geometry
3D geometry
inequalities
MATHCOUNTS
number theory
prime numbers

Ok..I took the Sprint round, and got a really fail score(23)... I made some dumb mistakes as usual:
12/48=1/48 (:wallbash_red:)(Thats only one of my mistakes :maybe: )

Here are some problems I need help on:

1.The two positive integer solutions of the equation: $ x^2\minus{}mx\plus{}n\equal{}0$ are $ k$ and $ t$, where $ m$ and $ n$ are both prime numbers $ k$>$ t$. What is the value of $ m^n\plus{}n^m\plus{}k^t\plus{}t^k$?
I keep getting fail answers when I try this... :(

2.How many distinct pairs of perfect squares differ by 35?
Why do I keep getting $ 4$, when the answer key says its $ 2$?!

3.Blah Blah class is made up of 20% 6th graders, and 30% 7th graders. The rest are 8th graders. The average age of the 6th graders is 11.4. The average age of the 7th graders is exactly one year greater than that of the sixth graders. If the average age of the eight graders is exactly 2 years greater than that of the 6th graders, what is the average of the entire class?
Idk why I got the wrong answer...my work was all messy, so I couldnt find my solution :blush:

4. The variable $ x$ varies directly as the cube of $ y$, and $ y$ varies directly as the square root of $ z$. If $ x$ equals 1 when $ z$ equals 4, what is the value of $ z$ when $ x$ equals 24?

Bleh..thats it..phew!

Also, my hand is aching now...I dont have the PDF version of this so I had to type it all out...uggh..does anyone have the PDF version?
Also, I know that chapter and states tests' solutions can be found online, but does anyone know where I can find the solutions to the nationals tests?

[hide=" 1"]

We know that $ k$ and $ t$ are both positive integers, and $ k > t$, so $ k \equal{} \frac{m \plus{} \sqrt{m^2 \minus{} 4n}}{2}$ and $ t \equal{} \frac{m \minus{} \sqrt{m^2 \minus{} 4n}}{2}$. Now, in order for these to be real, the discriminant $ m^2 \minus{} 4n$ must be greater than or equal to 0. $ m^2 \minus{} 4n \ge 0 \implies (m \minus{} 2n)(m \plus{} 2n) \ge 0$.

Since the product of these two expressions is greater than or equal to 0, we know they are either both 0, both negative, or both positive. It's impossible for them to both be 0, because than $ m \equal{} 2n$, and they are prime numbers. If they are both negative, we have $ m < 2n$ and $ m < \minus{}2n$. Adding gives us $ 2m < 0$ or $ m < 0$. But if $ m < 0$, then how can it be prime? Therefore, they must both be positive. We have $ m > 2n$ and $ m > \minus{}2n$. Obviously, for this to occur, $ m > n$.

For this to be valid, we need a pair of prime numbers that satisfy those inequalities and so that $ m \minus{} \sqrt{m^2 \minus{} 4n}$ and $ m \plus{} \sqrt{m^2 \minus{} 4n}$ are both positive and divisible by 2. In order for this to happen, $ m$ must be larger than $ \sqrt{m^2 \minus{} 4n}$. Thus we hav $ m > \sqrt{m^2 \minus{} 4n}$ or $ m^2 > m^2 \minus{} 4n \implies 0 > n$...which confuses me...
[/hide]

[hide=" 2"]

We have $ (a \plus{} b)(a \minus{} b) \equal{} 35$, so those two expressions must be factors of 35. If $ a \plus{} b \equal{} 1$ and $ a \minus{} b \equal{} 35$, then $ 2a \equal{} 36 \implies a \equal{} 18, b \equal{} \minus{}17$, which makes the first pair, and the second pair results when $ a \plus{} b \equal{} 5$ and $ a  \minus{} b \equal{} 7$. Thus, we have two pairs.

[/hide]

[hide=" 3"]

We know that 50% of the class is 8th graders, 30% is 7th, and 20% is 6th. We can just go ahead and assume there are 100 students. That means the total age of the 6th graders is $ 20 \times 11.4 \equal{} 228$, that of the 7th graders is $ 12.4 \times 30 \equal{} 372$, and for the 8th graders, it's $ 13.4 \times 50 \equal{} 670$. The total sum of the ages of everyone is 1270, so our answer is $ 1270 \div 100 \equal{} 12.7$.
[/hide]

[hide=" 4"]

We know that $ x \equal{} ky^3$, $ y \equal{} l\sqrt{z}$, and $ x \equal{} m\sqrt{z}^3$. Since $ x \equal{} 1$ when $ z \equal{} 4$, we have $ 1 \equal{} \sqrt{4}^3m \implies m \equal{} \frac{1}{8}$, so $ x \equal{} \frac{1}{8}z$. Thus, when $ x \equal{} 24$, we have $ z \equal{} 192$.
[/hide]

[hide="Hint for Izzy"] If $ m$ and $ n$ are prime numbers and $ k > t$, then $ t = 1$ and $ k = n$. Also, we see that $ m + 1 = n$. Also, the problem stated that they were all positive integers.[/hide]

[hide="2)"] Let the two perfect squares be $ a^2$ annd $ b^2$. We are given that $ a^2 - b^2 = 35$ and are supposed to find all $ (a,b)$ satisfying the above. Rewriting the equation,
\[ a^2 - b^2 = (a + b)(a - b) = 35\]
Let us take the case when $ a + b = 1$.
\[ a + b = 1\]
\[ a-b=35\]
\[ 2a = 36\implies{(a,b) = (18, - 17)}\]
Now for the second case
\[ a + b = 35\]
\[ a-b=1\]
\[ 2a = 36\implies{(a,b) = (18,17)}\]
Let us do the third one
\[ a + b = 5\]
\[ a-b=7\]
\[ 2a = 12\implies{(a,b) = (6, - 1)}\]
Lastly,
\[ a + b = 7\]
\[ a-b=55\]
\[ 2a = 12\implies{(a,b) = (6,1)}\]
Either the answer key was incorrect, or you mistakenly left out the word "positive" or something of that sort. [/hide]

[hide=" 1"]

Ah, so $ m \equal{} 3$ and $ n \equal{} 2$, and from there you can easily find the answer.
[/hide]

For the last one it didnt say positive..thats why I got 4 i think.. :?:  :o

The issue here is, $ (18, \minus{} 17)$, $ (6, \minus{} 1)$, etc. aren't the perfect squares.  It asks for the distinct perfect squares which have a difference of $ 35$.  So actually, we would have $ (18^2, (\pm 17)^2)$ and $ (6^2, (\pm 1)^2)\implies$ $ (324, 289)$ and $ (36, 1)$, our two distinct pairs.

[hide="4"]$ \frac{x}{y^3}\equal{}k \implies x\equal{}ky^3, \frac{y}{\sqrt{z}}\equal{}l \implies y\equal{}l\sqrt{z}$, so substituting yields $ x\equal{}k(l\sqrt{z})^3$.  When $ x\equal{}1, z\equal{}4$.  Substituting yields $ 1\equal{}8kl \implies kl\equal{}\frac{1}{8}$.  Thus, $ x\equal{}27 \implies 27\equal{}\frac{1}{8}\sqrt{z^3} \implies 27 \cdot 8\equal{}\sqrt{z^3} \implies z\equal{}2^2 \cdot 3^2\equal{}\boxed{36}$.[/hide]

This is confirmed by the answer key.

Ok, so I took the 2007 national Target Round and got a 6/8, and I guess thats all right...
These are the two problems I got wrong:

John noticed that the angle formed by the minute hand and the hour hand on a standard 12-hour clock was $ 110$ degrees when he left home after $ 6$ p.m, and when he came back before $ 7$ p.m, the angle formed was also $ 110$ degrees! How long did he leave if he left for more than $ 5$ minutes?

What I did:

Hour Hand Moves $ \Longrightarrow$ $ \frac {1}{2}$ degrees per minute.
Minute Hand Moves $ \Longrightarrow$ $ 6$ degrees per minute.

So they move apart at the rate of $ 6 \minus{} \frac {1}{2} \equal{} 5.5$ degrees per minute.

So $ \dfrac{110}{5.5} \equal{} 20$....he left at $ 6: 20$...

And he will come back when they are $ 110$ degrees apart again, or when they travel $ 250$ degrees apart.

So $ \dfrac{250}{5.5} \equal{} \dfrac{500}{11}$...he comes back $ \dfrac{500}{11}$ minutes after $ 6$...

So that means he left for $ \dfrac{500}{11} \minus{} 20 \equal{} 280/11$...and thats the answer I got, but the answer key says the answer is $ 40$....

Help? :maybe:

The other one I got wrong:

Using each of the digits $ {1,2,3,4}$ exactly once and using zero or more + signs, how many different totals can be obtained?

Also, BTW, I had to type this whole thing $ 2$ TIMES! I was revising through my solution, and I saw a typo so I pressed "Back Space" and the whole thing was gone...I am really angry at myself....
Hopefully I'll get some good solutions :)

[hide="Solution to Clock Question"]
The formula for the angle formed by the minute hand and the hour hand is $ |5.5m \minus{} 30h|$ where $ m$ is the minutes past the hour and $ h$ is the hour.  We know that $ h \equal{} 6$.  So, now we simply solve for the two values of $ m$.
\[ |5.5m \minus{} 180| \equal{} 110\implies m \equal{} \frac {140}{11} \text{ or } \frac {580}{11}\]
So we have $ \frac {580}{11} \minus{} \frac {140}{11} \equal{} \boxed{40}$.

NOTE:  If you try doing $ |5.5m \minus{} 30h| \equal{} 250$, you'll get a negative value for $ m$ and also a value of $ m$ that is greater than or equal to $ 60$, which can't happen.  [i]If [/i]something similar happened when solving the $ |5.5m \minus{} 180| \equal{} 110$ equation (if one or both of the answers were negative/greater than or equal to 60), then you would also need to have solve for values in the $ |5.5m \minus{} 30h| \equal{} 250$ equation.  Since we got two "good" answers, we didn't have to do that.[/hide]

Does anyone know where the answers to the nationals can be found? I have the handbook problems, but is there a place where all the answers to the nationals exist? Thanks in advance if you can recommend anything.

You won't find them online because MATHCOUNTS doesn't put national material online. You can get them from trading though. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189&start=660]here.[/url]

[quote="generalmotors"]Sorry, but I dont think I explained the last problem clearly :blush:

You can even have the following possibilities:

1234
123+4
321+4
12+34
etc.

Also, can someone tell me what mistake I made in my solution for the first problem?[/quote]

Help? :(

[quote="isabella2296"]You could have just hit "forward".  :P

The second one:

We have 2 choices, nothing or a +, for each space. Since there are 3 spaces, this is simply $ 2^3 \equal{} 8$.[/quote]

Unless it was incorrect?

I am pretty sure only one person got this correct in competition (that would be Justin Ahmann, who apparently guessed 3 random numbers, 2,2 and 11, and multiplied them together).

We use a happy casework type thingy.

Cases:

4 digit
3+1
2+2
2+1+1
1+1+1+1

4 digit=24 numbers (note all of these are distinct 4 digit)

3+1=4 choices for hundreds, 3 for tens. Note that 123+4=124+3 YAY! so 12 (all 3 digit)

2+2= 4c2 choices for tens so 6 total (2 digit)

2+1+1=4 choice for tens so 4 (2 digit)

1+1+1+1=1 choice (10)

This makes 24+12+6+4+1=47.

Could there be repeats?

4 digits- No.

3 digits- No.

2 digits- Now we must list

2+2

Note that the sum of the digits is ten, and it ranges from 37-73, so we have

Tens digits:

1,2 so 37

1,3 46

1,4 55

2,3 55

2,4 64

3,4 73 so one duplicate so far.

Now 2+1+1

This will be 19,28,37, and 46, so 2 more duplicates.

Thus, 3 duplicates total so 47-3=44

Covoluted, eh?

Thanks Mewto! Also, you mean  [b]convoluted[/b] right? :wink:

No, this was actually so convoluted it lost the n.

[quote="generalmotors"]
Also, can someone tell me what mistake I made in my solution for the first problem?[/quote]

I still dont get what I did wrong :(

@Mewto: Ha Ha :P


limit induction inequalities logarithms integration function calculus

Few years in a row the same problem appears in "Mathematics I" course on my faculty:

$ \lim_{n\to \infty} \left(\frac {1}{2}\cdot\frac {3}{4}\cdot\cdots\cdot\frac {2n \minus{} 1}{2n}\right) \equal{} 0$

The solution includes the induction based proof of the inequality

$ \frac {1}{2}\cdot\frac {3}{4}\cdot\cdots\cdot\frac {2n \minus{} 1}{2n} < \frac {1}{\sqrt {2n \plus{} 1}}$

And every year, students ask the same question: how could we in God's name know what to put on RHS. As a student I accepted it as a matter of intuition or so... But now as a UTA I feel I should give them a better answer than that (matter of practice, intuition etc.). Any comments?

Stirling's approximation.  I suppose if you wanted to be intuitive about it you could note that

$ \frac {1}{2} \cdot \frac {2}{3} \cdot .... \cdot \frac {2n \minus{} 1}{2n} \equal{} \frac {1}{2n}$

and then argue that asymptotically $ \frac {1}{2} \cdot \frac {3}{4} \cdot ... \cdot \frac {2n \minus{} 1}{2n}$ should approach $ \frac {2}{3} \cdot \frac {4}{5} \cdot ... \cdot \frac {2n \minus{} 2}{2n \minus{} 1}$.  So if their product is exactly $ \frac{1}{2n}$...

I don't know if my post is what you are looking for, but here goes anyway.

Let $ P$ be the product you have written.  In some sense, that product is missing the in-between factors.  Namely, it is missing the factors in the product
\[ Q \equal{} \frac{2}{3} \cdot \frac{4}{5} \cdot \cdots \cdot \frac{2n}{2n \plus{} 1} .
\]
The product $ PQ$ cancels nicely, evaluating to $ \frac{1}{2n \plus{} 1}$.  Also, if we look factor-by-factor, we see that $ P \le Q$.  Putting those two facts together, we have $ P^2 \le \frac{1}{2n \plus{} 1}$, which is what you wanted to show.

Edited: beaten by t0r.

A couple of tries at possible intuition:

First try: We're looking at $ \prod_{k \equal{} 1}^{n}\left(1 \minus{} \frac1{2k}\right).$ Take the logarithm of that to get $ \sum_{k \equal{} 1}^n\ln\left(1 \minus{} \frac1{2k}\right).$ What really matters is the behavior for large $ k,$ in which we say that $ \ln\left(1 \minus{} \frac1{2k}\right)\approx \minus{} \frac1{2k}.$ Then $ \minus{} \sum_{k \equal{} 1}^n\frac {1}{2k}\approx \minus{} \frac12\ln n \plus{} C.$ Exponentiate that back to get that original product probably behaves like $ \frac {C}{\sqrt {n}}.$ From here you might have to tinker with the constant to get it to work.

---------

Second try: It would be nice if that product were to telescope. Oh, too bad, it doesn't telescope. Well, what if it did? Maybe we can try supplying the missing factors:

$ \left(\frac12\cdot\frac34\cdots\frac {2n \minus{} 1}{2n}\right) \left(\frac23\cdot\frac45\cdots\frac {2n}{2n \plus{} 1}\right) \equal{} \frac12\cdot\frac23\cdot\frac34\cdots\frac {2n}{2n \plus{} 1} \equal{} \frac1{2n \plus{} 1}$

Now, both of the two products in the parenthesis - they're about the same size? Maybe. It's worth a shot to act as if they were and see where it leads.

-------------

Neither of these is fully accurate; neither is the final polished argument. But either way, you get a sense of where the $ \frac {C}{\sqrt {n}}$ might be coming from.

Edit: t0rajir0u and Ravi both posted while I got sidetracked by a conversation partway through. So my second try has come up three times now.

How about trying to estimate the product in a fairly naive way?

First, take a logarithm. Sums are easier to handle than products.
$ S_n\equal{}\ln\left(\frac12\right)\plus{}\ln\left(\frac34\right)\plus{}\cdots\plus{}\ln\left(\frac{2n\minus{}1}{2n}\right)$
$ \equal{}\ln\left(1\minus{}\frac12\right)\plus{}\ln\left(1\minus{}\frac14\right)\plus{}\cdots\plus{}\ln\left(1\minus{}\frac{1}{2n}\right)$
Now, we compare this to an integral. $ 2S_n$ is a midpoint Riemann sum for $ \int_1^{2n\plus{}1}\ln(1\minus{}\frac1x)\,dx$. Since that function is concave (its second derivative is $ \minus{}\frac{2x\minus{}1}{(x^2\minus{}x)^2}$), $ 2S_n>\int_1^{2n\plus{}1}\ln(1\minus{}\frac1x)\,dx\equal{}\Big[(x\minus{}1)\ln(x\minus{}1)\minus{}x\ln x\Big]_1^{2n\plus{}1}$
$ \equal{}2n\ln(2n)\minus{}(2n\plus{}1)\ln(2n\plus{}1)\equal{}2n\ln\left(\frac{2n}{2n\plus{}1}\right)\minus{}\ln(2n\plus{}1)>\minus{}1\minus{}\ln(2n\plus{}1)$
Dividing by 2 and taking an exponential, we get that the product is greater than $ \frac1{\sqrt{e}}\frac1{\sqrt{2n\plus{}1}}$. That's the wrong direction, but it's close enough to be interesting. Looking closer at that $ \frac1{\sqrt{2n\plus{}1}}$, we find that $ \ln\frac{2n\minus{}1}{2n}<\frac12\ln(2n\minus{}1)\minus{}\frac12\ln(2n\plus{}1)$ and we have the induction.

All three preceding posts went up while I was writing this. Oh well, it's different enough to be worthwhile.

Hehe, it's funny that three of us came up with the same idea at the same time.  I guess jmerry wins the prize for originality.

Here's yet another approach.  The reciprocal of the original product is
\[ \left(1 \plus{} \frac{1}{1}\right) \left(1 \plus{} \frac{1}{3} \right) \cdots
  \left( 1 \plus{} \frac{1}{2n \minus{} 1} \right).
\]
By expanding out this product, we get a sum that is at least
\[ \frac{1}{1} \plus{} \frac{1}{3} \plus{} \cdots \plus{} \frac{1}{2n \minus{} 1}.
\]
That sum is well-known to approach infinity (harmonic series).

I should add that Stirling's formula is actually stronger than necessary to understand this statement; it gives an asymptotic of $ \frac{1}{ \sqrt{ \pi n } }$.

Nevertheless, I'd like to support the logarithmic argument as a nice general-purpose method.  I was about to discuss it when the Merryfields both posted about it :P  In general, we have many more tools to attack sums with than to attack products, so since the logarithm of a product becomes a sum...

The problem with the main approaches is that arguing that the two products are asymptotically equivalent is a bit of a leap of faith and most students are not Euler..

I call for finding an entirely simpler approach than what has been suggested in the OP.

The most important presumption for all these 'intuitive' approaches is that the limit exists. So let us presume that outright (since as far as I know, there is no intuition, aside from Ramanujan's, that can tell us that at a glance).

1) We know that the limit cannot be nonzero by a cursory glance (this is a simple result in the theory of infinite products, in this case having to do with the divergence of harmonic series).

2) The only possibility is that the limit must be zero since it is a limit of non-negative terms. So how about using the squeeze theorem since showing that the series tends to 0 directly looks daunting? Assume the existence of some infinitesimal $ b_n$ for which $ \prod\limits_{k \equal{} 1}^n {(1 \minus{} {\textstyle{1 \over {2k}}}} ) < b_n$. We would like a case where $ \prod\limits_{k \equal{} 1}^{n \plus{} 1} {(1 \minus{} {\textstyle{1 \over {2k}}}} ) < b_n \frac {{2n \plus{} 1}}{{2n \plus{} 2}} < b_{n \plus{} 1}$.

3) In choosing $ b_n$, it would be nice if $ \mathop {\overline {\lim } }\limits_{n \to \infty } \frac {{b_{n \plus{} 1} }}{{b_n }} < 1$ since the majority of classroom infinitesimal sequences satisfy that property but clearly that is impossible in this case. But this is in fact a hint about where to go next. We are dealing with series and so we recall immediately that the standard example of infinitesimals for which this inequality fails when we learned the d'Alembert ratio test is $ \frac {1}{{n^a }}$, where $ a > 0$. This is starting to look promising since the suggested sequence fits the bill.

4) Intuition tells us to try $ a \equal{} {\textstyle{1 \over 2}}$. Why? Simple. $ \frac {1}{{n^{{\textstyle{1 \over 2}}} }} > \frac {1}{{n^a }}$ for $ a > {\textstyle{1 \over 2}}$ so we know that when we choose the exponent closer to 0, even if a value greater than  1/2 works, we are safe. Put another way, you can see from the inequality that if an exponent greater than 1/2 works then 1/2 works as well so might as well choose a as small as possible to hedge our bets. Square roots just happen to be nicer than powers of $ {\textstyle{1 \over \pi }}$, for example.

5) As is common in mathematics, our first guess of $ b_n \equal{} \frac {1}{{n^{{\textstyle{1 \over 2}}} }}$ fails horribly. However, rewriting $ \frac {{2k \plus{} 1}}{{2k \plus{} 2}} < \frac {{\sqrt k }}{{\sqrt {k \plus{} 1} }}$ as $ \frac {{2k \plus{} 1}}{{\sqrt k }} < \frac {{2k \plus{} 2}}{{\sqrt {k \plus{} 1} }}$, you can see immediately that the denominator in $ b_n$ should have been $ {\sqrt {2k \plus{} 1} }$.

And hence you get the result.  :P

I hope I was not too brash..  :oops:

Thank you all for detailed and quick answers. It was interesting to see that the topic attracted pretty much attention  :)

I'm sure it wouldn't take long for me to find one of these arguments on my own, but it's really hard to take enough time for each problem planned for tutorials... And to find balance between epsilon delta proofs and computation problems.


This is just to see how confident you guys are.

I don't think my opinion matters much, since I am not so good at math as you guys are.

I am pretty confident I can get into top 10. I really want to get into the top 4 and go to state, though :]

... Why chapter countdown? it counts for pretty much nothing.  As long as your team gets to state, you go to state reve. So just do well on the team round.

Too bad our team sucks.

Lol, you probably don't care about chapter. After all, 27th at nationals pretty much guarantees you 1st/2nd in chapter.

[hide="i am worried"]i actually have to care this year because to kids on a nother team got perfect AMC8 scores.

this was last years results
p just refers to people on our team
our team
captain-30
p1-28
p2(this years captain)-32
me-29
team round score-16

their team
captain-36
p1-25
p2-24
p3-21
team round score-18

one of the kids that got a perfect last year was in 5th grade so he didnt take these tests. im expecting both of them to average like a 40 or something and same with me and our captain. so its basically down to the other two kids on our team and their bests are somewhere around 37.[/hide]

[quote="mathcrazed"]...... who are you... I failed at nats last year... one of my teammates turned down the temp in my room (just for fun, of course) to 59-60 degrees. Caught a cold that night lol... I don't think top 4 from chapter is the cutoff anyways. I'm not exactly sure what it is though ~_-... I just looked it  up... Winning team, and the top 2 individuals not on the winning team. So... Aim for top 6... Assuming that the winning team's members are amazing...[/quote]
abtou catching a cold, i did at staes last year. it wasnt fun. i couldnt concentrate on the sprint round partially because the first 5 minutes i was working and anticipating the guy to bring me a tissue box, which he kinda forgot to.

O.O Oh ya. There's this 6th grader in indiana who's beast. I think... He's half asian (WHY ARE THEY ALL SO GOOD?!?!) like the benjamins -_-... argh. And there's the benjamin's little bro T.T. I swear, I AM DEAD at state.  I'll probably get like 10th -_-. and then I'll go into the restroom and cry, and then not care xD.

Now we're posting results o.O?

Just to brag:

Our Team

Captain-44
P1-44
P2-42
P3-I forget... 38? 40?

Team- 20? xD
How do you remember!!

The 6th and 7th graders are so beast this year.

Haha, I just happen to know your name.

For my region, its top 2 teams plus top 4 individuals go to state.

in my state, its top team and top 2 individuals not on the team. wild card teams are 2 highest scoring 2nd place teams. the team we beat was the 1st wild card team

Hehe, just wondering, What state do you live in?

If I don't make it, that means I suck.
...
Our chapter countdown is unofficial though.

our team last year chap

p1. 39
p2. 34
p3. 33
p4. 32(team cap)
Team round-20

wow, and we thought we were good. no wonder our capter always sucks at states

AHAHAHA.  You're at the best school for mathcounts in Virginia right? And you're beast

I'm pretty confident about getting 1st written, but there's this other kid who has a 50-50 chance of beating me in countdown and therefore being the champion of New Hampton town. XD

[quote="abacadaea"][quote="xpmath"]If I don't make it, that means I suck.
[/quote]
if you suck, that means i suck even more. :P

i live in pennsylvannia. what about you?[/quote]
Ohio, which is really easy.
Our scores last year:
p1=39
p2=38
p3(me)=37
p4=34
we got 18 on team

EDIT-But that was our beastly team which I left behind in MI.My new team doesn't have too much of a chance of making state.

does it really matter who the captain is as long as they have good handwriting?

Anyways, these are our scores on the 2005 State test we took this weekend.
(doing a harder test to prepare.)

Me- 24 sprint 14 target (38)
Other- 21 sprint 12 target (33)
Other- 22 sprint 12 target (34)
Other- 19 sprint 14 target (33)

Team-18

So on a state competition which is harder than what is at chapter, we got 52.5 which isn't a bad score. Especially for state.

BTW, this is all of our first years with the exception of one guy who has done it for 2. We are all 8th graders in Southwest Region of Connecticut.

I think I have a great chance of taking first in Chapter and possibly state as this is a pretty easy area, and I have fast countdown.


calculus
derivative
function
integration
absolute value
calculus computations

derivative of |x|^4

is d/dx |x|^4 = 4 |x|^3 ?

No calculus please.  Mods, move this to a higher level.

Note that $ \forall x \in \Re$; $ |x|^4\equal{}x^4$
Hence, $ \frac{d}{dx}|x|^4\equal{}4x^3$
In general, for all (otherwise differentiable) functions containing $ |x|$, we only have to check differentiabilty at $ x\equal{}0$

You also have to treat $ x < 0$ differently from $ x > 0.$

For instance, $ \frac {d}{dx}|x|^3 \equal{} \begin{cases}3x^2, & x\ge0 \\
\minus{} 3x^2, & x < 0\end{cases}.$

In often doesn't pay to try to handle $ |\cdot|$ as a symbol in its own right to be carried along. Treat the absolute value as a piecewise defined function, and handle each piece separately.

$ \frac {d}{dx}|x| \equal{} \frac {|x|}{x}$ for any $ x\ne 0$ ; over any interval, $ \int |x|\ \mathrm {dx} \equal{} \frac 12\cdot x|x| \plus{} \mathcal R\ .$

Generally, for any odd natural number $ n\ge 3$  we have :

for any $ x\ne 0$, $ \frac {d}{dx}|x|^n \equal{} nx^{n \minus{} 2}|x|$ ;

over any interval, $ \int |x|^n\ \mathrm {dx} \equal{} \frac {1}{n \plus{} 1}\cdot x^n|x| \plus{} \mathcal C\ .$

[b]Problem.[/b] $ F \in \int f(x)\ \mathrm {dx}\implies$ $ \int f(|x|)\ \mathrm {dx} \equal{} \mathrm {sign}\ (x)\cdot\left[F(|x|) \minus{} F(0)\right] \plus{} \mathcal C\ .$

I'm not very sure. Is the indefinite integral $ F(x)\equal{}\int f(x) dx$ defined by $ f(x)\equal{}\frac{dF(x)}{dx}$? In that case you can get the result by differentiating both sides using chain rule.

[quote="Virgil Nicula"][color=darkred][b]Problem.[/b] $ F \in \int f(x)\ \mathrm {dx}\implies$ $ \int f(|x|)\ \mathrm {dx} = \mathrm {sign}\ (x)\cdot[F(|x|) - F(0)] + \mathcal C\ .$[/color][/quote]
[b]Proof.[/b] Let $ G: \mathrm R\rightarrow\mathrm R$ be a primitive of the function $ g(x) = f(|x|)$. Thus,

$ g(x) = \{\begin{array}{ccc} f( - x) & \mathrm {,} & x < 0 \\
 \\
f(x) & \mathrm {,} & x\ge 0\end{array}$ $ \implies$ $ G(x) = \{\begin{array}{ccc} - F( - x) + k & \mathrm {,} & x < 0 \\
 \\
F(x) & \mathrm {,} & x\ge 0\end{array}$

From $ G(0 - 0) = G(0) = G(0 + 0)$ obtain $ k = 2F(0)$. Therefore,

$ \int g(x)\ \mathrm {dx} = \{\begin{array}{ccc} - F( - x) + F(0) & \mathrm {,} & x < 0 \\
 \\
F(x) - F(0) & \mathrm {,} & x\ge 0\end{array}$ $ = \mathrm {sign\ }(x)[F(|x|) - F(0)] + \mathcal C\ .$

[b]Remark.[/b] $ \mathcal C=\mathcal C\pm F(0)\ !$


When you first moved to college did you bring a lot of things with you? how was the first day? How was your roomate?

How much you bring usually depends a lot on how close you are to school. If you can drive, you can bring a lot more stuff. I have to fly to school, so I usually don't bring very much. My mom just moved this year, though, and I had to do something with my possessions, so this year I have a bit more stuff in my dorm.

I'd say don't bring anything unless you're sure you'll need it (really need, not just vaguely want). If you later realize you're missing something, you can buy it or get it from home during a break.

---

You never know whether you and your roommate will get along. My roommates freshman year were cool. But my experience in no way predicts what yours will be like.

---

The first few days are usually pretty boring, with lots of unpacking and orientation activities. It's when you start choosing classes, and then going to classes, that things start getting interesting. Or if you're at a place like Rice and you like to drink, it's when upperclassmen come back to campus that things get interesting. :P

(Disclaimer: I do not condone the excessive drinking activities that are popular at almost every university these days. Freshmen, don't let upperclassmen pressure you into making yourself sick! You can have fun and be safe at the same time.)


LaTeX
function
quadratics
algebra
quadratic formula

[quote="sharkman"]If a polygon of n sides has (n/2)(n-3) diagonals, how many sides will a polygon with 65 diagonals have?  Is there a polygon with 80 diagonals?

My work:

I equated (n/2)(n-3) to 65.  Is this the correct way to solve for n?

How do I answer the second question?

To my understanding, (n/2)(n-3) is our diagonal, right?

The letter n represents the number of sides, right?[/quote]

PLEASE use $\LaTeX$ next time.
[hide="1"]Solve for $n$

$\frac{n(n-3))}{2}= 65$
$n^{2}-3n-130 = 0$
$(n+10)(n-13) = 0$
$n=13,-10$
$n>0 \Rightarrow n=13$[/hide]

[hide="2"]Suppose there exists an integer $n$ such that $\frac{n(n-3)}{2}= 80$. Then:
$n(n-3) = 160$
$n^{2}-3n-160=0$
$D = 9+640 = 649$

$D$ is not a perfect square, so $n$ is not an integer.

$\textrm{NO SUCH POLYGON EXISTS}$
[/hide]

let $f(d) = \pm\sqrt{8d+9}$
[hide="I need help finishing the set of all solutions"]$\frac{3\pm\sqrt{8d+9}}{2}= n$
$n$ is a whole number, so the numerator must be even. Since 3 is an odd number, $f(x)$ is any odd number that satisfies $f(x)+3>0$
-step function?
That being said: $f(x)^{2}= 1, 3^{2}, 5^{2}, 7^{2}... = 9+8d$
-series?[/hide]Anyone wanna finish that off for me?

EDIT: missing money signs.

[quote="sharkman"]The letter n represents the number of sides, right?[/quote]
Try reading the question, it's often very enlightening.

[quote="sharkman"]I read the question (as I do every question here) more than once but nothing (to me) indicated that a square root should be used as part of the equation.

What word(s) indicated this to you?

Thanks[/quote]
I don't think you read what I quoted...

[hide="1"]
$\frac{n}{2}(n-3) = 65 \implies n(n-3) = 130$
$\implies n^{2}-3n-130 = 0$  Use the quadratic formula to find that its a 13-gon.[/hide]

[hide="2"]Set it up the same way:
$n(n-3) = 2\cdot 80$.  Simplify the equation and put it into the form $ax^{2}+bx+c = 0$.  If there is no positive integer solution then there is no such $n$-gon[/hide]

bro take the quad formula, then plug in the quadratics a,b,c
note that c = -2d

[quote="igiul"]let $f(d) = \pm\sqrt{8d+9}$
I need help finishing the set of all solutions]$\frac{3\pm\sqrt{8d+9}}{2}= n$
$n$ is a whole number, so the numerator must be even. Since 3 is an odd number, $f(x)$ is any odd number that satisfies $f(x)+3>0$
[/quote]

Take the quad formula, then plug in quadratic coefficients $a$, $b$, $c$.
[hide="see here"]
$\frac{n(n-3)}{2} = d$
$n^{2}-3n = 2d$
$n^{2}-3n-2d = 0$
THEREFORE $a = 1$ ; $b =-3$ ; $c =-2d$
[hide="next step"]
Then we do what my 7th grade teacher referred to as the [i]PLUG N' CHUG[/i].

$n = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}= \frac{3\pm\sqrt{9+8d}}{2}$
See it now? If you're still having trouble, pm me and I can help you work through it.
[/hide]
[/hide]
Of course, I'm no expert. People like torajirou probably know exactly what I'm trying to say (terminology and phrasing).


Princeton
college
calculus
calculus computations

$\frac{dy}{dx}=2y-4x$

Initial condition $f(0)=1$

Using step sizes of $\Delta{x}=.1$, find $f(.2)$

I am terrible at these find of approximations. The exam is Tuesday and please show your exact work for this and how to do it analytically and not using the calculator. Thanks in advance for the help!

ah i finally understand....sorry guys. mods can delete this now...

Well, just in case:

With Euler's method all you really need to know is that:

\[x_n=x_{n-1}+h\]

and

\[y_n=y_{n-1}+h(y_{n-1}')\]

So we start with $x_0=0$, $y_0=1$.

Then we have $x_1=.1$ and $y_0=1+.1(2(1)-4(0))=1.2$.

Next, $x_2=.2$ and $y_2=1.2+(.1)(2(1.2)-4(.1))=1.4$.

I was wondering if anyone knows if there is a program built into the 89 that does this automatically?  I looked but could not find one.  It would be helpful to save even a little time on the exam.

i wrote one for 83-plus... :D, and by the way i dont' think that ther is a high chance that they ask about it.

It's in the curriculum.  It is covered in the Princeton Review.  So I think there is a fair chance I'll see it once (probably not more than once though).  I'm sure I can afford the time but it would still be nice to have the program...

i agree. its usually on there in one way or another. sometimes its on FRQ and sometimes im multiple choice


请牛人帮忙想想，最近做论文时碰到的一问题，命题应该是没有错的，但就是证不出，请帮忙解答一下，万分感谢。

n * n的矩阵$ A_n$，其中的元素$ a_{ij}$的定义是f的第i阶导数与第j阶导数的乘积在[0, 1]上的积分，即$ \int_0^1f^{(i)}f^{(j)}dx$，其中i, j取值0, 1, 2, ... , n - 1，f属于$ C^{\infty}[0,1]$。

现在想证明的是$ A_n$的最小特征值$ \lambda_n$随矩阵维数n的增加而趋于0，想了很长时间，实在做不来，请指教。

如果无法证明上述的命题，那么能否证明$ A_n$的行列式随n的增加而趋于0？谢谢！

下面是我的一些基本想法，但最后还是无法证明。请牛人指教。

Suppose $ \lambda_1 > \lambda_2 > \cdots > \lambda_n$ is the n eigenvalues of $ A_n$. Since $ \lambda_1$ might be much larger than $ \lambda_n$,

$ \lambda_n$ could be "covered" by the trace of $ A_n$, i.e. tr(A).

So let's consider tr($ A^{\minus{}1}_n$) which equals to $ \frac{1}{\lambda_1} \plus{} \cdots \plus{} \frac{1}{\lambda_n}$. Thus we have

$ tr(A^{\minus{}1}_n) \equal{} \frac{1}{\lambda_1} \plus{} \cdots \plus{} \frac{1}{\lambda_n} < \frac{n}{\lambda_n}$.

So $ \lambda_n < \frac{n}{tr(A^{\minus{}1}_n)}$.

And then comes our problem, who can give an estimate of tr($ A^{\minus{}1}_n$)?

Why $ \lim_{n\rightarrow\infty} \frac{n}{tr(A^{\minus{}1}_n)} \equal{} 0$?

Hope someone can help me. Thanks a lot.


A rectangular box is 8 cm thick, and its square bases measure 32 cm by 32 cm. What is the distance, in centimeters, from the center point $ P$ of one square base to corner $ Q$ of the opposite base? Express your answer in simplest terms.

[asy]import three;
draw((0,0,1/4)--(1,0,1/4)--(1,1,1/4)--(0,1,1/4)--(0,0,1/4)--cycle3,linewidth(2));

draw((0,1,0)--(1,1,0),linewidth(2));
draw((1,1,0)--(1,0,0),linewidth(2));
draw((0,1,0)--(0,1,1/4),linewidth(2));
draw((1,1,0)--(1,1,1/4),linewidth(2));
draw((1,0,0)--(1,0,1/4),linewidth(2));

dot((1/2,1/2,1/4));
dot((0,1,0));

label("P",(1/2,1/2,1/4),W);
label("Q",(0,1,0),E);[/asy]

Drop a perpendicular from P to the bottom of the box, and call this point R.

PR = height box = $ 8$

RQ = half length of base's diagonal = $ \frac {32 \sqrt {2}}{2}$ = $ 16 \sqrt {2}$

Use the Pythagorean Therorem on this triangle to find PQ, the hypotenuse.

PQ = $ \sqrt {8^2 \plus{} (16 \sqrt {2})^2}$

PQ = $ \sqrt {576}$

PQ = $ 24$


Let a;b;c>-1.Prove that:
   $ \sum\frac{1\plus{}a^2}{1\plus{}b\plus{}c^2}\geq 2$

[quote="karis"]Let a;b;c>-1.Prove that:
   $ \sum\frac {1 \plus{} a^2}{1 \plus{} b \plus{} c^2}\geq 2$[/quote]
We have : $ a,b,c>\minus{}1$ therefore $ 1\plus{}b\plus{}c^2$, $ 1\plus{}c\plus{}a^2$, $ 1\plus{}a\plus{}b^2 \geq 0$
So $ b^2\plus{}1 \leq 2b$
$ \Rightarrow 2(1\plus{}b\plus{}c^2) \leq (1\plus{}b^2)\plus{}2(1\plus{}c^2)$
i.e $ 2(1\plus{}c\plus{}a^2) \leq (1\plus{}c^2)\plus{}2(1\plus{}a^2)$
     $ 2(1\plus{}a\plus{}b^2) \leq (1\plus{}a^2)\plus{}2(1\plus{}b^2)$
So,
$ \sum\frac {1 \plus{} a^2}{1 \plus{} b \plus{} c^2}\geq \sum\frac{2x}{y\plus{}2z}\equal{}\sum\frac{2x^2}{xy\plus{}2xz}$
where $ x\equal{}1\plus{}a^2$, $ y\equal{}1\plus{}b^2$, $ z\equal{}1\plus{}c^2$. Clearly $ x,y,z>0$.
Other, by [b]Cauchy-Schwarzt[/b] we have :
$ \sum\frac{2x^2}{xy\plus{}2xz} \geq\frac{2(x\plus{}y\plus{}z)^2}{3(xy\plus{}yz\plus{}zx)} \geq 2$
We are done.


[quote="236factorial"]The positive four-digit integers that use each of the four digits 1, 2, 3, and 4 exactly once are ordered least to greatest. What is the 15th integer in the list?[/quote]

[hide]
4x3x2x1=24

24/4=6

15/6=2 r3

124,142,214

$3214$?[/hide]


In quadrilateral $ ABCD$, $ m\angle B \equal{} m\angle C \equal{} 120^{\circ}$, $ AB \equal{} 3$, $ BC \equal{} 4$, and $ CD \equal{} 5$. Find the area of $ ABCD$.

$ \textbf{(A) } 15 \qquad\textbf{(B) } 9\sqrt{3} \qquad\textbf{(C) } \frac{45\sqrt{3}}{4} \qquad\textbf{(D) } \frac{47\sqrt{3}}{4} \qquad\textbf{(E) } 15\sqrt{3}$

drop perpendiculars from $ A$ and $ D$ to line $ BC$, the feet of these perpendiculars being $ P$ and $ Q$, respectively. Find $ AP$ an $ DQ$ (this is basic, because you have 30-60-90 triangles). Find the area of $ APQD$ (it's a trapezoid whose bases and height you know, so basic. then subtract the areas of $ APB$ and $ CQD$ (whose areas are found treeevially) to get the area of $ ABCD$.

all in all, this problem is [i]soooo[/i] basic.

Do you have a picture for the proof. It's kind of hard to imagine what you are talking about.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14819[/img]

 :(

On an arbitrary line through the point $ O,$ we define two points $ P,\ Q,$ such that $ OP \equal{} AB$ and $ OQ \equal{} BC$ $ ($ $ O,$ between $ P$ and $ Q$ $ ).$

Through the points $ P,\ Q,$ we draw two lines and parallel to the ray $ (b),$ which intersect the rays $ (a),\ (c),$ at points $ A',\ C'$ respectively and we denote the point $ B'\equiv (b)\cap A'C'.$

Through the point $ P,$ we draw the line parallel to the line segment $ A'B'C',$ which intersects the rays $ (a),\ (b),\ (c),$ at points $ A'',\ B'',\ C'',$ respectively.

It is easy to show that $ \frac{A''B''}{B''C''} \equal{} \frac{A'B'}{B'C'} \equal{} \frac{OP}{OQ} \equal{} \frac{AB}{BC}$ and the solution is completed.

Kostas vittas.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14823[/img]

 :thumbup:


Prove that:
$\frac{1}{a}+\frac{1}{b}-(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}})^{2}\geq \frac{3}{2}$
where $a;b$ are positive real numbers satisfying $a^{2}+b^{2}=1$


linear algebra
matrix
algebra
polynomial
Conjugate

Let $A$ be complex matrix dimensions $n \times n$, and $\overline{A}$ be matrix with conjugate elements.
Prove that $\det (A \overline{A} + I) \geq 0$.

thx

[quote="alekk"]yep, it's true that the set of diagonalizable matrices with distinct eigenvalues is dense. (just trigonalize a matrix $A$ and change a little bit the diagonal). But are you sure that $A$ and $\overline{A}$ commute ?[/quote]

I'm sorry - I obviously wasn't thinking well and didn't have time to check over my "proof" (didn't even have time to finish it, and I usually dislike posting incomplete results).

I can show that $A\overline{A}$ has a real characteristic polynomial, so the determinant you're looking at is real:
For any square $A,B$, $AB$ and $BA$ have the same characteristic polynomial. Now $\det(\lambda I-A\overline{A})=\det(\lambda I-\overline{A}A)=\det(\overline{\overline{\lambda} I-A\overline {A}})=\overline{\det(\overline{\lambda} I-A\overline{A}})$
If $f$ is the characteristic polynomial, this gives $\overline{f(\lambda)}=f(\overline{\lambda})$, and $f$ is real on the real axis. This makes $f$ a real polynomial, with all the baggage that comes with that.

The tough part: showing that all negative roots of $f$ must have even multiplicity.

It has already been posted but no solution were found. I've already seen an ugly solution but I don't rememeber the ideas ... Anyone with a nice approach ?

I am not sure not how much this approach helps, and it might already have been considered earlier on. But here goes:

Write $A = C + iD$ for real $n \times n$ matricies $C$ and $D$. Then we must have $\overline{A} = C - iD$. Then $Z = A\overline{A} = C^2 + D^2 + i(-CD + DC)$. Suppose if $\lambda$ is a real eigenvalue of $A\overline{A}$ with a complex $n$ dimensional eigenvector $v =x + iy$ (with x and y real n dimension vectors), and denote $\Re[Z] = C^2 + D^2$ and $\Im[Z] = -CD + DC$, for the sake of ease. Then $Zv^T =$ $\lambda v^T$ implies the following:
\[ \Re[Z]x^T - \Im[Z]y^T = \lambda x^T  \]
\[ \Im[Z]x^T + \Re[Z]y^T = \lambda y^T  \]
thus
\[ \left( \begin{array}{*{20}c} \Re[Z] - \lambda I_n & -\Im[Z] \\ \Im[Z] & \Re[Z] -\lambda I_n \end{array} \right) \left( \begin{array} {*{20}c} x \\ y \end{array} \right) = 0  \]

here $0$ represents the $2n$ dimensional zero vector. The left $2n \times 2n$ block matrix $F$ must be singular, since the $2n$ dimension vector doesnt have all zero entries. So all the real eigenvalues of $Z$ are precisely the real eigenvalues of $F$, and vice-versa.

Perhaps simplifying $det(F)$ as a function of the given matricies, and restricting $\lambda < 0$  could lead somewhere, since a positive or complex eigenvalue of $Z$ is not a problem.

I'd really appriciate some ideas! :lol:

Also on a related note, I ran some few numerical tests using Mathematica on the eigenvalues of $A\overline{A}$, and so far haven't ran accross a negative eigenvalue yet, and naturally have gotten positive and complex eigenvalues.

Negative eigenvalues aren't too hard to create in pairs; try a rotation matrix (by 90 degrees) for $A$.

This is a continuation of my previous post. Please comment (on possible/probable incorrectness):

Write $A = C + iD$ for real $n \times n$ matricies $C$ and $D$. Then we must have $\overline{A} = C - iD$.
Then $Z = A\overline{A} = C^2 + D^2 + i(-CD + DC)$. Suppose if $\lambda$ is a real eigenvalue of $A\overline{A}$ with a complex $n$ dimensional eigenvector $v =x + iy$ (with x and y real n dimension vectors), and denote $\Re[Z] = C^2 + D^2$ and $\Im[Z] = -CD + DC$, for the sake of ease. Then $Zv^T =$ $\lambda v^T$ implies the following:
\[ \Re[Z]x^T - \Im[Z]y^T = \lambda x^T \]
\[ \Im[Z]x^T + \Re[Z]y^T = \lambda y^T \]
thus
\[ \left( \begin{array}{*{20}c} \Re[Z] - \lambda I_n & -\Im[Z] \\ \Im[Z] & \Re[Z] -\lambda I_n \end{array} \right) \left( \begin{array} {*{20}c} x \\ y \end{array} \right) = 0_{2n} \]
So all the real eigenvalues of $Z$ are precisely the real eigenvalues of $F$ the $2n$ square block matrix above (without the $\lambda I_n$ expression), and vice-versa.

With this observation we conjecture the following: If $\lambda < 0$ is an eigenvalue of $Z$, then it occurs with an even multiplicity in $F$ and therefore in $Z$. We needn't worry about the positive eigenvalues of the complex ones (they come in conjugate pairs). If therefore we show that the characteristic polynomial of $F$ (which has all real coefficients) is always non-negative on the real line, then we are done. To this end, we note that if the matrix $\Re[Z] - xI_n$ is invertible for real $x$ (it is singluar for atmost $n$ values of $x$) then we may rewrite $F - xI_{2n}$ as:

\[ \left( \begin{array}{*{20}c} \Re[Z] - xI_n & 0 \\ 0 & \Re[Z] - xI_n \end{array} \right)\left( \begin{array}{*{20}c} I_n & -(\Re[Z] - xI_n)^{-1}\Im[Z] \\ (\Re[Z] - xI_n)^{-1}\Im[Z] & I_n \end{array} \right) \]

Then the characteristic polynomial of $F$ which we denote as $p(x) = \det(F-xI_{2n})$ as a function of the given matrix gives ${p(x)= \det(\Re[Z]-xI_n})^{2}\det(I_n+((\Re[Z]-xI_n)^{-1}\Im[Z])^2) \geq 0$ as the individual determinants are always non-negative. With the continuity of the determinant function, we must have $p(x) \geq 0$ as desired. Thus it follows that $\det(A\overline{A}+I_n) \geq 0$.

thanks for the feedback but something about that proof bothers me, to me I dont think the argument i posted above does justice in showing that the $F$ and $Z$ have the same list of real eigenvalues,ie" if $\lambda$ is a real eigenvalue and it occurs in $Z$ with some multiplicity $m$, in its characteristic polynomial then i dont think my proof implies that it occurs with multiplicity $m$ in the characteristic polynomial of $F$.

..Either that or its way to late


Can someone tell me what kind of problems I'm supposed to post here?

Edit: I guess the earlier ones were too easy.

Ok then AMC questions it will be  :roll:


[u][b]The author of this posting is : grodij (edited by Darij Grinberg)[/b][/u]
____________________________________________________________________

In the following, the abbreviation "circle P(p)" will mean the circle with center P and radius p.

Given two circles B(b) and C(c) and a line l.

[b](a)[/b] Find the point A on the line l such that the sum of the powers of the point A with respect to the circles B(b) and C(c) is minimal.

[b](b)[/b] For every point A on the line l, let a tangent from A to the circle B(b) meet this circle B(b) at Y, and let a tangent from A to the circle C(c) meet this circle C(c) at Z. Find the point A on the line l such that the sum AY + AZ is minimal.

[u][b]The author of this posting is : Myth (edited by Darij Grinberg)[/b][/u]
____________________________________________________________________

[i]Solution of problem [b](a)[/b]:[/i]

The powers of the point $A$ with respect to the circles $B\left(b\right)$ and $C\left(c\right)$ are $AB^2-b^2$ and $AC^2-c^2$, respectively. Thus, we have to minimize $AB^2+AC^2$. Let $B^{\prime}$ and $C^{\prime}$ be the projections of $B$ and $C$ on $l$. We have $AB^2+AC^2 = B^{\prime}B^2 + C^{\prime}C^2 + AB^{\prime}^2 + AC^{\prime}^2$. From AM-QM it is easy to see that optimal $A$ is the midpoint of $B^{\prime}C^{\prime}$.

[i]Solution of problem [b](b)[/b]:[/i]

I will only consider the case when the none of the circles B(b) and C(c) intersects the line l. In fact, this is the only case which seems really worth considering (in the other cases, the sum AY + AZ is not real for some positions of A; of course, one can still ask for the minimal of all real values of this sum, but this is a rather boring question).

Let B' and C' be the orthogonal projections of the points B and C on the line l. Since none of the circles B(b) and C(c) intersects the line l, then $b\leq b^{\prime}$ and $c\leq c^{\prime}$, where $b^{\prime} = BB^{\prime}$ and $c^{\prime} = CC^{\prime}$ are the distances from the points B and C to the line l. Now, since $b\leq b^{\prime}$, we have $b^2 \leq b^{\prime}\ ^2$, and thus $b^{\prime}\ ^2 - b^2 \geq 0$. Let $B_1$ be the point on the ray B'B such that $B^{\prime}B_1 = \sqrt{b^{\prime}\ ^2 - b^2}$. Then, $B^{\prime}B_1\ ^2 = b^{\prime}\ ^2 - b^2$.

Thus, in the right-angled triangle $AB^{\prime}B_1$, the Pythagoras theorem yields

$AB_1\ ^2 = AB^{\prime}\ ^2 + B^{\prime}B_1\ ^2 = AB^{\prime}\ ^2 + b^{\prime}\ ^2 - b^2 = AB^{\prime}\ ^2 + BB^{\prime}\ ^2 - b^2$.

But from Pythagoras, applied to the right-angled triangle AB'B, we have $AB^{\prime}\ ^2 + BB^{\prime}\ ^2 = AB^2$. Thus, $AB_1\ ^2 = AB^2 - b^2$. Now, since the line AY touches the circle B(b) at the point Y, we have $AY\perp BY$ and BY = b. Thus, in the right-angled triangle AYB, Pythagoras yields $AY^2 = AB^2 - BY^2 = AB^2 - b^2$. Hence, $AY^2 = AB_1\ ^2$, and $AY = AB_1$. Similarly, we can find a point $C_1$ such that $AZ = AC_1$. Now, instead of minimizing AY + AZ, we are left with the question of minimizing $AB_1 + AC_1$. But since $B_1$ and $C_1$ are fixed points, this is a well-known problem (one has to take A lying inside the segment $B^{\prime}C^{\prime}$ and satisfying the relation $\frac{B^{\prime}A}{C^{\prime}A} = \frac{B^{\prime}B_1}{C^{\prime}C_1}$).

  Darij


needs techno music to go along with it.  ;)

AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
RUN FOR YOUR LIVES!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 :rotfl:

:spam: :spam: :spam:

This should go in G&FF or something. More like in the test forum. Or your blog.

my head is hurting now :P
i agree with mathnerd314....this should go in your blog.


For a given point $A$ let us define a transformation $F_A$ that for any set of points $(B)$ gives a set of all the points equidistant to $A$ and $B$. For $n\geq 3$ and two distinct points $A$ and $B$, $\mathop{F_A \circ F_A \circ \dots \circ F_A}\limits_{n}(B)$ is a closed curve forming an angle at one point. Given $n$ find the value of the angle.

The answer to this problem is obviously:
$\alpha = \pi \left( 1 - \frac{2} {n} \right)$
Waiting for proofs...


Have you been accepted? My friend just got the email, 14 minutes ago.

I haven't gotten anything yet.  Do they send out rejection letters as well?

[quote="Klebian"]Not sure. But he says titu is still signed onto gmail. So I'm guessing they're still being sent out.[/quote]
I'M IN!!!!!!!!
I JUST GOT MY ACCEPTANCE LETTER
THE ONLY THING STOPPING ME FROM SCREAMING THIS OUT IS THAT MY PARENTS WOULD KILL ME
I'm going to start a poll

I got mine too. :P  I told my mother that I didn't get in, then told her that I did.  Then she said that I had self-esteem issues.  Anyway, congratulations to everyone who got in!

[quote="Aegor"]I told my mother that I didn't get in, then told her that I did.  Then she said that I had self-esteem issues.[/quote]
:rotfl: :rotfl:
I should do that this morning :D


trigonometry
geometry
circumcircle
trig identities
Law of Sines
geometry unsolved

From the vertex $ A$ of the equilateral triangle $ ABC$ a line is drown that intercepts the segment $ [BC]$ in the point $ E$. The point $ M \in (AE$ is such that $ M$ external to $ ABC$, $ \angle AMB \equal{} 20 ^\circ$ and $ \angle AMC \equal{} 30 ^ \circ$. What is the measure of the angle $ \angle MAB$?

M stay on a circle with radius AB and center B 'cause $ \angle ABC \equal{} 2\angle AMC$. So $ \triangle BMA$ is an isosceles triangle and $ \angle MAB \equal{} \angle BMA \equal{} 20$
But it seems too simple  :oops:

The solution of[b] ¬[ƒ(Gabriel)³²¹º]¼[/b] is not tru.
Because if $ \angle MAB \equal{} 20^0$ $ \Longrightarrow$ the suggestion.
$ \bullet$ $ \angle KBC \equal{}\angle CBM \equal{} 30^0$ $ \Longrightarrow$ $ BKCM$ is circums cribed quadriangle.
$ \bullet$ if $ \angle MAB \equal{} 20^0$ $ \Longrightarrow$ $ \angle BKM \equal{} 130^0$ but $ \angle BKM \equal{} \angle MCB$ $ \Longrightarrow$ $ \angle MCB \equal{} 130^0$.
But $ \angle MCB \plus{}  \angle MBC \equal{} 130^0$
$ \Longrightarrow$ the suggestion.
real $ \angle MAB > 20^0$

Real $ \angle MAB > 20^0$
Who can fine it. I will try and post it soon.

[quote="thanhnam2902"]$ \bullet$ if $ \angle MAB \equal{} 20^0$ $ \Longrightarrow$ $ \angle BKM \equal{} 130^0$[/quote]
Recheck that statement.

thanhnam2902 on your drawing where exactly is point M? i assume it's at the second A, but it makes your solution hard to follow

let $ \angle MAB\equal{}\alpha$, then $ \angle MAC\equal{}60\minus{}\alpha, \angle BMA\equal{}100\minus{}\alpha, \angle BCM\equal{}30\plus{}\alpha$.

from law of sines we can get
$ \sin \alpha\cdot \sin 20\cdot \sin 30\cdot \sin 60\equal{}\sin(60\minus{}\alpha)\cdot\sin 60\cdot \sin (100\minus{}\alpha)\cdot \sin (30\plus{}\alpha)$

this can be rewritten as
$ \sin \alpha\sin 20\equal{}\sin(2\alpha\plus{}60)\sin(100\minus{}\alpha)$

or
$ \cos(\alpha\minus{}20)\minus{}\cos(\alpha\plus{}20)\equal{}\cos(3\alpha\minus{}40)\minus{}\cos(\alpha\plus{}160)$

since $ \cos(\alpha\minus{}20)\plus{}\cos(\alpha\plus{}160)\equal{}0$, it follows that $ \cos(\alpha\plus{}20)\plus{}\cos(3\alpha\minus{}40)\equal{}0$, from where it's easy to conclude that $ \alpha\equal{}50$  :D

let $ D$ be on line $ BC$ such that $ B$ lies between $ C$ and $ D$, and $ AB \equal{} BD$. since $ \angle ADC \equal{} \angle AMC \equal{} 30$ it follows that $ ADMC$ is cyclic. let $ O$ be the circumcenter of $ ADMC$. note that $ \angle AOC \equal{} 2\angle AMC \equal{} 60 \equal{} \angle ABC$, so $ A,B,O,C$ are cyclic. note that $ \angle AOD \equal{} 2\angle ACD \equal{} 120 \equal{} \angle ABD$, so $ A,O,B,D$ are cyclic. since the circumcircles of $ A,B,O,C$ and $ A,O,B,D$ cannot coincide it follows that $ B \equal{} O$, from where it follows easily that $ \angle BAM \equal{} 50$  :D

edit: there's a much more easier solution... $ \angle CAD\equal{}90$, and since $ B$ is the midpoint of $ CD$ it follows that $ B$ is the circumcenter of $ ACD$, from where we conclude that $ B\equal{}O$... the rest follows easily..

Sory! My picture is wrong, solution of [b]campos[/b] is very nice. Thank you very much.

Oh! Problem of [b]Valentin Vornicu [/b]is very nice. Thanh you very much!

Actually this year where proposed very weak problems for 9th grade.This is also an easy one.

[quote="campos"]

from law of sines we can get
$ \sin \alpha\cdot \sin 20\cdot \sin 30\cdot \sin 60 \equal{} \sin(60 \minus{} \alpha)\cdot\sin 60\cdot \sin (100 \minus{} \alpha)\cdot \sin (30 \plus{} \alpha)$

[/quote]

can you explain this ? thank you ! :(

I think you solution {CAMPOS} is right but yor result is wrong <<angle BAM=20>>

I solve this problem with Ceve theorem .  (angle BAM=x)

sin(60-x)/sin(x) * sin(160-x)/sin(100-x)  * sin(30+x)/sin(90+x) =1

& we now  sin(60-x)=cos(30+x)  & sin(90+x)= cos(x)

 & cos(30+x) * sin(30+x)=1/2 * sin(60+x)  &   cos(x) * sin(x)=1/2 * sin(2x)

& sin(160-x)=sin(20+x)  &   sin(100-x)=sin(80+x)

then we have sin(60+x) * sin(20+x) =sin(2x) * sin(80+x)

then we have cos(40+x)=cos(80-x) then x=20

[color=darkblue][b][u]Atention[/u][/b], I modified the enunciation giving a degree of freedom to the point $ M$ ![/color]

[quote="Valentin Vornicu"] [color=darkred]Let  $ ABC$ be a equilateral triangle and let $ M$ be a point for which the line $ BC$

 separates the points $ A$ , $ M$ and $ \{\begin{array}{c} m(\widehat {AMB}) = 20^{\circ} \\
\ m(\widehat {AMC}) = 30^{\circ}\end{array}$ . Ascertain $ m(\widehat {MAB})$ .

[b][u]Remark.[/u][/b] Denote $ E\in BC\cap AM$ . [b]CAUTION ![/b] Appear two cases : $ E\in (BC)$ or $ B\in (EC)$ .[/color] [/quote]
[color=darkblue][b][u]Proof I.[/u][/b] Denote the point $ N\in BC$ for which $ B\in (NC)$ and $ BN = BC$ . Observe that $ M(\widehat {ANC}) = m(\widehat {AMC}) = 30^{\circ}$ .

Thus, $ ANMC$ is inscribed in the circle with the center $ B$ . Hence $ BM = BA$ , i.e. $ m(\widehat {MAB}) = m(\widehat {AMB}) = 20^{\circ}$ .

[b][u]Proof II.[/u][/b] $ \{\begin{array}{c} m(\widehat {BAC}) = m(\widehat {ACB}) = m(\widehat {CBA}) = 60 \\
\ m(\widehat {CMA}) = 30\ \ ,\ \ m(\widehat {AMB}) = 20\end{array}$ . Denote $ m(\widehat {BAM}) = x$ .

[b]Case[/b] $ 1\blacktriangleright\ E\in (BC)$ . Observe that $ \{\begin{array}{c} m(\widehat {MAC}) = 60 - x \\
\ m(\widehat {BCM}) = 30 + x \\
\ m(\widehat {MBC}) = 100 - x\end{array}$ .

Apply the well-known property in the quadrilateral $ ABMC$ :

$ \boxed {\boxed {\ \begin{array}{c} \\
\ \sin\widehat {BAM}\cdot\sin\widehat {MBC}\cdot\sin\widehat {CMA}\cdot\sin\widehat {ACB} = \sin\widehat {MAC}\cdot\sin\widehat {CBA}\cdot\sin\widehat {AMB}\cdot\sin\widehat {BCM} \\
 \\
\sin x\sin (100 - x)\sin 30\sin 60 = \sin (60 - x)\sin 60\sin 20\sin (30 + x) \\
 \\
\sin x\sin (80 + x) = 2\cos (30 + x)\sin 20\sin (30 + x) \\
 \\
\sin x\cos (10 - x) = \sin 20\sin (60 + 2x) \\
 \\
\sin x\cos (10 - x) = \sin 20\cos (30 - 2x) \\
 \\
\sin 10 + \sin (2x - 10) = \sin (50 - 2x) + \sin (2x - 10) \\
 \\
\sin 10 = \sin (50 - 2x) \\
 \\
10 = 50 - 2x\ \ \vee\ \ 10 + (50 - 2x) = 180 \\
 \\
x = 20\ \ \vee\ \ x = - 60\not\in \\
 \\
\boxed {\ x = 20\ } \\
 \\
\end{array}\ }}$

[b]Case[/b] $ 2\blacktriangleright\ B\in (EC)$ . Observe that $ m(\widehat {BCM}) = 30 - x$ and $ m(\widehat {BMC}) = 10$ .

Apply the trigonometrical form of the [b]Ceva's theorem[/b] to the interior point $ B$ for the triangle $ AMC$ :

$ \boxed {\boxed {\ \begin {array}{c} \\
\ \sin\widehat {BAM}\cdot\sin\widehat {BMC}\cdot\sin\widehat {BCA} = \sin\widehat {BAC}\cdot\sin\widehat {BMA}\cdot\sin\widehat {BCM} \\
 \\
\sin x\sin 10\sin 60 = \sin 60\sin 20\sin (30 - x) \\
 \\
\sin x\sin 10 = \sin 20\sin (30 - x) \\
 \\
\cos (x - 10) - \cos (x + 10) = \cos (x - 10) - \cos (50 - x) \\
 \\
\cos (x + 10) = \cos (50 - x) \\
 \\
x + 10 = 50 - x \\
 \\
\boxed {\ x = 20\ } \\
 \\
\end{array}\ }}$[/color]

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension.[/u][/b] Let $ ABC$ be an $ B$ - isosceles triangle, i.e. $ BA = BC$ and let $ M$ be a point for which

the line $ BC$ separates the points $ A$ , $ M$ and $ m(\widehat {AMC}) = 90^{\circ} - m(\widehat {BAC})$ . Prove that $ \widehat {AMB}\equiv\widehat {BAM}$ .[/color][/quote]

[color=darkblue][b][u]Proof .[/u][/b] Let $ N\in BC$ for which $ B\in (NC)$ and $ BN = BC$ . Observe that $ m(\widehat {ANC}) = m(\widehat {AMC}) = 90^{\circ} - m(\widehat {BAC})$ .

Thus, $ ANMC$ is inscribed in the circle with the center $ B$ . Hence $ BM = BA$ , i.e. $ \widehat {AMB}\equiv\widehat {BAM}$ .

[b]Remark.[/b] Now you can generate and another similar problems ![/color]

Thank you M. Nicula ! Multsam fain ! Your proof is of a "biblical" simplicity !
I've made a figure for Proof 1 to thank you again !


geometry
circumcircle
incenter
geometric transformation
homothety
ratio
geometry proposed

Let $ O,I,N,Na$ be the circumcenter, incenter, nine-point center, and the Nagel point of triangle $ ABC$, respectively. Prove that $ ONa\parallel IN$.

it's very simple:

The Nagel point is the incenter of the anticomplementary triangle and the nine-point center is the circumcenter of the medial triangle so and homothety with center the centroid of ABC and ratio $ \minus{} \frac {1}{2}$ take O on N and Na on I, from wich $ ON_a \parallel IN$


function
vector
algebra unsolved
algebra

(a) Does there exist an injective function from $ R^2$ to $ R$?
(b) Does there exist an injective function from $ Z^2$ to $ Z$? (where $ Z$ is the set of all integers?)
(c) Does there exist any integer $ k$ so that there exists an injective function from $ R$ to $ Z^k$?

My only idea for (a) is to use the fact that a linear map from vector space $ R^2$ to $ R$ is not injective, since the dimension of $ R^2$ is greater than that of $ R$. But not all functions from $ R^2$ to $ R$ are linear maps, so I'm stuck.
Perhaps using countability arguments would help, but I know very little about countability.

it's just an issue of cardinality, since you're simply talking about maps in the category Set. so the answers are yes, yes, no.

[quote="epitomy01"](a) Does there exist an injective function from $ R^2$ to $ R$?[/quote]
Yes since it always exists bijection between $ A$ and $ A^n$ as soon as $ A$ is an infinite set. For the fun, an example :

Let $ f(x)$ be an injective function from $ \mathbb{R}$ in $ (0,1)$ (for example $ f(x)\equal{}\frac{\arctan(x)}{\pi}\plus{}\frac{1}{2}$)

Let $ g(x,y)$ be the injective function from $ (0,1)\times(0,1)$ in $ (0,1)$ defined (using decimal representations) as :
$ g(\overline{0.abcdef\ldots},\overline{0.a'b'c'd'e'f'\ldots})$ $ \equal{}\overline{0.0aa'0bb'0cc'0dd'0ee'0ff'0\ldots}$

Then $ g(f(x),f(y))$ is an injective function from $ R^2$ to $ R$.

[quote="epitomy01"] (b) Does there exist an injective function from $ Z^2$ to $ Z$? (where $ Z$ is the set of all integers?)[/quote]
Yes since it always exists bijection between $ A$ and $ A^n$ as soon as $ A$ is an infinite set. For the fun, an example :

Let $ f(x)\equal{}\frac{|4x\minus{}1|\minus{}1}{2}$. $ f(x)$ is a bijective function from $ \mathbb{Z}$ in $ \mathbb{N}\cup\{0\}$
Let $ g(x,y)\equal{}y\plus{}\frac{(x\plus{}y)(x\plus{}y\plus{}1)}{2}$. $ g(x,y)$ is a bijective function $ (\mathbb{N}\cup\{0\})^2$ to $ \mathbb{N}\cup\{0\}$

Then $ g(f(x),f(y))$ is a bijective function from $ Z^2$ to $ \mathbb{N}\cup\{0\}$, and so an injective function from $ Z^2$ to $ Z$.

[quote="epitomy01"] (c) Does there exist any integer $ k$ so that there exists an injective function from $ R$ to $ Z^k$?[/quote]
No, since $ R$ is uncountable and $ Z^k$ is countable.

[quote="greentreeroad"]why g(x,y)=y+(x+y)(x+y+1)/2 is a bijective function from N^2 to N(includes 0).
Please explain[/quote]

It is a rather well known bijection from from $ N^2$ to $ N$(including 0).

$ 1)$ non-mathematical view :
You "count" the grid $ N^2$ in the following order :
(0,0)
(1,0),(0,1)
(2,0),(1,1),(0,2)
(3,0),(2,1),(1,2),(0,3)
(4,0),(1,3),(2,2),(3,1),(4,0)
and so on

$ 2)$ mathematical view

Let ${ g(x)=\frac{x(x+1)}2}$. $ g(x)$ is a strictly increasing fonction (from $ N$ in $ N$)
We have then $ f(x,y)=y+g(x+y)$ and we clearly have $ g(x+y)\leq f(x,y)<g(x+y+1)$

Assume now $ f(x,y)=f(x',y')$. Then, since $ g(x+y)\leq f(x,y)<g(x+y+1)$ and $ g(x'+y')\leq f(x',y')<g(x'+y'+1)$ and $ g(x)$ strictly increasing, $ f(x,y)=f(x',y')$ implies $ x+y=x'+y'$, and then $ y+g(x,y)=y'+g(x'+y')$ implies $ y=y'$ and so $ x=x'$. So $ f(x,y)$ is injective.

Now, let $ a$ be any non negative integer.
Since $ g(x)$ is a strictly increasing fonction (from $ N$ in $ N$) and since $ g(0)=0$, it exists a value $ b$ such that $ g(b)\leq a<g(b+1)$
Let then $ y=a-g(b)$. We have $ 0\leq y<g(b+1)-g(b)=b+1$. So $ 0\leq y\leq b$
Let then $ x=b-y$. We have $ x\geq 0$

So we have $ x\geq 0$ and $ y\geq 0$ such that $ a=g(b)+(a-g(b))=g(x+y)+y=f(x,y)$ and so $ f(x,y)$ is surjective.

And so $ f(x,y)$ is bijective.


A school bus has a maximum capacity of 44 passengers. Clemente school is taking a field trip, and the school rules say that the ratio of students to adults must be at most 10 to 1. Given that 121 students are going on the trip, what is the minimum number of buses needed?

The most students we can fit onto a bus is $ 40$, because we must have at least $ 4$ adults.  Thus, we simply take:

$ \left \lceil \frac{121}{40} \right \rceil \equal{}\boxed{4}$.

Since there are $121$ students going on the trip, there would be $12.1$ adults, rounding to $13$. This means there are $121+13=134$ passengers going on the field trip. Since one bus can hold $44$ passengers, we need to compute $134/44$ to get the number of buses. This is equal to $4$ when rounded up.


i)Let $ \sqrt{2+\sqrt{2+\sqrt{2+x}}}+\sqrt3\sqrt{2-\sqrt{2+\sqrt{2+x}}}=2x$ when $ x>0$, find all answers of this equation.
ii)Let a and b be integers that makes $ (\sqrt[3]{a}+\sqrt[3]{b}-1)2=49+20\sqrt[3]{6}$, find $ a+b$.

[quote="snowyowl"]i)Let $ \sqrt{2+\sqrt{2+\sqrt{2+x}}}+\sqrt3\sqrt{2-\sqrt{2+\sqrt{2+x}}}=2x$ when $ x>0$, find all answers of this equation.
[/quote]

In order the secund summand be real, we need $ 2\geq\sqrt{2+\sqrt{2+x}}$ and so $ x\leq 2$; So we have $ 0<x\leq 2$ ans it exist $ y\in[0,\frac{\pi}{2})$ such that $ x=2\cos(y)$.
Then :
${ \sqrt{2+x}=\sqrt{2+2\cos(y)}}=2\cos(\frac{y}{2})$ and $ \sqrt{2+\sqrt{2+x}}=2\cos(\frac{y}{4})$.
$ \sqrt{2+\sqrt{2+\sqrt{2+x}}}=2\cos(\frac{y}{8})$.
$ \sqrt{2-\sqrt{2+\sqrt{2+x}}}=2\sin(\frac{y}{8})$.

And the equation becomes $ 2\cos(\frac{y}{8})+2\sqrt3 \sin(\frac{y}{8})=4\cos(y)$ which can be written $ \sin(\frac{\pi}{6})cos(\frac{y}{8})+\cos(\frac{\pi}{6})\sin(\frac{y}{8})=\cos(y)$ and so :

$ \sin(\frac{y}{8}+\frac{\pi}{6})=\cos(y)$ and so $ \frac{y}{8}+\frac{\pi}{6}=\frac{\pi}{2}-y$ (since $ y$ and $ \frac{y}{8}+\frac{\pi}{6}$ $ \in[0,\frac{\pi}{2}))$

So $ y=\frac{8\pi}{27}$ and $ x=2\cos(\frac{8\pi}{27})$


geometry
3D geometry
inequalities
algebra
factorization
sum of cubes

Let [tex]n \geq 2[/tex] and suppose [tex]k_1,\;k_2,\;...,\;k_n[/tex] are natural numbers such that [tex]\sum_{j = 1}^nk_j^3 \leq 7n[/tex]. What is the biggest possible value of [tex]\sum_{j = 1}^nk_j[/tex]?

Is it : 2*n-1 + [(n-1)/7] ?

[hide]I won't give details because my reasoning is very complicated (and not 100% rigourous) but the main ideas are :

we must choose as small k_i (to maximise the sum) as we can but not too small (to reach 7n).

From n*a :^3: >= 7n, we must take as many '2' as we can.

Hence the following first values (n : ki :  :Sigma: :  :Sigma:  :^3: ) and the formula given :

2 : 1  2 : 3 : 9

3 : 1  2  2 : 5 : 17

4 : 1  2  2  2 : 7 : 25

5 : 1  2  2  2  2 : 9 : 33

6 : 1  2  2  2  2  2 : 11 : 41

7 : 1  2  2  2  2  2  2 : 13 : 49

8 : 0  2  2  2  2  2  2  2 : 14 : 56

9 : 1  1  2  2  2  2  2  2  2 : 16 : 58

10 : 1  1  2  2  2  2  2  2  2  2 : 18 : 66 [/hide]

0 isn't considered a natural number, just a whole number.  But even so, That solution works, just by using

8: 1 1 2 2 2 2 2 2 instead of 0 2 2 2 2 2 2 2

I think that is what he meant.

Is that solution the same as n+ floor(6n/7) which is what I got? It does match with the maximum values belenos posted...

Well I'm about to go to bed, but my first thought on the problem:

[hide]

Lets try to find the sequence with maximal sum, but given that, minimal sum of cubes. Consider two of the terms, a and b:

(a+1)^3 + (b-1)^3 = a^3 + b^3 + 3(a+b)(a-b+1). So if we swap a to a+1 and b to b-1, we keep optimal sum but the sum of cubes would decrease if a<b-1. So we can assume that all terms of the sequence are either x or x+1 for a fixed x.

[/hide]

Another approach (in spoiler): [hide]Use the inequality k :le: (1/7)k3 + 6/7[/hide].

Now that I've woken up I'll finish it off:

[hide]

Above I proved all numbers were x or x+1: obviously x=1 as if all were >=2, then the sum of cubes > 8n. So all numbers are 1 and 2: to maximise sum, we maximise number of 2s. If there are k twos, the sum is n+k and the sum of cubes is 7k+n. So we have k<=6n/7, and thus the maximum sum is n+[6n/7].

[/hide]


linear algebra
matrix
linear algebra unsolved

$ \left( \begin{array}{cccc} 0 & 0 & 0 & a_{1,n} \\
0 & 0 & ... & 0 \\
0 & a_{n \minus{} 1,2} & 0 & 0 \\
a_{n,1} & 0 & 0 & 0 \end{array} \right)$ $ \in$ $ M_{n}(\mathbb{C})$

how to find the Jordan canonical form of the above matrix?
 :D

 :roll:
Assume n is even, I think we can consider the determinantal divisor of it

we denote $ A\equal{}\left( \begin{array}{cccc} 0 & 0 & 0 & a_{1,n} \\
0 & 0 & ... & 0 \\
0 & a_{n \minus{} 1,2} & 0 & 0 \\
a_{n,1} & 0 & 0 & 0 \end{array} \right)$

Clearly
$ det(\lambda I_{n}\minus{}A)\equal{}(\lambda^{2}\minus{}a_{n,1}a_{1,n})(\lambda^{2}\minus{}a_{n\minus{}1,2}a_{2,n\minus{}1})...(\lambda^{2}\minus{}a_{\frac{n}{2}\plus{}1,\frac{n}{2}}a_{\frac{n}{2},\frac{n}{2}\plus{}1})$

and to look at all minors of order $ n\minus{}1$, we can see its determinantal divisor of order $ n\minus{}1$ is integer $ 1$ :o

so....

The Jordan canonical form is :
$ \left(\begin{array}{cccccc}\sqrt{a_{n,1}a_{1,n}}&&&&& \\ &\minus{} \sqrt{a_{n,1}a_{1,n}}&&&&\\&&...&&&\\&&&...&&\\ &&&&\sqrt{a_{\frac{n}{2}\plus{}1,\frac{n}{2}}a_{\frac{n}{2},\frac{n}{2}\plus{}1}}&\\&&&&&\minus{} \sqrt{a_{\frac{n}{2}\plus{}1,\frac{n}{2}}a_{\frac{n}{2},\frac{n}{2}\plus{}1}}\end{array}\right)$

but I can't work it clearly,and i don't think about the details, like if $ a_{n,1}a_{1,n}\equal{}a_{n\minus{}1,2}a_{2,n\minus{}1}$,
then the Jordan canonical form is the same one? I just assume all the $ a_{j,k}$, $ a_{k,j}$ are different , where $ j\plus{}k\equal{}n\plus{}1$  :oops:

$ A^2$ is a diagonal matrix. Thus if $ A$ is invertible then $ A$ is diagonalizable.

$ 2$ examples where $ A$ is not invertible:
$ n\equal{}5$: If the false diagonal of $ A$ is $ [*,0,0,*,*]$ then the Jordan form is $ diag(J,0,*,*)$
$ n\equal{}4$: if the false diagonal of $ A$ is $ [0,0,*,*]$ then the Jordan form is $ diag(J,J)$ where
$ J\equal{}\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and where $ *$ denotes a non zero element.


Let $ A$ be a $ n\times n$ matrix with complex elements. Prove that $ A^{\minus{}1} \equal{} \overline{A}$ if and only if there exists an invertible matrix $ B$ with complex elements such that $ A\equal{} B^{\minus{}1} \cdot \overline{B}$.

This is a very useless problem, IMHO: no use of matrices at all, just take $ B\equal{}af(A)\plus{}f(a)I_n$ for suitable $ a$ (to make $ B$ invertible) of modulus $ 1$. Here, $ f$ means complex conjugate. The converse is obvious. Bad problem.

Still you should see the results (very bad). And actually after thinking about it $ f(A)\equal{}\overline A$ I think and $ f(a)\equal{}\overline a$, and of course just add the condition for $ B$ invertible. Haven't worked out the last condition (B invertible) but I think this might work, because there are finite values of $ a$ for which $ B$ is not invertible, and we just take one that is good.

Of course, it's what I said. And of course the problem can be misleading, because people would believe that it's smart and uses advanced knowledge. I still cannot believe that a problem like the 4-th one, which is completely trivial with high-tech and very challenging otherwise, appeared in the RMO. It's the sign of a serious lack of originality, I think. As for this problem, I think it comes from higher mathematics and is probably used in computations of some cohomology groups, at least the statement makes me believe that.

Dunno, let's see the TST, though I am also kind of disappointed with the questions this year (overall).

Can anyone tell me why they use IMC 2002 problem B5 in a national contest? If I were participant, this would have saved me an hour or more of thinking, as I don't find the problem trivial at all (though some may say it is just misleading). Does really nobody in the comitee read past IMC exams (and 2002 is not that long gone, but yet old enough to be on Kalva)?

  darij

Well, look also at the discussion started on problem 4 and maybe you will understand why such kind of problems appear. Funny enough, I see that the "creator" of this problem also proposed some problems to IMC, but this one does not appear in his "creations". I find it very hard to believe he has not seen this problem at IMC. The problem with our olympiad is that there are very few "creators", who "create" everything, every year. This makes that horrible situations like this year appear: with "memory" or "experience" or "training" one can solve two problems in a few minutes, problems which otherwise are really not easy. But that's a matter of choice of course, as is my opinion that this is the worst choice of a set of linear algebra problems. However, it's not my job, this is just a comment on how imperfect world is.  :D


[quote="Svejk"]Is this true ? If $abc=1$ , $a+b+c\geq ab+bc+ca$[/quote]
If $a;b;c$ negative,this inequality is false

a, b and c can't all be negative since abc = 1. However the statement is false. This equality is not satisfied when a = b = 2 and c = 1/4.

the inequality is equivalent to $0\geq (1-a)(1-b)(1-c)$... i think it's easy to conclude...


factorial
number theory unsolved
number theory

Find x;y;z if x;y;z are natural numbers and x!y!=z!
Thank for your help. Sorry all of you because i am not good at English.

Well i believe that the only solns of this could be x or y being 1 and the other  being = z :)

Well this can be easily proved ( i think) :)

I guess you are looking the general solns.

The obvious general solution is $ (x,y,z) \equal{} (1,n,n)$ where n is a natural number
another solution is $ (x,y,z) \equal{} (n, (n! \minus{} 1), n!)$ where n is a natural number

Are there finite number of solutions to this equation??

There is one more solution missing. Find that particular solution :)

I could observe this.

$ 6! \equal{} 720 \equal{} 8.9.10$
$ 7!.6! \equal{} 10!$

The general idea is that, if I am able to split a given factorial as the product of k consecutive numbers (with 1x2 not included), then I can generate a solution for this.

But how do I generalize the above observation?

This topic   has  already been posted.
Have a look here

http://www.mathlinks.ro/Forum/viewtopic.php?t=191139


logarithms
number theory
number theory unsolved

Ok first off hello all, this is my first post and I am new to number theory and am trying to learn more as time goes on. My question will be probably easy if not simple for most gurus and math heads here. I have looked and searched the internet for my solution and have found none yet. There is also probably a name for my problem but any help is greatly appreciated.

1+x where x approaches infinity (or continues to grow 10,11,12.....100.....1000000), how many times will (1+x)/2 before it equals 1

To be more precise how many times will it take to equal the value of 1 if you add 1 to another number then divide by 2
over and over.After each division you will get another number which you will then add to 1 and then again divide by 2 until you reach 1.

Any number with decimals from this division are to be rounded with one exception; if the 1-(the answer to the last division you just did) < 1 then consider your problem complete.

For example, the number 1 and 10

1+10=11
11/2=5.5 (rounds to 6)

1+6=7
7/2=3.5 (rounds to 4)

1+4=5
5/2=2.5 (rounds to 3)

1+3=4
4/2=2

1+2=3
3/2=1.5 (consider your problem complete because 1-(1.5)<1

So for this example between 1 and 10 it took you 5 times which you came up with the numbers (6,4,3,2,1.5). What I want to know is as the number 10 in this example gets larger, what is the relationship between how many times it will take to get to one and the difference between 10 and 1. I would like to know the equation of this.

I have constructed a table to the best of my ability to give an example.That should be attached and by heart I am a programmer and have java code if requested to understand if I have not been clear enough. Any help is greatly appreciated. Thanks to you all.

Rounding?

Well, the problem is much easier without rounding, so let me solve that version first.

Let $f(x) = \frac{x+1}{2}$.  Then

$f^{n}(x) = \frac{x}{2^{n}}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^{n}}= 1+\frac{x-1}{2^{n}}$

Given $x$ you want to find $n$ such that $f^{n}(x) = 2$.  Well, then

$x-1 = 2^{n}\implies \boxed{ n = \log_{2}(x-1) }$.

Are there any known names for these relationships? Or is this type of relationship have a name? I'm sorry its just that i'm not a math major or number theory major. I am just trying to learn.

Well, it's [i]approximately[/i] geometric.


1)  Five hundred people attended the recital.  How
many children and adults attended the recital?

2)   The auctioneer said, "Take them both for 20 dollars!"
"I'll take one for 10 dollars,"  I said.
He said, "No, each one is 20 dollars."
The auctioneer said, "No, you misunderstood me."
Who's correct?

3)  There are four brothers and sisters in one family.
There are four brothers and sisters (all different people
from the first group) in another family.  How many total
males and females** between the siblings are there?

**  Assume everyone is clearly either a male or a female
 for this problem (as opposed to an asexual person or a
hermaphrodite.)

[hide="I think..."]1) 500

2) "I"

3) 0 (there's no one given standing between the two families)[/hide]

[hide="my answer"][u]Answer to puzzle 1[/u]:

children + adult = total number of persons

Therefore, precisely 500 children and adults attended the recital.

[u]Answer to puzzle 2[/u]:

The auctioneer is selling house numbers like “1”, “17”, “245” etc, with each house number priced at 20 dollars.  When the auctioneer said, "Take them both”, he referred to a two digit house number, with both the digits together available at 20 dollars.

Therefore, the auctioneer is correct.

[u]Answer to puzzle 3[/u]:

Assuming that no sibling of a given family is a half sibling of the other, the total males and females between the siblings in a family is 4.[/hide]

[hide="For #1..."]The answer is 0 because nobody is both a child and an adult.[/hide]

[hide="#2"]The use of the word "He" implies that a third person (a person other than "the auctioneer" and "I") is speaking.  The auctioneer's second statement then is likely directed at "he", not "I".  Even if we assume that the auctioneer is correct in saying that "he" misunderstood him/her (the auctioneer), there is not enough information to determine whether or not "I" can buy only one of the item being described or if "I" must purchase both.[/hide]

[quote="t0rajir0u"][quote="Arrange your tan"]1Assume everyone is clearly either a male or a female
 for this problem (as opposed to an asexual person or a
hermaphrodite.)[/quote]
Asexuality is an orientation, not a gender.[/quote]

But biologically, it [i]can[/i] mean having no gender if you define gender a certain way.

As applied to species that reproduce asexually, yes.  I've never heard the term used to refer to a human without sexual organs, if that's what you're implying.  And sex chromosome abnormalities usually aren't thought of as the absence of gender.

Interesting digression...

[quote="Oxford English Dictionary"]Biol. a. Not sexual, without sex. In Bot. formerly applied to cryptogams; cf. agamic.
b. In general contexts: without sexuality.[/quote]

So generally the words means "without sexuality," particularly since this is not a biology forum.

Anyway, the last two puzzles are quite confusing...

On the digression: see [url=http://www.asexuality.org]here[/url]. It could use the exposure. I never had a name for it until I stumbled across a link myself.
Asexuality is an overloaded word with other meanings, but we have plenty of those already, and this usage isn't likely to conflict with others.

[quote="Arrange your tan"]1)  Five hundred people attended the recital.  How
many children and adults attended the recital?[/quote]

Problems like this make me wish we had parentheses and order of operations in the english language for clarity.


There are n cars waiting at distinct points of a circular race track. At the starting signal each car starts. Each car may choose arbitrarily which of the two possible directions to go. Each car has the same constant speed. Whenever two cars meet they both change direction (but not speed). Show that at some time each car is back at its starting point.

please try this one, its nice, you are going to have fun with it!

consider instead that the balls on collision just-i.e. instead of rebounding,they just  'go through each other'.then since all the balls are moving at the same speed,after a certain time all the balls,clearly will be at exactly where they started after one rotation.
in the present situation, since the collisions are perfectly elastic(which we need to assume) instead of exactly the same balls we have a permutation of the balls but in exactly the same spots as the inital starting point,and also moving in the same direction as the balls initially went. it then follows that after some time,the initial situation will repeat.[/list]

to clarify:i was thinking in terms of balls colliding instead of cars moving round.....

(i think i have solved this problem before.....!)

Ok, I know, every period, the cars form a permutation, How do you guarantee that the initial permutation is going to appear again?, i know it is just a detail and my solution is the same as yours but obviously with this detail solved....

I've also seen this problem before (and its solution; I didn't solve it though, it was shown to me :(), and I didn't want to spoil the pleasure of those who hadn't :).

When the labels of the cars reach the initial positions, it's just as if the whole thing
started all over again, so by the next time they reach the initial position, the exact same permutation $\pi$ will have been performed on them. This means that we have to wait for $k$ turns until the cars are all in their initial positions, where $\pi^k=e$ ($e$ is the identity permutation; for any $\pi$ there is such an $k$).

We solve this problem through a few observations.
Label the cars in integers from $0$ to $n-1$ counterclockwise starting from an arbitrary car.

[i]Observation 1.[/i] The collision condition given in the problem statement is equivalent to two cars passing through each other and switching numbers, continuing in the same speed and direction.
Since every car has the same angular speed, they will all return to their original position regardless of direction after a full revolution. Thus, it is shown that the cars will eventually end in the same position at the same time, possibly having different numbers (but not necessarily). $\square$

[i]Observation 2.[/i] A car labeled $k$ for $0 \leq k \leq n - 1, k \in \mathbb{Z}$ must be located between the cars labeled $k + 1 \pmod n$ and $k - 1 \pmod n$ at all times in the same order. (In other words, a car cannot somehow pass through the cars directly adjacent to it while keeping the same number, using the the artificial collision condition used in Observation 1.)
This implies that the permutation of the numbers of the cars after a revolution will be a rotation of the original permutation of the numbers. $\square$

In this manner, after the cars travel one revolution we can generate that new permutation/rotation by adding some integer $a$ to every car's number, taking $\pmod n$ if a car's number exceeds $n - 1$. After $n$ revolutions we will have added $na$ to every car's number. Since $na + k \equiv k \pmod n$, after $n$ revolutions the cars will be back in their original permutation and location, which completes the proof. $\blacksquare$


inequalities
geometry
inequalities proposed

[color=purple] Let triangle ABC with three altitudes $h_{a},h_{b},h_{c}$  and two radii R,r.
         Prove that
    $\frac{3\sqrt{3}R}{4r}\le \frac{a}{h_{a}}+\frac{b}{h_{b}}+\frac{c}{h_{c}}\le \frac{\sqrt{3}R^{2}}{2r^{2}}$
     Welcome and I hope you'll have fun ! :D [/color]

[color=purple]It's not hard problem I think. But any one can give me the proff or the stronger ? [/color]

for the right inequality use:
i) a/ha + b/hb +c/hc= a^2 + b^ 2 + c^2 / 2rp    ( S = rp)
ii)the inequality of Mitrinovic: p>= 3r*sqrt3
iii) a^2+b^2+c^2<=9*R^2

S=area of the triangle
p=semiperimeter
r= the radius of the incircle

[color=purple]Thanks but can you write clearly ?  :) [/color]


vector
linear algebra
matrix
linear algebra open

A $simple$ $vector$ is a vector which all of it's entries are {1,-1,0}.
A $simple$ $base$ is a base that all of it's vectors are simple
Let $A$ be adjacency matrix of a graph.Find all(some)of classes of graphs,for which the nullspace of $A$ has a simple base.

I am really shocked :o .No class of graphs have been found with desired property?


Does there exists a function f defined on the positive integers with values in the same set such that f(f(n))=s(n) for all n, where s(n) is the sum of digits of n? I don't know the answer, though I put this problem long time ago.
  Harazi

I think the answer is yes.  Unfortunately I don't know how to write the solution rigorously, so I would very much appreciate it if somebody else could tighten up the gaps.

A couple of (obvious) things to note:
1) If N is partitioned into sets Ai such that Ai = {x: s(x) = i}, then each of these sets has an infinite number of elements.
2) s(n) < n for n > 9

Let a <- b mean f(b) = a

Let f(i) = i for i = 1, 2, ..., 9

Construct sequences as follows:
1 <- ___ <- 10 (smallest x such that s(x) = 1) <- ___ <- 19 (smallest x such that f(x) = 10) <- ___ ...
1 <- ___ <- 100 (2nd smallest x such that s(x) = 1) <- ___ <- 199999999999 (smallest x such that f(x) = 100) <- ___ ...

Now pair up the above 2 sequences.

1 <- ___ <- 1000 (3rd smallest x such that f(x) = 1) <- ___ <- (smallest x such that f(x) = 1000) <- ___ ...
1 <- ___ <- 10000 (4th smallest x such that f(x) = 1) <- ___ <- (smallest x such that f(x) = 10000) <- ___ ...

Pair up the above 2 again...etc.

Now we do
1 <- ___ <- 10 (smallest x such that s(x) = 1) <- ___ <- 28 (2nd smallest x such that f(x) = 10) <- ___ <- 1999 (smallest x such that f(x) = 28)
1 <- ___ <- 10 (smallest x such that s(x) = 1) <- ___ <- 37 (3rd smallest x such that f(x) = 10) <- ___ <- 19999 (smallest x such that f(x) = 37)

Pair those 2 up.

In general it should look like this:

1st <- 1st <- 1st <- 1st <- ...
1st <- 2st <- 1st <- 1st <- ...

1st <- 3rd <- 1st <- 1st <- ...
1st <- 4th <- 1st <- 1st <- ...
...

1st <- 1st <- 2nd <- 1st <- ...
1st <- 1st <- 3rd <- 1st <- ...

1st <- 1st <- 4th <- 1st <- ...
1st <- 1st <- 5th <- 1st <- ...

...

1st <- 1st <- 2nd <- 2nd <- ...
1st <- 1st <- 2nd <- 3rd <- ...

1st <- 1st <- 2nd <- 4th <- ...
1st <- 1st <- 2nd <- 5th <- ...

And on and on and on it goes.  Now do the same beginning with 2 <- ___, 3 <- ___ etc.

I sincerely hope the idea is correct or this will just be an extremely long and useless post. ;)

Very interesting idea. I will have to check if it verifies, but it seems that it works.

It doesn't.  Obviously there is a problem in combining this already:

1st <- 1st <- 2nd <- 1st <- ...
1st <- 1st <- 3rd <- 1st <- ...

1st <- 1st <- 4th <- 1st <- ...
1st <- 1st <- 5th <- 1st <- ...

Never mind then.

I have only now realised this. Too bad. Anyway, the idea of construction is beautiful.

Another (hopefully not so useless this time) thought:

Consider the sequence of numbers n, s(n), s(n), ..., sk(n), sk+1(n), where sk(n) > 9 and sk(n) \leq 9.  Obviously this sequence of numbers is unique for each n.  Also, let si-1(n) be the aith smallest number in the set {x: s(x) = si(n)}.  If a1, a2, ak, ak+1 and sk+1(n) are known, then the original n is unique.

Now consider the sequence backwards, i.e. sk+1(n), sk(n), ..., s(n), s(n), n.  If one of the numbers is changed, then all the numbers after that number will be different from the original sequence.

If we let sk+1(n) = 1, 2, ..., 9, and ai = 1 to \infty, then the union of all possible sequences = N.

Now does this pairing work?
1 <- ___ <- 1st <- ___ <- x1th <- ___  <- x2th <- ___ <- x3th <- ___
1 <- ___ <- 2nd <- ___ <- x1th <- ___  <- x2th <- ___ <- x3th <- ___

1 <- ___ <- 3rd <- ___ <- x1th <- ___  <- x2th <- ___ <- x3th <- ___
1 <- ___ <- 4th <- ___ <- x1th <- ___  <- x2th <- ___ <- x3th <- ___

...

And let xi vary from 1 to \infty for each i.

I forgot to say that in my opinion this is a good construction. But since last time something escaped to my attention, I ask everyone to think about this construction and say if it's ok.

Oh. It is not easy to understand.
I have a stupid question. Do we need all sequnces contain distict natural numbers? But this pairing contains continuum sequences.


The USAMO qualifiers for 2005 were announced earlier this week on the AMC web site.  There were 6 qualifiers from New York City.  Here they are:
[code]
Long Cheng       11  Stuyvesant
Hooyoung Chung   11  Stuyvesant
He Cui           11  Cardozo
Geehoon Hong     12  Cardozo
Theodore Spaide  12  Dalton
Danny Zhu         9  Stuyvesant
[/code]
There were also 6 qualifiers from other parts of New York State:
[code]
Michael Ding     10  Glen Cove
Sunny Kam        10  Great Neck South
Jennifer Lin     11  Arlington HS (La Grangeville)
David Rush        8  Scarsdale
John Schulman    11  Great Neck South
Neal Wadhwa      12  Ward Melville (Setauket)
[/code]
Congratulations to all of them!

In a topic on  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31725]Top USAMO High Schools[/url], I posted a list of schools in the country that have produced the most USAMO qualifiers over the last 10 years.  Stuyvesant is at number 3 on the list, and Hunter is tied for 12th.


1x1
2x2
3x3
4x4
5x5
sqrt2xsqrt2
sqrt5xsqrt5
sqrt10xsqrt10

8?

No, because there are 5 points on each side, you can only make up to a 4x4 square. Other than that you are correct, so it is 7.

...which...makes $ 8$! LOL! But don't we also have $ \sqrt{17}\times\sqrt{17}$?

Oh! I see. Yeah, that's right. Thanks for catching that!


In my AP Chemistry test a few days ago, we were asked to find the number of structural isomers of heptane. I found 6, but it was long and tedious. Is there a general formula for number of structural isomers of alkanes; $C_nH_{2n+2}$? If there is, I'd like to see the derivation.

Sloane has some information on this.  http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000602

[quote="Rushil"]I believe it can be proved that one cannot find the formula for the no. of isomers of a specific compound. Somebody did prove it as I read in a discrete Math book although it didn't mention the actual guy who did this and I haven't found hid name still. Anyone else have any ideas???[/quote]

I remember reading in some book the derivation for the general number of isomers for some type of hydroxyl. I think it was a very simple class of molecule; single bonds all around and then the one hydroxyl group somewhere.


geometry
incenter
angle bisector
geometry proposed

Given a triangle $ ABC$.
Suppose that a circle $ \omega$ passes through $ A$ and $ C$, and intersects $ AB$ and $ BC$ in $ D$ and $ E$.
A circle $ S$ is tangent to the segments $ DB$ and $ EB$ and externally tangent to the circle $ \omega$ and lies inside of triangle $ ABC$.
Suppose that the circle $ S$ is tangent to $ \omega$ at $ M$.
Prove that the angle bisector of the angle $ \angle AMC$ passes through the incenter of triangle $ ABC$.

I seem to remember it from somewhere: it was weither posted before or I've struggled to solve it. Anyway, what I wanted to say was this: isn't $S$ tangent to $\omega$, $BE$ and $BD$? (instead of $BD$ you wrote $ED$).

yes I posted it before and noone manage to give a sol.official sol is really floundering but I know really elementary one.

Last year,I heared from Gharakhani (One of the people in iranian team for IMO2004) that it has an elementary solution.But I don't know that solution.
So send that solution. ;)  ;)  :P

I proved a lemma that can solve quite strightforward the problem. I need some time to translate from italian to english the statment and the solution. In the meantime, if you can understand some italian you can have a look at

http://olimpiadi.sns.it/modules.php?op=modload&name=Forums&file=viewtopic&topic=2635&forum=5

I solved this problem some time ago using harmonic conjugates and some properties of projective geomtry........if i remember i will post the solution which was rteally nice.....
 (I remember that g7 of isl 2002 can be solved similary)

Iran99

Let E and F be the intersections of c(O) with AC and BC resp.; M and N be the midpoints of AE and BF; G and L be the inters. Of circle  c(O) with lines MO and NO resp.
If I is the incenter of ABC lets call P the inters. Between LA and BI and Q the inters. Between GB and AI. From the lemma RLxxxx we have that c(API) and c(BQI) intersect c(O) in the same point Y. Lets call W and X the intersection between c(API) with AC and c(BQI) with BC.
Let H and K be the intersection of c(O) with AI and BI. It is not too difficult to prove that LG//HK then LH=GK from which < IAP = < IBQ. Then ABPQ is cyclic and it easy to prove that P,W,X,Q are collinear and WX//LG.
If R and S are the inters. Of YG with AI and YL with BI from lemma RL250704 we have that YSIR is cyclic and tangent to c(O). Then SR//LG.
From < IYR = <ISR = <IKH = < BAH = < IAW it follows that W, Y and G are collinear. Similarly one can prove that X, Y and L are collinear too. So c(XYW) is tangent to c(O).
From < AWY = < RIY = < RSY = <YXW it follows that c(XYW) is tangent to AC and similarly one can prove that is tangent to BC too.
To conclude the proof observe tha < API = < AYI and < BQI = < BYI, but we know that < API = < BQI from which we have the thesis.

lemma RL250704

Dato un cerchio c e un triangolo ABC inscritto e siano R ed S due punti su c(ABC) e I un punto interno tale che < RIS sia supplementare ad < ABC. Se T = AC.SI provare che, al variare di C (fissati A e B) c(CTI) ha un punto fisso.

sprmnt21 I don't understand the Lemma RLXXXX that you mean can you give me a proof

Given D, C and F,E on a circe c(A) and I inside so that <FIE + <DGC = 180, where G is a generic point on the circle c(A). Then all the circles c(GJI), where J=GD.EI, as G varies on the circle,  pass through a fixed point.

Proof.

If X is the (second) common point to c(A) and c(GJI) for a generic G and let L be a generic different position of G on c(A) and M=LD.EI, we have that <IML = <MJG-<MDJ. But given that <MDJ=<LXG and <IJG = <IXG then <IML=<IXG-<LXG=<IXL ==> X  c(LMI) too.
As a corollary of this lemma we have that, when L=X, c(XOI) is tangent to c(A)

I post my solution here but i write in chinese, anybody help to translate?

Dear Mathlinkers,
you can see my proof of this problem on my website http://perso.orange.fr/jl.ayme  vol. 2  Un remarquable résultat de Vladimir Protassov.
Sincerely
Jean-Louis

My solution:
We cyclically rename ABC as BCA and rename M as T. Denote by M as the midpoint of the arc $DCB$. We prove the problem statement when $D, C , M , B$ are in this order. The other cases can be handled similarly.
Let $S$ be tangent to $AB, AC$ at $X, Y$ resp.
By the negative homothety sending $(S)$ to $\omega$, we have $X, T, M$ collinear. Let $CM $ intersect $YX$ at $Z$. As $\angle MDX =\angle MTD$ we have that $MDX \sim MTD$ and so $MD^2 = MX.MT$. By simple angle chasing we have $TCYZ$ to be a cyclic quadrilateral from $ \angle TYZ + \angle TCZ = 180$. From this conclusion, $\angle TZC = \angle TXZ$ and so $MZ^2 = MX.MT$. This gives that $MZ = MD$ and so $Z$ is the $B$ excenter of BCD. If I is the incenter of ABC, then from $\angle CTZ = \angle CIZ$, ITCYZ is cyclic. Thus $\angle ITC = \angle IZC = \angle CTB/2$ so $IT$ bisects $\angle CTB$ as required.


What is the 2004th positive odd integer?

its kinda odd. my problemon alcumus keeps staying this one. it wont go OFF.

_______________________________

GO AOPS!!!!  :!:

The $ n$th positive odd integer can be determined with $ 2n\minus{}1$. Thus, the $ 2004$th positive odd integer is $ 2(2004)\minus{}1\equal{}\boxed{4007}$.

The odd integers are $ 1,3,5,\cdots$ This is an arithmetic series with first term $ 1$ and common dif 2, so the 2004th odd integer is $ 1\plus{}2(2004\minus{}1)\equal{}4007$

To start out, you must know the correct formula. To find an "odd positive integer formula" is 2x-1,(try it with any number, it works!) such that x is the integer. In this case, 2004 is. Substitute, and you get 2(2004)-1. This equals 4007.

__________________________
AOPS RULES!!
 :lol:

We tried to submit this through Alcumus, but it wouldn't take the input. So for people who don't know the formula and need to have it explained from scratch, my 11yo daughter came up with this solution --- which we prefer and think you may wish to include as an alternative.

When we start counting, the first odd number comes before the first even number. The odds and evens come in pairs, but in each pair, the odd member comes before the even member:

first odd = 1, first even = 2,
second odd = 3, second even = 4,
third odd = 5, third even = 6, etc.

In each case, the nth even number = 2n, so the nth odd number = 2n - 1.
Therefore, the 2004th odd number = 2 x 2004 - 1 = 4007.

What I did was I tried finding the arbitrary odd numbers.  I found the second odd number (3) and the fifth (9) and the 16th (31) (which was a lot of legwork but proved my conjecture).  I found that the nth odd number was 2n - 1 just because of the pattern.  2*2-1=3, 2*5-1=9, etc.  (Sorry, I'm unfamiliar with Latex formatting.)

[quote="DeniseIL"][hide="DeniseIL's solution"]In each case, the nth even number = 2n, so the nth odd number = 2n - 1.
Therefore, the 2004th odd number = 2 x 2004 - 1 = 4007.[/hide][/quote]

This was my approach as well. The 2004th positive odd integer will come immediately before the 2004th positive even integer, therefore...
[hide="hidden solution"]The 2004th positive even integer is $2 \times 2004 = 4008$, so the 2004th positive odd integer is $4008 - 1 = \boxed{4007}$.[/hide]

For the record, you'll be amazed by how many facts you can discover about odd and even numbers by letting your even numbers equal $2n$, your odd numbers equal $2n+1$, and then playing around with some algebra. For example, why is the sum of two odd numbers always even? Let $2n+1$ be one odd number, and let $2m+1$ equal another odd number. $(2n+1)+(2m+1)=2n+2m+2=2(n+m+1)$. Therefore, the sum of any two odd numbers is always even.

Tim

 Solution:
The first positive odd integer is 1. To get to the 2004th positive odd integer, we must add 2 exactly 2003 times. Make sure you see why it's not 2004 times -- that would give us the 2005th positive odd integer, just like adding one 2 to 1 gives us the second positive odd integer, not the first. Therefore, our answer is $1 + 2003\cdot 2 = \boxed{4007}$.

I don't think so, since not everything is in $LaTeX$

The alcumus solution isn't always in LaTeX.
I just thought that because he wrote
[quote]... Make sure you see why it's not 2004 times -- that would give us the 2005th positive odd integer, just like adding one 2 to 1 gives us the second positive odd integer, not the first.  ... [/quote]
which sounds a bit more Alcumus-solution type than something a user would write

[b]But I might be wrong! :)[/b]

Sounds like Alcumus to me....
There's always that one person who copies and pastes the Alcumus solution :P


i need help in solving this area problem. please reply if you have any idea how to solve it, or the solution if possible. thanks.

a closed rectangular box, having a square base, has a total surface area of 16 sq. ft. If a main diagonal of the box has a length of 3 ft, what are the dimensions of the box?

thanks very much.

If $a$ is the side of the square, and $b$ is the other side, $S$ the surface area and $D$ the main diagonal, we get

$S=2a^2+4ab=16\Rightarrow \frac{S}{2}=a^2+2ab=8$.
$D=\sqrt{2a^2+b^2}=3\Rightarrow D^2=2a^2+b^2=9$.
$D^2-\frac{S}{2}=a^2-2ab+b^2=(a-b)^2=1\Rightarrow b=a\pm 1$.
$8=a(a+2b)=a(3a\pm2)=3a^2\pm2a$.
$3a^2-2a-8=0\Rightarrow a=2$ (since $a>0$).
$3a^2+2a-8=0\Rightarrow a=\frac{4}{3}$ (since $a>0$).

Substituting back to get $b$, we get dimensions of $2\times2\times1$ and $\frac{4}{3}\times\frac{4}{3}\times\frac{7}{3}$.

you've been very helpful. thank you so much....


I give you a solid triangle and ask you to bisect each of its sides.
You can see that the lines connecting these points of bisection then form a triangular hole with $3$ solid triangles bordering the hole.

Then I tell you to do the same operation of bisection on each of the remaining solid triangles, opening up new holes similar to the first. each of the new triangles has dimensions that are exactly half that of the original and a quarter of the area

After $n$ (where 'n' is any postiive integer) such operations what
will be the area of the resulting bizzare figure.

?

Am I misunderstanding the problem??
[hide] $(\displaystyle\frac{3}{4})^n$?? [/hide]


inequalities geometry circumcircle inradius

Prove that in $ \triangle ABC$ exists the relation ${{ \boxed {\ p\ \le\ \frac {a^2}{b + c} + \frac {b^2}{c + a} + \frac {c^2}{a + b}\ \le\ \frac {2p(R - r)}{R}}}\ }\ .$

[b]Remark.[/b] The left inequality is easily and is equivalently with $ \boxed {\ \sum\frac {a}{b + c}\ \ge \ \frac 32\ }$ ([u]Nesbitt[/u]).

The right inequality is more interesting and is equivalently with $ \frac rR + \sum\frac {a}{b + c}\ \le\ 2\ .$

[quote="Potla"]What is $ p$? I understand that $ R$ is the circumradius and $ r$ is the inradius of $ ABC$.
Thanks[/quote]
[b]p[/b] is the semiperimeter ; $ p\equal{}\frac{a\plus{}b\plus{}c}{2}$

chebychev:
suppose that $ a\geq b\geq c$
$ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{c \plus{} a} \plus{} \frac {c^2}{a \plus{} b}\geq\frac{1}{3}(a\plus{}b\plus{}c)\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\geq \frac{a\plus{}b\plus{}c}{2}\equal{}p$


geometry
circumcircle
geometric transformation
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14793[/img]

 :(

Draw parallels $ a, b$ to $ CD$ through $ A, B.$ Lay the given segment $ s$ on $ a$ with one endpoint $ A$ and the other enpoint $ E$ inside the given circle $ (O).$ $ M$ is the midpoint of $ AB.$ Line $ EM$ cuts $ b$ at $ F$ and $ BF \equal{} AE \equal{} s.$

Pick arbitrary point $ Q$ on the arc $ CD$ opposite to $ A, B$ and Draw the $ \triangle ABQ.$ Parallel to $ AQ$ through $ E$ cuts $ BQ$ at $ H$ and parallel to $ BQ$ through $ F$ cuts $ AQ$ at $ G.$ $ \angle AQB \equal{} \angle AGF \equal{} \angle BGE \equal{} \text{const},$ $ GH \parallel CD,$ $ GH \equal{} AE \equal{} BF \equal{} s$ and $ A, B, E, F$ are fixed $ \Longrightarrow$ locus of $ G, H$ are circles, circumcircles $ (X), (Y)$ of the $ \triangle AGF, \triangle BHE.$ (The 2nd triangle and its circumcircle are just translations of the 1st triangle and its circumcircle by the distance $ AE \equal{} s$ in the direction of the ray $ (AE.$ This can be used in the construction.) The circle $ (X)$ intersects $ CD$ at $ K_1, K_2$ and the circle $ (Y)$ at $ L_1, L_2,$ such that $ K_1L_1 \equal{} K_2L_2 \equal{} AE \equal{} BF \equal{} s.$ At the same time $ AK_1, BL_1$ meet at $ P_1 \in (O)$ and $ AK_2, BL_2$ meet at $ P_2\in (O).$ Points $ P_1, P_2$ solve the problem.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14818[/img]

 :thumbup: