inequalities geometry trigonometry symmetry

No, because that's not true. Consider $ (x,y,z) \equal{} (1/2 \minus{} \epsilon,1/2 \minus{} \epsilon,2\epsilon)$ for some very small positive number $ \epsilon$ -- you will have approximately $ \frac {1/4}{\sqrt {1/4}} \equal{} \frac {1}{2} > \frac {\sqrt {2}}{6}$

You really need to consider all three terms together.

ah yea thanks.

Anyways, I think I have a bunch of useless things but I can get out of that;

$ x^2\plus{}y^2\plus{}z^2\equal{}1\minus{}2(xy\plus{}yz\plus{}zx)$  Dont know if that helps but it seems to involve a lot of its terms.

I try to make a lot of substitutions, like $ x\equal{}1\minus{}y\minus{}z$ and same with the other two terms but I cant seem to get it to a more known type of inequality.

[quote="the future"]Let $ x$, $ y$, and $ z$ be positive real numbers with $ x \plus{} y \plus{} z \equal{} 1$.  Prove that

$ \frac {xy}{\sqrt {xy \plus{} yz}} \plus{} \frac {yz}{\sqrt {yz \plus{} zx}} \plus{} \frac {zx}{\sqrt {zx \plus{} xy}} \leq \frac {\sqrt2}{2}$

Does this basically boil down to proving that $ \frac {xy}{\sqrt {xy \plus{} yz}} \leq \frac {\sqrt2}{6}$?[/quote]

[hide="hold on"]
Warning: crazy idea ahead

$ \frac{xy}{\sqrt{xy\plus{}yz}}\le\frac{xy}{\sqrt{xy\plus{}yz\plus{}xz}}$
now we need to prove $ \sqrt{xy\plus{}yz\plus{}xz}\le \frac{1}{2}$
we know $ (x\plus{}y\plus{}z)^2\equal{}x^2\plus{}y^2\plus{}z^2\plus{}2xy\plus{}2yz\plus{}2xz\equal{}1$
so $ xy\plus{}yz\plus{}xz\equal{}\frac{1}{2}\minus{}x^2\minus{}y^2\minus{}z^2$
therefore $ xy\plus{}yz\plus{}xz\le \frac{1}{2}$ and we are done

i dunno, the first step looks kinda odd to me
[/hide]

I don't think that first step is right: simplify it, it says $ xz\le0$. Or notice that the RHS is dividing by more than the LHS, so it has to be smaller.

How do we know to consider $ (x,y,z) \equal{} (1/2 \minus{} \epsilon,1/2 \plus{} \epsilon,2\epsilon)$ for some very small positive number $ \epsilon$?

How do we do calculations with episilon representing a small number? Is there any difference between normal computation? Would would something like $ \frac{\epsilon}{5}$ turn into?

What was our initial inspiration to state this $ \frac{xy}{\sqrt{xy\plus{}yz}}\le\frac{xy}{\sqrt{xy\plus{}yz\plus{}xz}}$?

[quote="aufha"]How do we know to consider $ (x,y,z) \equal{} (1/2 \minus{} \epsilon,1/2 \plus{} \epsilon,2\epsilon)$ for some very small positive number $ \epsilon$?

How do we do calculations with episilon representing a small number? Is there any difference between normal computation? Would would something like $ \frac {\epsilon}{5}$ turn into?

What was our initial inspiration to state this $ \frac {xy}{\sqrt {xy \plus{} yz}}\le\frac {xy}{\sqrt {xy \plus{} yz \plus{} xz}}$?[/quote]

Whatever came after it, actually. :/

I'm not surprised it was wrong.  The question is how to fix it.

Meh, we might as well abandon the idea.

What about substitution? like a=xy, b=yz, c=zx? and uh, how to relate that to x+y+z=1 uh  :maybe: .

I'm not asking questions to criticize or say anything is wrong.

I'm asking because I'm pretty dumb and I don't understand a bunch of things I asked about in my above post...  :(

Those were showing that the two "solutions" above don't work.

Well I manage to get that the the equality occurs if and only if $ x\equal{} y\equal{} z \equal{}\frac{1}{3}$

This souds silly, but could you apply a geometric interpretation of the problem where x,y, and z are sides of the triangle?  And the area can be represented by 1/2xysinC or something like that to get the xy?  Probably not though since its too stretched.

[hide]This problem is slightly hard and tricky!But I think I found a nice solution! :wink:
We have:square root(xy+xz)=square root[x(1-x)]>=square root(1/9+x/3)
Doing similarly for 2 more pairs we have: LHS=< xy/square root(1/9+x/3)+yz/square root(1/9+z/3)+zx/square root(1/9+x/3)(2)
Now using Cauchy-Schwartz we have: RHS of(2)=<3(xy+yz+xz)[xy/(1/3+y)+yz/(1/3+z)+xz/(1/3+x)](3)
We need to prove RHS of(3)<=1/2
Using AM-GM we have: 3(xy+yz+xz)=<(x+y+z)^2=1
Therefore, we just need to prove: xy(1/3+y)+yz(1/3+z)+xz(1/3+x)=<1/2
Clearing all denominator and we will get the equivalent inequality:
81xyz+18(x^2y+y^2z+z^x)<=4+3(xy+yz+xz)
The most important step here is using hypothesis (x+y+z)=1 to make the inequality become homogenous!(The degree of the LHS are all third degree!)
This step will be: 81xyz+18(x^2y+y^2z+z^2x)<=4(x+y+z)^3+3(x+y+z)(xy+yz+zx)
Expansion of the RHS we need to prove: 48xyz+3(x^2y+y^2z+z^2x)=<4(x^3+y^3+z^3)+15(xy^2+yz^2+zx^2)
Now using AM-GM we have: 3/2*(x^3+y^3+z^3+xy^2+yz^2+zx^2)>=3(x^2y+y^2z+z^2x)
The rest is: 5/2*(x^3+y^3+z^3)+27/2*(xy^2+yz^2+zx^2)>=48(xyz)(AM-GM for two pairs of three variables,respectively)(Done!) :) [/hide]

[quote="aufha"]How do we know to consider $ (x,y,z) \equal{} (1/2 \minus{} \epsilon,1/2 \plus{} \epsilon,2\epsilon)$ for some very small positive number $ \epsilon$?

How do we do calculations with episilon representing a small number? Is there any difference between normal computation? Would would something like $ \frac {\epsilon}{5}$ turn into?[/quote]

I actually meant to have $ x\equal{}y\equal{}1/2 \minus{} \epsilon$. What I meant was that we are approaching $ (1/2,1/2,0)$. Because we are on an open interval, we cannot have that exact point, but we can be arbitrarily close to that, so the value can be arbitrarily close to the value produced by that point.

Really, $ \epsilon \approx 0$, so $ k \epsilon \approx 0$ too. I only used the coefficient $ 2$ to show that together, the three terms added to $ 1$.

[quote="the future"]Well I manage to get that the the equality occurs if and only if $ x \equal{} y \equal{} z \equal{} \frac {1}{3}$

This souds silly, but could you apply a geometric interpretation of the problem where x,y, and z are sides of the triangle?  And the area can be represented by 1/2xysinC or something like that to get the xy?  Probably not though since its too stretched.[/quote]

You don't know that x,y,z will form a triangle.

it was surprising at first when i thought it was straight forward from this famous one :
a,b,c >0 then :
$ \frac{a}{\sqrt{a\plus{}b}}\plus{}\frac{b}{\sqrt{b\plus{}c}}\plus{}\frac{c}{\sqrt{a\plus{}c}} \leq \frac{5}{4}\sqrt{a\plus{}b\plus{}c}$
However , the equality doesn't occur  :blush:

The difficulty in all the high school forums have gone up significantly in the last few years...

[quote="Temperal"]Note that if each individual term on the LHS is taken to the $ n$th power and the resulting inequality is proven, then the original inequality holds due to power mean. I'm not sure if this does anything, but maybe second powers could get rid of those square roots and then after writing with a common denominator and simplifying, cauchy could be used? I dunno.

By the way, shouldn't this be in pre-oly?[/quote]
Well, you have the same idea as I did above.However, you should do more specifically to prove that indeed it can lead to the solution since this inequality is a little bit strict :) .Doesn't anybody see my solution above and check it? :|

Ah, yes, I forgot, it's the reciprocal of the average, not the reciprocal.

Hm, I might check it someday...  :P

Piano, if you really want $ a\equal{}xy,b\equal{}yz,c\equal{}xz$, then $ \sqrt{abc}\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)\equal{}1$ (substitute in and check it!)

[quote="ghjk"]This problem is slightly hard and tricky!But I think I found a nice solution! :wink:
We have: $ \sqrt{xy\plus{}xz}\equal{}\sqrt{x(1\minus{}x)}\ge \sqrt{\frac{1}{9}\plus{}\frac{x}{3}}$ [/quote]
If this is true, then $ x(1\minus{}x)\ge \frac{1}{9}\plus{}\frac{x}{3}$
$ x^2\minus{}\frac{2x}{3}\plus{}\frac{1}{9}\le 0$
$ (x\minus{}\frac{1}{3})\le 0$, which is obviously not true unless $ x\equal{}\frac{1}{3}$

So shouldn't it be $ \sqrt{x(1\minus{}x)}\le \sqrt{\frac{1}{9}\plus{}\frac{x}{3}}$?

Also, simply checking with $ x\equal{}\frac{1}{2}$ disproves that.

[hide="Hint"]Substitute: $ x \equal{} \sin^2{a}$, $ y \equal{} \sin^2{b}$, $ y \equal{} \sin^2{c}$. Ain't too hard after that...[/hide]

Alternatively, I think mixing variables pretty much destroys it...

The expression $ \frac {xy}{\sqrt {xy \plus{} yz}} \plus{} \frac {yz}{\sqrt {yz \plus{} zx}} \plus{} \frac {zx}{\sqrt {zx \plus{} xy}}$ is equal, up to symmetry, to $ \frac {yz}{\sqrt {xy \plus{} yz}} \plus{} \frac {zx}{\sqrt {yz \plus{} zx}} \plus{} \frac {xy}{\sqrt {zx \plus{} xy}}$.

Is it possible to add this to the former and then reason that the sum of both expressions must be $ \leq \sqrt {2}$?

By the way, I wouldn't think of trig here  :D

Not too sure if this is right...

[hide="maybe right?"]
\[ \displaystyle \sum_{cyc}\dfrac {xy}{\sqrt {xy \plus{} yz}} \equal{} \displaystyle \sum_{cyc}\dfrac{xy}{\sqrt {y(x \plus{} z)}} \equal{} \displaystyle \sum_{cyc}\dfrac{xy}{\sqrt {y(1 \minus{} y)}}
\]
Thus, let [EDIT] $ z \equal{} min(x,y,z)$
Then the max of the expression occurs when the denominators are all $ \sqrt {z(1 \minus{} z)}$, so we substitute this in...
Then, the expression becomes:
\[ \dfrac{\displaystyle \sum_{cyc} xy}{\sqrt {z(1 \minus{} z)}}
\]
But by amgm, $ \displaystyle \sum_{cyc} xy \ge \dfrac{1}{3}$
But, also, since we assumed $ z \equal{} min(x,y,z)$, and all are positive, we know that $ z\le \dfrac{1}{3}$ Therefore, the expression becomes:
\[ \dfrac{\sqrt {2}}{2} \ge \dfrac{\dfrac{1}{3}}{\dfrac{\sqrt {2}}{3}}
\]
Which is clearly true, so we are done... :S[/hide]

You can't assume $ x\ge y\ge z$ as the ineq. is cyclic (not symmetric). You can assume $ z$ is $ \min(x,y,z)$ though. Otherwise nice :)


Bob asked his wife to accompany him to a war movie being featured that night. During a scene when grenades were exploding and guns were firing, Bob decided the time was right; he pulled out a gun and shot his wife. He then took her out of the theatre without anybody trying to stop him. Why not?

[hide="a guess..."]It was a dream.[/hide]

if uve heard problem b4 plz dont say answer

[hide="what about..."]Bob and his wife are actors in the movie and Bob kills his wife's character in the movie. When the movie's in the theater, bob and his wife go watch it. Then in the movie bob kills his wife. after the movie, they walk out of the theater and live happily ever after[/hide]

Yeah I agree with Funcia...

Anyway:

[hide="stupid answer"] Maybe Bob had a divorce at that second and then when his wife sat next to him, Bob refused to have his ex-wife sit next to him and shot his wife.[/hide]

Lol :rotfl:  :rotfl:  :rotfl:

[quote="randomdragoon"][hide="Classic..."]They were at a drive-in theater. He shot her in the car so no one heard him, and then he drove out with her body.[/hide][/quote]
Yeah, I agree with this answer.
[/hide]

I sort of do...

But then if they were at a drive- in, the original problem really states that Bob walks out, not drives.

Does this make sense?  :huh:  :huh:

[quote="Ihatepie"] He then took her out of the theatre without anybody trying to stop him. Why not?[/quote]

[b]he does not walk out, the  :?:  says that he too her out of the theater...so i think that the drive in makes perfect sense...hope that helps you! [/b]

[quote]He then took her out of the theatre without anybody trying to stop him. Why not?[/quote]

But he took her out of the theatre. So that means its inside a building.

well drive in theatres are also theatres right? :maybe:

[hide="a really stupid answer"]there was nobody else in the theater.[/hide]

or [hide]they have a home-theatre[/hide]

[hide="haha...jk"]
His wife was in the movie and he shot the screen, but then he walked out with his real wife![/hide]

lol nice one quevvy

[hide="hilarious"] how about Bob and his wife are very very rich and have their own private theater?[/hide]

I agree with Junggi but here's[hide="another possibility"]Everyone hated his wife and so they were happy she was shot[/hide]

he shot her accidentally and is taking her to the hospital

Although that doesn't go with "time is right" bit.

Maybe it's one of those faked injuries for  $\$$  from the insurance stuff.


[quote="Learner"]I've found (with the help of a friend who answers questions) two solutions that were wrong or incomplete. One I posted and got responses bashing the book, which wasn't my intention. I'd like to do it constructively. Is there a place?[/quote]I do not really understand what you want. Could you please be more explicit? Thanks.

As in where do we post suggestions for the AoPS books, not the web site.

[quote="Valentin Vornicu"]In that case I think it's best to address them to Richard Rusczyk directly, throught PM or email (his email is rrusczyk@artofproblemsolving.com) rather than posting them somewhere on the forum.[/quote]

Why not post them on the forum? I think community members should get a chance to see and discuss it. But remember that many of us here are loyal readers of the AoPS books, so be careful with word choice if you think there's something wrong with it.

[quote="MithsApprentice"]But remember that many of us here are loyal readers of the AoPS books, so be careful with word choice if you think there's something wrong with it.[/quote]That's exactly why I think that comments about this should go private. We don't want to scare anyone away. In plus chances are the Richard will look into his email more often than in every topic of this forum ;)
Maybe it's best to let Richard answer this.

[quote="TripleM"]I remember someone posting about this a long time ago - and I think it was agreed that it was a reasonable idea posting them on the forum first, in case you had just made a mistake. I think you did the right thing Learner - the responses weren't exactly responding to what you had asked! I think usually the responses will either confirm or deny that there is a mistake, and then you could PM rrusczyk.[/quote]He hasn't posted any thing concrete. Just that he found a couple of possible mistakes or incomplete proofs. I think (as an author myself) that it is wise to contact the author in such cases. As Richard is very busy at this time, he is probably not aware of this topic (that's one reason for which I think this is not such a good idea - it will take time for him to find this topic :) ). But this is my opinion.

[quote="Valentin VornicuHe hasn't posted any thing concrete. Just that he found a couple of possible mistakes or incomplete proofs.
[/quote]
(he had posted one of the possible mistakes in one of the other boards. Thats what he was referring to.)

I think (as an author myself) that it is wise to contact the author in such cases. As Richard is very busy at this time, he is probably not aware of this topic (that's one reason for which I think this is not such a good idea - it will take time for him to find this topic :) ). But this is my opinion.[/quote]

Yup, I agree. And for the same reason you should be very careful to make sure you haven't made a mistake before telling Richard, so its worthwhile posting it on the forum.


hint: [hide]Apply formula for Rate x Times = Distance. Find downstream and upstream rates by solving for one variable.[/hide]

[quote="Interval"]I do not get your hint.[/quote]

Well, I do.

In the future, please provide us with an idea of your mathematical background (i.e. what courses you have taken, etc.). Thank you.

And about the original problem: Let x equal the speed of the boat. Have the boat travel a given distance, and solve some equations. Remember to add and subtract the speeds respectively.

The assumption means that the motorboat always has the same speed.  If you try the problem again using now a ranger's and mathnerd 314's hints, but can't solve it, here is my solution:
[hide] Say the distance that the motorboat travels is d and its speed is m.  This means that its speed upstream is m-3 (since the stream decreases its speed by 3 mph when it is going in the opposite direction) and its speed going downstream is m+3 (since the stream adds 3 mph to its speed when it is going in the stream's direction).  Now, using the distance formula, as now a ranger's hint said, we have that d/(m-3)=5 and d/(m+3)=2.5  Thus, we have that d=5m-15 (multiplying both sides by m-3 on the first equation) and d=2.5m+7.5 (multiplying everything by m+3 on the second equation).  Notice that both equations have something equal to d.  This means that the "somethings" are equal to each other.  Therefore, 2.5m+7.5=5m-15.  Therefore, 2.5m=22.5 and thus $m=9 mph$, which is our answer [/hide]

[quote="Interval"]A motorboat heads upstream on a river that has a current of 3 mph. The trip upstream takes 5 hours, while the return trip takes 2.5 hours. What is the speed of the motorboat?

Assume that the motorboat keeps a constant speed relative to the water.

NOTE: What is the meaning of the assumption made above?

I've seen upstream and downstream questions before. However, I always get confused when it comes to setting up the right table or chart that leads to the correct equation that leads to the answer.[/quote]
[hide]Let $r$ be the speed of the motorboat.

$2.5(r+3)=5(r-3)$

$2.5r+7.5=5r-15$

$22.5=2.5r$

$r=\boxed{9\text{ mph}}$[/hide]
About the assumption: If the speed of the motorboat varied throughout it's "trip," the answer would be different. To simplify things, it is assumed that the motorboat maintained a constant speed.


Find the number of equivalence relations on A = {0, 1, y, n}.

Here's my logic: An equivalence relation must be reflexive, symmetric, and transitive. Thus, an equivalence relation R must have {(0,0), (1,1), (y,y), (n,n)}, to the very least. We must then consider all possible ordered pairs we can attach to this basic arrangement. For each ordered pair (x, y), (y, x) must also be in R because we need symmetry. We can attach (0,1), (0,y), (0,n), (1,y), (1,n), (y,n) and their inverses for 6 more equivalence relations. Or we can attach 2 of these choices and their inverses in 6C2 = 15 more ways for 15 more equivalence relations. Or we can attach 3 of these choices and their inverses in 6C3 ways, so on... So there should be 1 + 6 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = [b]64 [/b]possible equivalence relations?

Could someone please verify this?

Thanks.

Every disjoint decomposition of a set (i.e., breaking the set into disjoint subsets) defines an equivalence relation on the set. You simply say that the elements in each disjoint subset are equivalent to each other and this relation is obviously transitive, reflexive and symmetric. A finite set has a finite number of disjoint decompositions and this number has to be found. A 4-element set has the following disjoint decompositions:

1. One 4-element subset (the set itself), $C_4^4 = 1$ possibility.
2. One 3-element subset + one 1-elemet subset, $C_3^4 = 4$ possibilities.
3. Two 2-element subsets, $\frac 1 2 C_2^4 = 3$ possibilities.
4. One 2-element subset + two 1-element subsets, $C_2^4 = 6$ possibilities.
5. Four 1-element subsets, $\frac 1 4 C_1^4 = 1$ possibility.

The total number of disjoint decompositions of the 4-element set $A = \{0, 1, y, n\}$, equal to the number of equivalence relations on this set, is

$C_4^4 + C_3^4 + \frac 1 2 C_2^4 + C_2^4 + \frac 1 4 C_1^4 = 1 + 4 + 3 + 6 + 1 = 15$

Thx Yetti, I didn't realize that equivalence relations form a bijection with the number of partitions.


Could someone please post a solution to this seemingly very nice inequality? i've tried long and hard with fruitless results.

Can anyone help? I had no luck in solving this problem. :)

[quote="Parkdoosung"]Let $ a,b,c,d > 0$.
Prove that $ (\frac {a \plus{} b}{a \plus{} b \plus{} c})^{2} \plus{} (\frac {b \plus{} c}{b \plus{} c \plus{} d})^{2} \plus{} (\frac {c \plus{} d}{c \plus{} d \plus{} a})^{2} \plus{} (\frac {d \plus{} a}{d \plus{} a \plus{} b})^{2}\geq\frac {16}{9}$.[/quote]
I think this can be done with the S.M.V. but i cant prove it because i'm new to this method.
Can anybody else post solution with mixing variables???

[quote="Parkdoosung"]Let $ a,b,c,d > 0$.
Prove that $ (\frac {a \plus{} b}{a \plus{} b \plus{} c})^{2} \plus{} (\frac {b \plus{} c}{b \plus{} c \plus{} d})^{2} \plus{} (\frac {c \plus{} d}{c \plus{} d \plus{} a})^{2} \plus{} (\frac {d \plus{} a}{d \plus{} a \plus{} b})^{2}\geq\frac {16}{9}$.[/quote]
See also here: http://can-hang2007.blogspot.com/2010/02/inequality-117-v-q-b-can.html

[quote="minhhoang"]Oh, i can't read your solution.  Can you upload your solution for me, can_hang2007 ? Thanks !  :maybe:[/quote]
Use a djvu reader program, then you can read it!

[quote="minhhoang"]I can't use your link, Can, can you upload your solution ?  :blush:[/quote]
See attachment.


IMO Shortlist
combinatorics proposed
combinatorics

Prove that it is not possible to write numbers $ 1,2,3,...,25$ on the squares of $ 5$x$ 5$ chessboard such that any neighboring numbers differ by at most $ 4$.

@ Raja
Are numbers diagonal to each other considered neighbours?

no, it seems like neighbors = adjacent... (?) sorry for my bad english

[quote="snain"]Suppose A an B are in the same line (res. Column or diagonal). Line (res. Column or diagonal) contains maximum of 5 numbers so | A - B | <= (5-1)* 4 = 16.

1 and 25 can not be in the same line, Column or diagonal. (I)

1 can not have more than 4 neighbours so 1 is in the corner. (II)

25 can not have more than 4 neighbours so 25 is in the corner. (III)

(II) and (III) => 1 and 25 are in the same line or Column or diagonal witch is in contradiction with (I).[/quote]

Twice wrong-
1)Numbers on diagonal don't need to differ by less than 5
2)1 and 25 can have 4 neighbours

As I can see, I think this problem is just the special case of [url=http://www.mathlinks.ro/resources.php?c=1&cid=17&year=1988]IMO shortlist 1988 #4[/url]
However, it's still hard even today!

I have solved this, but with massive caseworks. :blush:  I think I cannot prove the general problem with my method, so I'll just wait for someone to write a better proof.

I'm going to prove general case.Namely,

[b]Theorem[/b]
Write numbers $1$~$n^2$ on the squares of $n$×$n$ chessboard.Then there are two neighbouring squares whose numbers differ by at least $n$.

proof)We write $1$~$n^2$ numbers in turn.It is obvious that If the number of empty squares which are neighbor to written squares is more than(or equal) $n$ a moment, theorem certainly holds.We consider the moment which some row(or column) is completed(all squares on some row are written numbers).

[b]Case1[/b]:some row is completed
On each column, there is a empty square which is neighbor to written squares.So theorem certainly holds.

[b]Case2[/b]:some column is completed
It is the same as Case1.

[b]Case3[/b]:some row and some column are simultaneously completed
We suppose that this case is occurred when the square A is written.Just before the square A is written, there is a empty square which is neighbor to written squares on each column.So theorem certainly holds.

From case1,2,3, we prove the theorem.$\blacksquare$


A four-digit positive integer $ \overline{a(a\minus{}1)(a\minus{}1)a}$ can be wrriten as a sum of three consecutive squares (for example,$ 1^2\plus{}2^2\plus{}3^2$).
Find the number.

This got a little ugly, but...

[hide="Solution"]
Expanding it out, we have
\[ a\plus{}10a\minus{}10\plus{}100a\minus{}100\plus{}1000a\equal{}1111a\minus{}110\equal{}k^2(k\plus{}1)^2\plus{}(k\plus{}2)^2\\
1111a\minus{}110\equal{}3k^2\plus{}6k\plus{}5\\
1111a\minus{}115\equal{}3k^2\plus{}6k\\
1111a\minus{}124\equal{}3k^2\plus{}6k\minus{}9\equal{}3(k\minus{}1)(k\plus{}3)
\]

Take this $ \mod 3$:
$ a\equiv 1\pmod 3\implies a\equal{}3q\plus{}1\implies a\equal{}1,4,7$.

Now simply bash.
$ 1111a\minus{}115\equal{}3k^2\plus{}6k\implies 3k^2\plus{}6k\plus{}(115\minus{}1111a)\equal{}0$.
$ k$ must be an integer, so the discriminant of this thing must be a perfect square, for starters.
That is, $ \sqrt{36\minus{}12(115\minus{}1111a)}$ is an integer.
Plugging in $ 4$ gives $ 51984$, which is $ 228^2$. The others give only non-perfect squares.
$ k\equal{}\frac{\minus{}6\plus{}228}{6}\equal{}37$.

So the answer is $ \boxed{4334}\equal{}37^2\plus{}38^2\plus{}39^2$
[/hide]

If anyone has a cleaner way, I'd like to see it.

[quote="Anavel_Gato"]A four-digit positive integer $ \overline{a(a \minus{} 1)(a \minus{} 1)a}$ can be wrriten as a sum of three consecutive squares (for example,$ 1^2 \plus{} 2^2 \plus{} 3^2$).
Find the number.[/quote]

Let $ \overline{a(a \minus{} 1)(a \minus{} 1)a} \equal{} k^{2} \plus{} (k \plus{} 1)^{2} \plus{} (k \plus{} 2)^{2}$ .  The numbers $ k^2$,$ (k \plus{} 1)^{2}$,$ (k \plus{} 2)^{2}$ have diferent remainders when are divided by $ 3$, then $ k^{2} \plus{} (k \plus{} 1)^{2} \plus{} (k \plus{} 2)^{2} \equiv 2 {mod3}$.So $ a \plus{} a \minus{} 1 \plus{} a \minus{} 1 \plus{} a \equal{} 4a \minus{} 2 \equiv 2 {mod 3}\Rightarrow 4(a \minus{} 1)\equiv0 {mod3}$ so $ a \equal{} 4$ or $ a \equal{} 7$ but $ a \equal{} 7$ there's no solution ,so $ a \equal{} 4$

                                                                                [b]ee"[/b]


inequalities
search
inequalities unsolved

Let a, b, ,c are real numbers. Find the min value of k such that
$ a^4\plus{}b^4\plus{}c^4\plus{}k(ab\plus{}bc\plus{}ca)^2\geq \sum_{sym}ab(a^2\plus{}b^2)$
If k=6, prove that this inequality is true. :D

[quote="EUCLA"]I'll prove the case $ k \equal{} 6$.
Let's prove it for positive reals. If it is true for positive, it is also true for some of $ a,b,c$ negative since $ LHS\ge 0$
\[ a^4 \plus{} b^4 \plus{} c^4 \plus{} 6(ab \plus{} bc \plus{} ca)^2\ge 2ab(a^2 \plus{} b^2) \plus{} 2bc(b^2 \plus{} c^2) \plus{} 2ca(c^2 \plus{} a^2)
\]

\[ \sum_{cyc}{a^4} \plus{} 2\sum_{cyc}{a^2b^2} \plus{} 12\sum_{cyc}{a^2bc}\ge 2ab(a \minus{} b)^2 \plus{} 2bc(b \minus{} c)^2 \plus{} 2ca(c \minus{} a)^2
\]

\[ (a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} 12abc(a \plus{} b \plus{} c)\ge 2ab(a \minus{} b)^2 \plus{} 2bc(b \minus{} c)^2 \plus{} 2ca(c \minus{} a)^2
\]
By am-gm we get : $ RHS\le \sum_{cyc}{\bigg(\frac {a^2 \plus{} b^2}{2}\bigg)^2}$

Then:
\[ 2(a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} 24abc(a \plus{} b \plus{} c)\ge a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2
\]
which is clearly true.[/quote]
think it's not enough to prove it for positives only :wink:

:huh: Isn't LHS just sum of squares, or am I missing something evident?

edit: you edited, now..why?

edit2: understood  :wink:

[quote="EUCLA"]:huh: Isn't LHS just sum of squares, or am I missing something evident?

edit: you edited, now..why?[/quote]

it is but allowing some of $ a,b,c$ to be negative we may achieve $ ab\plus{}bc\plus{}ca$ to be $ 0$, and it's not quite obvious then that the inequality holds, at least not to me..

I'll try, this time, in finding $ k$. Sorry if I'm mistaking again. :oops:

My guess is $ \boxed{k\equal{}\minus{}\frac{2}{3}}$

\[ \sum_{cyc}{a^4}\plus{}k\sum_{cyc}{a^2b^2}\plus{}2k\sum_{cyc}{a^2bc}\ge \sum_{cyc}{2ab(a^2\plus{}b^2)}\]
\[ (a^2\plus{}b^2\plus{}c^2)^2\plus{}2abck(a\plus{}b\plus{}c)\plus{}(k\minus{}6)\sum_{cyc}{a^2b^2}\ge \sum_{cyc}{2ab(a\minus{}b)^2}\]

We now have to minimize lhs and maximize rhs.

By am-gm $ 2ab(a\minus{}b)^2\le \bigg(\frac{a^2\plus{}b^2}{2}\bigg)^2$

\[ 2\sum_{cyc}{a^4}\plus{}6\sum_{cyc}{a^2b^2}\plus{}2abck(a\plus{}b\plus{}c)\plus{}(k\minus{}6)\sum_{cyc}{a^2b^2}\ge 0\]
\[ 2\sum_{cyc}{a^4}\plus{}2abck(a\plus{}b\plus{}c)\plus{}k\sum_{cyc}{a^2b^2}\ge 0\]

Since it is homogeneus (of degree 4) one can set $ abc\equal{}1$

Then, by am-gm we get \[ 6\plus{}6k\plus{}3k\ge 0\]

So we get the wanted result.

[quote="thegod277"]Let a, b, ,c are real numbers. Find the min value of k such that
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} k(ab \plus{} bc \plus{} ca)^2\geq \sum_{sym}ab(a^2 \plus{} b^2)$
If k=6, prove that this inequality is true. :D[/quote]
By ABC method
We have to prove ineq in two case
1.$ c \equal{} 0$
if $ a$ or $ b \equal{} 0$ ineq true
if $ a,b\neq 0$.Ineq <=>
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{a^2} \plus{} k\ge \frac {a}{b} \plus{} \frac {b}{a}$
$ t \equal{} \frac {a}{b} \plus{} \frac {b}{a}$
ineq <=> $ k\ge \minus{} t^2 \plus{} t \plus{} 2$
<=> $ k\ge Max \minus{} t^2 \plus{} t \plus{} 2 \equal{} 0$ ($ |t|\ge 2$)
2.$ b \equal{} c\neq 0$ ($ b \equal{} c \equal{} 0$ easy)
ineq <=> $ a^4 \plus{} k(b^2 \plus{} 2ab)\ge 2ab(a^2 \plus{} b^2)$
$ a \equal{} 0$ easy :D
$ b \equal{} ax (a\neq 0)$
ineq <=> $ k\ge \frac {2x(x^2 \plus{} 1) \minus{} 1}{(x^2 \plus{} 2x)^2}$
$ f'(x) \equal{} 0$ < = > $ x \equal{} 1,x \equal{} \minus{} 1,x \equal{} s$ (in there $ s^3 \minus{} s^2 \plus{} 2s \plus{} 2 \equal{} 0$)
So $ Max f(x) \equal{} Max(f(1),f(s)) \equal{} f(1) \equal{} \frac {1}{3}$
so $ k\ge f(s)$
$ k\ge Max (0,\frac {1}{3}) \equal{} \frac {1}{3}$
Sorry if I am false :blush:

[color=green]@thegod277, use Search tool plz, u can find more about ABC method.  :roll: [/color]

I should have done something wrong, in using am-gm on negatives, i suppose  :(

[quote="thegod277"]Sorry, but I don't know ABC method :( . What is ABC method? :huh:[/quote]
See here:
http://www.mathlinks.ro/viewtopic.php?t=165132

Oh, thank you.  :D
Send EUCLA,vodanh_tieutot: Thank you for your proof. You can see a general problem in Inequalities Proposes & Own problem,a post "very very hard" of me. And the minximum value of k of my inequality is $ \frac{1}{3}$, because if we choose a=b=c, then $ 3\plus{}9k\geq 6$, so $ k\geq \frac{1}{3}$.  :wink:

[quote="thegod277"]
the minximum value of k of my inequality is $ \frac {1}{3}$, because if we choose a=b=c, then $ 3 \plus{} 9k\geq 6$, so $ k\geq \frac {1}{3}$.  :wink:[/quote]
In some case min,max k isn't with a=b=c :D
Ex:
Find k Max ($ a,b,c\ge 0$)
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}25\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\ge k$
:D

Thank you. But I can prove the inequality is true for $ k\equal{}\frac{1}{3}$. So, kmin=1\3.
 :D


superior algebra
superior algebra solved

[tex]\frac{6563} { \sqrt{1-\frac{3*(\frac{0.5}{1})}{10000} }}[/tex]

How do I figure out how many significant digits should be in the final answer?

The 0.5 and 1 in the the equation are inputs.
This equation computes how much shift is in a hydrogen line due to gravitational redshift.

I get 6563.49 as the answer, but 0.5 & 1 in the (0.5/1) , and the 3  are only 1 significant digit.  The "1-" is a pure 1 (1.000000000).

I'd hate to give an answer of 7000.  That would ruin everything.

You're going about this the wrong way: the shift is tiny compared to the original value. You don't want to add it on to the original value, because the inaccuracy in that value will swamp you.
$a\cdot\left(\frac1{\sqrt{1-x}}-1\right)=a\cdot\left(\frac x2+O(x^2)\right)$
Since your value of $x$ is tiny and has low precision, the $O(x^2)$ term can be ignored. The shift is then $\frac{ax}2=\frac{6563\cdot3\cdot.5}{10000\cdot2}\approx.5$. You only have one significant figure here for the obvious reason.

This doesn't mean you can conclude that the shifted value is $6563.5$; you don't have enough significant figures in the original value to say that.

Thank you, jmerry.  That's about what I expected.


[quote="4everwise"]Find all integers $x$ such that $x^2-5x+6$ is a positive prime number.

[hide="Hint for those who want it"]When good mathematicians are stuck, they should always say the f word.

Factor!!  :D

Am I funny or what? (If you're thinking about saying no, don't)  :) [/hide][/quote]

[hide]

  Factoring $x^2 -5x + 6$ yields:

$(x-3)(x-2)$

  A prime number has two factors, "1" and itself.  Hence, we simply set each equation as 1, and then solve.

$x-3=1$
$x=4$

  The result is 2.

$x-2=1$
$x=3$

  The result is 0, which is not a prime number.

  Hence, the only value that works is $\boxed{2}$[/hide]  By result I meant value of the equation $x^2 - 5x + 6$

[hide]$x^2-5x+6=(x-2)(x-3)$.

We want one of these to be one, so the result may still be prime. Either x-2=1 or x-3=1.

If x=3, we get the answer 0, which doesn't work. If we try x=4, the result is 2, which is prime.

x can only be 4. [/hide]

Ok, 236factorial has determined that one of the answers is 4. There is one more.  ;)

[hide="Hint"]We know that x=2 and x=3 don't work. [hide="Continuation of Hint"] Oviously, 4 is the only positive value that works. Are there any negative values that work?  :roll: [/hide][/hide]

Are negative numbers considered prime? I thought that primes only referred to positives... or is it just that every time a problem talks about primes, they mean "positive primes"?

[b]EDIT:[/b] No, there is no such thing as a negative prime. All numbers that are negatives of primes actually have four factors: itself, one, negative one, and negative itself. And $-1$ doesn't count. ;)

[hide]Well either $(x-2)$ or $(x-3)$ has to be $-1$.
If $x-3=-1$ then $x=2$, but that doesn't work.
So $x-2=-1$ and $x=1$, which yields $(-1)(-2)=2$.
So $x$ can be $1$ too.[/hide]

Oh, that's what 4everwise meant. He didn't quite word it correctly, though...

[quote="Melissa"]Well either $(x-2)$ or $(x-3)$ has to be $-1$.
If $x-3=-1$ then $x=2$, but that doesn't work.
So $x-2=-1$ and $x=1$, which yields $(-1)(-2)=2$.
So $x$ can be $1$ too.[/quote]Good job!
 :)
I accidentally worded my second hint incorrectly. :blush:


My school offered amc 10a for the first time this year but i only got a 129 so even though it qould qualify me for AIME i want to take 10b to get a better USAMO index and my school refused to offer amc 10b so i am going to like tennessee or kentucky to take amc 10b so i was just wondering what i need to do i mean do i just show up at the test site or what :?:

you need to contact the director of this program in that school or college.


function
number theory proposed
number theory

For the positive integers written in base $10$ we define the function $T$ such that
the final figure  is moved to the front of the number  : for ex. $T(21345) = 52134$ .

Find the smallest $n$ such that $T(n) = 6n$ .

 :cool:

[quote="Misha123"]n=10a+b with b as the last digit. T(n)=b.10^(k-1)+a with k as the number of digits of n. T(n)=6n is equivalent to       b(10^(k-1)-6)=59a. b<10 and 59 is prime so 59|(10^(k-1)-6). The least possible k is 58 because 59|(10^58-1) (Little Fermat) and 6 is the inverse of 10 modulo 59. a must have k-1=57 digits so the least possible value of b is 6. n=10a+b=(10^57-6).60/59+6=(10^57-6)/59.

Misha[/quote]

How do you know that 58 is the smallest k such that 59|(10^k-1) ? True that 10^58 = 1 mod 59, but it could be that 10^29 = 1 mod 59 too.  We are looking for the smallest possible k.

Also, once you get k, how do you know that b = 6?


does any body know a good book for mobius functions? thanks

Please read the posting guidelines.  The Getting Started for is for basic high school problem solving math.  Mobius functions are rarely studied before the undergraduate level.

According to Mathworld...

http://mathworld.wolfram.com/MoebiusFunction.html

frt thanks alot,...

but i have seen it many years ago... ;)

no no,

i mean i have seen the link that he wrote...


counting
derangement
function
combinatorics unsolved
combinatorics

Any one have single idea to simply this ?

$\sum^n_{k=0}(-1)^k\binom{n}{k}(n-k)!$

[hide="source"]
I come across this when trying to solve the problem which is : "There are $n$ boxes each with one different ball in them . Then you rearrange them so that no ball is in their original boxes (each boxes must have only one ball) . Find the number of way to do so . " I get the above expression but I think it can be further simplified which I dont know how .[/hide]

[quote="zanttrang"]This is the classic "derangement" problem.  The sum you found is usually expressed as

$n! \sum^n_{k=0}\frac{(-1)^k}{k!}$

I've never seen any simpler analytic expression for this.  However, an expression involving the nearest integer function is

$\left[\frac{n!}{e}\right]$

This can be seen by examining the McLaurin series for $e^{-1}$:

$e^{-1} = 1 - 1/1! + 1/2! - 1/3! + ...$.  So,

$n! \sum^n_{k=0}\frac{(-1)^k}{k!} = n!(e^{-1} + \epsilon)$

where $| \epsilon | < 1/(n+1)!$

I don't recall how to make the final leap to the solution, but you can see why this would be a reasonable approximation for the sum.

For more information, check out Mathworld's site on the topic:

http://mathworld.wolfram.com/Derangement.html[/quote]

wow , thats cool . Thanks  :D


$(a)$  Prove that for all primes $p > 5$, $p^{4}\equiv 1 \mod 240$
$(b)$  If $p^{4}\equiv 1 \mod 240$ and $p \in \mathbb{Z}$, is $p$ necessarily prime?

A note for the first, it only works for $7$ and up.
[hide="Solution"]
All primes except $2$ and $3$ can be written as $6k \pm 1$

Note that $\frac{(6k+1^{4})-1}{24}=k+9k^{2}+36k^{3}+54k^{4}$ and that $\frac{(6k-1^{4})-1}{24}=-k+9k^{2}-36k^{3}+54k^{4}$

Since $k^{N}\equiv k \pmod{2}$, the two above are clearly even.

Taking them in mod 5 gives:
$k-k^{2}+k^{3}-k^{4}$ and $-(k+k^{2}+k^{3}+k^{4})$
Testing over 0 through 4,

-the first is not divisible by 5 only for $k\equiv 4$, which is not a problem because if we substitute $k=5z+4$, then $6k+1=30z+24+1=5(6z+5)$, which is not prime.

-the second is not divisible by 5 only for $k\equiv 1$, which is also not a problem because if we substitute $k=5z+1$, then $6k-1=30z+6-1=5(6z+1)$ which is not prime for all $z>0$. If $z=0$, it gives the prime number 5, which does not work.

Thus for all $p \in \text{Primes}-\{ 2,3,5 \}, \; p^{4}\equiv 1 \pmod{240}$

[/hide]

[hide="Second"]
Dunno if 241 is prime, but $p=241^{2}$ is definitely a counterexample.
[/hide]

[quote="Allan Z"](B) No definately not, the easiest counterexample is $240*2+1$, which can be factored into $13*37$[/quote]

In fact, I believe the smallest nontrivial counterexample is $p = 49$.  Generally, $p^{4}\equiv 1 \bmod 240 \Leftrightarrow 2, 3, 5 \not | p$.  Of course, the easiest counterexample is $240 \times 0+1$.

[quote="t0rajir0u"][quote="Allan Z"](B) No definately not, the easiest counterexample is $240*2+1$, which can be factored into $13*37$[/quote]

In fact, I believe the smallest nontrivial counterexample is $p = 49$.[/quote]Yes never forget that 2401 :P

[hide="(a)"]$p^{4}-1\equiv 0\pmod{24}$ if $p^{4}-1\equiv 0\pmod{16,3,5}$.

Notice that $p^{4}\equiv (-p)^{4}\pmod{n}$.
For p greater than 5, $p\equiv 1,(-1),2,(-2) \pmod{5}\implies p^{4}\equiv 1 \pmod{5}$.
For p greater than 3, $p\equiv 1,(-1) \pmod{3}\implies p^{4}\equiv 1 \pmod{3}$.
For p greater than 2, $p\equiv 1 (-1),3 (-3),5 (-5), 7 (7-)\pmod{16}\implies p^{4}\equiv 1\pmod{15}$.
Thus, $p^{4}\equiv 1\pmod{240}$, for primes greater than 5.[/hide]

Sorry, everybody.  A couple of typos have been fixed.  For (B), $p$ has to be integral and for (A), $p$ has to be $> 5$.


The point $ (2, 3)$ is reflected about the $ x$-axis to a point $ P$. Then $ P$ is reflected about the line $ y \equal{} x$ to a point $ Q$. What is the $ x$-coordinate of $ Q$?

When a point is reflected over the $ x$-axis, the $ y$-coordinate changes to its additive inverse.  Point $ P$ is $ (2,\minus{}3)$.  When reflecting over the line $ y\equal{}x$, I believe both the coordinates change to its additive inverse.  So point $ Q$ is $ (\minus{}2,3)$.  So the $ x$-coordinate is $ \boxed{\minus{}2}$

Reflecting $ (2,3)$ over the $ x$-axis, we change the sign of the $ y$-coordinate, so we get $ (2, \minus{} 3)$. Then, reflecting across $ y \equal{} x$, which flips the coordinates, we get $ ( \minus{} 2,3)$. The $ x$-coordinate of $ ( \minus{} 2,3)$ is $ \fbox{ \minus{} 2}$.

EDIT: Beaten.

Actually when you're reflecting a point across $y=x$, the coordinates are flipped i.e. $(x,y)$ becomes $(y,x)$. Thus the answer is actually $\boxed{-3}$.


it is well-known that when $ \nu(X) < \infty\,\,$  and $ p\leq q$ then $ \L^{q}\subset L^{p}$
Could any one prove the converse that if $ p\leq q$ and $ \L^{q}\subset \L^{p}\,\,$ then $ \nu(X) < \infty$?

Remark : we just consider $ X\subset\mathbb{R}$
thanks so much

Let $ X$ be an arbitrary $ \sigma$-finite measure space and $ 1\le p<q\le\infty$. Choose $ r$ strictly between $ p$ and $ q$.
If $ X$ has infinite measure, we can construct something that looks like $ x^{\minus{}1/r}$ as $ x\to\infty$. It's in $ L^q(X)$ but not $ L^p(X)$
If $ X$ has subsets of arbitrarily small positive measure, we can construct something that looks like $ x^{\minus{}1/r}$ as $ x\to 0^\plus{}$. It's in $ L^p(X)$ but not $ L^q(X)$.

This should answer your questions.


geometry
circumcircle
geometric transformation
reflection
incenter
cyclic quadrilateral
geometry proposed

Let $ ABC$ be a triangle with $ BC > AC > AB$. Let $ A',B',C'$ be feet of perpendiculars from $ A,B,C$ to $ BC,AC,AB$, such that $ AA' \equal{} BB' \equal{} CC' \equal{} x$. Prove that:
a) If $ ABC\sim A'B'C'$ then $ x \equal{} 2r$
b) Prove that if $ A',B'$ and $ C'$ are collinear, then $ x \equal{} R \plus{} d$ or $ x \equal{} R \minus{} d$.

(In this problem $ R$ is the radius of circumcircle, $ r$ is radius of incircle and $ d \equal{} OI$)

$ H$ is the orthocenter. $ A', B', C'$ are not feet of perpendiculars from $ A, B, C$ to $ BC, CA, AB,$ but points on the A, B-, C-altitudes $ AH, BH, CH,$ respectively.

[color=blue][b](a)[/b][/color] Assuming $ \triangle A'B'C' \sim \triangle ABC$ are oppositely similar, otherwise $ AA' \equal{} BB' \equal{} CC' \equal{} x \equal{} 0$ satisfies the conditions. $ \triangle A'B'C' \sim \triangle ABC$ $ \Longrightarrow$ $ A'B'C'H$ is cyclic. Assume $ AA' \equal{} BB'.$ Locus of circumcenters $ O'_c$ of the $ \triangle B'HA'$ is a line $ p_c \parallel CI$ parallel to the internal bisector of the $ \angle BCA.$ Likewise, assuming $ BB' \equal{} CC',$ locus of circumcenters $ O'_a$ of the $ \triangle C'HB'$ is a line $ p_a \parallel AI$ and assuming $ CC' \equal{} AA',$ locus of circumcenters $ O'_b$ of the $ \triangle A'HC'$ is a line $ p_b \parallel BI.$ Let $ P_a, P_b, P_c$ be reflections of the circumcenter $ O$ in $ BC, CA, AB.$ Then $ P_a \in p_a, P_b \in p_b, P_c \in p_c.$ $ \triangle P_aP_bP_c \cong \triangle ABC$ are centrally congruent with similarity center $ N$ (common 9-point circle center). Lines $ p_a, p_b, p_c$ are internal bisetors of the $ \triangle P_aP_bP_c,$ concurrent at its incenter $ F.$ Circle $ (F)$ with radius $ FH$ is the only circle through $ H$ cutting $ AH, BH, CH$ again at $ A', B', C',$ such that $ AA' \equal{} BB' \equal{} CC' \equal{} x > 0.$ The Fuhrmann circle of $ \triangle ABC$ goes through $ H$ and cuts $ AH, BH, CH$ again at $ A', B', C',$ such that $ AA' \equal{} BB' \equal{} CC' \equal{} 2r$ (see the last reply at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=101687[/url]) $ \Longrightarrow$ $ (F)$ is the Fuhrmann circle and $ x \equal{} 2r.$

[color=blue][b](b)[/b][/color] Let $ X, Y, Z$ be reflections of $ A', B', C'$ in the internal angle bisectors $ AI, BI, CI.$ $ AX \equal{} BY \equal{} CZ \equal{} x$ $ \Longrightarrow$ $ X \in AO, Y \in BO, Z \in CO$ are on a circle $ \mathcal P(O, x \minus{} R)$ (radius can be positive or negative). Let $ k \equiv A'B'C'$ be the collinarity line and $ k_x, k_y, k_z$ reflections of $ k$ in $ AI, BI, CI.$ Let $ k_x, k_y$ intersect at $ Q$ (different from $ X, Y$). By angle chase for the $ \triangle B'HA'$ and quadrilateral $ XOYQ,$ $ \angle B'HA' \equal{} \angle XQY \equal{} \frac {\angle XOY}{2} \mod \pi$ $ \Longrightarrow$ $ Q \in \mathcal P(O, x \minus{} R).$ Similarly, $ k_y, k_z$ intersect at a point on the circle $ \mathcal P$ different from $ Y, Z$ $ \Longrightarrow$ $ k_x, k_y, k_z$ concur at $ Q \in \mathcal P.$ Let $ U, V, W$ be reflections of $ Q$ in $ AI, BI, CI.$ The incenter $ I$ is the circumcenter of the cyclic quadrilateral $ QUVW,$ because perpendicular bisectors $ AI, BI, CI$ of its 3 sides $ QU, QV, QW$ concur at $ I.$ $ U, V, W \in k$ are collinear $ \Longrightarrow$ this is possible only when $ I \equiv Q$ ($ \equiv U \equiv V \equiv W$), therefore $ I \in \mathcal P,$ $ x \minus{} R \equal{} \pm OI$ and $ AA' \equal{} BB' \equal{} CC' \equal{} x \equal{} AX \equal{} BY \equal{} CZ \equal{} R \pm OI.$

We can solve part a easier .
We know $A'B'C'H$ in cycle .
so $ A'H . B'C' + HB' . A'C'  = HC' . A'B' $
 $(x-AH) . B'C' +(x-BH) . A'C'  = (CH - x) . A'B' $
 $\triangle A'B'C' \sim \triangle ABC$ so  $ AB = k A'B' $ and ....
so $(x-AH) . BC +(x-BH) . AC  = (CH - x) . AB $
so $ 2xp  = BC.AH + AC.BH + AB.CH = 4S = 4rp $
so $x=2r$

I mean $2p = AB + BC + CA $ and $S$ is the area of the triangle .

Is there a strict proof for $A'B'C'H$ being cyclic? Angle chasing might work only we presume a specific configuration of those points.


algebra polynomial probability function geometry circumcircle inradius

Here's some easy but nice problems I made up if you guys wanna do them... :whistling:

1)a) If the roots of the polynomial $2x^3+18x^2-12x+8$ are $a,b,c$, compute

$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$
b) find a polynomial who's roots are $a+3,b+3,c+3$
c) find a polynomial whos roots are $3a,3b,3c$

2) What is the probability that if we pick two numbers out of $0$ to $100$, they'll differ by at most $4$?

3) How many solutions are for $a+b+c+d+e\leq 70$?

4) How many ways can we distribute $20$ burgers among $5$ students so that no student gets less than $2$ or more than $6$?

I don't know what the answer to this problem is, so can someone give a detailed explanation of generating functions?

5) Find the distance between the centers of the circumcircle and incircle of a $12,13,14$ triangle

 number 5's answer will be a little messy though so you can probably leave the answer unsimplified in the $d^2$ formula

Enjoy :)

Are these problems too hard or does nobody want to do them?? I thought people in the Carolina's Board wanted people to post problems??? ;)  :diablo:

well, i guess since no one wants to do them...i'll answer my own questions..
[hide="1a"]
The polynomial with the roots reciprocal to $2x^3+18x^2-12x+8$ is $8x^3-12x^2+18x+2$, and we know that $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac) \Longrightarrow a^2+b^2+c^2=(\frac{3}{2})^2-2(-\frac{9}{4})=\boxed{\frac{27}{4}}$[/hide]
[hide="1b"]
For this, we have to take $f(x)$ and convert it into $f(x-3)$, so we use synthetic division and have $\boxed{2x^3-66x+152}$
[/hide]
[hide="1c"]
For this, we are multiplying the coefficients from left to right by $3^0,3^1,3^2,3^3$, so we have $\boxed{2x^3+54x^2-108x+216}$[/hide]
[hide="2"]
We can set this problem up geometrically, thus making a square with sides of $100$, the area of our desired numbers is $100^2-96^2 \Longrightarrow (100+96)(100-96)=784$, so our probability is $\frac{784}{100^2}$[/hide]

[hide="3"]
Since $\binom{n+r-1}{r-1}$ is applied and since it's an equality, we can fill another number in to give the extra sum left over, so we have $\boxed{\binom{75}{5}}$[/hide]

Number 4, I'm not exactly sure how to do, [hide="but"]using generating functions, we can have $(x^2+x^3+x^4+x^5+x^6)^5$ and we should be looking for the value of the coefficient of $x^{20}$, which I can't do because I don't know how to simplify the thing above...[/hide]
[hide="5"]
I'm not going to solve this right now because it's messy and takes too long, but just use Heron's formula to find the area of the triangle which will get us the inradius using $rs=A$ and also the circumradius using $R=\frac{abc}{4A}$ and use $d^2=R(R-2r)$ to find the distance.[/hide]

[quote="prybarczyk"]I can give you an explanation of how to solve number four if you need it.  Let me know.[/quote]

sure, i'm lost on generating functions  :?

Open the attachment.  I hope this makes sense.  I did not have a lot of time to write it up.  If you have any questions, let me know.


foll all $n$ is a number positive integer
a function $g(x)$ give by:
$g(x)$=-$n$ if $x$ :leq -$n$
$g(x)$=$x$ if -$n$<$x$ :leq $n$
$g(x)$=$n$ if $n$<$x$

let a function $f(x)$:R-R  such that:
for all $n$ in N*  then $g(f(x))$ is continous
prove that :$f(x)$ is a continous fuction

Let \[g_{m}\left( x\right) =\left\{ \begin{array}[c]{l}-m,-m<x\\ x,-m\leq x<m\\ m,m<x \end{array}\right. .\] If $f$ is locally bounded, then it is continuous. Indeed, suppose that $\left\vert f\left( x\right) \right\vert \leq M,$ for all $x$ in some interval $I$  and pick some $m>M.$ Then $g_{m}\left( f\left( x\right) \right) =f\left( x\right)$ for all $x\in I,$
hence the conclusion.

Now, suppose that $f$ is discontinuous at $a$ and not bounded from above.
There exists a sequence $\left( x_{n}\right)$ such that $x_{n}\rightarrow a$ and $f\left( x_{n}\right) \rightarrow\infty.$ Pick $m$ such that $m>\left\vert f\left( a\right) \right\vert .$ Then $\underset{n\rightarrow \infty}{\lim}g_{m}\left( f\left( x_{n}\right) \right) =m$ and $g_{m}\left( f\left( a\right) \right) =f(a) <m.$ It follows that $g_{m}\left( f\left( x\right) \right)$ is not continuous at $a,$ a contradiction.


I don't think previous threads have directly addressed this question. What is the point of taking an AP and SAT II in the same subject? I know that many colleges require 3 SAT IIs, but it seems redundant to take, say, AP Chemistry and SAT II Chemistry, since the AP is just a harder version of the same test.

Also, I was wondering, is it possible to take AP Physics B and AP Physics C in the same year? I looked at the schedule and it looks like the B exam and the mechanics part of the C exam overlap. Are the mechanics and E&M part of the C exam considered two different exams, or are they both "AP Physics C"?

[quote="aspiringchemlete"]What is the point of taking an AP and SAT II in the same subject? I know that many colleges require 3 SAT IIs, but it seems redundant to take, say, AP Chemistry and SAT II Chemistry, since the AP is just a harder version of the same test.[/quote]

The two tests are graded with a rather different philosophy and are used for different purposes.  The subject correspondence is not exact for other subjects, either - the SAT II US History test, from what I've heard, actually has a lot of material in it not in the AP curriculum.

[quote="aspiringchemlete"]Also, I was wondering, is it possible to take AP Physics B and AP Physics C in the same year? I looked at the schedule and it looks like the B exam and the mechanics part of the C exam overlap. Are the mechanics and E&M part of the C exam considered two different exams, or are they both "AP Physics C"?[/quote]

I wouldn't recommend it.  I think you can take the Mechanics test on a different date, but it's just a lot of hassle.  Mechanics and E&M are considered separate exams, although some colleges will not give you credit unless you have taken both.

[quote="aspiringchemlete"]I don't think previous threads have directly addressed this question. What is the point of taking an AP and SAT II in the same subject?[/quote]

SAT II tests are specific admission requirements at certain colleges, in various numbers and combinations, so you take those just to get into those colleges. You can take the AP tests because some colleges also ask about AP scores as part of the admission process.

Caltech:

[url=http://www.admissions.caltech.edu/uploads/File/CommonApp/application2008/Supplement08.pdf]Caltech supplement to Common Application[/url]

Harvard:

[url=http://www.admissions.college.harvard.edu/utilities/electronic_resources/download/Rollo0708App_onlinesupplement.pdf]Harvard supplement to the Common Application[/url]

Yale:

[url=http://www.yale.edu/admit/freshmen/application/pdf/yale_supplement.pdf]Yale supplement to the Common Application[/url]

Princeton:

[url=http://www.princeton.edu/pr/admissions/u/appl/pdf/PU_0708_comsupp.pdf]Princeton supplement to the Common Application[/url]

Duke:

[url=http://www.admissions.duke.edu/jump/applying/PDF/FormA_modifiable.pdf]Duke supplement to the Common Application[/url]

AP test scores can be used after admission for class placement, credit toward graduation, or for advanced standing, depending on differing policies at each college.

Thanks for the clarification!

I'm primarily using APs to stand out a bit more in the admission process. I have decided to take APs that I think I'll score well in, and then take SAT IIs to cover the other subjects so that I can demonstrate a versatile knowledge.

What's an SAT II?

What's an AP?

I've heard these words before. They are tests of some sort, right?
Where can I find out what these are...

[quote="aufha"]What's an SAT II?[/quote]

The first result when you Google that term.

[quote="aufha"]What's an AP?[/quote]

The third result when you Google that term (unless results are personalized in some odd way on your computer).

Google is your friend.

http://www.google.com/

Are the percentiles the same for a given score?

For example, I got an 800 on Math Level 2, but I was only in the 90th percentile. Does everyone who gets an 800 have 90th percentile on their score report?

Yes, everyone with the same scaled score receives the same percentile.  About 10% of students who take the Math Level 2 test receive an 800, hence the 90% percentile.

[quote="Γιώργος"]Greetings!
Im a student aboyt 16 years old. Il ive in Greece iand im interested about maths and international competions. I have a question. In order to get to M.I.T Cambridge or somewhere else you need to get to thwe I.M.O first?[/quote]
You should post a new thread if you have a new topic to discuss. The answer to this is "no", with some reservations, as I explained in the other thread where you asked this question.


(a_n) ,(b_n) two sequences decreasing to 0 and the sum of whose terms are infinite. Prove that the serie of min(a_n, b_n) diverges or give an exemple to disprove it.
Thank you very much.

it was already posted here.
at first construct positive  sequance (not necessary monotone) $ a_{n}$ and $ b_{n}$
such that $ \sum a_{n}$ and $ \sum b_{n}$  is equal to $ \infty$.  and $ \sum \min\{ a_{n}, b_{n} \} < \infty$
 then try to do them monotone decreasing by this way:
consider serie $ a_{1} \plus{} ... \plus{} \underbrace{\frac {a_{k}}{n_{k}} \plus{} ... \plus{} \frac {a_{k}}{n_{k}} }_{n_{k}}$ similary for $ b_{n}$ after all   let $ a'_{n }$  be $ n$ -th term of the serie $ a_{1} \plus{} ... \plus{} \underbrace{\frac {a_{k}}{n_{k}} \plus{} ... \plus{} \frac {a_{k}}{n_{k}} }_{n_{k}}$. so by choosing $ n_{k}$ sufficiently large you will get sequance $ a'_{n}$ and $ b'_{n}$ with required property

Did you know you reach this thread with
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314159
using digits of pi?


Consider an $n \times n$ array of numbers $a_{ij}$ (standard notation). Suppose each row consists of the $n$ numbers $1,2,\ldots n$ in some order and $a_{ij} = a_{ji}$ for $i , j = 1,2, \ldots n$. If $n$ is odd, prove that the numbers $a_{11}, a_{22} , \ldots a_{nn}$ are $1,2,3, \ldots n$ in some order.

If $n$ is odd, then all numbers outside the diagonal appear in pairs $(i,j),(j,i)$, so there is one of each number left on the diag.

[color=#666666]Hi..........[/color]

You can solve this problem at very way (and easy way). I now write one of the fun solution below:
suppose $ 2$ matrix $ n*n$ in the right and the left of first matrix. Put $ 1,2,3,...,n$ in the original  diagonal(in the first matrix) and then draw the parallel line by another diagonal. each line that transit each cells (that have a number) put same number in the other calls. in the end you can move $ 2$ matrix side over the original matrix. you will see this matrix is very you want.

[color=#666666]Bye........[/color]

I did it by contradiction.

first of all, we see that the parts of the arrays on both sides of the diagonal are mirror images of each other. so, there must be equal numbers of each element on both sides of the diagonal.

suppose that the 'diagonal boxes' do not contain all numbers from 1 to n.

consider the 'missing numbers' ( numbers not in the diagonal positions). there are n copies of each number. since these numbers are not in the diagonal, they must be on both sides of it. also, there must be equal numbers of each element on each side. so, there must be n/2 copies of the missing numbers on each side.

but n/2 is not an integer. contradiction.

therefore, the a_11, a_22, ..., a_nn is a permutation of 1, 2, 3......, n.  :)

[quote=murgi]IF THE NO IN THE ROWS ARE IN INCREASING THEN THOSE IN THE COLOUMNS ARE ALSO INCREASING.................AND VICE VERSA

THIS SIMPLIFIES THE PROBLEM. :)[/quote]

how does that 'simplify the problem'?

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$a_{11}$&$a_{12}$ & .& .& .& .&. &. & .& .& .& .& $a_{1n}$\tabularnewline
\hline
$a_{21}$&$a_{22} $& .&. & .& .& .&. & .&. & . & .& $a_{2n}$\tabularnewline
\hline
.&. & .& .& .&. & .& .& . &. &. & . &. \tabularnewline
\hline
.& .& .& .& .&. &. &. &. & . & .&. & .\tabularnewline
\hline
& & & & & & & & & & & & \tabularnewline
\hline
& & & & & & & & & & & & \tabularnewline
\hline
& & & & &  &  & &  & & & & \tabularnewline
\hline
& & & & & & & & & & & & \tabularnewline
\hline
& & & & & & & & & & & & \tabularnewline
\hline
& & &  & & & & & & & & & \tabularnewline
\hline
.&  .& .& .&  .& .&. &. &. &. & .& .&. \tabularnewline
\hline
.&.& . &. & .& .& .&. &. & .& .& .& .\tabularnewline
\hline
 $a_{n1}$&$a_{n2}$ &. &. &. & .&. &. &. & .&. &. &$a_{nn}$ \tabularnewline
\hline
\end{tabular}
[hide=Lemma 1]
The ith row and ith column are exactly the same.(because $a_{ij}=a_{ji}$)
[/hide]
[hide=Lemma 2]
Every number in a row or a column are distinct.(direct from lemma 1 and condition given in the question)
[/hide]
Hence, from these lemmas we can say that every number from 1 to n will occupy
the diagonal position atleast once. But there are only n diagonal positions.

So, each number will exactly occupy the fiagnoal position exactly once and hence,
we can say that numbers $a_{11}, a_{22} , \ldots a_{nn}$ are $1,2,3, \ldots n$ in some order.


I'm new to this site and not at all sure if it is alright to post this and if it is in the right section/forum. Here's my problem anyway:

If you roll one dice, the mean value of many rolls will be 3.5 (or (1+2+3+4+5+6)/6). Now, my question is what the formula is for the mean value of many rolls of n dices is if you always take the greatest value of the dices. This probably makes no sense at all (I'm not a native English speaker by the way) so I'll try with an example: Two dices, many rolls: If you get 1, 2 then 2 is the number to add. If 2, 3 then 3. If 1, 6 then 6 etc. Add all the highest numbers and divide by number of possibilities. For 1 dice the mean will be 3.5, for many dices the mean will approach 6. Working on the problem I came up with the following formula, is it correct?:

(1^n+2^n+3^n+...+6^n) / (1^(n-1)+2^(n-1)+....6^(n-1)) = (n (above) € k=1 (below) k^n) /  (n-1 (above) € k=1 (below) k^(n-1)) The formula to the left of the equal sign is a sum that I don't know how to write on a computer.

Thanks for any help. :)

sum in latex is $\sum^n_{k=0} a_i$ or alike.

The english part of your question is perfect, but the long summation looks a bit awkward :) I think you meant to write twice "n" above the summation instead of once n-1? :)

In any case, for n=2 we have that 11/36*6+9/36*5+7/36*4+5/36*3+3/36*2+1/36*1=4.472222222 while your sum yields 4.333333333. But it is a quite good approximation! (after 2 the difference decreases and converges to 0 in the end. but it is not the exact formula.)

I think this may be a more clear way: chance on the maximum being ...
1 : 1/6^n
2 : (2/6)^n - (1/6)^n
3 : (3/6)^n - (2/6)^n
4 : (4/6)^n - (3/6)^n
5 : (5/6)^n - (4/6)^n
6 : 1 - ((5/6)^n

so we multiply the i-th with i to get the desired result. for notation, (k/6)^n = ak

a1 + 2(a2 -a1) + 3(a3 -a2) + ... + 6(1 -a5), which results in our asked thing being 6 - (5/6)^n - (4/6)^n - (3/6)^n - (2/6)^n - (1/6)^n

Can you post how you did it? You may have a very good formula just missing one little piece or something :)

Thanks a lot Peter! I understand your solution perfectly. :) I was thinking in the same way myself but went wrong somewhere. Only now do I understand how simple the problem was. Thanks again. :) Have a nice weekend!


induction
combinatorics unsolved
combinatorics

In a wagon, every $m \geq 3$ people have exactly one common friend. (When $A$ is $B$'s friend, $B$ is also $A$'s friend. No one was considered as his own friend.) Find the number of friends of the person who has the most friends.

Generalized friendship theorem........

anyone has a proof to this one?

I don't get it.

If three people have exactly one friend $X$, then won't four people have the exact same friend X?  Because if four people have a common friend $Y \neq X$, then the three people in that set of four that have a common friend $X$ also have a common friend $Y$, contradicting that they have exactly one common friend.

But this is obviously too easy for chinese TST.. am I missing something?

---

(edit)  oops:  m is fixed, i get it.

[quote=grobber]Assuming the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=friendship&t=48500]Friendship Theorem[/url], it's easy to find the structure of such graphs (i.e. graphs in which every $m$ vertices have exactly one common neighbor). This is because amazingly enough, the problem gets easier when $m$ exceeds $2$ ($m=2$ is the classical version of the theorem).

The graphs look like this: there are $m-1$ vertices $x_1,\ldots,x_{m-1}$, each of which is connected to all other vertices, and every other vertex has exactly one neighbor except for $x_1,\ldots,x_{m-1}$. We assume there are at least $m+1$ vertices, of course; otherwise the problem isn’t interesting, since the hypothesis is vacuously satisfied. In fact, it follows from this that all vertices must have degree $\ge m$. Otherwise we could find a common neighbor of a set formed by a vertex and all its neighbors. This common friend would have to be a neighbor of the vertex, and also of itself, but this ontradicts the fact that we do not consider a vertex to be a neighbor of itself.

Suppose we’ve shown this for all values smaller than $m$ (for $m\ge 3$). Let $x_1$ be one of the vertices with maximal degree. We apply the induction hypothesis for $m-1$ to the graph induced on the set of neighbors of $x_1$. We will then find neighbors $x_2,\ldots,x_{m-1}$ of $x_1$, each of which is connected to all other neighbors of $x_1$. Because of the maximality assumption, the neighbors of $x_i$ are the neighbors of $x_1$ for all $i=\overline{1,m-1}$.

Now let $x$ be a neighbor of $x_i$ different from all the $x_i$. Since $y,x_1,\ldots,x_{m-1}$ have exactly one common neighbor, $x$ is connected to exactly one other neighbor of $x_i$, so the graph $G$ induced on the set formed by $x_i$ and their neighbors has the form described two paragraphs above. Suppose now we can find a vertex $y$ not in $G$. Notice that there must be at least one neighbor $x$ of $x_i$ different from all $x_i$; otherwise $y,x_i$ can have no common friend. Take such a neighbor $x$. The vertices $x_1,\ldots,x_{m-2},x,y$ have exactly one common friend, and this friend can only be the unique neighbor of $x_i$ linked to $x$ (call it $\bar x$). Now repeat the procedure with $\bar x$ instead of $x$. We find that $y$ is linked to $x$ as well. But we now see that $y,x_i$ have at least two common friends ($x$ and $\bar x$), contradiction. This means that $x_i$ and their neighbors are all the vertices in the graph, and we have what we wanted.[/quote]

How do you use induction hypothesis if $x_1,x_2,\ldots,x_{m-1}$ have two common neighbours ?

Another solution:
Consider a graph $G$ where vertices represent the people in the wagon and edges their friendship relations. Let $x$ be the vertex with the largest neighborhood. Note $|\Gamma(x)|=n$. Obviously, $n \geq m$ (otherwise, there will be less than $m$ people in $G$ and the hypothesis is trivially satisfied) and we need to determine the value of $n$. Pick $m-1$ people from $\Gamma(x)$ and $x$. This $m$-tuple has exactly one common friend in $\Gamma(x)$ (since $x$ has neighbors only in $\Gamma(x)$). If a vertex $v \in \Gamma(x)$ has at least $m$ neighbors in $\Gamma(x)$, then these $m$ neighbors have two common friends in $G$, namely $x$ and $v$, contradiction. Hence, $v$ has at most $m-1$ neighbors in $\Gamma(x)$, which means that it can be the common friend of at most 1 $m$-tuple consisting of $m-1$ vertices in $\Gamma(x)$ and $x$. We have $\binom {n}{m-1}$ $(m-1)$-tuples in $\Gamma(x)$, hence:
$$\binom {n}{m-1} \leq n \Longleftrightarrow$$
$$\frac{n \cdot (n-1) \cdot \dots (n-m+2)}{(m-1)!} \leq n \Longleftrightarrow$$
$$ (n-1) \cdot \dots (n-m+2) \leq (m-1)!$$
If $n \geq m+1$, then:
$$ (m-1)! \geq (n-1) \cdot \dots (n-m+2) \geq (m+1-1) \cdot \dots (m+1-m+2)=\frac{m!}{2} \Longleftrightarrow$$
$$ m \leq 2$$
which is rejected by hypothesis. Therefore $n \leq m$ and since $n \geq m$, we conclude $n=m$.


inequalities
function
combinatorial geometry
inequalities proposed

Prove the following inequality:
$e^{\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}}{\leq}\frac{e^{x_{1}}+e^{x_{2}}+\cdots+e^{x_{n}}}{n}{\leq}\frac{e^{\alpha}}{1+\alpha}{\cdot}e^{\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}}$,
where $x_{i}\in[0,1](i=1,2,\ldots,n)$ and $\alpha=\frac{1}{e-1}$.
P.S.
Of course the first part of the inequality is very easy(it is just written to make the inequality symmetric and good-looking :))  and can be justified by numerous famous inequalities.But what about the second part I haven't seen it in literature yet,but I won't be surprised if you tell that you have seen it.Anyway I will be glad to see new ideas and solutions.  :)

It follows from one converse of Jensen:
\[\displaystyle e^{\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}}{\leq}\frac{e^{x_{1}}+e^{x_{2}}+\cdots+e^{x_{n}}}{n}
\\
\le(e-1)\cdot\frac{x_1+\ldots+x_n}{n}+1
{\leq}\frac{e^{\alpha}}{1+\alpha}{\cdot}e^{\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}}.\]
I mean if f is convex in $[a,b]$ then for $x_i\in[a,b]$:
\[f\left(\frac{x_1+\ldots+x_n}{n}\right)
\le\frac{f(x_1)+\ldots+f(x_n)}{n}
\\
\le\frac{f(a)-f(b)}{a-b}\cdot\left(\frac{x_1+\ldots+x_n}{n}-a\right)+f(a).\]

[quote="siuhochung = Soarer"][quote="dmitin"]
I mean if f is convex in $[a,b]$ then for $x_i\in[a,b]$:
\[f\left(\frac{x_1+\ldots+x_n}{n}\right)
\le\frac{f(x_1)+\ldots+f(x_n)}{n}
\\
\le\frac{f(a)-f(b)}{a-b}\cdot\left(\frac{x_1+\ldots+x_n}{n}-a\right)+f(a).\][/quote]

How to prove it??[/quote]

This is very easy:

Since the number $x_i$ lies in the interval [a; b], there exists a nonnegative $t_i$, with $0\leq t_i\leq 1$, such that $x_i=t_ib+\left(1-t_i\right)a$. Since the function f is convex in the interval [a; b], we have

$f\left(x_i\right)=f\left(t_ib+\left(1-t_i\right)a\right)\leq t_if\left(b\right)+\left(1-t_i\right)f\left(a\right)=t_if\left(b\right)+f\left(a\right)-t_if\left(a\right)$
$=f\left(a\right)-t_i\left(f\left(a\right)-f\left(b\right)\right)$.

Thus,

$\frac{1}{n}\sum_{i=1}^{n}f\left(x_i\right)\leq\frac{1}{n}\sum_{i=1}^{n}\left(f\left(a\right)-t_i\left(f\left(a\right)-f\left(b\right)\right)\right)=\frac{1}{n}\left(\sum_{i=1}^{n}f\left(a\right)-\sum_{i=1}^{n}t_i\left(f\left(a\right)-f\left(b\right)\right)\right)$
$=\frac{1}{n}\left(nf\left(a\right)-\left(f\left(a\right)-f\left(b\right)\right)\sum_{i=1}^{n}t_i\right)=f\left(a\right)-\frac{1}{n}\left(f\left(a\right)-f\left(b\right)\right)\sum_{i=1}^{n}t_i$.

On the other hand, since $x_i=t_ib+\left(1-t_i\right)a=t_ib+a-t_ia=a-t_i\left(a-b\right)$, we have

$\sum_{i=1}^{n}x_i=\sum_{i=1}^{n}\left(a-t_i\left(a-b\right)\right)=\sum_{i=1}^{n}a-\sum_{i=1}^n\left(t_i\left(a-b\right)\right)=na-\left(a-b\right)\sum_{i=1}^{n}t_i$,

so that

$\left(a-b\right)\sum_{i=1}^{n}t_i=na-\sum_{i=1}^{n}x_i$,

and

$\sum_{i=1}^{n}t_i=\frac{1}{a-b}\left(na-\sum_{i=1}^{n}x_i\right)$.

Thus,

$\frac{1}{n}\sum_{i=1}^{n}f\left(x_i\right)\leq f\left(a\right)-\frac{1}{n}\left(f\left(a\right)-f\left(b\right)\right)\sum_{i=1}^{n}t_i = f\left(a\right)-\frac{1}{n}\left(f\left(a\right)-f\left(b\right)\right)\cdot\frac{1}{a-b}\left(na-\sum_{i=1}^{n}x_i\right)$
$=f\left(a\right)-\frac{f\left(a\right)-f\left(b\right)}{a-b}\cdot\frac{1}{n}\left(na-\sum_{i=1}^{n}x_i\right)=f\left(a\right)+\frac{f\left(a\right)-f\left(b\right)}{a-b}\cdot\frac{1}{n}\left(\sum_{i=1}^{n}x_i-na\right)$
$=f\left(a\right)+\frac{f\left(a\right)-f\left(b\right)}{a-b}\cdot\left(\frac{1}{n}\sum_{i=1}^{n}x_i-a\right)$,

qed..

  Darij

Now the generalization of [b]Kantorovich Inequality[/b].
Given the function $g$ defined on the segment $[a,b]$.Let it be positive,convex and differentiable on $[a,b]$.Then for every $a=x_{1}<x_{2}<\ldots<x_{n}=b$:
$\frac{\lambda_{1}g(x_1)+\lambda_{1}g(x_2)+\cdots+\lambda_{n}g(x_n)}{g(\lambda_{1}x_{1}+\lambda_{2}x_{2}+\cdots+\lambda_{n}x_{n})}{\leq}\frac{L(c)}{g(c)}$,where
,$\lambda_{i}{\geq}0$ with $\sum_{i=0}^{n}{\lambda_{i}}=1$ and $c$ is a point from $[a,b]$ s.t.
[i]a)[/i]$g'(c)=\frac{g(c)}{c-\frac{ag(b)-bg(a)}{g(b)-g(a)}}$,when $g(a){\neq}g(b)$.
[i]b)[/i]$g'(c)=0$,when $g(a)=g(b)$.
$L(x)$ is the equation of the line passing through points $(a,g(a))$ and $(b,g(b))$.The [b]Kantorovich Inequality[/b] is just a simple corollary from the inequality above when $g(x)=\frac{1}{x}$.
For concave $g$ the sign of inequality is inversed.
P.S.
I got this when I was trying to prove the [b]Kantorovich Inequality[/b] posted in the section Computations and Tutorial.The original inequality posted by me is the case $g(x)=e^{x},[a,b]=[0,1]$.
How do you like it guys? ;)

[quote="siuhochung"][quote="dmitin"]
I mean if f is convex in $[a,b]$ then for $x_i\in[a,b]$:
\[f\left(\frac{x_1+\ldots+x_n}{n}\right)
\le\frac{f(x_1)+\ldots+f(x_n)}{n}
\\
\le\frac{f(a)-f(b)}{a-b}\cdot\left(\frac{x_1+\ldots+x_n}{n}-a\right)+f(a).\][/quote]

How to prove it??[/quote]

Well I used to think about this inequality from geometric point of view. You may plot the graph of convex function and take convex hull of this graph arc. Vertical section of this set at $x=\frac{x_1+\ldots+x_n}{n}$ yields this inequality.


Let $ ABC$ be an $ A$ - isosceles triangle with $ A=40^{\circ}$ . Denote the points

$ \{\begin{array}{ccc}
M\in (AB) & , & MA=MC\\\\
N\in (MC) & , & AN\perp BC\\\\
P\in (AN) & , & PN=NC\end {array}$ . Prove that $ PM=PA$ .

[quote="Virgil Nicula"]Let $ ABC$ be an $ A$ - isosceles triangle with $ A = 40^{\circ}$ . Denote the points

$ \{\begin{array}{ccc} M\in (AB) & , & MA = MC \\
 \\
N\in (MC) & , & AN\perp BC \\
 \\
P\in (AN) & , & PN = NC\end {array}$ . Prove that $ PM = PA$ .[/quote]
[hide="Solution"]
Notice that $ P$ is a point on the interior of $ \triangle ACM$ such that $ \angle PAM = 20$ and $ \angle PAC = 20$.  Since $ AM = CM$, we have that $ \angle ACM = \angle CAM = 40$.  Since $ AB = BC$, we have that $ \angle ABC = \angle ACB = \frac {180 - \angle BAC}{2} = 70$.  Thus, $ \angle MCB = \angle ACB - \angle ACM = 70 - 40 = 30$.  Also, $ \angle ANC = 90 + \angle NCB = 120$.  Since $ PN = NC$, we have that $ \angle PCN = \frac {180 - \angle PNC}{2} = 30$.  Thus, $ \angle PCA = \angle ACM - \angle PCN = 40 - 30 = 10$.  Therefore, by http://www.artofproblemsolving.com/Forum/viewtopic.php?t=188349, we have that $ AP = PM$.  [/hide]

[hide="Alternate Solution"]
Let the midpoint of $ BC$ be $ Q$.  Thus, we can see that $ A$, $ P$, $ N$, and $ Q$ are collinear.  Since $ AM = CM$, we have that $ \angle ACM = \angle CAM = 40$.  Since $ AB = BC$, we have that $ \angle ABC = \angle ACB = \frac {180 - \angle BAC}{2} = 70$.  Thus, $ \angle MCB = \angle ACB - \angle ACM = 70 - 40 = 30$.  Now, notice that $ \angle PNC = \angle NQC + \angle NCQ = 90 + 30 = 120$.  Since $ PN = NC$, we have that $ \angle PCN = \frac {180 - \angle PNC}{2} = 30$.  Therefore, $ \angle PCB = \angle PCN + \angle PCB = 30 + 30 = 60$.  Also, $ P$ lies on $ AQ$, which is the perpindicular bisector of $ BC$, so $ PC = PB\implies \angle PCB = \angle PBC\implies \angle PBC = 60$.  Thus, $ \triangle PBC$ is equilateral, which shows that since $ \angle MCP = \angle BCM$, we have that $ CM$ is the perpindicular bisector of $ PB$.  This means that $ MP = MB\implies \angle MBP = \angle MPB$.  Notice that $ \angle MBP = \angle ABC - \angle PBC = 70 - 60 = 10\implies \angle MBP = \angle MPB = 10$.  Hence, $ \angle AMP = \angle MPB + \angle MBP = 10 + 10 = 20 = \angle MAP\implies AP = MP$. [/hide]


AMC
AIME
geometry
trigonometry
AMC 12
AMC 10
MATHCOUNTS

I've read in other posts on this forum of underclassmen debating whether it would be better to take the AMC 10 or 12. The general consensus seemed that it would be better to take the 12 since that would give you a better chance of qualifying for the AIME; though people said that there were of course exceptions in which case it would be better to take the 10. What are those exceptions? Is it just students looking for a perfect score or becoming school winner (knowing that they can't win that distinction taking the 12) that are recommended to take the 10, or are there any other cases?

Well, if you're really good at geometry, or if you don't know 11th and 12th grade math that well, I think it'd be better to take the AMC 10.

[quote]I belive trig also shows up on the AMC12 but not the AMC10 (or less anyway). So if you're not really familiar with that you'd probably be better off sticking with the AMC10.[/quote]
As I understand it, the AMC10 was introduced in an attempt to broaden the appeal of the AMC exams and stem erosion in the number of participants. The AHSME was intimidating. There were probably students who took the AHSME in the 9th or 10th grade and found the amount of material that they'd never seen (and their results) to be discouraging. The AMC organizers didn't want to drive such students away, and didn't want teachers of 9th and 10th grade students discouraging participation because it was "too hard."

So the intended audience for the AMC10 was really this: younger students without particularly extensive or advanced-for-high-school coursework (trigonometry being an example of a relatively advanced topic), seeking a slightly gentler competitive opportunity, and merely seeking to do well by their own standards, with no particular concern about qualifying for additional rounds.

It's fairly clear to me that serious competitors - students with real problem-solving experience, with realistic ambitions for advancing at least to the AIME - should take the AMC12. Even lacking the coursework, those who did well in previous competitions ([i]e.g.[/i], Mathcounts) are probably still better off with the more difficult exam.

At the time this was instituted, I wrote a letter to someone (can't remember who - Harold Reiter, perhaps?) expressing the concern that there might be schools in which rigid rules would be set up that only allowed 9th and 10th graders to take the AMC10, even if they were the aforementioned serious competitors. My concerns may have been overblown.

Thanks for all the advice. I think I will take the 12, since last year I only barely qualifed for the AIME through the 10 (was like 125 or something), and I don't want all of the time I spent preparing for the AIME to be for naught if I don't even qualify to take it again this year.

Wait so then wouldn't you take the 10?

And why don't you just take both?

Yeah I heard at Phillips Exeter 9th and 10th graders are not allowed to take AMC12, which is really strange.

My school doesn't offer both days, which is why I must choose between the two. I thought that my barely qualifying score would mean that I'd probably be better off taking the 12 since I have spent some time preparing for AMCs/AIME - that's the impression I got from reading the other responses.

[quote="dasherm"]That's an excellent point.  I'm probably taking both, but no matter what, I'm taking the 10.  Why?  Because I can!

BTW, I would say that making it to the AIME is easier from the 10, because the easier problems overcompensate for the higher cutoff.[/quote]

Hmm that depends, if you make a lot of careless errors in practices, then DON'T take AMC10. I have never taken AMC10 in my life and I'm not sure if I should take it(at another school) this year since this is my last chance.

But AMC's is only the first step, so yeah I don't care it that much (especially I don't need to qualify for USAMO through index).

Find practice AHSME/AMC12 problems on these forums and see how you do on them. They range all the way from the first 5, normally posted in places like the MathCounts forum or occasionally Getting Started, to the mid-range problems normally posted in Getting Started and sometimes Intermediate, and the hardest, posted in Intermediate and this forum. If you think that there's any noticeable chance that you might not qualify through the AMC12, take the 10 instead, unless you're also completely sure that you won't qualify through that, in which case you might want to take the 12 just for the practice.

My thought on this is that, its not much more difficult to qualify for the AIME through the USAMO by going through the AMC 12.  Although I didn't personally take this route, I regretted it, because I originally didn't make the cutoff for the AMC 10 the first time.  Plus it would prepare you for later years in preparation.

There's advantages to taking the 10. It's easier in general, no really advanced topics and usually 1 or 2 geometry problems that are 20-25 that you can completely bs with trig. But it's easier to make mistakes, so check over answers a ton.

[quote="Pakman2012"]My thought on this is that, its not much more difficult to qualify for the AIME through the USAMO by going through the AMC 12.  [/quote]

I would have to completely agree with this. In fact, the 12 and the 10 are equivalent here, both giving you a probability of 0 (not 1.01) since the USAMO does not qualify you for the AIME.

I think that in general, assuming that you're not extremely unbalanced in your ability at different topics, the AMC 12 is easier to qualify with. Why? Because you only need to answer about 15 questions, most of which are actually identical to the AMC 10. That means if you could have gotten 120 on the AMC 10, you can get 100 on the AMC 12, and you don't have to bother doing those pesky easy problems that might cost you due to stupid mistakes. There isn't usually any trig in problems 1-15 anyway, so if you're worried about qualifying for AIME, then the level of your math education does not play significant role here. If you're not worried about qualifying for AIME, then AMC 12 is better anyways as it can only help you qualify for USAMO.

Wow, I have no idea what I was trying to say there  :blush:

Go with the AMC 12 in my opinion.  You're "free" to attempt more problems and give answers.

Try a couple of each.  I personally think I have a 90% to qualify from the AMC 12, and only like a 60% on the AMC 10.

If you can consistently do 15 questions common to both, you're better off taking the 12.

[quote="kyyuanmathcount"]Its really hard to decide.  Ive taken two practice AMC 10's, one had a load of algebra, the other had a bunch of geometry.  The testmakers really vary the material a lot.  I personally feel that I would be better taking the AMC 10, because there's ample time to check your answers for stupid mistakes and i "qualified" on both of the practices.[/quote]

My guess is that the questions are varied to encourage studying all areas and not leaving weak spots for possible future mathematicians. Also, an idea: try taking a practice AMC 12, and see if you qualify through that.

I'd take the AMC 12. Contrary to how it may seem, it's actually much easier to qualify because you only need 100 points, which could just be 11 right and 14 skipped, and there are at least 11 AMC 10-level questions on the AMC 12, so yeah. Also it's hard to be "comfortably" above 120 pts on AMC 10 because even 20 right and the rest skipped would be only 12.5 pts above, and you never know, you could get some wrong.

My best suggestion:

Take what you're confident to get you AIME and hardcore for AIME. If you can score decently on AIME, it won't matter much which one (AMC-10 or AMC-12) you take.

[quote="JesusFreak197"]If you have the option to take the 10, it won't matter at all what you scored, because it means you're in 10th or younger and therefore qualify by AIME score only, right? Or does index actually still have effect?[/quote]

Theoretically, I think there is that extremely slim possibility that you set the floor score yourself, but yeah pretty much the floor score is how you're going to qualify if you're in xth grade, x $\le$ 10.

man I really want to take AMC10 should I???

Oh What if you misbubble your grade by accident (hint hint :P )?

the amc 10 is easier to get a good score unless you know how to solve all the amc 10 questions that will be on the amc 12.  If you try the amc 12, be sure you can answer 11 correctly.

[quote="Zellex"]I have another question....would a perfect on the AMC 10 be something better than a 120+ or a 130+ on the AMC 12?[/quote]

Depends on what you mean by better... Famewise, a perfect AMC 10 would be a lot rarer, so yeah. Indexwise, I'm under the assumption that you're an underclassman, so it wouldn't matter at all because you have the benefit of the floor (except maybe USAMO indices, but that's mostly USAMO-weighted anyway).

My school didnt offer the A test, so i only get to take one :( And i dont know which to take (though im leaning towards the 12)... A friend of mine that did take the 12A commented that it was unusual in that normally there are maybe 8-9 "easy " questions, 8-9 moderate questions, and 7-9 hard ones (not those exact numbers, but the difficulty of questions is distributed equally), but this one had maybe 15 doable or easy questions and the rest were impossible (this coming from a 10th grader). I got a copy of the test afterwards and i tried the first 15 problems.. i got 13 but i easily couldve gotten 15 if i spent enough time. My concern is that the 12B may be more like the usual ones, and i dont know if im able to do most of the moderate problems and on...

so should i take the 10 or the 12?? i barely qualified for aime through the ten last year (126/5), but ive deproved a bit, so any thoughts?

nvm i took the 12a from last year.. just looked at this years though and i know i can do the first 16, so does it seem like 12B is the right choice?

If you're a sophomore, take the 12. It'll give you more experience, since you can only take the 12 in 11th and 12th. I'm doing the same thing.

Also, there are definitely more than 15 easy/doable questions, and none of the questions were "impossible." I know what your friend meant ;), but, while I may be better at math than he is, I could quite easily do most of the problems from the 12A, and the ones that weren't easy were just a little more difficult to approach. Once you find the right method, they're all straightforward.

i think the AMC 10 might be only good for getting a 150....

for example, this year's AMC had 13 repeat problems between the 12 and 10.
if you got all 13 of those, that's a solid AIME qualifier

k thx for the replies.. i signed up for 12 :D


What makes one a 'Mathematician?' Is it:

      1)To have a 'college degree."(With Honor,perhaps.)
      2)To be a 'full-Pleadge' School/Colege Faculty Professor?(But don't have any 'New Idea' to contribute.)?
      3)To be a "COACHED" of school/colleges,or universities mathematics department with gusto?
      4)Spending more time in editing,revising,criticizing submitted
mathematics works for publication in reputable journals??
      5)To be a "Mathematics Department Chairman,"yet without
any 'contributive ideas?'(Mathematics,of course.)
      6) To count the 'Medals' in mathematical competition(Ex.'Putnam,'etc.),and its recognition?

                  I'm no 'mathematical novice.'I'm just beginning
mathematics enthusiast,with little knowledge of 'practical mathematics.

                                                Respectfully,
                                              A.Eddington.

[quote="Xevarion"]I don't think titles are important. A mathematician is someone who studies mathematics. :D[/quote]
Nice :D :D

I think a mathematician is someone who [i]does[/i] mathematics.  By "doing" mathematics, I mean creating mathematics which is new to the creator.  Saying that one just needs to study mathematics renders "mathematician" a meaningless term, as college students across the country study calculus... I would even say that there are lots of people who study mathematics seriously but passively, and I don't count them as mathematicians.  (I'm thinking in particular of many beginning mathematics majors, who will be mathematicians in a year or so but aren't yet.)

You're being nitpicky about semantics to mess up my nice little aphorism. :P By studying I didn't mean "preparing for a test" kind of studying.  :wink:

Actually, this was a question that came up on Wikipedia not to long ago.  What makes someone a [i]notable[/i] mathematician?

Anyone who loves math and devotes his whole life to it.

ET Bell (a math historian) numbered mathematicians up to his time in the thousands, but how many are [i]notable[/i] today? they all devoted their lives to math.

we should clarify what is meant by notable. most of these thousands were probably outstanding and notable mathematicians in their time to their peers.[/b]

Mathematician is a man who makes many jokes to his friends and understands their jokes too.

that's a good one  :lol: . no sarcasm intended  :D

[quote="Xevarion"]I don't think titles are important. A mathematician is someone who studies mathematics. :D[/quote]

correct  :lol:

[quote="Xevarion"]I don't think titles are important. A mathematician is someone who studies mathematics. :D[/quote]

Yes,

You can see for example the biography of an unknown mathematician, but hes the one I prefer : [size=150][color=red]RAMANUJAN[/color][/size]

http://en.wikipedia.org/wiki/Srinivasa_Ramanujan

He started [size=150]to discover theorems from the age of 13[/size]!! :lol:
And he Left schooling very soon beacause he was [size=150]studying just Mathematics[/size]  :lol:
He was veery[size=150] poor[/size] and was often near the point of [size=150]starvation[/size]!! :D

Yes, quite inspiring, not unlike the story of Michael Faraday, who discovered important laws regarding electromagnetic theory.

A mathemathician is a mathemathician for him/herself when they are happy at doing math, curious to discover and a young soul, like a child, but with the knowledge that they dont know really a thing, but trying to discover what is the truth...like a philosopher making philosophy in things that might not even exist in reality....

A mathemathichian for the world is one with the carachteristics you mentioned above, but this isn't really that interesting, because these are only credits and titles, fame, and ego. But a mathemathician is not that, not olny that. Something deeper that no one can understand, in our different world, stated but not existing in real world, but in real mind, and in real thought.


Princeton
college
MIT
Putnam
number theory

I am trying to get myself into the University of California branches. But I wanted to know whether pure mathematics courses exist in the UCs. Because, I want to choose mathematics as a major in college. But the site of UC Irvine doesn't say anything about there existing the option of choosing pure mathematics as a major.
I would be thanful for any reply to my post.

From my knowledge, UC Berkeley's math department is one of the best in the world. It regularly ranks with the top private colleges (harv, princeton, mit, stanf) for the top 5 in the US. It definitely has a pure mathematics section. My dad obtained his degree in Number Theory there. Berkeley also has an applied math and statistics department. I'm sure that most bigger UCs (LA) would have pure math also.

Yes, they do. You can be a math major practically anywhere.

In particular, here's the [url=http://www.math.uci.edu/undergraduate/major.html]UCI program[/url].

Incidentally, I've got a math degree from Berkeley- that's definitely the best math program in the system.

Thanks for your responses. I want to apply to Berkeley, San Diego
and Irvine branches of the University of California. Does someone with a bad GPA and a score of 2250 on the SAT( Math800, Reading
760, Writing 690) stand a chance of being accepted by these universities? One more thing. If one is born in California is he treated as an in-state student after the first year at the University?

Number of times a student placed in the top 500 on the Putnam in the last four years:

UC Irvine: zero
California State University, Long Beach: 3

Come where you can really get to know the faculty!

I think I will apply to CSULB too. But I need a lot of financial aid. Without it, I cannot go to any college. However, I won't mind if I can have federal loans. Are all US citizens eligible for federal loans?

What about CSULB? Does it provide enough financial aid? I hope I will be treated as a domestic student.

Check this out:
http://www.csulb.edu/divisions/students2/presidents_scholars/
It's really nice. I would recommend applying there no matter what your GPA is.

The program in shbmsc90's link is open only to graduates of California high schools.


function
real analysis
real analysis unsolved

$K$ a compact, $x_0 \in K$, $C_{x_0}(K, \mathbb{R})$ the set of continuous functions of $K$ towards $\mathbb{R}$ that vanish at $x_0$.
Suppose that $E$ is an algebra of function of $C_{x_0}$ such that:
if $x \neq y$ are in $K$, there exist $f \in C_{x_0}$ such that $f(x) \neq f(y)$ and $f > 0$ on $K - \{x_0\}$

Prove that $E$ is dense in $(C_{x_0}(K, \mathbb{R}), |.|_{\infty})$

$E$, together with the constant functions on $K$, forms an algebra $\widetilde E$ which separates points and which contains a constant function, hence is dense by Stone-Weierstrass. Any good enough approximation in $\widetilde E$ of a function which vanishes at $x_0$ will have to have to involve a small constant function, which we can ignore, thus getting a good approximation in $E$.

Sorry if it sounds unclear, but I think it's pretty obvious what's going on.

P.S.

Why do we need that positivity condition? :?