AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 10 A
AMC 12

I count 6. Random.org says 3=ZzZ... asks first.

Pythag chooses truth.

Btw, all dares must be able to be done online, with some degree of verifiability.

Post all your past AMC, AIME, USAMO scores.

Yawn. Ez.

5th grade:
AMC 12A 94.5
AMC 12B 94.5

6th grade

AMC10A 144
AMC12B 118.5
AIME I 9
USAMO 10

7th grade (this year)

AMC 10A 144
AMC 12B 138
AIME I 11
USAMO 42 (jk)

Dude what do you mean 42? That is a horrible score. The best score is obviously -0. (where -0 approximately =0)


Five congruent squares are placed side-to-side as shown to make a single rectangle whose perimeter is 168 inches. How many square inches are in the area of one of the squares?
[asy]draw((0,0)--(5,0)--(5,1)--(0,1)--cycle);
draw((1,0)--(1,1));
draw((2,0)--(2,1));
draw((3,0)--(3,1));
draw((4,0)--(4,1));[/asy]

Let the side length of the square be $ x$. Then, $ x\plus{}x\plus{}5x\plus{}5x\equal{}168 \implies 12x\equal{}168 \implies x\equal{}14$. The area is then $ x^2\equal{}14^2\equal{}\boxed{196}$.


geometry
circumcircle
perpendicular bisector
geometry unsolved

It is given a convex quadrilateral $ ABCD$ in which  $ \angle B\plus{}\angle C < 180^0$.
Lines $ AB$ and $ CD$ intersect in point E. Prove that
$ CD*CE\equal{}AC^2\plus{}AB*AE \leftrightarrow \angle B\equal{} \angle D$

From  $ \angle{B]+\angle{C}<180}$ we conclude that $ EA<EB,ED<EC$.Let O be the center of the circumcircle of  $ ADE$ with radius  R.

   $ CD.CE=AC^2+AB.AE$.So $ CD.CE=AC^2+AB(BE-BA)$.$ CD.CE-AB.AE=AC^2-AB^2$.$ CO^2-R^2-BO^2+R^2=AC^2-AB^2$

  Finally $ OC^2-OB^2=AC^2-AB^2$.Now it is easy to conclude that $ OA$ perpendicular to  $ BC$.

   Let $ OM$ perpendicular bisector of $ AE$.Then  $ BNMO$ cyclic and $ \angle{B}=\angle{AOM}$.

   Obviously $ \angle{ADC}=\angle{AOM}$.

   Done.

Let $ AC$ and circumcircle of triangle $ EBC$ intersects at point $ F$.
$ m(B) \equal{} m(D) \Longleftrightarrow$
$ |CD|.|CE| \equal{} |CA|.|CF| \equal{} |AC|^2 \plus{} |AE|.|AB|$ and solution is done.  :wink:

Let $F=\overline{AD}\cap\overline{BC}$ and $M$ be the midpoint of $\overline{AC}$. Let $\Omega$ be the circumcircle of $\triangle EDB$. Let $f(P)=PC^2-\mathcal{P}_\omega(P)$. It is straightforward to see $f$ is linear.  Then, $f(M)=f\left(\frac12 A+\frac12 C\right)=\frac12f(A)+\frac12f(C)$. Now, since $f(A)=AC^2+AB\cdot AE$ and $f(C)=CD\cdot CE$, we have $AC^2+AB\cdot AE=CD\cdot CE\iff f(M)=0$. Now, $f(M)=0$ iff $M$ lies on the radical axis of $\Omega$ and the circle of radius $0$ centered at $C$, $\odot C$. Since their radical axis is the line connecting the midpoints of the tangents from $C$, it follows $f(M)=0\iff A$ lies on the polar of $C$. From Brokard, this happens iff $EDBF$ is cyclic, which happens iff $\angle B=\angle D$. $\blacksquare$


looks like A is a diene(yield might be low)
B-diels alder reaction product
C-Then allylic bromination

This is a side question.

How exactly does nomenclature with bridged compounds work? This would be very useful when dealing with diels alder reaction products...

NBS also introduces bromine alpha to carbonyl groups.

so C would be with Bromine attached to the tertiary C next to either of the carbonyl gps . right?
rather than the one shown.

Yes, that would be another possibility. Probably, a mixture of products is to be expected. Next step: column cromatography.

which would be major? the one alpha to carbonyl or alpha to double bond. which one one is more reactive . I think that the H alpha to double bond is more stable bcoz it can more easily undergo homolytic cleavage to resonance stablize the radical . than the CO double bond. Right?


Let $ a,b,c\in R^\plus{}$. Prove that $ \sum_{cyc}\frac{ab(a\plus{}c)}{b\plus{}c}\geq \sqrt{3abc(a\plus{}b\plus{}c)}$

Yes, it's an exercise there. Can you find a solution?

Thanks, csbonny.

Can you solve the other two: http://www.mathlinks.ro/viewtopic.php?t=187378, http://www.mathlinks.ro/viewtopic.php?t=187380 ?

yes..put $ a\equal{}\frac {{\mathop a\nolimits^2 \mathop c\nolimits^2 }}{b}$
$ b\equal{}\frac {{\mathop a\nolimits^2 \mathop b\nolimits^2 }}{c}$
$ c\equal{}\frac {{\mathop b\nolimits^2 \mathop c\nolimits^2 }}{a}$
and $ x\equal{}b^2c^2$
$ y\equal{}a^2c^2$
$ z\equal{}a^2b^2$

[quote="easternlatincup"]Wow, and what about $ \sum_{cyc} \frac {a^k \plus{} b^k}{a \plus{} b}\geq \sqrt {\frac {8(a \plus{} b \plus{} c)(a^kb^k \plus{} b^kc^k \plus{} c^ka^k)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}$?[/quote]

theory2
$ x\equal{}\frac{1}{a\plus{}b},y\equal{}\frac{1}{a\plus{}c},z\equal{}\frac{1}{b\plus{}c}$
$ a\equal{}a^{k},b\equal{}b^{k},c\equal{}c^{k}$


For my physics class i have to build a device using 5 sheets of regular paper and some tape.  The device is supposed to make an eggs survive a 2 meter drop without cracking. Any ideas on how i should make the device?

You could make a parachute, which is probably what your teacher is expecting people to do.
Or, for a slightly more off-the-wall idea, you could make a cord out of paper strips, taped to the egg and to something at the top. But tape the paper in such a way that the paper rips as the egg falls. The energy from the falling egg is (as much as possible) used to rip the paper, slowing the egg down enough for it to make a safe landing. Kind of like that scene in the James Bond movie where Bond grabs onto the top of a giant banner and it tears as he falls, slowing him down. You might need to make a few holes in the paper for it to tear correctly though, like a pack of postage stamps.


Hi,
Can someone give me some help with this reaction?

Potassium dichromate solution is added to an acidifed solution of sodium sulfite.

For the unbalanced NIE I get:
$Cr_2O_7^{2-} + SO_3^{2-} \rightarrow Cr^{3+} + SO_2$

But when I balance this, I get:
$Cr_2O_7^{2-} + SO_3^{2-} + 16H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + SO_2 + 8H_2O$

Don't the $e^{-}$'s have to cancel though?

It is not balanced.

For the full redox reaction to be balanced you have to balance atoms, charge and you have to get rid of electrons.

[hide="balanced reaction with proper products"]
$Cr_2O_7^{2-}+3SO_3^{2-}+8H^{+}\rightarrow2Cr^{3+}+3SO_4^{2-}+4H_{2}\textrm{O}$

[/hide]

oh your right.. i have both being reduced at the moment.  Whoops.

Can something like $SO_4^{2-}$ be oxidized?  If so, what would it form?  I'm so terrible at figuring out the products of these.

Google peroxydisulfate. But that's very rare compound.

[quote="SnowStorm"]

Can something like $SO_4^{2-}$ be oxidized?  If so, what would it form?  I'm so terrible at figuring out the products of these.[/quote]

yes, $SO_4^{2-}$ can be oxodized...it will turn into $S_2O_8^{2-}$ i believe...
notice that the oxidation is not due to Sulfur, but because of Oxygens...

same thing happens with $NO_3^-$

[quote="amirhtlusa"]same thing happens with $NO_3^-$[/quote]

Do you have any reference for that? Never heard about such reaction.

IIRC it happens with sulphuric and phosphoric acid, perhaps others too, but not with nitric.

Exactly...as a matter of fact it was an olympiad problem...

NO3 DOES get oxadized...but notice that in the oxadization of NO3, nitrogen will not be oxodized, rather Oxygen will, as it goes to N2O5 and O2...

[quote="amirhtlusa"]NO3 DOES get oxadized...but notice that in the oxadization of NO3, nitrogen will not be oxodized, rather Oxygen will, as it goes to N2O5 and O2...[/quote]

Please elaborate.

Do you know anything about conditions this reaction takes place in? Oxidizer?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let G be a group, and let a be in G. The set C(a)= { x in G| xa=ax} of all elements of G that commute with a is called the centralizer of a.
a) compute C(a) if G=S3 and a=(1,2,3)
b) compute C(a) if G=S3 and a=(1,2)

I know that There are only 6 elements in S3, but how do I find xa and ax for each of the 6 choices of x. Do I really only need to do this for 4, 2 elements that will obviously be in C(a).

What's the problem about calculating some products of permutations?

I got part a but I'm not sure about part b

You can observe that $(12)$ don't commute with $(13)$. So $C_{(12)}=\{(1),(12)\}$.
Infact $C_*$ is a subgroup of $S_3$, thus...


i already posted this but in the wrong place ...
I am currently preparing a report in which the general theme is "time" , i ll have to present it with a question in the introduction (or any curiosity for which i am supposed to give a solution by observing a phenomenon (probably with an experience) and be able to defend all the initiatives i choosed. For example , my first ideas was to expose a way how to understand time, but it seems that the relativity theory is too advanced for me (i am at the second year of university , maths and physics section) , i am really determined to do something as original as possible that's why i am asking now for your ideas, any ideas :) ?

Why is it only few people had a straight[b] pipeline[/b] to creativity? To be '[b] original[/b] thinker[b]'[/b] is not really
that hard in terms of conceiving original[u],[b] good[/b] ideas[/u].Would you believe that that genius '[b]Albert[/b] Einstein[b]'became creative only after so many '[/b]mathematics[b] problems[b]'[/b] encountered by him? If you get what i mean,then,perhaps,becomimg original[b] on your part within your
reach. Just[b] think[/b] of[b] any[/b] ideas[b] through[/b] question[b] asking[/b].[b] Remember,mathematics[/b] are just a
handmaiden[b] of scientific methodology.Once a problem is identified(Assuming its really 'original/novel one.),then that's the time to search a right[/b] mathematical[b] tools[/b] so as to solve a given problem. Its easy to be creative,and original.


I will give out many questions to see who is the best mathematician. Anyone can participate.

Round 0:

Due on August 1, 2005

if f(x)=$\sqrt{x+2x}\cdot4x$, to what value of x would the equation equal 288?

Please post the answers, so it will be easier to see who won!

You're supposed to pm the solutions, not post them!!!

oops... :blush: sorry. i've never done a contest before. thanks for reminding me.

Yeah, that's right, but, nevertheless, he should edit/remove his post... :)

Make this Round 0 because it is already spoiled  :(

Yes...that's sad.

When we pm you, do you want us to show you our work? Or doesn't that make any difference in how many points you could earn?

In Other Words: If I showed my work and copied the answer wrong, would I still get full credit?

Well, I understand it is quite hard to show your work sometimes, but It doesn't really matter.

Round 1:

Due August 1

What is the sum of $1+2^2+2^3+2^4...2^{20}$?

Round 2:

Simplify:

$\displaystyle {\frac{\frac{2}{3}x}{\frac{(x+2)(x+2)}{x+1}}}$

Round 3

Question 1: Simplify:

$a+ax+ax^2+ax^{20}$.

Question 2: 1 quid = 5 quads.
10 quads = 7 quidquads.
91 quidquads=1 master

How many quids are 1 master?

Question: What is $1\cdot2\cdot...9\cdot10$.

No calculators on any problems.

People who may participate so far:
236 Factorial
10000th User

It means private message.  The feature is located in the top toolbar when you log in.

I don't think so. He says due August 1

Iknowwhereyoulive: funny name, please fix your signature. It is having problems

[quote="math92"]sorry, 1-5th is elementary and 6-8th is middles school :)[/quote]

That's what dark knight said. :gathering:

Well, anyway I have changed the contest, since I can't post hard problems here. I will post the problems in the MATHCOUNTS forum in the middle school section.

Not in SC!  But it differs, because in LA 8th grade is "called" high school.  For me, it's still middle school!

well, if i don't receive any more ansers by tommorow, i will just conduct a duel between

Xantos C. Guin
plinius
236factorial
Drunken_Math

The other contest or this one? I thought this one was over a while ago  ;)

Sorry. This one is over. I meant the one inthe mathcounts forum.

These problems are of variable level.  Please don't do this again in what is supposed to be a strictly curricular forum.

This should not be in the the classroom. Moved to MC.

Lock this thread. the contest is over and there shouldn't be anymore discussion on a ended competition.  ;)


Can anyone sort of list some basic properties of the incentre of an triangle?


inequalities
vector
inequalities proposed

this might have been posted before, but anyway, i was thinking of a general inequality as follows, but i have no clue how i could prove it:

for all $\alpha,\beta \in\left(\mathbb R^+\right)^n$ and , $x \in\left(\mathbb R^+\right)^m$ where $a\succ b$:

$\prod_{i=1}^n\left(\sum_{j=1}^m x_j^{\alpha_i} \right)\ge \prod_{i=1}^n\left(\sum_{j=1}^m x_j^{\beta_i} \right)$

Note:
$a\succ b\Leftrightarrow \sum_{i=1}^k a_{(i)} \ge \sum_{i=1}^k b_{(i)}$ and $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i$, where $a_{(i)}$ notes the i-th greatest "number" of the vector $a$.

Oh, sorry Peter.  :blush:  :blush:  I am just seeing that this one is equivalent to Vasc's inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=48136 post #3. Actually, I saw this post of Vasc some days ago and I proved the inequality for myself (well, in exactly the same way as Vasc did in post #7), but then forgot about it...

  darij

thx though darij, that shows to me that i get some kind of feeling for ineqs :)
but i always forget to apply log and if i think of it i dont think it to the end  :roll:


so what did you guys think?
post your answers.

yo carbon edit out your answers

they can still send in solutions by email
its best to wait at least a couple days till you guys start discussing

Please refrain from posting solutions until April 15. We'd like to receive them first!

Thanks,
Zac


geometry
trigonometry
geometric transformation
analytic geometry

[img]http://i29.tinypic.com/euq7sy.jpg[/img]

Can someone tell me how to get this started? To find x, where should I look?

the last 30, is on plane $ xy$ or $ xz$??

 however you should try to analize the movements(translations) that the point $ (0,0,0)$ had to do in order to get in to the finish, so try to follow the path and analize independently the translations parallel to the $ x$ axis the $ y$... $ z$...

For example to get into the point $ (5,8,13)$ you can sum the movements $ (1,1,2)$, $ (3,5,8)$ and $ (1,2,3)$, something like that, note that the $ x$, $ y$ and $ z$ coordinates are independent from each other

 :)


$z^2-x^2=zx=y^2\geq0$
So $z^2-zx-x^2=0$,$z=x=0$or $z=\frac{1+\sqrt{5}}{2}x$ or $z=\frac{1-\sqrt{5}}{2}x$
Easy to see $z=x=0$ impossible.
And $z^2>x^2,|z|>|x|$
So $z=\frac{1+\sqrt{5}}{2}x$,$x\neq0$,$y=\pm\sqrt{\frac{1+\sqrt{5}}{2}}$
Then take these in the first one.
Got it!


This year will be my first year taking the AMC12, and considering some of the topics are more advanced than those on the AMC10, I'm wondering how much I should prepare.

Is the material in AOPS I enough? Or are the problems harder than that?

Thanks!

Usually, AoPS Volume 1 is good for everything up until the AMC 10. Since the AMC 10 and AMC 12 have some overlap, I would say it covers about the first half of the AMC 12. For the rest, the Intermediate series or Volume 2 might be appropriate. On the other hand, if you didn't want to buy books, you could do lots of practice tests and prepare that way. Keep in mind that you should always understand the solution of a problem you got wrong when practicing.

I would also say to practice AIME questions.

So do you think volume 1 is good enough for mostly all amc 10 questions (except for maybe the last 5?)  Which chapters in volume 1 and volume 2 are most important to do for amc 10/12 prep?

I have the volumes 1 and 2, and I'd say the chapters to prepare for the AMC 10 in volume 1 are part of 1, 3 - 16, 18, 20 - 26, and 29. For the AMC 12, all of them in volume 1 and the easy to medium-easy (yes, I know I'm rigorous) parts of volume 2.

EDIT: (this is way in the future) You have found the 2013 Late Easter Egg #2. PM me with the number 86274995 to complete this step.

For the AMC 12 in Volume 2, I would Prioritize: 1,2,3,4,6,9,13,17,19,23

[quote="Poincare"][quote="batteredbutnotdefeated"]Alternatively, if you don't have much time, you could look into FSFMO as well.[/quote]
What's that?[/quote]
First Steps for Math Olympians.


Find all pairs of positive integers m and n for which 4mn - m - n + 1 is a perfect square...

[hide="no idea"] since there is a plus 1, im guessing when factored, its in the form (x+1)(x-1)[/hide]

abacadaea, that equation cannot be factored.

Are you sure the problem isn't $ 4mn \minus{} 2m \minus{} 2n \plus{} 1$?  As written, I don't think it's Mathcounts-level at all.

If it is factored as stated in the original problem, I would set it equal to n^2, where n is an integer.  I would then subtract m+n (-m-n) from both sides, and then factor the left side, getting (2m-1)(2n-1)=n^2-m-n.  However, I have to agree with t0----- in how this would be a crazy hard number theory problem.

You're using the variable $ n$ twice.

Anyway, the SFFT idea [hide="actually does work!"] Set $ 4mn \minus{} m \minus{} n \plus{} 1 \equal{} k^2$.  Then

$ 16mn \minus{} 4m \minus{} 4n \plus{} 4 \equal{} 4k^2 \implies$
$ (4m \minus{} 1)(4n \minus{} 1) \equal{} 4k^2 \minus{} 3$.

Given any value of $ k$, any divisor $ d$ of $ 4k^2 \minus{} 3$ that's congruent to $ 3 \bmod 4$ produces the valid solution

$ (m, n) \equal{} \left( \frac{d \plus{} 1}{4}, \frac{4k^2 \minus{} 3 \plus{} d}{4d} \right)$. [/hide]


[quote="GameBot"]What is the product of the sum of the first three primes the the sum of the reciprocals of the first three primes? Express your answer as a mixed number.[/quote]
Shouldn't it be "What is the product of the sum of the first three primes [i]and [/i]the sum[$ \dots$]"?  Anyways...
The first three primes are $ 2, 3, \text{and }5$.  Their sum is $ 10$, and the sum of their reciprocals is $ \frac {1}{2} \plus{} \frac {1}3 \plus{} \frac {1}5 \equal{} \frac {5}6 \plus{} \frac {1}5\equal{} \frac {31}{30}$.  So our product is $ 10\cdot\frac {31}{30} \equal{} \frac {31}3 \equal{} \boxed{10\frac {1}3}$


1) Prove that if lim $ s_n \over{n}$=L(non zero) then $ s_n$ is not bounded.Is the converse of this statement also true?Justify your answer.

Can you find a sequence of real numbers $ s_n$ which has no convergent subsequence and yet for which $ mod s_{n}$ converges?Justify your answer.

Ah, sweet homeworks again :). The rule is that when posting a homework question, you should show what you've done on the problem. Then you may expect a correction or a hint as to what to do next.


[quote="hendrata01"]If $a,b,c$ positive numbers such that $abc=1$ then prove that:

$3(a+b+c+8(ab+bc+ca)+12)\left( 1+2\left( \frac{1}{|7a-20|}+\frac{1}{|7b-20|}+\frac{1}{|7c-20|} \right) \right) \geq 7\left( \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}} \right)^2$[/quote]
are you sure ?
this is not hard:
$3(a+b+c+8(ab+bc+ca)+12)\left(1+2\left( \frac{1}{|7a-20|}+\frac{1}{|7b-20|}+\frac{1}{|7c-20|}\right)\right)>3\cdot 7(ab+bc+ca)=7\cdot 3\left[(\frac{1}{\sqrt{a}})^2+(\frac{1}{\sqrt{b}})^2+(\frac{1}{\sqrt{c}})^2\right]\geq 7(\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}})^2$
  :lol:  :lol:

oh sorry.... I was wondering for a long time why is that so thick (weak). I meant it to have an equality where $a=b=c=1$.
So I must've made an algebraic error.
Anyway, the real problem is this (sorry again :P)

$3(a+b+c+8(ab+bc+ca)+12)\left( \frac{1}{13}+2\left( \frac{1}{|7a-20|}+\frac{1}{|7b-20|}+\frac{1}{|7c-20|} \right) \right) \geq 7\left( \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}} \right)^2$


function
superior algebra
superior algebra unsolved

my own problem, I have no idea for it:

for every nonempty set $ A$ there exists an operation $ \circ: A\times A\to A$ such that $ (A,\circ)$ is a group.

(prove or disprove)

Posted before. For finite $ A$ take cyclic groups. Otherwise we can even find a field of that cardinality: take $ B\equal{}\mathbb Q(A)$ (rational functions with variables $ \in A$). Then $ |B|\equal{}|A|$, thus we got a field of this cardinality. If you just want a ring/group, take $ \mathbb Z^{(A)}$ (functions $ A \to \mathbb Z$ with coordinatewise addition).

Another interesting question is, by how many ways ban a set become a group...

Happy new year!

By the way, why should a set be non-empty, in order to be a group?

Probably because some definitions require it? It is rather pointless to have the "empty group."

[quote="MellowMelon"]If $ G$ is a group, the identity axiom says "There exists an element $ e \in G$ such that..." and you can stop there. There is an element in $ G$. No empty sets can be groups.[/quote]
Why even say that?
By definition a group is a set with a special binary operation $ G\times G\mapsto G$, how can this make sense if $ G$ is empty? I never seen anyone define Cartesian products for empty sets.

$ S \times \emptyset \equal{} \emptyset$ for all sets $ S$, there is no problem in applying the standard definition of "pairs".


inequalities
geometry
inradius
circumcircle
geometric transformation
reflection
rectangle

[b]Lemma.[/b] $ \triangle ABC$ is acute with altitudes $ h_a,h_b,h_c,$ such that $ h_a \ge h_b \ge h_c.$ Let $ P$ be a point on $ \overline{BC}$ and $ X,Y$ the feet of the perpendiculars from $ P$ to $ AB,AC.$ Then
\[ h_b \ge PX \plus{} PY \ge h_c\]
[b]Proof.[/b] Let $ M$ and $ N$ be the projections of $ P$ onto the altitudes $ h_b \equiv BH_b$ and $ h_c \equiv CH_c.$ Hence, $ \angle XPB \equal{} 90^{\circ} \minus{} \angle B$ and $ \angle H_bBC \equal{} 90^{\circ} \minus{} \angle C.$ Since $ \angle C \ge \angle B,$ due to $AB \ge AC$  $\Longrightarrow$ $ \angle XPB \ge \angle H_bBC.$

Let $ \triangle PM'B$ be the reflection of $ \triangle PMB$ about the perpendicular bisector of $PB.$ $ M'$ lies  on the semicircle with diameter $ PB$ and $ PM'$ is an internal ray of $ \angle XPB,$ due to $ \angle XPB \ge \angle H_bBC \ge \angle M'PB.$ Then chord $ PM'$ is greater than the chord $ PX,$ i.e. $ PM' \equal{} MB \ge PX.$ Since $ H_bMPY$ is a rectangle, we have then

$ MB \plus{} PY \ge PX \plus{} PY \ , \ PY \equal{} MH_b  \Longrightarrow  MB \plus{} MH_b \equal{} h_b \ge PX \plus{} PY.$

Similarly, we get $ PX \plus{} PY \ge h_c$ $\Longrightarrow$ $ h_b \ge PX \plus{} PY \ge h_c.$
____________________________________

Let $ O$ be the circumcenter of $ \triangle ABC.$ WLOG assume $ h_a \ge h_b \ge h_c.$ Let the parallel to $ BC$ through $ O$ intersects $ AB,$ $AC$ at $ B',C'$ and the altitude $ h_a$ at $ D.$ Let $ M_a,M_b,M_c$ be the projections of  $ O$ on $ BC,AB,AC.$ Since $ \triangle AB'C' \sim \triangle ABC,$ we use the above lemma for $\triangle AB'C'$ with shortest altitude $AD.$ Thus

$ AD \ge OM_a \plus{} OM_b \Longrightarrow AD \plus{} OM_a \equal{} h_a \ge OM_a \plus{} OM_b \plus{} OM_c.$

By Carnot theorem for the acute triangle $ \triangle ABC,$ we get then

$ h_a \ge OM_a \plus{} OM_b \plus{} OM_c \equal{} R \plus{} r.$  Equality holds when $ \triangle ABC$ is equilateral.

[quote="Moonmathpi496"]A triangle has inradius $ r$ and circumradius $ R$. Its longest altitude has length $ H$. Show that if the triangle does not have an obtuse angle, then $ H\geq r\plus{}R$. When does equality hold?[/quote]
It is equivalent to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1380206&sid=c6fc2ca60bcccef9dd9f8e74cfed3e89#p1380206]INMO 2008 Problem 5[/url]


quadratics
algebra
number theory unsolved
number theory

[quote="ovidius"]$ a_0 \equal{} a_1 \equal{} 1$ and $ \forall \ n\geq 1, \ (n \plus{} 3)a_{n \plus{} 1} \equal{} (2n \plus{} 3)a_n \plus{} (3n)a_{n \minus{} 1}$.

Prove that $ a_n\in \mathbb{Z},\ \forall\ n$[/quote]

Compute the first values and look up in Sloane - sooner or later you will see that your numbers are the [url=http://www.research.att.com/~njas/sequences/A001006]Motzkin numbers[/url]. An alternative recurrence making clear the integrality:

$ a_{n \plus{} 2} \minus{} a_{n \plus{} 1} \equal{} a_0a_n \plus{} a_1a_{n \minus{} 1} \plus{} ... \plus{} a_na_0$.

The only question remaining is how to prove the equivalence between these two recurrent equations in a halfways aesthetic way. The straightforward proof defines a formal series $ P \equal{} \sum_{i \equal{} 0}^{\infty}a_iX^i$, then notices that
$ \left(a_0 \equal{} 1,\ \ \ \ \ \ \ \ \ \ a_1 \equal{} 1,\ \ \ \ \ \ \ \ \ \ \left(n \plus{} 3\right)a_{n \plus{} 1} \equal{} \left(2n \plus{} 3\right)a_n \plus{} 3na_{n \minus{} 1}\right)$
is equivalent to the differential equation $ P^{\prime} \plus{} 2\frac {P \minus{} 1}{X} \equal{} 3P \plus{} 2XP^{\prime} \plus{} 3X^2P^{\prime} \plus{} 3XP$, while
$ \left(a_0 \equal{} 1,\ \ \ \ \ \ \ \ \ \ a_{n \plus{} 2} \minus{} a_{n \plus{} 1} \equal{} a_0a_n \plus{} a_1a_{n \minus{} 1} \plus{} ... \plus{} a_na_0\right)$
is equivalent to the quadratic equation $ P^2X^2 \plus{} PX \plus{} 1 \equal{} P$. Solve the quadratic equation and (let a slave) verify that the solution satisfies the differential equation. (Without Sloane, this would be even more work: solve a differential equation.) But there must be something more behind the two equivalent recurrent equations, or was this problem meant to be solved on two pages of computations???

  darij


Here's one that wasn't written in this century:

A powerful unvanquished excellent black snake which is 80 angulas in length enters into a hole at the rate of 7 1/2 angulas in 5/14 of a day, and in the course of 1/4 of a day its tail grows 11/4 of an angula. O ornament of arithmetricians, tell me by what time this serpent enters fully into the hole?

If you've got an answer, post it in spoiler and leave it for your younger brothers and sisters. Feel free to speculate on who, when, and where this problem comes from.

Okay, I am not sure if this is correct, but:

[hide]

Rate = Work / Time

Let t=time it takes for the snake to enter the hole

Length of snake = 80 A + t*( 11/4 A / 1/4 day)

Length of snake = time it takes for it to get into hole x Rate

Lenth of snake = t*(7.5 A / 5/14 day)

80 A + t*( 11/4 A / 1/4 day) = t*(7.5 A / 5/14 day)

80 + t*11 = 21*t

80 = 10*t

8 = t

So after 8 days (?!) it will have gotten into the hole.

[/hide]

As for the origin of the question:

[hide]Angulas probably has something to do w/fish, since if you are an "angler" you like to fish.  I take (took) French, so I don't know any Spanish, but that's what this probably is.  Since it has to do with fish, I am going to guess it was from a place where there is water, maybe around the mediterranean sea.  Probably from an ancient but important mathematician like Euclid.  Just a guess.[/hide]

Julie:

As for the answer:

[hide]That is correct.[/hide]

As for the origins:

[hide]It's not European. And it's most certainly not Euclid. Euclid would never have used language like that, and it's not his kind of problem.[/hide]

Here's my try at the solution....

[hide]

First off, I just simplify all the important notes ...so the snake would be going in the hole at a rate of 15/2 angulas in 5/14 of a day.  Each 1/4 of a day, 11/4 angulas grow.  Sanke is 80 angulas long.

In one day, the snake would go into the hole [tex](15/2)*(14/5)[/tex] 21 angulas.  But since it grows [tex](11/4)* (4)[/tex] 11 angulas each day, it means that the snake only enters into the hole 10 angulas per day.

Since the snake is 80 angulas long, it would take 8 days.

[/hide]

For the second part, I got no clue heh...

I think this problem is from the same source:

"Inside a forest, a number of apes equal to the square of one-eighth of the total apes in the pack are playing noisy games. The remaining twelve apes, who are of a more serious disposition, are on a nearby hill and irritated by the shrieks coming from the forest. What is the total number of apes in the pack?"

Yes, tokenadult, that's the same source. Of course, the ape problem, being a quadratic equation, does require at least some algebra. I like solutions of the black snake problem, like kool_dudy's, that use no algebraic notation. One thing about the ape problem: the mathematical tradition from which this sprang was never all that concerned about solutions being unique.

Since tokenadult is probably no likelier to tell you where this came from than I am, where is that? (Hints in another day or two.)

I'll try tokenadult's problem:

[hide]

Let t be the total number of apes.  (t/8)62 = (1/64)t^2 apes are playing games, and twelve apes are on the hill.  Thus (1/64)t^2 -t + 12 = 0.  Multiplying both sides by 64, t^2 - 64t + 768.  From the quadratic formula, the two solutions are 48 and 16.

[/hide]

As for where and when this was made, I have no clue...

I will venture a guess as to the national origin of the questions:

[hide]I would expect that they are Egyptian.[/hide]

nr1337 is correct. As for the origins - not Egypt. Where in Egypt are you going to find apes (monkeys?) in a forest?

Here's one last problem, and this time there are some more specific cultural hints:

Arjuna, exasperated in combat, shot a quiver of arrows to slay Carna. With half his arrow he parried those of his antagonist; with four times the square root of the quiverful he killed his horse; with six arrows he slew Salya (Carna's charioteer); with three he demolished the umbrella, standard, and bow; and with one he cut off the head of the foe. How many were the arrows which Arjuna let fly?

Kent Merryfield wrote:Feel free to speculate on who, when, and where this problem comes from.

A guess: [hide]Brahmagupta?

Although Euclid did pose problems somewhat similar to these. I actually posted one not too long ago. But what's interesting is that the Greeks solved most algebraic problems geometrically, because they did not have an efficient number system or the concept of abstract numbers that characterized the development of algebra in the East.[/hide]

gauss202: Good guess. Right nationality, but a few hundred years too early.

Since I'm a googler...I found this...

[hide] (India, ca. 1150)[/hide]

...which is the date of the problem

kool_dudy said "(India, ca. 1150)", which is correct. These problems are taken from the  [i]Lilivati[/i] written by the Indian mathematician Bhaskara, also known as Bhaskara II (to disitinguish him from an earlier Bhaskara) or Bhaskarachaya.

Everyone with an interest in the history of mathematics should know about the St. Andrews University website:

[url]http://www-history.mcs.st-andrews.ac.uk/history/[/url]

There you can read about Bhaskarachaya, including the charming (if probably false) legend of how the [i]Lilivati[/i] came to be written.


I found the second problem of the third round very interesting.
I have managed to find a better upper bound.
I just wondering if it is know which is the best possible value for this, as well as for the general case of having m integers instead of 16.

Thanks

I think you should not discuss such things during the round.

I don't uncover any inforamation needed for the solution, i just want to know if this problem has been solved in general.

DO NOT discuss about solutions of the problems during the competition!


Let have three number $ a,b,c \in (1,+\infty), a\not=c $. Discuss the equation:

\[
\frac{{\log \left( {ax} \right)}}{{\log \left( {bx} \right)}} = \frac{{\log \left( {bx} \right)}}{{\log \left( {cx} \right)}}
\]

Let $\alpha = \log(a), \beta = \log(b), \gamma = \log(c)$. Let $X = \log(x).$
Then the equation reduces to
$\alpha \gamma - \beta^2 +(\alpha + \gamma - 2\beta) x = 0$.

Since $\alpha \neq \gamma$ it follows that $\alpha \gamma - \beta^2$ and $ \alpha + \gamma - 2\beta$ cannot be both $0$.
Thus :
If $ac = b^2$ then $\alpha + \gamma - 2\beta= 0$ and $\alpha \gamma - \beta^2 \neq 0$ so that the eqation has no solution.
If $ac \neq b^2$ then the equation has exactly one solution.

Pierre.

The solution is with  $b^2\not=ac \wedge x\not= 1/b \wedge x\not= 1/c$
\[
x=10^{\displaystyle\frac{log^2 b-log a\cdot log c}{log a+log c-2log b}}
\]


my guess is:

for all reals $ {a_{1}},{a_{2}}... ,{a_{n}}$ we have the identity:

$ (\sum_{1\leq i\leq n}a_{i})(\sum_{1\leq i\leq n}a_{i}^{2}\minus{}\sum_{1\leq i < j\leq n}a_{i}a_{j}) \equal{}\sum_{1\leq i\leq n}a_{i}^{3}\minus{}3\sum_{1\leq i < j < k\leq n}a_{i}a_{j}a_{k}$

for $ n \equal{} 1$ we have the simple equality: $ {a}(a^{2}) \equal{}{a^{3}}$
for $ n \equal{} 2$ we get another well-known identity: $ (a\plus{}b)(a^{2}\plus{}b^{2}\minus{}ab) \equal{} a^{3}\plus{}b^{3}$
for $ n \equal{} 3$ we get Euler's Identity itself: $ (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}ac\minus{}bc) \equal{} a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc$
and for $ n \equal{} 4$ we get this identity which can be easily proven:
$ (a\plus{}b\plus{}c\plus{}d)(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\minus{}ab\minus{}ac\minus{}ad\minus{}bc\minus{}bd\minus{}cd) \equal{}$
$ a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}d^{3}\minus{}3(abc\plus{}abd\plus{}acd\plus{}bcd)$

...

is it true for $ {n}$ in general?


factorial
number theory proposed
number theory

Prove that no number of the form $10^{-n}$, $n\geq 1,$ can be represented as the sum of reciprocals of factorials of different positive integers.

It is obviosly, because $\sum_{i}\frac{1}{a_{i}!}=\frac{P}{a_{n}!}$, were $a_{n}$ - maximal of them and $(P,a_{n}!)=1.$

It's hardly so obvious: 1/2+1/6=2/3, so your main lemma is wrong :)


Find all possible primes so that (5^p) + (4p^4) is a perfect square.

Shouldn't be too hard for number theory admirers like myself.

Good Luck!  :)

Email me the answer to marcospertierra@hotmail.com.

[hide]
$ 5^p \plus{} 4p^4 \equal{} k^2 \Longrightarrow 5^p \equal{} (k\minus{}2p^2)(k\plus{}2p^2)$, so let $ k\plus{}2p^2\equal{}5^s,\ k\minus{}2p^2 \equal{} 5^r$. Subtracting, $ 4p^2 \equal{} 5^s\minus{}5^r \equal{} 5^{r} \cdot (5^{s\minus{}r}\minus{}1)$. Thus $ r \equal{} 0$, or otherwise $ 5|p \Longrightarrow p \equal{} \boxed{5}, k \equal{} 75$. So we have $ 4p^2 \equal{} 5^{p}\minus{}1 \Longrightarrow p\equal{}0,1$.
[/hide]

A similar problem:Find all solution triplet to the equation-
4^x+4^y=z^2.

[hide]wlog let y=x+a for some nonnegative a. Thus:

$ 4^x (1\plus{}4^a) \equal{} z^2$

We can break solutions into families based on their $ a$ values, since $ 4^x\equal{}(2^x)^2$ is already a square number. It then suffices to solve the equation:

$ 1\plus{}4^a \equal{} z^2$
$ z^2\minus{}(2^a)^2\equal{}1$

Thus, the difference of the squares of two integers is $ 1$. Thus, we must have $ z^2\equal{}1$ and $ 4^a \equal{} 0$. However, the latter condition has no solutions, so the original equation has no solutions.[/hide]

Sorry I made a mistake.The equation is:
1+4^x+4^y=z^2

I'll try.

[hide]$ 4^x\plus{}4^y\equal{}z^2\minus{}1 \rightarrow 4^x\plus{}4^y\equal{}(z\plus{}1)(z\minus{}1)$.  Considering $ \pmod {4}$, $ 4^x\plus{}4^x \equiv 3 \pmod 4$ if $ z$ is even.  But the only odd sum with $ 4^x\plus{}4^y \equiv 1 \pmod 4$, so z is odd.  Noticing that the product for any two consecutive even integers plus one is a perfect square, the only problem is which products can be made with powers of 4.  If $ x>1, y>1$, then the expression is equivalent to $ 4^n(4^{x\minus{}n}\plus{}4^{y\minus{}n})$.  $ 4^{x\minus{}n}\plus{}4^{y\minus{}n} \pm 2 \ne 4^n$ because powers of four grow too quickly, as seen in $ 4^1> 2$, and using $ 4^0$ once will make the sum odd.  The only solution occurs when $ n\equal{}0$, so $ x\equal{}1, y\equal{}1, z\equal{}3$.[/hide]

[quote="azjps"][hide]
$ 5^p \plus{} 4p^4 \equal{} k^2 \Longrightarrow 5^p \equal{} (k \minus{} 2p^2)(k \plus{} 2p^2)$, so let $ k \plus{} 2p^2 \equal{} 5^s,\ k \minus{} 2p^2 \equal{} 5^r$. Subtracting, $ 4p^2 \equal{} 5^s \minus{} 5^r \equal{} 5^{r} \cdot (5^{s \minus{} r} \minus{} 1)$. Thus $ r \equal{} 0$, or otherwise $ 5|p \Longrightarrow p \equal{} \boxed{5}, k \equal{} 75$. So we have $ 4p^2 \equal{} 5^{p} \minus{} 1 \Longrightarrow p \equal{} 0,1$.
[/hide][/quote]

sorry. the correct answer is p=5. besides, 0 and 1 are not prime.

[hide="Answer"](x,y,z)=(1k,2k-1,2^(2k-1)+1)
Actually this was one of our IMO TST problems this year.
[/hide]


Find all functions $ f: \mathbb{R}^{\plus{}}\rightarrow \mathbb{R}$ that satisfy:

$ f(x)f(y)\equal{}f(xy)\plus{}2005 \left( \frac{1}{x}\plus{}\frac{1}{y}\plus{}2004 \right)$ for all $ x,y>0.$

[quote="moldovan"]Find all functions $ f: \mathbb{R}^{ \plus{} }\rightarrow \mathbb{R}$ that satisfy:

$ f(x)f(y) \equal{} f(xy) \plus{} 2005 \left( \frac {1}{x} \plus{} \frac {1}{y} \plus{} 2004 \right)$ for all $ x,y > 0.$[/quote]

$ y\equal{}1$ $ \implies$ $ f(x)(f(1)\minus{}1)\equal{}2005(2005\plus{}\frac 1x)$. So $ f(1)\neq 1$ and $ f(x)\equal{}a(2005\plus{}\frac 1x)$

Plugging back in the original equation, we get $ a\equal{}1$ and the unique solution $ \boxed{f(x)\equal{}2005\plus{}\frac 1x}$


Take 3 lines $ l$, $ l_{1}$ and $ l_{2}$ in general position in the space. Consider a following map $ F$ from the line $ l_{1}$ to the line $ l_{2}$. By definition for each point $ A$ (which belongs to $ l_{1}$), the point $ F(A)$ is the point of intersection of the line $ l_{2}$ with the plane passing through the line $ l$ and the point $ A$. Prove that the map $ F$ is projective, i.e. $ F$ preserves the cross ratio of each 4 points.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=266658[/url]


In the attatched diagram, $PR=1, \ PS=2, \ PQ=4, \text{ and }\angle QPT = 45^\circ$. What is the length of $QT$?

(Edited diagram--removed angle measure so that more problems can be made from it)

[quote="i_like_pie"][quote="cincodemayo5590"]In the attatched diagram, $PR=1, \ PS=2, \ PQ=4, \text{ and }\angle QPT = 45^\circ$. What is the length of $QT$?[/quote]
Are you sure $2PR=PS?$ They look pretty equal.[/quote]

The diagram isnt to scale  :P

And actually.. the original problem is pretty trivial and can be solved easily without my original intended idea. You can solve the first one if you want, my second (harder, i think) question is in post #4.

[hide="Solution"]By power of a point, $PR\cdot PQ=PS\cdot PT\implies PT=\frac{PR\cdot PQ}{PS}=\frac{1\cdot 9}{2}={9\over 2}$. Now apply Cosine Law to $\triangle PQT$: $QT=\sqrt{9^{2}+\left({9\over 2}\right)^{2}-2\cdot 9\cdot{9\over 2}\cdot{\sqrt{2}\over 2}}={9\over 2}\sqrt{5-2\sqrt{2}}$[/hide]

[u]A Harder Problem[/u]:

In the diagram, $PR=1, \ PS=2, \ QR=7$, and the segment $PT$ does not extend into the sector defined by chord $QR$. If $QT$ is equal to the average of $PQ$ and $PT$, find the length of $QT$.

($\angle QPT$ is [u]not[/u] given in this problem like it was in the original)


hi, i have solved the last problem from my last post.

I have another one:

Question:
Customer brings $2 to bank for change into 20, 10 and 5 cents coins in such a way that the number of 10 cents coins is two thirds the number of 20 cent coins and the number of 5 cent coins is two more than the number of 20 cent coins. How many coins of each type should the teller give this client?

I let X be number of 20 cents coins.

this is my equation but its wrong??

(10 * 2/3X) + ((X+2) * 5) = 200

Anyone please help me

thank you

your equation was
    (10 * 2/3X) + ((X+2) * 5) = 200
you added the value of 10 cent and 5 cent coins but by mistake didnt put the value of 20 cent coins itself.
  equation would be
    (10 * 2/3X) + ((X+2) * 5) +20X = 200
SOLVING IT, THIS WOULD GIVE THE ANSWER AS
20CENTS=6,
10 CENTS=4
5CENTS=8  :roll:

oh i see i was wondering where to put the 20 cents in.

thank you i got the ansswer now


geometry
3D geometry
sphere
calculus
integration
perimeter
rotation

i am asking why does the integral from -r to r of 2:pi: :sqrt:( r 2 -t  2  )dt isn't equal to the surface of a sphere.
 r is the ray of the sphere.
  2 :pi:  (:sqrt:r 2 -t 2 ) is the perimeter of a cercle whose ray is (:sqrt:r 2 -t 2.
  I AM NOT ASKING TO CALCULATE THIS INTEGRAL.BECAUSE I AM SURE THAT IT ISN'T EQUAL TO 4 :pi: R 2 . SO WHERE IS THE ERROR???/ ;)

might  wanna post this in the Calculus Forum ;)
and do you want to prove the integral??

[quote="JBL"]Moderator should bump this up to a calculus forum.  I'm not at all clear what question the "why" is supposed to be asking, though.[/quote]

uh, yeah...I would if I could.  But an admin has yet to respond to my thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37321]here[/url] :?

A better reason: consider your integral.  You are approximating a sphere with small cylinders of radius $\sqrt{r^2 - t^2}$ and height $dr$.  Near $t = 0$, this reasonably approximates the shape of the sphere.  However, near $t = r$ or $t = -r$, your surface of approximation isn't even remotely close in shape to the surface of the sphere -- indeed, they will be nearly perpendicular to each other.  So the integral is calculating a limit of some surface that doesn't actually approach the surface of the sphere.

For an integral that works, try calculating the surface area as that of a rotated circle.

[quote="Bictor717"]Ummm... because that's not the surface area formula?
$A=2\pi\int^b_a f(x)\sqrt{1+(f'(x))^2}\;dx$[/quote]

and f(x) is what???
do you think that the integrale you gave is the surface of a sphere...
i don't see any ray or.....
try to explain clearly.
thanks

[quote="Bictor717"]Ummm... because that's not the surface area formula?
$A=2\pi\int^b_a f(x)\sqrt{1+(f'(x))^2}\;dx$[/quote]

is there someone who will clarify to me the meaning of this integral.f(x) is the function of what?????/
other question
the stars under your usernames .what do they mean and how to get them.
thanks.

Let $y=f(x)$ have a continous derivative on the interval $[a,b]$. the area $S$ of the surface of revolution formed by revolving the graph of $f$ about a horizantal or vertical axis is:
$S=2\pi\int_{a}^{b}r(x)\sqrt{1+(f'(x))^2} dx$
where $r(x)$ is the distance between the grph of $f$ and the axis of revolution...
the stars under the user name, is your rank.
the more posts you have, the more stars you get.
i think you'll get the first one around your 20th post...


In a cube $ABCD-A_1B_1C_1D_1$, $N$ is the midpoint of $AB$, $M$ is the midpoint of $B_1B$, $O$ is the intersection of $BC_1$ and $B_1C$.

Find a line through $O$ and intersets both $MA$ and $CN$ and prove that there exists only one of this kind of line.

Let $\alpha$ be the plane $NB_1C$. Clearly $O$ and $CN$ are in $\alpha$. $\alpha$ cuts $AN$ at $K$. In $\Delta ABB_1$ $K$ is medicenter so $\frac{B_1K}{KN}=2$ as $O$ is the midpoint of $B_1C$ $OK$ is not parallel to $CN$ so $OK$ is the line.
Assuming a different line cuts AM and CN leads to AM and CN are complanar.


function
real analysis
real analysis solved

Here's a cute one :).

Give an example of a continuous surjection from the real line to the unit square, such that the preimage of each point has continuum cardinality.

I know you guys will solve it fast :), but I was wondering how many nice examples we would see.

[quote="Moubinool"]Peano curve[/quote]

 There is a simpler example. The construction looks very much alike the Weierstrass function. :)

There are examples even simpler, I think. When I first saw this, I considered the image of a real in $(0,1)$ with binary expansion $0.b_1b_2\ldots$ to be $(0.b_1b_4b_7\ldots,0.b_2b_5b_8\ldots)$. In order to avoid the cases when we have two binary representations, make a convention; for instance, always take the expansion with infinitely many $1$'s in the end.

Does it work?

Looks like it is, since if a sequence tends to a number, then the terms of the sequence tend to that number in their binary expansion (in the mentioned conditions, i.e. if we choose a convention for representing the numbers with more thn one binary expansion).

[quote="grobber"]There are examples even simpler, I think. When I first saw this, I considered the image of a real in $(0,1)$ with binary expansion $0.b_1b_2\ldots$ to be $(0.b_1b_4b_7\ldots,0.b_2b_5b_8\ldots)$. In order to avoid the cases when we have two binary representations, make a convention; for instance, always take the expansion with infinitely many $1$'s in the end.[/quote]

Consider the following convergent sequence and its limit: $0,00100...0001111.. \to 0,0001111....$.
We see that $0,00100...0001111..$ maps to $(0,00...0111...;0,00...0111...)\to (0,0)$ and limit points maps to $0,00111...;0,00111...)\neq (0,0)$ (something like that).

It was in my head from the beginning, but it have came outside only now.

Maybe I am missing something as always :(

I see. It seems like that thing about the convention for numbers with two representations is deemed not to work. Sorry about that. :blush: I don't know if I can find a simple example.

P.S. You're not the one who's always missing something here, Myth, I am  ;)

Ok, I figured we could do this:

The Peano curve springs into mind immediately when you see this, but I had a problem with it: it was not clear to me (maybe it should have, but it wasn't :?) that the Peano curve takes each point of the square as a value $c$ times ($c$ is the continuum cardinality). That's why I modified that a bit.

Let $I=[0,1]$. Assume we have a continuous function $f:I\to I$ which takes every value of $I$ $c$ times. Assume then that $\phi:I\to I^2$ is a continuous surjection. Then $\phi\circ f:I\to I^2$ is a function satisfying our requirements.

Now, for $\phi$ take the Peano curve, and for $f$ take the projection of the Peano curve on $[0,1]$, i.e. $f=\pi\circ\phi$, where $\pi:I^2\to I$ is the projection on the $x$ axis.
Dos it work now?


hi, i cant solve it... is a peg game...

[img]http://img301.imageshack.us/img301/7356/peggame7tu.th.jpg[/img]

pls help me... thx for all.
bye.

Is the rule that pegs can jump over one another, and that the jumped over peg is removed and the goal is to have one peg left?


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $ A$ and $ B$ be two symmetric positive definite matrices in $ \mathcal M_n\left(\mathbb R\right)$. Prove that for each $ m\in \mathbb N$, the polynomial $ g\left(t\right) \equal{} \text{tr}\left(A \plus{} Bt\right)^m$ is a polynomial with non-negative coefficients.

Can we use the fact that there exists a basis in which both $ A,B$ are diagonal? Then $ A\plus{}Bt\equal{}TT^{\minus{}1}(A\plus{}Bt)TT^{\minus{}1}\equal{}T(D_1\plus{}D_2t)T^{\minus{}1}$,
 $ (A\plus{}Bt)^m\equal{}T^{\minus{}1}(D_1\plus{}D_2t)^m T$.

[quote="berti"][quote="Lippilus"]Can we use the fact that there exists a basis in which both $ A,B$ are diagonal?[/quote]So $ A,B$ would commute. I think this is not true in general.[/quote]
I think it is true.Because you may assume:
$ TAT^{\minus{}1}\equal{}E_n$,
And then consider $ TAT^{\minus{}1},TBT^{\minus{}1}$.
To find an orthangal matrix such that:
$ OTBT^{\minus{}1}O^{\minus{}1}$ is diagonal.

The Lippilus 's result is absolutely false as Berti said. The Zhaobin's formula is false; the exact result is :$ \exists{T}$ an invertible matrix s.t. $ TAT^T\equal{}I$.

Clearly $ g$ is a polynomial of degree $ \leq{m}$. The $ g$-coefficients are positive; to prove this fact we show that $ \forall{k\leq{m}},g^{(k)}(0)>0.$
For example let $ m\equal{}3$: $ g'(t)\equal{}tr(B(A\plus{}Bt)^2\plus{}(A\plus{}Bt)B(A\plus{}Bt)\plus{}(A\plus{}Bt)^2B)\equal{}3tr((A\plus{}Bt)^2B)$ and $ g'(0)\equal{}3tr(A^2B)$. More generally $ g^{(k)}(0)\equal{}\lambda{tr(A^{m\minus{}k}B^k)}$ where $ \lambda>0$. Now it's sufficient to prove that if $ C,D$ are positive definite symmetric matrices then $ tr(CD)>0$. We may assume that $ C\equal{}diag(c_1,\cdots,c_n)$; then $ tr(CD)\equal{}\sum{c_id_{i,i}}$ and we conclude with $ c_i>0,d_{i,i}>0$.

I have written too quickly my last post. My proof is uncorrect.Let $ m\equal{}4$: $ g'(t)\equal{}4tr(A\plus{}Bt)^3B)$. $ g''(t)/4\equal{}tr(B(A\plus{}Bt)^2B\plus{}(A\plus{}Bt)B(A\plus{}Bt)B\plus{}(A\plus{}Bt)^2BB)\equal{}2tr((A\plus{}Bt)^2B^2)\plus{}tr((A\plus{}Bt)B(A\plus{}BT)B)$ and $ g''(0)\equal{}2tr(A^2B^2)\plus{}tr(ABAB)$.
Thus the formula giving $ g^{(k)}(0)$ is false. In fact $ g^{(k)}(0)$ is a sum of traces of matrices in the form $ C_1,\cdots,{C_m}$ where we find in the $ C_i$: $ m\minus{}k$ $ A$-copies and $ k$ $ B$-copies.

If $ A_1,\cdots,A_k$ are definite positive symmetric matrices then $ trace(A_1\cdots{A_k})>0$ can be false. Yet experiments with computer show that, for our matrices $ A,B$, $ trace(A^uB^vA^wB^t\cdots)>0$ is true. I havn't any proof of this result but if it's true, then the Hatami's problem is solved.

Hmm, guys, are you sure this problem is solvable? Just try to type Bessis-Moussa-Villani conjecture on google and you will see why I have some doubts.

Thanks Harazi.
In fact $ trace(ABABBA)$ can be negative!!

The problem in the exam is for $ n \equal{} 2$. I thought it may be correct for larger $ n$.


algebra
polynomial
function
calculus
Ross Mathematics Program
real analysis
abstract algebra

I) For $ f : [0,1] \to R $, let  $ B_n (f,x) $ be the Bernstein polynomial of order n of the function f, defined by
\[
B_n (f,x)=\displaystyle \sum_{k=0}^{n} \displaystyle{n\choose k} f \left (\frac {k} {n}\right )x^k (1-x)^{n-k}
\]
Prove that if f is continuous on [0,1], then $ \{B_n(f)\} $ converges unifomly on [0,1] to f.

II) Use the risult of point I) to prove the approximation theorem of Weierstrass. If
$ f : [a,b] \to R $ is continous in [a,b], then for every $ \epsilon >0 $ there is a polynomial P such that
\[
|f(x)-P(x)|< \epsilon \ ; \ \ \ \ \forall x \in [a,b]
\]

Very nice theorem. I found a proof in my probabiliy book. Just note that

$p_k =  \displaystyle{n\choose k}x^k (1-x)^{n-k}$ is the binomal distrobution.

fyi, another proof: http://planetmath.org/?op=getobj&from=objects&id=6145

p/s: how to make hidden content in this forum?

I'm currently using Ross's textbook for a course I'm teaching:

[i]Elementary Analysis: the Theory of Calculus[/i] by Kenneth A. Ross, in the Springer UTM series.

Ross devotes an entire section to exactly this proof of the Weierstrass approximation theorem.

Fourier analyst that I am, I have a different favorite proof, but this is a nice one.

[quote="Kent Merryfield"]Fourier analyst that I am, I have a different favorite proof[of Weierstrass approximation theorem][/quote]

What Fourier proof do you have in mind?

I know the following "soft" argument suitable for quickly convincing people that the result is true.
1. It suffices to approximate polygonal lines (piecewise linear functions).
2. To do that, it suffices to approximate $|x|$ on some interval containing 0
3. Truncate the Taylor series of $|x| = \sqrt{1 - (1-x^2)}$ to get a polynomial approximation.

The Bernstein polynomial approximation is very slowly converging, but represents a positive function as a sum of positive polynomials.

[quote="fleeting_guest"]What Fourier proof do you have in mind? [/quote]
Focusing first not on ordinary polynomials but trigonometric polynomials.

If $f$ is continuous and periodic, then the Cesaro means of the Fourier series for $f$ converge uniformly to $f$. We prove this directly and constructively, exhibiting the Cesaro means as convolution of $f$ with the Fejer kernel, which is a "nice" positive approximation to the identity. Since the Cesaro means of a Fourier series are trigonometric polynomials, we have that a continuous periodic functions can be approximated uniformly arbitrarily closely by trigonometric polynominals.

If you start with a general function continuous on an closed bounded interval, you could just splice in a little connector piece to make it continuous and periodic with period slightly larger than the original interval. To get approximation by ordinary polynomials, just note that any trigonometric polynomial, which is a finite linear combination of exponentials, has a power series with an infinite radius of convergence. Hence the partial sums of that power series converge uniformly on any bounded set.

[quote="Kent Merryfield"]
If $f$ is continuous and periodic, then the Cesaro means of the Fourier series for $f$ converge uniformly to $f$. We prove this directly and constructively, exhibiting the Cesaro means as convolution of $f$ with the Fejer kernel, which is a "nice" positive approximation to the identity. Since the Cesaro means of a Fourier series are trigonometric polynomials, we have that a continuous periodic functions can be approximated uniformly arbitrarily closely by trigonometric polynominals. [/quote]

That reminds me of another way.  Convolution with normalized (unit mass) Gaussians $e^{-cx^2}$ will converge to $f$, and one can truncate the power series for $e^x$ to do the same with a polynomial.   Presumably this also has a probabilistic interpretation (heat kernel at time $t \to 0$) and so maybe a relation to the Bernstein polynomials.


:D Congaratulations [b]khashayar jan [/b]however the result was terrible for Iran. :(  :(

I'm so sorry. I didnt want to post a new topic. what should I do now?  :blush:  :oops:

[quote="sisioyus"]may i ask a question?
it is about that:
could girls in iran participate in mathematical competitions?
if girls was selected to participate in imo, could she show her face to others?[/quote]
 :maybe: Yes.They can participate in mathematical competitions, but with Hejab, that means that according to the religion and the law they can show their facec but they have to cover their hairs and other parts of their bodies.There was only 3 girls from Iran who participate in IMO,I think! But the reason of this matter is it:The highschools for the boys are better than the highschools for the girls in Iran! just this!
Do you know Maryam Mirzakhani? May be she is the best iranian scientist! She was one of that above three girls!!!
excuseme for my poor English!!!
Bye :)

[quote="boxedexe"]There seems to be a lot of misunderstanding about Iran; Iran is very different from the other countries in the middle east.  Regarding Maryam Mirzakhani, I think (I hope I am not making a mistake) that she has been (or used to be) the best IMO performer in Iran's history.  About Iran's performance this year, I can only say that I hope Iran performers better in the coming years.[/quote]
 :lol: You're right!


number theory
least common multiple
inequalities
number theory proposed

Let  be a positive integer with . Prove that $ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$

is not an integer

Solution:
Let $ 2^{k}\parallel n!$ and $ 2^{l}\parallel \sum_{i=2}^{n}(\frac{n!}{i})$,so i want to prove that $ k>l$.It is easy to understand,because if we denote $ 2^{k-k_{i}}\parallel \frac{n!}{i}$ then $ l=min\{k-k_{i},i=2,3\dots n\}$.But for even $ i$ we have $ k_{i}>0$.So $ k>l$. :lol:

A solution for why for $ m,n\in\mathbb{N}$, $ \frac{1}{n}\plus{}\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}m}$ cant be an integer:

each positive integer can be written in the form: $ 2^{r}(2t\plus{}1)$ where $ r,t\geq0$, now if we write each of the numbers of the set $ \left\{n,n\plus{}1,\cdots,n\plus{}m\right\}\equal{}\left\{2^{r_{1}}(2t_{1}\plus{}1),\cdots,2^{r_{m\plus{}1}}(2t_{m\plus{}1}\plus{}1)\right\}$, first we'll prove that only one of the $ r_{i}$'s has maximum between all $ r_{i}$'s. For the sake of contradiction lets assume there exist $ k$ and $ j$ that $ k>j$ and $ r_{k}\equal{}r_{j}$ are both maximum value between $ r_{i}$'s. So we must have: $ 2^{r_{j}}>m\minus{}j\plus{}1$ (because if not then $ n\plus{}j\minus{}1\plus{}2^{r_{j}}\equal{}2^{r_{j}}(2t_{j}\plus{}2)$ will be divisible by $ 2^{r_{j}\plus{}1}$ and $ n\plus{}m\geq n\plus{}j\minus{}1\plus{}2^{r_{j}}$ which is in contradiction with the maxmumity of $ r_{j}$ :D ), but:
$ 2^{r_{j}}|n\plus{}k\minus{}1\minus{}(n\plus{}j\minus{}1)\equal{}k\minus{}j$
as a result $ k\minus{}j\geq2^{r_{j}}$ and $ m\minus{}j\geq k\minus{}j$ as a consequence $ m\minus{}j\geq2^{r_{j}}$ but we had $ 2^{r_{j}}>m\minus{}j\plus{}1$ and this is contradiction, this shows that only one of the $ r_{i}$'s can have the maximum value between all of $ r_{i}$'s!
Now let $ r_{j}$ be the maximum between them. Let $ B\equal{}\textbf{lcm}\left[n,n\plus{}1,\cdots,n\plus{}m\right]$, so:

$ A\equal{}\frac{1}{n}\plus{}\frac{1}{n\plus{}1}\plus{}\cdots\plus{}\frac{1}{n\plus{}m}\equal{}\frac{a_{0}\plus{}a_{1}\plus{}\cdots\plus{}a_{m}}{B}$ where $ a_{i}\equal{}\frac{B}{n\plus{}i}\;\;\;(0\leq i\leq m)$. now its obvious that $ 2^{r_{j}}\parallel B$. as a consequence $ a_{i}\;(i\neq j)$ is even because $ r_{i}<r_{j}$ but $ a_{j}$ is odd, so $ a_{0}\plus{}a_{1}\plus{}\cdots\plus{}a_{m}$ is odd but $ B$ is even, and this fact says that $ A$ cant be an integer! :)

Another similar Problem:

Prove that $ \frac{1}{3}\plus{}\frac{1}{5}\plus{}\cdots\plus{}\frac{1}{2n\plus{}1}$ cant be an integer!

[quote="sinajackson"]
Prove that $ \frac{1}{3}\plus{}\frac{1}{5}\plus{}\cdots\plus{}\frac{1}{2n\plus{}1}$ cant be an integer![/quote]
Obviosly. use that exist prime $ n\plus{}1\le P\le 2n\plus{}1$.

[quote="sinajackson"]A solution for why for $ m,n\in\mathbb{N}$, $ \frac{1}{n}\plus{}\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}m}$ cant be an integer[/quote]well let $ S \equal{}\frac{1}{n}\plus{}\frac{1}{n\plus{}1}\plus{}...\plus{}\frac{1}{n\plus{}m}$ then we have:
$ \frac{m\plus{}1}{m\plus{}n}\leq S\leq\frac{m\plus{}1}{n}$
now assume the contrary,assume that $ S\in\mathbb{N}$ then we must have:
$ 1 <\frac{m\plus{}1}{n}\minus{}\frac{m\plus{}1}{m\plus{}n}$
$ \Rightarrow mn\plus{}n^{2}< m^{2}\plus{}m$
now if $ m < n$ then we have $ m^{2}\plus{}m^{2}< m^{2}\plus{}m\rightarrow m < 1$ which is a contradiction...
so $ n\leq m$ thus $ 2n\leq m\plus{}n$,thus there exists a prime number $ p$ between numbers $ n,m\plus{}n$ now note that:
$ S\in\mathbb{N}\Rightarrow\frac{(n\plus{}1)...(m\plus{}n)\plus{}...}{n(n\plus{}1)...(m\plus{}n)}\in\mathbb{N}$
$ \Rightarrow n\mid (n\plus{}1)...(m\plus{}n)$
now let $ p_{0}$ be the greatest prime number between $ n,m\plus{}n$ we have:
$ p_{0}\mid\frac{n(n\plus{}1)...(m\plus{}n)}{p_{0}}$
now we claim that $ 2p_{0}\not\in\{n,n\plus{}1,...,m\plus{}n\}$,because in this case we get that there exists a prime number $ q$ s.t $ p_{0}< q < 2p_{0}$ which contradicts the choice of $ p_{0}$...
so $ 2p_{0}\not\in\{n,n\plus{}1,...,m\plus{}n\}$
hence $ p_{0}\not\mid\frac{n(n\plus{}1)...(m\plus{}n)}{p_{0}}$ which is a contradiction...
so $ S\not\in\mathbb{N}$...

[quote="othman"]a related problem but really hard is :for all integers $ a_{n\ge 0}$ in arithmetical progession (ie:$ a_{n}\minus{}a_{n\plus{}1}$ is a constant ) we have the same result  :D[/quote]

Let $ a_{0}\equal{} a,\ a_{i\plus{}1}\minus{}a_{i}\equal{} d$. We can WLOG assume that $ (a,d) \equal{} 1$. Now $ \frac{1}{a_{1}}\plus{}\frac{1}{a_{2}}\plus{}\dots\plus{}\frac{1}{a_{n}}\equal{}\frac{a_{2}a_{3}\cdots a_{n}\plus{}a_{1}a_{3}\cdots a_{n}\plus{}\dots\plus{}a_{1}a_{2}\cdots a_{n\minus{}1}}{a_{1}a_{2}\cdots a_{n}}$. This implies $ a_{1}\mid a_{2}a_{3}\cdots a_{n},\ a_{2}\mid a_{1}a_{3}\cdots a_{n},\ a_{3}\mid a_{1}a_{2}a_{4}\cdots a_{n},\dots$ and so on. Thus $ a_{i}^{2}\mid P$ for $ 1\le i\le n$ where $ P \equal{} a_{1}a_{2}\cdots a_{n}$. So $ [a_{1}^{2},a_{2}^{2},\dots,a_{n}^{2}]\mid P$.

Now let $ p$ be a prime such that $ 1\le p\le n$ and let $ g \equal{} (a_{k},a_{k\plus{}p})$. We will prove that $ g \equal{} 1$. As $ g\mid a_{k\plus{}p}\minus{}a_{k}\equal{} dp$, if $ g \equal{} m p$,($ m\mid d$) we get $ p\mid g\mid a_{k}\equal{} a\plus{}kd\forall k\Rightarrow p\mid a$. Thus $ p \equal{} (a_{k},a_{k\plus{}p})$ for all $ k$ implies $ p\mid (a_{k},a_{k\plus{}1})$. But $ (a_{k},a_{k\plus{}1})\mid a_{k\plus{}1}\minus{}a_{k}\equal{} d$. So $ p\mid d$, which is a contradiction because $ (a,d) \equal{} 1$.

Now let $ g\mid d$. But this is not possible either because then by the same reasoning we get $ g\mid a$. Thus $ g \equal{} 1$.

To be continued...

[quote="TTsphn"]The first problem I thinks can use chebyshev law :There exist a prime number between n and 2n[/quote]
if you read the solutions carefully you can see that they use it too...  :wink:

My teacher say that there is 15 solution for this problem and give may student in my school but I solve only 2 solution.use or don't use Chebyshev law


function
algebra
polynomial
advanced fields
advanced fields unsolved

If $X = V(x_1x_4 - x_2x_3) \subset \mathbf{A}_\mathbf{C}^4$ and $U = \{(x_1,x_2,x_3,x_4) \in X: x_2x_4 \neq 0\}$, then $x_1/x_2$ if $x_2 \neq 0$ and $x_3/x_4$ otherwise defines a regular function on $U$.

Show that this function can't be written as a quotient of two polynomials globally (on the whole of $U$).

This is a standard exercise in the appropriate class, and you've garbled the statement. The only points we should be excluding from $U$ are those for which $x_2=x_4=0$.

Since the function is homogeneous of degree zero, we can convert any quotient expression for it into a quotient of homogeneous polynomials of equal degree.

Lemma: Every homogeneous polynomial $f$ of positive degree has a zero on $U$.
Actually, the set of zeros in $X$ includes a variety of dimension at least two, since the zeros of $f$ include the zeros of some irreducible factor, and the intersection of two varieties of dimension three in four-dimensional space has dimension at least three.

Using this, the denominator of any possible expression has zeros on $U$, and no such expression can define our regular function everywhere.

[quote]Actually, the set of zeros in X includes a variety of dimension at least two, since the zeros of f include the zeros of some irreducible factor, and the intersection of two varieties of dimension three in four-dimensional space has dimension at least three. [/quote]Could you explain this a bit or give a more elementary proof without dimension theory?


probability
college
probability and stats

What is the probability of getting a two pair hand in 5 card poker?
(answer: 4.75%)

I could not solve by counting each of the card's probabilities.

Here is my approach:
2-pair occurs at the very beginning as 1*3/51*48/50*3/49*44*48 but I reach 1/15 of the answer.

Thanks for any help/suggestions.

Finding the probability of each ranked hand in 5-card poker is a classic counting exercise. And if a student were to get all but one answer correct, the most likely missed answer would be two pair - but the usual mistake is a factor of 2 rather than a factor of 15.

The total number of poker hands is $ \binom{52}{5}\equal{}\frac{52\cdot51\cdot50\cdot49\cdot48}{1\cdot2\cdot3\cdot4\cdot5}\equal{}2598960.$

How many of those are two pair?

First pick the two denominations that will be paired out of the 13 possible denominations. We can do that in $ \binom{13}{2}\equal{}78$ ways.

Then choose which two out of four possible cards of that denomination form our pair - and do that twice, independently. We can do that in $ \binom{4}{2}\binom{4}{2}\equal{}36$ ways.

Finally, choose the card that forms the fifth card of the hand; it must be of a different denomination (since we don't have a full house), which leaves 44 possibilities. We can choose that in $ 44$ ways, which we can consider as $ \binom{44}{1}.$

The total number of two pair hands is $ 78\cdot 36\cdot 44 \equal{} 123552.$

The probability of two pair is $ \frac{123552}{2598960}\approx 0.04754.$

I don't completely understand the rationale of your explanation, but it looks like you may be trying to compute the probability of being dealt the two-pair hand [i]in a particular order[/i]. But order doesn't matter in a poker hand.

Thank you very much for the explanatory solution.
Regards

The other weekend I played about 50 hands of Texas Holdem. I never paired with anything on the flop. I saw the flop roughly 65% of the hands. What is the probability that happens? :) (this is a [b]real[/b] story)

Edit: by I never paired, I mean my highest card (with the flop) was A-high. No pairs, no straights, no nothing.

The number of hands valued at less than a pair is
\[ \binom{13}{5}\cdot 4^5\minus{}\binom{13}{5}\cdot 4\minus{}9\cdot 4^5\plus{}9\cdot 4\equal{}1303560\]
The first term is the number of ways of getting five different denominations; from that we subtract the flushes (including straight flushes), then the straights (including straight flushes), then add back in the straight flushes to compensate for the double counting.

The two hole cards plus the three cards of the flop constitute a single 5-card "hand". The probability of a bust hand is $ \frac{1303560}{2598960}\approx .50157.$ (Pretty close to a coin flip.)

The probability of, say, 32 hands in a row being "bust" hands is then $ (.50157)^{32}\approx 2.6\times 10^{\minus{}10}.$

Now we get to decide whether we [i]really[/i] believe Valentin.

[quote="orl"]Okay, this a bit off-topic. But [url=http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.bams/1183522375]here (pdf)[/url] is an interesting article of John von Neumann on simplified poker variants. [url=http://www.swiss-press.com/newsflashartikel.cfm?key=297648]John Ferguson[/url] is known for his joint work on poker with his father [url=http://www.math.ucla.edu/~tom/]Thomas Ferguson, statistics professor at UCLA.[/url][/quote]

you mean Chris Ferguson.  But the 'real' authority on mathematics in poker is Bill Chen, who has a Ph.D from Berkeley, two WSOP bracelets, and has written 'The Mathematics of Poker', which is the closest you'll come to a poker textbook anywhere.

Jaja, I meant Chris Ferguson. John von Neumann is guilty of implying this mental association!  :D I guess $ \$7.5$ millions in price money is not that bad. Maybe this post should get into the career forum.  :wink:

ironically Chris Ferguson is probably worth hundreds of millions solely because of his role in starting full tilt poker, and his career winnings (which are largely a product of variance) are a drop in the bucket compared to that.

[quote="Kent Merryfield"]The probability of, say, 32 hands in a row being "bust" hands is then $ (.50157)^{32}\approx 2.6\times 10^{ \minus{} 10}.$
Now we get to decide whether we [i]really[/i] believe Valentin.[/quote]It actually happened. I didn't want to complicate the math problem anymore, but out of those flops I either chased a few hands on the river, or bluffed them. Anyhow I never did make a pair (or more). It was literally my worst night in poker.


geometry
geometric transformation
reflection
trigonometry
ratio
circumcircle
parallelogram

Let $ ABCD$ be a convex quadrilateral such that $ AB\cdot CD\equal{}AD\cdot CB$. Let $ P$ be the reflection of $ A$ in $ BD$. Prove that $ \angle PCD\equal{}\angle ACB$.

I derive it using the [b] Ceva's Theorem (sine form) [/b].

[hide] Applying the theorem to triangle $ CBD$ with concurrent lines $ CA, BA, DA$ and $ CP, BP, DP$ and note that $ \angle{ABP} \plus{} \angle{ABC} \equal{} \angle{ADP} \plus{} \angle{ADC} \equal{} \pi$. (rewrite the hypothesis as $ \frac{BA}{BC}\equal{}\frac{DA}{DC}\equal{}k$ to get the fact that $ B, D$ lies on the [b] Appolonius circle [/b] of $ A, C$ with respect to ratio $ k$) [/hide]

[quote="mathangel"]

[hide] and note that $ \angle{ABP} \plus{} \angle{ABC} \equal{} \angle{ADP} \plus{} \angle{ADC} \equal{} \pi$.[/quote]

Why is this?

Also, any other solutions?

Please check your statement again...I think that there is a typo

Probably is better to restate the problem, in order to ask for the proof that the line segment $ CP,$ is the isogonal conjugate of $ CA,$ with rerspect to the angle $ \angle BCD$ $ ($ instead of $ \angle PCD \equal{} \angle ACB$ $ ).$

Kostas Vittas.

PS. Some other problems dear [b][size=100]Carlez Tevos[/size][/b], nice like your, are waiting for a long time to be discussed. So, give please more time.

As it was mentioned by [b]vittasko[/b],your problem is incorrectly stated. :wink:
Your statement will be correct only if we suppose $ \angle BAC\leq\frac {\pi}{2}$,so i'll solve the problem only for that case.
Proof:
Denote $ M$ as the midpoint of $ BD$ and $ E$ as the point on circumcirlce of $ ABCD$,such that $ AE\| BD$.
Then obviously $ BEDP$ is parallelogram,thus $ E,M,P$ are collinear.In other hand,we know that $ ABCD$ is harmonic and its well-known property is that diogonal is symmedian.So $ CE$ is the symmedian of $ \angle BCD$,hence $ C,M,E$ are collinear,and finally $ C,P,E$ are collinear so we are done.

C is located on the Appollonius circle of triangle $ ABD$ passing through $ A$. If $ \angle BCD > \angle BAD$, then $ \angle PCD \plus{} \angle ACB \equal{} 180^\circ$.

Best regards & A Happy New Year!
sunken rock


find all integer solution to
$a^{2}+b^{2}+c^{2}=a^{2}b^{2}$

I don't see any other solutions than $a=b=c=0$.

$a=b=c=0$ istheonly solution which you can see by looking the equation modulo 4.


inequalities
trigonometry
inequalities unsolved

WLOG suppose $ab+bc+ca=1$ , denote $a=\tan{\frac{A}{2}}$,etc...,where $A,B,C$ are angles of a triangle. We only prove $\prod\cos{\frac{A}{2}}\leq \frac{3\sqrt{3}}{8}$. This is easy problem.

[quote="chien than"]Let $a;b;c$ are positive real numbers.Prove that:
\[\frac{\sqrt[3]{(a+b)(b+c)(a+c)}}{2}\geq \sqrt{\frac{ab+bc+ca}{3}}\]
[/quote]
\[\Longleftrightarrow\frac{(a+b)^{2}(b+c)^{2}(c+a)^{2}}{64}\ge\frac{(ab+bc+ca)^{3}}{27}\]

\[\Longleftrightarrow (a+b)^{2}(b+c)^{2}(c+a)^{2}\ge\frac{64}{27}(ab+bc+ca)^{3}\]
But we have: $(a+b)(b+c)(c+a)\ge\frac{8}{9}(a+b+c)(ab+bc+ca)$, then we have to prove that:
$(a+b+c)^{2}\ge 3(ab+bc+ca)$ and it's simple.
[hide] I think it's well-known problem and it's not of you, [b]chien than[/b][/hide]

[quote="chien than"]Let $a;b;c$ are positive real numbers.Prove that:
$\frac{\sqrt[3]{(a+b)(b+c)(a+c)}}{2}\geq \sqrt{\frac{ab+bc+ca}{3}}$[/quote]
It's Carlson's inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27766


I'm trying to get a one inch bottom margin (not too hard because that's the default) and then I'd like another inch between the page number and the bottom of my text. I've tried a variety of combinations of

%\setlength{\voffset}{0 in}
\setlength{\topmargin}{0 in}
%\setlength{\headheight}{0 in}
%\setlength{\headsep}{0 in}
%\setlength{\topskip}{0 in}
%\setlength{\textheight}{5.0 in}
\addtolength{\textheight}{-1.0 in}
\flushbottom
\addtolength{\footskip}{1.0 in}

Any suggestions?
Thanks
Leah

To know exactly what settings you need to change, look in the section 6.4 in "The not so short introduction to $\LaTeXe$".  If you don't have this document, just look for it on google, it's free.

btw, I think it's a must to go trough to document.

ledurt


Let $A,B\in\mathcal M_{2}(R)$. Show that:

(a) $\det(A^2+B^2)\geq\det(AB-BA)$

(b) If $\det(A^2+B^2) = \det(AB-BA)$, then $\det A=\det B$

[color=blue][b] Problem.[/b]

  Let $A, B\in M_{2}(\mathbb{R})$. Prove that

     a) $det(A^{2}+B^{2})\geq det(AB-BA)$.

     b) if $det(A^{2}+B^{2})= det(AB-BA)$, then $det A=det B$.

   [/color]

   [i] Solution. [/i]

     a)  We have \[ (A+iB)(A-iB)=(A^{2}+B^{2})-i(AB-BA)  \] and \[ (A-iB)(A+iB)=)=(A^{2}+B^{2})+i(AB-BA)  \]
       We have \[ d=det[(A+iB)(A-iB)]\geq 0.  \] On the other hand we have \[ 2d=det[(A+iB)(A-iB)]+det[(A-iB)(A+iB)]=2det(A^{2}+B^{2})+2det[i(AB-BA)]=2det(A^{2}+B^{2})-2det(AB-BA)\geq 0.  \]
         The conclusion follows.

      b) Now, let's study when equality does occur. Welll , if $det(A^{2}+B^{2})=det(AB-BA)$ then $d=0$ so
          \[ det(A+iB)=det(A-iB)=0.  \] Let's consider $f(x)=det(A+xB)$ with $f(i)=f(-i)=0$ of course. It follows easily that $det A=det B=2$. Done! :)  ;)

(a) $(A+iB)(A-iB)=(A^2+B^2)-i(AB-BA)=C-iD$. Now $C=\left(\begin{array}{cc}c_{1}&c_{2}\\ c_{3}&c_{4}\end{array}\right)$ and $D=\left(\begin{array}{cc}d_{1}&d_{2}\\ d_{3}&d_{4}\end{array}\right)$.

[i]First[/i]  we have $\det(C-iD)=\det(A+iB)(A-iB)=\det(A+iB)\det(A-iB)$ $=\det(A+iB)\overline{\det(A+iB)}=|\det(A+iB)|\geq 0$ and also a real number.

On the other hand $\det(C-iD)=\det(C)-det(A)-i(d_{4}c_{1}+d_{1}c_{4}-d_{3}c_{2}-d_{2}c_{3})$ $=\det(C)-\det(D)-i\alpha$. Being a real number it's clear that $\alpha=0$. Hence $\det(C)-\det(D)\geq 0 \Leftrightarrow \det(C)\geq\det(D)$ :)

(b) If $\det(A^2+B^2)=\det(AB-BA)\Rightarrow\det(A-iB)=0$ or $\det(A+iB)=0$. But like we did in a), $/det(A-iB)=\det(A)-\det(B)-i\beta=0\Leftrightarrow \det(A)=\det(B)$.

Is this right :-?

Yes, it is almost the same with my solution. :)


In a 3d space, I have to find a way to apply a value based on the distance from another point.

Imagine, a camera, an image (flat 2d plane) in a cartesian space. The camera and the layer are represented by single coordinates. I am only concerned with the Z distance, and based on the Z distance, I will apply an effect to the image (a blur, for instance).

For example, when the layer is 14000 units distant from the camera, I need the blur value to equal 4700. Then when the layer is 2500 units from the camera, the blur value must equal 2200.

In addition, the things I can test for (variables I can input) are camera position (cameraZ), layer position (layerZ), near blur (nearX) and far blur (farX).

This defines a line, and I can draw it out on paper (roughly), but do not know how to convert this to a formula that will return values for all positions inbetween and beyond.

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

nearX = 2200;
farX = 4700;

cameraZ = <<look up the Z position of the camera>>;
layerZ = <<look up the Z position of the layer>>;
deltaZ = Math.abs(Math.abs(cameraZ - 1000000) - Math.abs(layerZ - 1000000)); // find the distance between the camera and layer, whose positions can be positive or negative
deltaX = Math.abs(Math.abs(nearX - 1000000) - Math.abs(farX - 1000000)); // find the distance between nearX and farX

slope = (deltaX) / (deltaZ);

(slope * deltaZ) + nearX; // this is about where I gave up. I am getting a tangential graph, not a linear one. I don't understand why.


Is there any way to get my practice proof writing graded by someone?

Well, I've never actually graded a USAMO (i.e. I don't know exactly what they would take off points for), but I'd be happy to comment on your practice proof writing, and perhaps make a rough guess as to what you might get for it on a real USAMO.  Post it here or PM me.

(I went to MOP a long, long time ago... over 10 years ago... )

Yakko

I took it last year, and I remember thinking that I did much better than I really did.  You have to make sure that you PROVE every assertion you make and that you don't make any assumptions.  As long as you have a solid chain of logic leading to the answer, you should get full or nearly full credit.

Good luck preparing for the test this year.

You could certainly post a proof for critique in the relevant Olympiad section.  There are many knowledgeable people there.


Was working on something and I got over 100 errors over this chunk of code, all because of a "Missing \cr" command or a "Misplaced \cr" command,
[code]\begin{eqnarray*}
\displaystyle\sum_{k=1}^{2004}\dfrac{1}{1+\tan^2\left(\dfrac{k\pi}{2\cdot 2005}\right)}&=&\displaystyle\sum_{k=1}^{2004}\dfrac{1}{\sec^2\left(\dfrac{k\pi}{2\cdot 2005}\right)} \\
&=&\displaystyle\sum_{k=1}^{2004}\cos^2\left(\dfrac{k\pi}{2\cdot 2005}\right)} \\
&=&\cos^2\left(\dfrac{\pi}{2\cdot 2005}\right)+...+\cos^2\left(\dfrac{1002\pi}{2\cdot 2005}\right)+\sin^2\left(\dfrac{1002\pi}{2\cdot 2005}\right)+... \\
&=&\cos^2\left(\dfrac{\pi}{2\cdot 2005}\right)+\sin^2\left(\dfrac{\pi}{2\cdot 2005}+...+\cos^2\left(\dfrac{1002\pi}{2\cdot 2005}+\sin^2\left(\dfrac{1002\pi}{2\cdot 2005}\right) \\
&=&1+1+1+...+1 \\
&=&1002(1) \\
&=&1002
\end{eqnarray*}[/code]

1. You have an extra } at the end of line 3 which shouldn't be there (TexnicCenter shows the } in red if the cursor is placed just after it)
2. Later on you have plenty of \left( but no \right)
3. eqnarray automatically uses display mode so there's no need for \displaystyle or \dfrac (fractions inside fractions shouldn't be in display mode)
4. you should not be using the deprecated eqnarray - see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ 5[/url]

Try[code]
\begin{align*}
\sum_{k=1}^{2004}\frac{1}{1+\tan^2\left(\frac{k\pi}{2\cdot 2005}\right)}&=\sum_{k=1}^{2004}\frac{1}{\sec^2\left(\frac{k\pi}{2\cdot 2005}\right)} \\
&=\sum_{k=1}^{2004}\cos^2\left(\frac{k\pi}{2\cdot 2005}\right) \\
&=\cos^2\left(\frac{\pi}{2\cdot 2005}\right)+...+\cos^2\left(\frac{1002\pi}{2\cdot 2005}\right)+\sin^2\left(\frac{1002\pi}{2\cdot 2005}\right)+... \\
&=\cos^2\left(\frac{\pi}{2\cdot 2005}\right)+\sin^2\left(\frac{\pi}{2\cdot 2005}\right)+...+\cos^2\left(\frac{1002\pi}{2\cdot 2005}\right)+\sin^2\left(\frac{1002\pi}{2\cdot 2005}\right) \\
&=1+1+1+...+1 \\
&=1002(1) \\
&=1002
\end{align*}[/code]which gives
\begin{align*}\sum_{k=1}^{2004}\frac{1}{1+\tan^{2}\left(\frac{k\pi}{2\cdot 2005}\right)}&=\sum_{k=1}^{2004}\frac{1}{\sec^{2}\left(\frac{k\pi}{2\cdot 2005}\right)}\\ &=\sum_{k=1}^{2004}\cos^{2}\left(\frac{k\pi}{2\cdot 2005}\right) \\ &=\cos^{2}\left(\frac{\pi}{2\cdot 2005}\right)+...+\cos^{2}\left(\frac{1002\pi}{2\cdot 2005}\right)+\sin^{2}\left(\frac{1002\pi}{2\cdot 2005}\right)+... \\ &=\cos^{2}\left(\frac{\pi}{2\cdot 2005}\right)+\sin^{2}\left(\frac{\pi}{2\cdot 2005}\right)+...+\cos^{2}\left(\frac{1002\pi}{2\cdot 2005}\right)+\sin^{2}\left(\frac{1002\pi}{2\cdot 2005}\right) \\ &=1+1+1+...+1 \\ &=1002(1) \\ &=1002 \end{align*}

thanks for the help...
I might not have known about the align environment...


function
search
algebra proposed
algebra

Find all functions $ f: R^{ + } \to R^{ + }$ that satisfy
\begin{align} \nonumber x^2(f(x) + f(y)) & = (x + y)f(yf(x)) \end{align}
for all $ x,y \in R^{ + }$.

Besides I've got a question: how can I find problems with the search query?

If $ f(x_1)\equal{}f(x_2)$, we have $ \frac{x_1\plus{}y}{x_1^2}\equal{}\frac{x_2\plus{}y}{x_2^2}$, which implies $ x_1\equal{}x_2$. Hence, $ f(x)$ is injective.

$ P(x,x): f(xf(x))\equal{}xf(x)$. Suppose $ f(1)\equal{}a$, then we have $ f(a)\equal{}a$. But $ f(x)$ is injective, so $ a\equal{}1$.
$ P(1,y): 1\plus{}f(y)\equal{}(1\plus{}y)f(y)$.

Hence, $ f(y)\equal{}\frac1y$.


Find all pairs of positive integers m,n s.t. $ mn\minus{}1$ divides $ n^3\plus{}1$


A rectangle inscribed in a triangle has its base coinciding with the base $b$ of the triangle. If the altitude of the triangle is $h$, and the altitude $x$ of the rectangle is half the base of the rectangle, then:

$\textbf{(A)}\ x=\dfrac{1}{2}h \qquad
\textbf{(B)}\ x=\dfrac{bh}{b+h} \qquad
\textbf{(C)}\ x=\dfrac{bh}{2h+b} \qquad
\textbf{(D)}\ x=\sqrt{\dfrac{hb}{2}} \qquad
\textbf{(E)}\ x=\dfrac{1}{2}b$

[hide]There is a smaller triangle, sharing its vertex with the original triangle, and sitting “on top” of the rectangle. What’s its altitude? How are those two triangles related?[/hide]

[hide="My Solution"]
[asy]defaultpen(linewidth(.8pt));
unitsize(4cm);

pair A = origin;
pair B = (1,0);
pair C = (0.5,0.75);
pair D = midpoint(A--C);
pair E = midpoint(B--C);
pair F = foot(E,A,B);
pair G = foot(D,A,B);
pair H = foot(C,A,B);
pair J  = intersectionpoint(D--E, C--H);

draw(A--B--C--cycle);
draw(C--H);
draw(D--E--F--G--cycle);

label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",D,NW);
label("$E$",E,NE);
label("$F$",F,S);
label("$G$",G,S);
label("$H$",H,S);
label("$J$",J,NW);

draw(rightanglemark(G,D,E,2));
draw(rightanglemark(D,E,F,2));
draw(rightanglemark(E,F,G,2));
draw(rightanglemark(F,G,D,2));[/asy]

Let the base of the rectangle be $ q$. We have $ x \equal{} \frac {1}{2}q$. We also know that $ \overline{DE} \parallel \overline{AB}$. Therefore, $ \triangle CDE \sim \triangle CAB$. Because $ DG \equal{} JH \equal{} EF \equal{} x$, we know that $ JC \equal{} CH \minus{} JH \equal{} h \minus{} x$. We also have $ GF \equal{} q \equal{} 2x$. Therefore, by similar triangle ratios,

$ \frac {h \minus{} x}{h} \equal{} \frac {2x}{b}\implies bh \minus{} bx \equal{} 2xh\implies x(2h \plus{} b) \equal{} bh\implies \boxed{x \equal{} \frac {bh}{2h \plus{} b} \text{ (C)}}$.
[/hide]


topology
geometry
3D geometry
parameterization
invariant
vector

Dear all, help me to find mistake in my writing.

" This paper consists of two parts. In first part, authors gave an approach to inertial manifolds for nonautonomous dynamical systems. Their result on the existence of inertial manifolds rely on two geometrical assumptions, called cone invariance and squeezing property. Beside that, conditions which ensure the global pullback attractor is contained in the inertial manifold are also given.

In this second part of the paper, they considered special nonautonomous dynamical systems, namely processes (or two-parameter semi-flows). This is a first application in approach and comparison with known results they verify the assumptions for semilinear nonautonomous evolution equations whose linear part satisfies an exponential dichotomy condition and whose nonlinear part is globally bounded and globally Lipschitz."

I am looking forward to listening your comments.

Thank you very much.

With best wishes,

Hoc.

You want grammar comments, right?

"This paper consists of two parts. In [b]the[/b] first part, [b]the[/b] authors give an approach to inertial manifolds for nonautonomous dynamical systems. Their result on the existence of inertial manifolds rel[b]ies[/b] on two geometrical assumptions, called cone invariance and [b]the[/b] squeezing property. Beside[b]s[/b] that, [b]the authors[/b] give conditions which ensure the global pullback attractor is contained in the inertial manifold.

In the second part of the paper, the [b]authors[/b] consider special nonautonomous dynamical systems, namely processes (or two-parameter semi-flows). This is a first application in approach and comparison with known results they verify the assumptions for semilinear nonautonomous evolution equations whose linear part satisfies an exponential dichotomy condition and whose nonlinear part is globally bounded and globally Lipschitz."

I'm having trouble with the last sentence.

Further to JRav's changes:

"This paper consists of two parts. In the first part, the authors give an approach to inertial manifolds for nonautonomous dynamical systems. Their result on the existence of inertial manifolds relies on two geometrical assumptions, called cone invariance and the squeezing property. [i]In addition[/i], the authors give conditions which ensure [i]that[/i] the global pullback attractor is contained in the inertial manifold.

In the second part of the paper, the authors consider special nonautonomous dynamical systems, namely processes* (or two-parameter semi-flows). This is [i]the[/i] first application [i]of this[/i] approach, and comparison with known results [i]verifies[/i] the assumptions for semilinear nonautonomous evolution equations whose linear part satisfies an exponential dichotomy condition and whose nonlinear part is globally bounded and globally Lipschitz."

I'm guessing somewhat about the intended meaning of the last sentence.

* Can this really be the right word?

Dear everybody,

Thank you very much for your comments.

Best regards,

Hoc.

Hello everybody,

I have two paragraphs, help me to prepair them, please. Thank you very much.

1. "This paper talk about normal form of duffing-van der Pol oscillator under nonautonomous parametric perturbations. The author gave a new generalization of Poincarés normal form theory to nonautonomous differential equations applied to the Duffing-van der Pol oscillator under a nonautonomous bounded parametric perturbation. The nonautonomous normal form is calculated for parameter values corresponding to the pitchfork scenario. In this paper, the author used the computer algebra program Maple to effort accomplish the enormous computational. Finally, the author also indicate application of the normal form, which can be used to the study nonautonomous bifurcation phenomena".

2. "This paper introduced about the dichotomy spectrum for linear dunamic equations $ x^\delta\equal{}A(t)x$ in $ \mathbb{R}^N$ on a time scale T (e. g. T=Z or $ T\equal{} \mathbb{R}$). This new spectrum consists of at most N closed intervals of the real line. In the autonomous cases with $ T\equal{}\mathbb{R}$ these intervals reduce to the real parts of the eigenvalues of A. In any cases the spectral intervals are associated with invariant vector boundles comprising solutions with a common exponential growth rate. A spectral theorem, which desribes all possible forms of the dichotomy spectrum, is the main result of this paper".

Have a nice day to you.

Hoc.

"This paper talk[b]s[/b] about [b]the[/b] normal form of [b]the[/b] duffing-van der Pol oscillator under nonautonomous parametric perturbations. The author [b]gives[/b] a new generalization of Poincarés normal form theory to [i] nonautonomous differential equations applied to the Duffing-van der Pol oscillator under a nonautonomous bounded parametric perturbation[/i]. The nonautonomous normal form is calculated for parameter values corresponding to the pitchfork scenario. In this paper, the author uses the computer algebra program Maple to effort[b]lessly[/b](?) accomplish the enormous computation. Finally, the author [b]gives an[/b] application of the normal form, which can be used to the study nonautonomous bifurcation phenomena".

The italicized part essentially repeats the first sentence so you should change that.

"Talks about" should be "discusses" or some other synonym: paper doesn't talk.  "Poincarés" should be "Poincaré's" (it's a possessive).  Be consistent and correct in your capitalization of things like "duffing-van der Pol."  I agree with JRav that the first sentence seems unnecessary.

Also, while we're on the subject, "prepair" is not a word, and if it were, it would mean "to put things in pairs beforehand." ;)

Hi, Guys

Thank you very much for your comments.

Best regards,

Hoc.


hameye taabe haye $F$ ra biyabid:

$F: N\rightarrow N$
$mf(f(m))=(f(m))^2$
$gcd(m,n)=d ,f(m.n).f(d)=d.f(m).f(n)$
$f(m)=1\Longleftrightarrow m=1$

Braye anoya ke ketabe :

OLAMPIYADE RIYAZI DAR IRAN ,Neveshteye Ebadolahe mahmodiyan

Ra daran ,

Jelde 2 ,Safaye 174 and 175 ,

Javabe masale :

$f(x)=x$ .

rahe hal 2 safas ke man ,hoseleye copy kardane ono nadaram

soale beautifylie,man az harchi moadele tabeie khosham miad akhe shabihe bazy mimoone va lezzatbakhshe,nazar shoma darbareye inmozoo chie? :D


For all of you USABO semifinalists, post your results here after April 20th.   Oh yeah, and whoever gets here first, post the cutoff too.

yeah im a USABO semifinalist. hopefully I made finalist...but the questions were easy this year on the semifinals so id put money on it that the cutoffs are high. hopefully finalist on the 20th!

[quote="GoBraves"]What is USABO?[/quote]US Biology Olympiad I presume. USAMO for Biology  :?

yeah pretty much, then theres IBO. too bad no BOSP (as opposed to MOSP) but USABO is pretty much a summer program in itself.

Hey adid, you're on this forum too?  I'm onomotopia121.

hey onomatopoeia! yeah im a CC guy too...been here for a while. yeah with the addition of this biology thread, i might make a permanent move over to these forums...lol. im "biomaster" at CC now because i got frozen (again. ha ha!)

[quote="GoBraves"]What is USABO?[/quote][url=http://www.cee.org/usabo/index.shtml]USA Biology Olympiad[/url], the qualifying test for the US team for the International Biology Olympiad.

yeah you can take it your senior year too, all through high school. that isnt cool your teacher forgot. the exams were fun this year; the open exam was especially challenging this year

Just got the call from CEE.  I'm in finals!  YEAH!

lol yeah ... apparently there're "21" finalists this year because I'm an intl student who can't go to the finals though I scored in top 20. ah w/e. congrats justdudxit. =)

(Is there)/(Will there be) a list of  qualifiers?

[quote="furious"]i was really upset when i realized that my biolody teacher forgot to give us the usabo. she had it on her computer but forgot about it. i know i could have qualified (or at least i will think that i would have  :lol: ) maybe i can take it next year.[/quote]
I know how you feel. Happened to me this year, too, and I know I would at least be a semifinalist . . . but oh well. Chem olympiad was fun, anyway. :P

sheiza...123 does NOT feel good right now. lol

Hmm. 113 as a junior, 125 was the cutoff, I think. How much work to make the camp and the team next year? Adidasty, any word?

well, i got a 123. i was pretty mad because its only 2 points off!!!!! well how much work.....to make USABO? i got pretty close and i *just* read campbells. its a lot to read, but thats the way to do it. buy the 6th edition (next year they might change the definitive source to 7th) and just read read read read read. memorize all that you can. then, make sure you are careful during the open exam. for the semifinals, all i can say is be sure you do well on the free response part.


superior algebra
superior algebra unsolved

A fifteen puzzle consists of fifteen small square tiles, numbered 1 to 15, which are mounted in a 4 x 4 frame in such a way that each tile can slide vertically or horizontally into an adjacent square (if it is not already occupied by another tile), but the tiles cannot be lifted out of the tray. On the packet in which the puzzle was sold, it is asserted that it is impossible to maneuver the tiles from the first to the second of the configurations shown below. Prove this assertion.

1   2   3   4        15  14  13  12
5   6   7   8        11  10   9    8
9  10  11 12        7    6    5   4
13 14  15  -        3    2    1    -

Oh, this is such an old chestnut. Puzzle was invented by Sam Lloyd, and is completely solved theoretically - it's all about the parity of the permutations we want to achieve.


[hide]
Let $x=1/a$ etc, then it's equivalent to
\[\sum_\text{cyc}\frac{yz}{x^{2}+2yz}\leq 1\]
After multiplying by 2 and subtracting 1 from each term, it's equivalent to
\[\sum_\text{cyc}\frac{x^{2}}{x^{2}+2yz}\geq 1\]
But this is clearly true from
\[2yz\leq y^{2}+z^{2}\Rightarrow \frac{x^{2}}{x^{2}+2yz}\geq \frac{x^{2}}{x^{2}+y^{2}+z^{2}}\]
[/hide]

[quote="arqady"][quote="K81o7"]
$\frac{9}{2x+2y+2z+3}\ge \frac{9}{6+3}$

[/quote]
I think, it's wrong. :wink:[/quote]

I wonder what I was thinking when I typed this...was I asleep?

[quote=arqady]$a,$ $b$ and $c$ are positive numbers. Without expanding prove that:
$\frac{a^{2}}{2a^{2}+bc}+\frac{b^{2}}{2b^{2}+ac}+\frac{c^{2}}{2c^{2}+ab}\leq1.$[/quote]
Middle school mathematics magazine N0.1(2019):

Mike, can you verify this one?
\\
For the positive real numbers $a, b, c$ with $a+b+c=1$, or $ a+b+c=abc=2 $
\\
\\
$ \frac{a^{2}}{2a^{2}+bc}+\frac{b^{2}}{2b^{2}+ac}+\frac{c^{2}}{2c^{2}+ab} \ge k \ \ ; $
\\
\\
Thank you!
Greetings!

[quote=Lonesan]
For the positive real numbers $a, b, c$ with $a+b+c=abc=1$, then:
[/quote]
\[1 = a+b+c \geqslant 3\sqrt[3]{abc} = 3.\]

[quote=arqady]$a,$ $b$ and $c$ are positive numbers. Without expanding prove that:
$$\frac{a^{2}}{2a^{2}+bc}+\frac{b^{2}}{2b^{2}+ac}+\frac{c^{2}}{2c^{2}+ab}\leq1.$$[/quote]
Let $a,$ $b$ and $c$ are positive numbers. Prove that:
$$\frac{b^2+c^2}{2a^2+bc}+\frac{c^2+a^2}{2b^2+ac}+\frac{a^2+b^2}{2c^2+ab}\geq2.$$

[quote=mudok][quote=sqing] Let $a,$ $b$ and $c$ are positive numbers. Prove that:
$$\frac{b^2+c^2}{2a^2+bc}+\frac{c^2+a^2}{2b^2+ac}+\frac{a^2+b^2}{2c^2+ab}\geq2.$$[/quote]

$\sum_{cyc}\frac{b^2+c^2}{2a^2+bc}\ge \sum_{cyc}\frac{2bc}{2a^2+bc} , \ \ \ \sum_{cyc}\frac{2bc}{2a^2+bc}\ge 2 \iff $

[quote=arqady]$a,$ $b$ and $c$ are positive numbers. Without expanding prove that:
$\frac{a^{2}}{2a^{2}+bc}+\frac{b^{2}}{2b^{2}+ac}+\frac{c^{2}}{2c^{2}+ab}\leq1.$[/quote][/quote]
Thank you very much.
Let $a,$ $b$ and $c$ are positive numbers. Prove that:
$$\sqrt{\frac{b^2+c^2}{2a^2+bc}}+\sqrt{\frac{c^2+a^2}{2b^2+ac}}+\sqrt{\frac{a^2+b^2}{2c^2+ab}}\geq\sqrt{6}.$$

Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive numbers . Prove that $$\sum_{k=1}^{n}\sqrt{\frac{S-a^{n-1}_k}{(n-1)a^{n-1}_k+\frac{T}{a_k}}}\geq\sqrt{n(n-1)}.$$
Where $S=a^{n-1}_1+a^{n-1}_2+\cdots+a^{n-1}_n,T=a_1a_2\cdots  a_n.$
[quote=Nguyenhuyen_AG][quote=sqing]Let $a,$ $b$ and $c$ are positive numbers. Prove that:
$$\sqrt{\frac{b^2+c^2}{2a^2+bc}}+\sqrt{\frac{c^2+a^2}{2b^2+ac}}+\sqrt{\frac{a^2+b^2}{2c^2+ab}}\geq\sqrt{6}.$$[/quote]
\[\sqrt{\frac{a^2+b^2}{2c^2+ab}} \geqslant \frac{\sqrt6(a^2+b^2)}{2(a^2+b^2+c^2)}.\]
See [url='https://drive.google.com/file/d/0ByIABDU-2YwJbmM3WjkwTGhOMHM/view']epsilon_vol13_2017[/url], page [b]257[/b].[/quote]
Thank you very much.


find [tex]A^2[/tex] where [tex]A[/tex] is the sum of the absolute values of all roots of the following equation:
[tex]x=\sqrt{19} +\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+x}}}}[/tex]

the solution provided by AMC doesn't seem too intuitive.  can someone provide a full solution to the problem?

If x satisfies that equation, then it will also satisfy
x= :sqrt: 19+91/x.  Multiplying by x leads to a quadratic whose roots are  1/2*(:sqrt: 19 :pm:  :sqrt: 383).

that's part of what i don't understand.

why does being a root of the former equation imply that it is also a roots of the latter equation?

I think in the original question, it should be 91/x instead of 91/( :sqrt: 19 + x).  Then if you keep plugging in x into the denominator there, you'll see that x can be represented as an infinite continued fraction.

Let [tex]f(k)=\sqrt{19}+\frac{91}{k}[/tex]
Let [tex]f(x)=a, f(a)=b, f(b)=c, f(c)=d[/tex]. Sub in value of [tex]a[/tex] first, then [tex]b[/tex] then [tex]c[/tex] then [tex]d[/tex] we have[tex] f(d)=x[/tex],

If [tex]f(m)=n[/tex], then [tex]f(n)=m[/tex]

[tex]d=f(x)=f(c)[/tex] gives us [tex]x=c [/tex]   [tex]f(c)=b=f(x)=f(a)[/tex] yields [tex]a=x[/tex],
or [tex]f(x)=x[/tex]
or
[tex]x=\sqrt{19}+\frac{91}{x}[/tex]

Many problems with a pattern smeared through a couple iterations stem from ideas like that.  I always look for a simple function [tex]f(x)[/tex] so that the given can be written as [tex]x = f^{(m)}(x)[/tex], since we may extract the solutions [tex]c[/tex] with the property that [tex]f(c) = c[/tex].  Sometimes the  function [tex]f[/tex] isn't as obvious.  Alternatively, you could substitute the given into itself. Here are a few related problems:

1. Solve for real [tex]x[/tex] given [tex]x = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+x}}}}[/tex].

2. Solve for real [tex]x[/tex] that satisfy the infinite equation [tex]2x = \frac{1}{x + \frac{1}{x + \frac{1}{x + \frac{1}{\cdots}}}}[/tex].

3. Solve for real [tex]x[/tex] for which [tex]x = \frac{-1 + 5\left(\frac{1 - 5x^2}{2}\right)^2}{2}[/tex].

4. (Jae Bae) Find a real solution to [tex](x-2)^2(x-6)^2 = x[/tex].


I would like to know how to draw an arc in Asymptote (it's not necessairly part of a circle).

I have 3 points and I would like an arc going through them.

Furthermore, how do you include radicals in Asymptote? For example, I have a coordinate with $ \sqrt{6}$

Many examples about arc are here: [url]http://www.piprime.fr/developpement/arc_asy/?posts_per_page=-1[/url]

[quote="pytheagle"]
Furthermore, how do you include radicals in Asymptote? For example, I have a coordinate with $ \sqrt {6}$[/quote]
I believe it is:
[code]
sqrt(6)
[/code]