus{}1)\minus{}2$ and so  $ (E3)$ : $ 4f(0)\equal{}2f(1)\plus{}2f(\minus{}1)\minus{}4$
$ P(2,\minus{}2)$ $ \implies$ $ f(f(0))\equal{}f(2)\plus{}f(\minus{}2)\minus{}4$ and so  $ (E4)$ : $ f(2)\plus{}f(\minus{}2)\equal{}2f(0)\plus{}4$

Adding the four equalities $ (E1)$ to $ (E4)$ implies $ 0\equal{}\minus{}4$

So, no such function exists.


[quote="satsun"](i) how many odd six-digit palindromic numbers are  there?
(ii) how many odd seven-digit palindromic numbers are there in which every digit appears at most twice?[/quote]

(i)[hide]If the last digit is odd, then the first digit is odd also, so for the first digit we have 1,3,5,7,9 which means 5 chocies. Second and third digits we can have any numbers, so 10 digits each. Once we choose the first three, the last three digits have already been chosen by our definition of a palindrome.

So we have $5\cdot10\cdot10 = \fbox{500}$[/hide]

{ii}[hide]Ok, so we're still odd, but now we have to add in the fact that we can't have a digit appearing twice. Like last scenario, we have 5 choices for the first number. Next we have to remember our restriction of "ever digit appearing at most twice", and we have to think about how that's going to work with palindromic numbers.. if I choose 1 for the first digit, that means 1 is already going to have a place in the last spot, taking care of our restriction. So for the second digit we have 9 choices (all numbers except for the first number). The third digit we have 8 choices (all numbers except the first and second numbers). The fourth digit we have 7 choices (all numbers except the first, second, and third.) And, just like the last problem, once we choose a certain number of digits (in this case, 4) the rest are chosen for us.

This gives us $5\cdot9\cdot8\cdot7 = \fbox{2520}$[/hide]

Number 1: Odd digits are $1,3,5,7,9$ so the first digit has to be like the last so $(5)(10)(10)(1)(1)(1)$, which is $500$.

Number 2: Don't really know :?

$(5)(10)(9)(8)(1)(1)(1)$, which is $3600$?

[quote="Iversonfan2005"][quote="nr1337"][quote="Iversonfan2005"]Number 2: Don't really know :?

$(5)(10)(9)(8)(1)(1)(1)$, which is $3600$?[/quote]

How about 5*9*8*7 = 2520?[/quote]

Oh yes, because you already used one number in the first digit, sorry about that :blush:[/quote]

No problem.  Dumb mistakes happen to me all the time.


geometry
combinatorial geometry
IMO Shortlist
combinatorics
IMO Longlist

Given $5$ points in the plane, no three of them being collinear. Show that among these $5$ points, we can always find $4$ points forming a convex quadrilateral.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]

This is the famous result of Esther Klein which introduced Erds and Szekeres to Ramsay theory....

Clearly, there is nothing to prove if the convex hull of the five points is a quadrilateral or a pentagon. Since the points are not collinear, we only have to deal with the case that the convex hull is a triangle, say $ABC$, with two of the points, say $D,E$ in its interior.
Now, the line $DE$ meet two side of the triangle $ABC$ (recall that no three points are collinear), say $AB$ and $AC$. Thus $B,C,D,E$ are in some order the vertices of a convex quadrilateral, and we are done.

Pierre.

Isn't this rather combinatorics? I just can't believe I'd solve a [i]Geometry[/i] longlisted question so easy :)

If point E is outside ABCD, the problem is trivially solved. If it's inside, draw the diagonal AC. Since no 3 are collinear, E is on one side of AB, say WLOG on the side towards D.

Then it follows immediately that ABCE is convex :)

Peter

Ah darn, Pierre is once again faster.

And this time with a <1 page solution this must be really easy! :D

Let the $5$ points be $A$, $B$, $C$, $D$, and $E$. We first consider $A$, $B$, $C$, and $D$. If each of these is in the exterior of the triangle determined by the other $3$ points, we are done by $ABCD$. If one of these is in the interior, however, WLOG let it be $A$. We now proceed with casework based on the position of $E$.

If $E$ is in the exterior of triangle $BCD$, the lines $AB$, $AC$, and $AD$ split the plane into $3$ regions. If $E$ is in the region determined by $AB$ and $AC$, then $ABEC$ is convex, and analogous reasoning applies to the other $2$ regions.

If $E$ is in the interior of triangle $BCD$, extend $BA$ bast $A$ to intersect $CD$ at $E$, and define $F$ and $G$ similarly. Then, if $E$ is in the interior of triangle $ACG$ or $ADF$, $AECD$ or $AEDC$ is convex respectively. Analogous reasoning applies to the other $2$ pairs of triangles. $\blacksquare$


geometry
induction
combinatorics solved
combinatorics

Hi!
This is the first (but certenly not the last) time I bother you with an own problem.
Let D be a disc of radius R>1. Infinitely many segments with endpoints in D are drawn such that every point of D is the endpoint of exactly one segment. For any X belonging to D, let X' be the other endpoint of the the segment which has the endpoint X. You are given that AB<1 iff A'B'<1. Prove that one can find A in D such that AA'<1.
I know it's not a nice problem, but I'm interested in other solutions becaues mine is rather strange.

adrian you are welcome to bother us anytime you want, in fact we encourge that.

PS be sure not to miss the head start for the MathLinks Contest this time, ok? :D

Sure, Valentin.
But what about the problem? Is it so hard? Or just too ugly?

No it looks nice....
I thought having a solution but my crucial lemma revealed to be false (aaargh...), so....

Pierre.

I don't know if it's Ok, but here it goes:

We can look at X' as f(X), with f: D->D a bijection.

From the condition AB<1 iff A'B'<1 we get that the image of an open disk of radius 1 is also an open disk of radius 1. Moreover, the image of the center of the first disk is the center of the image disk.

Let's assume there exist disks O and O' (they're the image of one another) s.t. O and O' intersect. The image of one of the points in the intersection region must be in both O and O', so it's also in the intersection region (which looks sort of like a lens). We assume XX'>=1 for all x in D. This means that the distance between any point in the intersection of the 2 disks above and its image is also >=1. Let's look at the point in the "center" of the region. We can easily see that in order for its image to be at least 1 unit apart, the intersection must be the whole disk, so the 2 disks O and O' must coincide. This gives us 2 cases:

(1) For a certain open disk O of radius 1 its image is the same as O, in which case the center of O is the same as its image, so the distance between them is 0<1;

(2) For all open disks of radius 1 their images do not intersect them, which is equivalent to saying that the centers of the disk and its image are at least 2 units apart or, in other words, fdor all points X in D, XX'>=2 (notice that if a disk of radius 1 centered at X doesn't fit in D, we only need to consider the intersection between the disk of radius 1 and D). In order to finish the proof, I want to prove that the sentence [b]"the image of an open disk of radius 1 is also an open disk of radius 1. Moreover, the image of the center of the first disk is the center of the image disk."[/b] is also true if we replace 1 with n \in N*. This can be done easily by induction. By using this new sentence with 2 instead of 1 we get back to the initial problem, and we can't continue to say [b]"for all X in D XX'>=n"[/b] for indefinitely large n, so the proof will eventually end.

Think it's correct? It might be flawed, since I thought it over during a train ride, and I can't concentrate that well on trains :D.

this may be due to my noobness but I'd think this is rather geometry than combinatorics :?

This is usually called 'combinatorial geometry' :D

>  is also true if we replace 1 with n  N*. This can be done easily by induction.

Well it may be true but could you give a complete proof of that?

Pierre.

Here's what I did (if I understand correctly what you're asking me to do :)):

Every point inside the disk (O1) of radius 1 goes to a point inside (O1'), right? Take A to be such a point, and let (A) be the disk of center A and radius 1. All the points inside (A) are turned into the points of (A'), another disk of radius 1, with center A' inside (O1). When A moves inside (O1), the disks (A) cover the surface of a disk (O2), centered at O1=O2 and of radius 2, and since the points of the disks (A') are inside a disk of center O2'=O1' and radius 2, it means that all the points of an open disk of radius 2 and center O2 go to an open disk of center O2' and radius 2. To prove that this is a bijection, all we have to do is think conversely, from (O2') to (O2).

If I understand correctly, this means that we can say [b]"AB<n <=> A'B'<n"[/b] for all positive integers n. I'm not sure about the proof above, maybe I'm missing something. Please share your opinions with me :).

You are right about "AB < n iff A'B' < n".
But, what is the problem about the fact that XX'  \geq n will eventually become untrue? You will have XX' < n for some X but how will you get XX' < 1?
As you mentioned, the image of an open disk is not necessarly an open disk (it is the intersection of an open disk with D).

Pierre.

I don't think the fact that the disks may not fit inside D is crucial. In fact, I don't even see why it is important for my initial "proof" (since I'm not sure it's Ok yet :D).

Ok, let me say it clearer (n-disk means disk of radius n):

We assume that XX'>=1 for all X. Then I have showed that any 1-disk and its image are disjoint. This, in turn, means that XX'>=2 for all X, because the center of any 1-disk is turned into the center of its image disk. From here we get (by the same reasoning as b4) that every 2-disk and its image are disjoint. This means that XX'>=4 for all X, and so on.

It can't go on forever, because XX' is a segment included in D, so XX'<=2R for all X, so we have a contradiction arising from the assumption that XX'>=1 for all X.

Ok, nice job Grobber!

Is it your solution too, Dzeta?

Pierre.

No, my solution is different.
Supose XX’>=1 for every X. Let O be the center of the disc. Take S(X) the simetrical of X related to O (I mean S(X), X and O colinear and OX=OS(X)). Define a sequence of points An(n>=0) by A0=O and An+1=An’ if n is even and An+1=S(An) if n is odd.
I prove by induction on n that for any 0<=i<j<=n, AiAj>=1. For n=1 this is obvious. Supose it’s true for any smaller n. I will prove it now for n. If j!=n (j is not n), then we already now that AiAj>=1. So, supose j=n.
Case 1. n is even. Supose there is one i such that AiAn<1. If i is even, then Ai-1An-1<1 (since Ai=S(Ai-1) and An=S(An-1)), imposible (because the asertion is true for n-1). If i is odd, then take two cases. If i=n-1, then A0An-1<=1/2<1, false. If i<n-1, then S(Ai)S(An)=Ai+1An-1<1, false (i+1!=n-1 since they are diferent mod 2).
Case 2. n is odd. Supose there is one i such that AiAn<1. If i is odd, then Ai’An’<1 means Ai-1An-1<1, false. If i is even, take again the 2 cases. If i=n-1,  An-1An<1 means An-1An-1’<1. If i<n-1, Ai’An’<1 means Ai+1An-1<1, false again and the claim is proved.
Now take n>(R+1/2)^2-1 (R is the radius). Take the points A0, A1, …, An and draw n+1 disc of radius ½ such that every Ai is the center of one disc. The claim I proved shows that the discs are disjoint. But the sum of their areas is bigger than the area of of a disc D’ with center O and radius R+1/2, which contains them all, imposible and the problem is solved.
Anyway, nice solution, Grobber!


Functional Analysis
advanced fields
advanced fields unsolved

Let 'X' be a Banach space and S,T $\in BL(X)$. $(T_{n})$ and
$(S_{n})$ be a sequence in BL(X). Suppose that $T_n\rightarrow T$
and $S_n\rightarrow S$ STRONGLY as $n\rightarrow \infty$ then show
that $S_nT_n\rightarrow ST$ strongly. Suppose that $T_n\rightarrow T$ and $S_n\rightarrow S$ WEAKLY as $n\rightarrow \infty$ then can
we say that $S_nT_n\rightarrow ST$ weakly ?

For the first part (about strong convergence):

Since $(T_n)_n$ is bounded for all $x\in H,\ (\|T_n\|)_n$ is bounded by the Banach-Steinhaus theorem. Fix an $x$. We have $\|T_nS_nx-TSx\|\le\|T_nS_nx-T_nSx\|+\|T_nSx-TSx\|$. The left summand goes to $0$ because if $\|T_n\|\le M,\ \forall n$, then $\|T_nS_nx-T_nSx\|\le M\|S_nx-Sx\|$, and $S_nx\to Sx$. The right summand also goes to $0$, so $(T_nS_n)x\to (TS)x,\ \forall x\in H$, i.e. $(T_nS_n)_n$ converges to $TS$ strongly, as desired.

For the second part, the answer is negative, I think:

Let $X$ be a separable Hilbert space with an orthonormal basis $(e_i)_{i\ge 1}$. Let $S,T$ be the trivial operators (i.e. they send everything to zero), and let $S_n,T_n$ be the following diagonal operators: $S_n$ sends $e_n$ to $e_1$ and $e_j$ to $0$ for $j\ne n$, and $T_n$ sends $e_1$ to $e_n$ and $e_i$ to $0$ for $i\ne 1$. Obviously, $(S_nx,y)\to(Sx,y),\ (T_nx,y)\to(Tx,y)\ \forall x,y\in X$, so the sequences $(S_n)_n,(T_n)_n$ converge weakly to $S$ and $T$ respectively. However, $(STe_1,e_1)=0$, while $(S_nT_ne_1,e_1)=(e_1,e_1)=1,\ \forall n$, so $[(S_nT_ne_1,e_1)]_n$ does not converge to $(STe_1,e_1)$.


algebra
system of equations
algebra unsolved

Solve the system of equations:
$\left\{ \begin{aligned} x^4+y^2-xy^3-\frac{9}{8}x = 0 \\ y^4+x^2-yx^3-\frac{9}{8}y=0 \end{aligned} \right.$

You didn't find all the solutions. Though I didn't analyse your solution carefuly but I think you lost some in the case $y=\frac{1}{2x}$.

I'm making stupid mistakes, as always. I hope it's correct right now.

That is right.

Alternatively we can multiply the first equation by $y$, second equation by $x$, subtract one from another and factor. We get same conditions:
$2xy-1=0$ or $x-y=0$.

For the second condition, we can substitute $y=\frac{1}{2x}$ into the first equation, multiply it by $8x^2$, and get $8x^6-9x^3+1=0$, which yields $x=1$ or $\displaystyle\frac12$. So there are another two solutions: $(1,\frac12)$ and $(\frac12,1)$.

This problem comes from the Austrian-Polish MO 1985   :D

Why do they all comes from the Austrian-Polish MO?I mean they can't find some new problem?

[quote="TomciO"]$8x^4 + 8y^2 - 8xy^3 - 9x = 8y^4 + 8x^2 - 8yx^3 - 9y$ is equivalent to:
$(x-y)(8x^3 + 16x^2y + 16xy^2 + 8y^3 - 8x - 8y - 9)=0$
Consider the case when $x \not = y$
Then we have
$8x^3 + 16x^2y + 16xy^2 + 8y^3 - 8x - 8y - 9 = 0$ or $8x^4 + 16x^3y + 16x^2y^2 + 8xy^3 - 8x^2 - 8xy - 9x = 0$
From this equation substract the first from the system to obtain:
$16x^3y + 16y^3x + 16x^2y^2 - 8x^2 - 8y^2 - 8xy = 0$ or
$2xy(x^2 + y^2 + xy) = x^2 + y^2 + xy$
$x^2 + y^2 + xy$ is $0$ iff $x=y=0$ so we can divide by it to get:
$y=\frac{1}{2x}$

Put this into the equation $8x^3 + 16x^2y + 16xy^2 + 8y^3 - 8x - 8y - 9 = 0$ to see it's equivalent to:
$x^6 = 1$ so $x=1$ and $y=\frac{1}{2}$ or $x=-1$ and $y=-\frac{1}{2}$. It's easy to check that the first pair is indeed a solution.
We are left with the case $x=y$. It's not hard to verify that this condition yields 2 solutions. The final answer is:
$(x, y) \in \{(0,0), (\frac{8}{9}, \frac{8}{9}), (1, \frac{1}{2})\}$[/quote]
your solution is rigth, but I think 2nd solution is $(\frac{9}{8}, \frac{9}{8}$)$

The solution of his line too long!!!!!!!!!!.
Solve the system of equations:
$ \left\{\begin{aligned}x^{4}+y^{2}-xy^{3}-\frac{9}{8}x = 0          (1)\\ y^{4}+x^{2}-yx^{3}-\frac{9}{8}y=0               (2)\end{aligned}\right. $
My solution:
$(1)*x^2$,$(2)*y^2$ then subtract side, hence $x^6-y^6-\frac{9}{8}x^3+\frac{9}{8}y^3=0$
equivalent to $(x^3-y^3)(x^3+y^3-\frac{9}{8})=0$
equivalent to $x=y or x^3+y^3=\frac{9}{8}$
with x=y (easy)
$x^3+y^3=\frac{9}{8}$     (3)
With $x^3+y^3=\frac{9}{8}$ then (1):x,(2):y then add side, we get
$\frac{x^3+y^3}{xy}=\frac{9}{4}$ equivalent to $xy=1/2$, hence $x^3y^3=1/8$      (4)
Since (3) and (4)............

1° case: If [size=150]x,y≠0[/size] and [size=150]x/y[/size] isn't a cube root of unity:

From the first equation, we have:
[size=150]x³ - y³ - 9/8 = -y²/x[/size]
[size=150]x³ + y³ - 9/8 = -y²/x + 2y³[/size]

Analogously, from the second equation,[size=150] x³ + y³ - 9/8 = -x²/y + 2x³[/size]

Then,[size=150] -y²/x + 2y³ = -x²/y + 2x³[/size]
[size=150](x³ - y³)/xy = 2(x³ - y³)[/size]
[size=150]xy = 1/2[/size]

Multiplying the first equation[size=150] x²[/size] :
[size=150]x^6 - 9x³/8 + 1/8 = 0[/size]
[size=150]x³ =  1 or 1/8[/size]

Analogously ,[size=150] y³ = 1 or 1/8[/size]

w.l.o.g. , [size=150]x³=1[/size] ,[size=150] y³=1/8[/size]
[size=150]x=w^i[/size] ,[size=150] y=w^j/2 [/size],[size=150] i,j[/size] in[size=150] {0,1,2}[/size] ,[size=150] w = cis(2π/3) = cos(2π/3) + i sin(2π/3)[/size]
[size=150]3 l i+j[/size] , because[size=150] xy=1/2[/size]
Checking the solutions,[size=150] (x,y) = (1,1/2) , (1/2,1) , (w, w²/2) , (w²/2, w) , (w², w/2) , (w/2, w²)[/size] are solutions

2º case: If [size=150]x,y≠0[/size] and[size=150] x/y[/size] is a cube root of unity:
Then [size=150]x³=y³[/size]. From the equations:
[size=150]y² - 9x/8 = 0[/size]  , [size=150] x² - 9y/8 = 0[/size]

Easily can check [size=150](x,y) = (9/8, 9/8) , (9w/8, 9w²/8) , (9w²/8, 9w/8)[/size] are solutions in this case

3° case: If [size=150]x=0[/size] or [size=150]y=0[/size]
In this case[size=150] (x,y)=(0,0)[/size] is solution.


A man enters bank 1 with a certain sum of money,he gives Rs 20 to the Cleark at the entrance and enters the bank, the money doubles in the bank and he returns out with the money and on his way he again gives Rs 20 to the cleark at the entrance.
He Now enters bank 2 with the remaining money and at the entrance he gives  Rs 20 to the cleark and
enters the bank, the money doubles in the bank and he returns out with the money and on his way he again gives Rs 20 to the cleark at the entrance.
He now enters bank 3 with the remaining money and at the entrance he gives  Rs 20 to the cleark and
enters the bank, the money doubles in the bank and he returns out with the money and on his way he again gives Rs 20 to the cleark at the entrance. He has no money now left with him when he comes out of Bank 3 .How much did he originally have. :lambda  :blush:

[quote="NoSoupForYou"][hide="Da answer..."]$300$[/hide][/quote]

i got something different [hide]$ 52.50 $[/hide]

[quote="NoSoupForYou"]That won't work because then he only has 10$ as he leaves the second bank. Not enough to pay the clerk at the third bank.[/quote]
I also think the answer is Rs 52.5.
BTW, what does Rs mean?

Oops... misread the problem. Heh heh, I solved for an interest rate of $\text{-50}$%.  :D  Yes, $52.50$ is correct.

Rs = rupees, the Indian national currency.


A cube with two faces ABCD, BCEF, has side 12. The point P is on the face BCEF a perpendicular distance 5 from the edge BC and from the edge CE. A beam of light leaves A and travels along AP, at P it is reflected inside the cube. Each time it strikes a face it is reflected. How far does it travel before it hits a vertex?

hmmm...i could do this by just making a lot of equations and reflecting them through faces and whatnot until the photon hits a vertex again. That would work, but it would mean a lot of mindless computation. Better way, anyone?

Can you think of any conditions that must be true when it finally hits a vertex? See if you can do it without computation.


inequalities
number theory
prime numbers
relatively prime

If   $ P\equal{}p_{1}p_{2}...p_{n}$ ,  $ p_{1},p_{2}..p_{n}$  are the first  $ n$  prime numbers than prove that:

$ p_{n\plus{}1}<P$

[quote="Elio (n)"]If   $ P \equal{} p_{1}p_{2}...p_{n}$ ,  $ p_{1},p_{2}..p_{n}$  are the first  $ n$  prime numbers than prove that:

$ p_{n \plus{} 1} < P$[/quote]

Observe number $ A \equal{} p_1 p_2 ...p_n \minus{} 1$ , which  is prime ,because it is not dividable by any $ p_i , (i \equal{} 1,2, ...,n)$

Obviously $ A \geq p_{n \plus{} 1}$ , where $ p_{n \plus{} 1}$ is $ (n \plus{} 1)$th prime number.

Since $ P > A$ and $ A \geq p_{n \plus{} 1}$ then $ \Rightarrow P > p_{n \plus{} 1}$ ,as desired.

Here is another one (similary) :

Let be $ p_{k}$ any prime number which divides $ A\equal{}p_{1}p_{2}..p_{n}\minus{}1$ therefore $ p_{k}\le A$

Since A and P are relative primes , it comes that $ p_k$ is defferent from $ p_{1},..p_{n}$ , so it must be $ p_k\ge p_{n\plus{}1}$

Finally we get $ p_{1}p_{2}..p_{n}> p_{1}..p_{n}\minus{}1 \ge p_{k}\ge p_{n\plus{}1}$       done .

[quote="enndb0x"]Observe number $ A \equal{} p_1 p_2 ...p_n \minus{} 1$ , which  is prime ,because it is not dividable by any $ p_i , (i \equal{} 1,2, ...,n)$ [/quote]

not true, $ 2\cdot 3\cdot 5 \cdot 7 \minus{} 1 \equal{} 209 \equal{} 11\cdot 19$

Thank you ocha,

Observe number $ A\equal{}p_1 p_2 ...p_n \minus{}1$ ,which is relatively prime with $ p_i  ,\forall i\equal{}1,2,...,n$

If $ A$ is prime number ,then it's  greater or equal to $ p_{n\plus{}1}$ .

Then $ P >A \geq p_{n\plus{}1}$

If $ A$ is composite ,then it's prime factors are greater than $ p_n$ ,which means that $ A \geq p_n$  ,then $ P>A\geq p_n$

Is it correct now ?  :maybe:


induction
number theory unsolved
number theory

hi!

here's a good q

simple enough for many..but nt fer me

We call a positive integer n special if every positibe integer less than or equal to n can be written as the sum of distinct divisors of n.

Prove that the product of two special nos is also special...
eg of a special no

6
since 1=1,2=2,3=3,4=1+3,5=2+3,6=1+2+3

need a reply fast

We can prove the following general form:

If $ n$ is a special number and $ m$ is a positive integer, then $ mn$ is a special munber.

Proof

$ n$ is a special number,there is $ n\equal{}n_1\plus{}n_2\plus{}\cdots\plus{}n_e$ such that $ n_i$'s are all distinct and $ n_i|n.$

And we have $ mn\equal{}mn_1\plus{}mn_2\plus{}\cdots\plus{}mn_e$ such that $ mn_i$'s are all distinct and $ mn_i|mn.$

So $ mn$ is a special munber.

(End of proof)

(1) This is from Canada, 2002:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77778[/url]

(2) To Makoto Kohno: You are not using the correct definition.  It is not enough to express $ n$ itself as the sum of distinct divisors.  You must express every number from 1 to $ n$ as the sum of distinct divisors.

[quote="nsato"]You must express every number from 1 to $ n$ as the sum of distinct divisors.[/quote]

Thank you for your note.
I misunderstood the definition. :o

hi nsato,

could you please get the proof in a much better way

i could not get the thing that well.

could you put it for me in a form of a formal proof. :maybe:  :maybe:  :maybe:

need your help fast...

sfsreporter

I have a question.

For example 1=1 , 2=2 and 1|2.
For example 1=1 , 2=2 , 3=1+2 , 4=4 and 1|2|4.

So, are '2'  and '4'  special numbers ?

More generaly , is 2^n a special number ?

[quote]1=1,2=2,3=3,[/quote]

made such a question.

no
2^n cannot b said as the general form as the example given by me is 6 which is not 2^n..

1=1
2=2
3=1+2
4=4
5=4+1
6=2+4
$ \cdots$ using binary representation.

So $ 2^n$ are special numbers , I think.

to elaborate the proof in that other link

we want to prove that every number smaller or equal to $ mn$ can be represented as a sum of some different devisors of $ mn$. ($ mn$ obviously can be represented in such a way) every integer $ x < mn$ is of a form $ am \plus{} b$ where $ a < n$ and $ b < m$.($ a,b$ not both equal to $ 0$)

because $ n$ is a special number $ a$ can be represented (if not equal to 0) as a sum of some different devisors of $ n$ denote $ a \equal{} z_1 \plus{} z_2 \plus{} ...z_k$ where $ z_i$ are these devisors. therefore $ am$ can be represented as a sum of some devisors of $ mn$ because $ am \equal{} mz_1 \plus{} mz_2 \plus{} ...mz_k$ and $ mz_i|mn$.

because $ m$ is a special number $ b$ can be represented (if not equal to zero) as a sum of some different devisors of $ m$. denote $ b \equal{} w_1 \plus{} w_2... \plus{} w_l$. an can also be represented consequently as a sum of some different devisors of $ mn$ because $ w_i|m|mn$

$ min(mz_i)\geq m > max(w_i)$ therefore none of the $ mz_i$s can be equal to any of $ w_i$s. therefore $ am \plus{} b$ is also representable as sum of different devisors of $ mn$. and with this the proof is finished.

as for the question if $ 2^n$ is a special number. sure it is! $ 2$ is and because the product of 2special numbers is again a special number $ 2\cdot2$ is also a special number, also $ 2\cdot4$, $ 2\cdot8$ etc.

if u don't like this argument u can also do an induction. $ 2$ is a special number. assume $ 2^n$ is also. than because every devisor of $ 2^n$ is also a devisor of $ 2^{n \plus{} 1}$ all numbers up to inclooding $ 2^n$ can be represented due to the induction hypothesis. and all numbers of the form $ 2^n \plus{} k$ where $ 0 < k < 2^n$ can be represented like $ 2^n \plus{} x_1 \plus{} x_2 \plus{} ..$ where these $ x_i$ are representation of $ k$. therefore every number smaller or equal to $ 2^{n \plus{} 1}$ can be represented in the desired way and induction is complete


Given $x,y,z\ge 0$ prove that
$3(xy+yz+zx)(x+y)(y+z)(z+x)\ge 8xyz(x+y+z)^{2}$ :lol:

Sorry I have some mistake I have edited it!  :)

[quote="gemath"][quote="Soarer"]Use
$(x+y)(y+z)(z+x) \ge \frac{8}{9}(x+y+z)(xy+yz+zx)$
[/quote]
Could you please help me to solve it ?[/quote] \[\iff x^{2}y+xy^{2}+y^{2}z+yz^{2}+z^{2}x+zx^{2}\ge 6xyz.\]


(ABCD)=-1

prove that:  1/BC + 1/AB +  1/AD + 1/CD

well whatever it is i simplified it to -[(B+D)(A+C)]

Haha, demji, this is geometry, not algebra ;) A, B, C and D are points, not variables.

Arne,thanks to u information about the problem ;)

and zhaoli az u know i didnt use latex to write this problem. :blush:

i mean:(1:BC) +( 1:AB) + (1:CD) +( 1:AD)

But what do you want us to prove?

1/BC + 1/AB + 1/CD + 1/AD is equal to what?

I've never heard of a geometric figure being harmonic. I can't even think of one now.
I assumed that the definition ashegh gave in his other harmonic post applies symmetrically to the other sides.
$BC\cdot AD=AC\cdot BD \to AB\cdot CD=AC\cdot BD$
$\dfrac{1}{AB}+\dfrac{1}{BC}+\dfrac{1}{CD}+\dfrac{1}{AD}=\dfrac{AB+BC+BD+AD}{AC\cdot BD}$

That's not a very interesting result. I guess that's why it's called harmonic.


LaTeX
inequalities proposed
inequalities

[quote="Maverick"]Let ABC be a triangle. Prove that:

(la+lb+lc)/R \sqrt (2p)  \leq  \sqrt p (1/ \sqrt bc + 1/ \sqrt ca + 1/ \sqrt ab )[/quote]

Explain..... :?

Difficult to unterstand.....

 :)  :)

Indeed. Maverick can you write this problem in Latex, please? :)

[quote="Maverick"]Let ABC be a triangle. Prove that:

(la+lb+lc)/R \sqrt (2p)  \leq  \sqrt p (1/ \sqrt bc + 1/ \sqrt ca + 1/ \sqrt ab )[/quote]
I think I understud this problem  :lol:
$\frac{(l_a+l_b+l_c)}{R\sqrt{2p}}\leq\sqrt{p} (\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}} + \frac{1}{\sqrt{ab}} )$ :?
But if I am right,why hi didn't write this problem in this form
$\frac{(l_a+l_b+l_c)}{R}\leq\sqrt{2}p (\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}} + \frac{1}{\sqrt{ab}} )$ :?


There are six clearly distinguishable frogs sitting in a row. Two are green, three are red, and one is blue. Green frogs refuse to sit next to the red frogs, for they are highly poisonous. In how many ways can the frogs be arranged?

The solution could be a bit more detailed about the part why the frogs have to be placed in groups; Something like: because green ones don't want to be near reds, every red must be adjacent to red frog or blue frog or be in the end. But the only way to do this is to put all the reds to one corner with blue between them and the greens. And then it could conclude the fact that greens are also together, and show the calculation.

IMO.

[hide=Solution]We can examine the number of possible ways to arrange the frogs.

There are 6 spots:

_____ ______ ______ ______ ______ ______

Clearly, the reds must all sit together in a block of 3, or else at least one green frog will be poisoned (:(). Therefore, they have the choice of RRR_ _ _ or _ _ _ RRR. That gives two choices. Then there is only one spot for the blue frog that is valid.

There are $3!$ ways to arrange the red frogs. There are $2!$ ways to arrange the green frogs. The total is $2\cdot 2!\cdot 3!=\boxed{24}$ ways to arrange the frogs.[/hide]

[quote=Equinox8]I feel like levans created this problem.[/quote]

How? Just because there is a frog?

[font=TIMES NEW ROMAN]As far as I know, levans mostly takes care of the code and site support but it's possible.

Why doesn't this work?
[hide=bad, wrong sol]
We use complementary counting. The total number of possibilities is 6!. To create a bad possibility, let there be a chunk where the green and red frogs are together. We can generate 5*2!*4! of these possibilities. Finally, 6!-5*2!*4! = 480 (wrong).
[/hide]

[quote=programmeruser][quote=ProbablyNot]@above the error is 5*2!*4!. Can you explain how you got that?[/quote]

There's 5 ways to place the chunk, 2! ways to order the frogs inside the chunk, and 4! ways to order the frogs outside the chunk[/quote]

there can be two pairs of red and green frogs next to each other so you need PIE
also its easier not to calculate the complement its


geometry
search
AoPS Books
number theory

How much does the Art of Problem Solving book Volume 1 help on a scale of 1-10? 1 being the worst and 10 being the best.

I'd say around 2, unless you're not reading it upside-down.

Are you sure Treething?

I mean, c'mon, is reading upside down really that difficult (after a few hours)? I'm sure you could get used to it, although it would be a nuisance.

The AoPS books are very helpful. I didn't expect there too be that much information.

For example, one can start the books with basic algebraic knowledge (Algebra I)...
and by the end know all of classroom Geometry, Algebra II, and Pre-Calculus/Trig with problem solving background and a touch of number theory, analytical geometry, etc.

It takes perseverance though, for people who already have at least an alg II background to go through the early chapters of AoPS 1 (it seems redundant at first), but I believe they're helpful.

Wrong forum.  Countless threads created for this topic.  Try Search feature.  Locked.


linear algebra
matrix
linear algebra unsolved

Let $A$ and $B$ be two complex square matrices of the same size such that $A^{2005}=B^{2005}=I$ and $AB=BA$. Show that $A+B+I$ is invertible

A and B are simultaneously diagonalizable; it remains to show that if a and b satisfy a^2005=1 and b^2005=1 then a+b+1<>0.
It is easy to show that a+b+1=0  and norm(a)=norm(b)=1 implie {a,b}={j,j^2} where j=exp(2iPI/3).
j^2005<>1 because 3 is not a divisor of 2005.


AMC
AIME
geometry
3D geometry
sphere
USA(J)MO
USAMO

I answered 1-16 and number 18 and got them all right.  I didn't get to any of the other problems though cuz i'm slow and i ran out of time.

I got 143 (#24 and #25 blank, all others right), my best ever.  Despite this, I thought it was on the whole just a bit harder than the A exam (on which I got 132), especially towards the end.   #24 and #25 were not just a matter of running out of time--I probably wouldn't have been able to solve them in any reasonable amount of time.  I mainly scored better than on the A because I didn't make any careless mistakes.  Among teammates with whom I compared answers, it seemed like the most missed problems were #11 and #14, and most of them left #21-25 blank.

[color=indigo]Thank you! I took the 10A this year. I'll take one of each for the next few years. Problem #14 was hard (I didn't try it) but I got 11 and 21.  [/color]

[quote="LadyKn1ght"][color=green]I got 101 by getting 11 right and leaving the others blank.
And I'm only in seventh grade!  :D   [/color][/quote]

u did the EXACT same thing as me! except i answered #1-11 (got them right) and then gave up because i hadn't learned some of the stuff... one of my friends said that amc 12A was a lot easier than 12B... he took both! (i'm in 8 grade)

132.5 (20 right, 5 blank)

Happy with this score. I've answered more questions before (like on the A test) but this is the first time I've managed to avoid making stupid mistakes.

133.5. This is a 16 points improvement from last year! I answered 22. I made a stupid mistake on #12...also, I can't believe how easy #16 is now that I think about it again.
Overall, quite happy.

Same here, 133.5. And also made a dumb reading mistake on number 12 - thought it asked for n/m, not n/p.

112.5  Ridiculously low, I know, but it was also with a 102 degree fever and the flu, so while not ecstatic, I'm content.

111.5, answered 14, and they were all correct. I could've gotten 3 more I think, except my brain just got frozen all of a sudden.

Oh well, qualified for AIME, haha. That'd be 3 hours of procrastination.  :)

Edit: typos

My guess is that it was harder to make a 150 on the B.  Not many students are going to take the nice approaches to both 24 and 25 (which we'll discuss at the Math Jam today).  Not seeing the nice approach on one could cost 10 minutes.  Not seeing the nice approach on both would mean you're probably better off checking all your other answers -- unless you're very fast.

I didn't take the B test, but am I still allowed to join the math jam?

I had done something like 125
18 rights and 7 none

Its my first time entering is it really hard to get a really good score in AIME

yeah, I got a 108.... it was after a long day of school and of only a few hours of sleep..
but still, It was my first ever AMC (well, I guess other than the A) and I'm a sophmore.

I got a 136 on the 12B, and a 143 on the 12A

i was kind of disappointed by the 136 because i didn't feel like i got any hard problems, just managed to correctly answer the easy ones. (first 21)

i doubt i gave it my all because i didn't think i could surpass my previous score on the A.

oh well, i'm satisfied overall.

139.5 (22 right, omitted 18, 23, 25).  I made a stupid mistake on #18 and kept on getting an answer that wasn't among the choices...*sigh*

A and B tests are almost equal in difficulty, i think.  B MAY have been a little harder (i did better on B, though).

137 on the B (down 3.5 pts from the A, but a 137 is a pretty cool score, and I get to use my A score anyways)

I got 100.5.   I answered 1-14, 16, and 18, but got 10 and 11 wrong because of stupid mistakes, and 18 wrong because of a miscalculation due to time pressure.

[quote="MCrawford"][quote="DPopov"][quote="MCrawford"]That's a great score. Just so that you understand the system in the future -- you might consider taking the AMC 10 at least once each of the next few years because that could cut down on the score you need on the AIME to qualify for the USAMO. While that might not matter as much this year, it could in 1, 2, or 3 years.[/quote]

Does this mean that I should have taken the 10 for a better chance at qualifying for the USAMO from the AIME? (I am in the 10th grade with 126.5 AMC 12 A score)[/quote]

I'm honestly not certain.  I have no idea where the AMC system is posted and there was just one exam for everyone back when I was in school.[/quote]The information is posted [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2004-5pub/05TM-USAMO/html-05U/04-05usamo-text.html]here[/url].  You'll probably have to read it at least twice to understand it though (I know that I did).

Their site also states:
[quote="AMC"]In advising students who desire to be selected for the USAMO whether to take the AMC 12 level contests or the AMC 10 level contests, past statistics show that it is to their advantage to take the AMC 12 level contests.[/quote] So there you go.

The other option is to move to Wyoming:
[quote="AMC website"]The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.[/quote]

[quote="MCrawford"][quote="LadyKn1ght"][color=green]I got 101 by getting 11 right and leaving the others blank.
And I'm only in seventh grade!  :D   [/color][/quote]

That's a great score.  Just so that you understand the system in the future -- you might consider taking the AMC 10 at least once each of the next few years because that could cut down on the score you need on the AIME to qualify for the USAMO.  While that might not matter as much this year, it could in 1, 2, or 3 years.[/quote]

Can you explain in more detail why taking both AMC10 and AMC12 will cut down on the score needed on the AIME to qualify for the USAMO?

But volliebal12, you don't know what the lowest AIME score will be. Its possibe it is quite high (ie those with good AMC12 scores dont do badly on AIME.) In that scenario, you don't get the equivalent of 150 from the 10.

[quote="volliebal12"]Now, Im not sure if when they say freshman and sophomores they mean AMC 10 takers or just freshman and sophomores, but on their site they say that the scores (to make USAMO) work like this:

AMC12+AIME score, highest make it.

but, if you took the amc10, you get an automatic 150 on the amc, because what you have to do is get the same or better score of the worst score of the 12 takers that make it.

however, im not sure if sophomores (or under) who took the 12 count as a sophomore or a junior/senior.[/quote]

Sophomores and under who took the 12 do get the benefit of the floor value (as far as I know).


geometry parallelogram trapezoid rhombus rectangle

ABCD is a parallelogram with perpendicular diagonals. Take points E, F, G, H on sides AB, BC, CD, DA respectively so that EF and GH are tangent to the incircle of ABCD. Show that EH and FG are parallel.

If it has an incircle, would it be a square?

ABCD does not have to be a square, just a rhombus. AEFCGH is a hexagon with an incircle, therefore its diagonals AC, EG, FH concur at a point X (Brianchon theorem). As $AE \parallel CG$, triangles $\triangle AEX \sim \triangle CGX$ are centrally similar with similarity center X and similarity coefficient $\frac{AX}{CX},$ having equal angles. As $HA \parallel CF,$ triangles $\triangle AHX \sim \triangle CFX$ are also centrally similar with the same similarity center X and the same similarity coefficient $\frac{AX}{CX},$ having equal angles. Therefore the quadrilaterals $AEXH \sim CGXF$ are centrally similar as well, which means that their corresponding diagonals $HE \parallel FG$ are parallel. See another solution [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=58842[/url] with points at infinity.

[color=darkblue]Let $ABCD$ be a convex quadrilateral. Define :

$\boxed T$- the set of the all [b]trapezoids[/b] $(\ AB\parallel CD\ \ \vee\ \ AD\parallel BC\ )$ ;

$\boxed P$- the set of the all [b]parallelograms[/b] $(\ AB\parallel CD\ \ \wedge\ \ AD\parallel BC\ )$ ;

$\boxed{Rh}$- the set of the all [b]rhombus[/b] $(\mathrm{\ parallelogram}\ \ \wedge\ \ AB=AD\ )$ ;

$\boxed{Re}$- the set of the all [b]rectangles[/b] $(\mathrm{\ parallelogram}\ \ \wedge\ \ A=B\ )$ ;

$\boxed S$- the set of the all [b]squares[/b].

Then, in the my opinion, there are the following inclusion relations : $Rh\cup Re\subset P\subset T\ \ ;\ \ S=Rh\cap Re\ .$

For example, the rhombus is a (particular) parallelogram ![/color]


I am going to try to create another game.
Spell arena is a game that if you sign up, you can play. When your rond comes up, I will give you spells. Some are attack, and some are defense. Each attack that you recieve lowers your magical power. Each defense you deploy lowers the attack power. Once you have no magical power left, you are out. You also lose all of your spells. If you have no more spells, you are out.

I will PM you your spells, and you must keep track of them, too  ;)

Only two people can play at one time.

No one accept me and the players in the current round may talk during rounds.

If one of these rules are broken(or if someone cheats) the round restarts. ;)

The first two players that respond and say that they want to play will be in round one. If there are already two players, please do not request to play until that round is finished.

To deploy a card, you must notify everyone that viewed this thread(reply to this thread) that you are deploying a card. You must tell the attack and defense numbers that come with the card too.

You can only deploy a card once.

I will give each person ten cards.

If one person deploys an attack card and the other person doesn't respond in three days, the attack is deployed.

If the amount of attack and the amount of defense is the same, nothing happens.

If a player responds for more than 5 days, he will be disqualified, meaning that the other person wins.

Please respect the fact that I am taking pains to create this game. Great! Now my fingers hurt!

Also, tell me if you like this game.

Don't be shy to sign up!

Good! We need one more player. After the first two rounds, I will alter the game a little. You can also participate twice. Even thrice. You can participate as many times as you like.

I'll play, though I don't completely understand the rules.

ok. I'll clear it up for you. I give you and Ignite ten cards, with either attack, or defense. A strategy is to play a defense card with an attack card. If the defense is greater than the attack, the attack is eliminated. If the attack is greater than the defense, the defense is subtracted from the attack, and the magic meter will go down. There are 10 points per magic meter. If the magic meter reaches zero, or if the players run out of cards, the person with the least magic loses, and the other player wins. I will pm your cards, and their powers, randomly. I am PMing NOW!

Round one begins. No one but I, daine and Ignite168 may post in this thread until round one is done.
I guess that this is a warlock vs. witch round.
You may enter the arena now, warlock Ignite 168 and witch daine(I couldn't find any other suitable names(witches and warlocks can be good)).

This oddly reminds me of Yu-gi-oh/Pokemon TCG/ some other random TCG.....

I REPEAT, no one accept me, ddaine and Ignite can talk UNTIL the END of round one.

Daine, you have three days to reply, or else the attack card layed down will be played. If you don't reply the next day, you will forfeit to ignite.

The cards are played. The outcome:
[hide]nothing happend, accept that the cards were used up[/hide]

A player should play another attack card, before this coming Thursday.

same outcome as before. The forfeit time is Wed. when I say so.

I tell daine to go.

Hey daine, is this a list of your cards or no?

Defense: 1
Defense: 2
Defense: 5
Defense: 5
Attack: 1
Attack: 1
Attack: 2
Attack: 3
Attack: 3
Attack: 5

I play attack 3.

Edit: no to your question.

NO more lynching information! This is NOT an information stealing game!

P.S. Remember, you can't use a card twice! And it's time for Ignite to go.

I play defense 3.

I think this is a pointless game, why? All it comes down to is what cards you were dealt and who got better ones. The only thing that could change that is activity and the activity rule is pretty lax.

daine, you have posted a card that you no longer have. I'll let that go, but your weakest card will go. From now on, PM me your card choices.

[quote="DarkKnight"]I know I'm not supposed to post, but the people shouldn't be able to see what the other person's card was. They should PM you the cards they want to play and you should just post the outcome.[/quote]
MUCH better idea.
mbabbit, you should just post when someone has sent you a card, asking for the other person to send a defense card.

[quote="mbabbitt2003"]daine, you have posted a card that you no longer have. [/quote]
When did I post a card that I no longer had?

"attack 3" ;)  I'm keeping track of the game ;)

Nothing happend, AGAIN???!!! daine, did you know what card Ignite was pm ing???

P.S. Ignite has until Saturday to put down a card.


induction
combinatorics unsolved
combinatorics

A word is a finite sequence of two symbols a and b. The number of the symbols in the word is said to be the length of a word. A word is called 6-[i]aperiodic[/i] if it does not contain a subword of the form cccccc for any word c. Prove that f(n) > (3/2)^n, where f(n) is the total number of 6-aperiodic words of length n.

[quote="A1lqdSchool"]Hint :
By Induction to prove :f(n)>3/2 f(n-1), n \geq 2[/quote]
It is time to explain your hint!  ;)


Hi, I couldn't seem to find this in the LaTeX guide on the website. Say I have
$aa \ldots a,$
but I want to indicate that there are $n$ total $a$'s in the previous expression. On paper normally I write something underneath the $a$'s, this something looks like a { rotated 90 degrees counterclockwise, and then put an $n$ underneath it. How can I do this in LaTeX?

I hope this description was clear enough. Thanks.

\underbrace is probably what you are looking for:
\underbrace{aa \ldots a}_n gives $\underbrace{aa \ldots a}_n$

Great. That's exactly what I wanted, thanks!


geometry
geometric transformation
rotation
symmetry

A unit square is rotated about a vertex by an acute angle, a, such that cos(a)=4/5.

what is the shaded area?

The key here is noting the symmetry, i.e. the angles with $ a$ degrees.

It is easy to see that the answer is $ 2\cdot\frac12\cdot1\cdot\frac12\equal{}\boxed{\frac{1}{2}}$.

[quote="math154"]The key here is noting the symmetry, i.e. the angles with $ a$ degrees.

It is easy to see that the answer is $ 2\cdot\frac12\cdot1\cdot\frac12 \equal{} \boxed{\frac {1}{2}}$.[/quote]

how did you get your final answer? I understand how you got all the dimensions, but just not the final step.

Split the shaded region into two right triangles down the long diagonal.


Today I encountered the formula $B=\mu_{0}nI$ being used, where $B$ is the magnetic field, and $I$ is the current. I've never seen it before. Can you tell me what formula this is, and what is $n,\mu_{0}$ (give me the link, etc.)?


???????????????????????????????????????????????

I seem to use R2CuLi everywhere................................

1. Convert to pentane
2. Reaction with lithium di-(1'-methyl butyl) copper
3. Bromination to get Br attached to the carbon attached to the methyl group.
4. Elimination with alcoholic KOH

u get 2 compounds now (doesn't matter)

5. Reaction with (CH3COO)2Ag, THF, NaBH4, H2O (oxymercuration-demercuration)

Your sequence doesn't make sense. For example, from where come from the Gilman reagent in 2.?

What do you mean from where does it come????

I cant use gilman reagent directly??
Do you want to say that in a synthesis, i can use gilman reagent only if i have prpared it with available material?????

The purpose of the problem is to synthesize 4-methylnonan-4-ol using pent-1-ene as the [i]only organic starting material[/i].

i am useless. :( ....dont know anything about chemistry...

Here is my proposal:

1. hydroboration/oxidation to get 1-pentanol
2. oxidation with PCC in dichloromethane to get pentanal
3. self-condensation using sodium hydroxide
4. oxymercuration/demercuration
5. Wolf-Kishner reduction

I think I may need acidic workup at the end too, but I'm not sure.  I'm thinking that the carbanion that is formed in the 5th step could pick up a proton intramolecularly or from one of molecules that hasn't undergone the reduction yet, rather than from the ethylene glycol that the reaction would be carried out in.

i think your syntheseis will work HIDE away..

Would the oxymercuration/demercuration step be specific? I don't think so.

The water will only add in a Markovnikov sense in the oxymercuration/demercuration reaction will it not?

And in this case to what does Markovnikov sense corresponds to?

Ah, I think I see the problem.  Using NaBH4 in the presence of an carbonyl group for the demercuration step will yield a diol rather than a hydroxyaldehyde.  So I suppose I just need to switch around steps 4 and 5, to get around this problem?

[quote="Carcul"]Propose a synthesis of 4-methylnonan-4-ol from pent-1-ene.[/quote]

Take one equivalent of the 1-pentene and oxymercurate (using Hg(OAc)_2 and NaBH_4, for example) to form pentan-2-ol. Oxidize with, for example, chromic acid to form the ketone. Now, take another equivalent of 1-pentene and react with HBr in the process of peroxides so that homolysis and radical reaction takes place to yield 1-bromopentane. Insert Mg metal oxidatively to obtain the Grignard reagent and attack on the carbonyl group of the previously formed ketone; acidic workup yields the product (which is, of course, racemic).

Yes, that's perfectly correct.

[quote="MysticTerminator"]Insert Mg metal oxidatively[/quote]

Why oxidatively?

sorry, I meant the magnesium is being oxidized ... our book calls the reaction of R-Br to R-MgBr "oxidative insertion of magnesium"


http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME_1_2006-2007/Problem_7

That figure would make triangle AEC have sides 1, 5, and 6.

but thats impossible.

so the figure doesn't exist.

am i wrong?

EDIT: whoops, hehe could be on the same line XD

[hide="hint"]
because i'm too dumb to finish it.
the ratio of ab to bd is the same as the ratio of areas of acf to cfd
we know aef is 5/6 of acf

[/hide]

[hide="Legitimate Hint"]Note that F is on the circumcircle of AEB.[/hide]
[hide="Bogus Hint"]Assume that AE is perpendicular to EB[/hide]


induction
Putnam
real analysis
real analysis solved

Let $\{a_n\}$ be a sequence of positive numbers. Is it possible that \[\limsup_{n\to\infty}\left(\frac{1+a_{n+1}}{a_n}\right)^n<e.\]

Let's see what happens if we want to have the contrary.

First of all, we would need $\limsup \frac{1+a_{n+1}}{a_n}\le 1$, so that we don't have arbitrarily large terms of the form $\left(\frac{1+a_{n+1}}{a_n}\right)^n$. Second of all, we cannot have $\limsup\frac{1+a_{n+1}}{a_n}< 1$, because then we would have $a_{n+1}<a_n-1$ for all sufficiently large $n$, which would make some of the $a_n$'s negative. This means that $\limsup \frac{1+a_{n+1}}{a_n}=1$. It's rather easy to show from here that $a_n\to\infty$.

The condition we want to prove is impossible is now translated to $n\cdot\left(\frac{a_{n+1}+1}{a_n}-1\right)< t<1$ for sufficiently large $n$. In particular, we have $n\cdot\left(\frac{a_{n+1}+1}{a_n}-1\right)<1$. For simplicity, we may assume this holds for $n\ge 1$ (it holds for sufficiently large $n$). We thus get $a_n<n\left(a_1-\left(\frac 12+\frac 13+\ldots+\frac 1n\right)\right)$, which goes to $-\infty$ as $n\to\infty$. This is a contradiction.

Did I do anything wrong? :?

There are some unnecessary arguments.

My solution was based on the fact that $\left(\frac{1+a_{n+1}}{a_n}\right)^n\leq \left(1+\frac{1}{n}\right)^n$ for all large $n$. So we immediately obtain $n\cdot\left(\frac{a_{n+1}+1}{a_n}-1\right)<1$.

[quote="grobber"]In particular, we have $n\cdot\left(\frac{a_{n+1}+1}{a_n}-1\right)<1$. For simplicity, we may assume this holds for $n\ge 1$ (it holds for sufficiently large $n$). We thus get $a_n<n\left(a_1-\left(\frac 12+\frac 13+\ldots+\frac 1n\right)\right)$.[/quote]
This conclusion is unclear for me. :?
I obtained contradiction in another way.

Since Grobber ignored my question, I will explain it myself. It was unclear not only for me...

Suppose $\frac{1+a_{n+1}}{a_n}\leq 1+\frac{1}{n}$ holds for all $n\geq N$. We can write it as $a_{n+1}\leq \frac{n+1}{n}a_n-1$.
Let's prove by induction that $a_n\leq n\left(\frac{a_N}{N}-\frac{1}{N}-\frac{1}{N+1}-...-\frac{1}{n}\right)$ for all $n>N$.
Indeed, for $n=N+1$ it is true. Then $a_{n+1}\leq \frac{n+1}{n}a_n-1\leq (n+1)\left(\frac{a_N}{N}-\frac{1}{N+1}-...-\frac{1}{n}\right)-1=(n+1)\left(\frac{a_N}{N}-\frac{1}{N+1}-...-\frac{1}{n+1}\right)$.

[quote="Ravi B"]By the way, Putnam 1949 had essentially the same problem.[/quote]

Hmmm...
$a_n$ is a sequence of positive reals. Show that $\varlimsup_{n\to\infty} \left(\frac{a_1+a_{n+1}}{a_n}\right)^n \geq e$.

[quote="Myth"][quote="Ravi B"]By the way, Putnam 1949 had essentially the same problem.[/quote] I am sure that Centaurus inhabitants invented it many millions years ago.[/quote]

Don't be sarcastic. But appreciate that other people point to earilier sources.  :)


Can anyone prove that:
$ \sum_{i\equal{}1,2,..,n}\binom{n}{i}i (\frac{5}{6})^{i} (\frac{1}{6})^{n\minus{}i}\equal{}\frac{n}{6}$

or more simply
$ \sum_{i\equal{}1,2,..,n}\binom{n}{i}i (\frac{1\minus{}p}{6})^{i} (\frac{p}{6})^{n\minus{}i}\equal{}np$

The second part looks wrong.

If you're trying to prove that the expected value of a random variable with a binomial distribution is $ np$, I believe it's an induction on $ n$. The trick is $ i\binom{n}{i}\equal{}n\binom{n\minus{}1}{i\minus{}1}$ and some change of index.

The second part was meant to be something equivalent to
\[ \sum_{i \equal{} 0}^n \binom{n}{i}\cdot i \cdot (1 \minus{} p)^{i} p^{n \minus{} i} \equal{} np.
\]
No need for induction: just use $ i \cdot \binom{n}{i} \equal{} n \cdot \binom{n \minus{} 1}{i \minus{} 1}$ (an easy algebraic or combinatorial fact) and the binomial theorem.

Wow! You guys are fast.

I was thinking of generating functions.

You don't need anything special...suppose that we have an event that happens with probability $ p$. What is the expected number of times that the event happens if we run the experiment $ n$ times. We could do it the $ k$ successes occurs $ \binom{n}{k}\cdot p^k\cdot (1\minus{}p)^k$, and sum it up, or we could note that we can sum the expected value for each turn and sum, or $ n\cdot p$, proving the identity.


Sprint level:

The area of a rectangle with length (x + 5) and width (x – 8) is 237.  What is the area of a rectangle with length (x – 10) and width (x + 7)?

[hide]Lowering the length of one side by 2 and increasing the length of one side by 2 decreases the volume by 4.
[b]233[/b]
[/hide]

Treething I think that only works whe you start with a square.

8x6=48

increase one by 2 and decrease the other by 2

10x4 = 40

But it goes down by 8

You're going to need another method, since that one doesn't work

[hide]237- 30 gives you 207. I hope that's right.[/hide]

[quote="cj"][hide="my simple way"] (x+5)(x-8)=237, then x^2-3x=277;
(x-10)(x+7)=x^2-3x-70 = 277-70=207.[/hide][/quote]

I would say yours is by far the fastest way.  Good job!


Let $a,b,c$ be the length of the sides of triangle. Prove that:
$\displaystyle a^2b^2+b^2c^2+c^2a^2 \geq abc(a+b+c)$

isn't this just rearrangement?
WLOG a >= b >= c.
Applying rearrangement on the two sets {ab,ac,bc} and {ab,ac,bc},
we get (ab)(ab) + (ac)(ac) + (bc)(bc) >= (ab)(ac) + (bc)(ab) + (ac)(bc) which is
a^2*b^2 + b^2*c^2 + a^2*c^2 >= abc(a+b+c).

Hxtung, sorry, are you joking? The inequality $\displaystyle a^2b^2+b^2c^2+c^2a^2 \geq abc \left( a+b+c \right)$ evidently holds for any three real numbers a, b, c, since, using the inequality $u^2 + v^2 \geq 2uv$ for two arbitrary real numbers u and v, we have

$\displaystyle a^2b^2+b^2c^2+c^2a^2 = \frac12 \cdot \left( a^2 \left(b^2 + c^2 \right) + b^2 \left( c^2 + a^2 \right) + c^2 \left( a^2 + b^2 \right) \right) \geq \frac12 \cdot \left( a^2 \cdot 2bc + b^2 \cdot 2ca + c^2 \cdot 2ab \right) = a^2 bc + b^2 ca + c^2 ab = abc \left( a + b + c \right)$.

  Darij

Are you sure the $a,b,c$ must be the sides of a triangle??
using the inequality $x^2+y^2+z^2\ge xy+yz+zx$ we find that
$a^2b^2+b^2c^2+c^2a^2\ge abbc+bcca+caab=abc(a+b+c)$
Did I miss something?


[quote="Virgil Nicula"][color=darkred]Let $ ABCD$ be a convex quadrilateral , $ AC\not\perp BD$ and $ X\in (BD)$ so that

$ XB^2\cdot (DC^2 \minus{} DA^2) \minus{} XD^2\cdot (BC^2 \minus{} BA^2) \equal{} AB^2\cdot CD^2 \minus{} AD^2\cdot BC^2$ .

If $ R$ is the reflection of  $ A$ w.r.t. $ BD$ then prove that $ ACRX$ is concyclically.

Particularly, $ XA\equal{}XR\implies$ $ \widehat {XCA}\equiv\widehat {XCR}$ .[/color][/quote]

[color=darkblue][b][u]Remark.[/u][/b] See the particular case $ AB\cdot CD \equal{} AD\cdot BC$ , i.e.

$ \frac {XB}{CB} \equal{} \frac {XD}{CD}$ at http://www.mathlinks.ro/Forum/viewtopic.php?t=63420 [/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABCD$ be a convex quadrilateral , $ AC\not\perp BD$ and $ X\in (BD)$ so that
$ XB^2\cdot (DC^2 - DA^2) - XD^2\cdot (BC^2 - BA^2) = AB^2\cdot CD^2 - AD^2\cdot BC^2\ \ (*)$ .
If $ R$ is the reflection of  $ A$ w.r.t. $ BD$ then prove that $ ACRX$ is concyclically.
Particularly, $ XA = XR\implies$ $ \widehat {XCA}\equiv\widehat {XCR}$ .[/color][/quote]
[color=darkred][b][u]Lemma[/u][/b] (well-known or you can easily it). Are given two fixed points $ B$ , $ D$ and three real constants $ m$ , $ n$ , $ p$ so that $ m+n\ne 0$ . Then the geometrical locus of the point $ X$ for which $ m\cdot XB^2+n\cdot XD^2=p$ is a circle $ C(O)$ , where $ O\in BD$ .[/color]

[color=darkblue][b][u]Proof of the proposed problem.[/u][/b] Denote $ \{\begin{array}{c}
m=DC^2-DA^2\ ,\ n=BA^2-BC^2\\\\
p=AB^2\cdot CD^2-AD^2\cdot BC^2\\\\
f(X)=m\cdot XB^2+n\cdot XD^2\end{array}$ .

Observe that $ \{\begin{array}{c}
AC\not\perp BD\Longleftrightarrow AB^2+CD^2\ne AD^2+BC^2\Longleftrightarrow m+n\ne 0\ .\\\\
f(X)=fC)=f(A)=f(R)=p\ .\end{array}$

Using the upper lemma results that $ ACRX$ is cyclically.

[b][u]Remark.[/u][/b] $ XA=XR$ $ \implies$ the ray $ [CX$ is the bisector of the angle $ \widehat {ACR}$ .

[b][u]Particular case.[/u][/b] $ AB\cdot CD=AD\cdot BC$ , i.e. $ \frac {BA}{BC}=\frac {DA}{DC}$ . The relation $ (*)$ becomes

$ \frac {XB^2}{XD^2}=\frac {BC^2-BA^2}{DC^2-DA^2}=(\frac {BC}{DC})^2\cdot\frac {1-(\frac {BA}{BC})^2}{1-(\frac {DA}{DC})^2}$ $ \Longrightarrow$ $ \frac {XB}{XD}=\frac {BC}{DC}$ ,

i.e. the ray $ [CX$ is the bisector of the angle $ \widehat {ACR}$ .Therefore, $ \widehat {RCB}\equiv\widehat {ACD}$ .[/color]


Hi all,
this is the link of IMC-site.

http://www.imc-math.org.uk/

For the day-1 Pb 6 ,day-2 Pb-6 and day-2 Pb 3
are solutions that don't seem the best I think :blush:

..I mean that here, in the mathforum, are many and better ideas!!! :wink:

They are just the same as they were in the longlist (i.e. as authors proposed).


calculus
integration
derivative
trigonometry
real analysis
real analysis unsolved

1/(1+cos(2x)^2)    < ---i just need to know how to find the anti-derivative of this algebraicly

1/(x^4-1)         <----and this one using partial fraction decomposition....

holy crap man you're smart....

thank you very much :lol:

nevermind i know how to solve it now. thanks man.


ratio
geometry
incenter
geometric transformation
reflection
circumcircle
trigonometry

Consider the family of  nonisosceles triangles $ABC$ satisfying the property $AC^2 + BC^2 = 2 AB^2$. Points $M$ and $D$ lie on side $AB$ such that $AM = BM$ and $\angle ACD = \angle BCD$. Point $E$ is in the plane such that $D$ is the incenter of triangle $CEM$. Prove that exactly one of the ratios
\[ \frac{CE}{EM}, \quad \frac{EM}{MC}, \quad \frac{MC}{CE}  \]
is constant.

Here's a sketch of the proof:

Try to prove (by using the relation and the median bisector) that $\frac {GA}{GB}=\frac {CA}{CB}$, where $G$ is the centroid. This means that $G$ is on the Apollonius circle of $ABC$ corresponding to $C$. It's prety obvious that if we take a point $X$ on the apollonius circle corresponding to $C$, then $X'$, the reflection of $X$ in $AB$ will be situated on the cevian which is isogonal to $CX$. We do this for $X=G$ and find that the reflection of $G$ in $AB$ is on the symmedian from $C$. It is now clear that $E$ is exactly this reflection of $G$ in $AB$. This means that $ME=MG=\frac {MC}3\Rightarrow \frac {EM}{MC}=\frac 13$, and this shows that $\frac {EM}{MC}$ is constant.

I didn't spend time thinking about how to show that the others aren't constant, but that can't be too hard (if another of those ratios were constant it would mean that all triangles $EMC$ are similar). Come to think of it, it would mean that the angle $\angle CMD=\angle CMA$ is constant, and this is obviously false.

Further properties of such triangles are discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=28877 . I am planning to find and post further proofs of these when I will find time (however, probably not too soon).

  Darij

Hrmmm ok well here was a pretty straightforward proof:

Reflect C across AB to C'. Let P be the point such that PA and PB are tangent to the circumcircle of ABC. From the definitions and the fact that CP is a symmedian, it follows that the intersection of CP and C'M is E. In order to prove that $\frac{EM}{MC}$ is constant, it suffices to show that $\frac{C'E}{EM}= \frac{CC'}{MP}$ is constant.

From the condition, after subtracting the law of cosines equation, we get

$2 \cdot AC \cdot BC \cos C = AB^{2}$
$2 = \frac{c^{2}}{ab \cos C}$.

Now the ratio $\frac{CC'}{MP}$ is equal to

$\frac{[CAC'B]}{[ABP]}= \frac{ab \sin C}{\frac{c^{2}\tan C}{4}}$
$= \frac{4ab \cos C}{c^{2}}= 2$

so we win. The other part of the problem, as grobber noted, is necessarily routine.

We do not use barycentric coordinates.

Consider the point $X$ on the C-symmedian such that it lies on the circle of $ABC$. We know that $ACBX$ is a harmonic quadrilateral, thus $XC$ is the symmedian of triangle $AXB$ (and $ACB$). Furthermore, $\frac{AX}{BX}=\frac{AC}{BC}=\frac{AD}{DB}$, thus $\angle CXD=\angle DXM$ (as angle bsiector with symmedian) and we know $\angle XCD=\angle MCD$. Therefore $E=X$ and notice by the ratio-lemma that

\[\frac{EM}{MC}=\frac{\sin\angle EAB}{\sin\angle BAC}\cdot\frac{AX}{AC}=\frac{\sin\angle ACM}{BAC}\cdot \frac{\sin\angle MCB}{\sin\angle ABC}=\frac{AM^2}{CM^2}\]

We now apply Stewart's theorem, and $AM(BC^2+AC^2)=2AM(CM^2+AM^2)$, or $AM(8AM^2)=2AM(CM^2+AM^2)\rightarrow CM^2=3AM^2$ and we have a fixed ratio for $\frac{EM}{CM}$.

If the other ratios were fixed, then all such $CEM$ would be similar and this is easy to see is not true.

We do not use barycentric coordinates.

Since $CE$ is a symmedian, we intersect it with the tangents to the circumcircle at $A$ and $B$ and call this point $X$.  Let $CE \cap AB=F$.  Since $\angle CMF=\angle EMF$ and $\angle XMA=\dfrac{\pi}{2}$, we get that (CFEX) is harmonic, implying that $E$ is on the circumcircle.

Then we have $\dfrac{EM}{MC}=\dfrac{EF}{FC}=\dfrac{FB}{AF}\cdot\dfrac{FE^2}{FB^2}=\dfrac{a^2}{b^2}\cdot\dfrac{\sin^2 \angle MCB}{\sin^2 \angle CAB}$.  Applying Stewart's Theorem, we get that $CM=\dfrac{\sqrt{3}c}{2}$, so we get $\sin \angle MCB=\dfrac{\sin\angle ABC}{\sqrt{3}}$.  Thus we have $\dfrac{EM}{MC}=\dfrac{1}{3}$.

On the other hand, a quick computation using the fact that $\triangle BCM \sim \triangle ECA$ gives that $\dfrac{MC}{CE}=(\dfrac{3}{8})(\dfrac{a^2+b^2}{a^2b^2})$, which is clearly not constant.  Multiply this by $\dfrac{EM}{MC}=\dfrac{1}{3}$ to see that $\dfrac{CE}{EM}$ is not constant as well.  Thus, we are done.

QED


calculus
integration
function
complex analysis

For an "appropriate" meromorphic function $f$ prove
\[\sum_{n=0}^\infty f(n)-\int_{0}^\infty f(x)dx=\frac{f(0)}{2}+i\int_{0}^\infty\frac{f(it)-f(-it)}{e^{2\pi t}-1}.\]
What does it mean 'appropriate'?

It means that you have to figure out what exactly is needed to make the formula work yourself. How to do that? Start with some simple function like $\frac 1{(z+1)^{2}}$ when you are pretty sure that, at least, each sum and each integral in the formula makes sense. Next, prove the formula for this particular function. Then just look at what properties of the function you have used in that proof and call a function "appropriate" if it has the same properties. Then, if you want to really master the question, show that each of the properties on your list is necessary by constructing corresponding counterexamples. That's the general idea. I'll leave it to you to figure out what to do with this particular problem now.


function
combinatorics unsolved
combinatorics

[quote="Rushil"]In how many ways can a $2\times n$ grid be covered by
(a) 2 monominoes and $n-1$ dominoes
(b) 4 monominoes and $n-2$ dominoes.[/quote]

[color=blue][b]Solution.[/b][/color]
[color=blue](a)[/color] Let $x_n$ be the number of ways a $2\times n$ grid be covered by 2 monominoes and $n-1$ dominoes. The sequence $x_n$ satisfies the recurrence relation:

\[ \begin{array}{l} x_n = 2x_{n - 1} - x_{n - 3} \\ \\ x_1 = 1,{\rm{ }}x_2 = 4,{\rm{ }}x_3 = 9 \\ \end{array}  \]

With the method of charateristic equations (or by generating functions) we can find the closed formula

\[ x_n = \frac{{15 - 7\sqrt 5 }}{{10}}\left( {\frac{{1 - \sqrt 5 }}{2}} \right)^n + \frac{{15 + 7\sqrt 5 }}{{10}}\left( {\frac{{1 + \sqrt 5 }}{2}} \right)^n - 4  \]

The first ten values of $x_n$ are 1,4,9,17,30,51,85,140,229,373.

[color=blue](b)[/color] Let $y_n$ be the number of ways a $2\times n$ grid be covered by 4 monominoes and $n-2$ dominoes. The sequence $y_n$ satisfies the recurrence relation for $n \ge 2$

\[ \begin{array}{l} y_n = 4y_{n - 1} - 4y_{n - 2} - 2y_{n - 3} + 4y_{n - 4} - y_{n - 6} \\ \\ y_2 = 1,{\rm{ }}y_3 = 7,{\rm{ }}y_4 = 35,{\rm{ }}y_5 = 103,{\rm{ }}y_6 = 259,y_7 = 585 \\ \end{array}  \]

The closed formula for $y_n$ is

\[ y_n = 33 + 8n + \frac{{203\sqrt 5 - 375}}{{25}}\left( {\frac{{1 - \sqrt 5 }}{2}} \right)^n + \frac{{18 - 8\sqrt 5 }}{5}n\left( {\frac{{1 - \sqrt 5 }}{2}} \right)^n - \frac{{203\sqrt 5 + 375}}{{25}}\left( {\frac{{1 + \sqrt 5 }}{2}} \right)^n + \frac{{18 + 8\sqrt 5 }}{5}n\left( {\frac{{1 + \sqrt 5 }}{2}} \right)^n  \]

The first ten values of $y_n$  ($2 \le n \le 11$) are 1,7,35,103,259,585,1237,2495,4863,9235

Leon


I don't know how to prove this theorem:
For every positive integer n,there exists a prime p such that n≤p≤2n.


calculus
vector
trigonometry
AoPS Books
real analysis
real analysis theorems

Hello all,

I am a sophomore in high school taking Algebra II.  I am teaching myself precalculus with ease.  Nothing seems difficult, except for vectors which are more boring than difficult.  However, over the summer at Rutgers University I will be taking a class for precalclus to cement in my skills.

I am new to AoPS, but I love ATTEMPTING(lol) to solve problems.  However, most problems are out of my domain.

What would you suggest to do to prepare for Calculus.  Will doing AoPS problems help me?  How about the books?

My teacher says that Calculus is mainly algebra and trigonometry. If so, what can I do to sharpen those skills to  the point that those operations become trivial.

I appreciate any and all input.

--Potential

PS: I know most of precalculus trigonometry.

[quote="Potential"]\

My teacher says that Calculus is mainly algebra and trigonometry.

\[/quote]
I don't agree.
I love Calculusand for me it' :wink: s most interesting than Algebra. Algebra is language of math.

[quote="Potential"]I am teaching myself precalculus with ease.[/quote]

That's good. Very good.

[quote="Potential"]Vectors are more boring than difficult.[/quote]

That's almost the exact same problem I had, except that it was 2 or 3 variable equations. What I did was I found some difficult versions of the problems and(tried to(at least)) solve those. Once I solved those, I could dare say that I had mastered those.

[quote="Potential"]Most problems are out of my domain.[/quote]

Try looking at them closely. Ask yourself some questions like:

"What do I know?"

"What do I want, and how can I express it?"

"What does this problem look like?"

The "What do I know?" is explained in the problem.
The first part ot "What do I want, and how can I express it?" is in the problem, and then the second part can be left up to you.
The "What does this problem look like?" is a good question to ask, because you can employ some *insert problem type here* tactics to solve the problem once you answer the question.

[quote="Potential"]What would you suggest to do to prepare for Calculus?  Will doing AoPS problems help me?  How about the books?[/quote]

For the first problem, doing precalculus is part of preparing for Calculus. But then there are some other problems, like

"Compute

$\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}$"

that you *might* not know how to do. Then you should learn how to do them(AND FAST!!!).

For the second problem, the answer is undefined, because specific people have specific needs.

For the third, the AoPS books will definitely help you. Volume 1 will prepare you for Highschool basics, and help you get into Intermediate problems.

Volume 2 is Intermediate problems, and some Pre-Olymipad problems. They can really deapen(I can't spell)your understanding in mathematics.

Volume 3 does not exist. We are sorry for any inconveniance.

[quote="Potential"]My teacher says that Calculus is mainly algebra and trigonometry.[/quote]

Then you have the right to tell your teacher that he/she is wrong. :) Knowing Trig and Algebra won't get you that far in Calculus.

Hope this helps!!!

Thank you for your response.

I appreciate your tips.

Does anyone else have any suggestions?tips?

Thanks


$ a^2(b\plus{}c)\equal{}b^2(a\plus{}c) \implies a^2b\plus{}ca^2\equal{}ab^2\plus{}b^2c$
$ a^2b\minus{}ab^2\equal{}b^2c\minus{}ca^2 \implies ab(a\minus{}b)\equal{}c(b\plus{}a)(b\minus{}a)$

since $ a\not\equal{} b$, we can divide each side by $ a\minus{}b$.

$ ab\equal{}\minus{}c(b\plus{}a) \implies ab\equal{}\minus{}bc\minus{}ca \implies ab\plus{}bc\plus{}ca\equal{}0 \minus{}\minus{}\minus{}\alpha$

and we are looking for the value of $ c^2a\plus{}bc^2.$

$ a\times \alpha \equal{} a^2b\plus{}abc\plus{}a^2c\equal{}0 \implies a^2b\plus{}a^2c\equal{}\minus{}abc\equal{}2008$

$ c\times \alpha \equal{} abc\plus{}bc^2\plus{}c^2a \implies c^2a\plus{}bc^2\equal{}\minus{}abc\equal{}2008$

answer : 2008

$ ab \equal{} \minus{} c(b \plus{} a) \implies ab \equal{} \minus{} bc \minus{} ca \implies ab \plus{} bc \plus{} ca \equal{} 0 \minus{} \minus{} \minus{} \alpha$

and we are looking for the value of $ c^2a \plus{} bc^2.$

$ a\times \alpha \equal{} a^2b \plus{} abc \plus{} a^2c \equal{} 0 \implies a^2b \plus{} a^2c \equal{} \minus{} abc \equal{} 2008$

$ c\times \alpha \equal{} abc \plus{} bc^2 \plus{} c^2a \implies c^2a \plus{} bc^2 \equal{} \minus{} abc \equal{} 2008$

answer : 2008[/quote]

i dont understand after line $ ab \equal{} \minus{} c(b \plus{} a) \implies ab \equal{} \minus{} bc \minus{} ca \implies ab \plus{} bc \plus{} ca \equal{} 0 \minus{} \minus{} \minus{} \alpha$
thanks  :blush:

I'm not sure, but I think this is what he means:

We're looking for the value $ ac^2\plus{}bc^2.$

$ c\times\alpha\equal{}abc\plus{}bc^2\plus{}ac^2\implies ac^2\plus{}bc^2\equal{}\minus{}abc$

So now we're trying to find the value of $ \minus{}abc$

$ a\times\alpha\equal{}abc\plus{}a^2b\plus{}a^2c\implies \minus{}abc\equal{}a^2b\plus{}a^2c$

so $ c^2{(a\plus{}b)}\equal{}ac^2\plus{}bc^2\equal{}\minus{}abc\equal{}a^2b\plus{}a^2c\equal{}\fbox{2008}$


Two mail carriers meet on their routes and have a conversation.
A: "I know you have three sons. How old are they?"
B: "If you take their ages, expressed in years, and multiply those numbers, the result will equal your age."
A: "But that's not enough to tell me the answer!"
B: "The sum of theese three numbers equals the number of windows in that building."
A: "Hmm[pause]. But it's still not enough!"
B: "My middle son is red-haired."
A: "Ah, now its clear!"

How old are the sons?

He was distinguishing that there were no twins...

I think there are more info to be given...

Here's a similar problem:
I recently overheard a conversation that went roughly as follows:
Bob: "here's a problem that might interest you. On my bus this morning there were only three other passengers, all of whom I knew, We discovered that the product of their ages was 2450, and that the sum was exactly twice your age. How old are they?"
Jim:"Hang on. You haven't given me enough info."
Bob: "Oh sorry. I forgot to mention that one of the passengers on the bus was someone older than me, and I am--"
"Jim: "I know how old you are. And I now know the passengers' ages, too"
How old are Jim, Bob, and each of the three other passengers?

[quote="funcia"][quote="modeler"][quote="funcia"]other than what kstan said, how can you conclude anything?
[hide]2,3,5
3,4,5
571
there are a lot of possibilities![/hide][/quote]

funcia can you please explain your examples?
I cannot think of any example yet :(
The product of the number is equal to someone's age, this is a great restriction (you can't find a 2450-yrs old man, can you?)[/quote]
well the windows don't say anything...the two men are able to get a job and from the way they talk i concluded(maybe a little too quickly) they are 20-70 years old
2,3,5,=30 years old; 10 windows
and i suppose you can do the rest
3,4,5= 60 year old,12 windows
..
.
.
you just have to know that the sons are pretty young and red hair is a pretty good clue(i guess) though i think there is more to the red hair thing because i heard something sort of...oh well... i forgot it[/quote]

Are you sure that there is another possible choice other than 2,3,5 for the case 30 years old and 10 windows? If it is the only choice, the person should have guess the answer without the red-hair clue

that isn't the only choice i was showing how they work read kstan's post for a way that could help....

[hide]My answer is $(1,5,8)$ since the next solution will give a big value of dad's age. I assume that the 90 years old or older persons is too old ti be the mail barrier. :lol:
I suggest that the two solutions $(a,b,b)$ and $(x,y,z)$ without red hair clue statisfies:
$xyz=ab^{2}$
$x+y+z=a+2b$

we may assume $x<y<z$, and chek what happen if one of the $x$, $y$, or $z$ equal to $a$ or $b$.
The other clue is that one of numbers $x,y,z$ must have divisor to be share so that we obtain $a,b,b$ with statisfies $xyz=ab^{2}$.
for example in $1,5,8$  we have $8$ which can share its factor $2$ to $1$ so that we obtain $2,2,10$.
[/hide]

why can't it be 3,4,5? I'm sure there is someone family like that in the world.

no it can't be 3,4,5 .. $(3)(4)(5)=60$ and $3+4+5=12$, there is no other integers triple  whose sum is $12$ and product is $60$.

[hide]
Let investigate the conversation..

The product of the three numbers is equal to $B$'s age, since $B$ knows his own ages, he guest what are these three numbers? then he realize that the information is still not enough to determine the solution. So he ask another clue by saying "But, it is not enough..."  then he get a new clue that is "the sum of three desire numbers is equal to the number of windows in that building",
from this conversation we know that $B$ can count the number of  windows (but of course we don't know it), then he'll guest:
"what are the three numbers, whose product is my age and the sum is the number of windows in that building?

From the $B$ point of view, he know much information than us, he know his age and he know the number of windows.. BUT he say that it is still not enough to tell him what the ages??

Its because there are two or more triple say $(a,b,c)$ and $(x,y,z)$ such that $abc=xyz$ and $a+b+c=x+y+z$.

 After $A$ said that "my middle son...", then:
-$B$ know the solution, why? it because there are just only two possibility after the preceding clues. And the answer is that triple whose number has "middle"  that is triple $(x,y,z)$ with $x<y<z$ and the other posibility that is removed after red-haired clue must be the triple $(a,b,c)$ with at least one of its component is equal to other (i.e in the form $(a,a,b)$).

Now we need some algebra to determine $(x,y,z)$ with $x<y<z$ statisfies $xyz=a^{2}b$ and $x+y+z=2a+b$.
Don't forget to assume that $B$ isn't very old (up to 100)..
[/hide]

[quote="moogra"]same reason as it can't be 3,4,5 because it's product is 24 and sum is 9. No other set of numbers meet that requirement

btw I think 1,5,8 is right
the last clue would work (it could have been 2,2,10)[/quote]
why does another set of numbers need to meet it?

Because he only gets it when the man tells him that there is a middle child, saying that there is more than one possibility originally
So we agree that it's 1,5,8?

[quote="modeler"]Two solutions by my computer program
[hide]
Age= 1, 5, 8, Window = 14, dad = 40 (The other choice without red-hair clue is 2, 2, 10)
Age= 2, 5, 9, Window = 16, dad = 90 (The other choice without red-hair clue is 3, 3, 10)

The next solution is 3, 6, 8, but the father will be 144 years old :P

[/hide][/quote]

it can also be 2,5,9 but how can a 88 year old person still reproduce without issues?

Guys, the product is the OTHER mail carrier's age, not the father's.

Yes $1,5,8$ is the answer. there are infintely many possibilities but the solution keep large. You can't find a mail barrier whose age is up to 90 years don't you.... :lol:

I'm sure there is/was a carrier that was 90... lot's of weird things happen in the world, but the chance of having kids is off

[quote="kstan013"]funcia, for the 3,4,5, I hope a 60-year-old dad wouldn't have a 5-year old as his oldest kid! I think they would have to be between 20 and 40 with kids that young, so 2,4,5 at the oldest and 2,3,4 at the youngest[/quote]
arranged mail carrying marriage inj a world where mail carriers are royalty


trigonometry
inequalities
geometry
incenter
perimeter
Euler
geometric inequality

Given triangle $ABC$ with three bisector $l_{a},l_{b},l_{c}$ prove that
$\frac{1}{l_{a}}+\frac{1}{l_{b}}+\frac{1}{l_{c}}\le\frac{3}{2r(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2})}$

[color=purple]
     I don't think it's a nice and strong problem gemath.I can prove more and by a easy proff :P
  Here it's is
   $\frac{1}{l_{a}}+\frac{1}{l_{b}}+\frac{1}{l_{c}}\le \frac{2}{3r}(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2})$
     Because $\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2})\le \frac{3}{2}$ so mine is stronger than yours :lol:
    And here is the proff: Let I is the inscribe center of triangle ABC
 It's easy to check that $\sin\frac{A}{2}=\frac{r}{IA}$ thus $\frac{\sin\frac{A}{2}}{r}=\frac{1}{IA}$ similarly for $\frac{\sin\frac{B}{2}}{r},\frac{\sin\frac{C}{2}}{r}$
    So the above inequality can rewrite in the form
   $\frac{1}{l_{a}}+\frac{1}{l_{b}}+\frac{1}{l_{c}}\le \frac{2}{3}(\frac{1}{IA}+\frac{1}{IB}+\frac{1}{IC}) (*)$
     In triangle it's not hard to show
 $\frac{1}{l_{a}}=\frac{b+c}{2\sqrt{bcp(p-a)}}$ and similarly for$\frac{1}{l_{b}},\frac{1}{l_{c}}$
    Hence $\frac{1}{l_{a}}+\frac{1}{l_{b}}+\frac{1}{l_{c}}=\frac{(b+c)\sqrt{a(p-b)(p-c)}+(c+a)\sqrt{b(p-c)(p-a)}+(a+b)\sqrt{c(p-a)(p-b)}}{2\sqrt abcp(p-a)(p-b)(p-c)}$
  And similaly we can show that
    $\frac{2}{3}(\frac{1}{IA}+\frac{1}{IB}+\frac{1}{IC})=\frac{2(a+b+c)}{3}.\frac{\sqrt{a(p-b)(p-c)}+\sqrt{b(p-c)(p-a)}+\sqrt{c(p-a)(p-b)}}{\sqrt{abcp(p-a)(p-b)(p-c)}}$
  Therefore $(*)$ is equipvalent to
$\frac{(b+c)\sqrt{a(p-b)(p-c)}+(c+a)\sqrt{b(p-c)(p-a)}+(a+b)\sqrt{c(p-a)(p-b)}}{2\sqrt abcp(p-a)(p-b)(p-c)}\le \frac{2(a+b+c)}{3}.\frac{\sqrt{a(p-b)(p-c)}+\sqrt{b(p-c)(p-a)}+\sqrt{c(p-a)(p-b)}}{\sqrt{abcp(p-a)(p-b)(p-c)}}$
   $\leftrightarrow (b+c)\sqrt{a(p-b)(p-c)}+(c+a)\sqrt{b(p-c)(p-a)}+(a+b)\sqrt{c(p-a)(p-b)}\le \frac{4}{3}p(\sqrt{a(p-b)(p-c)}+\sqrt{b(p-c)(p-a)}+\sqrt{c(p-a)(p-b)}) (*')$
   Without loss of genenrality asumme that $a\geb\gec$ and using Chebysev inequality we can get $(*')$ easily .The proff is completed
    On the same occasion  the problem of JiChen is
  $cosA+cosB+cosC\ge\frac{3}{2r}(\frac{1}{m_{a}}+\frac{1}{m_{b}}+\frac{1}{m_{c}})$
    I think it's real nice problem, gemath can you show me the way you used to make your inequality or the stronger  :D .Thank alot
 @ gemath Have you voted for me :D? I think with my proff you can vote 6* for me :D  [/color]

[quote="evarist"]
 $\frac{1}{l_{a}}=\frac{b+c}{2\sqrt{bcp(p-a)}}$ and similarly for$\frac{1}{l_{b}},\frac{1}{l_{c}}$
    Hence $\frac{1}{l_{a}}+\frac{1}{l_{b}}+\frac{1}{l_{c}}=\frac{(b+c)\sqrt{a(p-b)(p-c)}+(c+a)\sqrt{b(p-c)(p-a)}+(a+b)\sqrt{c(p-a)(p-b)}}{2\sqrt abcp(p-a)(p-b)(p-c)}$
[/quote]
Hey are you sure I don't think so  :D
[quote="evarist"]    $\frac{2}{3}(\frac{1}{IA}+\frac{1}{IB}+\frac{1}{IC})=\frac{2(a+b+c)}{3}.\frac{\sqrt{a(p-b)(p-c)}+\sqrt{b(p-c)(p-a)}+\sqrt{c(p-a)(p-b)}}{\sqrt{abcp(p-a)(p-b)(p-c)}}$
 [/quote]
And I think you lost number $2$ but however after all small mistakes you had true  :lol:  solution but it isn't nice for this mine, mine has very easy solution but you want stronger inequalities ? Ok there are many stronger  :D
$\frac{1}{l_{a}}+\frac{1}{l_{b}}+\frac{1}{l_{c}}\le\frac{1}{h_{a}}+\frac{1}{l_{b}}+\frac{1}{l_{c}}\le\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{l_{c}}\le\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{1}{r}\le\frac{3}{2r(\sum\sin\frac{A}{2})}$
And strongest  $\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{1}{r}$   :D
Anyway I rate below average that's all  :D

[color=purple]You are very humorous :D and your post really funny! :rotfl: .The stronger  :P i want to talk here are the similar or... something as the following problem :) [/color]
 $1-\frac{2[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]}{(a+b+c)^{2}}\leq3r^{2}(\frac{1}{{l_{a}}^{2}}+\frac{1}{{l_{b}}^{2}}+\frac{1}{{l_{c}}^{2}})\leq 1$
[hide="author"] [color=purple]Evarist_WaltherJanous  :lol:[/color] [/hide]

[quote="gemath"]Given triangle $ ABC$ with three angle-bisectors $ w_{a},w_{b},w_{c}$, prove that
$ \left(\frac {1}{w_{a}} \plus{} \frac {1}{w_{b}} \plus{} \frac {1}{w_{c}}\right)\left(\sin\frac {A}{2} \plus{} \sin\frac {B}{2} \plus{} \sin\frac {C}{2}\right)\le\frac {3}{2r}$.[/quote]
[quote="evarist"][color=purple]I can prove more and by a easy proof. :P Here it's is
$ \frac {1}{w_{a}} \plus{} \frac {1}{w_{b}} \plus{} \frac {1}{w_{c}}\le \frac {2}{3r}\left(\sin\frac {A}{2} \plus{} \sin\frac {B}{2} \plus{} \sin\frac {C}{2}\right)$.
Because
$ \sin\frac {A}{2} \plus{} \sin\frac {B}{2} \plus{} \sin\frac {C}{2}\le \frac {3}{2}$,
so mine is stronger than yours. :lol:
Let $ I$ is the inscribe center of triangle $ ABC$. It's easy to check that
$ \sin\frac {A}{2} \equal{} \frac {r}{IA}$.
So the above inequality can rewrite in the form
$ \frac {1}{w_{a}} \plus{} \frac {1}{w_{b}} \plus{} \frac {1}{w_{c}}\le \frac {2}{3}\left(\frac {1}{IA} \plus{} \frac {1}{IB} \plus{} \frac {1}{IC}\right)$.[/color][/quote]
If $ I$ is the incentre of a triangle $ ABC$, $ g_{a},g_{b},g_{c}$ are the Gergonne cevians, then

$ \frac {2}{IA} \plus{} \frac {2}{IB} \plus{} \frac {2}{IC}\geq \frac {3}{g_{a}} \plus{} \frac {3}{g_{b}} \plus{} \frac {3}{g_{c}}$,

with equality if and only if $ \triangle ABC$ is equilateral.

[hide="Algebraic form."]Let $ a,b,c,$ are sides and $ s$ is the semi-perimeter, then

$ IA \equal{} \sqrt {\frac {bc(s \minus{} a)}{s}}$,

$ g_{a} \equal{} \sqrt {(s \minus{} a)\left[s \minus{} \frac {(b \minus{} c)^2}{a}\right]}$.

Put $ s \minus{} a \equal{} \frac {1}{x},s \minus{} b \equal{} \frac {1}{y},s \minus{} c \equal{} \frac {1}{z}$, the geometric inequality is equivalent to the following algebraic inequality

$ 2\sum{x\sqrt {\frac{yz \plus{} zx \plus{} xy}{(x \plus{} y)(x \plus{} z)}}}\geq 3\sum{x\sqrt {\frac {y \plus{} z}{4x \plus{} y \plus{} z}}}$,

for positive real numbers $ x,y,z$.[/hide]
[b]Remark[/b] $ w_{a}\geq g_{a}$.

[quote="Ji Chen"]$ 2\sum{x\sqrt {\frac {yz \plus{} zx \plus{} xy}{(x \plus{} y)(x \plus{} z)}}}\geq 3\sum{x\sqrt {\frac {y \plus{} z}{4x \plus{} y \plus{} z}}}$.[/quote]
If  $ x,y,z$ are nonnegative real numbers, then

$ 90(x \plus{} y \plus{} z)\sum{x\sqrt {(y \plus{} z)(4x \plus{} y \plus{} z)(4y \plus{} z \plus{} x)(4z \plus{} x \plus{} y)(yz \plus{} zx \plus{} xy)}}$

$ \geq 180[x^4(y \plus{} z) \plus{} y^4(x \plus{} z) \plus{} z^4(x \plus{} y)]$

$ \plus{} 1079[x^3(y^2 \plus{} z^2) \plus{} y^3(z^2 \plus{} x^2) \plus{} z^3(x^2 \plus{} y^2)] \plus{} 2533(x^3yz \plus{} y^3zx \plus{} z^3xy)$

$ \plus{} 4669(xy^2z^2 \plus{} yz^2x^2 \plus{} zx^2y^2)$

$ \geq 135(x \plus{} y \plus{} z)\sum{x(y \plus{} z)\sqrt {(x \plus{} y)(x \plus{} z)(4y \plus{} z \plus{} x)(4z \plus{} x \plus{} y)}}$.

I will prove [u][b]evarist[/b]'s stronger inequality[/u] i.e. $\boxed{\ \frac 1{l_a}+\frac 1{l_b}+\frac 1{l_c}\, \le\, \frac {2\cdot(\sin\frac A2+\sin\frac B2+\sin\frac C2)}{3r}\ }$ .

Using the identity $\boxed{\ \sum\, \frac 1{l_a}=\frac {s\sum\sin\, \frac A2+r\sum\, \cos\frac A2}{2rs}\ }$ (one can see its proof in the beginning of my post from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=327619][b][color=red]here[/color][/b][/url])

the above inequality reduces to: $\frac {3r}s\sum\, \cos\frac A2\, \le\, \sum\, \sin\frac A2\iff \boxed{\ 3\prod\, \tan\frac A2\sum\, \cos\frac A2\, \le\, \sum\, \sin\frac A2\ }$ . Using

the substitutions: $\{\begin{array}{llll}
A=180^{\circ}-2X \\
B=180^{\circ}-2Y \\
C=180^{\circ}-2Z\end{array}$ where $ \|\begin{array}{cccc}X,Y,Z\in(0,90^{\circ})\\ \\ X+Y+Z=180^{\circ}\end{array}\| $ we will transform the previous inequality (in [b]any[/b]

triangle) into one confined to an [b]acute-angled[/b] triangle : $\boxed{\ 3\prod\, \cot X\sum\, \sin X\, \le\, \sum\, \cos X\ }\ \ (\ast)$ . For conveniece,

we will denote by $R$ , $r$ , $s$ the classical elements of $\triangle\, XYZ$ (therefore, $r$ and $s$ are different from those appearing above)

Thus, by taking into account the well known set of identities : $\|\begin{array}{llll}
\sum\, \sin X=\frac sR \\ \\
\sum\, \cos X=1+\frac rR \\ \\
\sum\, \cot X=\frac {s^2-(2R+r)^2}{2rs}\end{array}\|$ the inequality $(\ast)$ becomes:

$3\cdot\frac {s^2-(2R+r)^2}{2rs}\cdot\frac sR\, \le\, \frac {R+r}R\iff 3\cdot(s^2-(2R+r)^2)\, \le\, 2r(R+r)$ which is true since by [u][b]Gerretsen[/b]'s inequality[/u]

one has $s^2-(2R+r)^2\le 2r^2$ while the inequality $3\cdot 2r^2\le 2r(R+r)$ reduces to the famous [u][b]Euler[/b]'s inequality[/u] i.e. $R\ge 2r$ .

[quote="quykhtn-qa1"]Furthermore, we have: $\boxed{\, \frac 1{l_a^2}+\frac 1{l_b^2}+\frac 1{l_c^2}\, \le\, \frac 1{h_ah_b}+\frac 1{h_bh_c}+\frac 1{h_ch_a}\, }\, \le\, \frac 1{3r^2}\ \ \ .$[/quote]
Nice one, [b]quykhtn-qa1[/b]. However, the above "boxed" inequality is not strong at all and it is trivialized after using the identities:
\[\left\{\begin{array}{llll}
\sum\, (\frac 1{l_a})^2 & = & \frac {(2R+r)s^2+(4R+r)r^2}{8Rr^2s^2} \\ \\
\sum\, \frac 1{h_bh_c} & = & \frac {s^2+r^2+4Rr}{4r^2s^2}\end{array}\right.\ \ \ .\]
Indeed, it is easy to see that it reduces to $s^2\, \le\, (4R+r)(2R-r)$, which is obviously weaker than [u][b]Gerretsen[/b]'s inequality[/u] i.e. $\boxed{\, s^2\, \le\, 4R^2+4Rr+3r^2\, }$ .