MATHCOUNTS Euler AMC USA(J)MO USAMO geometry trigonometry

Kevin Lei, Lewis Chen, some other guy from hopkins, me.

Yay I made nats.

So... now what?

[quote="nikeballa96"]which one's james4l? and the other ones some random yury dude.[/quote]

Confusing ca with tx?

[quote="r15s11z55y89w21"]ca=/=tx

However, just adding a letter makes it true:

ca=tax

Okay, I just made another lame "joke"...

Anyway, where's Kevin Lei from? (just wondering)[/quote]

Kevin Lei is Euler_Apprentice, and he's from miller.

So are there any training sessions or something?

james4l=Lewis Chen

5th place :mad:

If any one of you four get cancer, can I take your spot?

Or, if pythag011 wants those extra 3-4 days to study for MOP, he can give up his place to me :D

hm, i wouldn't have made nats there.
i wuold've gotten top seven or so, but not nats.

This is the frickin stupid thing about MATHCOUNTS.

I can have a score that's double the score of say, someone in Guam, yet they win their state and I don't get a thing except a 1st place Countdown trophyand a 4th place trophy, yet I still don't get to go to nats.

Why don't they just take the top 228 scores throughout the nation!?!?!?!?!?!

And, [b]one arithmetic error[/b] cost me nats, as well as first place, as well as <insert something here after we can discuss scores>.

[size=67]I keep making excuses.  Sigh.[/size]

that's a good point

i think it'd be better if every state was balanced between good and bad.
but it wouldn't be good to take top 228 scores because other people don't get a chance

...well, in that case, good job pythag011...and wow, three from the Bay Area?

[quote="Kent Merryfield"]Of course, I wouldn't mind if they split California along the same lines as the splitting of the state championship meet, and had two teams. Note that pythag011 is the first Southern Californian to make the team in the last three or four years.[/quote]

Last two years actually; someone from southern California made it in 2007.

[quote="r15s11z55y89w21"]...and wow, three from the Bay Area?[/quote]
I don't know this for sure, but I'd guess that the last non-Bay Area northern Californian to make the state team was Austin Shapiro in - what would that be - 1997? Or did David Stolp ever make the state team?

hi guys

NC has at least one person who would be guaranteed to make CA nats

because he got a[color=indigo]n AHEM[/color]?

not_trig, no scores. please edit that out.

Rankings - Norcal

(if there is a space I think they are not in AOPS)

1. Eulers_Apprentice
2. james4l
3. Michael2Choi
4. AIME15
5. EcstaticPotter
6. Matthew Chow
7. miller4math
8. Spencer Yee
9. Charles Liu
10. Nikhil Buduma

Ok, here is where things get a bit more unpredictable, seeing as they only called up the top 10. (I used the fact that the countdown was a top-bottom style)

11. Matthew Lin
12. Lexing Tong
13. Kevin Chen
14. Julia Huang
15. Grace Lin
16. some person who beat Kevin Lei

pythag011 would have been around first to third, anyways making the team.

BTW, does anyone know anyone about the coach?

According to Eulers_Apprentice on ftw, I got 1st in ca... but he's not very sure...

Oh darn Jordan only had Grace Lin in the top 16?
And oh man, Calvin from JLS did not place, nor did anyone from Terman :O

[quote="serialk11r"]Err....Lin=林.
As for your other interests...I don't know who she is, other than that she is the little sister of some person who was in my math class a while ago.[/quote]

lol sister of Albert Lin? I know her, family friends. She seriously did not know any math until recently. She's not really hot as least of what I can remember of her but I haven't seen her in years -- if I thought she was hot though, I would be a pedophile then as she's in like 6th/7th grade.

No JLS people? I'm disappointed.

Actually today the 7th grade kid who won Peninsula Chapter told me he placed in top 16, so I'm assuming he's the 16th place person. (from JLS)

[quote="Eulers_Apprentice"]wow... i just discovered this topic.  :o
why are you guys talking about grace lin? i dont even know who she is...
and ur not really a pedophile because u have to be like 50 to be a pedophile. wow this is random.
anyways

i forgot, but either
miller4math got 7th and EcstaticPotter got 6th
or
miller4math got 6th and EcstaticPotter got 5th

i think its the second one. i take that back.
i know its the second one. im not gonna erase what i just typed cause im too lazy  :)
wow this post is getting really long so im just going to stop.[/quote]

no AIME15 was 5th.

Also, I get free plane tickets to bay area, yay.

errr pythag actually
AIME15 got 4th in [b]norcal[/b]
which is what were talking about. not CA in total

So the countdown was not top-bottom entirely? I remember Johnny beat Nikhil in cd.

Oh well - results are out. http://www.mathcounts-ca.org/uoptop25and402009.pdf

[quote="bowser"]good job!!
I didn;t even make states....ruined the history of my school
we were tied with another school and they used some tiebreaker which they won so they went to states :mad:
o well even if i went to states I wouldn't have gotten top 4  :|[/quote]

 :ninja: its okay, i was so stupid i didnt take math counts

Just to clarify, in 2007 someone from SoCal called Daichi something (he's Japanese, and went back to Japan after nats) made the nats team.

also these rankings are totally different from last year, things fluctuate FAST as people graduate middle school.

[quote="bookaholic"]If you had grown up in Guam, I'd imagine you would have had far less opportunity to get interested in and learn about math. There are huge advantages to living in a competitive state (or at least the regions of a competitive state that do the competing). It means a school-wide and/or regional network of people with the same interest and the same or higher ability. I think this is probably the motivation behind state USAMO qualifiers as well.[/quote]
Meh

What sucks the most, then, is living just outside of a competitive region?

Yeah, fo sho that's definitely the worst =(


number theory
relatively prime
number theory unsolved

Prove that n is prime iff $\sigma(n)+\phi(n)=nd(n)$.

$\phi (n)$ is the number of positive integers less than $n$ which are relatively prime with $n$.
$d(n)$ is the number of positive divisors of $n$ (including 1 and $n$).
$\sigma (n)$ is the sum of the positive divisors of $n$.

I guess it must be moved to number theory section.

Clearly, if $n$ is prime then $\sigma (n) = n+1, \phi (n) = n-1$ and $d(n)=2$ so that the equality holds.

Now, assume that $n$ is a composite number.
Note that $\phi (n) \leq n-1.$
Let $1=d_1 < d_2 < \cdots < d_k = n$ be the positive divisors of $n$. Then $k \geq 3.$
Therefore :
$nd(n) - \sigma (n) = kn - (d_1 + \cdots + d_k) = (n-d_1) + \cdots + (n-d_k) \geq (n-1) + (n-d_2) > n-1 > \phi (n).$
Thus $n$ does not satisfy the desired equality, and we are done.

Pierre.


geometry
geometric transformation
reflection
function
analytic geometry

the inverse of the relation $ {(2,\minus{}3),(5,\minus{}4),(8,\minus{}2)}$ is?

(the answer to this is just $ {(\minus{}3,2),(\minus{}4,5),(\minus{}2,8)}$ right? or am i missing something :huh: )

Sorry, I can't understand the problem? What do you mean "inverse" in this case?

that's exactly what i was stuck on.

Verbatim, the problem reads:

"The inverse of the relation $ {(2,\minus{}3),(5,\minus{}4),(8,\minus{}2)}$ is

A. $ {(\minus{}2,3),(\minus{}5,4),(\minus{}8,2)}$

B. $ {(2,3),(5,4),(8,2)}$

C. $ {(\minus{}2,\minus{}3),(\minus{}5,\minus{}4),(\minus{}8,\minus{}2)}$

D. $ {(\minus{}3,2),(\minus{}4,5),(\minus{}2,8)}$

E. None of the above

Oh, maybe the reflection of a point across the line $ y\equal{}x$, $ inverse$ comes from the inverse fuction?

i mean, i guess, but this is supposed to be a $ very$ easy test, so thinking is not supposed to be involved in the process....

so would B be right?

an inverse of a function $ f$ is the function $ g$ such that $ g(f(x))\equal{}x$.for instance, the inverse of
$ f(x)\equal{}2x\plus{}3$ would be $ g(x)\equal{}\frac{x\minus{}3}{2}$

so, to put it all in a nutshell...

the inverse of given coordinates is  just the x and y's flipped?(as in: switch the x's and the y's)


question is-- find last two digits of $ 7^{7^{1000}}$
[hide="approach 1"]$ \phi 100 \equal{} 40$. By euler's theorem, $ a^{\phi(n)}\equiv1\pmod{n}$. So, here if we divide the exponent $ 7^{1000}$ by 40, we get an equivalent power less than "n". Again, we hav to solve $ 7^{1000}$ in mod(40). taking $ \phi$ again , $ \phi 40 \equal{} 16$ .dividing 1000 by 16 , remainder=8. We are left with $ 7^8$ .But, $ 7^2\equiv1\pmod{16}$. so, $ 7^8$ is 1(mod 16). Expression left is$ 7^1 \pmod{100}$. Hence, [b]answer is 07.[/b][/hide]
[hide="approach 2"]
first take $ 7^7$, break into power of 4 and 3. bt, $ 7^4 \equiv 1 \pmod{100}$.so,we are left with $ 7^{3^{1000}} \equal{} 7^{9^{500}}\equiv7^{500}\equiv7^{20}\equiv1\pmod{100}$ so, [b]answer is 01[/b][/hide]
[hide="approach 3"]break mod inot 4 and 25. Afetr calc. we obtain $ x\equiv1\pmod{4} and x\equiv1\pmod{25}$, where x is given expression. solving here, i got by chinese remainder [b]01 as answer.[/b] [/hide]
where is the error?

In approach 1 you applied $ \bmod 16$ to the wrong level of exponentiation.  Write your work done more clearly.

Approach 2 is completely invalid.  First of all, $ 7^{7^{1000}} \neq (7^7)^{1000}$, and second of all it is not at all valid to "break into powers of $ 4$ and $ 3$" except by the binomial theorem.

You haven't shown any work in approach 3, but $ 7^{7^{1000}} \equiv \minus{}1 \bmod 4$, so something is wrong.

The fastest check here is [hide="fixing approach 1"] $ \lambda(100) \equal{} 20$ where $ \lambda(n)$ is the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url] and $ \lambda(20) \equal{} 4$, so $ 7^{1000} \equiv 1 \bmod 20$ hence $ 7^{7^{1000}} \equiv \boxed{07} \bmod 100$.  [/hide]

notice that the last 2 digits in the powers of 7 repeat in groups of 4
(01,07,49,43...)
and since $ 7^2 \equiv 1 \pmod{4}, (7^2)^{500} \equiv 1 \pmod 4$
so since the exponent $ 7^{{1000}}$ is 1 mod 4, $ 7^{7^{1000}} \equiv 7 \pmod{100}$
$ \boxed{07}$ is the answer

In approach 1, wut i tried to do was that in $ a^p\pmod{n}$, we can split 'p' into $ (k\cdot%Error. "phin" is a bad command.
+r)$. And as $ a^{\phi(n)}\equiv{1}\pmod{n}$, r is the resultant exponent power which is less than n and also equivalent to the initial expression in $ Z_n$. Hence, to obtain this we have to divide the exponent by $ \phi(n)$, so as to obtain 'r'. This means same as finding equivalent power in $ Z_{\phi(n)}$. So, i solved this and got 1 as resultant exponent.
I think nothing shud be wrong with this approach.

[quote="analytic"]we hav to solve $ 7^{1000}$ in mod(40). taking $ \phi$ again , $ \phi 40 \equal{} 16$ .dividing 1000 by 16 , remainder=8. We are left with $ 7^8$ .But, $ 7^2\equiv1\pmod{16}$. so, $ 7^8$ is 1(mod 16). [/quote]
This step is incorrect.  You are evaluating $ 7^8 \bmod 40$, not $ \bmod 16$.


Dr. Alan visits 9 patients on his ward on Monday. On Tuesday, and each day thereafter, one patient is discharged and two more are admitted to the ward, except on Saturday and Sunday, when no one is discharged. How many patients are on the ward at the end of the following Tuesday?

[quote="aleph0"]9+8+2=19.The message is too small. Please make the message longer before submitting.

[color=blue]Please read the sticky on writing solutions well. -Isabella2296[/color][/quote]
...

Well, the solution is, add one patient to the group every day, except Saturday and Sunday, when there are two patients added. Therefore, we have two weekend days and six weekday days... (sounded weird). Therefore, $2\cdot2+6\cdot1=4+6=10$, but remember, this isn't the answer. This is the number of patients increased from the first Monday to the next (week) Tuesday. Since there were nine patients, we add: $9+10=\boxed{19}$.


geometry
geometric transformation
reflection
homothety
AMC
USA(J)MO
USAMO

Let ABCD be a convex quadrilateral, the diagonals of which intersect orthogonally at O.
Prove that the reflections of O with respect to the sides of the quadrilateral are concyclic.

(I use the code window in order to make the ASCII graphics below.)

[code]
        B       Y     C
        /-------|-----\_
      X/__'-._  | __-'__\_Z
      /   --__'-O---''    \_
     /     __--'/'-_        \_
    /   __-    |    '-_       \_
   /__--      /        '-_      \_
 A/-___      |            '-_     \_
       ----_/__              '-._   \_
            W  ----____          '-_  \_
                       ----____     '-_ \_
                               ----____'-_\_
                                       ----_\D

The reflections of the point O in the sidelines AB, BC, CD, DA are the images of the orthogonal projections X, Y, Z, W of the point O on these sidelines AB, BC, CD, DA under the homothety with center O and factor 2. Hence, it is enough to show that the points X, Y, Z, W are concyclic (since the images of concylic points under a homothety are concyclic points again).

Since < AWO = 90° and < AXO = 90°, the points W and X lie on the circle with diameter AO; hence, < XWO = < XAO, or, equivalently, < XWO = < BAO. Similarly, < ZWO = < CDO, thus

  < XWZ = < XWO + < ZWO = < BAO + < CDO,

and similarly < XYZ = < ABO + < DCO.

Now, in order to establish the concyclicity of the points X, Y, Z, W, it is enough to show < XWZ + < XYZ = 180°. In other words, we have to prove

  (< BAO + < CDO) + (< ABO + < DCO) = 180°.

However, since triangles AOB and COD are right-angled, we have < BAO + < ABO = 90° and < CDO + < DCO = 90°, and thus

  (< BAO + < CDO) + (< ABO + < DCO)
  = (< BAO + < ABO) + (< CDO + < DCO)
  = 90° + 90° = 180°,

qed.[/code]

  darij

Again, this problem was intended for the beginners Darij :)

There is also a solution using inversion :) Is anyone interested?

By the way, here's my solution before I looked at the above solution.

Let cyclic quadrilateral $ABCD$ have projections of $O$ onto $AB$,$BC$,$CD$, and $AD$ as $Y$,$W$,$Z$, and $X$, respectively.
If $\angle XZW+\angle XYW=180^{\circ}$, quadrilateral $XYZW$ is cyclic (if we have a qudrilateral $PQRS$, then $\angle P+\angle R=\frac{arcQRS+arcQPS}{2}=\pi rad=180^{\circ}$ implies that if two opposite angles of a quadrilateral sum up to $180^{\circ}$, then the quadrilateral is cyclic).

Quadrilaterals $OXDZ$,$OZCW$,$OWBY$, and $OXAY$ are cyclic since they all have 2 opposite 90 degree angles.

$\angle ODA=\angle OZX$, $\angle OCD=\angle OZW$, $\angle OYW=\angle OBC$, $\angle OYX=\angle OAD$, $\angle OAD+\angle ODA=90^{\circ}$, and $\angle OCB+\angle OBC=90^{\circ}$.

Hence,

$\\\angle XZW+\angle XYW=\angle OZX+\angle OZW+\angle OYX+\angle OYW\\ =\angle ODA+\angle OCB+\angle OBC+\angle OAD\\ =(\angle ODA+\angle OAD)+(\angle OCB+\angle OBC)\\ =90^{\circ}+90^{\circ}=180^{\circ}$

And the conclusion follows.

Masoud Zargar

Actually, this was on an old USAMO, but it's not hard. What about that inversion solution? I have one--invert twice about O... :D

I suppose the inversion maps the points to a line and turns it into a collinearity problem.

[quote="mathwizarddude"][quote="Differ"]http://www.amazon.com/Geometry-Revisited-Mathematical-Association-Textbooks/dp/0883856190

If you don't have it, you should get it.[/quote]
hmm, is there a site where I can read that book for free (or website with similar articles on inversions), as I don't feel like buying it (too expensive...I'm not rich at all... moreover, even if i have that book, spending time finishing that book would not give me time to do other problems posted on this site and learn other strategies - right now i don't even think i have seen at most 1% of the problems posted on this site)? But the book is pretty cool, though (only gotta read 6 pages... so stingy, lol)
Thanks for any info...  :D[/quote]

Borrow it from your friends! One of them is bound to have it. (everyone from Southern California has a copy (or two))

Oh really? Perhaps not for the place I'm living here (I know CA has a lot of Asians, but not for UT)... i bet less than 1% of the people even *like* math here...

[quote="boxedexe"]Seeing that this thread has been revived, for those of you who are interested in a solution via inversion, look for Inversion in Mathematical Excalibur (google!).[/quote]

I can't be bothered reading the solution in Excalibur, but it is different to this:

As noted previously, by a clear homothecy it suffices to prove $X,Y,Z,W$ is cyclic where $X,Y,Z,W$ are the orthogonal projections from $O$ to the sides $AB,BC,CA,AD$. Invert with centre $O$. Let the line passing through $A,X,B$ be mapped to the circle $\Gamma$. Since $OX$ is the shortest distance from $O$ to the line $AB$, it follows $OX'$ is the longest distance from $O$ to $\Gamma$ again. Then $X'$ is diametrically opposite $O$ in $\Gamma$ and then $\angle X'B'O=\angle X'A'O'=90^{\circ}$. But clearly also $\angle A'OB'=90^{\circ}$ since $A'B'$ lie on the fixed rays $OA,OB$. So $OA'X'B'$ is a rectangle. Repeat for $Y',Z',W'$ to conclude $X'Y'Z'W'$ is a rectangle which implies $XYZW$ is cyclic.


Given $ a,b,c \geq0$.Prove that
$ \frac1{5a^2\plus{}bc}\plus{}\frac1{5b^2\plus{}ca}\plus{}\frac1{5c^2\plus{}ab} \geq \frac3{a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca}$

[quote="dduclam"]Given $ a,b,c \geq0$.Prove that
$ \frac1{5a^2 \plus{} bc} \plus{} \frac1{5b^2 \plus{} ca} \plus{} \frac1{5c^2 \plus{} ab} \geq \frac3{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca}$[/quote]
$ \sum_{cyc}\frac {1}{5a^2 \plus{} bc} \equal{} \sum_{cyc}\frac {b^2c^2}{5a^2b^2c^2 \plus{} b^3c^3}\geq\frac {(ab \plus{} ac \plus{} bc)^2}{15a^2b^2c^2 \plus{} a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3}.$
Id est, it remains to prove that $ (ab \plus{} ac \plus{} bc)^2(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} ac \plus{} bc)\geq3(a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3) \plus{} 45a^2b^2c^2,$
 which obviously true.
I think, the following similar inequality nicer:
For psitive $ a,$ $ b$ and $ c$ prove that
\[ \frac {1}{a^2 \plus{} bc} \plus{} \frac {1}{b^2 \plus{} ac} \plus{} \frac {1}{c^2 \plus{} ab}\geq\frac {9}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} c \plus{} bc}
\]

by Cauchy we have :
$ \frac{1}{a^{2}\plus{}bc} \plus{} \frac{1}{b^{2}\plus{}ac} \plus{} \frac {1}{c^{2}\plus{}ab} \geq  \frac{(1\plus{}1\plus{}1)^2}{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca}$

OK,khashi70!  What about the following Vasc's inequality:
\[ \frac {1}{a^2 \plus{} bc} \plus{} \frac {1}{b^2 \plus{} ac} \plus{} \frac {1}{c^2 \plus{} ab}\geq\frac {3}{ ab \plus{} ac \plus{} bc}
\]
:wink:

[quote="arqady"]\[ \frac {1}{a^2 \plus{} bc} \plus{} \frac {1}{b^2 \plus{} ac} \plus{} \frac {1}{c^2 \plus{} ab}\geq\frac {3}{ ab \plus{} ac \plus{} bc}
\]
:wink:[/quote]

Nice inequality ... Here is what I did ... The inequality is equivalent to

$ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\plus{}abc(a^3\plus{}a^2b\plus{}b^2a\plus{}b^3\plus{}c^2a\plus{}a^2c\plus{}c^3\plus{}b^2c\plus{}c^2a\plus{}3abc)\geq 0$ which is true  :)

The following more general inequality holds
If $ \frac{2}{3}\leq r\leq 3\plus{}\sqrt{7}$  then

$ \frac {1}{ra^2 \plus{} bc} \plus{} \frac {1}{rb^2 \plus{} ac} \plus{} \frac {1}{rc^2 \plus{} ab}\geq\frac {r\plus{}2}{r( ab \plus{} ac \plus{} bc)}$

the proof is similar  :wink:

Given $ a,b,c\geq0$.Prove that
$ \frac a{b^3\plus{}8c^2a}\plus{}\frac b{c^3\plus{}8a^2b}\plus{}\frac c{a^3\plus{}8b^2c}\geq\frac2{a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca}$


calculus
integration
trigonometry
inequalities
calculus computations

For $ f(x)\equal{}1\minus{}\sin x$, let $ g(x)\equal{}\int_0^x (x\minus{}t)f(t)\ dt.$

Show that $ g(x\plus{}y)\plus{}g(x\minus{}y)\geq 2g(x)$ for any real numbers $ x,\ y.$

hi kunny-kun  :D

$ g(x) \equal{} \int_0^x (x \minus{} t) f(t) \, dt$

It's easy to see $ g(z) \equal{} g( \minus{} z)$

making a small substituition $ x \equal{} \frac {a \minus{} b}{2}\land y \equal{} \frac {a \plus{} b}{2}$

we get the inequality $ \frac {g(a) \plus{} g(b)}{2}\geq g\left(\frac {a \minus{} b}{2}\right)$

making $ b \to \minus{} b$ ,  $ \frac {g(a) \plus{} g(b)}{2}\geq g\left(\frac {a \plus{} b}{2}\right)$

I think there is a Famouse Inequality like above, but i cannot remember it.

:( could someone help me?

thanks Carcul-kun

So $ g$ is convex, according to the inequality

so $ g''(z)  \geq 0$

Using Cauchy Integration formula to differentiate this in a easy way

$ g''(x)\equal{}f(x)\equal{}1\minus{}\sin (x)\geq 0$

So $ g$ is convex and Jesen is true and we are donne


it's aspecial cas of hardy inequality just see:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=inequality+hardy&t=29756


Q- Prove that when three consecutive even numbers are squared and the results are added, that the sum always has at least 3 divisors.

Thank you for your help.

[quote="tµtµ"][hide]$2n + (2n+2) + (2n+4) = 2 \cdot 3 \cdot (n+1)$[/hide][/quote]

How does that show when they're squared?

Considering the numbers to be 2n, 2n+2, and 2n+4, squaring and adding them gives us $4(3n^2+2n+5)$, which I think has three factors (2, 4, and $3n^2+2n+5$). Don't forget 1 either  :D

Let the numbers be x-2, x, and x+2.

When the numbers are squared and added, we get (3x^2 + 8), where x mod 2 == 0.

So, we can replace x with 2x, giving us 3(2x)^2 + 8, yielding 12x^2 + 8, which is only divisible by certain numbers. When factored, this gives us 4(3x^2+2), which is obviously divisible by all factors of 4, which are 1, 2, and 4. Thus, all consecutive even numbers when squared and added are divisible automatically by 1, 2, and 4. And finally, for the last divisor, the number is divisible by itself, yielding four divisors total (including 1 and the number itself). This also suggests that when the result is obtained, there have to be [b]at least [/b] four divisors for the number given.[/b]

[quote="Karth"]Let the numbers be x-2, x, and x+2.
[/quote]
Here you make a mistake, because x could be an odd number ;)

The correct notation (used by tµtµ and 236!) would be : 2n, 2n+2, 2n+4

Let the numbers be $2(n-1), 2n, 2(n+1)$ where $n$ is any integer.  Squaring and adding we get $4(n^2 -2n + 1 + n^2 + n^2 +2n + 1) = 2^2 (3n^2 + 2)$.  To find the divisors of this, we do the prime factorization of number, add one to the exponents and multiply.  No matter what the factorization of the second part is, it will always be multiplied by the 3 from the $2^2$, therefore the minimum number of divisors is 3.

@mathmanman:

That's why on the last step, I replaced x with 2x, cuz no matter what x is (when it is an integer), 2x is always even.  :D

Yeah, I saw that :)
But, I was just talking about the notation ;)
I mean, your proof is good, but.. you do not begin very well :P
Like, for example, an integer is usually denoted by "n"


search
inequalities
inequalities unsolved

Where do you use z in your problem, and what exactly are you summing?

It's symmetric sum notation, do a search for the Inequalities thread, and it should become clear what is intended.

no answer for this ,agree or disagree :(

yes, it comes from a books

i mean that it must be $3\sqrt{11}$ except for $4\sqrt{6}$

you mean the right hand side of the inequality should be $3 \sqrt{11}$ instead of $4 \sqrt{6}$?


trigonometry
vector
trig identities
Law of Cosines

If O is the origin, A is the point (4,-2) and B is the point (7,-5), find m<AOB to the nearest tenth of a degree.

I get about 8.9 degrees

Could someone check it for me?

THanks!!

[hide="solution"]We graph the points and use the Law of Cosines to get

AOB≈8.972626615

So your estimation is very close but I would say about 9.[/hide]

Using a vector approach:
[hide]We need to find the angle between $<4,-2>$ and $<7,-5>$. The cosine of the angle between two vectors is equal to the dot product divided by the product of the magnitudes of the vectors; thus calling the angle $x$, $cosx = \frac{38}{\sqrt{20 \times74}}$ and it follows that $x\approx 8.973$.[/hide]


Equipe brasileña:

BRA1: Gustavo Lisbôa Empinotti
BRA2: Marcelo Tadeu de Sá Oliveira Sales
BRA3: Matheus Araújo Marins
BRA4: Matheus Secco Torres da Silva

Digam suas equipes!!

PERUVIAN TEAM:
PER1.- CESAR CUENCA .
PER2.- FERNANDO MANRIQUE.
PER3.- IVAN ALEXANDER.
PER4.- SALAS HUAMAN.

                                            ESTE ES EL PODEROSISIMO!!! EQUIPO PERUANO...

PD:
 Yo ya toy viejo pa esto ..:P..

[color=blue][b]Paraguay:[/b]

1. Ariel Schvartzman.
2. Marcos Martínez.
3. Edgar Ávalos.
4. Edgar Elizeche.

Saludos de parte de un nuevo ex-olímpico.[/color]

Chle

1 Anibal Veloso
2 Benjamin Baeza
3 Daniel Contreras
4 Matias Garrido


number theory
greatest common divisor
Gauss
Euler
number theory unsolved

Probably this is an easy one but so far i havent solved any gcd problem so its hard for me...  :(
$1.$ Prove that if $ab+cd$ is devisible by $a-c$ then $ad+bc$ is also devisible by $a-c$
$2.$ Prove that if $ad-bc$ and $a-b$ are devisible by $k$ and $(a,k)=1$ then $c-d$ is also devisible by $k$

Notice that \[ (ab + cd) - (ad + bc) = (a - c)(b - d) \] so, if $a - c$ divides $ab + cd$, then $a - c$ must divide $ad + bc$ as well.

Do you see why?

And do you see how to do the second part?

[quote="Gohan"]
$2.$ Prove that if $ad-bc$ and $a-b$ are devisible by $k$ and $(a,k)=1$ then $c-d$ is also devisible by $k$[/quote]
$k|(ad-bc)$  and  $k|(a-b)$  then $k|(ac-bc)-(ad-bc)$
$k|a(c-d)$
$k|(c-d)$ (gauss)
:)

[quote="Gohan"]10x Arne and Fermat-Euler...  :D[/quote]
thanx
Arne  :winner_first:
Fermat-Euler  :winner_second:


Consider a row of $2n$ squares colored alternately black and white. A legal move consists of choosing any contiguous set of squares and inverting their colors. What is the miniimum number of moves required to make the entire row be one color?

The answer is pretty obvious, but be careful with your proofs!

so can we pick one square only..like lets say u change the one square from black to white..or do u have to pick more then one square in a row

ya thats true...and i guess that solution is also valid..nicely done

[quote="t0rajir0u"][hide="Well..."] Say we want to turn the squares black.  If we invert an even number of squares, we have done nothing.  If we invert an odd number of squares, we can increase the number of black squares by at most $1$, so it will take at [b]least[/b] $n$ moves.

If we are allowed to flip one square at a time, we can flip the $n$ white squares, so it takes at [b]most[/b] $n$ moves.

Ergo it takes [b]exactly[/b] $n$ moves. QED.

(We must be allowed to flip one square at a time, or else the case $n = 1$ is insoluble.) [/hide][/quote]

The argument is a little trickier, I think. It's not always true that you can only increase the number of black squares by at most $1$. For example, say you invert one black square. Now you have 3 white squares; you can invert all of those, thus increasing the number of black squares by $3$. Obviously, this isn't "efficient," but it makes the proof a little trickier.

Hmm, true.  I hadn't considered that.   :|

[quote="pianoforte"][hide="induction? :maybe: "]I think the minimum is $n$.

Base case.
For $n=1$, the figure looks like $\text{BW}$, and clearly it takes only 1 move to change to all black or all white.

Inductive step.
Assume that it takes only $n$ moves turn $2n$ squares of BW's into one color. For $2(n+1)$ squares, the figure looks like $BW\cdots BW(BW)$. By assumption, it takes $n$ moves to turn it into $BB\cdots BB(BW)$, and $1$ more move to make the entire $2(n+1)$ row into W's.

So it takes a minimum of n moves.(?)[/hide][/quote]

That makes the assumption that if you have $2(n+1)$ squares the optimal solution is to change the first $2n$ squares before the last $2$. Which is nontrivial and not even completely true since there exist many strategies for reaching the optimal solution of $n+1$ moves.

Another inductive "solution": After applying a single move to $2(n+1)$ squares (assuming you don't affect the end squares), combine all contiguous white and black squares. In one move, we have reduced it to the $2n$ case. The problem is that we don't know that the optimal solution doesn't affect only some of the squares we combined (e.g. only the left square in a pair of white squares). This induction seems a little more promising, though.

I am under the impression that there should exist a "slick" and simple solution- the result is, after all, intuitively obvious.

[quote="Ravi B"]How about the extremal principle?

[hide="Hint"]Define the [i]weight[/i] of a single move to be the number of squares flipped by it.  Define the [i]weight[/i] of a sequence of moves to be the sum of the weights of all of its moves.  Among all the sequence of moves that result in a unicolor row, consider those that use the minimum number of moves.  Among all these minimum-move sequences, consider one that has the minimum weight.  In such a "minimum minimum" sequence, show that its moves are disjoint.[/hide][/quote]

[hide="Neat Idea!"]Well, by contradiction, suppose two moves overlap. If instead of using these two moves we invert each of the parts affected by only one of these two moves, we achieve the same result, since the squares affected by both end up unchanged anyway, and have a corresponding weight loss (haha) of $2 \cdot (\text{number of overlapped squares})$, a contradiction to the minimum weight assumption.

Then, we may apply the originally "flawed" arguments to achieve the result.[/hide]


Hi,

Is there an easy way to evaluate $ \sum_{i\equal{}1}^{\infty}\frac{1}{i^2}$?

Sorry if this is too hard/easy for this forum. :maybe:

Thanks,
ThinkFlow

I don't think it can get easier than [url=http://en.wikipedia.org/wiki/Basel_problem]this[/url].

And [url=http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf]this[/url] is also useful.


I have gotten worse at math because I have gotten much more careless since last year. Is there any way to practice avoiding them?

basically, just check over your work. double check your arithmatic as you go along, and later during the test, perhaps at the two-hour mark or so, check your answers again. read the questions several times, especially lengthy ones. make sure you give what they asked for (for example, if you solve for x and y, and the problem asks for x, don't give y)

[quote="davidyko"]Yeah, check.
I rather went overboard with checking this year, and ran out of time 7/8 of the way through doing #11. And I STILL didn't catch my mistake on #2 which plagued half the population.  :D[/quote]

i had a bit of time to briefly check 1 through 5. I did catch my mistake in #2 because from my interpretation there could be two answers.

Be [b]really[/b] careful the first time you do the problem. That ensures that you'll spend enough time on each problem and sometimes even eliminates the need to check back later. I usually like to run through my work right after doing a problem, but this year it seemed like checking was taking a really long time and I needed to make sure I got to all the problems. So instead of the usual procedure, I decided to mark a few answers I wasn't confident about and to check on them again later.

[quote="bpms"]Chess 64's method of using mod 9 to check arithmetic works pretty well[/quote]
what's Chess64's method?

for multiplication, i do standard multiplication and then check with this algorithm that i discovered 4 years ago in pre-algebra class :P
the product of numbers  equals the square of their arithmetic mean subtracted by the square of half their difference.
$xy = (\frac{x+y}{2})^{2}-(\frac{x-y}{2})^{2}$
it's useful for checking products of two integers that have an even sum.
to check a square:
$x^{2}= (x+y)(x-y)+y^{2}$
useful when you're trying to check $48^{2}$
$48^{2}= 46*50+2^{2}$

[quote="fractalz"][quote="bpms"]Chess 64's method of using mod 9 to check arithmetic works pretty well[/quote]
what's Chess64's method?[/quote]
Mod 9

I heard that people should just do work very fast in practice, even if you make more mistakes. Then, when you are used to working so fast, then working just a little bit slowly  will greatly reduce mistakes under time constraint.

Write NEATLY on your scrap paper. Put only one question per sheet of scrap paper. Number each scrap paper with the question number.

Try to do things a different way when you check over your work. If you have a sequence of numbers to add, try adding them in a different order and see if you get the same answer. Plug in your answers to the original problem and make sure they work (this is how I caught the planets question -- when I was moving the planets to the 210 position I realized they would be collinear at 105  too). Underline the question. Put a box around key phrases of the question that aren't italicized, but should be.


geometry
geometric transformation
homothety
geometry open

Construction Trapezoidal know two diagonal measure of length be m and n , $ \widehat{BAI} \equal{} \frac{1}{2}  \widehat{BIA}$ (I be intersection two diagonal)

here's a solution to your nice problem
construct a line $ (L)$ and put on it the lenght $ m\plus{}n \equal{} A'B'$ then draw a circle with radius $ m$ and center $ R$
now draw a lenght egal to $ \sqrt{(m\plus{}n)n}$ it's a well known technic
draw a circle with center $ B'$ and radius $ \sqrt{(m\plus{}n)n}$ so we get $ I$
you can see now that  $ ABI$ is deduced from $ A'B'I$ by homothety


[quote="great math"]It is a double- application of Cauchy-Schwartz! :)[/quote]

i don't think so.( also use cauchy Schwart).Will you show me your proof. I think i's not really easy

Well, if you want to know the answer, here is my solution

From the given condition, we have

$ 2 \ge \sum_{cyc}\frac{a^2\plus{}b^2}{a^2\plus{}b^2\plus{}1}$

Applying Cauchy- Schwartz as follows

$ 2\left(2(a^2\plus{}b^2\plus{}c^2)\plus{}3\right) \ge \left(2(a^2\plus{}b^2\plus{}c^2)\plus{}3\right)\left(\sum_{cyc}\frac{a^2\plus{}b^2}{a^2\plus{}b^2\plus{}1}\right) \ge (\sum_{cyc}\sqrt{a^2\plus{}b^2})^2$

It implies us to the following evaluation

$ 2\left(2(a^2\plus{}b^2\plus{}c^2)\plus{}3\right) \ge (\sum_{cyc}\sqrt{a^2\plus{}b^2})^2$

Amazingly, by long expansion, we get

$ a^2\plus{}b^2\plus{}c^2 \plus{}3 \ge \sum_{cyc}\sqrt{(a^2\plus{}b^2)(a^2\plus{}c^2)} \ge \sum_{cyc} (a^2\plus{}bc)$

In conclusion, we obtain

$ 3 \ge ab\plus{}bc\plus{}ca$
 as desired.

We complete our proof.


Develop a mathematical procedure by which you might judge when one function fits a data set better than another.

??? How would you do that?

Well, think about it.  Given a set of data and some possible functions, what makes some of them closer approximations than others?

they have to go through or almost go through most points in the data. but is the question looking for something more?  :maybe:

[quote="t0rajir0u"]A more concrete example:  Suppose I have the data points

$(0, 0.1), (1, 1.1), (2, 4.1), (3, 9.1), (4, 16.1), (5, 25.1)$

Clearly the function $f(x) = x+0.1$ goes through two of these points, whereas the function $f(x) = x^{2}$ goes through none of them - yet which one is the better fit?[/quote]

The $f(x) = x^{2}$ one because it follows the trend of the data points.

What do you mean by that?

You probably mean that if you use the two rules to generate data points, the first rule produces

$(0, 0.1), (1, 1.1), (2, 2.1), (3, 3.1), (4, 4.1), (5, 5.1)$

Whereas the second rule produces

$(0, 0), (1, 1), (2, 4), (3, 9), (4, 16), (5, 25)$

And clearly that second set of points is much "closer" than that first set of points.  So can you encapsulate this "closeness" mathematically?


This is a problem that I saw on a Japanese TV show and I'm pretty stumped
(I'll use "times" for the multiplication sign):

abc times cba = 2002 times x

Find x if a, b, c, and x are different integers and abc and cba are two 3-digit integers which have reversed digits.

i don't know if this helps, but 2002=2*7*11*13.

then we know either a or c is divisible by 2 without loss of generality let it be a, and let a=2d, 2d-b+c is divisible by 11. actually, 2d-b+c is either 11 or 0.

another thing you could do is say that abc=100a+10b+c, cba=100c+10b+a.

abc x cba=10001ac+1010bc+1010 ab+100a^2+100b^2
+100c^2. all of this we know is equal to 0 mod 2, 7, 11, 13, which give us the equations:

ac=0 mod 2.
5ac+2bc+2ab+2a^2+2b^2+2c^2=0 mod 7
2ac+9bc+9ab+a^2+b^2+c^2=0 mod 11
etc. you get the point. not sure if this is helpful tho...

this becomes quite trivial if you know the divisibility rules.

It's well explained [url=http://www.m-a.org.uk/xplusyfiles/home/issue_11/multiple_tricks/index.htm]here[/url]:

and use the fact that a,b,c are integers and try and set a bound for x.
Then we can also find x as with the remainder rules we can determine the l.c.f. of x


[quote="AIME15"]$ \frac {3 \times 5}{9 \times 11} \times \frac {7 \times 9 \times 11}{3 \times 5 \times 7} \equal{}$
\[ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ \frac {1}{49} \qquad \textbf{(E)}\ 50
\]
[/quote]

[hide]Cancel, cancel, cancel, cancel, and then cancel.

$ \boxed{1}$[/hide]

[hide="Solution"]
Notice that all terms cancel and we are left with $\boxed{\text{(A)} 1}$
[/hide]

...why did you revive a 5.5 year old thread that had already been resolved?

Lol, since the threads were already bumped by [b]@competitivecoder[/b] ( :o ), I decided to give nice additional solutions that were $\LaTeX$ified for future thread viewers :)

[quote=AIME15]$ \frac{3 \times 5}{9 \times 11} \times \frac{7 \times 9 \times 11}{3 \times 5 \times 7}\equal{}$

\[ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 0 \qquad
\textbf{(C)}\ 49 \qquad
\textbf{(D)}\ \frac{1}{49} \qquad
\textbf{(E)}\ 50
\][/quote]

It cancels out

[quote=Posymk25]Love your solution but I don't think we should bump over 10 year old threads  :-D[/quote]

it was only 5 years since the last post before him

Wasn't this the first problem from the first AMC 8 ever?

Link to original problem: [url]https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_1[/url]

The people who did this problem on the real test are probably in their 40's. That's crazy.... :huh:

I am trying to imagine a smart kid tricking everyone in his class to multiply all that out...

[hide]1 of course. Because they all cancel out[/hide]

[quote=pubulous][hide]everything eliminates: 3/3, 5/5, 7/7. 9/9, 11/11. the answer is A. [/hide]
also, why are we bumping threads that are older than some of you?[/quote]

lol. interesting bump tho

The answer is $1$, or a, everything is cancelled

[quote=Quentissential][quote=AIME15]$ \frac{3 \times 5}{9 \times 11} \times \frac{7 \times 9 \times 11}{3 \times 5 \times 7}\equal{}$

\[ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 0 \qquad
\textbf{(C)}\ 49 \qquad
\textbf{(D)}\ \frac{1}{49} \qquad
\textbf{(E)}\ 50
\][/quote]
[quote=Quentissentia[l][hide=:D, sol]A[/hide][/quote]
thats my answer[/quote]

why revive a 11 yr old forum

[quote=motorfinn][quote=mathapple101]Here is a solution with a motivation, to help future viewers  :coolspeak:

We firstly must understand the meaning of $ \times $. This problem is on the [b]AMC 8[/b], so this is meant for [i]8th[/i] graders to solve. Thus, we may hypothesize that $ \times $ means a [i]variable[/i], specifically the variable $X$.

[color=#f00]Lemma 1 : 7 < 8[/color]
[i]Proof:[/i]
Note that if we subtract $7$ from both sides, we get that:
$$ 0 < 1 $$
Now, we use binomial theorem to solve: note that $0^2 + 1^2 = 1^2$, so we have a right triangle with side lengths $0,1,$ and $1$. Since $1$ is the hypothenuse, and $0$ is a leg, and since a leg is always smaller than the hypothenuse, we find that $0 < 1$. $\Box$
[color=#f00]Lemma 2 : The $\times$ is a variable[/color]
[i]Proof:[/i]
We refer to this postulate:
[quote = my 7th grade teacher]
From now on, we use dots for multiplication and the "x" is a variable, not the multiplication sign.
[/quote]
Using Lemma 1, we get that $7<8$, and since this is the AMC 8 (proved earlier), which is meant for 8th graders, and because we must use dots for multiplication, we can conclude that the $x's$ are variables. There is an alternative proof at the end. $\Box$

Now, since the $\times$ are variables, we can make the expression:
$$ \frac{3x5}{9x11}x\frac{7x9x11}{3x5x7}$$
By communative property and algebra, this equal
$$\frac{15x^2}{99x}\frac{693x^2}{105x^2} =   \frac{10395x^4}{10395x^3} = x$$
To find $x$, we guess that it equals $0$, so the answer is $\boxed{A}$.

[hide=alternative proof]
Note that the main confusion is the $x's$. Since my ex is a [i]pig[/i] and is annoying, and since multiplication is also pretty annoying, the $\times$ is multiplication. Therefore, everything cancels out, and we get the answer of $\boxed{1}$.
[/hide][/quote]

wow you're so cool and funny! bumping a 5 year old thread to make a joke?!?! i never would become so funny in my wildest dreams![/quote]

actually, you are being much funny right now ! you so sarcastic it make me laugh ! you must be in wildest dreams !

[quote=PerfectPanther13][quote=franzliszt]Imagine some poor kid actually multiplying all that out :noo:[/quote]

I almost did xD[/quote]

Me too then i realized whats the point you can just cancel out everything.

[quote=mathapple101]Here is a solution with a motivation, to help future viewers  :coolspeak:

We firstly must understand the meaning of $ \times $. This problem is on the [b]AMC 8[/b], so this is meant for [i]8th[/i] graders to solve. Thus, we may hypothesize that $ \times $ means a [i]variable[/i], specifically the variable $X$.

[color=#f00]Lemma 1 : 7 < 8[/color]
[i]Proof:[/i]
Note that if we subtract $7$ from both sides, we get that:
$$ 0 < 1 $$
Now, we use binomial theorem to solve: note that $0^2 + 1^2 = 1^2$, so we have a right triangle with side lengths $0,1,$ and $1$. Since $1$ is the hypothenuse, and $0$ is a leg, and since a leg is always smaller than the hypothenuse, we find that $0 < 1$. $\Box$
[color=#f00]Lemma 2 : The $\times$ is a variable[/color]
[i]Proof:[/i]
We refer to this postulate:
[quote = my 7th grade teacher]
From now on, we use dots for multiplication and the "x" is a variable, not the multiplication sign.
[/quote]
Using Lemma 1, we get that $7<8$, and since this is the AMC 8 (proved earlier), which is meant for 8th graders, and because we must use dots for multiplication, we can conclude that the $x's$ are variables. There is an alternative proof at the end. $\Box$

Now, since the $\times$ are variables, we can make the expression:
$$ \frac{3x5}{9x11}x\frac{7x9x11}{3x5x7}$$
By communative property and algebra, this equal
$$\frac{15x^2}{99x}\frac{693x^2}{105x^2} =   \frac{10395x^4}{10395x^3} = x$$
To find $x$, we guess that it equals $0$, so the answer is $\boxed{A}$.

[hide=alternative proof]
Note that the main confusion is the $x's$. Since my ex is a [i]pig[/i] and is annoying, and since multiplication is also pretty annoying, the $\times$ is multiplication. Therefore, everything cancels out, and we get the answer of $\boxed{1}$.
[/hide][/quote]

btw, 6th graders can also take this...

[quote=eevee_12][quote=royliang]Why is it so overcomplicated?[/quote]

no u :love:[/quote]

what a thing to say to someone thats just asking a question!

[hide=bruh.... :dry: ]its $1$ cause everything cancels out![/hide]

[quote=NInjas25]Ok no one needs the answer anymore. The answer is one. We all now know. :roll:[/quote]

haha yea


simplify \[ \sqrt [3]{2\plus{}\sqrt{5}}\plus{}\sqrt [3]{2\minus{}\sqrt{5}}\].

[quote="DaReaper"]simplify
\[ \sqrt [3]{2 \plus{} \sqrt {5}} \plus{} \sqrt [3]{2 \minus{} \sqrt {5}}
\]
.[/quote]

If, $ \sqrt [3]{\sqrt {5} \plus{} 2} \plus{} \sqrt [3]{\sqrt {5} \minus{} 2}$,
$ \sqrt [3]{\left(\frac{\sqrt {5}}{2} \plus{} \frac{1}{2}\right)^3} \plus{} \sqrt [3]{\left(\frac{\sqrt {5}}{2} \minus{} \frac{1}{2}\right)^3} \equal{} \left(\frac{\sqrt {5}}{2} \plus{} \frac{1}{2}\right) \plus{} \left(\frac{\sqrt {5}}{2} \minus{} \frac{1}{2}\right) \equal{} \sqrt{5}$.

But, $ \sqrt [3]{2 \minus{} \sqrt {5}} \not \in \mathbb{R}$.
Cf. [url=http://www.mathlinks.ro/viewtopic.php?t=243230]a nice eqution 1![/url]

[hide="identity"]If a+b+c=0, then $ a^3\plus{}b^3\plus{}c^3\equal{}3abc$.[/hide]
[hide="proof"]Consider the cubic with roots a,b, and c. It is $ x^3\minus{}(a\plus{}b\plus{}c)x^2\plus{}(ab\plus{}ac\plus{}bc)x\minus{}abc$. Substituting x=a,b, and c, and then adding the resulting equations gives $ a^3\plus{}b^3\plus{}c^3\minus{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\plus{}(a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc)\minus{}3abc\equal{}0$, or $ a^3\plus{}b^3\plus{}c^3\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}ac\minus{}bc)\plus{}3abc$. Since a+b+c=0, we have $ a^3\plus{}b^3\plus{}c^3\equal{}3abc$, as desired.[/hide]

[hide="solution to problem"] Let $ x\equal{}\sqrt[3]{2\plus{}\sqrt{5}}\plus{}\sqrt[3]{2\minus{}\sqrt{5}}$, so $ x\minus{}\sqrt[3]{2\plus{}\sqrt{5}}\minus{}\sqrt[3]{2\minus{}\sqrt{5}}\equal{}0$. Using our identity, we now have $ x^3\minus{}4\equal{}3x\sqrt[3]{\minus{}1}$. If we're looking for only real solutions, which I assume we are, as otherwise it'll get messy, we have $ x^3\plus{}3x\minus{}4\equal{}0$. Noticing that the coefficients add to 0, we have x=1 as a root. Dividing (x-1) out, we have $ (x\minus{}1)(x^2\plus{}x\plus{}4)$. The quadratic clearly has no real roots, so we find that the only real solution is x=1.[/hide]

easy  [hide]\[ \[
\begin{array}{l}
 x = \sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }} \\
  \Rightarrow x^3  = 4 + 3x\sqrt[3]{{4 - 5}} \\
  \Rightarrow x^3  + 3x - 4 = 0 \\
  \Rightarrow (x - 1)(x^2  + x + 4) = 0 \\
  \Rightarrow x = 1 \\
 \end{array}
\]

 [/hide]


trigonometry
function
algebra
polynomial
algebra unsolved

Solve the eq
$(\sin 3x+4\sin 2x\cos 3x)^{2}=11\cos^{2}3x$

hey guys...

dont make theses impossibles problems..

is a lost of time

the solution is a arc-function of a polynomial of degree 5

this dont have a explicit formula.

so  the solution is only approximations...

to do this, use newton method or other thing

or just make a good problem that have a good answer...

humpt


floor function
function
modular arithmetic
inequalities
number theory
relatively prime

Let $n$ be a natural number and $f(n) = 2n - 1995 \lfloor \frac{n}{1000} \rfloor$($\lfloor$ $\rfloor$ denotes the floor function).

1. Show that if for some integer $r$: $f(f(f...f(n)...))=1995$ (where the function $f$ is applied $r$ times), then $n$ is multiple of $1995$.

2. Show that if $n$ is multiple of 1995, then there exists r such that:$f(f(f...f(n)...))=1995$ (where the function $f$ is applied $r$ times). Determine $r$ if $n=1995.500=997500$

[hide="1"]
Let $f^r$ denote $\underbrace{f\circ f\circ f\circ \cdots \circ f}_r$. Since $f(n)\equiv 2n\pmod{1995}$, $f^r(n)\equiv 2^rn\pmod{1995}$. Then if $f^r(n)$ is divisible by 1995, so is $2^rn$. But 1995 and 2 are relatively prime, so 1995 must divide $n$.
[/hide]
[hide="2"]
If $n$ is a multiple of 1995, then so is $f^r(n)$ for any $r$. Also, $\frac{a}{1000}-1< \left\lfloor \frac{a}{1000} \right\rfloor \leq \frac{a}{1000}$, so $\frac{a}{200}\leq f(a)< \frac{a}{200}+1995$ for any positive integer $a$.

Suppose there exists a positive integer $n$ which is a multiple of 1995 but $f^r(n)\neq 1995$ for any $n$. Let $m$ be the least positive value of $f^r(n)$. Then $m\geq 1995\cdot 2$ since $m$ is a multiple of 1995. But the inequality gives \[ 0<\frac{m}{200}\leq f(m)<\frac{m}{200}+1995\leq \frac{101m}{200}<m \] so $0<f(m)<m$ which is a contradiction. Therefore the sequence must eventually reach 1995.

When $n=1997\cdot 500$, $\frac 52 \cdot 1995\leq f(n)<\frac 72 \cdot 1995$ so $f(n)$ must equal $3\cdot 1995$. Since $\frac{3\cdot 1995}{200}$ is clearly less than 1995, $f^2(n)=1995$, so $r=2$ in this case.
[/hide]


When you add vectors together, is it necessary to draw a 'sum' vector as well?

[geogebra]38c9fff7c5195921e033ddc64b18a52f57cb6b60[/geogebra]

I'm not sure what you mean; if we add $ <a,b>\plus{}<c,d>\equal{}<a\plus{}c,b\plus{}d>$, we don't need to draw it to figure out the result, if that is what you mean.


modular arithmetic
linear algebra
matrix
combinatorics unsolved
combinatorics

we have $2n+1$ real numbers $a_{1},a_{2},a_{3}...a_{2n+1}$
for all $1\le i\le 2n+1$
we can divide numbers $a_{1},a_{2},...a_{i-1},a_{i+1},...a_{2n+1}$
into two groups (each with $n$ elements) with equal sums
prove that $a_{1}=a_{2}=...=a_{2n+1}$

[b]Solution.[/b]  Assume to the contrary that there exists a set $\{a_{i}\}_{i=1}^{2n+1}$ with all numbers not equal and with minimal sum of values.  Let $S=a_{1}+\hdots+a_{2n+1}$.  Since when excluding $a_{i}$, $S-a_{i}$ can be divided into two sets with equal sums, $a_{i}\equiv S\pmod 2$, that is, $a_{i}$ have equal parity since $S\mod 2$ remains invariant.

If $a_{i}\equiv 0\pmod 2\ \forall i\in [1,2n+1]\cap\mathbb{Z}$, then we obtain a set $\left\{ \frac{a_{i}}{2}\right\}_{i=1}^{2n+1}$ satisfying the properties of the problem with an even smaller sum of values. Contradiction.

Similarly, if all the $a_{i}$ are odd, then we have a contradiction since we can obtain a set $\left\{ \frac{a_{i}+1}{2}\right\}_{i=1}^{2n+1}$ with a smaller total sum, not all elements equal. Contradiction.

And the conclusion follows.$\blacksquare$

You get the result for $\mathbb R$ out of that for $\mathbb Q$ by choosing a (Hamel)basis of $\mathbb R$ over $\mathbb Q$ and applying the result for rationals.

Of course, we may avoid Hamel basis (and hence axiom of choice) by using uniform approximation theorem: consider N such that $Na_{i}$ differ from some integers by less then 0.0001. Then Na_i satisfy the same condition.

Or, we may say that linear system have only trivial solution over Q, hence it have only trivial solution over R (bacuse rank of a matrix does not depend on a field, over whiich it is considered. Of course, if the characteristics of field is 0)


number theory
prime factorization
prime numbers
number theory unsolved

Prove: For each positive integer is the number of divisors whose decimal representations ends with a 1 or 9 not less than the number of divisors whose decimal representations ends with 3 or 7.

Let $ n$ be the positive integer.
Let set $ A$ contain all the divisors of $ n$ whose decimal representations end with $ 1$ or $ 9$.
Let set $ B$ contain all the divisors of $ n$ whose decimal representations end with $ 3$ or $ 7$.
We will show that the cardinality of set $ A$ is greater than or equal to that of set $ B$.
The numbers in $ A$ and the numbers in $ B$ are not divisible by $ 2$ or $ 5$, so we only need to consider divisors of $ n$ that do not have $ 2$'s or $ 5$'s in their factorizations. The size of $ A$ and the size of $ B$ do not change if the exponent of 2 or 5 in $ n$'s prime factorization is changed to $ 0$. Thus we can let $ n$ not be divisible by $ 2$ or $ 5$.
$ n$ can be expressed as
$ n \equal{} \displaystyle\prod_{w \equal{} 1}^a p_w \cdot \displaystyle\prod_{x \equal{} 1}^b q_x \cdot \displaystyle\prod_{y \equal{} 1}^c r_y \cdot \displaystyle\prod_{z \equal{} 1}^d s_z$
where $ p_w$ is a prime ending with $ 1$, $ q_x$ is a prime ending with $ 3$, $ r_y$ is a prime ending with $ 7$, $ s_z$ is a prime ending with $ 9$, and $ a, b, c$ and $ d$ are the sums of exponents on prime numbers ending with $ 1$, $ 3$, $ 7$, and $ 9$ respectively.
Now consider a divisor of $ n$. Let $ m$ be that divisor. Then we can express $ m$ as
$ m \equal{} \displaystyle\prod_{w \equal{} 1}^e p_w \cdot \displaystyle\prod_{x \equal{} 1}^f q_x \cdot \displaystyle\prod_{y \equal{} 1}^g r_y \cdot \displaystyle\prod_{z \equal{} 1}^h s_z$
where $ 0 \le e \le a$, $ 0 \le f \le b$, $ 0 \le g \le c$, and $ 0 \le h \le d$.
Taking $ m$ mod $ 10$, we get
$ m \equiv 1^e \cdot 3^f \cdot 7^g \cdot 9^h \equiv 3^f \cdot ( \minus{} 3)^g \cdot ( \minus{} 1)^h \equiv 3^{f \plus{} g} \cdot ( \minus{} 1)^{g \plus{} h} \mod 10$.
By Euler's totient theorem $ 3^4 \equiv 1 \mod 10$, so if $ f \plus{} g \equiv 0 \mod 4$, $ 3^{f \plus{} g} \equiv 1$ so $ m \equiv \pm 1$ and $ m$ ends with a 1 or a 9.
If $ f \plus{} g \equiv 1 \mod 4$, $ 3^{f \plus{} g} \equiv 3$ so $ m \equiv \pm 3$ and $ m$ ends with 3 or 7.
If $ f \plus{} g \equiv 2 \mod 4$, $ 3^{f \plus{} g} \equiv 9$ so $ m \equiv \pm 9$ and $ m$ ends with a 1 or a 9.
If $ f \plus{} g \equiv 3 \mod 4$, $ 3^{f \plus{} g} \equiv 7$ so $ m \equiv \pm 7$ and $ m$ ends with a 3 or a 7.
If $ f \plus{} g$ is even, $ m$ is in set $ A$. If $ f \plus{} g$ is odd, $ m$ is in set $ B$.
For each $ e, h$ such that $ 0 \le e \le a$ and $ 0 \le h \le d$,
There are $ i$ values of $ f$ that are even and $ j$ of them that are odd, such that $ i \ge j$. There are $ k$ values of $ g$ that are even and $ l$ of them that are odd, such that $ k \ge l$. $ ik \plus{} jl \ge il \plus{} jk$ by rearrangement. Since $ ik \plus{} jl$ values of $ f \plus{} g$ are even and $ il \plus{} jk$ values of $ f \plus{} g$ are odd, the number of values of $ f \plus{} g$ that are even is greater than or equal to the number of values of $ f \plus{} g$ that are odd. Therefore, the cardinality of set $ A$ is greater than that of set $ B$, and we are finished.


Give a counting explanation for why (2n)!(2m)! / [m!n!(m+n)!] is an integer.

heya there!

we know that the exponent of a prime factor for the canonic form of  (2n)!(2m)! if

sum { [2n/p_i^k]+[2m/p_i^k]} , the sum if after k

and for m!n!(m+n)! is

sum {[n/p_i^k]+[m/p_i^k]+[(m+n)/p_i^k]} again the sum is after k.

we know that [2a]+[2b]>=[a]+[b]+[a+b] for any a and b from R+.
so the number is integer!!! :D

cheers!

[quote="Lagrangia"]I don't know!!! I just gave a solution.. didn't even read what he needed(sorry)! :D what do you mean by a counting explanation... I think that this is a counting explanation... :D

cheers![/quote]

I think he means a combinatorial argument... like counting something in two ways.

I think that such an explanation would be hard to find.. guess that the solution I have posted is good enough! :D


Two people take turns cutting up a rectangular $6\times 8$ chocolate bar. You are allowed to cut only along a division between the squares and the cut can only be in a straight line. The last player who can break the chocolate wins. Is there a winning strategy for either player? Generalize.

[quote="E^(pi*i)=-1"][hide="Seems too easy to be right . . ."]
Each time the chocolate is broken, the number of pieces increases by one.  Initially there is 1 piece and at the end there are 48 pieces.  After the first player breaks the chocolate , there are 2 pieces, after the second player goes there are 3, after the first player goes again there are 4, etc.  The first player always leaves an even number of pieces and the second player always leaves an odd number of pieces.  Since 48 is even, Player 1 wins.  For an $m \times n$ bar, Player 1 wins iff either $m$ or $n$ is even; Player 2 wins iff both are odd.
[/hide][/quote]

I agree that that sounds too easy to be right, but it seems right.

Oh yes that's right. The version I was thinking about is the normal cutcakes;

You have a 6x8 chocolate bar. The same rules apply but player 1 can only cut horizontally and player 2 can only cut vertically. Pieces always maintain their orientation. Who has a winning strategy?

Which direction is the $6$ in?

So this problem is basically a matter of survivor: outlast.

Hey, I don't understand the problem. For example, player 1 breaks the 6x8 into a 6x2 and a 6X6. If player 2 breaks the 6x6 into a 5x6 and a 1x6. This means we have 3 pieces of bars. What next? could player 1 choose to break any of them until what's left are 48 1x1 chocolates bars?

I think a player should break every remaining block (that's not $ 1\times 1$) once during a single turn, otherwise the problem is trivial.

hm, so, correct me if i'm wrong. once we have 48 1x1 bars, the game ends?


group theory
abstract algebra
linear algebra
matrix
algebra
polynomial
inequalities

Is a finite subgroup of $ \mathrm{GL}_n(\mathbb{C})$ necessarily nipotent?

Any finite group can be embedded as a finite subgroup of $ GL_n(K)$, for any field $ K$ and large enough $ n$.

I think you mean "unipotent". If so then we can say that there are no non-trivial finite unipotent subgroups $ G$ of $ Gl_n(\mathbb C)$. To see this let $ A\in Gl_n(\mathbb C)$ a unipotent matrix of finite order: $ A^m \equal{} E$. Then $ A$ is diagonalizable, since the minimal polynomial $ f_A(X)$ divides the polynomial $ X^m \minus{} 1$, thus the zeros of $ f_A$ are all simple. If $ A$ is unipotent, then $ A \equal{} E \plus{} N$ with some nilpotent matrix $ N$ which must be 0 since $ A$ is diagonalizable. Thus $ A \equal{} E$.

NB: This remains true if $ \mathbb C$ is replaced by a field of characteristic 0.

There's another meaning here: a nilpotent group is one in which the chain of central groups ($ Z_{n\plus{}1}/Z_n$ is the center of $ G/Z_n$) eventually reaches the whole thing.

For this meaning, well, finite groups that aren't nilpotent exist. $ S_3$ is the first, and we can embed that in $ GL_2(\mathbb{C})$, or $ GL_3(K)$ for any $ K$.

[quote="jmerry"]$ S_3$ is the first, and we can embed that in $ GL_2(\mathbb{C})$, or $ GL_3(K)$ for any $ K$.[/quote]

In case you are curious, S3 embeds in GL(2,K) for any (associative, unital) ring K.  This is because it has a 2-dimensional representation over the integers.  Representations of Sn are triply nice: (1) the trace of every matrix in a representation over a field of characteristic 0 is rational, (2) up to conjugation every representation over a field of characteristic 0 is actually over the field of rationals, (3) up to conjugation every matrix representation over the rationals is actually over the integers.  Once you have a representation over the integers, you get a representation over any associative unital ring.  Sn only has a few normal subgroups, so it is usually easy to check the representation is faithful.

#1 is studied in the papers mentioned in the "all elements of the order are conjugate".  #2 is related to the Schur index, and #3 is why combinatorics is so interested in the representation theory of Sn.

The representation over an arbitrary ring takes a specified element of order 2 to [0,1;1,0] and a specified element of order 3 to [0,1;-1,-1].  [0,1;-1,-1] is never the identity, so the representation is always faithful.


calculus
integration
logarithms
LaTeX
derivative
calculus computations

Hello everyone:

I have questions regarding the following two integrals.

Thank you very much.

---

1. Differentiate: $ \frac {d}{dx}\left[\int_{e^{2x}}^3 (y^2 \minus{} 5 \ln y) dy \right]$

I did not have much trouble calculating it, but I got an answer of $ \minus{} 2e^{6x} \plus{} 20xe^{2x}$ instead of the given [hide] $ 2e^{6x} \minus{} 20xe^{2x}$[/hide].

When I looked at the answer, my mistake took place here:

$ \frac {d}{dx}\left[\int_{e^{2x}}^3 (y^2 \minus{} 5 \ln y) dy \right] \equal{} \minus{} \frac {d}{dx}\left[\int_3^{e^{2x}} (y^2 \minus{} 5 \ln y) dy \right]$

= $ [e^{4x} \minus{} 5(2x)][2e^{2x}]$

Originally, I thought that the negative sign was to be carried over after the integration part. Why does it go away after integration?

---

2. Differentiate: $ \frac {d}{dx}\left(\int_{x \minus{} 2}^5 9b^2 db \right)$

After some steps, I arrived at:

$ 375x^3 \minus{} 3(x \minus{} 2)^3$.

I started to differentiate here, and I applied the Chain Rule to $ \frac {d}{dx} 3(x \minus{} 2)^3$.

However, this resulted in an incorrect answer as the given answer simplified $ 3(x \minus{} 2)^3$ before differentiation. Why is my method erroneous?

[quote="catalan"]-

1. Differentiate: $ \frac {d}{dx}\left[\int_{e^{2x}}^3 (y^2 \minus{} 5 \ln y) dy \right]$

I did not have much trouble calculating it, but I got an answer of $ \minus{} 2e^{6x} \plus{} 20xe^{2x}$[/quote]

That's the correct answer.

I assume you meant this: [quote="catalan"]$ \frac {d}{dx}\left(\int_{x \minus{} 2}^{5x} 9b^2 db \right)$[/quote]  (It's always a good idea when using LaTeX to preview your posts, or at least read over the final product, to make sure you haven't misplaced any {}s, etc.)

[quote="catalan"]I applied the Chain Rule to $ \frac {d}{dx} 3(x \minus{} 2)^3$.

However, this resulted in an incorrect answer as the given answer simplified $ 3(x \minus{} 2)^3$ before differentiation. Why is my method erroneous?[/quote]  Well, either you or the text made a computational error, and if you don't show your computations or your answer then it's impossible to tell who it actually is.  Any valid differentiation rule correctly applied will give the correct final answer, and the chain rule leads to the same thing as the product rule or as expanding everything out and applying the power rule; all three methods give something equal to $ 9(x \minus{} 2)^2$ (which will appear with a minus sign in the final answer).

$ Q\equal{}\int_{e^{2x}}^3 (y^2 \minus{} 5 \ln y) dy$

by parts (i know this is the hard way, instead of FTC)

$ u\equal{}y^2\minus{}5\ln y$, $ dv\equal{}dy$
$ du\equal{}2y\minus{}\frac{5}{y}dy$, $ v\equal{}y$

$ Q\equal{}(y^3\minus{}5y \ln y) |_{e^{2x}}^3 \minus{} \int_{e^{2x}}^3 2y^2\minus{}5 dy$
$ \equal{} 27\minus{}15\ln 3 \minus{}e^{6x}\plus{}10xe^{2x} \minus{} (\frac{2y^3}{3}\plus{}5y )|_{e^{2x}}^3$
$ \equal{} 27\minus{}15\ln 3 \minus{}e^{6x}\plus{}10xe^{2x} \minus{} 18\plus{}15\plus{}\frac{2e^{6x}}{3}\minus{}5e^{2x}$
$ \equal{}\frac{\minus{}e^{6x}}{3}\plus{}(2x\minus{}1)5e^{2x}\plus{}C$

so you are rightwhen you diff

In the well-known conditions, we have  $ \boxed {\ \int_{a(x)}^{b(x)}f(t)\ \mathrm {dt}\equal{}f[b(x)]\cdot b'(x)\minus{}f[a(x)]\cdot a'(x)\ }$ .

[quote="Virgil Nicula"]In the well-known conditions, we have  $ \boxed {\ \int_{a(x)}^{b(x)}f(t)\ \mathrm {dt} \equal{} f[b(x)]\cdot b'(x) \minus{} f[a(x)]\cdot a'(x)\ }$ .[/quote]

I think you forgot to put a $ \frac{d}{dx}$ in front.


anybody knows how to bold stuff in Latex?

Wrong forum but OK, I know how to bold letters

[code]{\bf Hello}[/code]

$\text{Hello}~{\bf Hello}$


algebra
polynomial
inequalities
algorithm
inequalities proposed

Let $ a,b,c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}1.$ Prove that for any $ 0<k \le 2\sqrt{6} \minus{}4,$ we have
$ \frac{a^3}{a\plus{}k(a\minus{}b)^2}\plus{}\frac{b^3}{b\plus{}k(b\minus{}c)^2}\plus{}\frac{c^3}{c\plus{}k(c\minus{}a)^2} \ge \frac{1}{3}.$
:)

Thank you very much for your result, but I think it is very hard for us to prove the inequality for the result you find if we dont have computer's helping. However, for $ k\equal{}2\sqrt{6} \minus{}4,$ we can prove it in a very simple way.

We have a similar result is:
If $ a,b,c >0,a\plus{}b\plus{}c\equal{}1,$ then for any $ 0 \le k \le \frac{3\plus{}\sqrt{6}}{6},$ the following inequality holds
$ \sum \frac{a^3}{a\plus{}k(b\minus{}c)^2} \ge \frac{1}{3}.$
In here, $ \frac{3\plus{}\sqrt{6}}{6}$ is the best constant. :)

Dear can_hang2007,
I don't know whether it can be proven by your simple way that for $ k\equal{}\frac{59}{65}\,$ and even $ k\equal{}\frac{9}{10}$ which both are stronger than $ 2\sqrt{6}\minus{}4$.


geometry
circumcircle
quadratics
3D geometry
tetrahedron
Pythagorean Theorem
geometry proposed

Let three circles (1),(2),(3) have the same center O and their radius are $ R_1,R_2,R_3$ respectively. Three points A,B,C move on (1),(2),(3) respectively. Known that $ R_1 \equal{} 1,R_2 \equal{} \sqrt {2},R_3 \equal{} \sqrt {5}$, find the greatest value of the area of triangle ABC.

$ O$ is point in the $ \triangle ABC$ with distances from the vertices $ AO : BO : CO \equal{} 1 : \sqrt 2 : \sqrt 5.$ Suppose $ A, O, B$ are given, then for max area, $ CO \perp AB,$ etc. $\Longrightarrow$ $ O$ is orthocenter of the $ \triangle ABC.$ Rename it $ H$ and let $ (O,R)$ be the circumcircle of this triangle. Let $ AH$ cut $ (O)$ again at $ D.$ The isosceles $ \triangle BDO \sim \triangle DHC$ are similar, having equal angles $\Longrightarrow$ $ BH \cdot CH = BD \cdot CH \equal{} HD \cdot OD \equal{} HD \cdot R.$ Power of $ H$ to $ (O)$ is $ R^2 \minus{} OH^2 \equal{} AH \cdot HD \equal{} \frac{AH \cdot BH \cdot CH}{R}.$ On the other hand,

$ AH^2 \plus{} BH^2 \plus{} CH^2 \equal{} 12R^2 \minus{} (a^2 \plus{}  b^2 \plus{}  c^2) \equal{} 3R^2 \plus{} OH^2 \equal{} 4R^2 \minus{} \frac{AH \cdot BH \cdot CH}{R}$

Substituting $ AH, BH, CH,$ we get a cubic for $ R$: $ 4R^3 \minus{} 8R \minus{} \sqrt{10} \equal{} 0.$
Substituting $ R \equal{} \varrho \sqrt{10},$ this becomes $ 40 \varrho^3 \minus{} 8 \varrho \minus{} 1 \equal{} 0.$ Looking for a rational root, $ \varrho \equal{} \frac{1}{2}.$ Factoring it out, the remaining quadratic $ 20 \varrho^2 \plus{} 10 \varrho \plus{} 1 \equal{} 0$ has 2 more real roots, both negative, not acceptable. With $ R \equal{} \varrho \sqrt{10} \equal{} \frac{\sqrt{10}}{2},$ the triangle sides are

$ a \equal{} \sqrt{4R^2 \minus{} AH^2} \equal{} 3,\ \ b \equal{} \sqrt{4R^2 \minus{} BH^2} \equal{} \sqrt{8},\ \ c \equal{} \sqrt{4R^2 \minus{} CH^2} \equal{} \sqrt{5}.$

$ S \equal{} \frac{abc}{4R} \equal{} 3.$

Another solution:
P is the point satisfying PO is the perpendicular to the plane (ABC) and OP=1.
Apply Pythagorean theorem, $ PA\equal{}\sqrt{2},PB\equal{}\sqrt{3},PC\equal{}\sqrt{6}$
The volume of the tetrahedron PABC is denoted by V.
$ \frac{1}{3}OP.S(ABC)\equal{}V\leq \frac{1}{6}PA.PB.PC\equal{}\frac{1}{6}.\sqrt{2}.\sqrt{3}.\sqrt{6}\equal{}1$
$ \Rightarrow S(ABC)\leq 3$
The equality occurs when $ \angle APB\equal{}\angle BPC\equal{}\angle CPA\equal{}90^o$, then $ AB\equal{}\sqrt{5},BC\equal{}3,CA\equal{}\sqrt{8}$.
We can choose O(0,0), A(0,1), B(1,-1), C(-2,-1).


calculus
integration
function
calculus computations

[quote="emjay285"]Can this integral be integrated? $ \int{\frac{e^{x}}{x}\left(1-\frac{1}{x}\right)dx}$[/quote]

Sure, you can integrate it as follow:
$ \int{\frac{e^{x}}{x}\left(1-\frac{1}{x}\right)dx}$
$ \Rightarrow\int{\frac{e^{x}}{x}dx-\int{\frac{e^{x}}{x^{2}}dx}}$
Integration by parts:
$ \Rightarrow\int{\frac{e^{x}}{x}dx-(-\frac{e^{x}}{x}+\int{\frac{e^{x}}{x}dx)}}$
$ \Rightarrow\frac{e^{x}}{x}$

it's a particular case of a well known theorem

[color=darkred][b][u]Theorem[/b][/u][/color]
Let $ f$ be a differentiable function then
$ \boxed{\int e^{x}(f(x)\plus{}f'(x))\text{dx}\equal{}e^{x}f(x)\plus{}c}$
the result is obvious just differentiate $ e^{x}f(x)$


[hide]only see that B,C,I,H are concyclic..  :)[/hide]

in fact $B,C,H,O,I$ are concyclic and that leads to an easy solution.
by the way i was not a contestent at apmo .

Here is my solution:

Easy angle chasing yields $\angle BIC=120^\circ$, while more angle chasing yields $\angle BHC=120^\circ$. Hence $BIHC$ is cyclic, and if we denote the foot of the perpendicular from $C$ to $AB$ as $D$, then $\angle IHD=\angle IBC$. Now, let $\angle ABC=2\beta$. It is easy to find $\angle DCB=90-2\beta$, hence $\angle AHD=2\beta$. Therefore, $3\angle ABC=6\beta=2\angle AHI$.

we first prove a lemma
lemma = let $ABC$ be a triangle with $D$ as midpoint of minor arc $BC$ of circumcircle of $ABC$
also let $I$ be incentre of triangle $ABC$ than we prove that $D$ is centre of circumcircle of $BIC$.

proof = it is evident that $A$ excentre of triangle $ABC$ lie on circumcircle of $BIC$
also $BD=DC$ and thus $\angle CBD=A/2 = \angle CAD=\angle BAI$
thus $A,I,D$ are collinear.
now $\angle IBD = 90-C/2=\angle JCB=\angle BID$ where $J$ is $A$ excentre.
and thus $BD=DI$
thus $D$ is circumcentre of $BIC$.

main proof = by this lemma and $\angle A = 60$ we get that $\angle BOC=\angle BIC=\angle BHC = 120$
and hence $B,H,O,I,C$ are concyclic.
by concyclicity of $HIJB$ we have $\angle HBJ = \angle AIH = 120-C-B/2=60+B/2$
in triangle $AHI$ we thus get $\angle AHI=3B/2$ since $\angle HAI=60-B$
and thus $2\angle AHI = 3\angle B$.

we are done :D

A nice problem .
We will show that points B,I, H and C are concyclic wich is too easy since angle<BIC = 90°+A/2 = 120°=180°-A=<BHC.
Let (w) the cercle ( BIHC). Let Y be the intersection point of line AH with (w);then angle <YBC = <ABC .
[u]Prof[/u]
By simple angle chasing <BAY = <HCB =<HYB = <AYB  and (AY) perpendicular to (BC) so triangle BAY is isoceles then the equality follows.
Now denote by X the center of (w) ,then :
2<AHI=360°-2<IHY = <IXY = 2<IBY =<ABC+ 2<CBY = <ABC + 2<ABC = 3<ABC , QED

[hide=Solution]
First, it is well known that $\angle BIC=90+\frac{1}{2}A=120=180-A=\angle BHC$, so $BIHC$ is cyclic with vertices in this order. Let the midpoint of minor arc $\overarc{BC}$ be $M$. Observe $\odot(BIHC)$ is centered at $M$ by the Incenter-Excenter Lemma.

Now, $\angle AHI + \angle IHC + \angle BHC = 360$.
$\angle AHC = 180-B$ is obvious.
We chase $\angle IHC = \angle BHC + \angle IHB = 120+\frac{C}{2}$ since $\angle IHB = \frac{1}{2} \angle IMB = \frac{1}{2}C$.
This means that $\angle AHI = 60+B-\frac{C}{2} = A+B-\frac{C}{2} = 180-\frac{3C}{2}$.

To finish, $B+C=120 \implies 180=\frac{3}{2}(B+C) \implies 180-\frac{3C}{2}=\frac{3B}{2}$. Hence, $\angle AHI = \frac{3B}{2}$ as desired.
[/hide]

 This problem is easy.
First we should know $B,I,H,C$ are concyclic.
$<BHC=<CIB=120$

Let altitudes intersect sides $a,b,c$ in points $F,D,E$. Now, $\angle IBH=\angle ABD-\angle ABI=30^\circ-\frac{\beta}{2}$.
$\angle ICH=\angle ACI-\angle ACH=30^\circ-\frac{\beta}{2} \implies \angle IBH=\angle ICH$. Using that quadrilateral $BCHI$ is cyclic we obtain:
$\angle BCI=\angle BHI=\frac{\gamma}{2}$ and $\angle FHB=90^\circ-\angle FBH=90^\circ-\beta+\angle ABH=120^\circ-\beta$
Finally, $\angle AHI=180^\circ-\angle BHI-\angle FHB=60^\circ-\frac{\gamma}{2}+\beta=60^\circ-\frac{120^\circ-\beta}{2}+\beta=\frac{3\beta}{2} \implies 2\angle AHI=3\beta$ :)

Note $\angle BHC=\angle BIC=\angle BOC=120^{\circ},$ so $H,O$ lie on $(BIC).$

We claim that $AHO$ is isosceles with $AH=AO.$ We use complex numbers. Put $ABC$ on the unit circle and note that $H=A+B+C=A+D,$ so $AH=OD=AO.$ Now note that $\angle CAH=\angle BAO=90^{\circ}-\angle C,$ so $\angle HAO=\angle A-(180^{\circ}-\angle 2C)=\angle C-\angle B=2\angle C-120^{\circ}.$ Thus $\angle AHO=150^{\circ}-\angle C=\angle B+30^{\circ}.$

Also note $\angle HIO=180^{\circ}-\angle CHO=180^{\circ}-\angle A+\angle B=120^{\circ}+\angle B,$ so $\angle IHO = 30^{\circ}-\frac{1}{2}\angle B.$ Thus $\angle AHI=\frac{3}{2}\angle B,$ as desired.

[hide=solution sketch]Key Observation B, H, I, C are concyclic. Note that $\angle AHI = =180- \angle IHB-\angle BHK$, where K is the foot of altitude to BC from A. Now angle chase with the cyclic quadrilateral, and we get the desired result[/hide]

Let $M$ be the midpoint of minor arc $BC$ of $(ABC)$ and $O$ be its center. We have that $\triangle COM$ is equilateral and hence $OM=AH$. Also evidently $AH || OM$, hence $AOMH$ is a parallelogram. But since $OM=OA$ it's a rhombus, whence $OM=MH$. Now from the excenter/incenter lemma we have that $CHIOB$ is cyclic. We have $\angle AHI=\angle AOI=\angle AOC-\angle COH=\angle AOC-\angle IBC=\angle AOC-\frac{1}{2}\angle ABC=\frac{3}{2}\angle ABC$, as desired.

Note that we have $$\angle BIC = 180^{\circ}-\angle IBC - \angle ICB = 180^{\circ} - \frac{\angle ABC + \angle ACB}{2} = 180^{\circ} - \frac{180^{\circ}-\angle A}{2}$$ and since $\angle A = 60^{\circ}$, then $\angle BIC = 120^{\circ}$. Now, it is well-known that $$\angle BHC = 180^{\circ}-\angle A = 180^{\circ}-60^{\circ}=120^{\circ}$$ so as a result we have $\angle BIC = \angle BHC$ so $BIHC$ is cyclic. Now, reflecting $A$ over $BC$ and letting that point be $A'$ we know $A'$ is on the circumcircle of $BIHC$, so $$\angle AHI = 180^{\circ}- \angle A'HI = \angle A'BI = \angle B + \angle CBI = \frac32\angle B$$ so $2\angle AHI = 3 \angle B$.

First note that $BIC$ are concyclic. I claim that $H$ lies on this circle. Let $\Gamma_1$ denote the circumcircle of $\triangle ABC$ and $\Gamma_2$ denote $(BIC)$. Then, arc $\overarc {BAC}$ has measure $240^{\circ}$. Then, since $\angle BIC=90^{\circ}+\frac{\angle A}{2}=120^{\circ}$, looking at $\Gamma_2$, $\overarc{BC}$ not containing $I$ has measure $240^{\circ}$ as well. Hence, $\Gamma_1$ and $\Gamma_2$ are reflections of each other across $BC$. Since the reflection of $H$ across $BC$ lies on $\Gamma_1$, $H$ must lie on $\Gamma_2$.

Now,
\[3\angle ABC=6\angle IBC=6(180-\angle IHC)=1080-6(180-\angle AHI+\angle ABC)=6\angle AHI-6\angle ABC,\]
which implies the desired result

Points $I, H, O, B, C$ Will lie on same circle..
And then by some angle chase we will get the desired results

We start with a lemma.

[color=red][b]Lemma:[/b][/color] In a triangle $ABC$ with orthocenter $H$, $AH=2R\cos A$.

[i]Proof.[/i] Consult problem Exercise 4.4.7 in the AoPS Precalculus Book. $\square$

Now note that $AH=2R\cos A=2R\cdot \frac12=R=AO$. Also, $AI=AI$ and $\measuredangle HAI=\measuredangle IAO$ since $H$ and $O$ are isogonal conjugates. Thus, $\triangle AHI\cong \triangle AOI$.

Now observe the following angle equalities: \begin{align*} \angle BHC&=180^\circ-\angle A=120^\circ\\ \angle BIC&=90^\circ+\frac{\angle A}2=120^\circ\\ \angle BOC&=2\angle A=120^\circ \end{align*}

so hexagon $CHIOBI_A$ is cyclic.

The rest is just angle chasing: $$\measuredangle AHI=\measuredangle IOA=\measuredangle IOC+\measuredangle COA=\measuredangle IBC+2\measuredangle CBA=4\measuredangle CBI- \measuredangle CBI=3\measuredangle CBI$$ and multiplying both sides by $2$ gives the desired $2\measuredangle AHI=6\measuredangle CBI=3\measuredangle CBA$. $\blacksquare$

[hide=Remarks]This problem was quite brutal for directed angles...one must write the condition as $2\measuredangle AHI=3\measuredangle CBA$ and NOT $2\measuredangle AHI=3\measuredangle ABC$.

Some cool properties about a triangle $ABC$ with $\angle A=60^\circ$ and $\overline{AI}\cap(ABC)=M$ are

[b]Property 1:[/b] $(ABC)$ and $(IBC)$ (which we will call $\omega_1$ and $\omega_2$ respectively) are reflections over $BC$.
[i]Proof.[/i] Clearly $H\in \omega_2$ and $H'\in\omega_1$ where $H'$ is the refllection of $H$ over $BC$. This is true by the Reflecting the Orthocenter Lemma. Now, for a point $P$ on $\omega_1$, we have $\measuredangle BPC=\measuredangle BAC=60^\circ$ and $\measuredangle BP'C=-\measuredangle BPC=120^\circ=\measuredangle BIC$ as needed.

[b]Property 2:[/b] $\triangle COM$ is equilateral.
[i]Proof.[/i] Clearly, $IM$ bisects $\angle BIC=120^\circ$, so $\triangle MIC$ is isosceles with $\angle MIC=60^\circ$, implying the result.

[b]Property 3:[/b] $CHIOBI_A$ is cyclic.
[i]Proof.[/i] Proven in the solution to the main problem.

[b]Property 4:[/b] $\triangle AHI\cong \triangle AOI$.
[i]Proof.[/i] Proven in the solution to the main problem.

[b]Property 5:[/b] $\triangle IMH\cong \triangle IMO$.
[i]Proof.[/i] We have $MH=MO$ and $MI=MI$. Also note that $\measuredangle HMI=\measuredangle IMO$ since $H$ and $O$ are isogonal conjugates and thus reflections across the line $\overline{AIM}$. Thus $\triangle IMH\cong \triangle IMO$. [url]https://artofproblemsolving.com/community/c6h1408296p7899381[/url]

[b]Property 6:[/b] $AOMH$ is a rhombus.
[i]Proof.[/i] Observe that $AH\parallel OM$. From the above, we have $MH=MO=AO=AH$, finishing. (Actually, the parallel condition was not even needed.)

[b]Property 7:[/b] Line $AIM$ is the perpendicular bisector of $\overline{OH}$.
[i]Proof.[/i] This follows from $AO=AH$ (or $MH=MO$) combined with the $O$ and $H$ isogonality.

[b]Property 8:[/b] If $N_9$ is the center of the nine-point circle of $\triangle ABC$, then it lies on the line $\overline{AIM}$.
[i]Proof.[/i] $N_9$ is the midpoint of $\overline{OH}$ which does the trick.

Problems to try: [url=2007 APMO Problem 2]https://artofproblemsolving.com/community/c6h141391p799589[/url]
[/hide]

[quote=APMO 2007 P2]Let $ABC$ be an acute angled triangle with $\angle{BAC}=60^\circ$ and $AB > AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$ . Prove that $2\angle{AHI}= 3\angle{ABC}$.[/quote]
[b]Solution.[/b] Since $\angle A=60^{\circ}$, we have that $BIHC$ is cyclic. Now just by angle chasing, we get that:
$$\angle AHI=180^{\circ}-\angle IHD=180^{\circ}-(\angle IHB+\angle BHD)=180^{\circ}-(\frac{\angle C}{2}+\angle C)=180^{\circ}-\frac{3\angle C}{2}$$
$$2\angle AHI=360^{\circ}-3\angle C=3(120^\circ-\angle C)=3\angle B.\blacksquare$$

[b][color=lightblue]Claim:[/color][/b] $BIHC$ is cyclic.

[i]Proof.[/i] Notice that \begin{align*}\angle HCI&=\tfrac{1}{2}\angle C-(90-60)\\&=\tfrac{1}{2}\angle C-30+\tfrac{1}{2}(\angle B+\angle C-120)\\&=\tfrac{1}{2}\angle B-(90-\angle C)\\&=\angle HBI.\end{align*} $\blacksquare$

Hence, $$\angle AHI=(180-\angle IHC)+(180-AHC)=\tfrac{1}{2}\angle B+180-\angle A-\angle C=\tfrac{3}{2}\angle B.$$ $\square$

Let AD,BE and CF be altitudes of triangle. ∠EHF = 180 - 60 = 120 = ∠BHC and ∠BIC = ∠A + ∠B/2 + ∠C/2 = 120 so BIHC is cyclic.
∠AHI = ∠IHF + ∠FHA = ∠IBC + ∠DHC = ∠IBC + ∠ABC ---> ∠AHI = 3/2 ∠ABC.
we're Done.

$BHIC$ is cyclic since $\angle BHC = 180^{\circ} - \angle A = 120^{\circ} = 90^{\circ} + \frac{\angle A}{2}$. $D$ be the foot from $A$ to $BC$. Consider $DHIC$ where $\angle HDC = 90^{\circ}$, $\angle IBC = \frac{\angle B}{2}$, $\angle HIC = 180^{\circ} - 90^{\circ} + \angle B = 90^{\circ} + \angle B$. Now, trivial calculation gives, $\angle AHI = \frac{3}{2}\angle B$ and check that, this is the result we wanted.

Let $\angle ABC=2\theta$ and $\angle ACB=120-2\theta$. We wish to show that $\angle AHI=3\theta$.\\

Claim: $HIBC$ is cyclic. This is because $$\angle BHC=180-\angle A=120$$ and $$\angle BIC=90+(\angle A/2)=120.$$

Then, $$\angle IHB=\angle ICB=60-\theta.$$ Let $E$ be the foot from $B$ to $AC$. Then, $$\angle AHE=\angle C=120-2\theta,$$ so $$\angle AHI=180-\angle AHE-\angle IHB=180-(120-2\theta)-(60-\theta)=3\theta,$$ hence done.

Observe that $B, C, H, I$ are concyclic, as $\angle BHC = 120^\circ = \angle BIC$ from some simple angle chasing.
Now we have $\angle AHC = 180^\circ - B, \angle CHI = 180^\circ - \frac{B}{2} \implies \angle AHI = \frac{3B}{2}$, and we are done. $\square$


Prove that

$\sqrt{\frac{m_a}{a}}+\sqrt{\frac{m_b}{b}}+\sqrt{\frac{m_c}{c}}\geq\sqrt{6}$

Nice inequality! It's really yours? :?  I saw this i think some time ago, but....i don't know if this yours?
 Anyway, inequalities connected to this one can be found in my book for inequalities. :D  ;)

Sorry, I discovered  that in 1976!

In Russian montly "Kvant" 1976, number 7, page 16 :
prove $\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\geq\frac{3\sqrt{3}}{2}$
author: V. Matizen.

Maybe what you saw is this?

When I proved this inequality, I thought of mine.

[quote="Ph-An"]I think that the enunciation is wrong. See you for $a=b=c=1\ (m_a=m_b=m_c=\frac{\sqrt 3}{2})$.[/quote]

no! a=b=1, c-->2

Indeed, but then must  $\sum \sqrt \frac{m_a}{a}>\sqrt 6$ (strict !). You are right, thus you misleaded me.

Which inequality? :?  The one that Ph-An posted? :)

[quote="cezar lupu"]Which inequality? :?  [/quote]
prove that:
$\sqrt{\frac{m_a}{a}}+\sqrt{\frac{m_b}{b}}+\sqrt{\frac{m_c}{c}}\geq\sqrt{6}$
Cezar, can you solve this? :)

[quote="Ph-An"]Indeed, but then must  $\sum \sqrt \frac{m_a}{a}>\sqrt 6$ (strict !). [/quote]
You are right. If  $\sum \sqrt \frac{m_a}{a}>\sqrt 6$ then  $\sum \sqrt \frac{m_a}{a}\geq\sqrt 6.$ :)

Okay, this is my solution:

First, let's write out the problem completely:

[color=blue][b]Problem.[/b] Let ABC be a triangle with medians $m_a$, $m_b$, $m_c$. Prove the inequality

$\sqrt{\frac{m_a}{a}}+\sqrt{\frac{m_b}{b}}+\sqrt{\frac{m_c}{c}}>\sqrt6$.[/color]

Now, the [i]solution[/i]: First, we will prove the inequality

$\sqrt{\frac{a}{m_a}}+\sqrt{\frac{b}{m_b}}+\sqrt{\frac{c}{m_c}}>2\sqrt2$.

In fact, we have $m_a\leq\frac12\left(b+c\right)$, $m_b\leq\frac12\left(c+a\right)$ and $m_c\leq\frac12\left(a+b\right)$ (these are well-known and easy inequalities; they are even true with an < instead of a $\leq$; a short proof of them was given in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=252755#p252755]http://www.mathlinks.ro/Forum/viewtopic.php?t=21462 post #16[/url]). Also, we have $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2$ (this is also an old inequality; see http://www.mathlinks.ro/Forum/viewtopic.php?t=48825 , http://www.mathlinks.ro/Forum/viewtopic.php?t=50390 , http://www.mathlinks.ro/Forum/viewtopic.php?t=48311 , http://www.mathlinks.ro/Forum/viewtopic.php?t=46428 , http://www.mathlinks.ro/Forum/viewtopic.php?t=27481 ). Thus,

$\sqrt{\frac{a}{m_a}}+\sqrt{\frac{b}{m_b}}+\sqrt{\frac{c}{m_c}}\geq\sqrt{\frac{a}{\frac12\left(b+c\right)}}+\sqrt{\frac{b}{\frac12\left(c+a\right)}}+\sqrt{\frac{c}{\frac12\left(a+b\right)}}$
$=\sqrt2\cdot\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)>\sqrt2\cdot 2=2\sqrt2$.

Thus, the inequality

$\sqrt{\frac{a}{m_a}}+\sqrt{\frac{b}{m_b}}+\sqrt{\frac{c}{m_c}}>2\sqrt2$

is proven. Now, applying this inequality to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59415]the medians triangle[/url] of triangle ABC, whose sidelengths are $m_a$, $m_b$, $m_c$ and whose medians are $\frac34a$, $\frac34b$, $\frac34c$, we get

$\sqrt{\frac{m_a}{\frac34a}}+\sqrt{\frac{m_b}{\frac34b}}+\sqrt{\frac{m_c}{\frac34c}}>2\sqrt2$.

After multiplication by $\frac{\sqrt3 }{2}$, this becomes $\sqrt{\frac{m_a}{a}}+\sqrt{\frac{m_b}{b}}+\sqrt{\frac{m_c}{c}}>\sqrt6$, and the problem is solved.

By the way, this inequality:

[quote="arqady"]In Russian montly "Kvant" 1976, number 7, page 16 :
prove $\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\geq\frac{3\sqrt{3}}{2}$
author: V. Matizen.[/quote]

is a particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=21016 .

  Darij

[quote="darij grinberg"]
Now, the [i]solution[/i]: First, we will prove the inequality

$\sqrt{\frac{a}{m_a}}+\sqrt{\frac{b}{m_b}}+\sqrt{\frac{c}{m_c}}>2\sqrt2$.
[/quote]
My proof:
$\sqrt\frac{a}{m_a}=\sqrt2\cdot\sqrt[4]{\frac{1}{\frac{2a^2+2b^2-a^2}{a^2}\cdot1\cdot1\cdot1}}\geq\frac{4\sqrt2}{3+\frac{2b^2+2c^2-a^2}{a^2}}=2\sqrt2\cdot\frac{a^2}{a^2+b^2+c^2}.$
[quote="darij grinberg"]
By the way, this inequality:
[quote="arqady"]In Russian montly "Kvant" 1976, number 7, page 16 :
prove $\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\geq\frac{3\sqrt{3}}{2}$
author: V. Matizen.[/quote]
is a particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=21016 .
  Darij[/quote]
My proof:
${\frac{m_a}{a}=\frac{\sqrt3}{2}\sqrt{\frac{2b^2+2c^2-a^2}{3a^2}}=\frac{\sqrt3}{2}\cdot\sqrt{\frac{1}{\frac{3a^2}{2b^2+2c^2-a^2}\cdot1}}\geq\frac{\sqrt3}{2}\cdot\frac{2}{\frac{3a^2}{2b^2+2c^2-a^2}+1}}=$
$=\frac{\sqrt3}{2}\cdot\frac{2b^2+2c^2-a^2}{a^2+b^2+c^2}.$
Rozenberg Michael (arqady). :)


Solve for the equation $ \sin (\sin x) \equal{} \cos (\cos x)$

Here's my nearly done solution:
[hide]
I first substituted $ x \equal{} \frac {\pi}{2} \minus{} t$.
So, the equation would now be: $ \sin (\cos t) \equal{} \cos (\sin t)$
From here, $ \sin t \equal{} \arccos (\sin (\cos t))$.
If we draw some triangles, we would arrive at $ \sin t \equal{} \frac {\pi}{2} \minus{} \cos t$

eventually, $ \sin x \plus{} \cos x \equal{} \frac {\pi}{2}$

Is this right?
[/hide]
thanks!:)

Seems right...
Of course, since $ \sin x \plus{} \cos x \leq \frac{\sqrt{2}}{2}<\frac{\pi}{2}$, this implies there are no solutions...

You should probably be more careful.

$ \sin x\equal{}\sin y$ doesn't imply that $ x\equal{}y$. In particular $ x\minus{}y$ is a multiple of $ 2\pi$ so you may want to include this.

[quote="Brut3Forc3"]Seems right...
Of course, since $ \sin x \plus{} \cos x \leq \frac {\sqrt {2}}{2} < \frac {\pi}{2}$, this implies there are no solutions...[/quote]
But isn't $ \frac{\pi}{2} <2$? This means that there could be solutions because the maximum value of $ \sin x \plus{} \cos x \equal{}2$?

A [i]slightly [/i]more straightforward approach would be to solve $ \sin A \equal{} \cos B$, and then substitute in $ \sin x, \; \cos x$ for $ A, \; B$, respectively.

Is correctly "for any real $ x$ we have $ \sin x\plus{}\cos x\le\sqrt 2$ " ...

Yes. $ |\sin x \plus{} \cos x| \equal{} \sqrt{\sin 2x \plus{} 1}\leq \sqrt 2$.

So basically, this has no solutions? i'm actually considering this as an approach to solve russia 1995 trig equation.

[quote="Brut3Forc3"]Seems right...
Of course, since $ \sin x \plus{} \cos x \leq \frac {\sqrt {2}}{2} < \frac {\pi}{2}$, this implies there are no solutions...[/quote]

No, $ \sin x \plus{} \cos x \leq \sqrt 2$.


[quote="nanren"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1,$ prove that

$ \frac {1}{a^2 \plus{} b} \plus{} \frac {1}{b^2 \plus{} c} \plus{} \frac {1}{c^2 \plus{} a} > \frac {13}{2}.$[/quote]Using the arithmetic-mean — geometric-mean inequality,

$ \sum_{cyc}{\frac {1}{a^2 \plus{} b}} \plus{} \sum_{cyc}{\frac {(3 \minus{} b \plus{} 3c)^2(a^2 \plus{} b)}{(3bc \plus{} 3ca \plus{} 3ab \plus{} 1)^2}} > 2\sum_{cyc}{\frac {3 \minus{} b \plus{} 3c}{3bc \plus{} 3ca \plus{} 3ab \plus{} 1}}$

$ \equal{} \frac {22}{3bc \plus{} 3ca \plus{} 3ab \plus{} 1},$

it remains to prove that

$ 44(3bc \plus{} 3ca \plus{} 3ab \plus{} 1) > 13(3bc \plus{} 3ca \plus{} 3ab \plus{} 1)^2 \plus{} 2\sum_{cyc}{(3 \minus{} b \plus{} 3c)^2\left(a^2 \plus{} b\right)}.$

This comes down to the following quartic homogeneous inequality:

$ 44(a \plus{} b \plus{} c)^2\left(a^2 \plus{} b^2 \plus{} c^2 \plus{} 5bc \plus{} 5ca \plus{} 5ab\right)$

$ \minus{} 13\left(a^2 \plus{} b^2 \plus{} c^2 \plus{} 5bc \plus{} 5ca \plus{} 5ab\right)^2 \minus{} 2\sum_{cyc}{(3a \plus{} 2b \plus{} 6c)^2\left[a^2 \plus{} b(a \plus{} b \plus{} c)\right]}$

$ \equiv F(a,b,c) \equal{} F(a,a \plus{} s,a \plus{} t)$

$ \equal{} 12a^4 \plus{} 16(s \plus{} t)a^3 \plus{} 8(16s^2 \minus{} 13st \plus{} 16t^2)a^2 \plus{} 2(51s^3 \plus{} 46s^2t \minus{} 71st^2 \plus{} 51t^3)a$

$ \plus{} \frac {(2s \minus{} t)^2(4s^2 \plus{} 84st \plus{} 19t^2) \plus{} (2s^2 \minus{} t^2)^2}{4} > 0,$

which is clearly true for $ a \equal{} \min\{a,b,c\}.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=120780


articles inequalities quadratics algebra

Some article names contain misprints (like Chebyshevs inequality, which should actually be Chebyshev's inequality, etc.) Is there any easy way to edit them? The standard edit option doesn't seem to allow you to edit the article title  :?

That's because the article title is an intrinsic and immutable property of the page. I assume you'd have to actually move the contents to the uncreated page titled Chebyshev's Inequality (http://www.artofproblemsolving.com/Wiki/index.php?title=Chebyshev%27s_inequality&action=edit).  I'd also suggest making the original page a redirect after that.

I've been gradually cleaning up article names.  Two common mistakes that people should avoid are as follows:

1. Don't just capitalize the first letter in every word.  Only do so when the article name is a proper noun.  Otherwise, users will only be able to link with the same incorrect capitalization.  I've changed about 10 of these already and don't want to change 100 more.

2. Don't pluralize entries without good reason.  I just changed "quadratic equations" to quadratic equation.  With internal links, you can still write "quadratic equations" and have it link to the appropriate articles by using this syntax:
[[quadratic equation | quadratic equations]]
where the first term is the article name, and the second shows up in the text.

You can even write [[quadratic equation]]s and it will still be a link with "quadratic equations". ;)

But how can we redirect the pages? For example Rearrangement inequality to Rearrangement Inequality.

See the topic called "Two names for the same topic". It says:

"#REDIRECT [[American Regions Mathematics League]] "

Using that as the content of any article that would redirect to the article titled American Regions Mathematics League

Edit (Gah, need to think out my posts): You might also want to wait for the ongoing capitalization issue to settle before you set it one way or another.


Prove, that equation $x^x=y^3+z^3$ has infinitely many solutions in positive integers

Set $y=2^k , z= 2^k$, then we need $x^x=y^3+z^3 = 2^{3k} \cdot (1^3+1^3) = 2^{3k+1}$.
Assume $x=2^m$, then we need $2^{2^m \cdot m } = x^x = 2^{3k+1}$, thus $3k+1 = 2^m \cdot m$.
Now any $m \equiv 4 \mod 6$ (and also $m \equiv 5 \mod 6$) fulfills $2^m \cdot m \equiv 1 \mod 3$, thus for these $m$ such $k$ exists, giving us infinetely many solutions.
This procedure should work for most other starting terms, too, as long as $3 \nmid y+z$ and best when $y^3+z^3$ is a prime power.


function
calculus
derivative
limit
real analysis
real analysis unsolved

Hi everyone,

Is it possible to find a sequence of differentiable function $ \{f_n\}$ over $ (0,1)$ which converges pointwise
to $ f$, but such that the sequence of derivatives $ \{f'_n\}$ over $ (0,1)$ converges pointwise to a function
$ g$, with $ g\not \equal{} f'$?

Moreover, what is the simplest example of a series $ \sum a_n$ that converges but such that $ \sum a_n^2$ diverges?

Thank you.

It's easy enough to construct examples of differentiable $ f_n\to f$ for which perhaps the sequence $ f_n'$ diverges at some point, or for which $ f'$ doesn't exist somewhere. To do exactly what you asked is going to be a bit trickier. I'll have to think about that.

As for the series: you can't have that for $ a_n>0$ or absolute convergence, so the example must involve cancellation and conditional convergence. I'd go with $ a_n\equal{}\frac{(\minus{}1)^n}{\sqrt{n}}$ as my example.

Thank you for your answer. For the first problem, if the interval were $ [0,1]$ instead of $ (0,1)$  then the sequence is

\[ f_n(x)\equal{}\frac{x^n}{n}\]

which converges to $ f\equal{}0$ on $ [0,1]$. Now $ \{f'_n\} \equal{}\{x^{n\minus{}1}\}$ converges to $ g\equal{}0$ for $ x\not\equal{}1$ and
to $ g\equal{}1$ for $ x\equal{}1$. Obviously $ g\not\equal{}f'$.

But, what about the case in which the interval is $ (0,1)$?

For the second problem, what does your series converge to?

[quote="vnca1"]For the second problem, what does your series converge to?[/quote]  No reason to think the value it converges to is particularly nice, but it's easy to see it converges by the alternating series test, and easy to see that the square series diverges.

The first question is not hard as stated: just take some differentiable function $ f$ that has derivative $ 1$ at $ 0$ and vanishes outside $ [\minus{}1,1]$ and put $ f_n(x)\equal{}\frac 1nf(nx)$. Then $ f_n\to 0$ on the entire real line but $ f_n'(0)\to 1$ (though $ f_n'(x)\to 0$ for all $ x\ne 1$).

Let's make the question more interesting. Does there exist a sequence of continuously differentiable functions $ f_n$ such that $ f_n\to 0$ pointwise on the entire real line and $ f'_n$ converge pointwise to a non-vanishing function $ g$ on the entire real line. ;)

I'm sorry but I don't understand how everything you talked about is related to the first question.

I'll be a little more formula-driven than fedja, although this is essentially his example. (Just without the compact support.)

Let $ f_n(x)\equal{}\frac{x}{1\plus{}n^2x^2}$ on $ \mathbb{R}.$ Then $ \lim_{n\to\infty}f_n(x)\equal{}0$ for all $ x\in \mathbb{R}.$

In fact, we even have that $ f_n\to0$ uniformly, since $ |f_n(x)|\le\frac1{2n}.$

So $ f_n\to f,$ where $ f(x)\equal{}0$ and thus $ f'(x)\equal{}0$ everywhere.

In the meantime, $ f_n'(x)\equal{}\frac{1\minus{}n^2x^2}{(1\plus{}n^2x^2)^2}$ Taking pointwise limits,

$ \lim_{n\to\infty}f_n'(x)\equal{}\begin{cases}1&x\equal{}0\\0&x\ne 0\end{cases}.$

So that is almost what you asked for - a sequence of functions whose derivatives converge, but to something other than the derivative of the limit.

The only objection is that you wanted this on the interval $ (0,1).$ All right then, just let $ g_n(x)\equal{}f_n\left(x\minus{}\frac12\right).$ Then the problematical point lies at $ \frac12,$ in the middle of your interval.

Thank you, I see it now. This is very clever.


Prove that there don't exist rational numbers $ x,\ y$ such that $ x^2\plus{}y^2\equal{}3$.

Then we'd have $ a^2 \plus{} b^2 \equal{} 3c^2$ for integers. So a and b must both be divisible by 3. So a=3n, b=3m, then $ 3n^2\plus{}3m^2\equal{}c^2$. Then c must be divisible by 3 and we have a contradiction by infinite descent.

Sorry, I can't understand your solution. :(

$ a^2 \equiv 0, 1 \bmod 3 \implies a^2 \plus{} b^2 \equiv 0, 1, 2 \bmod 3$ and $ 0 \bmod 3$ is possible if and only if $ a^2, b^2 \equiv 0 \bmod 3$, so $ a$ and $ b$ are both divisible by $ 3$.  Equivalently, $ \minus{}1$ is not a quadratic residue $ \bmod 3$.


trigonometry
analytic geometry
complex numbers

[b]For how many real values of $ K$ will $ (2 \plus{} Ki)^3$ be a real number?[/b]
[hide="my solution"]the trigonometric representation of $ 2 \plus{} Ki$ is $ r\cdot (\cos \theta \plus{} i\sin \theta)$ for now we can ignore the $ r$
$ \cos 3\theta \plus{} i\sin \theta \equal{} (\cos \theta \plus{} i\sin \theta)^3 \equal{} \cos^3\theta \plus{} 3i\cos^2\theta \sin \theta \minus{} 3\cos \theta \sin^2\theta \minus{} i\sin^3\theta$
since we want the solution to be all real,
$ 3i\cos^2\theta\sin\theta \minus{} i\sin^3\theta \equal{} 0$
we can solve for theta to get $ \cos\theta \equal{} \frac {1}{2} \Longrightarrow \theta \equal{} 60$ deg. $ z \equal{} 2 \plus{} Ki \equal{} (r,60)$ in the complex plane there is only one coordinate with those polar coordinates that will result in a completely real result. thus, there is only one real value of K.
does that work? i know i have the right answer, but is my solution valid?[/hide]

*sorry  :oops:

You don't really need a trigonometric representation for such small powers.. expanding would do just fine.

[quote="t0rajir0u"]Your question is incomplete.  Is it "for how many real values of $ k$ is $ (2 \plus{} ki)^3$ real?  You don't need de Moivre's; just expand.[/quote]

$ (2 \plus{} ki)^3 \equal{} 8 \plus{} 12ki \minus{} 6k^2 \minus{} k^3i \equal{} (8 \minus{} 6k^2) \plus{} k(12 \minus{} k^2)i$

So we simply find the values of $ k$ at which the imaginary part vanishes, which are $ k \equal{} 0, k \equal{} 2\sqrt {3}, k \equal{} \minus{} 2\sqrt {3}$.


Determine (with proof) the largest integer $ n$ such that $ n$ is divisible by all positive integers less than $ \sqrt[3]{n}$.


An integer sequence is defined by \[a_{n}= 2a_{n-1}+a_{n-2}\]
and $a_{0}=0, a_{1}=1$

Prove that $2^{k}| a_{n}\iff 2^{k}| n$

does this help at all..the fact that:
$a_{n}=2^{n}-1$

That my friend, is completely wrong! Please don't post if it is just a wild guess.

:blush:  that generalization is wrong i didnt know that
$a_{0}= 0$

..my bad

EDIT: Ya i clearly misread the sequence formula..it doesnt make sense how i got that....my apologies

[hide="This might be of some help..."]
I'm not sure how to proceed next, but the recurrence has the explicit formula of $a_{n}= \frac{ (1+\sqrt 2)^{n}-(1-\sqrt 2)^{n}}{2\sqrt 2}$.[/hide]

By means of generating functions, e.g., we obtain \[a_{n}= \frac{(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}}{2\sqrt{2}}\]By Newtons binomial, we've got \[\begin{aligned}(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}& = \sum_{i=0}^{n}(\binom{n}{i}(\sqrt{2})^{i}-\binom{n}{i}(-\sqrt{2})^{i}\\ & = \sum_{\begin{array}{c}i=0 \\ i \textrm{ even}\end{array}}^{n}(\binom{n}{i}(\sqrt{2})^{i}-\binom{n}{i}(-\sqrt{2})^{i})+\sum_{\begin{array}{c}i=0 \\ i \textrm{ odd}\end{array}}^{n}(\binom{n}{i}(\sqrt{2})^{i}-\binom{n}{i}(-\sqrt{2})^{i}) \\ & = 0+\sum_{\begin{array}{c}i=0 \\ i \textrm{ odd}\end{array}}^{n}2 \binom{n}{i}(\sqrt{2})^{i}\\ & = 2\sqrt{2}\sum_{j=0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n}{2j+1}2^{j}\end{aligned}\]So \[a_{n}= \sum_{j=0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n}{2j+1}2^{j}\]How to proceed...  :P I'm not even sure if this is the right way to tackle it  :maybe:

Well, I have to go to bed soon, I think it can be done by induction:

$a_{n}-a_{n-2}=2a_{n-1}$, so terms that are 2 terms apart differ by a multiple of 2.

$a_{n}=2a_{n-1}+a_{n-2}=4a_{n-2}+2a_{n-3}+2a_{n-3}+a_{n-4}$
$\implies a_{n}-a_{n-4}= 4(a_{n-2}+a_{n-3})$
, so terms that are 4 terms appart differ by a multiple of 4.

If you can induct this and show that terms that are $2^{k}$ apart differ by a multiple of $2^{k}$, and you know that $0$ is a multiple of all $2^{k}$, so you can have $2k|n\implies2k|a_{n}$.

Kurt Gödel: Come on, you are very close for 1 part of the question.  :wink:

The Zuton Force: Can you please elaborate your solution and write the general form? (You only treated the cases where the difference is $2$ or $4$) Splendid idea though  :)

Zuton Force has revealed a glimpse of the idea... Why doesn't anyone notice how funny the statement is?
Take a close look at that recursion... do you see why $a_{n}= a_{k}a_{n-k+1}+a_{k-1}a_{n-k}$? :)

It is given for $k=2$; but substituting the formula in itself gives the next $k$. Now the induction step is in hand reach.  :wink:


geometry
quadratics
Pythagorean Theorem
algebra
system of equations

Let D= a^2+b^2+c^2, where a and b are consecutive integers and c=ab. Then Sqrt(D) is
(A) Always an even integer
(B) Sometimes an odd integer, sometimes not
(C) Always an odd integer
(D) Sometimes rational, sometimes not
(E) Always irrational

I will try to show my work. I have three equations in the beginning. The two listed in the question and a+1=b. I can use the two equations above to substitute into the third equation to get a^2+(a+1)^2+(ab)^2=D. I don't see how factoring helps. I think I got to the fact that D always is odd but can not figure out what the sqrt(D) must always be.

Another problem that I don't even know how to attack is:
Quadrilateral ABCD is inscribed in a circle with side AD, a diameter of length 4. If sides AB and BC each have lenght 1, then CD has length
(A)7/2 (B) 5 Sqrt(2)/2 (C)Sqrt(11) (D) Sqrt(13) (E)2sqrt(3)

Let's write b = a+1 and c = a^2+a. Then a^2+b^2+c^2 = a^2+(a+1)^2+(a^2+a)^2 = a^4+2a^3+3a^2+2a+1 = (a^2+a+1)^2, so sqrt(D) = a^2+a+1. Now if a is even, then a^2+a+1 is odd. But if a is odd, then a^2+a+1 is also odd, so sqrt(D) is always an odd integer (C).

Thanks except I don't know how you spotted the factorization of a^4+2a^3+3a^2+2a+1 to get
(a^2+a+1)^2. Also, I hope the geometry problem is solvable without the picture.

You don't really need that.

Leave b as it is for a:^2:b:^2:

So you get

a:^2:+(a+1):^2:+a:^2:b:^2: = 2a:^2:+2a+1+a:^2:b:^2:
= 2(a:^2:+a) + a:^2:b:^2: + 1

Look at that. It's trivial that the first part is even. The second part is even, because either a or b is even, thus the square is. Adding one to an even number is always odd.

Here is a hint for the geometry problem:

All you need to use is the Pythagorean Theorem and Ptolemy's Theorem. Try to somehow use them to get a system of equations that you can solve.

Can you give me a bigger hint then saying Ptolemy's theorem and pythagorean theorem because I still can't find the solution.

Well, there is a right angle at ABD since it is inscribed in a circle and AD is the diameter, so we can apply the Pythagorean Theorem to find BD = Sqrt(15). Now using Ptolemy's Theorem would result in an equation with two variables AC and CD, both of which we do not know right now. Is there another way we can get an equation including AC and CD so that we can solve both of them simultaneously?

Sorry, I know I am giving away how bad I am in math but I still don't get what equations I can set up.

Well, ACD is also a right angle since AD is the diameter and the angle is inscribed in the triangle. Therefore, we can establish the Pythagorean Theorem to obtain

AC^2 + CD^2 = AD^2 = 4^2 = 16

Applying Ptolemy's Theorem to ABCD, we have

AB*CD + BC*AD = BD*AC
1*CD + 1*4 = Sqrt(15) * AC
CD = Sqrt(15) * AC - 4

Substituting, we have

AC^2 + (Sqrt(15) * AC - 4)^2 = 16

Simplifying and solving the quadratic, we obtain AC = 0 or AC = Sqrt(15)/2. It is clearly the latter since AC must be greater than 0. Substituting again, we have

CD = Sqrt(15) * (Sqrt(15)/2) - 4 = 15/2 - 4 = 7/2

So the answer is (A).

Thanks, know I get it. It was hard for me to realize that both angle ABD and BCD were right angles.