John has five children.  What is the probability that at least half of them are girls?

Since this is an odd number, the probability is 1/2.

The probability at least half are girls, $ g$, is equal to the probability that at least half equal boys, $ b$.

Since the probability that #girls = #boys is zero (odd number of children), the sum of $ g$ and $ b$ is $ 1$, we know that $ g \equal{} \boxed{\frac {1}{2}}$

MEWTO:
1: EXPLAIN CALCULATIONS IN YOUR POST
2: DON'T RUSH THROUGH EXPLANATIONS
3: DON'T REPEAT EXACTLY WHAT OTHER PEOPLE SAY
4: DON'T RESPOND TO THIS JUST TO RAISE YOUR POST COUNT

I'm not, but also by Binomial theorom, the sum of this probability for

1 child: 1c1/2^1=1

3 children 3c2+3c3/2^3=3+1/8=1/2

5 children 5c3+5c4+5c5/2^5=10+5+1/32=1/2

Notice a pattern?

[quote="CA Math"]You're not what?

And also, you could have explained that the FIRST time, which is what you should always do.[/quote]

Well it is intuitevely obvious, which is what matters for FTW. Says the hypocrite who keeps replying to these posts.

But this is not FTW.

It is about Problem Discussion, which means that someone didn't understand the problem or how to do it.  Posting a spew of numbers would definitely not help anyone trying to learn.

And the probability is symmetrical because the only probabilities involved, the probability that a child is a girl or a boy, are both $ \frac{1}{2}$

[quote="Name of this forum"][size=200][b]FTW Problems[/b][/size] Discussion[/quote]

Enough said. Why are we spamming thuis thread?

Exactly my point.

You [b]are[/b] spamming.  Post [b]better[/b] solutions with [b]explanations[/b].

I would take this as a warning.  Remember that I already made a thread on this.

[quote="isabella2296"]
Hello Everyone,

For a while, there was a severe reduction in the amount of spam in this forum. However, I have noticed that there is a large outbreak once more. Even though I can delete, edit, lock, etc., it would be much better if there was very little spam in the first place. So I have posted some guidelines, the same ones that I posted a while before.

1. Do not post a second solution to a problem unless

-- You arrived at a different answer.

-- You have a different way of getting to the answer.

-- You noticed a mistake someone made.

-- You wish to clarify something that was unclear in someone else's post.

2. If you think that someone wrote a great explanation,

-- Rate the post 4, 5, or 6 cubes.

-- Private message the person if you want to, and tell them you like the explanation.

-- Do not make a post saying, "Nice explanation, _____________!"

3. If someone posts spam,

-- Rate the post 1 cube.

-- Private message the person detailing why it is spam, in case they did not know, and ask them to kindly abstain from spamming in the future.

-- Do not make a post saying, "Hey! That's spamming! Don't do it, __________!"

4. Keep in mind that each thread is only, and I repeat, only, for purposes of solving the problem shown.

-- Be careful not to stray off topic.

-- Before posting, think hard whether your post is necessary or not. Remember, you do not have to press the "Submit" button.

-- Do not post for a higher post count. What's the point of having a 1000 posts if they're all spam?

5. When you see a problem that involves a "broad" concept, which you do not understand,

-- Try not to make a new post and ask. For example, if a problem involves, say, proportions, and you do not know anything about proportions, there are two things you can do. One, you can private message the person who answered the problem and ask for details on proportions, or better yet, two, start a new thread in Classroom Math on Proportions.

-- If somebody ignores this guideline, and asks anyway, and you wish to answer their question, private message them and answer their question, or bring this guideline to their attention.

Thank you for taking the time to read this. I have locked this, so you may private message me any questions or concerns you have.[/quote]

Read this.

[sarcasm]Hm AIME that was a productive post following the guidelines that you just posted![/sarcasm]

-- Do not make a post saying, "Hey! That's spamming! Don't do it, __________!" '

And with regard to the problem, that assumes the probability for each child are independent and 1/2, when in reality you forgot about babies born as both genders, and the fact that the probability isnt exactly equal. HOwever with the given information, 1/2 is the answer.


Individual Round, Number 28:
Factor $6(x-8)^2+7(x-8)-5$ completely.  One of the factors is:

a) $(3x+19)$
b) $(2x+19)$
c) $(3x+5)$
d) $(2x-17)$
e) None of the Above

Team Round, Number 5:
Solve the systems:
$3x-5y=1$
$2x+7y=-1$

$4m-3n=-7$
$5m-n=2$

What is the SUM of the denominators of the x-coordinate and m-coodrinate solutions?
(Attempt to solve this problem within one minute)

[quote="Farazy"]Individual Round, Number 28:
Factor $6(x-8)^2+7(x-8)-5$ completely.  One of the factors is:

a) $(3x+19)$
b) $(2x+19)$
c) $(3x+5)$
d) $(2x-17)$
e) None of the Above
[/quote]

Too lazy to learn latex so stufoo :/

[hide="1"]
y = x - 8
6y^2 + 7y - 5
(2y - 1)(3y + 5)
(2x - 17)(3x - 19)
[b]d[/b]
[/hide]

I don't think number is hard to solve under one minute, if you can think fast. Make sure you always cancel the y coordinates, leaving the x's left.


advanced fields
advanced fields theorems

http://www.math.ucsd.edu/~sbutler/ron/04_04_openproblems.pdf

In this paper, Graham mentions that for a sufficiently large n, any r-coloring of Euclidean
n-space contains a monochromatic unit square, but says that the necessary n was not known.

Does anyone know what the current best bound on minimum n(r) is, or if any bound is known at all?

I haven't heard anything on this for probably two years. But I vaguely recall that there is an asymptotic. I can't point you in the direction of any papers, though.


algebra
polynomial
linear algebra
matrix
Ring Theory
superior algebra
superior algebra unsolved

Let $ A \equal{} [x_{ij} ]$ be the n x n matrix of variables. Prove that the
determinant of A is an irreducible polynomial in the polynomial ring
$ Z[x_{ij} ]$.

A fun one.

Suppose $ det\ A \equal{}  pq$.  First, take the quotient by all $ x_{ij}$, $ i\neq j$, so $ \overline{p}\cdot\overline{q} \equal{} \prod_i x_{ii}$.  Without loss of generality, $ \overline{p} \equal{} \prod_{i\equal{}1}^k x_{ii}$, $ \overline{q} \equal{} \prod_{i\equal{}k\plus{}1}^n x_{ii}$, with $ 0\leq k < n$, so $ det\ A \equal{} (\prod_{i\equal{}1}^k x_{ii} \plus{} r)(\prod_{i\equal{}1}^k x_{ii} \plus{} s)$, where $ r$ and $ s$ do not depend on the diagonal variables.

Consider this as a polynomial in $ \prod_{i\equal{}1}^k x_{ii}$.  The leading coefficient is clearly $ det\ C$, where $ C$ is the $ (n\minus{}k)\times (n\minus{}k)$ block matrix found in the lower right corner of $ A$.  So we have $ q \equal{} det\ C$.  Similarly, $ p \equal{} det\ B$, where $ B$ is the $ k\times k$ block matrix found in the upper left corner of $ A$.

We have shown $ det\ A \equal{} det\ B \cdot det\ C$.  But the right side does not depend on $ x_{1n}$ if $ k > 0$, while the left side does.  So we must have $ k\equal{}0$, whence $ p \equal{} 1$, and $ det\ A$ is irreducible.


[quote="TienVinh"][quote="N.T.TUAN"]Cac ban cho hoi ai lam mod cua forum Viet Nam nay nhi?[/quote]NNT tính làm mod hả?[/quote]

Khong he.To muon nho ban ay ti viec thoi ma.

tôi là mod ở đây,có vấn đề gì mà bạn cần,cứ PM cho tôi nhé.

lâu rồi không trở lại, bây giờ thì có được 1 mod.  :) tuy nhiên cái sub-forum này vẫn lạnh lẽo như nhà ma

toi khong ngi la no lanh leo nhu ma ( boi vi cha biet ma la lanh hay nong )   ma chu yeu cac thanh vien Vietnam chua biet nhieu den forum Vietnam nay . No se rat hay boi vi o day chung ta cung 1 ngon ngu , moi trao doi se thuan tien hon .
analysis , minh to mo hoi ban dang o tai Vietnam hay dang o nuoc ngoai ?
      minh nghi ban rat co ca tinh song khong nen noi nhung dieu  theo 1 con mat phan dien nhu the . Boi vi minh thay cac forum o cac nuoc cung dau co soi dong dau nhi  , chi duoc 1 vai thoi !!!!!!!!!
 Khong biet co dung khong ? Boi vi minh vao forum cua cac bang nuoc Mi thay con "lanh "  hon ma` .

[quote="princenguyen"]:lol: Anh ơi. Như thế nào để được làm MOD. Em mún làm MOD ^^. 1 chân Admin cũng ok :D[/quote]
Please PM to Valentin :D

[quote="princenguyen"]:lol: Anh ơi. Như thế nào để được làm MOD. Em mún làm MOD ^^. 1 chân Admin cũng ok :D[/quote]
Phan dau di, thay bao la oai lam day   :ninja:   :D

Cho em hỏi là bác NNTuan ở VN hay ở nước ngoài ạ ^^


I know this has an answer but i dont understand it please help it is...

5/x-4/y+6/z=15
6/x+2/y+9/z=14
7/x+6/y-3/z=19

the / are fractions
thanks for the help

i realized i posted this in the wrong topic and put it in the highschool part
you may delete this post if you would like

You may put $x'=\frac{1}{x},y'=\frac{1}{y},z'=\frac{1}{z}$
then you'll get
$5x'-4y'+6z'=15$
$6x'+2y'+9z'=14$
$7x'+6y'-3z'=19$
which is easy to solve.
(Eliminating 1 unknown by the first two equations and sub it into the third, or simply use Cramer's rule)


ratio
email
advanced fields
advanced fields unsolved

This is my first post and if someone can answer this, I'll be long-time user of the forum :)

Situation:
Everyone knows as you twist a pair of wires together, their total lengths will shorten.  Also the number of twists and the diameter of the wire are crucial in how much it will shorten.

Problem:
I am searching for an equation relating final, shortened, length to the starting length, number of twists, and guage of the wire.

I have been looking at this post in particular http://mathforum.org/library/drmath/view/53155.html

The equations for the helix were helpful until I realized they did not take into consideration the fact that each wire has a certain width(gauge).

Givens:
Gauge, starting length of wires, and number of twists.

If it helps:
Gauge 16 AWG ~= .0805 inches in diameter
200 m long wire ~= 7874 inches
1 twist per 2 or 4 inches.

THANKS!

[quote="nbourg8"]
I have been looking at this post in particular http://mathforum.org/library/drmath/view/53155.html
The equations for the helix were helpful until I realized they did not take into consideration the fact that each wire has a certain width(gauge).
[/quote]
But they [b]did[/b] take the width into consideration! The radius of the helix is exactly half of the diameter of the wire.  The helix itself represents the deformed central axis of the wire and the assumption is that there is no noticable stretching or shrinking along the central axis, which is true if you don't bend your wire too much.

Alright, on another forum someone posted this reply which is a completely different approach.  Could you verify this:

My guess would be that the new length would be equal to the original length (200m) multiplied by the ratio of the twist length (2" or 4" -- whatever you're going to use) and the sum of the twist length and 2*pi*Diameter of the conductor (#16AWG).

NewL = OldL * [TwistL / (TwistL + 2*pi*D)]

For example, if you have 200m (200,000mm) to start with, and your wire has an outside diameter of 1mm and your going to have one complete twist for every 2" (50.8mm), then the new length would be:

New length = 200,000*[50.8 / (50.8 + 2*3.14159*1)]mm
= 177,986mm = 178m

I guess, you'd still need a design check on that calc, but that's the way I see it anyhow.

[quote="nbourg8"]  Could you verify this:
My guess would be that the new length would be equal to the original length (200m) multiplied by the ratio of the twist length (2" or 4" -- whatever you're going to use) and the sum of the twist length and 2*pi*Diameter of the conductor (#16AWG).
[/quote]
Well, as he said, it is just a guess, and it is a wrong one :). Note that no explanation for the formula was given, the result was just postulated. So there is nothing to verify.

Okay, well I am still a little confused as to how "Dr. Math" has come up with his result on the following link.  I see he talks about a 1 twist per inch however, I dont see how I could change it to a twist per 2 or 4 inches.

The general formula is $L_{twisted}^2+\pi^2D^2N^2_{twists}=L_{untwisted}^2$ where $D$ is the diameter of the wire, $L_{twisted}$ is the length of the twisted wire, $L_{untwisted}$ is the length of the untwisted one, and $N_{twists}$ is the total number of twists. Now, $N_{twists}=n L_{twisted}$ where $n$ is the number of twists per unit length. So, we finally get
\[ L_{twisted}^2(1+\pi^2D^2 n^2)=L_{untwisted}^2\,. \]
If you make one twist every 2 or 4 inches, this just means that you have $n=\frac 12$ or $\frac14$ correspondingly.

Last question, I have uploaded a spreadsheat showing the calculations made using both feet and meters.  They display different values but yet use the same information.  This is expected though because of the different metric systems, however, when calculating the percentage of the final to initial values, they are different still.  Please review this excel spreadsheet as this should be the last of my questions.

Thanks for all the help.

http://www.yousendit.com/transfer.php?action=download&ufid=4C68D02C00E27620

Note: if you want you can change the contents of the file and reupload it to yousendit.com

my email is nbourg8 at gmail.com

Your diameter of wire in millimeters is just huge compared to that in inches. $15\,mm=0.6\,in$ which is much greater than $0.05\, in$. I'm not sure about the other convertions, but this one is certainly wrong.

Yup, that was issue.  Now percentages work out.  Its strange that the spreadsheet shows a very small decrease in wire length when common logic should tell you it would decrease MUCH more than 2.5 meters with 1 twist per inch.  If anyone else can detect an error in this method, please let me know.

1 twist per inch doesn't sound like that much to me. The result seems reasonable. If you twisted it as much as you could, you'd get a lot of twists per inch, and then your wire would certainly decrease a lot in length.

[quote="nbourg8"] Its strange that the spreadsheet shows a very small decrease in wire length when common logic should tell you it would decrease MUCH more than 2.5 meters with 1 twist per inch. [/quote]
What "common logic"? :? Seriously, there is no error. It may be a bit countrintuitive, but the decrease is indeed very small when the diameter of the wire is small.

Hrm, I guess I never thought of it like that.  I had to hand twist a pair of 16 gauge wire at 1 twist per 2 inches and it was hella hard.  Was probably the shielding that made things harder.  Either way, I can now take into account for any changes in diameter due to the shielding thanks to you guys.  YOU RULE :)

Added to Post***

Wow, if you account for the sleeve thickness the wire sits in, it makes a TREMENDOUS difference in the results.


[quote="kunny"]Let $a,b,c$ be positive reals such that $a+b+c=2.$
Prove that $\sqrt{1-\frac{a^3}{bc}}+\sqrt{1-\frac{b^3}{ca}}+\sqrt{1-\frac{c^3}{ab}}\leq \sqrt{3}.$[/quote]
Try $a=b=0.9,c=0.2$ ;)


What is $ \sqrt[4]{16} \cdot \sqrt[3]{8} \cdot \sqrt{4}$ expressed as a positive integer?

This is $ 2 \times 2 \times 2$ or $ \boxed{8}$.


I've done a few of these puzzles online before, so I decided to make my own. It's really easy in comparison... Have fun! (Note: The answer to the question What is 1+1? A)2 B)3 C)4 D)5 is [b]A[/b], not 2.)

[hide="Fat Quiz Hidden Inside"]Question 1:
The first question with answer C is
[b]A)[/b]4 [b]C)[/b]2
[b]B)[/b]3 [b]D)[/b]1

Question 2:
The answer to this question is
[b]A)[/b]A [b]C)[/b]C
[b]B)[/b]B [b]D)[/b]D

Question 3:
The first question with answer D is
[b]A)[/b]13 [b]C)[/b]9
[b]B)[/b]11 [b]D)[/b]7

Question 4:
The next question with answer C is
[b]A)[/b]10 [b]C)[/b]12
[b]B)[/b]11 [b]D)[/b]13

Question 5:
The next question with the same answer as this is
[b]A)[/b]6 [b]C)[/b]8
[b]B)[/b]7 [b]D)[/b]9

Question 6:
The answer to 12 is
[b]A)[/b]D [b]C)[/b]B
[b]B)[/b]C [b]D)[/b]A

Question 7:
The answer to 8 is
[b]A)[/b]B [b]C)[/b]C
[b]B)[/b]A [b]D)[/b]D

Question 8:
The answer to 7 is
[b]A)[/b]B [b]C)[/b]C
[b]B)[/b]A [b]D)[/b]D

Question 9:
This is the first question with answer
[b]A)[/b]A [b]C)[/b]C
[b]B)[/b]B [b]D)[/b]D

Question 10:
The only two consecutive questions to have the same answer are
[b]A)[/b]7 and 8 [b]C)[/b]12 and 13
[b]B)[/b]8 and 9 [b]D)[/b]13 and 14

Question 11:
The amount of questions with answer A is
[b]A)[/b]6 [b]C)[/b]2
[b]B)[/b]4 [b]D)[/b]0

Question 12:
The answer to 6 is
[b]A)[/b]D [b]C)[/b]B
[b]B)[/b]C [b]D)[/b]A

Question 13:
The amount of questions with answer B is
[b]A)[/b]prime [b]C)[/b]neither A nor B
[b]B)[/b]odd [b]D)[/b]both A and B

Question 14:
The amount of questions with answer B is
[b]A)[/b]3 [b]C)[/b]5
[b]B)[/b]4 [b]D)[/b]6

Question 15:
The amount of questions in this quiz is
[b]A)[/b]12 [b]C)[/b]14
[b]B)[/b]13 [b]D)[/b]15[/hide]

I get hopelessly confused by these things. The answer to 15 is D!

[hide="I FIGURED IT OUT!!!"]
1.B
2.A
3.C
4.B
5.A
6.A
7.B
8.A
9.D
10.C
11.A
12.D
13.D
14.A
15.D
[/hide]

This took like my entire math class, but I got it!

Sorry, erabel, but I don't think that's right.
Your answer for question 4 was B, but you answered C on question 10, not question 11.

Thanks for pointing that out. I'll try again. Didn't notice that. TY, cf249

Since it's been over a month and no one has figured this out, can you post the solution, Brut3Forc3?

[hide="Answer"]
1  B
2  A
3  C
4  A
5  B
6  A
7  B
8  A
9  D
10 C
11 A
12 D
13 D
14 A
15 D[/hide]

Some women shake hands standing up, some shake hands standing down, and some kiss cheeks standing up, and some kiss cheeks standing down.


Given x:y:z = 4:5:7, find the value of  x + y + z / y + z

By setting values for x, y, and z, the problem becomes easy. Let x=4, y=5, and z=7. Thus $ x\plus{}y\plus{}z\plus{}\frac{z}{y}\equal{}4\plus{}5\plus{}7\plus{}\frac{7}{5}\equal{}\boxed{17\frac{2}{5}}$.


how did you guys get such high SAT verbal scores? Did you study something special? did you do formal SAT test prep?
Ryan

This really doesn't belong as a separate topic in the MC forum. I answered in the SAT thread.

Dedicated studying, 3 hours a day, 5 days a week, 2 or 3 months=730. ;) Do it for 8 more months and you'll get an 800 :)

Out of the MC forum...

is 630 high for 7th grade?

If so - a lot of reading.

The best way for me to learn is to lovingly obsess over something interesting, even if it isn't the most direct way.

During lunch, go to the school library, and make it a game to read all of the books with the coolest sounding vocabulary. Read Moby Dick, Paradise Lost, Julius Caesar, etc. Look up the words you like the best, and the rest you can gain from context. After a year or so you should be fine.

My mind would explode if I memorized for 3hrs. every day. If you've got the skilz though, it's faster.

Forget the 3 hours a day thing. The best way by far is to read everything that interests you. If you dont' have an interest, you are lying. Most published serious texts, whether from the Teen magazine or the New York Times will help improve your vocabulary and especially your understanding of words in context. You must read though. To develop vocabulary, staring into an SAT vocab list simply isn't enough to get you a 800.

Exactly. I'm done with standardized tests forever, but I always did very well on the verbal sections. That didn't come by studying- instead I read for pleasure, all the time.
I never read those "great books", either; newspapers and science fiction were good enough.

Also, I did better on the real thing than on the small amount of practice I did. That's what adrenaline does for me, on any test I consider "important".

Most of the top-level students on math contests do have very good verbal skills; playing Anagrams at MOP can get scary. Once you have to start writing proofs, verbal and writing skills matter quite a bit.


geometry
circumcircle
inradius
geometry theorems

Is there anybody who can show the Carnot theorem :

the sum of the directed distances between the circumcentre of a triangle and its sides is equal to the sum of the circumradius and the inradius of the triangle.

It's actually called the Carnot theorem and is pretty well-known. See http://www.cut-the-knot.org/proofs/carnot.shtml or http://agutie.homestead.com/files/Geoproblem_B.htm#Carnot .

  darij

Ptolemy gives:
\[ OD \cdot c + OF \cdot a = R \cdot b \]
\[ OD \cdot b + OE \cdot a = R \cdot c \]
\[ OE \cdot c + OF \cdot b = R \cdot a \]

Since: $r \cdot (a+b+c) = 2S = OD \cdot a + OE \cdot b + OF \cdot c$

Add the four equations togather, we get:
\[ OD \cdot (a+b+c) + OE \cdot (a+b+c) + OF \cdot (a+b+c)= R \cdot (a+b+c) + r \cdot (a+b+c) \]
Hence:
\[ OD + OE + OF = R+r \]


geometry
perimeter
vector
geometry solved

I have a question from my geometry teacher (im in 10th
grade), and he told us to prove that a triangle would be equilateral if
the perimeters of three triangles' (made by the intersection of 3
altitudes of the triangle (H) -orthocenter- and two of the vertices (AB, AC, BC) )
are equal. But, I'm getting stuck. Please help.

I know if I find out if the altitudes bisect the angle or the side,
I would be able to prove that it is equilateral, I have been trying
for three days, but I have yet not got how to make the "first move"
on it so I can continue.

I have posted in the "Pre-Olympiad" but, as it appears to be quite a hard problem (according to my friend), I have decided to post it here. This question, as he says, is quite similar to the ones from BMO.

[quote="mecrazywong"]A hint: AH=2RcosA. From here we can deduce that sinC-cosC=sinB-cosB=sinA-cosA. Now the problem should be rather easy.[/quote]

The problem is, if A is obtuse then the non-signed length of AH is -2R cos A, and I don't think that Krazy_Brian meant "signed" perimeters (moreover, in the case of an obtuse triangle, sin x - cos x is not injective anymore, so you need an additional argument). Here is a solution for the non-signed case:

[color=blue][b]Problem.[/b] Let ABC be a triangle with orthocenter H. If the triangles BHC, CHA, AHB have the same perimeter, then show that triangle ABC is equilateral.[/color]

[i]Solution.[/i] We start with a well-known lemma:

[b]Lemma.[/b] If X, Y, U, V are four points, then $XY\perp UV$ holds if and only if $XU^2-XV^2=YU^2-YV^2$.

[hide="Proof of Lemma"][i]Proof of Lemma.[/i] Using vectors and their scalar products, we see that $XY\perp UV$ is equivalent to $\overrightarrow{XY}\cdot\overrightarrow{UV}=0$. But

$\overrightarrow{UV}=\frac{\overrightarrow{UV}+\overrightarrow{UV}}{2}=\frac{\left(\overrightarrow{UX}-\overrightarrow{VX}\right)+\left(\overrightarrow{UY}-\overrightarrow{VY}\right)}{2}=\frac{\left(\overrightarrow{UX}+\overrightarrow{UY}\right)-\left(\overrightarrow{VX}+\overrightarrow{VY}\right)}{2}$,

so that

$\overrightarrow{XY}\cdot\overrightarrow{UV}=\overrightarrow{XY}\cdot\frac{\left(\overrightarrow{UX}+\overrightarrow{UY}\right)-\left(\overrightarrow{VX}+\overrightarrow{VY}\right)}{2}$
$=\frac12\cdot\left(\overrightarrow{XY}\cdot\left(\overrightarrow{UX}+\overrightarrow{UY}\right)-\overrightarrow{XY}\cdot\left(\overrightarrow{VX}+\overrightarrow{VY}\right)\right)$
$=\frac12\cdot\left(\left(\overrightarrow{UY}-\overrightarrow{UX}\right)\cdot\left(\overrightarrow{UX}+\overrightarrow{UY}\right)-\left(\overrightarrow{VY}-\overrightarrow{VX}\right)\cdot\left(\overrightarrow{VX}+\overrightarrow{VY}\right)\right)$
$=\frac12\cdot\left(\left(\overrightarrow{UY}^2-\overrightarrow{UX}^2\right)-\left(\overrightarrow{VY}^2-\overrightarrow{VX}^2\right)\right)=\frac12\cdot\left(\left(\left|\overrightarrow{UY}\right|^2-\left|\overrightarrow{UX}\right|^2\right)-\left(\left|\overrightarrow{VY}\right|^2-\left|\overrightarrow{VX}\right|^2\right)\right)$
$=\frac12\cdot\left(\left(YU^2-XU^2\right)-\left(YV^2-XV^2\right)\right)=\frac12\cdot\left(\left(YU^2-YV^2\right)-\left(XU^2-XV^2\right)\right)$.

Hence, $XY\perp UV$ is equivalent to $\frac12\cdot\left(\left(YU^2-YV^2\right)-\left(XU^2-XV^2\right)\right)=0$, what is equivalent to $XU^2-XV^2=YU^2-YV^2$; this proves the Lemma.[/hide]

Now, let's solve the problem: Since H is the orthocenter of triangle ABC, we have $AH\perp BC$, so that the Lemma yields $AB^2-AC^2=HB^2-HC^2$; in other words, $\left(AB-AC\right)\left(AB+AC\right)=\left(HB-HC\right)\left(HB+HC\right)$. On the other hand, since the triangles CHA and AHB have the same perimeter, we get HC + HA + AC = HA + HB + AB, what transforms into HB - HC = - (AB - AC). Thus, the equation $\left(AB-AC\right)\left(AB+AC\right)=\left(HB-HC\right)\left(HB+HC\right)$ becomes $\left(AB-AC\right)\left(AB+AC\right)=-\left(AB-AC\right)\left(HB+HC\right)$. Now, if $AB\neq AC$, then $AB-AC\neq 0$, so we can divide this equation by AB - AC and obtain AB + AC = - (HB + HC), what is nonsense, since AB + AC > 0 and HB + HC > 0. Therefore, we must have AB = AC. Similarly, AB = BC. Hence, the triangle ABC is equilateral, and the problem is solved.

  Darij

[quote="darij grinberg"][quote="mecrazywong"]A hint: AH=2RcosA. From here we can deduce that sinC-cosC=sinB-cosB=sinA-cosA. Now the problem should be rather easy.[/quote]

The problem is, if A is obtuse then the non-signed length of AH is -2R cos A, and I don't think that Krazy_Brian meant "signed" perimeters

  Darij[/quote]
Well, if the triangle is obtuse, the problem is even easier to tackle:

WLOG, assume A is obtuse. Then we have AB+AH=CA+CH, however, we must have CB+CH>AB+AH, by the triangle inequality. Contradiction!
Thus the case A is obtuse can be eliminated.

Anyway, nice from you noticing such minor mistake :)

Oh... ok my bad... The triangle is acute. Sorry guys...

Thanks a lot darij, and mecrazywong, I'd been thinking about this for days, without a solution. (I had some ideas,  but they didnt work out right)

Once again, thanks very much!


vector
superior algebra
superior algebra unsolved

Supposed $ f$ is an automorphism onto $ R[x]$. Prove that $ f(a)\equal{}ma\plus{}b$ for all $ a \in R[x]$,for some $ m$ and $ b$ in the set of complex  numbers.

I don't believe this is true if "automorphism" is taken to mean "ring automorphism."  Do you mean "vector space automorphism"?


algebra
function
domain
complex analysis
complex analysis unsolved

Let V be an analytic subvariety of D, which is a domain in the n-dimensional
complex space. Let f be a projection from V to (n-1)-dimensinal complex
space which maps z to its frist n-1 components. Then the counter-image
of each f(z) must be a finite set.
Can anybody help me?

I know the solution now.
And I forgot a condition that V is compact.


ARML
topology
geometry
trigonometry
Euler
modular arithmetic
rotation

Well, it's happening on Saturday.  Who's coming?  I'm looking forward to taking a test on which Harrison's been working.

I don't think I am, but I could be wrong if my coach has worked wonders

I will be there as an individual. I would have paid to be a team of 1, but $\$40$ is too much for me.

@ Boy Soprano II: I'll assume that you will also be present at tommorow's Fall Cobb County Tournament.

I'll be there as well. Be scared. (Just kidding  :P )

I'll be there as an individual :P

I might have 1 or 2 people from my school join me.

Yeah, I'm looking forward to the power round. I need to get better at proof competitions.

Do you think Harrison and Sean made the power round really hard?

[quote="drunner2007"]Do you think Harrison and Sean made the power round really hard?[/quote]
Well, it probably won't be of olympiad difficulty, because that wouldn't be feasible unless they wanted really low scores, but something tells me that Harrison may have had temptations to write difficult problems, and hopefully he will have had a hard time overcoming them. :diablo:

[quote="drunner2007"]Yeah, I also suspect that Harrison will want to write hard problems.
Boy Soprano II do you think you can handle anything he throws at you?[/quote]
I hope so :lol: .  But mostly it is because I spend most of my time doing hard problems, so I hope that hard problems will be to my advantage.  Also, it makes the order of the top few places less of a lottery.

yeah i hate tournaments that are so easy that simple mistakes determine the winner

Does anyone know what time the tournament ends and what time awards start/end?

i think awards are at 1:00

a professor told me we'll start at 9:00, go for 1 hr. on the written test. after that, 15 (maybe 20) minute break, and then power round for 20 minutes. to finish off competition, ciphering for approximately 1 hr.

i don't understand why ciphering would take 1 hr and power round 20 minutes. should be interesting either way though

ciphering usually takes around that long, if not even longer, but thats a really short power round

[quote="drunner2007"]yeah. i mean, 1 ARML power round proof can take 20 minutes alone (near the end), so maybe the questions won't be that difficult.[/quote]

Either that, or Harrison could be conniving to make us all [i]think[/i] it is really easy, and when we're unsuspectingly starting the tournament... HE'LL GET YOU!  :rotfl:  :evilgrin:

[quote="drunner2007"]I'm trying to figure out #30.  I thought I found that it would have to be 0. My reasoning led me to think that all the terms involving an even number raised to a power would cancel, contributing 0 to the sum, and then all the odd numbers raised to a power would contribute 0 to the sum as well. I wonder what I'm doing wrong. What's a good general strategy?[/quote]
Well, each digit gives 2 outputs (which may be the same) (because $\phi(10) = 4$ and you add two two your exponent each time you repeat a digit, mod 4), and you have 100 of each of those, up to 2000.  So pretty much all you have to do is find the units' digit of $1^{1}+2^{2}+3^{3}+4^{4}+5^{5}+6^{6}$, which is not too difficult to do with the Euler-Fermat theorem.  (I hope that wasn't too dense.)

Take all the numbers mod 10 and add up: $1^{1}+2^{2}+3^{3}+\ldots+10^{10}$. This should give us some random last digit. We have 200 of such sums, so, we have to just add: $1^{1}+\ldots+6^{6}$, whose last digit is simply $9$.  :)

Ok makes sense. Not that bad when you look at it like that.

[quote="Potato Theory"]How do you get that it will give 200 of the same random digit for each set of 10?[/quote]

Hmm... could you rephrase your question? I don't understand it.  :maybe:

What I mean is, how do you know that the sum of the numbers $1^{1}+2^{2}+3^{3}+\ldots+10^{10}$ gives the same result as $11^{11}+12^{12}+13^{13}+\ldots+20^{20}$ mod 10?

I wrote a proof of Euler's Thm. using abstrabt algebra principles (groups and their properties).

in fact, the proof in AOPS 2 uses abstract algebra concepts without explicitly referring to them.

i might type it up

Well, of course, haha, every time you do algebra or number theory you use abstract algebra principles without explicitly referring to them.  Number theory, in particular, is easy to link to more fundamental algebraic principles because it uses several different structures, and it is all the more so because most students are underexposed to it, which I think is why they place such huge emphasis on it at programs like PROMYS and Ross.

Anyhow, what I found was a proof based on combinatorics and the fact that $\phi$ is a multiplicative function.  You have to use a stronger version of Fermat as a lemma.

My proof relied on the fact that $<U_{n},*>$ was a group, where $*$ is multiplication modulo $n$. Then I considered a certain set of elements, used the pigeon-hole principle to show that two of the elements are equal, and then used cancellation laws (which stem from the group properties).

[quote="Boy Soprano II"][hide="Theorem"](Fermat's Little Theorem) Let $p$ be a prime that does not divide $a$.  Then $a^{p}\equiv a \pmod{p}$.

Proof.  Suppose we are putting $p$ beads (which come in $a$ different colors) on a string.  There are clearly $a^{p}$ ways to do this.  If we throw out the $a$ ways to put all beads of the same color on the string, and now tie the  ends of the string together to make a necklace and divide by the $p$ ways we could rotate any given arrangement to give another that would look the same when the ends of the necklace are not tied together, we have $\frac{a^{p}-a}{p}$ possible arrangements.  But this implies that $p$ divides $a^{p}-a$, or \[a^{p}\equiv a \pmod{p}.\] [/hide]
Anyhow, that's Fermat's Little Theorem, which is pretty well known.

In fact, what you have posted is false.  What we do know is that $1^{1}$ gives the same units digit as $1^{21}$ (I'm ditching the extra digits), $2^{2}$ gives the same units digit as $2^{22}$, $3^{3}$ gives the same units digit as $3^{23}$, etc.  The reason is this :  in mod 2, $a^{k+20}\equiv (a^{(2-1)})^{20}(a^{k}) \equiv 1^{20}(a^{k})$, and similarly for mod 5; hence $a^{k}\equiv a^{k+20}\pmod{10}$ (the formal name for the fact that we can base mod 10 on mod 2 and mod 5 is the Chinese Remainder Theorem).  Anyhow, that gets us what we want.

[hide="For those who already know the Fermat Theorem"]I was going to post a proof of the Fermat-Euler theorem, but I thought it would probably be a bit over the top.  Anyhow, the proof (which I didn't realize existed until I tried to find it) was based on this proof of the Fermat Theorem, and I encourage you to find it.[/hide][/quote]

I did the Fermat's Little Theorem Proof by induction (it was quite interesting), but can you do the proof of Euler's generalization with induction as well? The totient function appears to be a bit intimidating  :maybe:

Induction, hahahahaha, that's funny.  That actually wouldn't work well for Euler, though, because first, you'd probably have to use some goofy rewording of it that doesn't require that the number you're raising to the $\phi(m)$ power to be relatively prime to $m$ (aye, there's the rub), and after that you'd probably have some strange divisibility lemmas on your hands.  If you felt a strong need to prove Fermat-Euler by induction, you would probably have to prove the strengthened version of Fermat to which I alluded earlier, and then use the Chinese Remainder Theorem and the convenient fact that $\phi$ is multiplicative.  Except I'm not sure how well proving the strong version of Fermat would go, either.

[quote="Boy Soprano II"][hide="Theorem"](Fermat's Little Theorem) Let $p$ be a prime that does not divide $a$.  Then $a^{p}\equiv a \pmod{p}$.

Proof.  Suppose we are putting $p$ beads (which come in $a$ different colors) on a string.  There are clearly $a^{p}$ ways to do this.  If we throw out the $a$ ways to put all beads of the same color on the string, and now tie the  ends of the string together to make a necklace and divide by the $p$ ways we could rotate any given arrangement to give another that would look the same when the ends of the necklace are not tied together, we have $\frac{a^{p}-a}{p}$ possible arrangements.  But this implies that $p$ divides $a^{p}-a$, or \[a^{p}\equiv a \pmod{p}.\] [/hide]
Anyhow, that's Fermat's Little Theorem, which is pretty well known.

In fact, what you have posted is false.  What we do know is that $1^{1}$ gives the same units digit as $1^{21}$ (I'm ditching the extra digits), $2^{2}$ gives the same units digit as $2^{22}$, $3^{3}$ gives the same units digit as $3^{23}$, etc.  The reason is this :  in mod 2, $a^{k+20}\equiv (a^{(2-1)})^{20}(a^{k}) \equiv 1^{20}(a^{k})$, and similarly for mod 5; hence $a^{k}\equiv a^{k+20}\pmod{10}$ (the formal name for the fact that we can base mod 10 on mod 2 and mod 5 is the Chinese Remainder Theorem).  Anyhow, that gets us what we want.
[/quote]

I'm sorry; I'm still sort of confused.  I see what you mean in that $a^{k}\equiv(10c+a)^{k}$ (that's how I sort of tried to work the problem), but I'm sort of lost after that.  For example, $3^{3}$ is not congruent to $3^{13}$ in mod 10.  I see your proof for intervals of 20, but what about the intervals in between (intervals of 20 offset by 10)?

I got that the digits alternate, for powers of 3: 3 and 7, for powers of 2: 4 and 6, and so on.

The intervals in between don't matter, because since we are dealing with the numbers 1 through 2000, there are 100 of each residue mod 20.  That is to say, it doesn't matter that $3^{3}\not\equiv 3^{13}\pmod{20}$; what matters is that $3^{3}\equiv 3^{23}\equiv \cdots \equiv 3^{1983}\pmod{10}$ and that $3^{13}\equiv 3^{33}\equiv \cdots \equiv 3^{1993}\pmod{10}$, because there are 100 numbers in each of those strings of congruences.

Ah!  I finally get it.  Thanks.  I don't know why I was so slow on this one.  :blush:


Are natural numbers of the form $ 2^{n}$ for $ n \equal{} 0, 1, 2, 3, ...$ the only natural numbers which cannot be expressed as a sum of two or more consecutive natural numbers? How do we prove this?

1+2=3
2+3=5
3+4=7

Every odd number is the sum of 2 consecutive numbers
And for more, another counter-example: 1+2+3=6

Sorry, I meant to type "cannot" but typed "can" by accident. I've edited it.

See my proof for this [b]exactly topic[/b] at [url]http://www.mathlinks.ro/viewtopic.php?search_id=1333233957&t=299164[/url]


$AB, BC, CD, DE$ are consecutive chords on a semicircle of unit radius with lengths $a, b, c, d$.
Prove that $a^2 + b^2 + c^2 + d^2 + abc + bcd < 4$.


Why is it that whenever I try to watch a video it plays for a few secs then it just shows an ad?
I have 1200+ karma

I've wasted like 100 karma trying to watch them...

Your download is too slow.  After clicking the video button, put it on pause and let it stream for awhile before viewing.

ok i figured it out now after switching to IE, there was no progress bar at the bottom of the video in firefox for some reason...


In a Trinomial ax^2 + bx +c, if a & c are prime nos., how many values can we have?

Notice the times$ \minus{}$it must have been cause by clicking submit multiple times or/and a slow internet connection.


The streets of the city of Duzhinsk are simple polygonal lines not intersecting each other in internal points. Each street connects two crossings and is colored in one of three colors: white, red, or blue. At each crossing exactly three streets meet, one of each color. A crossing is called positive if the streets meeting at it are white, blue and red in counterclockwise
direction, and negative otherwise. Prove that the difference between the numbers of positive and negative crossings is a multiple of 4.

(WLOG) Ignore all of the blue streets. Then the white and red streets form a series of alternating edge-colored polygons, as each intersection must have one white and one red street and the streets cannot cross. Also, the polygons are non-intersecting, and they each have an even number of edges.

Consider all polygons $ P$ which are not enclosed by any polygons, and going clockwise around the vertices, alternately label them $ A$ and $ B$, such that (wlog) each $ A$ is between a white and red street in that order.

Repeat this step on any polygons enclosed by the above polygons, except reverse the $ A$s and $ B$s such that each $ A$ is now between a red and white street in that order, and so forth continue labeling all of the vertices, alternating the order with each 'level'.

It is easy to see now that if 2 $ A$s or 2 $ B$s are connected by a blue street, then they have the same sign (consider if the blue path is within a polygon, between two polygons, or between a polygon and the polygon enclosing it), and if an $ A$ and $ B$ are connected they have opposite signs. We ignore the case with opposite signs because they do not change the difference.

There are the same number of $ A$s and $ B$s, and after removing the pairs of points with opposite signs, there still are the same number of $ A$s and $ B$s. Since the streets connect either 2 $ A$s or 2 $ B$s, then there are $ 2n$ of each type, with $ 4n$ total intersections. But the number of positive and negative intersections are even, so we are done.


Jim is selling a cell phone to Tommy. Tommy can bid once on Jim's cell phone and Jim will sell if the bid is above Jim's secret asking price, which is uniformly distributed form 0 to 100.  If Tommy gets the cell phone, he could then resell it for 80% more than his price; what should Tommy bid for the phone to maximize his expected return?

[quote="1234567890"]...80% more than his price...[/quote]

Do you mean Tommy's secret asking price?

Tommy can sell it for 80% above Jim's secret price.

Expected value isn't my thing, but I gave it a shot.  Tell me where I went wrong.

[hide]
Let the secret price be $ s$ and Tommy's bid be $ b$.  Then the profit $ p$ that Tommy makes (if $ s < b$) is
$ p = 1.8s-b$
So, we want to find the expected value of $ p$ in terms of $ b$.
Wouldn't this expected value just be the average of all $ p$?  Which is
$ \frac{ \int_{0}^{b}1.8s-b \; ds }{100}=-.001b^{2}$
Of course, this gives the best solution of $ b=0$, which doesn't seem like it'd be the correct answer.
[/hide]


$ m$ and $ n$ are an amicable pair if and only if $ \sigma{(n)}=\sigma{(m)}=n+m$.

Let $ p=3\cdot 2^{e-1}-1$, $ q=2p+1$, $ r=(p+1)(q+1)-1$. Prove that if all of $ p$, $ q$, and $ r$ are odd primes, then $ m=2^{e}pq$ and $ n=2^{e}r$ are amicable.


Let $ t_k$ be the number of natural numbers $ a$ for which $ \frac{n\plus{}2}{k\plus{}1}\le a\le \frac{n\plus{}1}{k}$ where $ n,k\in\mathbb{N}$.

Define $ A$ as $ A\equal{}\sum_{i\equal{}1}^{n\plus{}1} t_i$

Prove that $ t_k$ is equal to the number of ordered pairs which are solutions to $ kx\plus{}(k\plus{}1)y\equal{}n\minus{}k\plus{}1$, where $ x,y\in\mathbb{N}_0$, and evaluate $ A$.

$ kx \plus{} (k \plus{} 1)y \equal{} n \minus{} k \plus{} 1 \Leftrightarrow kz \plus{} y \equal{} n \plus{} 1$, where $ z \equal{} x \plus{} y \plus{} 1$. we need to count the number of integer solutions of $ kz \plus{} y \equal{} n$ with $ 0\leq y \leq z \minus{} 1$

under these conditions we have $ kz\leq kz \plus{} y \equal{} n \plus{} 1 \Rightarrow z\leq \frac {n \plus{} 1}{k}$. at the same time, we have $ n \plus{} 1 \equal{} kz \plus{} y\leq kz \plus{} z \minus{} 1$, so $ \frac {n \plus{} 2}{k \plus{} 1}\leq z$. this proves the first part.

And what about the second part?

Evaluate: $ \sum_{i\equal{}1}^{n\plus{}1} t_i$


Let f:N\{0}-N\{0} so that for all positive integers m,n we have:
f(f(m)+f(n))=m+n
Find all posible values of f(1988)

We did this in the Belgian IMO training sessions this year.
You can prove that f(n) = n for all n. Hence f(1988) = 1988.
I think this was a problem from the 1988 IMO LONGlist.
I'll post a full solution later.

It's even a shortlist problem. Solution at http://www.kalva.demon.co.uk/short/soln/sh8819.html .


It appears that there are 995 enrollments in the classes, and there are 220 in both, so775 are in some course.

Therefore, 75 students does not take anything.

I think the probability is 3/34.

[quote="colts18"]I got a weird answer... But then  I did the problem over again and got..
[hide]3/34[/hide][/quote]
What was your weird answer???
How did you get 3/34???

[quote="Silverfalcon"]This is from NCIML. Just like the title, if you draw a venn diagram, you can solve this problem easily. Good luck!

At Wassamata University, there are 850 students. There are 560 taking math, 435 taking science, and 220 taking both. If one student is chosen at random, find the probability that he or she will be taking neither math nor science.[/quote]
-
The total number of students that are taking math and science is $( 560+435-220)$
Thus,
$850-775$ students don't take math or science, in simpler terms 75 students

The probability that he or she will be taking neither math nor science is $\frac{75}{850}$ or $\frac{3}{34}$

Therefore,
 :rotfl:

[hide]3/34[/hide]

                                    :starwars: (I just like this one)

Yeah, I think everyone got it. It's a pretty easy question.

[hide]
There are 850 stucdents at the school, and 560 + 435 - 220 students taking at least math or science (there are some that will take both classes). $\frac{560 + 435 - 220}{850} = \frac{775}{850}$ So: $\frac{75}{850}$ is the probability of chosing a student that takes neither math nor science.
Simplify and you get: $\frac{3}{34}$[/hide]


Questo è un messaggio per i frequentatori dell'OliForum italiano. Come avrete forse notato è da un po' di tempo che non funziona per motivi di "web hosting" :(.
*Dovrebbe* tornare operativo nei primi giorni del 2007, il che vuol dire come minimo il 3 gennaio.

ci consoleremo x un po con mathlinks... :)

però l'oli è meglio  :lol:

Wow, grazie per l'informazione.
E anche per il lavoro! :wink:

p.s. sfratto?  :huh:

[quote="edriv"]p.s. sfratto?[/quote]
E' una lunga storia :( . Ne parliamo al WC (cioè al Winter Camp :lol: ).

Gia mi stavo disperando ma nessuno sa qualcosa di piu preciso?
Mi piaceva l oliforum  :lol:
Un abbraccio ragazzi
sastra

[quote="sastra81"]ma nessuno sa qualcosa di piu preciso?
[/quote]
Tornerà, ma la cosa è complicata...


Is there an incredibly quick way to turn huge (25-26 digit) base 2 numbers in base 10 numbers.

A girl at my school was saying someone could do it in seconds, but she wouldn't accept my explantion that figuring out 1 followed by 25 0's in base 2 is 2^25, and figuring that takes a LONG time.

well, just calculating 2^25 doesn't take that long . . . I mean, if you only want an estimate, it's really easy because 2^10 is just a bit more than 1000.  But if you want it exact, it still isn't that hard.  I mean, maybe "seconds" is a bit of an exaggeration, but I don't really think it should take more than a minute.  I have the powers of 2 up until 4096  = 2^12 memorized, and you can just square that and double it.

I mean, if you're talking about someone who claims to do this as a talent, if you just memorized the powers of 2 up until 2^25, you could do it as the person was writing it down.

I think it's definitely doable.  Impressive, but doable.


Γεια σας, είμαι φοιτητής και παρακολουθώ το φόρουμ εδώ και καιρό. Σας συγχαίρω για τις ωραίες ασκήσεις που προτείνετε και για το γεγονός ότι είστε τόσο δραστήριοι. Η άσκηση που σας στέλνω αφορά το κεφάλαιο των σειρών πραγματικών αριθμών, ασχολήθηκα αρκετά μαζί της αλλά δυστυχώς δε μπόρεσα να τη λύσω (μέχρι τώρα τουλάχιστον). Θα ήθελα να την κοιτάξετε
Ευχαριστώ πολυ

Υπόδειξη πάνω στην άσκηση
[hide]Το άθροισμα της σειράς είναι β/(β-α) όπου β>α>0
Έκανα τη σκέψη ότι β/(β-α)=[b]Σ[/b](α/β)^ν για ν=0,1,2,...
Ελπίζω αυτό να βοηθήσει
[/hide]

εισαι σιγουρος οτι υπαρχει? δεν πρεπει να συγκλίνει η σειρά ?

ΑΝ ΕΦΑΡΜΟΣΕΙΣ ΚΡΙΤΗΡΙΟ ΛΟΓΟΥ ΚΑΙ ΣΤΗ ΣΥΝΕΧΕΙΑ ΤΟ ΚΡΙΤΗΡΙΟ RAABE-DUHAMEL KATAΛΗΓΕΙΣ ΣΕ ΕΝΑ ΟΡΙΟ b-a+1...ΣΥΝΕΠΩΣ ΓΙΑ b>a Η ΣΕΙΡΑ ΣΥΓΛΙΝΕΙ ΚΑΙ ΑΠΟΔΕΙΚΝΥΕΤΑΙ ΟΤΙ ΚΑΙ ΓΙΑ b=a ΕΠΙΣΗΣ ΣΥΓΚΛΙΝΕΙ..ΣΤΗΝ ΠΕΡΙΠΩΣΗ ΠΟΥ a>b ΧΡΗΣΙΜΟΠΟΙΩΝΤΑΣ ΤΗΝ 1/n ΚΑΤΑΛΗΓΕΙΣ ΒΑΣΕΙ ΤΟΥ ΚΡΙΤΗΡΙΟΥ ΣΥΓΚΡΙΣΗΣ ΟΤΙ ΔΕΝ ΣΥΓΚΛΙΝΕΙ..

ΑΡΑ Η ΣΕΙΡΑ ΣΥΓΚΛΙΝΕΙ...ΤΩΡΑ ΤΟ ΠΟΥ ΣΥΓΚΛΙΝΕΙ ΕΙΝΑΙ ΑΛΛΟ ΘΕΜΑ!!

δεν ειδα την συνθηκη β>α ,
μονο τοτε συγκλινει (μαλιστα το κριτηριο λογου δεν ωφελει)
η συνεχεια λοιπον εδω http://www.mathlinks.ro/Forum/viewtopic.php?t=144619


Determine the pairs (m,n) of natural numbers for which the equation:

(1+(sin(nx))^2)/(1+(sin(mx))^2) = sin(nx) has real solutions.

[i]Kmal, 62, 1981[/i]

We have sinnx>0 and 1+sin^2mx=sinnx+1/sinnx>=2. Thus, we must have sin^2mx=1 and sinnx=1.

Please write the solution explicitely as pair (m,n), Harazi.  :)

Well, I didn't find all pairs, since there are a lot. It was just a suggestion. Sorry!

[quote="harazi"] we must have sin^2mx=1 and sinnx=1.[/quote]

Harazi,

If you solve sinnx=1 and you plug in  sin^2mx=1, will you not find the
require conditions on n, m for the problem ?

Yes, of course. Or you can see separately when the systems sinnx=1 and sinmx=1 or sinnx=1 and sinmx=-1 have a common solution. But I really don't like this job.

I think we may consider it as solved. Yes?


inequalities Euler geometry incenter algebra polynomial induction

Alright, here goes.

1.  2004!, which was fairly straightforward.  I've seen this question before, I'm sure.  Count the choices for which horizontal edges that you pass through.
2a.  Whoever said AM-GM is used to solve every CMO inequality was right.  You guys are psychic.
2b.  Not so bad.  If a and b are relatively prime then it all works out.  And you can assume that they're relatively prime by cancelling common factors.
3a.  Rather annoying to prove, but not bad.  And my final proof was somewhat elegant.
3b.  Oops.  This one was quite simple (perhaps even simpler than 1?) but I didn't go far enough with it.
4.  Yufei will have gotten this one, I'm sure.  27/4 (or was it 27/8, I forget what I put) is obtained from an equilateral triangle.  You could consider P/R and K/R^2 separately, couldn't you?  But I didn't prove my answer much.
5.  They explicitly said "positive integer" n, by the way - not natural number.  Put down some interesting stuff here, but no major headway.

No IMO chances for me.  Alas.

<geoffrey was a bit faster than me  :blush: >

1, 2 and 3a were relatively easy. 3b was tricky and i didn't get a full solution for it.

4 was definitely a surprise, although i did make some progress with it, playing around with the Euler line and the distance between the incentre and the circumcentre.

5, i read wrong initially. spent a ton of time trying to figure out which values of (a,b,c) work for which values of n (instead of looking for those that worked for all). I realized this a bit too late but still managed to put somewhat of a proof for part a and b. Part b, i think there are no solutions. Am i right?

I thought that #1 is harder than expected. I didn't get it on my first try (and I wasted at least 5 minutes thinking that the answers were binomial coefficients).

#2 was quite easy. I think the fact that the LHS was squared was give-away (the question would be essentially the same, though slightly more difficult if it weren't).

#3 was, well, at least I didn't find a nice solution to it (I don't even know if mine is correct). I doubt the jury will have an easy time marking this one.

#4 never thought that I'd see a day when my training in symmetrical polynomial inequalities could be so efficiently used. I used the [i]Ravi[/i] substitution $a=y+z$, $b=x+z$, $c=x+y$, and everything fell apart (well, almost) from Muirhead and AM-GM. In my opnion, this problem is an upgrade from 2002#3 (the latter was total jokes compared to the current one)

#5 I first misread the problem too, thinking that we need to find all solution sets $(a, b, c;n)$. But then I thought that would be way too difficult, so I reread the problem (whew). The problem (both parts) relied on [url=http://mathworld.wolfram.com/Newton-GirardFormulas.html]Newton's formula[/url]. Part b) was definitely easier than part a). For part a), I'm not sure about my solution, but it sufficies to just find triples that are 2,3-powerful. I got $(1, 1, 1)$, $(1, 1, 4)$ as answers (being the hothead I was that time, I wrote $(2, 2, 2)$ as an answer as well  :wacko:. That's gonna cost me a mark or two)

In my opinion, this test was harder than last year's (yeah, great unbiased opinion  :D), especially since I got off to a rough start (getting stuck on problem 1). I don't know, that kinda just broken my rhythm. I was also bit short on time, so my solutions were badly written (the one I especially fear for is #3).

My prediction is that because the questions are so divided (3 of 5 problems had parts a and b to it), partial marks play huge roles here. So don't be surprised if one gets more problems, but less points than another.

What do you guys think?

[edit:]
Did anyone notice an error on the blue sheet glued outside of the envelope? It says that IMO will be held in Cancun Mexico, but it has been long since changed to Merida, Mexico. Just hope that they don't order the wrong air tickets.

Also, I've managed to once again write one of my solutions in the wrong place. Oh well.

yeah, i got 3 and a bit, but probably can't shoot up to 21 points...

bah, no more imo for me...lol

btw, i believe #1 is from Michael, seems he made a lot of us pay... :)

Where can we find/post the solutions?

Would someone care to share their solution for #3?

Thanks in advance.

anyone want to post the questions on the forum?

I too would like to see the problems.  Gee, why couldn't there have been two inequalities last year  :?

for me, breezed through 1/2, then 3a was really challenging for me, i tried dividing up the circle, all sorts of sum arguments, etc... didnt work.
#4, i forgot to check if 81/4 worked out, turned out it didnt, i multiplied by 3 somewhere, OOPS.  #5a, messed up; #5b i wrote newton formula, but no time.

Problems and Solutions (kalva style!)

Edit: at the request of Yufei Zhao, I have clarified the problem statements.  I think they are _okay_ now.

$\hline$
$\hline$
1.  An equilateral triangle is composed of $n^2$ smaller equilateral triangles with $n$ on the base, and $2i-1$ triangles on the $i$th row.  Let $f(i)$ be the number of paths from the $1^{st}$ row triangle to the center one in the bottom row.  Find $f(2005)$.  [A path moves from one smaller triangle to the next, never goes from a lower (greater index'ed) row to a higher (lower index'ed) row, and moves to adjacent (connected by edges) triangles only.]

[hide="solution"]$f(i) = (i-1)f(i-1)$ and $f(2) = 1$, trivial induction shows $f(i) = (i-1)!$.[/hide]

$\hline$
2.  (a,b,c) are positive integers that satisfy (a^2 + b^2 = c^2), show a):  L = (c/a + c/b)^2 > 8 , and b): L is never an integer.

[hide="solution"]a)$ (c/a + c/b)^2 = (a^2 + b^2)(1/a^2 + 1/b^2 ) + 2(a/b + b/a) \geq 8$ with equality in (cauchy, amgm) when $a=b$; but $a=b$ implies $\sqrt{2}a = c$, a contradiction.
b) $c(a+b)/ab$ is rational, so we must have $k^2 = n$ a square, for some integer k.  Now, barring common factors which get cancelled when considering $c(a+b)/ab$, exactly one of $a, b$ is even.  Numerator = odd, Denominator = even, not an integer, done.[/hide]

$\hline$

3.  |S| > 2 consists of n points interior to a circle.  a): Show the existence of distinct $a,b,c \in S$ and A,B,C on the circle for which $Ap$ is smallest when $p = a$, and likewise for $B,b; C,c$, for all S, and that b): $\forall n$, we cannot guarantee that $a,b,c,d$ and $A,B,C,D$ exist (similarily defined.)

[hide="solution"]a) pick a,b,c s.t. abc = maximal area, and A where the altitude from a meets the circle such that Aa is shortest, similarily for B, C.  Now if there exists K s.t. KA < Aa, then the height from K to bc is larger than the height from a to bc, contradicting "maximal area."  (We define an altitude of a to be the perpendicular from the extension of bc meeting vertex a.)
b): inscribe an equilateral triangle in the circle, and consider the set of its midpoints and other points arbitrarily close to the center of the circle.  Then atleast one "center point" $x$ must be picked.  But no corresponding choice $X$ keeps $xX < xP$ for all $P \neq X$.[/hide]

$\hline$

4.  Let P, K, R be the perimeter, area, and circumradius of $\triangle ABC$.  Maximize $\frac{PK}{R^3}$.

[hide="solution"]We show (*) $P \leq 3\sqrt{3}R$ and (**) $K \leq \frac{3\sqrt{3}}{4}R^2$.  Put angles $AOB, BOC, COA$ as $2Z,2X,2Y$.  It is easy to show $P = 2(cos(X) + cos(Y) + cos(Z))$
and $K = \frac{R^2}{2}(sin(2X) + sin(2Y) + sin(2Z))$, and now to show (*) and (**), so combining,  $\frac{PK}{R^3} \leq 27/4$, now a=b=c works, so it must be the maximal value. [/hide]

$\hline$

5.  Let $(a,b,c)$ be positive integers such that $a \leq b \leq c$, $gcd(a,b,c) = 1$.  Let $(a,b,c) \in T_n \Leftrightarrow (a+b+c)|(a^n + b^n + c^n)$ for all positive integers $n$.  a) find the intersection of $T_n$; b) find the intersection of $T_{2004}$ and $T_{2005}$ that is not contained in $T_{2007}$.

[hide="solution"] Let $S_n = a^n + b^n + c^n$, $p,q,r = a+b+c, ab+bc+ca, abc$.  a) Now $s1 | s2 , s1 | s3 \Rightarrow s1 | 2q, s1 | 3r$.  Consider $0 = 6(x-a)(x-b)(x-c)$ (mod p).  Using $s1 | 2q, s1 | 3r$, we have
$0 = 6x^3$ (mod p).  So in paticular, $p | 6a^3, 6b^3, 6c^3$.   Observe $gcd(a,b,c) = gcd(a^3, b^3, c^3) = 1$.  Now $p | gcd(6a^3, 6b^3, 6c^3) = 6$, so $p = 1,2,3,6$.  We obviously can't have $p=1,2$.  $p=3$ gives us $(1,1,1)$; $p=6$ gives us $(1,1,4), (1,2,3), (2,2,2)$ of which only $(1,1,4)$ satisfies.  We can check that these two solutions satisfy for all n.

b) Observe the Newton Girard recursion for power sums: $S_{n+3} = S_{n+2}p - S_{n+1}q + S_nr$.  Put $n = 2004$.  The right side is clearly divisible by $S_1$, as given.  But it is also given that the left side is not.  Contradiction.  Thus there are no triples that satisfy.
[/hide]

$\hline$
$\hline$

edit:

[quote="billzhao"]Make sure that you write your solutions in the appropriate places (you'll know what I mean if you take the CMO, assuming they don't change the format).[/quote]
[quote="billzhao"]Also, I've managed to once again write one of my solutions in the wrong place. Oh well.
[/quote]
lol

[quote="Singular"]
4.  Let P, K, R be the perimeter, area, and circumradius of $\triangle ABC$.  Maximize $\frac{PK}{R^3}$.[/quote]

It's clear that $PK/R^3=\frac 14\cdot \left(\sum a\right)^3\cdot \left(\prod a+b-c\right)^2/(abc)^3$

Letting $u=a+b-c$ and so on, we need to prove

$(u+v+w)^\cdot (uvw)^2\leq 27/8^3\cdot [(u+v)(v+w)(w+u)]^3$

Taking out cubes, we get

$[(u+v+w)(uvw)^{\frac 23}]^3 \leq [3/8\cdot \sum u^2v]^3$

and this is just

$[\sum u^{\frac 53}v^{\frac 23}w^{\frac 23}]^3\leq [3/8\cdot \sum u^2v + 2uvw ]^3$

Now, $2uvw+3u^2v+3u^2w\geq 8u^{\frac 74}v^{\frac 58}w^{\frac 58}$, so after applying weighed AM-GM to the terms inside the bracket of the RHS of the last inequality, the result follows from muirhead.

Actually, to finish where Singular left off for #4, it's easy to show that:

sin(A)+sin(B)+sin(C) <= (3:sqrt:3)/2 and
sin(2A)+sin(2B)+sin(2C) <= (3:sqrt:3)/2.

Just apply Jensen's to sine, and you are done.  :)

Is sine concave over the interval 0, 180?

edit: thanks yufei, that's what i was going for.  ya.. 180, 360.

[quote="Singular"]Is sine concave over the interval 0, 180?[/quote]

Yeah, but not over [180, 360]. So the second claim $\sin(2A)+\sin(2B)+\sin(2C) \leq (3\sqrt{3})/2$. doesn't follow from Jensen when the triangle is obtuse.

oops, but once you get  sinA + sinB + sinC <= (3:sqrt:3)/2, then sinA * sinB * sinC <= (3:sqrt:3)/8 from AM-GM, then sin2A + sin2B + sin2C = 4sinA * sinB * sinC <= (3:sqrt:3)/2.

For me, this year's CMO actually felt easier than last years. (Maybe i screwed up too badly last year.)

Here are my solutions: (3a is a little fishy, i think)
1. I bijected the paths to ordered sequence $(a_1,a_2,\cdots a_{n-1})$ where $1<=a_i<=i$. Answer follows immediately
2. a) am-gm
b) Cancel out the gcd from a,b,c, show that the numbers are pairwise relative prime. Show that $c^2 | n$. Then, cancelling c^2, we get that an integer equals to $({a+b \over ab})^2$. For a or b =1 or $(a,b)=(2,2)$, show that it can't be a pythagorean tripple, otherwise $(a-1)(b-1)>1$ and we get that the integer is between 0 and 1, contradiction.

3. a) If that $S$ is in a line, then draw the perpendicular bisector of the segments containing each two adjacent points, these lines divide the circle into regions for each point. THen show, using pythagorean theorem, that each region on the circle would yield a valid point for each point in the region.
If it's not in a line, then the convex hull must contain at least 3 points. Choose 3 random vertices on convex hull, call them, $a$, $b$, $c$. For $a$, let the two sides of teh convex hull containing $a$ be $m$ and $n$. Draw rays from $a$ perpendicular to $m$, and $n$, away from the hull, then the region of the circle bounded by these two lines would yield a valid $A$. This is because when you draw the circle with centre $A$ and radius $aA$, for it to cut the convex hull, it must cut $m$ or $n$, but $m$ and $n$ angle away from the circle.
b) choose a small equalateral triangle (I wrote circumradius =1/4r) in the centre and other points infitestimally near the centre.

4. $Use R=2sinAsinBsinCR^2$, $K=2R(sinA+sinB+sinC)$, and $sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)$. Then we have $\prod{sinAcosA/2} \leq 27/16$. Which follows from jensen, since it's easy to show that lnsinXcosX/2 is concave, in [0,\pi]. Answer is 27/4.

5. This uses the same idea as last year's COMC last question. Basically, find a recursion and everything falls apart. My solution is same as the others.

[quote="pengshi"]since it's easy to show that lnsinXcosX/2 is concave, in [0,\pi]. Answer is 27/4[/quote]

I'm almost 100% sure that you're correct here, but how do you consider it 'easy'?.  My computations were fairly involved.  Although it's probably easier if you split it into ln sin x and ln cos x/2

[quote="blahblahblah"]
I'm almost 100% sure that you're correct here, but how do you consider it 'easy'?.  My computations were fairly involved.  Although it's probably easier if you split it into ln sin x and ln cos x/2[/quote]

Easy as in straightforward,no tricks involved.

Hi folks,

I knew the CMO was going to be written sometime soon (couldn't remember if it was yesterday or next Wednesday), so I did a little Google search, and quickly was referred to this site.  Oh wow, I see that many of you have posted solutions already, and congratulations to all of you on your performances.  I know many of you from past National Math Camps and such, and so I really enjoyed reading your posts today.  Also, I made up quite a few of the problems on this year's CMO (#1,2,4,5), and so it was nice to see so many new solutions to these problems - solutions that none of us committee members came up with ourselves.  Congratulations to all of you, and best wishes on future contests this year.

Regards,
Richard Hoshino.

Hey guys!
I haven't posted here before, but I was interested in seeing how people did and I though I might share some of my solutions.

I think I got essentially the same solutions as everyone else for questions 1 and 2... there should be a definite 14 marks there.

For 3a, I defined a "feasable region" for point Pn as being the set of points closest to Pn.  Since every point in the entire plane (excluding points ON perpendicular bisectors) must be on some point's region, then there must be at least 3 regions of infinite size.  These regions are continuous and contain points inside the circle, so they must contain points on the circle.  I had to put in a little argument for parallel lines cutting the plane up, but it wasn't too bad.

3b I just used analytic geometry with polar coordinates.  I said let the circle be r=4, and put the first 3 points at (1,0), (1,120degrees), (1,240 degrees).  Then put the 4th point at (0,0) and the 5th-nth points very very close to (0,0).  None of the regions will then touch the circle.  I had to prove a few intermittant steps, but the geometry was easy.  So #3 should hopefully bring me 7 marks as well.

For #4, I changed PK/R^3 to (1/2)*(P^2*r/R*3), where r is the inradius... since area = inradius*semiperimeter.  Now, since the expression is constant upon scaling the triangle (ie its degree is zero; it's homogenous), just let R=1 and suppose you have 3 points on a unit circle and you want to maximize the perimeter and the inradius of the triangle they formed.  Both are maximized when the triangle is equilateral (I proved both of them) so we are done.

I used Jensen's to do the proof for the perimeter.  For the inradius, some simple circle geometry worked.  Fix 2 of the points and move the 3rd one.  Take the angle bisectors of the 2 fixed points and find the incentre... you'll discover that it lies on another circle passing through the 2 fixed points, and clearly the maximum incenter occurs when it lies on the perpendicular bisector of the 2 fixed points.  Therefore with 2 fixed points, inradius is maximized when the 3rd point is positioned such that the triangle is isosceles, and thus with 3 movable points, inradius is maximied for an equilateral triangle.  27/4 is then the maximum value.

Hopefully 7 points unless I screwed up the Jensen.

#5 I didn't get very far on... I had no time; in fact I was rushing to complete the proofs for #4.  Probably 1 mark maximum there for the trivial discoveries I made.

So about 28 for me unless I did something stupid.

~ Elyot

Hey shaun here, I thought it was a really fun cmo this year... I didn't like how some problems were broken up into parts (a) and (b) because it required more reading and writing lol... but I understand since (a) and (b) seemed to be independent of each other... especially 5(b) which seemed easier than 5(a)...

I had the weirdest solution to problem 4... using euler's R >= 2r, I showed that the minimizing the given expression was equivalent to minimizing s/r... then I said that from Heron's, s(s-a)(s-b)(s-c) = s^2*r^2 which I rewrote as (s-a)/r * (s-b)/r * (s-c)/r = s/r.  So I had to minimize the left side... but since the sum of the 3 factors on the left side turns out to be s/r, I created the relationship x + y + z = xyz, so each is the tangent of an angle of a triangle... from which I think the desired result ensues... though I'm sure I messed up Jensen's lol...

No one in my school really solved problem 3... I think it confused people... and 1 and 2 were pretty straight forward... hopefully I got an honourable mention... but I didn't read gcd(a,b,c) = 1 in question 5(a) so I got (k,k,k) as an answer where k = 1,2,3... but the committee's nice right?

lol, see ya guys later... good luck on usamo...

nice committee? where did you get that idea from?  they marked my CMO papers last yr like the Putnam ones!!  I think I lost 6 marks on question 1 for losing one of the solutions! :(

[quote="Singular"]my solution on the contest was this:  K = abc/4R, and abc <= P^3 / 27, so PK/R^3 <= (P/R)^4 / 27*4 <= (3sqrt(3))^4 / 27*4 = 27/4.

Unfortunately somehow I multiplied an extra 3 in there.  But atleast the character of the soln was right[/quote]

But you didn't properly show where it was convex!  Jensens + failure to accurately state where f is convex should be heavily marked down.

[quote="Elyot"]When do they post/mail/tell us the results?

Also, when do they decide on the IMO team?[/quote]

Last year it was in the beginning of May.

You can check the [url=http://www.cms.math.ca/]CMS website[/url] once so often to find out whether they released the results.


Ok, I have a question, and I want to know if there is in fact a way to solve this rather than drawing it out, guess-and-check, plug-and-chug, and other guessing methods.

In chess, a knight moves in an L-shape manner—two spaces in one direction and one space in a direction perpendicular to the first direction—as shown. Beginning in and including the upper-left corner, what is the most number of squares in a 4x4 checkerboard that a knight can visit without visiting any square more than once?

[hide="Answer"]15[/hide]
Help!! Thanks! :D

For a completely rigorous proof, you need to show existence.  It's not difficult to construct one.

[hide="However..."]
if you don't need one, then note that it is impossible to achieve all 16 (this is a well-known result).  Guess 15. :D
[/hide]


Evaluate

$ \sin\left(\frac{17\pi}{4}\right) \cdot \cos\left(\frac{\minus{}15\pi}{4}\right) \cdot \tan\left(\frac{13\pi}{4}\right)$

No calculator!

[quote="Silverfalcon"]Evaluate

$ \sin\left(\frac {17\pi}{4}\right) \cdot \cos\left(\frac { \minus{} 15\pi}{4}\right) \cdot \tan\left(\frac {13\pi}{4}\right)$

No calculator![/quote]

[hide]
because $ \theta\equal{}\theta\plus{}2n\pi$

first term is equal to $ \sin{\frac{\pi}{4}}\equal{}\frac{\sqrt{2}}{2}$
2nd term is equal to $ \cos{\frac{\pi}{4}}\equal{}\frac{\sqrt{2}}{2}$
3rd term is equal to $ \tan{\frac{5\pi}{4}}\equal{}1$
so its $ \frac{1}{2}$

[/hide]


Four rectangular paper strips of length 10 and width 1 are overlapping perpendicularly in a square pattern such that each strip meets two other strips. How much area is covered?

Okay, so we have four $ 10 \times 1$ strips, for a total area of $ 40$.  In order to form a square region, we must have overlaps of $ 1 \times 1$ squares at the corners (can you see why?).  So we have $ 40 \minus{} 4 (1 \times 1)$, or $ \boxed{36}$.


trigonometry
real analysis
real analysis unsolved

Find a basis of solutions of the following equation:

$ x^{(4)}\minus{}x\equal{}0$

characteristic equation : $ \lambda^4\minus{}1\equal{}0 \ \iff \ \lambda \equal{} \pm 1, \pm i$
Thus, $ x\equal{}C_1e^x\plus{}C_2e^{\minus{}x}\plus{}C_3\cos x\plus{}C_4\sin x$


$ x_n: \equal{} \frac {( \minus{} 1)^n*n}{n \plus{} 1}$

I tried assuming they did converge to a number $ a$ and try to contradict it by showing a case where n is even and n is odd for some n[u]>[/u]k, but that wasn't working, any suggestions?

Try looking at convergence of x(2n) and x(2n-1)

Is my method correct, but just try it with $ x_{2n}$ and $ x_{2n \minus{} 1}$?


Given that $\frac{((3!)!)!}{3!} = k * n!$, where k and n are positive integers and n is as large as possible, find $k + n$.

This is well within range of getting started.

Keep in mind that $n! = n(n-1)(n-2) \ldots (3)(2)(1)$.

So, $3! = 3 * 2 * 1$.

That is an easy problem for AIME.  What problem number was it?

[hide]3! = 6, whose factorial euqals 720, whose factorial equals 720!.  Since 720 is divisible by 3!, you can just say $\frac{720!}{6} = 120\cdot 719!$.  When you add 719 and 120 up, you get 839, which is the answer.[/hide]


Let $ ABC$ be a balance triangle at $ A$ .Let $ O$ be a point inside $ ABC$ such that $ \large\hat{OBA} \equal{}\large \hat{OCB}$.Let $ M$ be the mid point of $ BC$.Prove that $ \large\hat{AOC}\plus{}\large\hat{BOM}\equal{}180^{o}$.

[quote="hoangclub"]Let $ ABC$ be a balance triangle at $ A$ .Let $ O$ be a point inside $ ABC$ such that $ \large\hat{OBA} \equal{} \large \hat{OCB}$.Let $ M$ be the mid point of $ BC$.Prove that $ \large\hat{AOC} \plus{} \large\hat{BOM} \equal{} 180^{o}$.[/quote]
We can prove that this holds if and only if $ \triangle ABC$ is isoceles with vertex $ A$.  Is this what you mean by "balanced at $ A$?"
[hide="Solution"]
Extend $ AO$ to meet $ BC$ at $ P$.  Consider the circumcircle of $ \triangle BOC$.  Notice that it is tangent to $ AB$ because $ \angle ABO\equal{}\angle OCB$.  Also, notice that $ \angle AOC \plus{} \angle POC \equal{} 180$ and we wish to show that $ \angle AOC \plus{} \angle BOM \equal{} 180$.  Thus, it suffices to show that $ \angle POC \equal{} \angle BOM$, but since $ OM$ is a median of $ \triangle OBC$, it suffices to show that $ OP$ is a symmedian of $ \triangle ABC$.  Yet, the tangent at $ B$ meets $ PO$ at $ A$, so we have that $ \angle AOC \plus{} \angle BOM \equal{} 180$ if and only if $ AC$ is tangent to the circumcircle of $ \triangle ABC$, so if and only if $ AB \equal{} AC$.  [/hide]


for any triplet of real numbers $ \left(x,y,z\right)$,
we define $ \left(x,y,z\right)^{*}$ to be a permutation of $ \left(x,y,z\right)$ that is arranged from more to less.
(i.e., non-increasing sequence)

Show that, for any $ a,b,c > 0$,
\[ \left(\frac {4}{3}a,\frac {4}{3}b,\frac {4}{3}c,0\right)^{*} \succ \left(a,b,c,\frac {a \plus{} b \plus{} c}{3}\right)^{*}
\]
:?:

P.S.(1) : I don't sure this is true or not?  :huh:
P.S.(2) : My English may be bad, apologize if I've done any mistake.  :blush:

Yes it's true.
http://en.wikipedia.org/wiki/Majorization


is there a Bijection from IR² to IR ?? ( i am sure there is :roll:  , but i need examples  :maybe: !!!! )

Yes, there is such a bijection, just not a continuous one.

The first thing I'll do is produce a surjective mapping from $\mathbb{R}$ onto $\mathbb{R}^{2}.$ Consider
$x\in\mathbb{R}$ to be given by the decimal expansion $x=\sum_{k=-m}^{\infty}d_{k}10^{-k}$ with the understanding that all numbers that can be represented by a terminating decimal be represented by a terminating decimal. Then let

$f(x)=\left(\sum_{j=-u}^{\infty}d_{2j}10^{-j}, \sum_{j=-v}^{\infty}d_{2j+1}10^{-j}\right)$

($u$ and $v$ are whatever numbers near $m/2$ needed to capture all of the original information.)

This mapping is surjective, but it's not quite injective. We can tinker with it to see if we could build a one-to-one correspondence based on the same idea, or we could just invoke the Cantor-Schröder-Bernstein Theorem to claim that such a one-to-one correspondence must exist.


K is positive integer non zero.
a(K) is real sequence such that for any K:
1) a(K)-2a(K+1)+a(K+2)  >= 0
2) a(1)+a(2)+.....  +a(K)  <= 1
prove that for any K >= 1:   0 <= a(K)-a(K+1) <= 2/(K 2 )

Maybe some conditions are missing? I think we have $a_k\ge 0,\ \forall k\ge 1$, for example, otherwise just take $a_k=1-k$. If this is indeed the case, I believe we can even show $0\le a_k-a_{k+1}\le\frac 2{k(k+1)}$.

YES YOU ARE RIGHT :blush:
a(K) >= 0
Sorry i transleted it from arabic and i am french educated


[quote="gauss202"]What is the grading rubric for the papers of the MMPC Part II? How many points is each question worth?
And what is the range of the top 10 scores usually like?[/quote]

Um

It is 10 points each.  So it is out of 50

However, MMPC has two parts (Part I and Part II).  part 1 out of 40, Part 2 out of 50.

However, your part two score is multiplied by 1.2 to get the scaled score.

So, if you get 37 on part 1 and 37 on part 2, your actual score is 37*1+37*1.2=81.4

It is really random the scoring.

Like for example, this year, first and second place were 99 and 98.8.  The previous year, the scores for first and second place were 68.6 and 67.2.

It it just random.


abstract algebra
number theory
least common multiple
group theory
superior algebra
superior algebra unsolved

The question is: Let G be an Abelian group with elements a and b with orders m and n respectively. Prove |ab| <= mn.  The second part is to find a better bound than mn.  I did the first part of the proof.  I determined now that the better bound would be the LCM of m and n.  But how would I go about stating a proof of that? Thanks. Also this is my first post but i have been lurking around a while.  There are tons of smart people around here. [/i]

If $a^{k}=b^{k}=e$, $(ab)^{k}=a^{k}b^{k}=e$. We have $a^{k}=e$ when $m|k$ and $b^{k}=e$ when $n|k$. When do we have both?


My Math Counts team is short one person and I was wondering if anyone who is in middle school and not registered for a Math Counts team would be willing to join it...just post here or PM me if so for more information. Our chapter competition is the Friday of the second weekend in February in Athens.

hi, I would like some more information
oh yeah... 7th grade in Geometry and Algebra 1

I PMed you and I am waiting for my PM to get out of the outbox.
Hopefully, everything is rather agreeable... :lol:

Oh and we have to check the person who isn't coming is def. still not coming but you are welcome to join our 3 person team anyway...(last year when i couldn't stay for the competition i practically let the whole team down.)

[quote="Arvind_sn"]I'm already signed on for a team (River Trail Middle School), but anyway, what's your school? And what grade are you in? (I thought you were in 9th grade)

But for future reference,

Arvind Narayan

7th grade

Algebra II

Hope I see u there!!!  :)[/quote]

You can't participate with two teams, you will get them both disqualified.

yes i know i just want to know which school he's (she's?) from.


abstract algebra
superior algebra
superior algebra solved

Let R= (Z/3Z) [x] / (x^2). Describe all isomorphic clases of R-modules M with |M|=27

It is proved   [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sum+direct&t=88843]somewhere[/url] in this site that if $k$ is a field, then every $k[x]/(x^{2})$ is a direct sum of modules isomorphic either to $k$ or to $k[x]/(x^{2})$.

Let $k=\mathbb Z/3\mathbb Z$ and $R=k[x]/(x^{2})$.
Then the $R$ modules with $27$ elements are
$k^{3}$ and $k\oplus R$.


let $P_{n}(x)$ is the sequence of polynomials such that
 i)   $P_{0}(x)=0$;
 ii)  $P_{1}(x)=x$;
 iii) $P_{n+2}(x)=xP_{n+1}(x)+(1-x)P_{n}(x).$

 Find all real roots of $P_{n}(x)$

[quote="ashwinrk_jain"]i get

                           Pn(x) = (2x)/(x^2-x-2)[/quote]

Good jokes!  ;)

oh i am sorry i have missed n in my problem


What is the greatest integer value for $ m$ such that $ 5m\minus{}3 > 8m\plus{}21$?

$ 5m \minus{} 3 > 8m \plus{} 21$

$ \minus{}3m > 24$

$ m < \minus{}8$

Since $ m$ must be less than $ \minus{}8$, the greatest possible value is $ \boxed{\minus{}9}$.

[hide=The Answer:]
\begin{align*}
& 5m - 3 > 8m + 21. \\
& 5m - 8m > 21 + 3. \\
& -3m > 24. \\
& m < -8. \\
\end{align*}

Therefore, the greatest integer value for $m$ such that $5m - 3 > 8m + 21$ is $\boxed{-8 - 1 = 9}$.
[/hide]

By the way, I found this by inputting random thread URLs. :)


Fing integer $ k>0$ so that the equation $ x^{2}+y^{2}+z^{2}= kxyz$ has integer roots, except $ (0,0,0)$.

Hey, is this in our Number Theory class? :lol:
I've already done this by considering the equation with $ x$ is a variable
$ x^{2}-kyz.x+y^{2}+z^{2}=0$

I prove that if $ (x,y,z)$ is one root of the equation and $ x\leq y\leq z$ then $ (kyz-x,y,z)$ is also a root of it.

With $ k$ even, we can easily prove there's no root (except $ (0,0,0)$) by applying descent.

With $ k=1$ and $ k=3$, it has infinite number of roots (Since we can find $ (3,3,3)$ or $ (1,1,1)$ is a root of it)

It remains with $ k\geq 5$ and $ k$ odd  :oops:

Have you tried my hint yet, bibobeo?

Let $ x_{0}, y_{0}, z_{0}$ be roots of the equation for which $ x_{0}+y_{0}+z_{0}$ is minimum
WLOG, we can assume that $ x_{0}\geq y_{0}\geq z_{0}$

Consider the equation with $ x$ as an variable
$ f(x)=x^{2}-ky_{0}z_{0}.x+y_{0}^{2}+z_{0}^{2}$

We can easily prove that $ f(y_{0})\geq0$

This will lead to $ kz_{0}\leq3$

Consider 3 cases: $ kz_{0}=1;kz_{0}=2; kz_{0}=3$

We can leave out the cases $ kz_{0}=1;kz_{0}=2$

So the only values of k are $ k=1$ and $ k=3$

[/code]

So simple algebra takes it to prove k is less than equal to 3, (Provided I'm not wrong).... :D


Gauss
geometry
3D geometry
prism
function
sphere

1) Is it important to do this one by Gauss' law? This is the easier way: the system you've described is a capacitor and its capacity is given by $C=\varepsilon_0{S\over d}$, where $S$ is the area of the plates, $\varepsilon_0 = 8.85 \cdot 10^{-12} {\rm F/m}$ and $d$ the distance between the plates. The voltage between the plates of a capacitor is $U = {q \over C}$ where $q$ is the charge on the plates. Furthermore, the electring field induced between the plates of a capacitor is given by $E = {U \over d} \Rightarrow E = {q \over \varepsilon_0 S} \Rightarrow q = \varepsilon_0 E S$. The charge on the plates is, thus, $q \approx 0.4 \cdot 10^{-9} {\rm C}$. Of course, the charge on the second plate is $-q$.

Now, let's solve this by applying Gauss' law: consider a four-sided orthogonal prism of very small width that has two sides very near and parallel to a plate, with the plate being between these sides. We'll take these sides to have the same surface as the plate - $S$. The charge encompassed by this prism is $q$, while the total flux through the surface of the prism will be $ES$. Now, Gauss' law states that ${q\over \varepsilon_0} = ES$, and now it's obvious. I hope the answer is clear.

2) We'll find the function $E(r)$ relating the electric field with the distance from the center. Consider a sphere of radius $r$. The total electric flux through the sphere is $\Phi = 4E(r)r^2\pi$, while the charge encompassed by the sphere is $q(r\leq R) = \frac{4}{3}\rho r^3\pi$ and $q(r > R)=q(R)$. By Gauss' law,  $\Phi = {q\over \varepsilon_0}$ and the rest is straightforward. You just need to plug in those values of $r$ you've stated.

[color=blue]You can find another solution [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=73047]here[/url][/color]


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

Show that the value of $\frac{d^{2}y}{dx^{2}}$ at the point $(x_{n},y_{n})$ on the curve $y=f(x)$ is given ,with usual notation , by the approximate relation

$y''_{n}\approx\frac{y_{n+1}-2y_{n}+y_{n-1}}{h^{2}}$

where $x_{r}=x_{0}+rh$

I copy out the problem from my book directly and my problem is I dont what is " usual notation " the question is talking about . Also how exactly we define $y'_{n}$ in recurrence relation as like what the question want us to prove . Thanks .

The "usual notation" is that $y_{r}=y(x_{r})$
Also, $y''_{n}=y''(x_{n})$.

When you get to actually proving it, Taylor's theorem is usually the best method for this sort of problem.

my attempt

from taylor expansion , we have

$y_{n+1}=y_{n}+hy'_{n}+\frac{h^{2}y''_{n}}{2}+\cdots$

so

$y''_{n}\approx\frac{2(y_{n+1}-y_{n}-hy'_{n})}{h^{2}}$

also $y'_{n}=\frac{y_{n+1}-y_{n-1}}{2h}$ , hence

$y''_{n}\approx\frac{y_{n+1}-2y_{n}+y_{n-1}}{h^{2}}$

is this ok ?

by the way , why is it $y'_{n}=\frac{y_{n+1}-y_{n-1}}{2h}$? (I just got this from book without proof..... )

[quote="shyong"]by the way , why is it $y'_{n}=\frac{y_{n+1}-y_{n-1}}{2h}$? (I just got this from book without proof..... )[/quote]

I believe that is equating the slope at a point and the slope of a secant near that point.


Prove that:
$\frac{a}{a^{2}-bc+1}+\frac{b}{b^{2}-ac+1}+\frac{c^{2}}{c^{2}-ac+1}\geq \frac{1}{a+b+c}$
where $a;b;c$ are positive real number satisfying $3(ab+bc+ca)=1$

let S the RHS.

by Cauchy, $\sum_{cyc}a(a^{2}-bc+1) . S \geq (a+b+c)^{2}$

but $\frac{(a+b+c)^{2}}{\sum_{cyc}a^{3}-abc+a}= \frac{1}{a+b+c}$

[b]Problem 2[/b]:Prove that:
$\frac{a}{a^{2}-bc+1}+\frac{b}{b^{2}-ac+1}+\frac{c^{2}}{c^{2}-ac+1}\leq 3$
where $a;b;c$ are positive real number satisfying $3(ab+bc+ca)=1$

To chien than
[hide]Tôi và Tuấn đã nhắc em nhiều lần lắm rồi mà sao không tuân thủ nội quy diễn đàn. Trong box Vietnam đã có cảnh cáo rồi mà sao em không nghe.[/hide]


Two men set out at the same time to walk towards each other from $ M$ and $ N$, $ 72$ miles apart. The first man walks at the rate of $ 4$ mph. The second man walks $ 2$ miles the first hour, $ 2\frac {1}{2}$ miles the second hour, $ 3$ miles the third hour, and so on in arithmetic progression. Then the men will meet:

$ \textbf{(A)}\ \text{in 7 hours} \qquad \textbf{(B)}\ \text{in }{8\frac {1}{4}}\text{ hours}\qquad \textbf{(C)}\ \text{nearer }{M}\text{ than }{N}\qquad \\
\textbf{(D)}\ \text{nearer }{N}\text{ than }{M}\qquad \textbf{(E)}\ \text{midway between }{M}\text{ and }{N}$

Let $ x$ be the number of hours it would take the men to meet.

Then we get the distance $ 4x$ for the $ M$ man and $ \frac{x}{2}[4\plus{}\frac{x\minus{}1}{2}]$ for the second man. They travel a total of 72 miles together when they meet. So we have the equation

$ 4x\plus{}\frac{x}{2}[4\plus{}\frac{x\minus{}1}{2}]\equal{}72$
$ 4x\plus{}2x\plus{}\frac{1}{4}x^2\minus{}\frac{1}{4}x\equal{}72$
$ 6x\plus{}\frac{1}{4}x^2\minus{}\frac{1}{4}x\equal{}72$
$ x^2\minus{}x\plus{}24\equal{}72$
$ x^2\minus{}x\minus{}48\equal{}0$

Using the quadratic formula, we get

$ x\equal{}\frac{1\plus{}\sqrt{1\plus{}48*4}}{2}$ (since $ x$ must be positive)
$ x\equal{}\frac{1\plus{}\sqrt{193}}{2}$

In that many hours, the man from $ M$ will have traveled $ 2\plus{}2\sqrt{193}$ miles, which is slightly under 30 miles, so that's the meeting point, closer to $ M$ than $ N$.

Answer: C

@gauss1181: I think you made an error, because I got a quadratic with integer roots. I started with the same equation as you though.

[hide="Continuing..."]
From step (3) of gauss's simplification of the equation, we have

$ 6x \plus{} \frac{1}{4}x^2 \minus{} \frac{1}{4}x \equal{} 72\implies 24x \plus{} x^2 \minus{} x \equal{} 288\implies x^2 \plus{} 23x \minus{} 288 \equal{} 0.$

Applying the quadratic formula and discarding the negative root, we find that they meet after $ x \equal{} 9$ hours. At $ x \equal{} 9$ hours, the first man travels $ 4\cdot 9 \equal{} 36$ miles and the other man travels $ 72 \minus{} 36 \equal{} 36$ miles. Therefore, they meet midway between $ M$ and $ N$ $ \implies \fbox{(E)}$.
[/hide]

oh, oops. sorry, I forgot to multiply 72 by 4 and thus my answer was wrong. :(
but thanks for the correction. :blush:


give x*y*z=1/8

prove that
a/(b+2c) + b/(c+2a)+ c/(a+2b)>=4/3
I have general problem but I can't prove it .

Sorry Hikaru,maybe you do a mistake here.What is x,y,z.The inequality is not right too.If a=b=c we just have
a/(b+2c) + b/(c+2a)+ c/(a+2b)=1....  :?

[quote="Hikaru"]
a,b,c>0;
prove that
a/(b+2c) + b/(c+2a)+ c/(a+2b)>=2/3
[/quote]

[quote="Master"]are you sure hikaru: when a=b=c the leftside=1
?????????????????[/quote] so what? [tex] 1 \geq \frac 23 [/tex] right ? :)
maybe the equality case is not in a=b=c=1 (it would seems so as the ineq is not symmetric in a,b,c).

I think this one can be proved by Cauchy.Let see:
[a(b+2c)+b(c+2a)+c(a+2b)][a/(b+2c)+b/(c+2a)+c/(a+2b)]>=(a+b+c)^2.
But a(b+2c)+b(c+2a)+c(a+2b)=3(ab+ac+bc)<=(a+b+c)^2.
So we have what we need.


calculus
integration
trigonometry
geometry
function
algebra
linear equation

Let $|\cos \theta| \neq |\sin \theta|$. Find the value of $k$ satisfying the following equation.

\[\int_{\cos ^2 \theta}^{\sin ^2 \theta} (x-\cos ^2 \theta)(x-\sin ^2 \theta)(x-k) dx=0\].

[hide]The answer has to be $\frac12.$

In general, $\int_u^v(x-u)(x-v)(x-k)dx=0$ when $k=\frac{u+v}2.$

It's all just a matter of symmetry. Integrated out, this is a linear equation for $k$ and has a unique solution. Letting $k$ be the midpoint of the interval works.[/hide]

Needless to say, you are right, Kent Merryfield.

The theme of this problem is the condition of the equality of two area between a cubic function and a line on condition that these graphs have three the intersection points. Let $\alpha,\beta,\gamma\ (\alpha <\beta <\gamma)$ be the abssisae of the intersection points,using beta function for Integral,

the condition can be described $\int_{\alpha}^{\gamma} (x-\alpha)(x-\beta)(x-\gamma)dx=0\Longleftrightarrow \frac{1}{12}(\gamma-\alpha)^3(2\beta -\gamma -\alpha)=0\Longleftrightarrow \beta =\frac{\alpha +\gamma}{2}$


nobody likes these ineqs?
or nobody can solve these ineqs? :)


If 2^n - 1 = ab and 2^k is the highest power of 2 dividing 2^n - 2 + a - b then k is even

Let $x=a-1 \geq 1$ and $y=b+1 \geq 3$. Then, it is given that $(x+1)(y-1)=2^n - 1$, that is $xy+y-x=2^n$. (1)
Moreover $2^n-2+a-b= ab-1+a-b = (a-1)(b+1) = xy$. Thus we want to prove that if $x=2^pq$ and $y=2^rs$ where $q,s$ are odd integers then $p=r$.

But from (1), if $p>r$, we have :
$2^n = 2^{p+r}qs + 2^rs - 2^pq = 2^r(2^pqs + s - 2^{p-r}q)$ where $2^pqs + s - 2^{p-r}q$ is odd and $2^pqs + s - 2^{p-r}q > 2^pqs  - 2^{p-r}q = 2^{p-r}q(2^rs - 1) \geq 2$. Such an odd number cannot divide a power of 2. A contradiction.

If $p < r$ then :
$2^n = 2^{p+r}qs + 2^rs - 2^pq = 2^p(2^rqs + 2^{r-p}s - q)$ where $2^rqs + 2^{r-p}s - q$ is odd and $2^rqs + 2^{r-p}s - q > 2^rqs  - q = q(2^rs - 1)$ which leads to the same contradiction.

Thus $p=r$ and we are done.

Pierre.

Dear Pierre

what if a=b=1 then how can you have x be greater than or equal to 1 and y be greater than or equal to 3.

one more question, how did you come up with this solution?

thanks for the solution.

Since the problem deals with the exponent of 2 in the prime decomposition of an expression, it is really classical to try to factorize the given expression, and that one must be recognized at first-sight.

If $a=b=1$ the considered expression is 0, thus the exponent of 2 is $+ \infty$ which you may consider as even but it does not make really sense.

Pierre.


The mean distance of Earth from the sun is 9.29 x 10^7 miles. Assuming that the orbit of Earth is circular around the Sun and that 1 revolution takes 365 days, find the linear speed of the Earth. Express your answer in miles per hour.

[hide]
$ r = \frac{d}{t}$
For this problem
$ d=2 \pi \left( 9.29 \cdot 10^{7}\ \text{miles}\right)$
$ t=365 \ \text{days}= 24 \frac{\text{hours}}{\text{day}}\cdot 365 = 8760 \ \text{hours}$
Then
$ r=21210.0457 \pi$.
[/hide]

The book's answer is approximately 2292 mile per hour.

Who do I get 2292 mph?

There is a similar problem posted on Yahoo! [url=http://sg.answers.yahoo.com/question/index?qid=20070711031118AAe7ipU]here[/url].
The answer there coincides with my previous post, so you book may be wrong.


trigonometry
analytic geometry
geometry
invariant
geometric transformation
rotation
induction

Post and discuss your round 2 solutions here!  My solutions were short this year - only two were longer than a page.

[quote="renatzu"]I think your solution to number 1 is wrong, it should be 2023, not 2033.[/quote]
You're right, but it's a very minor typo.  If you look at how I derived it, I get the right answer.  Thanks for pointing that out.

I thought mine were pretty good. My number 4 left a lot to be desired :D .

[quote="bpms"]Does anyone have a non-trig solution to number 5? Mine isn't too trig heavy since I used $\sin (x+y)= \sin x \cos y+\cos x+\sin y$. You guys don't think that I would need to derive it would you :maybe: .[/quote]
This identity is not correct (probably just a typo). Mine uses mainly coordinate geometry; I only used the Law of Cosines once.

you don't need any trig if you see that $\angle BAC=60^\circ$

My answers are now concealed for the rest of eternity   :diablo:

[quote="knexpert"]Here are my solutions. I made a generalization on number 4, but it probably isn't correct.[/quote]
I got the same generalization. :D Although, I just said "Similarly, the coloring to achieve the least $n$ is ...blah, blah..., so for $c$ colors, $n=k^{c}+k^{c-1}-1$."

For #5 I accidentally found BF instead of AF but I got the length of BF right (7.5).  Do you think the graders would take points off for that?

My solution for #1 was brute force-ish, here's basically what I did

1. Proved that all such numbers had to have four digits
2. Proved that the hundreds digit had to be less than or equal to 2
3. Proved that the hundreds digit could only be 0
4. Proved that the tens digit had to be even and the ones digit had to be congruent to -1 mod 4 (at least I think, I didn't have time to make a copy of my handwritten solutions)
5. Tried all the remaining possible numbers, and found the correct answer.

Do you think this will get full credit?

I think I might have gotten commended for one of either 2 or 5, most maybe both. Hoping for:

5 [b][color=red]5[/color][\b] 5 1 [b][color=red]5[/color][\b]. I only did a special case for 4.[/b]

actually 7.5 is wrong
... only a couple points
in any case, here are my solutions

[quote="rnwang2"]Please be nice and don't make fun of my solutions :yup:[/quote]
I had a similar solution for numbers 2 and 5. Good job not trig bashing either of them.

I "solved" problem #4 at 12 ish EDT so my solution obviously wrong.

My prediction:
min: 5 5 4 2 5
max: 5 [b]5[/b] 5 4 [b]5[/b]

These are probably high estimates.

Here are my solutions. I thought my number 1 will be the closest to a commended; I doubt any others will. I dont' have a number 4 because I couldn't understand the question, but hopefully I will get 20 points (which is better than the 13 I got on the last one :lol: )

I did more of a speculation for #4, although I did get $k^{2}+k-1$.

Anyway, I have a solution for #5 that seems pretty simple...

[hide="5"]
Since all three sides of $\triangle ABC$ are given, it is natural to use the Law of Cosines to determine $\angle BAC$; therefore, we can figure out $\angle BAE$ and $\angle EAD$.

Using the Law of Cosines, we get $\angle BAC = 60$. Thus, $\angle BAE = \angle EAD = 60$.

Drawing $\overline{EC}$ and $\overline{BE}$, we see that $\angle ECB = \angle BAE$ because they are both inscribed angles referencing to the same arc. Therefore $\angle ECB = 60$.

Likewise,
$\angle BEC = \angle BAC$

because they are both inscribed angles referencing to the same arc.

Therefore,
$\angle BEC = 60$

making $\triangle ECB$ equilateral.

Since $\triangle ECB$ is equilateral, $\overline{EC}= \overline{EB}= 7$. Now we have 2 diagonals and 3 sides of a cyclic quadrilateral. If we apply Ptolemy's Theorem, we see that $EA = 3$
$\triangle EFA$ is a $30-60-90$ triangle; therefore, $AF = \frac{EA}{2}= \frac{3}{2}$ [/hide]

EDIT: Wait...did you have to prove that the quadrilateral was cyclic?...I thought by the description of the problem, it was understood that it was cyclic...>.>

since the length for problem 5 is invariant with respect to BC, i suspect that commendations will only be given if you derive that the answer is $\frac{1}{2}*|AB-AC|$

number 1 I misread the problem and thought there's no solutions, but later I wrote a program and found two.  I use completing the squares with diophatine equations.  Hope I get commended.

number 2 is interesting: I have a rotation solution, hope to get commended on this one too.

Edit: everyone just extend the two sides?? com'on, be manly and use transformations :rotfl:  :P

number 3 is just brute force the first couple values, theorem of finite difference, and do induction on the real thing...

number 4 I wrote a program, guess the formula, and proved it :rotfl:

number 5 I try to draw a scaled diagram, so I'm like "let's find out the angles with law of cosines", then I found the 60 degree angle and the rest is easy.

Here's what I did to find the 60 degree angle:

"Note that a 5-7-8 triangle is just an $\frac{11}{2}$-$\frac{5\sqrt{3}}{2}$-7 right triangle joined to a $\frac{5}{2}$-$\frac{5\sqrt{3}}{2}$-5
right triangle by the $\frac{5\sqrt{3}}{2}$ side. Since the second right triangle is also a $30^\circ$-$60^\circ$-$90^\circ$
triangle, we can conclude that $m\angle BAC = 60^\circ$."

Lol... I wrote a program in C++ for #1 to find the answers and it returned no solution, so I went with that. That's the last time I trust my program-writing abilities (or lack thereof) after 2-3 years of not using them  :blush:

Then in writing my solution, I "verify" this faulty solution by making a critical arithmetic error (10-16=-4  :oops:).

Yay for writing solutions at 2 am (deadline is 3am EST).  :|

So, worst case scenario: 1 5 5 4 5
best case: 3 5 5 5 5.

[quote="bpms"]Unless I'm missing something, everyone used law of cosines on problem 5. Does any one have a pure synthetic solution?[/quote]

I do :)

Argh. I failed this round (and I will fail the rest of them). :oops:

[quote="ehehheehee"]You can figure out the angle is 60 degrees by applying Heron's and solving for the height and use the ratio (height)/AB which would be $\frac{\sqrt{3}}{2}$ and then prove the triangle is acute.

Or you can apply the pythagorean theorem by labling (AH) as x and (CH) as 5-x and equating the height...That is what I did in mathcounts before I knew heron's formula (which can find the amazing one's two triangles)

Or you can just plug in the law of cosines and be done with it...

But since the heron's formula can be derived through law of cosines and the second way can derive the law of cosines, there really isn't a difference.[/quote]
But heron's can be proven synthtically too. To Mr. Rusczyk, could you please post your solution, or are you saving it for the math jam?
Thanks,
BPMS

These are the correct solutions.  nr1337 has spoken.  I was too lazy to simplify #3, whatever.

[quote="knexpert"]Did anyone besides me do a solution to number 5 using coordinate geometry?[/quote]
I used coordinates to find the length of the segment as well.

[quote="bpms"]Unless I'm missing something, everyone used law of cosines on problem 5. Does any one have a pure synthetic solution?[/quote]

First off I use law of cosines.  The second time I FORGOT my solution ( :rotfl: ) so I came up with one non-coordinate non-trig solution.  But after I remembered the law of cosines solution I went with that one because it's shorter.  Now I forgot the second solution I came up with... :|


Find the number of degrees in the measure of the supplement of the complement of an angle with measure 27 degrees.

The compliment of 27 is 63, and the supplement of 63 is $ \boxed{117}$.

Alternatively:

Supplement of Compliment of an angle $ x$ is:

180-(90-$ x$) = 90+$ x$

therefore, plugging in 27 yields $ \boxed{117}$


Let $PQ$ be the diameter of a semicircle $H$. Some circle $O$ is internally tangent to $H$ and is tangent to $PQ$ at $C$. Let $A$ be a point on $H$, and let $B$ be a point on $PQ$ such that the line $AB$ is perpendicular to $PQ$ and is tangent to $O$. Prove that the line $AC$ bisects the angle $PAB$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=17446 and http://www.mathlinks.ro/Forum/viewtopic.php?t=51954 .

I heard the problem can be solved using inversion...

  darij

If J tangency point of H and O also K tangency point of AB and O, now J,K and Q are collinears, therefore PJQ and KBQ are similar:
PQ/KQ=JQ/BQ
BQ.PQ=JQ.KQ  (1)
Also:
QA^2=BQ.PQ  (2)
QC^2=JQ.KQ  (3)
From (1),(2) and (3): QA=QC.
Now: <ACQ=<CAQ
<APQ+<PAC=<CAB+<BAQ
where: <APQ=<BAQ, now: <PAC=<CAB i.e. AC bisects<PAB.


I searched around and I couldn't find a topic on this..so I posted it.

I found that Microsoft was giving away copies of its programming software (Visual Studio Express Edition) until sometime next year.

[quote]Visual Studio Express is free for one year

Until November 7, 2006, we are promotionally discounting the downloadable versions of Express to free. This doesn’t mean that the product turns off after a year, but rather that as long as you download the product before November 7, 2006, you can get it for free and you can use it forever.

[url=http://forums.microsoft.com/MSDN/ShowPost.aspx?PostID=126606&SiteID=1]http://forums.microsoft.com/MSDN/ShowPost.aspx?PostID=126606[/url]
[/quote]

:D

[url=http://msdn.microsoft.com/vstudio/express/]http://msdn.microsoft.com/vstudio/express/[/url]

I downloaded it [Visual Basic] and it was pretty good.

Cool! I downloaded Visual Studio Express a few months ago for free, but it was just a one-year version. I probably won't need it anymore now that I have Linux though.

Does anyone know what the difference is between the express edition and the regular version?


I think it is from crux...
$ \frac{a^2\plus{}bc}{b\plus{}c}\plus{}\frac{b^2\plus{}ca}{c\plus{}a}\plus{}\frac{c^2\plus{}ab}{a\plus{}b} \geq a\plus{}b\plus{}c$

[quote="Dr Sonnhard Graubner"]hello, what conditions must be fulfilled that the inequality is true?
Sonnhard.[/quote]
Oops i forgot: $ a,b,c>0$

hello, our given inequality is equivalent with
$ \frac{a^4\plus{}b^4\plus{}c^2\minus{}a^2b^2\minus{}b^2c^2\minus{}c^2a^2}{(a\plus{}b)(a\plus{}c)(b\plus{}c)}\geq0$,
the numerator is non negative because it is equivalent to $ (a^2\minus{}b^2)^2\plus{}(b^2\minus{}c^2)^2\plus{}(a^2\minus{}c^2)^2\geq0$.
Sonnhard.

Another solution:
$ \sum \frac{a^2\plus{}bc}{b\plus{}c}\plus{}a\equal{}\sum \frac{(a\plus{}b)(a\plus{}c)}{(b\plus{}c)}$
Seting $ a\plus{}b\equal{}x,b\plus{}c\equal{}y,c\plus{}a\equal{}z$ and our inequality become $ \sum\frac{xy}{z} \geq x\plus{}y\plus{}z$,which is true,because $ (xy)^2\plus{}(yz)^2\plus{}(zx)^2 \geq xyz(x\plus{}y\plus{}z)$ :wink: