How many four-digit numbers greater than 2999 can be formed such that the product of the middle two digits exceeds 5?

Case 1: 2nd digit is 1.

Case 1.1: 3rd digit is 6

$ 7 \cdot 1 \cdot 1 \cdot 10 \equal{} 70$ 4-digit numbers greater than 2999 with middle digits (1,6).

Case 1.2: 3rd digit is 7

$ 7 \cdot 1 \cdot 1 \cdot 10 \equal{} 70$ again.

Case 1.3: 3rd digit is 8

$ 7 \cdot 1 \cdot 1 \cdot 10 \equal{} 70$ once more.

We see so far the each possible combination of middle digits yields 70 more numbers. Thus we just need to find the total number of middle digit combinations that work.

2nd digit is 1: We can have $ 6 \sim  9$ for the 3rd digit, giving us 4 ways.

2nd digit is 2: We can have $ 3 \sim  9$, giving us 7 ways.

2nd digit is 3: We have $ 2 \sim 9$, giving us 8 ways.

2nd digit is 4: We can have $ 2 \sim  9$, giving us 8 ways.

2nd digit is 5: We can have $ 2 \sim  9$, giving us 8 ways.

2nd digit is 6: $ 1 \sim  9$, giving us 9 ways.

For each number thereafter, there will be 9 ways. So our answer is $ (4 \plus{} 7 \plus{} 3 \cdot 8 \plus{} 4 \cdot 9) \times 70 \equal{} \boxed{4970}$.

I think there is a faster way involving complementary counting...

@above yes

[hide="the same thing but faster"]

since we only care about the middle two digits, let's just find the valid possibilities for the middle digits then find out how many ways we can assign the other two.

if the two numbers are different, our possibilities are:

0 and any number 1-9 (9 numbers)
1 and any number 2-5 (4 numbers)
since the two numbers can be flipped, we have $(9+4) \cdot 2 = 26$

if they are the same, we can have $\{0,0\}, \{1,1\}, \{2,2\}$ for 3 arrangements.
since we have $26+3=29$ INVALID arrangements, there are $10 \cdot 10 - 29 = 71$ VALID arrangements.

from here, there are 7 possibilities for the first digit (3-9), and 10 for the last, so we have

$71 \cdot 7 \cdot 10 = \boxed{4970}$


calculus
integration
trigonometry
Euler
function
complex analysis
limit

using complex variables, or otherwise, show, for $ k>1$,  :maybe:

$ \textbf{(i)}\;\;\;\int_{0}^\infty\;\sin \left( y^{k}\right) \;\textbf dy\;=\;\boxed{\sin\left( \frac{\pi}{2k}\right)\cdot\Gamma{\left(\frac{k+1}{k}\right)}}$

$ \textbf{(ii)}\;\;\;\int_{0}^\infty\;\cos \left( y^{k}\right) \;\textbf dy\;=\;\boxed{\cos\left( \frac{\pi}{2k}\right)\cdot\Gamma{\left(\frac{k+1}{k}\right)}}$

Let the first integral Be $ A$ and the second $ B$.
Using euler formula and since $ y$ is real.

$ {\left(\cos \left(y^{k}\right)+i \sin \left(y^{k}\right)\right) dy=e^{i y^{k}}dy}$

$ {\left(A=\Im\left(\int_{0}^{\infty }e^{i y^{k}}\, dy\right)\right)\land \left(B=\Re\left(\int_{0}^{\infty }e^{i y^{k}}\, dy\right)\right)}$

$ {\left(y=t^{1/k}\right)\land \left(k=\frac{1}{s+1}\right)\land (-1<s<0)}$

$ {e^{i y^{k}}dy= (s+1) e^{i t}t^{s}dt}$

This is very simple to demonstrate

$ {\int_{0}^{\infty }\frac{t^{s}}{e^{t \alpha }}\, dt=\alpha^{-s-1}\Gamma (s+1)}$

Using

$ {(\alpha =-i)\land \left(s=\frac{1-k}{k}\right)}$

Making a substituition and takin the imaginary and the real part it's donne

[u]Outline[/u] ---> needs work.

Let $ \gamma$ be a contour starting on the origin moving to $ R$ and going counterclockwise with an angle of $ \pi/2k$ in a circular arch.

Consider the function $ f(z) = e^{iz^{k}}$ this function is holomorphic in an open set containing this contour. So $ \oint_{\gamma}f(z) dz = 0$.

Now, doing this peicewise we get,
$ \int_{0}^{R}\exp (ix^{k}) dx+\int_{0}^{\pi/2k}\exp \left( iR^{k}e^{ki \theta}\right) i Re^{i\theta}d \theta-\int_{0}^{R}\exp (it^{k}e^{i\pi/2}) e^{i\pi/2k}dt =0$

Now show that the middle term goes to 0.

[quote="sylow_theory"]
Now, doing this peicewise we get,
$ \int_{0}^{R}\exp (ix^{k}) dx+\int_{0}^{\pi/2k}\exp \left( iR^{k}e^{ki \theta}\right) i Re^{i\theta}d \theta-\int_{0}^{R}\exp (it^{k}e^{i\pi/2}) e^{i\pi/2k}dt =0$
[/quote]
I applogize, I didnt have so much time, I will try to be more explicit.

We will sent $ R\to \infty$. First look at the [b]third[/b] integral:
$ \int_{0}^{\infty}\exp \left( it^{k}e^{i\pi /2}\right) e^{i\pi/2k}dt = e^{i\pi/2k}\int_{0}^{\infty}e^{-t^{k}}dt$
Let $ u=t^{k}$ to get:
$ \frac{1}{k}e^{i\pi/2k}\int_{0}^{\infty}e^{-u}u^{k/(k-1)}dt = e^{i\pi / 2k}\cdot \left( \frac{1}{k}\Gamma \left( \frac{1}{k}\right) \right) = e^{i\pi /2k}\Gamma \left( \frac{k+1}{k}\right)$

Now look at the [b]second[/b] integral:
$ \int_{0}^{\pi / 2k}\exp \left( iR^{k}e^{ki \theta}\right) i Re^{i\theta}d \theta$
Note that,
$ \left| \exp( iR^{k}e^{ki\theta}) i Re^{i\theta}\right| = \left| R\exp \left( R^{k}(i\cos k\theta-\sin k\theta )\right) \right|= R \exp \left(-R^{k}\sin k\theta\right) \leq R\exp \left(-2kR^{k}\theta /\pi\right) *$
But then that means,
$ \left| \int_{0}^{\pi / 2k}\exp \left( iR^{k}e^{ki \theta}\right) i Re^{i\theta}d \theta \right| \leq \int_{0}^{\pi/2k}R\exp \left(-2kR^{k}\theta /\pi\right) d\theta = \frac{\pi}{2kR^{k-1}}\left( 1-e^{-R^{k}}\right)$
But,
$ \frac{\pi}{2kR^{k-1}}\left( 1-e^{-R^{k}}\right) \to 0$ as $ R\to \infty$.
Which means that,
$ \int_{0}^{\pi / 2k}\exp \left( iR^{k}e^{ki \theta}\right) i Re^{i\theta}d \theta \to 0$ as $ R\to \infty$.

Look at the original integral expression and take $ \lim_{R\to \infty}$ the above work shows that:
$ \int_{0}^{\infty}\exp (ix^{k}) dx = e^{i\pi / 2k}\Gamma \left( \frac{k+1}{k}\right)$
So that means,
$ \int_{0}^{\infty}\cos x^{k}dx+i \int_{0}^{\infty}\sin x^{k}dx = \cos \left( \frac{\pi}{2k}\right) \Gamma \left( \frac{k+1}{k}\right)+i \sin \left( \frac{\pi}{2k}\right) \Gamma \left( \frac{k+1}{k}\right)$
Equate real and imaginary parts.
Q.E.D.

By the way, did you get this killer integral from? Seriously you got to tell me.

$ *)$Because $ \frac{2}{\pi}\leq \frac{\sin \theta}{\theta}\leq 1$ on $ (0,\pi/2]$.

[quote="sylow_theory"]
Let $ u=t^{k}$ to get:
$ \lim_{R\to\infty}\;\frac{1}{k}\;e^{\frac{\imath\pi}{2k}}\int_{0}^{R^{k}}e^{-u}\;\boxed{u^{\frac{1}{k}-1}}\;du = e^{\frac{\imath\pi}{2k}}\left[ \frac{1}{k}\;\Gamma \left( \frac{1}{k}\right) \right] \;=\; e^{\frac{\imath\pi}{2k}}\;\Gamma \left( \frac{k+1}{k}\right)$

[/quote]

hey cool, sylow_theory (you really must like the [i]gamma [/i]function a lot... LOL !)   :coolspeak:

( PS: i made a slight/minor correction for $ \textbf I_{3}$ above... see "boxed" expression )

[ also, a nice job with showing $ \textbf I_{2}\to 0$ in the limit $ R\to\infty$]

thank you. :D  :P

$ I=\int_{0}^\infty \exp\left(-i x^{k}\right)$

Substituting:

$ u=ix^{k}$
$ x\frac{du}{dx}=kix^{k}=ku$

$ I=\left( \int_{0}^\infty \exp(-u)u^{1/k-1}\right) \frac{(-i)^{1/k}}{k}$

$ =\frac{1}{k}\Gamma[\frac{1}{k}]\exp\left( \frac{-i\pi}{2k}\right) = \Gamma[(k+1)/k]\exp\left( \frac{-i\pi}{2k}\right)$

Comparing Real and Imag parts gives the desired relations.


geometry
circumcircle
radical axis
power of a point
geometry proposed

Let $ ABC$ be a triangle.
There exist circle $ \Gamma_{A}$ such that through $ B, C$ and tangent to incircle of $ ABC$ at $ A_{1}$.
Similarly, we can define $ B_{1}$ and $ C_{1}$.
Then, prove that $ AA_{1}, BB_{1}, CC_{1}$ meet one point.

I think it's very nice problem if using step of solution of G7 :lol:
I think it can be posted before.

$ (O)$ is the triangle circumcircle and $ (I)$ its incircle, touching $ BC, CA, AB$ at $ D, E, F.$ Common tangent of $ (I), \Gamma_A$ at $ A_1$ cuts $ BC$ at the radical center $ P$ of $ \Gamma_A, (I), (O).$ Likewise, common tangents of $ (I), \Gamma_B$ and $ (I), \Gamma_C$ at $ B_1, C_1$ cut $ CA, AB$ at radical centers $ Q, R$ of $ \Gamma_B, (I), (O)$ and $ \Gamma_C, (I), (O).$ $ PQR$ is therefore radical axis of $ (I), (O).$ At the same time, $ DA_1, EB_1, FC_1$ are polars of collinear $ P, Q, R$ WRT $ (I),$ concurrent at the pole $ S$ of $ PQR,$ inside of $ (I)$ (Since $ IS \perp PQR,$ $ S \in OI \perp PQR.$)

Since $ S$ is inside of $ (I),$ a central projection exists, which takes the circle $ (I)$ into a circle $ (S)$ with center $ S.$ All projected points will be labeled with an asterisk. $ A_1^*, B_1^*, C_1^*$ become diametrically opposite to $ D^*, E^*, F^*$ in $ (S),$ $ A^*A_1^*, B^*B_1^*, C^*C_1^*$ become Nagel lines of $ \triangle A^*B^*C^*$ concurrent at its Nagel point $ M^*$, hence the lines  $ AA_1, BB_1, CC_1$ are also concurrent at a point $ M.$

Thank you yetti :lol:
I like these proofs that proof doesn't use trigonometric or any complex calculating.

My solution is using cevian nest theorem.
If $ DA_{1}, EB_{1}, FC_{1}$ are meet one point. Then, $ AA_{1}, BB_{1}, CC_{1}$ are concurrent, too.
But it's easy to show $ DA_{1}, EB_{1}, FC_{1}$ are concurrent by any strategy.

Really Thank you yetti^^


AMC
AMC 10
geometry
probability
number theory

Since AoPS has an entire forum for Virginians, and it is never used, I figured I might as well make a thread that will probably not be seen by anyone ever, but hey w/e.

This is for incoming freshmen or people who at some point hope to be incoming freshmen, with questions about math at TJ, or TJ in general. You can post questions here or probably get a quicker reply if you PM me, although I'll check this thread from time to time and maybe other people will too.

You can also just post here to let me/us know you'll be coming. I know a few of the incoming freshmen already, mostly siblings of past TJers, but I try to make it my business to know everyone.

~JSteinhardt

just wondering how long it'll take you to see this jacob  :)

I'm an incoming freshmen and I've had a lot of questions about math competitions and TJ in general.

Could you explain the eighth period and block scheduling?

Are there alot of group projects at TJ?

How is going to TJ a different experience than going to school in middle school? (For example, elementary school was a lot different from middle school
since in elementary school, I didn't have a locker, I didn't have to wear gym clothes, I had "home room teacher" who taught most things rather than having seven or so different teachers, and there was a magical 30 minute block of the day called recess)

What math clubs and activities are there during the eighth period?

How would I be able to join the TJ Varisty Math Team?

Could you explain lettering to me?

What are some other recommended math or science related clubs at TJ?

What are some of the math competitions students can take at TJ? How will find out how to take them?

How can I take both the AMC 10 and AMC 12?
OR
How can I take the AMC 10A and AMC10 B in one year?

What middle school did you go to?

What do you recommend a middle schooler applying to TJ to do? My younger brother might apply.

Thank you for your time.

I'm an incoming freshmen to TJ and I wanted to ask you a few questions about math and TJ in general.

I read that high school math contests test on "secondary school math topics" as it sometimes says on Aops wiki but I don't find all of these topics such as arithmetic, algebra, counting, geometry, number theory, and probability covered in school. Are math contests and math in the classroom very different?

What are group projects in math classes like?

Do I get to choose whether to take the AMC 10 or AMC 12? If I do, which should I take and why?

Could you explain the Big Sibs Program?

Are you going to be a big sib?

Before when I asked you questions, you said that if you were a Freshman, you'd do Math Team Wednesday A&B, Freshman Computer Team Friday A, and Senior Computer Team Friday B.

If freshmen can be in the Senior Computer Team, what does the Senior in "Senior Computer Team" mean?

Also, last time when I asked:
How would I be able to join the TJ Varisty Math Team?
You said:
"... just show up to meetings. If you want to be automatically registered for the mailing list and everything, you should come to the interest meeting which is during one of the first eighth periods of the year... "

It seems like if I try going to a few clubs to see what I like, I might miss interest meetings for the clubs I go to. What can I do to prevent that from happening?- or does it matter?

Will being in more than one club hurt my chances to do well in the clubs I choose to be in?

What did you do for the really big research project in IBET that you mentioned last time I asked you questions ?

What were some of the biggest changes in your life and new problems you faced when you went to high school? How do you think I should prepare for them and handle these challenges?

What do you think has been the most difficult part about going to T.J. or just high school in general?

Thank you for your time again.

-Chris Au

my family might move to north carolina before next summer, when i'll be a freshman. I heard there's an entrance exam? just wondering how hard the entrance exam is.

Also, What percentage of each class gets into the top colleges in the U.S.?

I hope I'm not misinterpreting your post here...

TJ is considered in Virginia a Governor's School and is thus state sponsored. To be able to even apply you have to live in the section of the state it's located in. (that would be northern Virginia) So I don't think you're going to be eligible. If you are, I'm going to be pretty surprised and annoyed, because I live in central VA and I certainly would have tried to go if I could.

oops i'm moving the Virginia. nto NC

if we move we're moving, near richmond is where we're going.

I have a question about the Data Analysis packets found on the TJHSST website for math classes.  I'm going to be in algebra 2 in the 07-08 school year. Is it important for me to go through the packets before school or will they teach them during school?

So at this point it's probably fastest actually to e-mail me if you need a quick response (jacob.steinhardt@gmail.com).

TJ is in northern VA, so if you live in Richmond you are probably out of luck, but you can apply to the governor's school in that area probably (talk to MellowMelon for more information about that, as he goes there).

Most other questions have been responded to in PM, and now Haitao knows how long it takes to respond (I think I stopped checking this after it was inactive for a long time, my mistake I guess). As for jnub, if I had to make a random guess I'd say you're Keeyoung, but I'm probably wrong.

[quote=Anniboy]lol, everyone who messaged here is probably an adult by now and working in corporate america[/quote]

:rotfl: I know!

[quote=Taco12]Hi, I'm going to TJHSST![/quote]

That is cool!

[quote=Taco12]Hi, I'm going to TJHSST![/quote]

Woah orz

[quote=Taco12]Hi, I'm going to TJHSST![/quote]

Hi, going to TJHSST! I am HIA2020, and I plan on going to TJHSST this fall if all goes well!

ok I emailed the Jacob Steinhardt dude, he didn't say anything yet, just emailed him just because saying "whats up dude"
"


modular arithmetic
LaTeX
number theory
relatively prime

I have seen a demonstration to show that $ 52517\equiv \minus{} 1\pmod{6}$. I am not sure of the steps, which involves drawing a table with 0,1 and 2 which has a few columns. I know that's vague, but hopefully descriptive enough. Tell me if you know where there is a description of the method.

Thanks!

First off, although $ 52517$ is $ \minus{}1 (mod 6)$, negative numbers in mods are generally not used. In the form $ b (mod n)$, $ b$ is a nonnegative integer less than $ n$.
To find what 52517 is in $ mod 6$, we use the results for $ mod 2$ and $ mod 3$.
52517 is $ 1 (mod 2)$ and $ 2 (mod 3)$, so it is $ 5 (mod 6)$.
This may be the table you were looking for:
$ 0 (mod 2)$ and $ 0 (mod 3)$ is $ 0 (mod 6)$
$ 1 (mod 2)$ and $ 1 (mod 3)$ is $ 1 (mod 6)$
$ 0 (mod 2)$ and $ 2 (mod 3)$ is $ 2 (mod 6)$
$ 1 (mod 2)$ and $ 0 (mod 3)$ is $ 3 (mod 6)$
$ 0 (mod 2)$ and $ 1 (mod 3)$ is $ 4 (mod 6)$
$ 1 (mod 2)$ and $ 2 (mod 3)$ is $ 5 (mod 6)$

how do you make use of the first two columns to get the third?

All he did was $ 0\pmod{3}\cdot0\pmod{2}\equiv0\pmod{6}$

[quote="cf249"]although $ 52517$ is $ \minus{} 1 (mod 6)$, negative numbers in mods are generally not used.[/quote]
This is not actually true -- it's frequently convenient to use negative values in mods (it can make multiplication easier, for instance), and there is literally no good reason [i]not[/i] to use them.  Also, to get the mod symbol in LaTeX, use "\mod mn" for $ \mod mn$ or "\pmod{mn}" for $ \pmod{mn}$.

[quote]how would you use the table method for $ 52517\pmod{8}$?[/quote] You wouldn't: this method is one version of the "Applied" Chinese Remainder Theorem, and the hypothesis of the Chinese Remainder Theorem requires that you are breaking the mod into relatively prime factors.  More specifically, it's obviously not possible to reconstruct a unique value modulo 8 from a value modulo 4 and a value modulo 2.


Give a triangle ABC,with  angle A is right.M is the midpoint of the segment BC.We construct a  right angle PMQ,with $ P \in [AB],Q \in [AC]$.
Prove the inequality:$ PQ^2 \geq AP.CQ\plus{}AQ.BP$

[quote="nkht-tk14"]Give a triangle ABC,with  angle A is right.M is the midpoint of the segment BC.We construct a  right angle PMQ,with $ P \in [AB],Q \in [AC]$.
Prove the inequality:$ PQ^2 \geq AP.CQ \plus{} AQ.BP$[/quote]

Since $ M$ is the midpoint of $ BC$ hence $ MP\parallel AC\implies$ $ P$ is the midpoint of $ AB$; similarly $ Q$ is also the midpoint of $ AC$.
So $ PQCM$ and $ PBMQ$ are $ \parallel^{\text{gm}}$'s
$ \implies MP\equal{}CQ\equal{}\frac 12 AC; MQ\equal{}AP\equal{}\frac 12 AB$.
So $ PQ^2\equal{}MP^2\plus{}MQ^2\equal{}\frac 14 (AB^2\plus{}AC^2)$
And $ AP\cdot CQ\plus{}AQ\cdot BP\equal{}\frac 14 (2AB\cdot AC)$.
So the given inequality rewrites as
\[ AB^2\plus{}AC^2\geq 2AB\cdot AC\]
This is true from AM-GM. Equality holds if and only if $ \triangle ABC$ is an isosceles right triangle.
Nice problem :)

Sorry ,but when I said the angle PMQ,it meant that PMQ is a right triangle at M not P.
So my problem is probably more difficult.


suppose five pirates found a treasure .now these pirates let me name them a,b,c,d,e are having their positions according to increasing form.like a is having highest position then it is b and so on.OK now a will decide how the money should be distributed among the them.
The game begins now. There will be voting .all the five will give their votes .If votes against a is more than 50%
a will be thrown out of the ship.
and then b will decide and so on.
now tell what decision should "a" take to have benefit as well as he could save his life.
solution tomorrow :)

Is the treasure continuous or discrete?  When this problem was given to me, there were $ 200$ gold pieces in the treasure and $ 400$ pirates.  :P

of course it is what u r thinking
there is no need of information of treasure.
 just consider it is 100 dollar

The discreteness of the treasure is important.  If the treasure can be distributed continuously, there are a continuum of strategies and there is no longer a best strategy.  The discreteness of the treasure is necessary to ensure the finiteness of the number of possible strategies.

[hide="The 100 dollar solution, if everyone gets an integer number of dollars, is..."] that the highest pirate can give himself $ 99$ dollars and the lowest pirate $ 1$ dollar.  Why? [/hide]

with my TX 2006 and 2007 national MC teams, we discussed this problem in many versions:
more people, you DONT want the money and you cant throw it overboard or anything, pirates are violent, pirates are peaceful, and in real life (this version phailed very well)

here is what i think the solution is, but i forgot the exact solution
[hide]
step 1: suppose there is 1 pirate- he gets all the money without resistance
step 2: suppose there are 2 pirates- the last pirate will definitely vote NO (except possibly if the first pirate gives all the money to the last pirate) because then there will be 1 pirate left and he gets all the money
step 3: the second pirate will vote yes as long as he gets 1 coin (because if he doesnt, the last person definitely wont, and then there will be 2 pirates left and eventually, the last pirate gets all the money), so give the second pirate 1 coin and keep the rest
step 4: the second pirate will vote yes if he gets at least 2 coins, and the last will vote yes as long as he gets 1 coin, so give them those coins and keep all the other coins (give nothing to the third person)
step 5: give the third pirate 1 coin and the last pirate 2 coins-- then keep al the rest
yeah-- so i stopped explaining near the end, but i hope you understand my reasoning
[/hide]

STEVEN when there r two guys a & b then if a has decide then b can't throw him out of ship bcoz a will of course not vote against his own decision.Thus he has only 50% vote against himself which is not enough.
I hope u get it now correctly but the way u are solving the problem is somehow.
u r closer to right answer.

hi
reply to answer (99,0,0,0,1)
just think a is saying he is going to take 99 dollars and going to give 1 dollar to e then why not c will take a chance that may be b decision will be somehow better for him.there r three pirates who will vote against him.
but i must say u r very near to correct answer.

Hmm.  I remember the $ (98, 0, 1, 0, 1)$ answer being wrong, but I can no longer remember why.  Possibly my memory is mistaken; following stevenmeow's induction with the correct conditions (and the additional condition that a pirate prefers to throw other pirates overboard given that he gets the same amount of treasure) gives me that answer.

i remember the correct answer was (98,0,1,0,1)only.
anyways kindly try to remember why it is wrong.
i ll wait for ur reply then only i will write the full solution.

oh... in the version i know of, the pirate splitting the money cannot vote for/against it

The solution :
lets first think about e .just think when there are just d and e are left whatever will be the decision of d would be e can't go against it .So e never wants that such situation should occur.
so if c 's decision is giving him at least one dollar to him e will vote for him
therefore situation become
c:99
d:0
e:1 dollar
now coming to b just want one person to vote for him
b:99
c:0
d:0
e:1
c don't want such situation
so if A's decision somehow better for him he will vote for him .
so the final scenario
a:98
b:0
c:1
d:0
e:1
going to post a new puzzle tomorrow.


we need more information here

I thought they ate kfc 6legged chickens, but why did the question ask about drumsticks?

well, what's drumsticks anyway? :?

Drum sticks are chicken legs. You have enough infomation you gotta read the prolem carefully.

[quote="A+MATH"]Albatross, Bobert, Caddlebot (now out of the hospital), and Frankinfueter are celebrating the Chinese New Year by feasting on KFC chicken (6 legged of course).

If Albatross ate 1/2 as many as Caddlebot
and Frankinfueter ate 3 more than Bobert
And Bobert ate 3 times as many as Albatross
How many drumsticks did Frankinfueter eat if Caddlebot ate 120?

Happy late New Year.  :D  :lol:[/quote]
[hide="solution"]
Caddlebot ate 120
Albatross ate 60
Bobert ate 180
so [b]Frankinfueter ate 183.[/b]
[/hide]

ooo...

I didn't know what a drumstick was...

Thanks Treething for that wonderful picture about the "DRUMSTICKS!!"
 :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D

[quote="A+MATH"]Albatross, Bobert, Caddlebot (now out of the hospital), and Frankinfueter are celebrating the Chinese New Year by feasting on KFC chicken (6 legged of course).

If Albatross ate 1/2 as many as Caddlebot
and Frankinfueter ate 3 more than Bobert
And Bobert ate 3 times as many as Albatross
How many drumsticks did Frankinfueter eat if Caddlebot ate 120?

Happy late New Year.  :D  :lol:[/quote]

[hide]
120/2=60 (Albatross ate 60)
60*3=180 (Bobert ate 180)
180+3=183 ( Frankinfueter ate [b]183[/b])[/hide]

[quote="A+MATH"]Bonus: If each chicken has 6 legs how many chickens were used to make the feast.[/quote]

[hide]
120/6=20 (20 chickens for Caddlebot)
60/6=10 (10 chickens for Albatross)
60*3/6=30 (30 chickens for Bobert)
183/6= 30.5 (30.5 chickens for Frankinfueter)
20+10+30+30.5= [b]90.5 chickens total[/b]
[/hide]

[quote="A+MATH"][hide="To DarkNight"]
Actually it would be 91 becaues you can't kill half a chicken
[/hide][/quote]

Oh... Ok.  :P

[quote="RC-7th"]But it asks how many chickens were used.

Maybe they ordered from KFC and the other half of the chicken was used on another order.[/quote]

Lol. YUMMY KFC. But why are the chickens six-legged? Don't chickens only have two?


Fie $a=\ln2,b=\ln3,c=\ln5$. Prima ecuatie se scrie
\[ x^2\frac{b+c}{a}+y^2\frac{a+c}{b}+z^2\frac{a+b}{c}=2(xy+xz+yz). \]
Adunand $x^2+y^2+z^2$, obtinem
\[ (a+b+c)\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)=(x+y+z)^2. \]
Restul e Cauchy-Schwartz...:)


LaTeX rotation geometry parallelogram ratio trigonometry

In triangle $ ABC$, $ M$ is the midpoint of side $ BC$. There is a point $ D$ on side $ AB$
for which $ CD$ intersects $ AM$ at $ E$ such that $ DE \equal{} AD$. Prove that $ CE \equal{} AB$.

Hah...this was my favorite...not a rigorous proof (actually, not vigorous at all), but it's pretty fun and it's what i wrote...

(don't speak Latex, i'm sorry...I'll learn someday)

<EAD is congruent to <AED because triangle AED is isosceles. <CEM is congruent to <AED because they're vertical angles, so <EAD is also congruent to <EAD.
m<EMC + m<EMB=180 degrees,CM=BM because M is the midpoint of BC.
Now, rotate triangle CEM 180 degrees, so that C is at the same point as B. Call the point where E is now E'. <E'CM is congruent to <EAD (since nothing has really changed about triangle CEM or CE'M) and therefore E'BA is an isosceles triangle, which means that AB=BE'=CE

It's really equivalent to following AltheMan's hint about building a parallelogram - for the parallelogram, draw line segments through C, parallel and congruent to BE, and, through B, parallel and congruent to CE.

As I said, not a rigorous proof, but a somewhat interesting one (for me, at least)

[quote="serialk11r"]That also works, and if I remember the problem correctly law of sines does the same? Ratio of areas is like 1 line as well.[/quote]
That's my solution when I did this problem one month ago :lol:

[quote="pythag011"][quote="serialk11r"]Bleh for this one you can just find the ratios of the areas and it gets killed pretty fast.[/quote]

Or, you could look at it and realize its a one-line application of menelaus.[/quote]

Could you please explain this to me? I'm still not very comfortable with using Menelaus, and every line I see in this diagram seems to go through a vertex. Would you care to elaborate? Thanks.

On that note, I solved this problem by extending AM to a point F such that BMF is isosceles (I first showed this was valid by showing that angle AFC had to be obtuse, which was a total pain.) The rest came from simple triangle congruences.


A sequence $ (a_n)$ of positive integers satisfies: $ a_n\equal{}\sqrt{\frac{a_{n\minus{}1}^2\plus{}a_{n\plus{}1}^2}{2}}$ for all $ n \ge 1$. Prove that this sequence is constant.

Let us suppose that x is the minor I finish of the succession, clearing we obtain that square x is the sum of the squares of I finish before x and after x, then if some of them is greater than x the other serious minor which serious a contradiction, therefore are equal and all the terms serious equal to x


The planes $3x+4y+7z = 9$ and $ax+by+cz = d$ are parallel. If $(2,1,6)$ is equidistant from these two planes, find $d/c$.

[hide="Hint"]The corner which two planes make is equal with a corner or the supplement that the normal vector generates.[/hide]

uhh... what... haha can you explain that again

[hide="Solution"]
This plane is $3x+4y+7z = k$ for $\forall k\in \mathbb{R}$.
And, Length of the perpendicular line which I took down from point $(x_{1}, y_{1}, z_{1})$ to plane $ax+by+cz+d=0$

\[\frac{|ax_{1}+by_{1}+cz_{1}+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}\]

So,
$\frac{|3 \cdot 2+4 \cdot 1+7 \cdot 6-9|}{\sqrt{2^{2}+1^{2}+6^{2}}}=\frac{|3 \cdot 2+4 \cdot 1+7 \cdot 6-k|}{\sqrt{2^{2}+1^{2}+6^{2}}}$
$\therefore k=9, 95$

$\therefore \frac{d}{c}=\frac{95}{7}$
[/hide]


number theory
prime numbers
prime factorization

Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will $ 20!$ be the resulting product?

[hide]$20!=2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19$

We have to find numbers $a,b$ such that $ab=20!$ and $(a,b)=1$.

Cosider of the set $S = \{2,3,5,7,11,13,17,19\}$

We can create the numbers $a$ and $b$ by choosing prime numbers from this set.
For every prime divisor $p$ of $a$ we can't have $p \mid b$.
If, for example $5 \mid a$, then we should have $5 \not{\mid} b\Rightarrow  5^4 \mid a$.

That is, once we have choosen the prime divisors of $a$, we must include also their exponents

So, we have only to choose the prime divisors of $a$.

Every prime number $p_i \in S$ has too possibilities: $p_i$ divides $a$ or not.
So there are $2^8=256$ ways to do that
(including the cases $a=1$ and $a=20!$)

$20!$ is not a perfect square (for example $17^2$ does not divide $20!$)
In the $256$ permutations, we have count two times any pair $(a,b)$.

i.e. $a=2^{18}$ and $b= 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19\Rightarrow$
there is also the pair
$a= 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19$ and $b=2^{18}$

So, half of these pairs have $a>b$ and the other half have $a<b$.

$\frac{a}{b}<1 \Rightarrow a<b$

So we have $128$ ways [/hide]

[quote="pontios"][So, half of these pairs have $a>b$ and the other half have $a<b$.

 [/hide][/quote]

Sorry to revive but how do you know this.

Definitely it a cannot be equal to b, or else the product of the numerator and denominator is 1. So a<b ,a=b(not possible), or a>b, since we can swap a and b between the numerator and denominator. Each of (a,b) and (b,a) will have one pair belonging to a<b and 1 to b>a

Just wanted to point out: We actually don't need to factorize $20!.$ In fact, this problem requires no computation.

The actual values of the exponents in the prime factorization of $20!$ are irrelevant--what we care about is how many primes there are. It is easy to count that there are $8$ primes between $1$ and $20.$ From here, the problem just requires observation, and not computation.

[hide="Intuitive Solution"]
For a fraction to be in lowest terms, a prime $p$ can only appear in the numerator or denominator, but not both (otherwise the fraction isn't in lowest terms).

Note that for the product of the numerator and denominator to equal $20!$, each of its $8$ unique prime factors (2, 3, 5, 7, 11, 13, 17, 19) must appear in either the numerator or denominator, but not both. This is equivalent to counting the number of ways we can split $8$ objects among $2$ sets.

However, we have a small catch: the fractions must be between $0$ and $1$. For every fraction that is in this range, there is another fraction that is greater than $1$. (To see this, consider any fraction $\frac{a}{b}$, with $a<b$. There is another fraction, $\frac{b}{a}$, which is outside the range.)

Thus, our answer is $\frac{2^8}{2} = 2^7 = 128$.
[/hide]


function
advanced fields
advanced fields unsolved

Show the impossibility of solving the general  seventh degree equation by functions of two variables.

What do you call a function of two variables¿


What positive integer is nearest to $ \sqrt[4]{1200}$?

We test numbers.  Since $ 5^4\equal{} 625$ and $ 6^4\equal{} 1296$, the given value is closest to 6.

6^4=1296

6 is the answer

[color=#FF80FF][hide="."][b][i][u]PLEASE RATE MY POST THANK YOU SO MUXH [/u][/i][/b][/hide][/color]


inequalities
LaTeX
inequalities proposed

1/a+ab+abc (1 is common) +1/b+bc+bca +1/c+ca+cab   =< 1/3sqrt3abc  (1/a+1/b+1/c)

Can you rewrite this question with LATEX?

I think right side is $ \frac{1}{3\sqrt[3]{abc}}.(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$

$ i) a \plus{} ab \plus{} abc \geq 3.\sqrt [3]{a^3b^2c}$

$ ii) x^3 \plus{} y^3 \plus{} z^3\geq x^2y \plus{} y^2z \plus{} z^2x$ And solution is done. :wink:


logarithms algebra function domain LaTeX

1. $ \frac{log p}{log 9}\equal{}\frac{log q}{log 12}\equal{}\frac{log (p\plus{}q)}{log 16}$

find $ \frac{q}{p}$.

2. $ \frac{log a}{log 9}\equal{}\frac{log b}{log 15}\equal{}\frac{log (a\plus{}2b)}{log 25}$

find $ \frac{b}{a}$.

hint?
i tried cross multiplying the first two expressions to get some weird expressions. and now im stuck.
i stink at logs.  :mad:

[hide="#1"]$ {\log a\over\log b}\equal{}\log_ba$, hence

$ \log_9p\equal{}\log_{12}q\equal{}\log_{16}(p\plus{}q)\equal{}l$, which gives

$ p\equal{}9^l,q\equal{}12^l,{q\over p}\equal{}\left({4\over 3}\right)^l$

Now from $ \log_{16}(p\plus{}q)\equal{}l$ we get

$ 9^l\plus{}12^l\equal{}16^l$

Divide the equation by $ 12^l$ to get

$ \left({3\over 4}\right)^l\plus{}1\equal{}\left({4\over 3}\right)^l$

Substituting $ x\equal{}\left({4\over 3}\right)^l$ we get $ {1\over x}\plus{}1\equal{}x\implies x^2\minus{}x\minus{}1\equal{}0\implies x\equal{}{1\plus{}\sqrt{5}\over 2}$ (negative solution is discarded). Now $ {q\over p}\equal{}x\equal{}{1\plus{}\sqrt{5}\over 2}$

Apply similar ideas to the second problem.[/hide]

Farenhajt nailed it in #1, and it's the same sort of method for #2, but I have time and, more importantly, you can never have too much logarithm practice!

[quote="unimpossible"]2. $ \frac {log a}{log 9} = \frac {log b}{log 15} = \frac {log (a + 2b)}{log 25}$

find $ \frac {b}{a}$.[/quote]
[hide="Hint for #2"]This question revolves around knowledge of the change of base identity, namely that $ \frac {\log_{a}{b}}{\log_{a}{c}} = \log_{c}{b}$. Simplify based on this.[/hide]

[hide="Worked Solution for #2"]Based on the change of base identity, we can simplify:
$ r = \log_{9}{a} = \log_{15}{b} = \log_{25}{(a + 2b)}$
Thus converting to exponential form $ a = 9^{r}$, $ b = 15^{r}$, $ a + 2b = 25^{r}$

We're after $ \frac {b}{a}$ so we write:
$ \frac {b}{a} = \frac {15^{r}}{9^{r}} = \left(\frac {15}{9}\right)^{r} = \left(\frac {5}{3}\right)^{r}$

We now have a formula for $ \frac {b}{a}$ as desired. However this is not finished: more information is provided that can simplify our answer and provide us with some concrete numbers.

From $ a + 2b = 25^{r}$ we can divide by $ b$ to get $ \frac {a}{b} + 2 = \frac {25^{r}}{b}$.
Now substituting our value for $ b$ into this we get:

$ \frac {a}{b} + 2 = \left(\frac {25}{15}\right)^{r}$
$ \frac {a}{b} + 2 = \left(\frac {5}{3}\right)^{r}$

Noting that $ \frac {b}{a} = \left(\frac {5}{3}\right)^{r}$ gives us the final simplification:
$ \frac {a}{b} + 2 = \frac {b}{a}$

We can now solve for this. Substituting $ u = \frac {b}{a}$ gives:

$ \frac {1}{u} + 2 = u$
$ 1 + 2u = u^{2}$
$ u^{2} - 2u - 1 = 0$
So $ u = 1 - \sqrt {2}, u = 1 + \sqrt {2}$. We disregard the negative solution $ 1 - \sqrt {2}$ by inspection, as this would imply one of $ a$ or $ b$ being negative which is not in the valid domain for $ \log$. Therefore $ \boxed{\frac {b}{a} = 1 + \sqrt {2}}$. Notice we never had to find the value of $ r$. Nice question.[/hide]

By the way, in $ \text{\LaTeX}$, the correct way to write the logarithm is:

[code]\log_{2}{8} = 3[/code]
Giving $ \log_{2}{8} = 3$

Not:
[code]log_{2}{8} = 3[/code]
Giving $ log_{2}{8} = 3$.

The difference in spacing and font style is subtle, but it does look better.

[quote="xpmath"]Eh? I remember questions similar to this from the first chapter of volume 2...[/quote]
Could be. I've never read the books (shame on me, etc).

yeah i did it the way fahrenhajt and kh- did.

anybody have a different method?


From a pile of 100 pennies, 100 nickels and 100 dimes, select 21 coins which have a total calue of exactly $1.00.  In your selection you must also use at least one coin of each type.  How many coins of the three types should be selected?

[hide]  we know there must be at least 1 nickel which means we need at least 5 pennies to make our total a nice and even 10 cents.  Now we have 21-6=15 coins left.  If all 15 of these coins were nickles we would only get 75 cents, 15 cents short.  However if we trade each nickel in for a dime (trading meaning replacing) we would get 5 more cents each time.  Thus we need 15/5= 3 dimes.  [b]So in all we need  5 pennies, 13 nickels and 3 dimes [/b] [/hide]

p=2
n=5
d=10

2x+5y+10a= 100
x+y+a=21

if a is 10 then x & y are 0 so a+x+y=10 not 21
if a is 9 then x can be 5 and y=0 or x=0 and y=2, so a+x+y=14 or 11 not 21
if a is 8 then x can be 5, 10 then y will be 2, 0 so a+x+y can be 15 or 18 not 21
if a is 7 then x can be 5, 10, 15 then y will be 4,2,0 so a+x+y will be 16, 19, 22 not 21
if a is 6 then x can be 5,10,15,20 then y will be 6, 4, 2, 0 so a+x+y will be 17, 20, 23 or 26 not 21
if a is 5 then x can be 5,10,15,20,25 then y  will be 8,6,4,2,0 so a+x+y will be 18, 21, 24, 27, 30
so a is 5, x is 10 and y is 6
So i need 5 dimes, 6 nickles and 10 pennies

@nn@, the total value of 5 dimes, 6 nickels and 3 dimes is 90 cents, and you need 100 cents, so that's incorrect - G-Unit's answer is the correct one.

$ d\equal{}dimes$
$ n\equal{}nickels$
$ p\equal{}pennies$

$ d\plus{}n\plus{}p\equal{}21$
$ 10d\plus{}5n\plus{}p\equal{}100$
$ d, n, p>1$

There are multiple answers.

[hide="Solution"]
We have the two equations
\[ d\plus{}n\plus{}p\equal{}21
\]
\[ 10d\plus{}5n\plus{}p\equal{}100
\]
Taking mod 5 of the second equation we recieve
\[ p\equiv 0\mod{5}
\]
We have 4 cases
\[ p\equal{}5,10,15,20
\]
[hide="p=5"]
\[ d\plus{}n\equal{}16
\]
\[ 10d\plus{}5n\equal{}95
\]
Multiplying by 5 and subtracting we receive
\[ 5d\equal{}15
\]
Dividing by 5 we receive
\[ d\equal{}3
\]
Subtracting from the first equation we recieve
\[ n\equal{}13
\]
[/hide]
[hide="p=10"]
\[ d\plus{}n\equal{}11
\]
\[ 10d\plus{}5n\equal{}90
\]
Multiplying by 5 and subtracting we receive
\[ 5d\equal{}35
\]
Dividing by 5 we receive
\[ d\equal{}7
\]
Subtracting from the first equation we recieve
\[ n\equal{}4
\]
[/hide]
[hide="p=15"]
\[ d\plus{}n\equal{}6
\]
\[ 10d\plus{}5n\equal{}85
\]
Multiplying by 5 and subtracting we receive
\[ 5d\equal{}55
\]
Dividing by 5 we receive
\[ d\equal{}11
\]
Subtracting from the first equation we recieve
\[ n\equal{}\minus{}5
\]
Therefore, there is no solution for $ p\equal{}15$
[/hide]
[hide="p=20"]
\[ d\plus{}n\equal{}1
\]
\[ 10d\plus{}5n\equal{}80
\]
Multiplying by 5 and subtracting we receive
\[ 5d\equal{}75
\]
Dividing by 5 we receive
\[ d\equal{}15
\]
Subtracting from the first equation we recieve
\[ n\equal{}\minus{}14
\]
Therefore, there is no solution for $ p\equal{}20$

The solutions are
[b]3 dimes, 13 nickles, 5 pennies[/b]
and
[b]7 dimes, 4 nickles, 10 pennies[/b]
[/hide]
[/hide]

@phr33k, um i think 5x10+6x5+30x1=110 maybe?  :)  :roll:


Draw $\triangle PQR$ and $\triangle PST$. Let $QV$ form the altitude of $\triangle PQR$ and $SV$ the altitude of $\triangle PST$.

If $\frac{QS}{QV}=\frac{1}{3}$ and $\frac{PT}{PR}=\frac{3}{4}$, what is the value of $\frac{\text{area of } \triangle PST}{\text{area of }\triangle PQR}$

No diagram. Solutions. :)

Solution:
[hide]
Note immediately that we are conspicuously given the two ingredients for finding the area of a triangle: bases and heights. All we have to do is work with the ratios given.

First, confirm the picture. QV and SV are both altitudes (given). QV must be an altitude to side PR, and thus QV is perpendicular to the line (not segment) defined by points P and R at V. Likewise, SV is perpendicular to the line defined by points P and T at V. Therefore P,V,T, and R are collinear because R and T are each on the same line PV.

Therefore, V, Q, and S are collinear, because only one unique line can be perpendicular to line PV at point V.

But here is the first significant ambiguity in the problem, one which leads to two possible solutions: Since V, Q, and S are collinear, and we are given 3QS=QV, we have two possible orders for points V, Q, and S: VQS, or VSQ. The order cannot be specified with the information given, so we must consider both.

In case VQS, VQ+QS=VS, so VS>VQ. Then Let VQ=3y and QS=y (given), so VS=4y. We are given PT=3x and PR=4x. Then the ratio of the areas is:
$\frac{\text {area of } \triangle PST}{\text {area of } \triangle PQR}=\frac{\frac{1}{2}(3x)(4y)}{\frac{1}{2}(4x)(3y)}=\frac{6xy}{6xy}=1$, and the areas are equal.

 in case VSQ, VQ-QS=VS, so VS<VQ. Then Let VQ=3y and QS=y (given), so VS=2y. We are given PT=3x and PR=4x. Then the ratio of the areas is:

$\frac{\text {area of } \triangle PST}{\text {area of } \triangle PQR}=\frac{\frac{1}{2}(3x)(2y)}{\frac{1}{2}(4x)(3y)}=\frac{3xy}{6xy}=\frac{1}{2}$.

These are the only two possibilities for the ratios of the areas, and they are both valid possibilities based on the given constraints.

[/hide]


A 15-digit positive integer,which consists of only 2s and 3s, is divided without a remainder by a 7-digit integer which consists of the same number,i.e aaaaaaa. Find the 15-digit integer without using any programming method but only purely mathematical ways.

Let $n$ be the 15-digit number and $m=222222222222222$.

Then $n=m+\sum_{k=0}^{14}{a_{k}\cdot 10^{k}},$ where $a_{k}\in \{0,1\}.$

That is,
if $a_{k}=0$ then the k-th digit of $n$ is $2$
if $a_{k}=1$ then the k-th digit of $n$ is $3$

Let $x=1111111$

$a\cdot x \mid n \Rightarrow  x \mid n.$
So, if there is a number $n$ which is divided exactly by $ax$ then it is divided exactly by $x$.
This means that it's enough to check the divisibility by $1111111$

We will find $m \mod{x}$

$m = 222222202222222+20000000$

The first number is a multiple of $x$.
The second can be written as $20000000 = 2 \cdot 10000000 = 2 \cdot (9999999+1) = 2 \cdot 9 \cdot x+2$

So $m \equiv 2 \mod{x}$

$n \equiv 2+\sum_{k=0}^{14}{a_{k}\cdot 10^{k}}\mod{x}$

$n \equiv 0 \mod{x}\Leftrightarrow \sum_{k=0}^{14}{a_{k}\cdot 10^{k}}\equiv-2 \mod{x}\ \ (1)$

So we have to find the remainders of $10^{k}$ in modulo $x$

$10^{7}= 9999999+1=9x+1 \Rightarrow  10^{7}\equiv 1 \mod{x}$

so for every $k$ we have $10^{k+7}\equiv 10^{k}\mod{x}$. This means that for $k\geq 7$ we have the same remainders

$10^{0}\equiv 10^{7}\equiv 10^{14}\equiv 1 \mod{x}$
$10^{1}\equiv 10^{8}\equiv 10 \mod{x}$
$10^{2}\equiv 10^{9}\equiv 100 \mod{x}$
$10^{3}\equiv 10^{10}\equiv 1000 \mod{x}$
$10^{4}\equiv 10^{11}\equiv 10000 \mod{x}$
$10^{5}\equiv 10^{12}\equiv 100000 \mod{x}$
$10^{6}\equiv 10^{13}\equiv 1000000 \mod{x}$

We have to choose $\{a_{k}\}$ such that $(1)$ is satisfied

$-2 \equiv 1111109 \equiv 2222220 \mod{x}$

From the above congruences we can see that the remainder of $\sum_{k=0}^{14}{a_{k}\cdot 10^{k}}$ in modulo $x$ is:

$(a_{6}+a_{13})10^{6}+(a_{5}+a_{12})10^{5}+(a_{4}+a_{11})10^{4}+(a_{3}+a_{10})10^{3}+(a_{2}+a_{9})10^{2}+(a_{1}+a_{8})10^{1}+(a_{0}+a_{7}+a_{14})$

So, clearly we cant' obtain the number $1111109$, because the last be cannot have a value grater of $3$

But we can obtain the number $2222220$, and there is only one way, when $a_{0}=a_{7}=a_{14}=0$ and all the rest $a_{k}=2$

So our number is $n=222222222222222+011111101111110 = \boxed{322222232222223}$

[color=brown][b]Note[/b]
By checking the divisibility of $n$ with every number $a \in \{2,3,...,9\}$ we can see that the 7-digit divisor aaaaaaa can be
either the number $x=1111111$ or the number $3x=3333333$,
because $3 \mid n$ and $3 \not{|}\ 1111111$[/color]

Woah... that's a long and complicated solution... I got the same answer but I just experimented with numbers divisible by 1111111... :oops:


I was wondering if there is a general formula for factorial, $P=1*2*3*...*n$, just like there is a formula for the sum of consecutive numbers , $\frac{n(n+1)}{2}$.  The answer to this problems eludes me.

Thank you, but I am aware that $n!=n(n-1)(n-2)....$, yet I was hoping for an answer.  There must be an answer to this, even though it might not be discovered.  I'll seek for an answer to this, and I would really appreciate any ideas.

[quote="JesusFreak197"]You multiply them out. Nothing more, nothing less. There is no formula for finding $n!$ other than saying $n!=n\cdot (n-1)\cdot (n-2) \cdot ... \cdot 2\cdot 1$.[/quote]
I don't think so. There is such formula called the gamma function with notation $\Gamma(x)$. For integer $n$, $\Gamma(n+1)=n!$. This gamma function is so powerful that it extends factorials real numbers and even complex numbers.
For more information on the gamma function, please refer to the link:
[url]http://mathworld.wolfram.com/GammaFunction.html[/url]
Besides, for approximation of factorials, we can use the Stirling's formula.
[url]http://mathworld.wolfram.com/StirlingsApproximation.html[/url]

I stand corrected, then. But it's not much help unless you know Calculus, and even then, it's very difficult for me to understand (I do know Calculus).

Well, there's also Stirling's formula, but it's just an equivalent of the factorial.

This can probably be found in leepakhin's links, but it is easy to show with integration by parts that \[n!=\int_0^\infty\frac{x^n}{e^x}dx.\]

Question: I just realized that if this is true for $n=0$ then $\int_0^\infty\frac{1}{e^x}dx=\int_0^\infty\frac{x}{e^x}dx$, but that couln't be, could it? It must just be true for $n\ge1$. It was a long time ago that I read about it and derived it.


function
trigonometry
calculus
calculus computations

[quote="kunny"]Find the maximum and minimum vale of $y=4\sin^{2}x-3\sin x\cos x+2\cos^{2}x.$[/quote]

$y=2\sin^{2}x-\frac{3}{2}\sin 2x+2\\=3-\cos 2x-\frac{3}{2}\sin 2x\\=3-\frac{1}{2}\sqrt{13}\sin (2x+a)$
where $\tan a=\frac{2}{3}$

So the max is $3+\frac{\sqrt{13}}{2}$ and min is $3-\frac{\sqrt{13}}{2}$


geometry
geometric transformation
algebra unsolved
algebra

Vanya has chosen two positive numbers, x and y. He wrote the numbers x+y, x-y, x/y, and xy, and has shown these numbers to Petya. However, he didn't say which of the numbers was obtained from which operation. Show that Petya can uniquely recover x and y.

I don't understand what the 'uniquely' means.
Anyway, Petra can squared each of the numbers and compare the differences of the results, with the given numbers to identify $xy$ and then $x+y$ and $x-y$ (and then $\frac x y$).
Thus $x = \frac 1 2 (x+y + x-y)$ from which we easily get $y$ too.

Pierre.

Pierre, I think something is wrong because if the first player took the number $-x$ and $-y$ you would get the same squares.....But not the same numbers....
Also, maybe I don't get something.....
By the way, the name of the second player is Petya (the russian equivalent for Pierre :D) , and not Petra :) , a woman's name  :lol:

[quote="Xixas"]Vanya has chosen two [b]positive[/b] numbers, x and y.[/quote]

[quote="DusT"]By the way, the name of the second player is Petya (the russian equivalent for Pierre :D) , and not Petra :) , a woman's name  :lol:[/quote]

:D :D

I think you are right....
Strange that at the actual competition I didn't even notice the condition about the positiveness of the numbers  :? I didn't solve the problem anyway (too many cases I think...)

[quote="pbornsztein"]I don't understand what the 'uniquely' means.
Anyway, Petra can squared each of the numbers and compare the differences of the results, with the given numbers to identify $xy$ and then $x+y$ and $x-y$ (and then $\frac x y$).
Thus $x = \frac 1 2 (x+y + x-y)$ from which we easily get $y$ too.[/quote]
How do you think: are there some cases?
For example, it is possible $(x+y)^2-(xy)^2=2(x-y)^2$  :?
Am I missing something?

Uniquely means "with no other possibilities", i.e. Petya can found numbers [tex]x[/tex] and [tex]y[/tex], that had been chosen by Vanya homologously (don't know, if this word is proper here).

Xixas! I think, your translation is fine.


trigonometry
geometry
circumcircle
inradius
geometry theorems

I can't find any good lists of geometric/trig identities anywhere, so I will start one here. If anyone finds a really good list of geometric/trig identities, please post the link here. And Mildorf seems to be the king of geometric/trig identities, so Mildorf, if you can post some that would be excellent!

I will try to restrict this to the more obscure identities. (It doesn't include sum-to-product or product-to-sum because they just take too long to type and cost lots of bandwidth/loading time.)

And if there are any mistakes, please tell me.

[hide]

$[ABC]$ is the area of the triangle, $R$ is the circumradius, $r$ is the inradius, $s$ is the semiperimeter.

$\tan \frac{1}{2}x = \frac{\sin x}{1+\cos x} = \frac{1-\cos x}{\sin x}$

$\cos A + \cos B + \cos C = 1 + \frac{r}{R}$

$\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A \cos B \cos C = 1$

$[ABC] = 2R^2 \sin A \sin B \sin C$

[/hide]

[quote="Philip_Leszczynski"]$\cos A + \cos B + \cos C = 1 + \frac{r}{R}$[/quote]

How do you (nicely) prove this?

this might be of interest:

http://myhome.personaldb.net/ideahitme/trig.pdf

Wow, thanks! Well, I think this thread is pretty much useless by now.  ;)

[b]Piece of poetry:[/b] \[ \underline {\overline {\left| \ cos A,\ \cos B,\ \cos C\ ...\ \right| }}  \] \[ \blacksquare  \] \[ \cos A+\cos B+\cos C=  \]
\[ 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+\cos C=  \]
\[ 2\sin \frac C2\cos \frac{A-B}{2}+1-2\sin ^2\frac C2=  \]
\[ 1+2\sin \frac C2\left( \cos \frac{A-B}{2}-\cos \frac{A+B}{2}\right)=  \]
\[ 1+4\sin \frac A2\sin \frac B2\sin \frac C2=  \]
\[ 1+4\sqrt{\frac{(s-b)(s-c)}{bc}\cdot \frac{(s-c)(s-a)}{ca}\cdot \frac{(s-a)(s-b)}{ab}}=  \]
\[ 1+\frac{4(s-a)(s-b)(s-c)}{abc}=  \]
\[ 1+\frac{4sr^2}{4Rrs}=  \]
\[ 1+\frac rR.  \] \[ \blacksquare  \] [u]Exercise (practice).[/u] Let $ABC$ be a acute triangle, its circumcircle $w=C(O,R)$ and the middlepoints $D\in [BC]$, $E\in [CA]$, $F\in [AB]$ of the sides of the triangle $ABC$. Then $OD+OE+OF=R+r$ (the [i]Euler's relation[/i]).

[u]1. A trigonometrical solution.[/u] We add the relations $OD=R\cos A$, $OE=R\cos B$, $OF=R\cos C$ and obtain $OD+OE+OF=R(\cos A+\cos B+\cos C)=R\left(1+\frac rR\right)=R+r$.

[u]2. A solution without the previous trigonometrical relation[/u]: We apply the [i]Ptolemeu's theorem[/i] in the cyclic quadrilaterals $AEOF$, $BFOD$, $CDOE$: $b\cdot OF+c\cdot OE=a\cdot R$; $c\cdot OD+a\cdot OF=b\cdot R$; $a\cdot OE+b\cdot OD=c\cdot R$. We add these relations to the wellknown relation $a\cdot OD+b\cdot OE+c\cdot OF=2sr$ and we will obtain the relation $(a+b+c)\cdot (OD+OE+OF)=(a+b+c)\cdot (R+r)$, i.e. $OD+OE+OF=R+r$.


The digits of a four-digit positive integer add up to 14. The
sum of the two middle digits is nine, and the thousands digit
minus the units digit is one. If the integer is divisible by 11,
what is the integer?

Since the middle two digits add up to 9, then the thousands and units digit add up to 5. The difference of those two digits is one, so the thousands digit is 3 and the units digit is 2. Since the number is divisible by 11, the sum of the odd digits would be equal to the sum of the even digits. (The divisibility rule for 11 is add the even and odd digits separately and compare the difference. If the difference is divisible by 11, then so is the number.) Since the thousands digit is 1 more than the units digit, the tens digit is one less than the hundreds digit. Thus, the number is 3542.


Let a,b,c be positive real numbers. Negative or prove that inequalities:
$ 3_\sqrt {a^2+b^2-ab} \  _+  3_\sqrt {b^2+c^2-bc}}  \ _+ 3_\sqrt {a^2+c^2-ac}  \geq\ \3_\sqrt {9(a^2+b^2+c^2)}$

I believe he means:

$ \sqrt [3] {a^2 \plus{} b^2 \minus{} ab} \plus{} \sqrt [3] {b^2 \plus{} c^2 \minus{} bc} \plus{} \sqrt [3] {a^2 \plus{} c^2 \minus{} ac} \geq \sqrt [3] {9(a^2 \plus{} b^2 \plus{} c^2)}$

My solution for it:
Applying Cauchy-Schwartz's inequality, we have:
LHS<=cube root(9)(a^2+b^2+c^2-ab-bc-ac)
Since a,b,c>0(in fact it can be zero) we will have: a^2+b^2+c^2-ab-bc-ac=<a^2+b^2+c^2
Therefore, we proved that: LHS<=RHS.So I disprove the sign of inequality above :)

You can write a key explicit (Sorry my English is badly)
If you do it I think the equal sign occur ab+bc+ca=0????

Well, just applying Cauchy-Schwartz for the LHS and considering that a,b,c>0, you will get my point! :)


Let $ P \in Z[x]$ be a polynomial.Prove that there exist two polynomial $ Q;R \in Z[x]$ such that $ P(x).Q(x)$ is the polynomial of $ x^2$ and$ P(x).R(x)$ is the polynomial of $ x^3$

This is INMO 2008 problem 6.
[url=http://www.isid.ac.in/~rbb/inmosol_08.pdf]Click here[/url] for the solution.


graph theory
superior algebra
superior algebra solved

Let $R$ be a ring and let $J \lhd R$. Define $\textrm{Rad}\ J = \left\{\left. a \in R \right| \exists n > 0: a^n \in J\right\}$. Prove that $\textrm{Rad}\ J \lhd R$.

I'm struggling with this and I don't know why, it shouldn't be too hard... :huh:  :(

The most difficult part is to prove that it's clused under addition:
Let $a^m,b^n \in J$. Then $(a+b)^{m+n} = \sum_{i=0}^{m+n} a^i b^{m+n-i} \binom{m+n}{i}$ is a sum of elements of $J$, so $\in J$.

Can I assume that $R$ is commutative? :huh:

The problem doesn't say that $R$ needs to be commutative, that's the thing I was struggling with... I could prove it easily for commutative $R$, you know. Confused...

Well, it's not true if $R$ isn't commutative- look at $n\times n$ matrices over some field $F$, and let $J=0$. The nilpotent matrices are certainly not an ideal; the only ideals are the whole ring and zero anyways.

I think you should assume that $R$ is commutative.

One of the cute things (just as with graph theory and stuff) is lack of true convention, in a commutative ring theory course it would be tedious to always include : 'it has unity element and is commutative', but this is one of the things I hate about ring theory, always confusion

Yes, I think your question about commutativity is  of importance, consider $R=M_{2}(\mathbb{Q})$ and simply $J=0$, so basically we will be speaking about nilpotent element in the radical

Take $A=\left[\begin{array}{cc} 0&1\\0&0\end{array}\right]$ and $B=\left[\begin{array}{cc} 0&0\\1&0\end{array}\right]$

$A$ is obviously nilpotent, however $AB$ is not, so the radical is not an ideal

I have a question too : what is the radical of the entire ring, intuition would say the total ring ? but my professor told me otherwise?

Of course the radical of the whole ring is the whole ring-  the radical of a set must contain itself, because that exponent is allowed to be 1.


induction
combinatorics unsolved
combinatorics

Suppose,45 people meet at a meeting.It is known that every two people
who [b]knows[/b] the same number of people do not know each other.
Find the maximum number of acquainted people.

" Find the maximum number of acquainted people"
What do you mean "acquainted people" ?

[quote="gnosis"]I think he means to find the maximum number of people(a group) such that everyone knows every other person in the group.[/quote]

No, no! :o I mean the maximum number of acquainted [b]pairs[/b]?

There is at most one person with 44 friends, at most two persons with 43 friends and so on. Since we want to maximize the number of pairs and $1+2+\dots+9=45$, we calculate (using $\sum k$ and $\sum k^2$, so that it could be generalized):

\[ 44+2\cdot43+3\cdot42+4\cdot41+5\cdot40+6\cdot39+7\cdot38+8\cdot37+9\cdot36=1740 \]

Half this number is 870, the maximal number of pairs.

And could you please provide a source for the problem?

Your solution is similar to mine. But, my one is more logical.
I began to consider 2-part graph, and maximized the number of pairs of acquainted people (the result is 23.22)
I thought that was a really little value, so I continued to consider the 3-part graph and so on.
I guessed the 9-part graph would bring me the result.
The next step was not simple. I solved completely your feeling: max=1/2(44+2.43+...)
I think that 45 is a specific number (=10 choose 2).
Improving this solution is not complex in induction but the result is not nice.


invariant
monovariant
search
combinatorics unsolved
combinatorics

A deck of 50 cards contains two cards labeled n for each n = 1, 2, . . . , 25. There are 25 people
seated at a table, each holding two of the cards in this deck. Each minute every person passes the
lower-numbered card of the two they are holding to the right. Prove that eventually someone has two
cards of the same number.

When you extract problems from a source like ACoPS, you should do a search to see if others have posted them.  I searched on the forum for the words "card pass" (without quotes) and this is the third hit:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=363083331&t=307580


The sides of an equiangular hexagon measure 10, 6, 12, 14, x, and y units, in the order given. Find x and y.

Can you find a solution? I don't really know if that is right of not.

Simply extend the three sides with length 6, 14 and y till they meet one another.
Then we can obtain an equilateral triangle with side length = 28.
Thus $ x\plus{}14\plus{}12 \equal{} 28$ -> $ x\equal{}2$,
$ x\plus{}y\plus{}10 \equal{} 28$ -> $ y\equal{}16$ and it's done :)

http://www.mathlinks.ro/viewtopic.php?t=212879

[hide="A general statement"]If an equiangular hexagon has side lengths $ a_1,a_2,a_3,a_4,a_5,a_6$, in this order, then we have
\[ a_1\minus{}a_4\equal{}a_2\minus{}a_5\equal{}a_3\minus{}a_6
\]
[i]Proof[/i]
Let the hexagon be $ ABCDEF$, where $ AB\equal{}a_1,BC\equal{}a_2,\ldots,FA\equal{}a_6$. Extend $ AF,DE,BC$ to obtain a triangle $ GHI$. Note that $ ABG,CDI,EFH,GHI$ are equilateral triangles. Now,
\[ GI\equal{}GB\plus{}BC\plus{}CI\equal{}a_1\plus{}a_2\plus{}a_3\\GH\equal{}GA\plus{}AF\plus{}FH\equal{}a_1\plus{}a_6\plus{}a_5\\HI\equal{}HE\plus{}ED\plus{}DI\equal{}a_5\plus{}a_4\plus{}a_3.
\]
$ GI\equal{}GH$ implies that $ a_2\minus{}a_5\equal{}a_3\minus{}a_6$, $ GH\equal{}HI$ implies that $ a_1\minus{}a_4\equal{}a_3\minus{}a_6$. Hence our proof is complete.

This is taken from Romanian TST.[/hide]

See also the (not yours) topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=174932 .

Kostas vittas.


[quote="rachid"]let x,y and z reel numbers and:
x>-1 ; y>-1 ; z>-1 ; t>-1 ; x+y+z+t=2
Prove that : x^3 +y^3 +z^3 +t ^3 >1/2[/quote]

Moved out of IMO 2008 forum.
Should be easy by RCF/LCF, but I hope someone finds something better (I have no time now).

  darij

hello , I now this inequality , I solved it months ago ( it should be solved without using any theorem)

we have:

$ \sum {(x \minus{} \frac{1}
{2})^2 (x \plus{} 1) \geqslant 0{\kern 1pt} {\kern 1pt} {\kern 1pt}  \Leftrightarrow {\kern 1pt} \sum {x^3  \geqslant \sum {(\frac{3}
{4}x \minus{} \frac{1}
{4})}  \equal{} \frac{1}
{2}} }$  :lol:

thank you mehdi ( I had the idea when  I was in a geography class  :D )

:D thanks geography class for this nice proof  :lol:

another way
$ 16(x^3\plus{}y^3\plus{}z^3\plus{}t^3) \geq (x\plus{}y\plus{}z\plus{}t)^3\equal{}8$

please how to prove that   (x+y+z+t)^3 <= 16(x^3 +y^3 +z^3 +t ^3).

[quote="Jure the frEEEk"]another way
$ 16(x^3 \plus{} y^3 \plus{} z^3 \plus{} t^3) \geq (x \plus{} y \plus{} z \plus{} t)^3 \equal{} 8$[/quote]
It does not work here because $ x$ can be negative. :wink:

my solution,
by caushy-shwarz we have :(x+y+z+t)²<4(x²+y+z²+t²)
1<x²+y+z²+t²
a second time caushy-shwarz:(x²+y+z²+t²)<(x^3 +y^3 +z^3 +t^3) (x+y+z+t)
so 1/2 <(x^3 +y^3 +z^3 +t^3)
true?.

[quote="rachid"]have you another way Mr arqady.[/quote]
My proof the same of the anas's proof. I don't see another way. :maybe:
I think, it's interesting to look for the minimal $ k$ for which the inequality
\[ x^3 \plus{} y^3 \plus{} z^3 \plus{} t ^3\geq\frac {1}{2}
\]
holds for all $ x,$ $ y,$ $ z$ and $ t$ such that $ x \plus{} y \plus{} z \plus{} t \equal{} 2$ and $ \{x,y,z,t\}\subset[k, \plus{} \infty).$ :wink:

[quote="anas"]we can also use chebychev  :)[/quote]
I don't see it.  :maybe:  Explain please. Thank you!

:blush:  sry, because the inequality is symmetric ( I guess  :roll: ), we assume that : $ x\geq y\geq z\geq t$ so by chebychev's inequality we have:
$ S \equal{} \sum{x^3} \geq \frac { {(x \plus{} y \plus{} z \plus{} t)(x^2 \plus{} y^2 \plus{} z^2\plus{}t^2)}}{4}$

[quote="anas"]:blush:  sry, because the inequality is symmetric ( I guess  :roll: ), we assume that : $ x\geq y\geq z\geq t$ so by chebychev's inequality we have:
$ S \equal{} \sum{x^3} \geq \frac { {(x \plus{} y \plus{} z \plus{} t)(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2)}}{4}$[/quote]
I think, you are wrong because it's no true that $ x^2\geq y^2\geq z^2\geq t^2.$ :wink:

[quote="rachid"]so ,is my solution true? :?:[/quote]
No! We can't  use "caushy-shwarz" here because $ x$ can be negative.


calculus
derivative
inequalities
inequalities proposed

if $ a,b,c >0$ prove that :

$ \sum \frac{a^2}{b^3\plus{}c^3} \ge  \frac{3}{2 \sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}}$

sorry if it has been posted before.

[quote="shoki"]if $ a,b,c > 0$ prove that :

$ \sum \frac {a^2}{b^3 \plus{} c^3} \ge \frac {3}{2 \sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}}$

sorry if it has been posted before.[/quote]

without loss of generality, we can assume that $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$, ineq becomes

$ \sum \frac {a^2}{3 \minus{} a^3} \ge \frac {3}{2}$

we can prove that

$ \frac {a^2}{3 \minus{} a^3} \ge \frac {7a^3 \minus{} 1}{12}$

actually

$ \frac {a^2}{3 \minus{} a^3} \minus{} \frac {7a^3 \minus{} 1}{12} \equal{} \frac {(a \minus{} 1)^2 \left(7 a^4 \plus{} 14 a^3 \plus{} 21 a^2 \plus{} 6 a \plus{} 3\right)}{12 \left(3 \minus{} a^3\right)} \ge 0$

so

$ \sum \frac {a^2}{3 \minus{} a^3} \ge \sum{\frac {7a^3 \minus{} 1}{12}} \equal{} \frac {3}{2}$

done.

my solution is somehow like yours ,

i have proved

$ \frac{a^2}{3\minus{}a^3} \ge \frac{5a^3\plus{}1}{12}$

$ \leftrightarrow 0 \le (a\minus{}1)^2(\frac{5}{12}a^4\plus{}\frac{10}{12}a^3\plus{}\frac{15}{12}a^2\plus{}\frac{1}{2}a\plus{}\frac{1}{4})$

which is right,
a question how did u find the expression  $ \frac{7a^{3}\minus{}1}{12}$?
thx for your answer :)

@shoki: your estimation is wrong. Take for example $ a\equal{} 1/2$.

[quote="shoki"]my solution is somehow like yours ,

i have proved

$ \frac {a^2}{3 \minus{} a^3} \ge \frac {5a^3 \plus{} 1}{12}$

$ \leftrightarrow 0 \le (a \minus{} 1)^2(\frac {5}{12}a^4 \plus{} \frac {10}{12}a^3 \plus{} \frac {15}{12}a^2 \plus{} \frac {1}{2}a \plus{} \frac {1}{4})$

which is right,
a question how did u find the expression  $ \frac {7a^{3} \minus{} 1}{12}$?
thx for your answer :)[/quote]

$ \frac {a^2}{3 \minus{} a^3} \minus{} \frac {5a^3 \plus{} 1}{12} \equal{} \frac{(a\minus{}1) \left(5 a^5\plus{}5 a^4\plus{}5 a^3\minus{}9 a^2\plus{}3 a\plus{}3\right)}{12 \left(3\minus{}a^3\right)}$
is not non-negative.

find the $ \frac {7a^{3} \minus{} 1}{12}$ you can think of tangentline

[quote="FelixD"]@shoki: your estimation is wrong. Take for example $ a \equal{} 1/2$.[/quote]
seems that u are right ,
i must have made a mistake in my computations...
sorry :(
@kuing:
well that's exactly what i have thought about but i found another equation for the tangent line  :huh: ...
thx :)  ,the idea of considering the tangent line is really great and nice...

[quote="FelixD"]Indeed it is^^ (but it uses derivatives... :P )[/quote]
yes,and that's why i said in my previous problems to avoid using derivatives because as u can see i used the tangent line idea....

I thought that you used jensen...^^ but the tangent method is in my opinion more powerful than jensen.

$ \frac{a^2}{3\minus{}a^3}$ is "Half concave and half convex" in $ (0,3)$ ...

[quote="kuing"]$ \frac {a^2}{3 \minus{} a^3}$ is "Half concave and half convex" in $ (0,3)$ ...[/quote]and it is even entirely convex in $ [0,\sqrt[3]{3})$ which is what we need here. So Jensen works. :) Fortunately, because I don't think "half concave and half convex" works when there is a pole in the middle.

[quote="spanferkel"][quote="kuing"]$ \frac {a^2}{3 \minus{} a^3}$ is "Half concave and half convex" in $ (0,3)$ ...[/quote]and it is even entirely convex in $ [0,\sqrt [3]{3})$ which is what we need here. So Jensen works. :) Fortunately, because I don't think "half concave and half convex" works when there is a pole in the middle.[/quote]

sorry, is $ \frac {x^{\frac {2}{3}}}{3 \minus{} x}$ half concave and half convex here...

let $ a^3 \equal{} x, b^3 \equal{} y,c^3 \equal{} z$, that $ x,y,z > 0, x \plus{} y \plus{} z \equal{} 3$, inequality becomes

$ \sum\frac {x^{\frac {2}{3}}}{3 \minus{} x} \ge \frac {3}{2}$

but $ \frac {x^{\frac {2}{3}}}{3 \minus{} x}$ is half concave and half convex in $ (0,3)$ so we can't just useing jensen's inequality to prove...

if for $ \frac {a^2}{3 \minus{} a^3}$ convex in $ [0,\sqrt [3]{3})$ we only get
$ \sum{\frac {a^2}{3 \minus{} a^3}} \ge 3\cdot\frac {(\frac {a \plus{} b \plus{} c}{3})^2}{3 \minus{} (\frac {a \plus{} b \plus{} c}{3})^3}$
but actually, $ 3\cdot \frac {(\frac {a \plus{} b \plus{} c}{3})^2}{3 \minus{} (\frac {a \plus{} b \plus{} c}{3})^3} \le \frac {3}{2}$ ...

oops you're right. So let's better stick with the tangent line method here  :lol:  ;)

[quote="shoki"]if $ a,b,c > 0$ prove that :

$ \sum \frac {a^2}{b^3 \plus{} c^3} \ge \frac {3}{2 \sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}}$

sorry if it has been posted before.[/quote]

The stronger Inequality seems also true for all $ a,b,c$ be positive real numbers
\[ \sum_{cyc} \frac {a^2}{b^3 \plus{} c^3} \ge \frac {3}{2 \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}}\]


$ n$ fair 6-sided dice are simultaneously rolled. The probability that exactly two of them show a number other than 1 is $ \frac{25}{216}$. Find $ n$.

We have $ \frac {25}{216} \equal{} \dbinom n2\left(\frac56\right)^2\left(\frac16\right)^{n \minus{} 2}$.

It is easy to see that $ n \equal{} \boxed4$ satisfies this.

[quote="aattss"]Were we supposed to notice that 5 squared is 25 and 6 to the power of four is divisible by 216?[/quote]

Yes, I think so.

How did they get the inequalities? I got the rest of the solution, I just don't get how they got the inequalities.

It can't be 3 for this reason:

It is easy to see that 5 squared is 25, and 6 cubed is 216. So people might think $\frac{1}{6}*\frac{5}{6}*\frac{5}{6}$ are the probabilities.
However, there are 3 ways the 1 could be rolled: first, second, or third. And that messes up the result.

[hide=Post #8 by vinayak-kumar][url=aops.com/community/user/112039][b]vinayak-kumar[/b][/url] · Jul 31, 2013, 7:15 PM [url=aops.com/community/p3164761](view)[/url][color=transparent]helo[/color]-----
Inequalities? I don't see any inequalities. And it cant be $3$ because $4^3\ne 256$.

-----
[color=#5b7083] #8[color=transparent]hellloolo[/color] [aops]x[/aops] 8 [color=transparent]hellloolo[/color]  [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

[tip=@vinayak-kumar][img]https://avatar.artofproblemsolving.com/avatar_112039.jpg[/img]
AoPS User[/tip] wdym you can't see the inequalities??

[hide=snippet of alcumus sol]Note that if $n \geq 5$, then $n < 3^{n-3}$ and $n-1 < 2\cdot 2^{n-3}$. We can multiply these inequalities to get that when $n \geq 5$, we have $n(n-1) < 2\cdot 6^{n-3}$, or $108n(n-1)<6^n$.[/hide]

see? inequalities

They said that the dice were rolled simotaniously. wouldn't we not need the n choose 2?

[hide=cheese] so I realized that there have to be 2 5/6’s and x-2 1/6’s. You have to multiply by Xc2 because of ordering. You can notice that the probability is noticeably large, and when you multiply by Xc2 it has to only effect the denominator by a power of 6. Therefore, the only logical answer for x would be 4, which is the right answer. [/hide]

[hide=sfd;gj gkl;sdjgl;sdkjf;gksdlfg jdl;ksfgjd;slfkgj sl;d]use guess and check[/hide]


I would assume that it ate integer amounts of the mutton:
so all we are doing is finding solutions to:
[hide]7u+11c=94 (where u is the number of cutlets and c is the number of chops)
so:
we just find value that work...
just start subtracting 7's from 94 and we get: 7(4) + 11(6) = 94. so turns out that 4 cutlets and 6 chops were eaten
[/hide]

[hide]We see that 94=3mod7 and 11=4 mod 7 so we determine when 3mod7=4xmod7 which occurs at x=6. We check: 94-6*11=28, so he ate 6 chops and 4 cutlets[/hide]

Pretty mods!

Solafide, sorry, but could you tell me you reasoning behind that? Why you chose to mod those perticular numbers? I can't seem to get it.

[quote="Gyan"]A Labrador  called Button
Ate Ninety-four rupees  worth of mutton
Each cutlet cost seven,
Each chop  cost eleven.
How much did he eat,
 That  old glutton?
:)

(ps a rupee is  just unit of  indian  money ..)
Level - Hard mathcounts level[/quote]
[hide="solution"]
x = cutlet
y = chop

7x + 11y = 94
You don't have any more info, so we plug and chug.

x = 4
y = 6

Greedy little dog :P .
[/hide]

Good work. All naswers are correct.
Button, if you are interested, is now  on  [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=25885]Diet ....:)[/url]

[hide="dwx does this help?"]Since 94 :equiv: 6(mod11) you know that when you buy 8 chope you are 6 rupees short.  For each chop less you buy, you assume you buy a cutlet instead which leaves you still 4 more short since 11 :equiv: 4(mod7).  Keep adding 4 to the six you are short until it is  :equiv: 0(mod 7).  That happens(14) after 2 trades of chop to cutlet.  So you bought only 6 chops, traded for 2 cutlets, and have money for 2 more cutlets.  Total : 6 chops, 4 cutlets.[/hide]


[color=darkblue]Let $ (O;R)$ is a circle. Let $ A, B, C$ are three points lie on circle $ (O;R)$. Let $ D$ is midpoint of arc $ AB$, let $ E$ is midpoint of arc $ AC$. Let $ DE$ meet $ AB$ and $ AC$ at $ H$, $ K$ respectively.

1) Prove that $ DE \perp AI$.

2) Prove that $ IK//AB$.[/color]

ADIE - kite, AHIK - rhombus, done.

Best regards,
sunken rock

Dear Mathlinkers,
(1) is well known traditionally, perhaps a new proof can be found
(2) note Td the tangent to (O) at D ; applied the Pascal degenerated theorem to Td EBACD.
Sincerely
Jean-Louis


Explain why the liquid in the two beakers behave the way they do.

Because the surface tensions of two liquids are different. :)

You are very right.
In fact in the first case the adhesion strenght, between the particles of the liquid and the particles of the container, is much more intense then the coesion strenght, between the particles of the liquid. So the contact surface tents to be maximum. This is the case of water.
In the second case  the adhesion strenght is less intense then the coesion strenght so the contact surface tents to be minimum. This is the case of Hg.


function
trigonometry
algebra unsolved
algebra

Is $ \{x\}\plus{}\sin{x}$ a periodic function?

$ \{x\}\equal{}x\minus{}[x].$

[quote="bilarev"]Is $ \{x\} + \sin{x}$ a periodic function?

$ \{x\} = x - [x].$[/quote]
No ,this function is not periodic . Suppose this function is periodic ,it means that there exist a real number T such that :
\[ f(x) = f(x + T),\forall x\in R
\]
Because this function is periodic T ,so we also have :$ f(x) = f(x + nT),\forall x\in R$
Take $ x = 0$ then we have :
$ \sin{nT} + \{nT\} = 0,\forall n\in N$ (1)
We have two case :
Case 1 : $ \frac {T}{\pi}$ is a rational .

Chose $ nT = 2k{\pi}$ then (1) can not hold .

Case 2 : $ \frac {T}{\pi}$ is an irrational number.

Then we have :
${ \{\frac {nT}{2\pi}}\}$ dense on $ (0,1)$
We only need to chose $ n$ such that :
${ \{\frac {nT}{2\pi}}\} < \frac {1}{4}$ then (1) also can't hold .
Therefore this function can't be  periodic .  :)


sorry
                $\sum\frac{1}{1-ab}\leq\ 8$

Is the statement correct? I see
\[2ab < a^{2}+b^{2}< 1\qquad ab<\frac{1}{2}\qquad 1-ab>\frac{1}{2}\qquad \frac{1}{1-ab}<2\]

is hard  :maybe:
and you can use SMV theorem


calculus
integration
function
real analysis
real analysis unsolved

Suppose that $(f_n)_{n\in N}$ be the sequence from all functions $f_n:[0,1]\rightarrow \mathbb{R^+}$ s.t. $f_0$ be the continuous function and $\forall x\in [0,1] , \forall n\in \mathbb {N} , f_{n+1}(x)=\int_0^x \frac {1}{1+f_n (t)}dt$. Prove that for every $x\in [0,1]$ the sequence of $(f_n(x))_{n\in N}$ be the convergent sequence and calculate the limitation.

This is in the wrong forum, I will move it.


1-Qingchun Ren [CHN]
2-Hangsheng Diao [CHN]
3-Ye Luo [CHN]
4-Xuancheng Shao [CHN]
5-Aliaksei Levin [BLR]
6-Rosen Dimitrov Kralev [BGR]
[color=blue]7-Peter Scholze [DEU][/color]
8-Aliakbar Daemi [IRN]
9-Toshiki Kataoka [JPN]
10-Tsukasa Kuribayashi [JPN]
[color=red]11-Iurie Boreico [MDA] (outstanding solution)[/color]
12-Cheng-Chiang Tsai [TWN]
13-Shinh-Yih Huang [TWN]
[color=blue]14-Andrey Gavrilyuk [RUS][/color]
15-Sergii Slobodianiuk [UKR]
[color=blue]16-Brian Lawrence [USA][/color]

-blue colored ones are Mathlinkers including Iurie Boreico.

How many of them are going to Slovenia??'

Does anyone have Iurie Boreico's solution????

I think Andrew and Brian will not come, Brian is involved in RSI, while Andrew (I might be mistakening) has graduated.

[quote="janieluvsmusic"]what are their mathlinker names?[/quote]

- Look for the previous posts in this section.

- I only know that, Magazinov from Russian team who scored 41 in IMO 2005 will be in IMO 2006, too.

Official solution :  :)

To prove the inequality it is enought to show this

$3 \geq \frac{x^2+y^2+z^2} {x^5+y^2+z^2} + \frac{x^2+y^2+z^2} {y^5+z^2+x^2} + \frac{x^2+y^2+z^2} {z^5+x^2+y^2}$

In view of the Cauchy-Schwarz inequality and the condition $xyz \geq 1$ we have :

$(x^5+y^2+z^2) (yz+y^2+z^2) \geq (x^\frac{5}{2} y^ \frac{1}{2} +y^2 + z^2)^2 \geq (x^2+z^2+y^2)^2$

or $\frac{yz+y^2+z^2}{x^2+y^2+z^2} \geq \frac{x^2+y^2+z^2}{x^5+z^2+y^2}$. Taking the cyclic sum

and using the fact $x^2+y^2+z^2 \geq xy+yz+zx$ we are done.

[quote="Davron"]Here is a Iurie's elegant proof :

To prove the inequality it is enought to show this

$3 \geq \frac{x^2+y^2+z^2} {x^5+y^2+z^2} + \frac{x^2+y^2+z^2} {y^5+z^2+x^2} + \frac{x^2+y^2+z^2} {z^5+x^2+y^2}$

In view of the Cauchy-Schwarz inequality and the condition $xyz \geq 1$ we have :

$(x^5+y^2+z^2) (yz+y^2+z^2) \geq (x^\frac{5}{2} y^ \frac{1}{2} +y^2 + z^2)^2 \geq (x^2+z^2+y^2)^2$

or $\frac{yz+y^2+z^2}{x^2+y^2+z^2} \geq \frac{x^2+y^2+z^2}{x^5+z^2+y^2}$. Taking the cyclic sum

and using the fact $x^2+y^2+z^2 \geq xy+yz+zx$ we are done.[/quote]

Cool!

[quote="Pascual2005"]Thats not his proof, that is the oficial solution, his solution is much more cooler!, He didnt used cauchy, he just completed some squares and do some manipulations... If I remember it i can post it later....[/quote]

OK

What is the username of the Brain? (Brian Lawrence)

If I am not mistaken, Darkseer on mathlinks is Cheng-Chiang Tsai [TWN]

[quote="mathmanman"]Actually, [b]Iura[/b]'s proof is already on the forum : http://www.mathlinks.ro/Forum/viewtopic.php?t=44480 . :)[/quote]

Sorry but I didn't find it... what number of post is it????

Did you read the thread?
Actually, his solution is on the first page (posted by [b]jpe[/b] if I remember well), but it is also said on the second page.. :)

See exactly here
http://www.mathlinks.ro/Forum/viewtopic.php?p=301848#p301848

[quote="Davron"]1-Qingchun Ren [CHN]
[/quote]
He shall be in Slovenia this year again.

Just out of curiosity, what are the usernames of the MathLinkers?

7 - Peter Scholze
11- iurie
14- GHAVR

but i don't know who is Brian in mathlinks.ro

davron

hmmm yes i am sure about that triple and i am not sure about Brian.

davron

Guys I just guess that Tosiki Kataoka from Japan,will be in Slovenia too.
So I think we can again expecte too much perfect scors this year :idea: .

[quote="shobber"]I think there was an awarding ceremony for the perfect scorers in IMO 2005. I have a photo of it which shows the four Chinese boys. But in the middle of them was a handsome boy which extremely looks like Peter.  So I wanna ask: Was it you, Peter?[/quote]
Finally, I am able to attach a picture.

What about the boy at the far left side? Is he also on the mathlinks?

It is Aliakbar Daemi from Iranian team.
I think he is not a mathlinker.

Wow, I didn't know we had any perfect scorers here on mathlinks! (as far I know Manolescu, Reid Barton, Reiher) all aren't a member.
Congratulations to all participants of course.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=101347[/url]
is the topic that made me found out about Peter Scholze's very good results (silver and perfect paper  :o ).  I am quite amazed that Germany has two strong players (remember Christian Reiher) in a row!

By the way, who is setout, and what exactly is his claim in that post?

[quote="fredbel6"]Wow, I didn't know we had any perfect scorers here on mathlinks! (as far I know Manolescu, Reid Barton, Reiher) all aren't a member.[/quote]It seems that you don't know far enough ;) Some of them are members of the site ;)

It seems that Peter is again in the middle of the Chinese!   :rotfl:

http://imo2006.dmfa.si/images/20060717/IMG_1205.JPG

[quote="setout"][quote="Peter Scholze"]Well, having quite the same scores as a few chinese guys, this isn't very surprising ;) (two chinese had 35 like me, one 34)...[/quote]

One thing is being a gold medailist

and to be gold medailist and play honestly and fair is the other one.

http://www.mathlinks.ro/Forum/viewtopic.php?t=101347[/quote]

One thing is to post a few insults to abreact after a bad test,

and to continue flaming and spam other topics is the other one.

  darij

[quote="setout"][quote="darij grinberg"]
and to continue flaming and spam other topics is the other one.
darij[/quote]

I am not trying to spam, I just try to protect all the future contestants from this kind of behavior.[/quote]
No you go one step further and claim that this was pure intention.
And you seem to write this claim and the rest everywhere (especially at places Peter posts to), which is spamming and helps noone.


induction
Functional Analysis
linear algebra
real analysis
real analysis unsolved

Let $ f, f_1, ... , f_n$ be such linear functionals defined on a linear space $ L$, that if $ f_1(x) \equal{} ... \equal{} f_n(x) \equal{} 0$, then $ f(x) \equal{} 0$. Prove that then there are such constants $ \alpha_1, ..., \alpha_n$ : $ f(x) \equal{} \sum_{i\equal{}1}^{n} \alpha_k f_k(x)$ for every $ x \in L$ (actually $ f(x)$ may be represented as a linear combination of $ f_k(x)$).

Thanks a lot if someone can help with this - I tried to prove this fact but failed; the only thing I know is that you have to use fact $ \cup Ker(f_k) \subset Ker(f)$ when proving this.

(found this problem at Functional analysis book, so I think it doesnt belong to linear algebra)

It is in fact a linear algebra problem. The trick is.. induction on n.

It seems to be a bit difficult for me because I thought this problem isnt so easy to solve - can you please tell me what's induction basis and step here? Оf course, those coefficients $ \alpha_1, .. \alpha_n$ are the same for every $ x$. Thank you.

For just $ f_1$, the base case is this. The co-dimension of the nullspace of a linear functional is 1, so write $ L \equal{} X \plus{} N_{f_1}$ where the sum is direct and $ X$ has dimension 1. Both $ f_1$ and $ f$ live only on $ X$, so the conclusion is clear.

edit: Well I suppose the above requires zorn's lemma. Probably an algebraic trick can do it:

$ f_1(x \minus{} \frac{f_1(x)}{f_1(z)}z) \equal{} 0$
So $ f$ applied to the same thing is zero, so
$ f(x) \equal{} \frac{f(z)}{f_1(z)}f_1(x)$

I don't remember if this particular proof requires Zorn's lemma; I probably won't be able to answer for a few days but if you can consult the linear algebra book by Hoffman, Kunze, the proof is in there.

Zorn lemma? :what?: Just notice that the values $ f_1(x),\dots,f_n(x)$ determine the value of $ f(x)$ uniquely and that the resulting mapping is linear. Now just recall that any linear mapping from a linear subspace of $ \mathbb C^n$ to $ \mathbb C$ can be written as $ y\mapsto \sum_j a_j y_j$.

The misguided zorn's lemma comment was just because I wanted to choose a complementary subspace to the nullspace of $ f_1$ - which in this case you can do easily by just choosing a non-zero element in the quotient $ L/N_{f_1}$.

The second thing I wrote gives another proof for the base case $ n\equal{}1$.

Any way, this is a standard problem which seems a little bit elusive because most authors just leave it as an exercise. But fedja gives a very neat proof.

[quote="Renevan"]Thanks for answers.
§outh§tar, I checked book you mentioned - unfortunately, I can't find exactly this fact there.
Well, also I think fedja's proof is enought obviously, but I found this problem at book : Kolmogorov, Fomin "Introductory Real Analysis", so I think proof should be based on the facts of functional analysis. Also this problem precedes paragraph about null spaces and hyperplanes, so I think Kalle's proof is most closely here - and likely Zorn's lemma is used here, because we used it a few times before when were proving other facts from that book. Can you please explain a bit more specifically what should we do here - prove that $ f(x)$ is maximal element in partially ordered set $ L$? Do we suspect here that if $ f(x)$ and $ g(x)$ are linear functionals on $ L$ then $ f < g$ if $ Ker(f) \subset Ker(g)$ ? Pff, I think I'm on a wrong way  :( . Maybe I can find full proof using facts listed above (I mean null spaces and co-dimensions) somewhere so I can read it and understand; dont know why but this seems to be too hard for me...Thanks in advance for all help and hints.[/quote]

See page 110 of the H&K book

Oops, it seems I didnt pay enought attention last time I checked this book  :blush:
Thanks to all, now it's clear.

Dear Renevan,

since all your posts are about Functional Analysis,
I hope the following book will be of use for you:
Kadets, V.M. A course of Functional Analysis /
Kharkov National University, 2006, 607 pp. (Russian)

One can download it from
http://page.mi.fu-berlin.de/~werner/kadetsbook/

In particular the fact you are asking about is
just Lemma 1 from Chapter 16.3.3 "Weak topologies"
(page 517).


Εστω ο διανυσματικος χωρος $ V$ διαστασης $ n$ και $ f: V \rightarrow V$ γραμμικη απεικονιση με $ f^n(x) \equal{} 0, f^{n \minus{} 1}(x) \neq 0$. Νδο υπαρχει διανυσμα $ x \in V$ τετοιο ωστε το συνολο $ \{x, f(x), f^2(x), ... , f^{n \minus{} 1}(x)\}$ να ειναι βαση του $ V$

Μάλλον εννοείς ότι υπάρχει $ x \in V$ ώστε $ f^n(x) \equal{} 0$ και $ f^{n \minus{} 1}(x) \neq 0$. (Διότι για όλα τα $ x \in V$ αποκλείεται να ισχύει.)

Τότε γι'αυτό το χ, τα $ \{x, f(x), f^2(x), ... , f^{n \minus{} 1}(x)\}$ είναι γραμμικώς ανεξάρτητα (και άρα βάση του $ V$). Πράγματι αν $ \lambda_0 x \plus{} \lambda_1 f(x) \plus{} \cdots \plus{} \lambda_{n\minus{}1}f^{n\minus{}1}(x) \equal{} 0$ τότε ...

[quote="Demetres"]Μάλλον εννοείς ότι υπάρχει $ x \in V$ ώστε $ f^n(x) \equal{} 0$ και $ f^{n \minus{} 1}(x) \neq 0$. (Διότι για όλα τα $ x \in V$ αποκλείεται να ισχύει.)

Τότε γι'αυτό το χ, τα $ \{x, f(x), f^2(x), ... , f^{n \minus{} 1}(x)\}$ είναι γραμμικώς ανεξάρτητα (και άρα βάση του $ V$). Πράγματι αν $ \lambda_0 x \plus{} \lambda_1 f(x) \plus{} \cdots \plus{} \lambda_{n \minus{} 1}f^{n \minus{} 1}(x) \equal{} 0$ τότε ...[/quote]

Τι να πω... μετεφερα την ασκηση οπως ακριβως μας την εδωσαν (βασικα το λεει ακριβως ετσι: η f εχει την ιδιοτητα $ f^n \equal{} 0, f^{n \minus{} 1} \neq 0$)... και εγω ετσι την ελυσα με την προυποθεση οτι τα παραπανω ισχυουν για καποια και οχι για ολα τα $ x$. Απο την αλλη ομως μου φανηκε πολυ απλο για να ειναι σωστο και γι αυτο και το ανεβασα εδω (το προβλημα αυτο μας το ειχαν δωσει για την seemous προετοιμασια)

[quote="Demetres"][quote="NickNafplio"]
βασικα το λεει ακριβως ετσι: η f εχει την ιδιοτητα $ f^n \equal{} 0, f^{n \minus{} 1} \neq 0$
[/quote]

Όταν λέει $ f^n \equal{} 0$, εννοεί ότι η $ f^n$ είναι η μηδενική συνάρτηση. Δηλαδή για κάθε $ x \in V$ ισχύει $ f(x) \equal{} 0$. Όταν λέει $ f^n \neq 0$, εννοεί ότι η $ f^n$ δεν είναι η μηδενική συνάρτηση. Δηλαδή υπάρχει $ x \in V$ για το οποίο ισχύει $ f(x) \neq 0$.[/quote]

Α οκ τοτε,
Ευχαριστω.


calculus
integration
algebra
function
domain
superior algebra
superior algebra unsolved

Let $ A\not \equal{}0$ be some finite dimensional $ \mathbb{C}$-algebra with no zero-divisors $ \not\equal{}0$.
(not necessarily commutative)
Then for $ n: \equal{}\dim_{\mathbb{C}}A$ and any $ a\in A$
we have $ 1,a,a^2,\ldots,a^{n\minus{}1}$ linear dependent over $ \mathbb{C}$.
So we find some non-zero $ f\in\mathbb{C}[X]$ of degree $ <n$ with $ f(a)\equal{}0$.
But $ \mathbb{C}$ is algebraically closed, so we find $ c,\lambda_1,\ldots ,\lambda_m\in\mathbb{C}$
with $ f\equal{}c(X\minus{}\lambda_1)\cdots (X\minus{}\lambda_m)$, and therefore $ 0\equal{}f(a)\equal{}c(a\minus{}\lambda_1\cdot 1_A)\cdots (a\minus{}\lambda_m\cdot 1_A)$.
But $ A$ has no zero-divisors $ \not\equal{}0$, which gives $ a\equal{}\lambda_k\cdot 1_A\in\mathbb{C}1_A$ for some $ k$ and all $ a\in A$.
So $ A\equal{}\mathbb{C}1_A\cong_{\mathbb{C}} \mathbb{C}$ is the only non-zero zero-divisor-free finite dimensional $ \mathbb{C}$-algebra.

[quote="puuhikki"]Does there exists a $ \mathbb{C}$-algebra which is an integral domain but not a field?[/quote]
$ \mathbb{C}[X]$?


If we keep a piece of ice in a sweater(not totally wrapped around by the sweater,just place it inside it),will it melt fster or slower?(And ofcourse,why)

The woolen material of a sweater prevents heat loss or heat gain from the
surroundings. Sweaters , by default are dumb - they don't care if the occupant is
a freezing blighter or a block of ice, the heat transfer is slow: so the ice melts slower,
provided the sweater is not wrapped around it, in which case, thetrapped air may melt the ice quickly.


Εστω κυκλος κεντρου $ O$ και σημειο $ P$ εκτος αυτου. Εστω τυχαια τεμνουσα $ PAB$ του κυκλου και η εφαπτομενη $ PC$ που βρισκεται προς το μερος της $ PO$ προς το οποιο κειται και η $ PAB$. Εστω $ D$ το αντιδιαμετρικο του $ C$ και $ M,N$ τα κοινα σημεια της $ OP$ με τις $ DA,DB$ αντιστοιχως. ΝΔΟ $ OM \equal{} ON$.

[quote="Dimitris X"]Η λύση μου είναι η εξής:
Φέρνουμε την άλλη εφαπτόμενη από το $ P$ προς τον κύκλο.Έστω $ PC'$
To $ BCAC'$ είναι αρμονικό τετράπλευρο.
Άρα η $ D(BCAC')$ είναι αρμονική τετράδα.
Για να δείξουμε λοιπόν ότι $ OM \equal{} ON$ ερκεί η $ DC'$ να είναι παράλληλη προς την $ PO$,το οποίο ισχύει γιατί εαν φέρουμε την $ CC'$ έχουμε $ CC' \bot OP$ και $ \angle CC'D \equal{} 90$ γιατί βένει σε ημικύκλιο..

Done!!![/quote]

Πολύ ωραία λύση, εξαιρετικό παράδειγμα για το αντίστοιχο μάθημα !

 Σάκη, έχεις μήπως και άλλη , διαφορετική  , λύση ;

Μπάμπης

Κ. Στεργιου η λυση που εδωσα στην ασκηση ειναι αρκετα πιο περιπλοκη και ως εξης:

Εστω η //λη απο το $ D$ προς την $ OP$ η οποια τεμνει την $ PC$ στο $ X$. Αρκει τοτε η δεσμη $ D(DX,DM,DO,DN)$ να ειναι αρμονικη. Τοτε θα ισχυει $ MO\equal{}ON$.
Εστω οτι η $ DN$ τεμνει την $ PC$ στο $ Q$ και η $ DM$ την $ PC$ στο $ S$. Τοτε αρκει $ (X,S,C,Q)\equal{}\minus{}1$.

Ομως $ OC\equal{}OD$ και $ OP//DX$. Αρα $ PC\equal{}PX$. Για να ειναι η παραπανω τετραδα αρμονικη αρκει απο θεωρημα Newton  $ PC^2\equal{}PX^2\equal{}PS.PQ$.
Ομως $ PC^2\equal{}PA.PB$. Οποτε αρκει $ PS.PQ\equal{}PA.PB$, δηλαδη αρκει το $ SABQ$ να ειναι εγγραψιμο.

$ \angle{SQB}\equal{}\angle{BCD}$ (καθετες πλευρες).
$ \angle{BCD}\equal{}\angle{BAD}\equal{}\angle{SAP}$. Αρα $ \angle{SQB}\equal{}\angle{SAP}$ και αρα $ SABQ$ εγγραψιμο.


Let $f(x)=mx+b$. For what values of $m$ and $b$ will $f(x)=f^{-1}(x)$ for all $x$?

[hide]
That's just reflecting it over the line $y=x$, so it's either $(m,b)=(1,0)$ or $(m,b)=(-1,b)$ for any $b$.[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $G$, $H$ and $K$ be finite groups such that $G \times H \cong G \times K$. Is it necessary that $H \cong K$?

Note that the corresponding result is not true for infinite groups, for instance, $G \times G^\omega \cong G^2 \times G^\omega$. However, it may be true for finitely generated groups.

I have managed to prove that $H$ and $K$ are fairly similar if $G \times H \cong G \times K$. That is, I have shown that:

\begin{itemize}

\item The number of homomorphisms from any finite group to $H$ equals the number of homomrphisms to $K$.

\item For an Abelian group, the number of homomorphisms from $H$ to that Abelian group equals the number of homomorphisms from $K$ to that Abelian group.

\item The Jordan Holder decompositions of $H$ and $K$ comprise the same simple quotient groups.

\end{itemize}

Hi

I 'm not presenting all the proof details for these statements. I have put up a document (that I'm still finishing) describing my knowledge state of the problem. The url is:

[url]http://www.cmi.ac.in/~vipul/unsolvedproblems[/url]

Vipul

This reminds me of http://www.mathlinks.ro/Forum/viewtopic.php?t=86575 ... it's a special case of your question.


i think this belongs in here... not entirely sure

what are the real solutions to [tex]2^{x}=x^{2}[/tex].

kind of hint: [hide]there are three solutions[/hide]

I could only find that [hide]2 and 4[/hide] were solutions. I can't find the third...

I'm pretty sure there's no way to find the third using elementary methods.

Yeah, although we can narrow it down to be somewhere between -1 and 0 fairly easily, (and between -1 and -1/2 with a bit more work), there's no good way to find the last root, which (mathematica tells me) has a value of about -0.766664695962123....

Can the third root be expressed in radicals?

I got 2 and 4 by simply looking at it.  All I knew about the last one was that it was negative.  So if anyone knows an elementary(with high school algebra or easier for example) way to find the last one I d be happy to know because I can't find one.

You can't :)

Well, I'd be very surprised if you could.

Using Maple, I got 2, 4, and [tex]-2{\frac {{\it LambertW} \left( \frac{\ln  \left( 2 \right)}{2}  \right)
}{\ln  \left( 2 \right) }}
[/tex]

Note: LambertW function is defined as [tex]LambertW(x)*e^{LambertW(x)}=x[/tex]

Yeah, the Lambert W-function sure sounds like "Getting Started" material to me...

Yeah, but it's the way Maple expressed the answer.

It's a trancendental function -- basically, just like the logarithm function (which solves the equation elog(x) = x), there's no elementary way of expressing it except in special cases.

i wasn't exactly sure about the third one.. i just knew it was there. the question showed up on a contest as "find the number of solutions".. and the popular answer was 2.

Visual inspection of the equation should yield 2 and 4; when you draw the graphs of both f(x) = 2x and f(x) = x2, they have to intersect somewhere in the 2nd quadrant. Thus there has to be another solution.

Hi - One way to intutively see this right away, is to look at the graph of the functions (plot it if necessary) . It is  fun to know  the shape of, at least the familier function , as welll as your familier friends..:)

$y=2^x$  is zero when  $x$ is  $-\infty$ ,  1 when  $x=0$ and rises rather fast (like $e^x$)
$y=x^2$ is familier parabola and  rises  slower than the above curve...

It is easy to see the two points (x=2, x=4) but obviously parabola has to cut at one more point when  x is negative, because first curve is going 'down' and the other is going up...

[hide]Another way to look at it (ant that is what i  did ) the function  $y = \ln (|x|) $ and $y=3 x$ ( 2/ln2 is about 3)  and again the line cuts that inverted volcano in three places...) [/hide]

[quote="shobber"][b]
Can someone give  me a solution to the first two roots in Gettign Started ways?
Don't tell me by saying the word $looking$![/b][/quote]

Nope.

You can only really find them by guessing.


The sum of the lengths of all the edges of a particular cube is 24 cm. When the cube is unfolded into a flat sheet (net) of six connected squares, each sharing at least one side with one another, what is the perimeter of the resulting polygon?

[hide]Any cube has 12 edges, so each one is long $ 2$.
When we unfold the cube on a plane, we get a sort of latin cross, whose perimeter is $ 2\cdot 14=28$[/hide]


Consider a triangle $ABC$. We have two points $D$ and $E$ such that $\measuredangle ADB=\measuredangle AEC=90^{\circ}$ and that triangle $ADB$ is inversely similar to triangle $AEC$. The points $D$ and $E$ both lie outside the triangle $ABC$. Let the lines $CD$ and $BE$ intersect at $R$. Prove that $AR$ is perpendicular to $DE$.

Ok guys, don't be mad at me or anything, but I'll use a homography yet again :).

Consider the points $A,D,E$ and the lines $DB,EC$ to be fixed and consider the mapping $B\mapsto C$ s.t. $ADB,AEC$ are similar and inversely oriented. This is clearly a homography, and it induces a homography from the pencil through $E$ to the one through $D$, $EB\mapsto DC$. Since $ED$ is mapped to itself, it means that this homography is a perspectivity, so the locus of $R$ is a line. All we need to do now is choose two particular positions of $R$, show that $AR\perp DE$ in those two situations, and the rest will follow.

As the two particular positions choose a) $B\to\infty$; in this case $C\to\infty$, and $R$ will be the intersection of two heights of $ADE$, meaning that it will also lie on the third one, so $AR\perp DE$; b) the case when $ADB,AEC$ are right isosceles triangles; in this case the problem is really well-known (it's probably even been discussed on the forum).

[quote="vinoth_90_2004"]Consider a triangle $ ABC$. We have two points $ D$ and $ E$ such that $ \measuredangle ADB \equal{}\measuredangle AEC \equal{} 90^{\circ}$ and that triangle $ ADB$ is inversely similar to triangle $ AEC$. The points $ D$ and $ E$ both lie outside the triangle $ ABC$. Let the lines $ CD$ and $ BE$ intersect at $ R$. Prove that $ AR$ is perpendicular to $ DE$.[/quote]

we can solve the problem in the following way.
Let F and G be the orthogonal projections of D and E on AB and AC respectively. If K=EB^DG and L=EF^DC then one can prove(*) that DLKE is cyclic. Then if M is the (second) point of intersection between c(ADF) and c(AGE), it holds, from radical axes theorem that, MA, DL and EK concurr at R. From this it follows quite strigthforward the claim.

(*) If K'=c(AGE)^c(ABD it is easy to see that K' lies on BE. If K=/=K' we would have that (let's suppose that K lies on the segment EK') <EAG lessthan <EKG = <DKB > <DAB which is a contraddiction. Then K==K' and <DKE = 180 - <GKE=180-<GAE. Similarly one can prove that <DLE=180 - <DFA  then we are done.

[quote="Armo"]Hey guys I have just proved that when $\angle{DBA}=\angle{ACE}=90^{\circ}$ then  $AR\perp{BC}$ . But only for acute angled triangle, yet. I"ll try to check it when one of the angles B or C is greater than 90. Any ideas?[/quote]

Actually is the same problem wrt the original one. Is enough to exachange the role of the triangle ABC and ADE. The above proof works for both configuration regardless the "acuteness" of triangle.

Yes,you are right sprmnt21.Now I have the proof for all triangles.But I used many   ratios in my solution and it isn't so nice as yours .So I shall not post it. :)


Obviosly $ f(\frac{k}{n})\plus{}f(\frac{n\minus{}k}{n})\equal{}1$, therefore sum is $ \frac{n\plus{}1}{2}.$


linear algebra
matrix
combinatorics unsolved
combinatorics

Let $ S_1,S_2,...,S_m \subseteq \{1,2,...,n\}$.For every  $ i \not \equal{} j$,$ S_i \not \equal{} S_j$,also $ \mid S_i \cap S_j \mid \equal{} 1$.Prove that $ m \le n$.

Re: "maybe well-known too," yes, it is.  In fact, there is another copy of the same problem on the front page of this very forum right now :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=256900

My apologize for double(maybe more than double) posting of this problem :blush:
But,Could somebody post "really" elementary solution for this one.'cause,By the link above i just found solution using some advanced level math(such as matrix) :blush:


Let x,y,z be positive real numbers such that z=x+y. Prove that

 (x^2+y^2+z^2)^3 \geq 54x^2y^2z^2


Let P be a point inside regular pentagon ABCDE, such that triangle APE is equilateral. Find the degree measure of obtuse angle BPD.


Circle $O$ is inscribed in quadrilateral $ABCD$. The tangent on $AB,BC,CD,DA$ are $E,F,G,H$. $EF$ and $GH$ intersects at $K$. $OK$ intersects $BD$ at $K'$. Prove that $OK$ is perpendicular to $BD$ and $OK = \frac{r^{2}}{OK'}$, where $r$ is the radius of the circle.

Both statements are equivalent to proving that $BD$ is the polar of $K$ wrt the circle.

Proof:

$EF$ is the polar of $B$ wrt the circle.
$GH$ is the polar of $D$ wrt the circle.

$\Rightarrow$ $K$ is the pole of $BD$ wrt the circle.
Done!  :P